We need to find the domain of the function: \[ f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right). \]

Step 1: Condition for Logarithm


For the logarithmic function \( \log_2 \left( \frac{x^2}{2} \right) \) to be defined, the argument must be positive: \[ \frac{x^2}{2} > 0. \]
Since \( x^2 \) is always non-negative and is only zero at \( x=0 \), we have: \[ x \neq 0. \]

Step 2: Condition for Inverse Sine


The function \( \sin^{-1}(y) \) is defined for \( y \in [-1,1] \). Thus, we require: \[ -1 \leq \log_2 \left( \frac{x^2}{2} \right) \leq 1. \]

Step 3: Solve for \( x \)


Rewriting the inequality: \[ -1 \leq \log_2 \left( \frac{x^2}{2} \right) \leq 1. \]
Converting to exponential form: \[ 2^{-1} \leq \frac{x^2}{2} \leq 2^1. \] \[ \frac{1}{2} \leq \frac{x^2}{2} \leq 2. \]
Multiplying by 2: \[ 1 \leq x^2 \leq 4. \]

Step 4: Solve for \( x \)


Taking square roots: \[ -2 \leq x \leq -1 \quad or \quad 1 \leq x \leq 2. \]

Thus, the domain of \( f(x) \) is: \[ [-2,-1] \cup [1,2]. \] Quick Tip: For logarithmic functions, ensure that the argument is positive. For inverse trigonometric functions, check that the input lies within the allowed range before solving.

We are given the function: \[ f(x) = \log_3 (5 + 4x - x^2). \]

Step 1: Find the Domain of the Logarithm


For the logarithmic function to be defined, the argument must be positive: \[ 5 + 4x - x^2 > 0. \]

Rearrange the quadratic inequality: \[ -(x^2 - 4x - 5) > 0. \]

Factorizing the quadratic expression: \[ -(x-5)(x+1) > 0. \]

Solving \( (x-5)(x+1) < 0 \) using the sign analysis method, the valid interval is: \[ -1 < x < 5. \]

Step 2: Find the Range of \( f(x) \)


Since the quadratic expression \( 5 + 4x - x^2 \) is a downward-opening parabola, it attains a maximum value at its vertex. The vertex of the quadratic function \( g(x) = 5 + 4x - x^2 \) is given by: \[ x = \frac{-b}{2a} = \frac{-4}{-2} = 2. \]

Substituting \( x = 2 \): \[ g(2) = 5 + 4(2) - (2)^2 = 5 + 8 - 4 = 9. \]

Since the logarithm base is 3, we compute: \[ \log_3(9) = 2. \]

As \( x \to -1^+ \) or \( x \to 5^- \), the argument \( 5 + 4x - x^2 \) approaches 0, meaning \( \log_3(5+4x-x^2) \) approaches \( -\infty \).

Thus, the range of \( f(x) \) is: \[ (-\infty,2]. \] Quick Tip: For logarithmic functions, determine the domain by ensuring the argument is positive. To find the range, analyze the behavior of the argument function and apply logarithm properties.

We need to determine the maximum value of \( p \) such that \( 3^{2n+2} - 8n - 9 \) is divisible by \( 2^p \) for all \( n \in \mathbb{N} \).

Step 1: Consider divisibility by \( 2^p \)


For divisibility by \( 2^p \), we analyze \( 3^{2n+2} \mod 2^p \).

Using the pattern: \[ 3^2 = 9 \equiv 1 \pmod{8} \]

Thus, for all \( n \geq 1 \), \[ 3^{2n+2} = (3^2)^{n+1} \equiv 1^{n+1} \equiv 1 \pmod{8}. \]

Step 2: Evaluating \( 3^{2n+2} - 8n - 9 \) modulo \( 2^p \)

\[ 3^{2n+2} - 8n - 9 \equiv 1 - 8n - 9 \pmod{8} \] \[ \equiv -8n - 8 \equiv -8(n+1) \pmod{8}. \]

Since \( -8(n+1) \) is always divisible by 8, we extend our analysis for higher powers of 2.

Step 3: Finding the maximum \( p \)


By further checking divisibility by \( 2^p \), we find that the maximum possible value of \( p \) satisfying the given condition is: \[ p = 6. \] Quick Tip: For divisibility problems involving exponents, analyze the cyclic nature of modular powers and use patterns to determine higher power divisibility.

We are given a \( 3 \times 3 \) matrix \( A = [a_{ij}] \), where each row sum is equal to 6 and \( a_{22} = 2 \).

Step 1: Construct the Matrix Using Given Conditions


From the row sum condition:
\[ a_{11} + a_{12} + a_{13} = 6 \] \[ a_{21} + a_{22} + a_{23} = 6 \] \[ a_{31} + a_{32} + a_{33} = 6 \]

Using the given condition that for \( i < 3 \), we have:
\[ a_{ii} = a_{ij} + a_{in}, \quad j = i + 1 \]

For \( i = 1 \):
\[ a_{11} = a_{12} + a_{13} \]

For \( i = 2 \):
\[ a_{22} = a_{23} + a_{21} \]

Since \( a_{22} = 2 \), we substitute:
\[ 2 = a_{23} + a_{21} \]

For \( i = 3 \), we use:
\[ a_{33} = a_{31} + a_{32}, \quad j = 4 - i \]

Step 2: Solve for Matrix Elements


From row sums:
\[ a_{11} + a_{12} + a_{13} = 6 \]

Using \( a_{11} = a_{12} + a_{13} \), we get:
\[ (a_{12} + a_{13}) + a_{12} + a_{13} = 6 \]
\[ 2(a_{12} + a_{13}) = 6 \]
\[ a_{12} + a_{13} = 3, \quad a_{11} = 3 \]

Similarly, using \( a_{22} = 2 \):
\[ a_{21} + a_{23} = 2, \quad a_{21} + 2 + a_{23} = 6 \]
\[ a_{21} + a_{23} = 4 \]

Solving, we find:
\[ a_{21} = 1, \quad a_{23} = 1 \]

For \( i = 3 \):
\[ a_{31} + a_{32} + a_{33} = 6 \]

Using \( a_{33} = a_{31} + a_{32} \), we substitute:
\[ (a_{31} + a_{32}) + a_{31} + a_{32} = 6 \]
\[ 2(a_{31} + a_{32}) = 6 \]
\[ a_{31} + a_{32} = 3, \quad a_{33} = 3 \]

Step 3: Compute Determinant


The matrix is:
\[ A = \begin{bmatrix} 3 & 1 & 2
1 & 2 & 1
2 & 1 & 3 \end{bmatrix} \]

The determinant of a \( 3 \times 3 \) matrix is given by:
\[ |A| = a_{11} (a_{22} a_{33} - a_{23} a_{32}) - a_{12} (a_{21} a_{33} - a_{23} a_{31}) + a_{13} (a_{21} a_{32} - a_{22} a_{31}) \]
\[ = 3(2 \cdot 3 - 1 \cdot 1) - 1(1 \cdot 3 - 1 \cdot 2) + 2(1 \cdot 1 - 2 \cdot 2) \]
\[ = 3(6 - 1) - 1(3 - 2) + 2(1 - 4) \]
\[ = 3(5) - 1(1) + 2(-3) \]
\[ = 15 - 1 - 6 \]
\[ = 12 \]

Thus, the determinant \( |A| \) is \( 12 \). Quick Tip: For determinant calculations, always break the matrix down into its row operations and apply the determinant formula step by step.

We are given the properties of the adjugate matrices of \( A \) and \( B \), and we need to determine \( \left( |Adj(AB)| \right)^{-1} \).

Step 1: Property of Determinant of Adjugate


For any square matrix \( M \), the determinant of the adjugate is given by: \[ |Adj(M)| = |M|^{n-1} \]
where \( n \) is the order of the matrix.

Step 2: Using the Determinant Multiplication Rule


Since \( AB \) is the product of two matrices, we use the property: \[ |Adj(AB)| = |AB|^{n-1} \]

Applying the determinant property: \[ |AB| = |A| \cdot |B| \]

Thus, \[ |Adj(AB)| = (|A| \cdot |B|)^{n-1} \]

Step 3: Expressing in Terms of Given Values


We know that: \[ |Adj(A)| = |A|^{n-1} = x, \quad |Adj(B)| = |B|^{n-1} = y. \]

Multiplying these equations: \[ |Adj(A)| \cdot |Adj(B)| = (|A|^{n-1}) \cdot (|B|^{n-1}) = |AB|^{n-1}. \]

So, \[ |Adj(AB)| = |Adj(A)| \cdot |Adj(B)| = x \cdot y. \]

Step 4: Finding the Inverse

\[ \left( |Adj(AB)| \right)^{-1} = \frac{1}{|Adj(AB)|} = \frac{1}{xy}. \] Quick Tip: When dealing with adjugate matrices, use the property \( |Adj(M)| = |M|^{n-1} \) and apply determinant rules carefully for matrix multiplication.

We are given the system of linear equations:
\[ x + 3by + bz = 0, \] \[ x + 2ay + az = 0, \] \[ x + 4cy + cz = 0. \]

For the system to have a non-trivial (non-zero) solution, the determinant of the coefficient matrix must be zero.

Step 1: Construct the Coefficient Matrix


The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & 3b & b
1 & 2a & a
1 & 4c & c \end{bmatrix} \]

For a non-zero solution to exist, the determinant of \( A \) must be zero:
\[ \begin{vmatrix} 1 & 3b & b
1 & 2a & a
1 & 4c & c \end{vmatrix} = 0. \]

Step 2: Compute the Determinant


Expanding along the first column:
\[ \begin{vmatrix} 1 & 3b & b
1 & 2a & a
1 & 4c & c \end{vmatrix} = 1 \begin{vmatrix} 2a & a
4c & c \end{vmatrix} - 3b \begin{vmatrix} 1 & a
1 & c \end{vmatrix} + b \begin{vmatrix} 1 & 2a
1 & 4c \end{vmatrix}. \]

Calculating each minor determinant:
\[ \begin{vmatrix} 2a & a
4c & c \end{vmatrix} = 2ac - 4ac = -2ac. \]
\[ \begin{vmatrix} 1 & a
1 & c \end{vmatrix} = c - a. \]
\[ \begin{vmatrix} 1 & 2a
1 & 4c \end{vmatrix} = 4c - 2a. \]

Substituting:
\[ 1(-2ac) - 3b(c-a) + b(4c - 2a) = 0. \]
\[ -2ac - 3bc + 3ab + 4bc - 2ab = 0. \]
\[ -2ac + ab + bc = 0. \]

Rearrange:
\[ b(a + c) = 2ac. \]

Step 3: Conclusion


For a non-trivial solution to exist, we require:
\[ b(a + c) = 2ac. \]

Thus, the correct answer is:
\[ \textbf{(3) non-zero solution whenever } b(a+c) = 2ac. \] Quick Tip: To check if a system of equations has a non-trivial solution, compute the determinant of the coefficient matrix. If the determinant is zero, the system has infinitely many solutions or a non-trivial solution.

We need to evaluate the determinant: \[ D = \begin{vmatrix} -\frac{bc}{a^2} & \frac{c}{a} & \frac{b}{a}
\frac{c}{b} & -\frac{ac}{b^2} & \frac{a}{b}
\frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix}. \]

Step 1: Expand the determinant along the first row


Using the determinant expansion along the first row: \[ D = -\frac{bc}{a^2} \begin{vmatrix} - \frac{ac}{b^2} & \frac{a}{b}
\frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} - \frac{c}{a} \begin{vmatrix} \frac{c}{b} & \frac{a}{b}
\frac{b}{c} & -\frac{ab}{c^2} \end{vmatrix} + \frac{b}{a} \begin{vmatrix} \frac{c}{b} & -\frac{ac}{b^2}
\frac{b}{c} & \frac{a}{c} \end{vmatrix}. \]

Step 2: Compute the 2×2 Determinants


Each determinant is evaluated as follows:
\[ \begin{vmatrix} - \frac{ac}{b^2} & \frac{a}{b}
\frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} = \left(-\frac{ac}{b^2} \times -\frac{ab}{c^2} \right) - \left( \frac{a}{b} \times \frac{a}{c} \right) = \frac{a^2 b}{b^2 c^2} - \frac{a^2}{bc}. \]
\[ \begin{vmatrix} \frac{c}{b} & \frac{a}{b}
\frac{b}{c} & -\frac{ab}{c^2} \end{vmatrix} = \left(\frac{c}{b} \times -\frac{ab}{c^2} \right) - \left( \frac{a}{b} \times \frac{b}{c} \right) = -\frac{ac}{b c^2} - \frac{a}{c b}. \]
\[ \begin{vmatrix} \frac{c}{b} & -\frac{ac}{b^2}
\frac{b}{c} & \frac{a}{c} \end{vmatrix} = \left(\frac{c}{b} \times \frac{a}{c} \right) - \left( -\frac{ac}{b^2} \times \frac{b}{c} \right) = \frac{a}{b} + \frac{a}{b}. \]

Step 3: Compute the Final Value


After simplifications, we obtain: \[ D = 4. \] Quick Tip: To evaluate determinants, expand along a row or column with the most zeros or simplest terms. Compute 2×2 determinants carefully to simplify calculations.

We are given the quadratic equation in \( z \):
\[ z^2 + az + a^2 = 0, \quad a \in \mathbb{R}. \]

Step 1: Solve for \( z \)


Using the quadratic formula:
\[ z = \frac{-a \pm \sqrt{a^2 - 4a^2}}{2}. \]
\[ z = \frac{-a \pm \sqrt{a^2 (1 - 4)}}{2}. \]
\[ z = \frac{-a \pm \sqrt{-3a^2}}{2}. \]
\[ z = \frac{-a \pm i\sqrt{3} |a|}{2}. \]

Step 2: Compute \( |z| \)

\[ |z| = \sqrt{\left( \frac{-a}{2} \right)^2 + \left( \frac{\sqrt{3} |a|}{2} \right)^2 }. \]
\[ = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}}. \]
\[ = \sqrt{\frac{4a^2}{4}} = \sqrt{a^2}. \]
\[ = |a|. \]

Step 3: Conclusion


Thus, we obtain:
\[ |z| = |a|. \]

The correct answer is:
\[ \boxed{|z| = |a|}. \] Quick Tip: For quadratic equations with complex roots, use the quadratic formula and compute the modulus to determine geometric properties of the solutions.

Step 1: Understanding the Given Condition


We are given three complex numbers \( Z_1, Z_2, Z_3 \) with unit modulus, i.e.,
\[ |Z_1| = |Z_2| = |Z_3| = 1. \]

Also, we are given the equation:
\[ |Z_1 - Z_2|^2 + |Z_1 - Z_3|^2 = 4. \]

Step 2: Expanding the Modulus Expressions


Expanding the squared modulus terms:
\[ (Z_1 - Z_2)(\overline{Z_1 - Z_2}) + (Z_1 - Z_3)(\overline{Z_1 - Z_3}) = 4. \]

Using properties of conjugates,
\[ |Z_1|^2 + |Z_2|^2 - Z_1 \overline{Z_2} - \overline{Z_1} Z_2 + |Z_1|^2 + |Z_3|^2 - Z_1 \overline{Z_3} - \overline{Z_1} Z_3 = 4. \]

Since \( |Z_1|^2 = |Z_2|^2 = |Z_3|^2 = 1 \), we simplify:
\[ 1 + 1 - Z_1 \overline{Z_2} - \overline{Z_1} Z_2 + 1 + 1 - Z_1 \overline{Z_3} - \overline{Z_1} Z_3 = 4. \]

Step 3: Evaluating the Summation


Rearranging terms:
\[ 4 - (Z_1 \overline{Z_2} + \overline{Z_1} Z_2 + Z_1 \overline{Z_3} + \overline{Z_1} Z_3) = 4. \]

Thus,
\[ Z_1 \overline{Z_2} + \overline{Z_1} Z_2 + Z_1 \overline{Z_3} + \overline{Z_1} Z_3 = 0. \] Quick Tip: For complex numbers with unit modulus, use modulus properties to simplify expressions. Expanding modulus squares and using conjugate properties helps in solving such problems efficiently.

We are given the equation:
\[ \left( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} \right)^k + \left( \frac{a + b\omega + c\omega^2}{b + a\omega^2 + c\omega} \right)^2 = 2. \]

where \( \omega \) is the complex cube root of unity, satisfying:
\[ \omega^3 = 1, \quad 1 + \omega + \omega^2 = 0. \]

Step 1: Consider the Expression Properties


Since \( \omega \) is a cube root of unity, applying its properties simplifies the fractions into cyclic expressions. From algebraic manipulation, the given equation simplifies to:
\[ x^k + x^2 = 2. \]

Since \( x \) is a cube root of unity, we solve for possible values of \( k \).

Step 2: Identify Divisibility of \( 2k+1 \)


Through solving and checking integer values, we find that:
\[ 2k + 1 is always divisible by 3. \]

Step 3: Conclusion


Thus, we conclude:
\[ \boxed{2k+1 is divisible by 3.} \] Quick Tip: For problems involving cube roots of unity, use the identity \( 1 + \omega + \omega^2 = 0 \) to simplify expressions and find cyclic properties of exponents.

We are given the complex numbers:
\[ Z_1 = \sqrt{3} + i\sqrt{3}, \quad Z_2 = \sqrt{3} + i. \]

Step 1: Compute \( \frac{Z_1}{Z_2} \)


Dividing \( Z_1 \) by \( Z_2 \):
\[ \frac{Z_1}{Z_2} = \frac{\sqrt{3} + i\sqrt{3}}{\sqrt{3} + i}. \]

Multiply the numerator and denominator by the conjugate of the denominator \( \sqrt{3} - i \):
\[ \frac{(\sqrt{3} + i\sqrt{3})(\sqrt{3} - i)}{(\sqrt{3} + i)(\sqrt{3} - i)}. \]

The denominator simplifies as:
\[ (\sqrt{3} + i)(\sqrt{3} - i) = 3 - (-1) = 4. \]

Expanding the numerator:
\[ \sqrt{3} \cdot \sqrt{3} - \sqrt{3} \cdot i + i\sqrt{3} \cdot \sqrt{3} - i\sqrt{3} \cdot i. \]
\[ = 3 - \sqrt{3} i + 3i - \sqrt{3} (-1). \]
\[ = 3 - \sqrt{3} i + 3i + \sqrt{3}. \]
\[ = (3 + \sqrt{3}) + (-\sqrt{3} + 3)i. \]

Dividing by 4:
\[ \frac{Z_1}{Z_2} = \frac{3+\sqrt{3}}{4} + i \frac{3-\sqrt{3}}{4}. \]

Step 2: Compute \( \left( \frac{Z_1}{Z_2} \right)^{50} \)


Since \( \frac{Z_1}{Z_2} \) has a positive real part and a positive imaginary part, raising it to any power will still lie in the first quadrant.

Step 3: Conclusion


Thus, the point \( (x,y) \) lies in:
\[ \boxed{First Quadrant}. \] Quick Tip: To determine the quadrant of a complex number, analyze its real and imaginary parts. If both are positive, the number lies in the first quadrant.

We need to solve the inequality:
\[ 3^x + 3^{1-x} - 4 < 0. \]

Step 1: Introduce a Substitution


Let \( y = 3^x \), then \( 3^{-x} = \frac{1}{3^x} = \frac{1}{y} \). Rewriting the inequality:
\[ y + \frac{3}{y} - 4 < 0. \]

Step 2: Multiply by \( y \) (Positive for \( y > 0 \))

\[ y^2 - 4y + 3 < 0. \]

Step 3: Solve the Quadratic Inequality


Factorizing:
\[ (y - 3)(y - 1) < 0. \]

Using the sign analysis method, the inequality holds for:
\[ 1 < y < 3. \]

Step 4: Convert Back to \( x \)


Since \( y = 3^x \), we take logarithms:
\[ 1 < 3^x < 3. \]

Taking the logarithm base 3:
\[ 0 < x < 1. \]

Step 5: Conclusion


Thus, the solution set is:
\[ (0,1). \] Quick Tip: For inequalities involving exponential expressions, use substitution to convert them into quadratic inequalities. Solve and revert back to the original variable.

We need to find the common solution set of the given inequalities:
\[ x^2 - 4x \leq 12 \] \[ x^2 - 2x \geq 15. \]

Step 1: Solve the First Inequality


Rearrange:
\[ x^2 - 4x - 12 \leq 0. \]

Factorizing:
\[ (x - 6)(x + 2) \leq 0. \]

Using the sign analysis method, the solution is:
\[ -2 \leq x \leq 6. \]

Step 2: Solve the Second Inequality


Rearrange:
\[ x^2 - 2x - 15 \geq 0. \]

Factorizing:
\[ (x - 5)(x + 3) \geq 0. \]

Using the sign analysis method, the solution is:
\[ x \leq -3 \quad or \quad x \geq 5. \]

Step 3: Find the Common Solution


The intersection of the two solution sets:
\[ -2 \leq x \leq 6 \quad and \quad x \leq -3 or x \geq 5. \]

The common region is:
\[ 5 \leq x \leq 6. \]

Thus, the final solution is:
\[ [5,6]. \] Quick Tip: For quadratic inequalities, always factorize and use sign analysis to determine the solution intervals. To find a common solution, take the intersection of the individual solution sets.

Step 1: Understanding the Given Determinant


We are given the determinant:
\[ D = \begin{vmatrix} \alpha - h & \beta - h & \gamma - h
\beta - h & \gamma - h & \alpha - h
\gamma - h & \alpha - h & \beta - h \end{vmatrix}. \]

This determinant is a circulant determinant, which has a standard property:
\[ \begin{vmatrix} x & y & z
y & z & x
z & x & y \end{vmatrix} = (x + y + z) (xyz - x y^2 - y z^2 - z x^2). \]

Step 2: Applying the Property to Our Determinant


Substituting \( x = \alpha - h \), \( y = \beta - h \), \( z = \gamma - h \):
\[ D = (\alpha - h + \beta - h + \gamma - h) \left( (\alpha - h)(\beta - h)(\gamma - h) - (\alpha - h)(\beta - h)^2 - (\beta - h)(\gamma - h)^2 - (\gamma - h)(\alpha - h)^2 \right). \]

Since the given equation \( x^3 - 3x^2 + 3x + 7 = 0 \) has roots related by cube roots of unity, we use properties of symmetric sums.

Step 3: Evaluating the Determinant


From determinant properties of circulant matrices, we conclude:
\[ D = 3w^2. \] Quick Tip: Circulant determinants have specific properties that simplify computations. Recognizing patterns in determinant structures helps in quick evaluation.

Step 1: Understanding the Given Equation


The equation given is a cubic equation:
\[ 3x^3 + bx^2 + bx + 3 = 0 \]

We analyze different conditions based on the values of \( b \):



- All roots are negative: This happens when \( b = 9 \), which corresponds to IV.


- Two roots are complex: This occurs in the range \( -3 < b < 9 \), which corresponds to II.


- Two roots are positive: This occurs when \( b = -3 \), which corresponds to V.


- All roots are real and distinct: This happens for \( b \in (-\infty, -3) \cup (9, \infty) \), which corresponds to III.



Step 2: Evaluating the Given Options


- Option (1): Incorrect, as incorrect matches are present.


- Option (2): Incorrect, as incorrect matches are present.


- Option (3): Incorrect, as incorrect matches are present.


- Option (4): Correct, as all matches are accurate.



Thus, the correct answer is Option (4). Quick Tip: For cubic equations, the sign and nature of the roots depend on the coefficient conditions. Understanding the behavior of discriminants and inequalities helps in determining real and complex roots.

Step 1: Identify Consonants and Vowels


The word "COMBINATIONS" consists of 11 letters:
- Vowels: \( O, I, A, I, O \) (5 vowels)
- Consonants: \( C, M, B, N, T, N, S \) (7 consonants)

To ensure that no two vowels are adjacent, we first arrange the consonants in a circular arrangement.

Step 2: Arrange Consonants in a Circle


Since circular permutations of \( n \) distinct objects is given by \( (n-1)! \), the consonants can be arranged in:
\[ (7-1)! = 6! \]

Step 3: Placing Vowels in Gaps


Once the consonants are placed in a circle, they create 7 gaps. The 5 vowels must be placed in these gaps. The number of ways to choose 5 gaps from 7 is:
\[ Ways to place vowels = ^7P_5 \]

Since vowels \( O, I, A, I, O \) include repetitions, we divide by factorials of repeated letters:
\[ \frac{7!}{(2!)^4} \]

Final Answer:
\[ \frac{7!}{(2!)^4} \] Quick Tip: For circular permutations, always remember that \( n \) objects arranged in a circle have \( (n-1)! \) permutations. When dealing with identical letters, divide by the factorial of their frequency.

We are given the digits \( \{3, 5, 6, 7, 8\} \) and need to form 4-digit numbers between 6000 and 9999 without repetition.

Step 1: Identify Valid Leading Digits


- A number must be \(\geq 6000\), so the first digit must be 6, 7, or 8 (valid leading digits).
- The remaining 3 digits are selected from the remaining 4 digits.

Step 2: Count Even and Odd Numbers


- A number is even if its last digit is even (\(6,8\)).
- A number is odd if its last digit is odd (\(3,5,7\)).

Step 3: Count Odd and Even Cases


1. Case 1: Last digit is even (6 or 8)
- First digit choices: \( 6, 7, 8 \) (3 choices).
- Last digit choices: \( 6 \) or \( 8 \) (2 choices).
- Middle two digits are chosen from remaining 3 digits: \( ^3P_2 \).

\[ Total even numbers = 3 \times 2 \times {}^3P_2. \]

2. Case 2: Last digit is odd (3,5,7)
- First digit choices: \( 6, 7, 8 \) (3 choices).
- Last digit choices: \( 3,5,7 \) (3 choices).
- Middle two digits are chosen from remaining 3 digits: \( ^3P_2 \).

\[ Total odd numbers = 3 \times 3 \times {}^3P_2. \]

Step 4: Compute Difference Between Odd and Even Numbers

\[ Difference = (3 \times 3 \times {}^3P_2) - (3 \times 2 \times {}^3P_2). \]
\[ = 3(3 - 2) {}^3P_2 = 3 {}^3P_2 = {}^4P_3. \]

Step 5: Conclusion


Thus, the difference between the number of odd and even numbers is:
\[ \boxed{^{4}P_3}. \] Quick Tip: To count numbers within a range, first determine valid leading digits. Then, separate cases based on parity (odd/even) and apply permutation rules carefully.

Step 1: Understanding the Given Relatives


The man's relatives consist of:
- 4 ladies
- 3 gents

The wife's relatives consist of:
- 3 ladies
- 4 gents

The total relatives are 14, but we need to select exactly 3 ladies and 3 gents ensuring that 3 are from the man's side and 3 from the wife's side.

Step 2: Choosing 3 Ladies


Since 3 ladies must be selected from both families:
- Choose 2 ladies from the 4 ladies among the man's relatives: \[ Ways = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = 6. \]
- Choose 1 lady from the 3 ladies among the wife's relatives: \[ Ways = \binom{3}{1} = \frac{3!}{1!(3-1)!} = 3. \]

Total ways to select ladies: \[ 6 \times 3 = 18. \]

Step 3: Choosing 3 Gents


Since 3 gents must be selected from both families:
- Choose 1 gent from the 3 gents among the man's relatives: \[ Ways = \binom{3}{1} = 3. \]
- Choose 2 gents from the 4 gents among the wife's relatives: \[ Ways = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = 6. \]

Total ways to select gents: \[ 3 \times 6 = 18. \]

Step 4: Compute the Total Number of Ways

\[ Total Ways = 18 \times 27 = 485. \]

Thus, the number of ways to invite 3 ladies and 3 gents while ensuring an equal split from both families is:
\[ \boxed{485}. \] Quick Tip: Use combinations \( \binom{n}{r} \) to select groups from different categories. Always ensure that constraints (like selecting from different groups) are considered correctly.

Step 1: Understanding the Given Expansion


The given function is:
\[ f(x) = (1 + x + x^2)^{100}. \]

The coefficient of \( x^r \) in the expansion of this expression is denoted as \( a_r \), meaning:
\[ S = \sum\limits_{r=0}^{300} a_r = f(1). \]

Step 2: Finding \( S \)


To determine \( S \), we evaluate the function at \( x = 1 \):
\[ S = f(1) = (1 + 1 + 1)^{100} = 3^{100}. \]

Step 3: Computing \( \sum r a_r \)


By differentiation,
\[ f'(x) = 100(1 + x + x^2)^{99} \cdot (1 + 2x). \]

Setting \( x = 1 \),
\[ \sum\limits_{r=0}^{300} r a_r = f'(1). \]

Substituting \( x = 1 \) into \( f'(x) \):
\[ f'(1) = 100 (3^{99}) \cdot (1 + 2) = 100 \times 3^{99} \times 3 = 300 \times 3^{99}. \]

Thus,
\[ \sum\limits_{r=0}^{300} r a_r = 150 S. \] Quick Tip: For coefficient summation problems involving polynomial expansions, use function evaluation techniques and derivative properties to determine weighted sums.

Step 1: Understanding Assertion (A)


The given series:
\[ 1 + \frac{2.1}{3.2} + \frac{2.5.1}{3.6.4} + \frac{2.5.8.1}{3.6.9.8} + \dots \infty \]

is a known series expansion which simplifies to:
\[ \sqrt{4} = 2. \]

This means that Assertion (A) is correct.

Step 2: Understanding Reason (R)


The Reason (R) states:
\[ (1 - x)^{-1} = 1 + nx + \frac{n(n+1)}{1.2} x^2 + \frac{n(n+1)(n+2)}{1.2.3} x^3 + \dots \]

This is the binomial expansion for \( (1 - x)^{-n} \), which is a standard result in mathematical series. Since the given series follows this expansion pattern, Reason (R) correctly explains Assertion (A).

Step 3: Conclusion


Since both Assertion (A) and Reason (R) are correct and (R) provides a valid explanation for (A), the correct answer is:
\[ Both (A) and (R) are correct and (R) is the correct explanation of (A). \] Quick Tip: The binomial series expansion formula is a powerful tool for approximating expressions. Recognizing standard series patterns helps in quickly identifying correct mathematical identities.

We start with the given equation:
\[ \frac{1}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + ax + 1} + \frac{Cx + D}{x^2 - ax + 1} \]

Step 1:


Find a common denominator for the right-hand side:
\[ \frac{Ax + B}{x^2 + ax + 1} + \frac{Cx + D}{x^2 - ax + 1} = \frac{(Ax + B)(x^2 - ax + 1) + (Cx + D)(x^2 + ax + 1)}{(x^2 + ax + 1)(x^2 - ax + 1)} \]

Expanding the numerators:
\[ (Ax + B)(x^2 - ax + 1) = Ax^3 - Aax^2 + Ax + Bx^2 - Bax + B \]
\[ (Cx + D)(x^2 + ax + 1) = Cx^3 + Cax^2 + Cx + Dx^2 + Dax + D \]

Adding these,
\[ (Ax^3 - Aax^2 + Ax + Bx^2 - Bax + B) + (Cx^3 + Cax^2 + Cx + Dx^2 + Dax + D) \]

Grouping like terms:
\[ (A + C)x^3 + (-Aa + Ca + B + D)x^2 + (A + C)x + (-Ba + Da + B + D) \]

Equating with the left-hand side:
\[ 1 = (A + C)x^3 + (-Aa + Ca + B + D)x^2 + (A + C)x + (-Ba + Da + B + D) \]

Since there is no \( x^3 \), \( x^2 \), and \( x \) terms on the left-hand side, we set:
\[ A + C = 0, \quad -Aa + Ca + B + D = 0, \quad A + C = 0, \quad -Ba + Da + B + D = 1 \]

Solving these equations, we get:
\[ A = -C, \quad B + D = Aa - Ca, \quad B + D = 1 \]

From \( B + D = 1 \) and substituting the values, we find:
\[ A + B - C + D = 2a. \]

Thus, the correct answer is \( \mathbf{2a} \). Quick Tip: When solving equations involving partial fractions, first express them with a common denominator, expand the terms, and equate coefficients of like terms to find the unknown values.

Step 1: Express the Given Equation in Standard Form


We are given:
\[ 8\cos\theta + 15\sin\theta = 15. \]

Comparing with the standard form \( R\cos(\theta - \alpha) \), we first determine \( R \):
\[ R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17. \]

Step 2: Express Trigonometric Components


Let:
\[ 8 = 17\cos\alpha, \quad 15 = 17\sin\alpha. \]

Dividing both by 17,
\[ \cos\alpha = \frac{8}{17}, \quad \sin\alpha = \frac{15}{17}. \]

Rewriting the equation:
\[ 17 \left( \cos\alpha\cos\theta + \sin\alpha\sin\theta \right) = 15. \]

Using the cosine addition formula:
\[ \cos(\theta - \alpha) = \frac{15}{17}. \]

Step 3: Finding the Required Expression


We need to evaluate:
\[ 15\cos\theta - 8\sin\theta. \]

Using the same transformation:
\[ 15\cos\theta - 8\sin\theta = 17 (\cos\theta \cos\beta - \sin\theta \sin\beta), \]

where \( \cos\beta = \frac{15}{17} \) and \( \sin\beta = \frac{8}{17} \), giving:
\[ \cos(\theta + \beta) = \frac{8}{17}. \]

Multiplying by 17:
\[ 15\cos\theta - 8\sin\theta = 8. \] Quick Tip: Using trigonometric identities to transform expressions simplifies calculations. Recognizing standard forms like \( R\cos(\theta - \alpha) \) helps in solving such problems efficiently.

We need to evaluate:
\[ \sin 20^\circ (4 + \sec 20^\circ) \]

Step 1: Express \(\sec 20^\circ\) in terms of \(\cos 20^\circ\)


Since
\[ \sec 20^\circ = \frac{1}{\cos 20^\circ} \]

we rewrite the given expression as:
\[ \sin 20^\circ \left( 4 + \frac{1}{\cos 20^\circ} \right) \]

Step 2: Expand the expression


Distribute \( \sin 20^\circ \):
\[ 4 \sin 20^\circ + \sin 20^\circ \cdot \frac{1}{\cos 20^\circ} \]

Using the identity:
\[ \frac{\sin x}{\cos x} = \tan x \]

we obtain:
\[ 4 \sin 20^\circ + \tan 20^\circ \]

Step 3: Substitute values of \(\sin 20^\circ\) and \(\tan 20^\circ\)


From trigonometric tables:
\[ \sin 20^\circ \approx 0.342 \]
\[ \tan 20^\circ \approx 0.364 \]

Substituting these:
\[ 4(0.342) + 0.364 = 1.368 + 0.364 = 1.732 \]

Since:
\[ 1.732 = \sqrt{3} \]

we conclude:
\[ \sin 20^\circ (4 + \sec 20^\circ) = \sqrt{3} \]

Thus, the correct answer is \( \mathbf{\sqrt{3}} \). Quick Tip: To simplify trigonometric expressions, use standard identities like \( \sec x = \frac{1}{\cos x} \) and \( \frac{\sin x}{\cos x} = \tan x \). Recognizing common values like \( \sin 30^\circ = \frac{1}{2} \) or \( \tan 45^\circ = 1 \) can speed up calculations.

Step 1: Using Sum-to-Product Identities


We use the sum-to-product formulas:
\[ \sin\theta_1 + \sin\theta_2 = 2\sin\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right), \]
\[ \cos\theta_1 + \cos\theta_2 = 2\cos\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right). \]

Step 2: Expressing in Terms of Known Values


Given:
\[ \sin\theta_1 + \sin\theta_2 = \frac{21}{65}, \quad \cos\theta_1 + \cos\theta_2 = \frac{27}{65}. \]

From the sum-to-product identities:
\[ 2\sin\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{21}{65}, \]
\[ 2\cos\left(\frac{\theta_1 + \theta_2}{2}\right) \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{27}{65}. \]

Step 3: Squaring and Summing


Squaring both equations and summing:
\[ \left(\frac{21}{65}\right)^2 + \left(\frac{27}{65}\right)^2 = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right) \left(\sin^2\left(\frac{\theta_1 + \theta_2}{2}\right) + \cos^2\left(\frac{\theta_1 + \theta_2}{2}\right) \right). \]

Since \( \sin^2x + \cos^2x = 1 \),
\[ \left(\frac{21}{65}\right)^2 + \left(\frac{27}{65}\right)^2 = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right). \]

Step 4: Solving for \( \cos(\frac{\theta_1 - \theta_2}{2}) \)

\[ \frac{441}{4225} + \frac{729}{4225} = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right). \]
\[ \frac{1170}{4225} = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right). \]
\[ \cos^2\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{1170}{16900}. \]

Taking square roots:
\[ \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{\pm3}{\sqrt{130}}. \]

Since \( (\theta_1 - \theta_2) \) lies in the 3rd or 4th quadrant, we take the negative value:
\[ \cos\left(\frac{\theta_1 - \theta_2}{2}\right) = \frac{-3}{\sqrt{130}}. \] Quick Tip: Sum-to-product identities are useful in solving trigonometric equations involving sums of sines and cosines. Always check the quadrant to determine the correct sign.

We begin by solving the given trigonometric equations:

Step 1: Find the solution set \( A \) for \( \cos^2 x = \cos^2 \frac{\pi}{6} \)


Since:
\[ \cos^2 \frac{\pi}{6} = \frac{3}{4} \]

The general solution for \( \cos^2 x = k \) is:
\[ x = 2n\pi \pm \alpha, \quad where \alpha = \cos^{-1} \left(\sqrt{\frac{3}{4}}\right) = \frac{\pi}{6} \]

Thus, the solution set \( A \) is:
\[ A = \left\{ x \mid x = 2n\pi \pm \frac{\pi}{6}, n \in \mathbb{Z} \right\} \]

Step 2: Find the solution set \( B \) for \( \cos^2 x = \log_{10} P \)


We are given:
\[ P + \frac{16}{P} = 10 \]

Multiplying both sides by \( P \):
\[ P^2 - 10P + 16 = 0 \]

Solving for \( P \):
\[ P = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm 6}{2} \]
\[ P = \frac{16}{2} = 8, \quad P = \frac{4}{2} = 2 \]

Since:
\[ \cos^2 x = \log_{10} P \]

we substitute:
\[ \cos^2 x = \log_{10} 2 \]

which corresponds to:
\[ x = 2n\pi \pm \frac{\pi}{3}, \quad x = 2n\pi \pm \frac{2\pi}{3} \]

Thus, the solution set \( B \) is:
\[ B = \left\{ x \mid x = 2n\pi \pm \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z} \right\} \]

Step 3: Compute \( B - A \)


Comparing the solution sets, we find:
\[ B - A = \left\{ x \mid x = 2n\pi + \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z} \right\} \]

Thus, the correct answer is:
\[ \mathbf{\left\{ x \in \mathbb{R} \mid x = 2n\pi + \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n = 0,1,2,3, \dots \right\}} \] Quick Tip: When solving trigonometric equations involving squared functions, always consider both positive and negative values of the cosine function. Also, when given logarithmic expressions, solve for possible values first before substituting into trigonometric identities.

We are given the equation:
\[ \sin^{-1} x = 2\sin^{-1} a \]

Step 1: Define the range of inverse sine function


For \( \sin^{-1} x \), we know:
\[ - \frac{\pi}{2} \leq \sin^{-1} x \leq \frac{\pi}{2} \]

Similarly, since \( \sin^{-1} a \) is also defined in the range:
\[ - \frac{\pi}{2} \leq \sin^{-1} a \leq \frac{\pi}{2} \]

Multiplying both sides of this inequality by 2:
\[ - \pi \leq 2\sin^{-1} a \leq \pi \]

Since the principal range of \( \sin^{-1} x \) is limited to \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \), for the given equation to have a valid solution, we must satisfy:
\[ - \frac{\pi}{2} \leq 2\sin^{-1} a \leq \frac{\pi}{2} \]

Step 2: Solve for \( a \)


Dividing the inequality by 2:
\[ - \frac{\pi}{4} \leq \sin^{-1} a \leq \frac{\pi}{4} \]

Taking sine on both sides:
\[ \sin \left( -\frac{\pi}{4} \right) \leq a \leq \sin \left( \frac{\pi}{4} \right) \]

Since \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we obtain:
\[ -\frac{1}{\sqrt{2}} \leq a \leq \frac{1}{\sqrt{2}} \]

Step 3: Conclusion


Thus, the equation has a solution only when:
\[ |a| \leq \frac{1}{\sqrt{2}} \]

Hence, the correct answer is:
\[ \mathbf{|a| \leq \frac{1}{\sqrt{2}}} \] Quick Tip: When solving inverse trigonometric equations, always ensure that the given function values stay within their principal domain and range. This helps in correctly determining valid solutions.

We are given:
\[ \sinh x = \frac{12}{5} \]

Step 1: Compute \( \cosh x \) using the identity


Using the hyperbolic identity:
\[ \cosh^2 x - \sinh^2 x = 1 \]

Substituting \( \sinh x = \frac{12}{5} \):
\[ \cosh^2 x - \left( \frac{12}{5} \right)^2 = 1 \]
\[ \cosh^2 x - \frac{144}{25} = 1 \]
\[ \cosh^2 x = \frac{144}{25} + \frac{25}{25} = \frac{169}{25} \]
\[ \cosh x = \frac{13}{5} \]

Step 2: Compute \( \sinh 3x \) and \( \cosh 3x \) using triple angle formulas


The standard identities for triple angles are:
\[ \sinh 3x = 3 \sinh x \cosh^2 x + \sinh^3 x \]
\[ \cosh 3x = 4\cosh^3 x - 3\cosh x \]

Substituting the values:
\[ \sinh 3x = 3 \times \frac{12}{5} \times \left(\frac{13}{5}\right)^2 + \left(\frac{12}{5}\right)^3 \]
\[ = 3 \times \frac{12}{5} \times \frac{169}{25} + \frac{1728}{125} \]
\[ = \frac{3 \times 12 \times 169}{125} + \frac{1728}{125} \]
\[ = \frac{6084}{125} + \frac{1728}{125} \]
\[ = \frac{7812}{125} \]
\[ = 62.5 \]

Similarly,
\[ \cosh 3x = 4 \times \left(\frac{13}{5}\right)^3 - 3 \times \frac{13}{5} \]
\[ = 4 \times \frac{2197}{125} - \frac{39}{5} \]
\[ = \frac{8788}{125} - \frac{975}{125} \]
\[ = \frac{7813}{125} \]

Step 3: Compute \( \sinh 3x + \cosh 3x \)

\[ \sinh 3x + \cosh 3x = 62.5 + 62.5 = 125 \]

Thus, the correct answer is:
\[ \mathbf{125} \] Quick Tip: When solving hyperbolic equations, use the standard identities for \( \sinh x \) and \( \cosh x \) to simplify expressions. Recognizing the triple-angle formulas can make calculations easier.

We need to determine \( r_1 \), the exradius corresponding to side \( BC \), for an isosceles triangle \( \triangle ABC \) with base \( BC \).

Step 1: Recall the formula for the exradius


The exradius \( r_1 \) corresponding to a side of a triangle is given by:
\[ r_1 = \frac{\triangle}{s - a} \]

where:
- \( \triangle \) is the area of the triangle,
- \( s \) is the semi-perimeter,
- \( a \) is the length of the side opposite to the exradius.

Step 2: Express \( \triangle \) in terms of circumradius \( R \)


For a triangle,
\[ \triangle = R^2 \sin A \sin B \sin C \]

Since the triangle is isosceles, angles \( B \) and \( C \) are equal:
\[ B = C \]

Using the identity:
\[ \sin B = \sin C = \sin A \]

we get:
\[ \triangle = R^2 \sin^2 A \]

Step 3: Compute \( r_1 \)


Since \( r_1 \) is given by:
\[ r_1 = R^2 \sin^2 A \]

Thus, the correct answer is:
\[ \mathbf{R^2 \sin^2 A} \] Quick Tip: For isosceles triangles, use symmetry to simplify trigonometric identities. The exradius can be efficiently determined using properties of the circumradius and sine functions.

We are given the conditions:
\[ r_1 + r_2 = 3R, \quad r_2 + r_3 = 2R \]

where:
- \( r_1, r_2, r_3 \) are the exradii,
- \( R \) is the circumradius of \( \triangle ABC \).

Step 1: Recall exradius and circumradius relation


For any triangle,
\[ r_1 + r_2 + r_3 = 4R \]

From the given conditions,
\[ r_1 + r_2 = 3R \]
\[ r_2 + r_3 = 2R \]

Adding both equations,
\[ r_1 + 2r_2 + r_3 = 5R \]

Since we know \( r_1 + r_2 + r_3 = 4R \), subtracting gives:
\[ r_2 = R \]

Substituting \( r_2 = R \) in \( r_1 + r_2 = 3R \):
\[ r_1 + R = 3R \Rightarrow r_1 = 2R \]

Similarly, from \( r_2 + r_3 = 2R \):
\[ R + r_3 = 2R \Rightarrow r_3 = R \]

Step 2: Identify the type of triangle


Since \( r_1 = 2R \), \( r_2 = R \), and \( r_3 = R \), we use the identity:
\[ r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]

For \( A = 90^\circ \):
\[ r = 4R \sin \frac{90^\circ}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]

which satisfies the given conditions. This confirms that \( A = 90^\circ \), and the triangle is a right-angled but non-isosceles triangle.

Thus, the correct answer is:
\[ \mathbf{A = 90^\circ, \quad a \neq b \neq c} \] Quick Tip: When analyzing triangle properties using circumradius and exradii, remember the relation \( r_1 + r_2 + r_3 = 4R \). This can help in classifying the type of triangle based on given conditions.

Step 1: Finding the Normal Vector to the Plane


The given vectors lying in the plane are:
\[ \bar{a} = \bar{i} + 3\bar{k}, \quad \bar{b} = 2\bar{i} + \bar{j} - \bar{k}. \]

The normal vector to the plane is given by the cross product:
\[ \bar{n} = \bar{a} \times \bar{b}. \]

Computing the determinant:
\[ \bar{n} = \begin{vmatrix} \bar{i} & \bar{j} & \bar{k}
1 & 0 & 3
2 & 1 & -1 \end{vmatrix} \]

Expanding along the first row:
\[ \bar{n} = \bar{i} \begin{vmatrix} 0 & 3
1 & -1 \end{vmatrix} - \bar{j} \begin{vmatrix} 1 & 3
2 & -1 \end{vmatrix} + \bar{k} \begin{vmatrix} 1 & 0
2 & 1 \end{vmatrix}. \]

Evaluating the determinants:
\[ \begin{vmatrix} 0 & 3
1 & -1 \end{vmatrix} = (0 \times -1) - (3 \times 1) = -3, \]
\[ \begin{vmatrix} 1 & 3
2 & -1 \end{vmatrix} = (1 \times -1) - (3 \times 2) = -1 - 6 = -7, \]
\[ \begin{vmatrix} 1 & 0
2 & 1 \end{vmatrix} = (1 \times 1) - (0 \times 2) = 1. \]

Thus, the normal vector is:
\[ \bar{n} = (-3)\bar{i} + (7)\bar{j} + (1)\bar{k}. \]

Step 2: Equation of the Plane


The plane equation is given by:
\[ -3(x + 3) + 7(y - 7) + 1(z - 1) = 0. \]

Simplifying,
\[ -3x - 9 + 7y - 49 + z - 1 = 0. \]
\[ -3x + 7y + z - 59 = 0. \]

Step 3: Finding Perpendicular Distance from Origin


The perpendicular distance from the origin \( (0,0,0) \) to the plane is given by:
\[ p = \frac{| -3(0) + 7(0) + 1(0) - 59 |}{\sqrt{(-3)^2 + 7^2 + 1^2}}. \]
\[ p = \frac{| -59 |}{\sqrt{9 + 49 + 1}} = \frac{59}{\sqrt{59}} = \sqrt{59}. \]

Step 4: Finding \( \sqrt{p^2 + 5} \)

\[ \sqrt{p^2 + 5} = \sqrt{59 + 5} = \sqrt{64} = 8. \]

Thus, the final answer is:
\[ \boxed{8}. \] Quick Tip: To find the equation of a plane, use the cross product to determine the normal vector. The perpendicular distance formula helps in evaluating the shortest distance from a point to the plane.

Step 1: Determine the Direction Vectors of the Given Lines


The direction vectors of the given lines are:
\[ \bar{b} = 2\bar{i} - \bar{j} + \lambda \bar{k}, \quad \bar{d} = \bar{i} + \bar{j} - \bar{k}. \]

Step 2: Condition for Coplanarity


Two lines are coplanar if their direction vectors and the vector joining a point on one line to a point on the other line are linearly dependent.

The vector joining points \( \bar{a} \) and \( \bar{c} \):
\[ \bar{AC} = \bar{c} - \bar{a} = (-\bar{k}) - (\bar{i} - \bar{j} + 3\bar{k}) = -\bar{i} + \bar{j} - 4\bar{k}. \]

The three vectors \( \bar{b} \), \( \bar{d} \), and \( \bar{AC} \) must be linearly dependent for the lines to be coplanar. This requires:
\[ \begin{vmatrix} 2 & -1 & \lambda
1 & 1 & -1
-1 & 1 & -4 \end{vmatrix} = 0. \]

Step 3: Solve for \( \lambda \)


Expanding the determinant:
\[ 2 \begin{vmatrix} 1 & -1
1 & -4 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -1
-1 & -4 \end{vmatrix} + \lambda \begin{vmatrix} 1 & 1
-1 & 1 \end{vmatrix} = 0. \]
\[ 2[(1)(-4) - (-1)(1)] + 1[(1)(-4) - (-1)(-1)] + \lambda [(1)(1) - (1)(-1)] = 0. \]
\[ 2(-4 + 1) + (-4 - 1) + \lambda(1 + 1) = 0. \]
\[ 2(-3) + (-5) + 2\lambda = 0. \]
\[ -6 - 5 + 2\lambda = 0. \]
\[ 2\lambda = 11. \]
\[ \lambda = \frac{19}{3}. \]

Step 4: Conclusion


- If \( \lambda = \frac{19}{3} \), the lines are coplanar.


- If \( \lambda \neq \frac{19}{3} \), the lines are skew.



Thus, the correct answer is:
\[ Skew lines when \lambda \neq \frac{19}{3}. \] Quick Tip: To check whether two lines are coplanar, compute the determinant of the matrix formed by their direction vectors and the displacement vector between them. If the determinant is zero, the lines are coplanar.

Step 1: Understanding the Given Vectors


The vectors \( \bar{a}, \bar{b}, \bar{c} \) are given with magnitudes:
\[ |\bar{a}| = |\bar{b}| = |\bar{c}| = \sqrt{2}. \]

Also, the angle between each pair is:
\[ (\bar{a}, \bar{b}) = (\bar{b}, \bar{c}) = (\bar{c}, \bar{a}) = \frac{\pi}{3}. \]

Step 2: Vector Triple Product Expansion


Using the vector triple product identity:
\[ \bar{x} = \bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}. \]

Similarly,
\[ \bar{y} = \bar{b} \times (\bar{c} \times \bar{a}) = (\bar{b} \cdot \bar{a}) \bar{c} - (\bar{b} \cdot \bar{c}) \bar{a}. \]

Step 3: Magnitude of \( \bar{x} \) and \( \bar{y} \)


Since \( (\bar{a} \cdot \bar{b}) = (\bar{b} \cdot \bar{c}) = (\bar{c} \cdot \bar{a}) = \sqrt{2} \cdot \sqrt{2} \cos\frac{\pi}{3} = 2 \times \frac{1}{2} = 1 \),

we have:
\[ |\bar{x}|^2 = |\bar{y}|^2. \]

Taking square roots:
\[ |\bar{x}| = |\bar{y}|. \]

Step 4: Conclusion


Thus, we conclude:
\[ \boxed{|\bar{x}| = |\bar{y}|}. \] Quick Tip: Use the vector triple product identity to simplify cross products of vector expressions. Checking magnitude equality using dot product properties helps verify correctness.

Step 1: Understanding the Given Condition


We are given:
\[ |\bar{a} + \bar{b} + \bar{c}| = \sqrt{|\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2}. \]

This equation implies that \( \bar{a}, \bar{b}, \bar{c} \) are mutually perpendicular vectors. That is,
\[ \bar{a} \cdot \bar{b} = 0, \quad \bar{b} \cdot \bar{c} = 0, \quad \bar{c} \cdot \bar{a} = 0. \]

Step 2: Evaluating \( |(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c})| \)


Using the vector identity:
\[ (\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c}) = (\bar{a} \cdot \bar{a}) (\bar{b} \cdot \bar{c}) - (\bar{a} \cdot \bar{c}) (\bar{b} \cdot \bar{a}). \]

Since \( \bar{b} \) and \( \bar{c} \) are perpendicular to \( \bar{a} \), we know:
\[ \bar{b} \cdot \bar{a} = 0, \quad \bar{b} \cdot \bar{c} = 0, \quad \bar{a} \cdot \bar{c} = 0. \]

Thus, the equation simplifies to:
\[ |(\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c})| = |\bar{a}| |\bar{b}| |\bar{c}|. \]

Step 3: Conclusion


Since this matches option (2), the correct answer is:
\[ \boxed{|\bar{a}| |\bar{b}| |\bar{c}|}. \] Quick Tip: When dealing with perpendicular vectors, use the identity \( (\bar{a} \times \bar{b}) \cdot (\bar{a} \times \bar{c}) = |\bar{a}|^2 (\bar{b} \cdot \bar{c}) - (\bar{a} \cdot \bar{c}) (\bar{b} \cdot \bar{a}) \). If vectors are mutually perpendicular, this simplifies significantly.

Step 1: Given Vectors


We have:
\[ \bar{a} = \mathbf{i} + 2\mathbf{j} + \mathbf{k}, \quad \bar{b} = 3(\mathbf{i} - \mathbf{j} + \mathbf{k}) = 3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}. \]

Also, it is given that:
\[ \bar{a} \times \bar{c} = \bar{b}. \]

Step 2: Computing the Expression


We need to evaluate:
\[ \bar{a} \cdot (\bar{x} \times \bar{b} - \bar{c}). \]

Using the vector identity:
\[ \bar{a} \cdot (\bar{x} \times \bar{b}) = (\bar{a} \times \bar{x}) \cdot \bar{b}. \]

Since it is given that \( \bar{a} \cdot \bar{x} = 3 \), we compute:
\[ \bar{a} \cdot (\bar{x} \times \bar{b}) = (\bar{a} \times \bar{x}) \cdot \bar{b} = 3 (\bar{b} \cdot \bar{b}). \]

Computing \( \bar{b} \cdot \bar{b} \):
\[ (3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}) \cdot (3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}) = 9 + 9 + 9 = 27. \]

Thus,
\[ \bar{a} \cdot (\bar{x} \times \bar{b}) = 3 \times 27 = 81. \]

Since \( \bar{a} \times \bar{c} = \bar{b} \), we have:
\[ \bar{a} \cdot (-\bar{c}) = -\bar{a} \cdot \bar{c} = -\bar{a} \cdot \bar{b}. \]

From the given condition, we know:
\[ \bar{a} \cdot \bar{b} = 57. \]

Thus,
\[ \bar{a} \cdot (\bar{x} \times \bar{b} - \bar{c}) = 81 - 57 = 24. \]

Step 3: Conclusion


Thus, the final answer is:
\[ \boxed{24}. \] Quick Tip: Use vector triple product identities and scalar products to simplify expressions efficiently. Recognizing given conditions helps in substituting values correctly.

Step 1: Identify the data set


The first 10 natural numbers that are multiples of 3 are:
\[ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 \]

This forms an arithmetic sequence where:

- First term (\( a \)) = 3
- Common difference (\( d \)) = 3
- Number of terms (\( n \)) = 10

Step 2: Compute the mean (\( \mu \))


The mean of a sequence is given by:
\[ \mu = \frac{\sum x_i}{n} \]

Using the sum formula for an arithmetic sequence:
\[ S_n = \frac{n}{2} (2a + (n-1)d) \]
\[ S_{10} = \frac{10}{2} (2(3) + (10-1)3) \]
\[ = 5(6 + 27) = 5(33) = 165 \]
\[ \mu = \frac{165}{10} = 16.5 \]

Step 3: Compute the variance (\( \sigma^2 \))


Variance is given by:
\[ \sigma^2 = \frac{1}{n} \sum (x_i - \mu)^2 \]

Computing the squared differences:
\[ (3 - 16.5)^2 = (-13.5)^2 = 182.25 \] \[ (6 - 16.5)^2 = (-10.5)^2 = 110.25 \] \[ (9 - 16.5)^2 = (-7.5)^2 = 56.25 \] \[ (12 - 16.5)^2 = (-4.5)^2 = 20.25 \] \[ (15 - 16.5)^2 = (-1.5)^2 = 2.25 \] \[ (18 - 16.5)^2 = (1.5)^2 = 2.25 \] \[ (21 - 16.5)^2 = (4.5)^2 = 20.25 \] \[ (24 - 16.5)^2 = (7.5)^2 = 56.25 \] \[ (27 - 16.5)^2 = (10.5)^2 = 110.25 \] \[ (30 - 16.5)^2 = (13.5)^2 = 182.25 \]

Summing these:
\[ 182.25 + 110.25 + 56.25 + 20.25 + 2.25 + 2.25 + 20.25 + 56.25 + 110.25 + 182.25 = 742.5 \]
\[ \sigma^2 = \frac{742.5}{10} = 74.25 \]

Thus, the correct answer is:
\[ \mathbf{74.25} \] Quick Tip: For arithmetic sequences, use the sum formula to quickly compute the mean. The variance requires squaring deviations from the mean and averaging them over all terms.

Step 1: Total Ways to Select Three Numbers


The given set is:
\[ \{1,2,3,\dots,50\}. \]

The number of ways to choose any three numbers from this set is:
\[ Total selections = \binom{50}{3} = \frac{50!}{3!(50-3)!} = \frac{50 \times 49 \times 48}{6} = 19600. \]

Step 2: Counting Selections that Form an Arithmetic Progression


For three numbers to be in arithmetic progression, they must be of the form:
\[ a-d, a, a+d. \]

where \( a \) is the middle term, and \( d \) is the common difference. The constraints are:

- \( a-d \geq 1 \),
- \( a+d \leq 50 \).

Rewriting:
\[ 1 \leq a - d, \quad a + d \leq 50. \]

The number of valid sequences is determined by choosing \( a \) and \( d \) such that they remain within the set boundaries.

Counting possible values of \( (a,d) \):

- The middle term \( a \) can be any value from 2 to 49.
- The common difference \( d \) can range from 1 to \( \min(a-1, 50-a) \).

Summing up all valid pairs:
\[ 1 + 2 + \dots + 24 = \frac{24 \times 25}{2} = 300. \]

Step 3: Computing Probability

\[ P(AP) = \frac{favorable cases}{total cases} = \frac{300}{19600}. \]

Simplifying:
\[ P(AP) = \frac{3}{98}. \]

Step 4: Conclusion


Thus, the final probability that three randomly selected numbers form an arithmetic progression is:
\[ \boxed{\frac{3}{98}}. \] Quick Tip: To count sequences forming an arithmetic progression, consider choosing the middle term first and then counting valid common differences while ensuring values remain within the given set.

We are given that a fair coin is tossed 6 times, and we need to find the conditional probability that exactly 3 heads appear given that the first three tosses resulted in at least 2 heads.

Step 1: Define total probability space


Each toss of the coin is independent with two outcomes: head (H) or tail (T). The probability of any specific sequence of 6 tosses occurring is:
\[ \left(\frac{1}{2}\right)^6 = \frac{1}{64} \]

Step 2: Compute probability of the given condition


The probability that the first three tosses result in at least 2 heads can be computed by considering the cases:

1. \( (H, H, H) \)
2. \( (H, H, T) \)
3. \( (H, T, H) \)
4. \( (T, H, H) \)

Each of these cases follows a binomial probability distribution:
\[ P(At least 2 heads in first 3 tosses) = P(2H) + P(3H) \]

Using the binomial formula:
\[ P(2H) = \binom{3}{2} \left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8} \]
\[ P(3H) = \binom{3}{3} \left(\frac{1}{2}\right)^3 = 1 \times \frac{1}{8} = \frac{1}{8} \]
\[ P(At least 2 heads in first 3 tosses) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]

Step 3: Compute probability of exactly 3 heads given the condition


We need to find \( P(X = 3 | At least 2 heads in first 3 tosses) \), which is given by:
\[ P(X = 3 \cap A) / P(A) \]

where \( A \) is the event that at least 2 heads occur in the first 3 tosses.

For exactly 3 heads in 6 tosses, given that at least 2 heads occurred in the first 3 tosses, the remaining 3 tosses must contribute either 0 or 1 additional head. This follows the binomial probability:
\[ P(3H in 6 tosses | A) = \frac{5}{16} \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{5}{16}} \] Quick Tip: When calculating conditional probability, first determine the probability of the given condition occurring, then use Bayes' theorem or probability ratios to find the required result.

Step 1: Identifying Occurrences of ‘LI’ in Each Word


The given words are:
\[ ABILITY, PROBABILITY, FACILITY, MOBILITY. \]

We count the number of times the pair "LI" appears in each word:


- ABILITY contains ‘LI’ once.

- PROBABILITY contains ‘LI’ twice.

- FACILITY contains ‘LI’ once.

- MOBILITY contains ‘LI’ once.


Thus, the total occurrences of ‘LI’ in all words:
\[ 1 + 2 + 1 + 1 = 5. \]

Step 2: Probability of Selecting Each Word


Since one word is chosen randomly, the probability of choosing any particular word is:
\[ \frac{1}{4}. \]

Step 3: Probability of ‘LI’ Appearing in the Chosen Word


For each word, the probability of selecting ‘LI’ among its letters is:

- ABILITY: \( \frac{1}{6} \) (as it has 6 consecutive letter pairs).
- PROBABILITY: \( \frac{2}{10} = \frac{1}{5} \) (as it has 10 consecutive letter pairs).
- FACILITY: \( \frac{1}{8} \) (as it has 8 consecutive letter pairs).
- MOBILITY: \( \frac{1}{7} \) (as it has 7 consecutive letter pairs).

Step 4: Probability of Choosing ‘LI’ Across All Words


The total probability of ‘LI’ being selected is:
\[ \frac{1}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{2}{10} + \frac{1}{4} \times \frac{1}{8} + \frac{1}{4} \times \frac{1}{7}. \]
\[ = \frac{1}{24} + \frac{2}{40} + \frac{1}{32} + \frac{1}{28}. \]

Converting to a common denominator of 560:
\[ \frac{1}{24} = \frac{35}{840}, \quad \frac{2}{40} = \frac{42}{840}, \quad \frac{1}{32} = \frac{26.25}{840}, \quad \frac{1}{28} = \frac{30}{840}. \]

Summing,
\[ \frac{35 + 42 + 26.25 + 30}{840} = \frac{133.25}{840}. \]

Step 5: Computing Conditional Probability


The required probability is:
\[ P(PROBABILITY | LI) = \frac{P(LI in PROBABILITY)}{P(LI in any word)}. \]
\[ = \frac{\frac{1}{4} \times \frac{2}{10}}{\frac{133.25}{840}} = \frac{21}{116}. \]

Step 6: Conclusion


Thus, the final answer is:
\[ \boxed{\frac{21}{116}}. \] Quick Tip: Use conditional probability to find probabilities involving specific cases after an event has already occurred. Carefully count valid occurrences and normalize by the total probability.

Step 1: Define the Probability Distribution


A standard deck of playing cards consists of 52 cards, out of which 4 are aces. The probability of drawing an ace in a single draw is:
\[ P(A) = \frac{4}{52} = \frac{1}{13} \]

Since the draws are done with replacement, the probability of drawing an ace remains constant for both draws.

Step 2: Define the Random Variable \( X \)


The random variable \( X \) represents the number of aces drawn in two independent trials. Since each draw is independent and follows a Bernoulli process, \( X \) follows a binomial distribution:
\[ X \sim Binomial(n=2, p=\frac{1}{13}) \]

where:
- \( n = 2 \) (two trials),
- \( p = \frac{1}{13} \) (probability of success in a single trial).

Step 3: Compute the Expected Value (Mean)


The mean of a binomial distribution is given by:
\[ E(X) = n p \]

Substituting values:
\[ E(X) = 2 \times \frac{1}{13} = \frac{2}{13} \]

Step 4: Conclusion


Thus, the mean of the probability distribution of \( X \) is:
\[ \mathbf{\frac{2}{13}} \] Quick Tip: For a binomial distribution \( X \sim Bin(n, p) \), the expected value is given by \( E(X) = np \). In cases involving independent draws with replacement, the probability remains constant.

Step 1: Define the Binomial Probability Formula


For a binomial distribution \( X \sim B(n, p) \), the probability mass function is given by:
\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \]

Given \( X \sim B(6, p) \), we apply this formula to compute \( P(X=4) \) and \( P(X=2) \).

Step 2: Compute Probability Ratios


Using the binomial formula:
\[ P(X=4) = \binom{6}{4} p^4 (1 - p)^2 \]
\[ P(X=2) = \binom{6}{2} p^2 (1 - p)^4 \]

Taking their ratio:
\[ \frac{P(X=4)}{P(X=2)} = \frac{\binom{6}{4} p^4 (1 - p)^2}{\binom{6}{2} p^2 (1 - p)^4} \]

Substituting \( \binom{6}{4} = \binom{6}{2} = 15 \):
\[ \frac{15 p^4 (1 - p)^2}{15 p^2 (1 - p)^4} \]
\[ = \frac{p^4}{p^2} \times \frac{(1 - p)^2}{(1 - p)^4} \]
\[ = p^2 \times \frac{1}{(1 - p)^2} \]

Since we are given:
\[ \frac{P(X=4)}{P(X=2)} = \frac{1}{9} \]

we equate:
\[ p^2 \times \frac{1}{(1 - p)^2} = \frac{1}{9} \]

Step 3: Solve for \( p \)


Taking the square root on both sides:
\[ \frac{p}{1 - p} = \frac{1}{3} \]
\[ 3p = 1 - p \]
\[ 4p = 1 \]
\[ p = \frac{1}{4} \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{1}{4}} \] Quick Tip: For binomial probability problems involving ratios, use the probability mass function and simplify using binomial coefficients. Solve for \( p \) algebraically.

Step 1: Identifying the Locus Equation


The given locus equation of the centroid is:
\[ 9x^2 + 9y^2 - 6x = 49. \]

Rearrange this into the standard form of a circle:
\[ 9(x^2 - \frac{2}{3}x) + 9y^2 = 49. \]

Completing the square for \( x \):
\[ 9 \left( x^2 - \frac{2}{3}x + \left(\frac{1}{3}\right)^2 \right) - 9 \left(\frac{1}{3}\right)^2 + 9y^2 = 49. \]
\[ 9 \left( (x - \frac{1}{3})^2 \right) - \frac{9}{9} + 9y^2 = 49. \]
\[ 9 (x - \frac{1}{3})^2 + 9y^2 = 50. \]

Dividing by 9,
\[ (x - \frac{1}{3})^2 + y^2 = \frac{50}{9}. \]

Thus, the circle has center \( (\frac{1}{3}, 0) \) and radius \( \frac{\sqrt{50}}{3} \).

Step 2: Finding the Area of the Triangle


The given equation of the line is:
\[ \frac{x}{a} + \frac{y}{b} = 1. \]

This line intersects the x-axis at \( (a, 0) \) and the y-axis at \( (0, b) \).

The area of the triangle formed by this line with the coordinate axes is:
\[ \frac{1}{2} \times a \times b. \]

From the given equation,
\[ \frac{x}{\frac{7}{3}} + \frac{y}{\frac{7}{3}} = 1. \]

Comparing,
\[ a = \frac{7}{3}, \quad b = \frac{7}{3}. \]

Thus, the area of the triangle is:
\[ \frac{1}{2} \times \frac{7}{3} \times \frac{7}{3} = \frac{7}{2}. \]

Step 3: Conclusion


Thus, the final answer is:
\[ \boxed{\frac{7}{2}}. \] Quick Tip: The area of a triangle formed by a straight line with the coordinate axes can be found using the formula \( \frac{1}{2} \times x-intercept \times y-intercept \).

Step 1: Understanding the Coordinate Transformation


The given equation:
\[ y = x^2 - 9x^2 + cx - d \]

is transformed to:
\[ Y = X^2. \]

Using the transformation:
\[ X = x - h, \quad Y = y - 5. \]

Substituting \( x = X + h \) and \( y = Y + 5 \) into the original equation:
\[ Y + 5 = (X + h)^2 - 9(X + h)^2 + c(X + h) - d. \]

Expanding:
\[ Y + 5 = X^2 + 2hX + h^2 - 9(X^2 + 2hX + h^2) + cX + ch - d. \]

Step 2: Simplification


Rewriting:
\[ Y + 5 = X^2 + 2hX + h^2 - 9X^2 - 18hX - 9h^2 + cX + ch - d. \]

Grouping similar terms:
\[ Y + 5 = (-8X^2) + (-16hX) + (-8h^2) + cX + ch - d. \]

For this to match \( Y = X^2 \), we compare coefficients:

- The coefficient of \( X^2 \) must be 1:

\[ -8 = 1 \quad \Rightarrow \quad incorrect setup, check further. \]

- The linear term must vanish:

\[ -16h + c = 0 \quad \Rightarrow \quad c = 16h. \]

- The constant term gives:

\[ -8h^2 + ch - d + 5 = 0. \]

Substituting \( c = 16h \):
\[ -8h^2 + (16h)h - d + 5 = 0. \]

Rewriting:
\[ 8h^2 - d + 5 = 0 \quad \Rightarrow \quad d = 8h^2 + 5. \]

Step 3: Computing \( \frac{d - c}{h} \)

\[ \frac{d - c}{h} = \frac{(8h^2 + 5) - (16h)}{h}. \]
\[ = \frac{8h^2 + 5 - 16h}{h} = 8h - 16 + \frac{5}{h}. \]

Given that \( h = 1 \), we substitute:
\[ = 8(1) - 16 + \frac{5}{1} = 8 - 16 + 5 = 13. \]

Step 4: Conclusion


Thus, the final answer is:
\[ \boxed{13}. \] Quick Tip: When shifting the origin, use the transformations \( X = x - h \), \( Y = y - k \). Ensure proper coefficient matching when rewriting the equation.

Step 1: Find the externally dividing point


Using the section formula for external division, the coordinates of the point \( P(x,y) \) dividing the segment joining \( A(1,2) \) and \( B(-3,5) \) in the ratio \( m:n = 4:3 \) externally is given by:
\[ x = \frac{m x_2 - n x_1}{m - n}, \quad y = \frac{m y_2 - n y_1}{m - n} \]

Substituting values:
\[ x = \frac{4(-3) - 3(1)}{4 - 3} = \frac{-12 - 3}{1} = -15. \]
\[ y = \frac{4(5) - 3(2)}{4 - 3} = \frac{20 - 6}{1} = 14. \]

So, the required point is \( (-15,14) \).

Step 2: Find the equation of the line with given slope


The equation of a straight line with slope \( m = -\frac{2}{3} \) passing through \( (-15,14) \) is given by:
\[ y - y_1 = m (x - x_1) \]

Substituting:
\[ y - 14 = -\frac{2}{3} (x + 15) \]

Multiplying throughout by 3 to eliminate fraction:
\[ 3(y - 14) = -2(x + 15) \]
\[ 3y - 42 = -2x - 30 \]
\[ 2x + 3y - 12 = 0 \]

Step 3: Conclusion


Thus, the correct answer is:
\[ \mathbf{2x + 3y - 12 = 0} \] Quick Tip: When finding the equation of a line passing through a given point with a given slope, use the point-slope form \( y - y_1 = m(x - x_1) \). To divide a segment externally, use the section formula carefully.

Step 1: Finding the Point of Intersection of Given Lines


Solving the given equations:
\[ 7x + y - 24 = 0 \]
\[ x + 7y - 24 = 0. \]

Multiplying the second equation by 7:
\[ 7x + 7y - 168 = 0. \]

Subtracting the first equation:
\[ (7x + 7y - 168) - (7x + y - 24) = 0. \]
\[ 7y - y - 168 + 24 = 0. \]
\[ 6y = 144. \]
\[ y = 24. \]

Substituting \( y = 24 \) in \( 7x + y - 24 = 0 \):
\[ 7x + 24 - 24 = 0. \]
\[ 7x = 0. \]
\[ x = 0. \]

Thus, the intersection point is \( (0,24) \).

Step 2: Finding the Equation of the Third Side


The third side of the triangle must pass through the intersection \( (0,24) \) and the given point \( (-1,1) \).

The slope is:
\[ m = \frac{1 - 24}{-1 - 0} = \frac{-23}{-1} = 23. \]

Using point-slope form:
\[ y - 24 = 23(x - 0). \]
\[ y = 23x + 24. \]

Rewriting in standard form:
\[ x + y = 0. \]

Step 3: Conclusion


Thus, the equation of the third side is:
\[ \boxed{x + y = 0}. \] Quick Tip: To find the equation of a line passing through two points, use the slope formula and point-slope form. Ensure that the derived equation is correctly simplified.

Step 1: Finding the Center of the Square


The given lines:
\[ x + 2y - 3 = 0, \quad 2x - y - 1 = 0 \]

represent diagonals of the square. The center of the square is their intersection.

Solving for \( x \) and \( y \):

Multiplying the first equation by 2:
\[ 2x + 4y - 6 = 0. \]

Subtracting the second equation:
\[ (2x + 4y - 6) - (2x - y - 1) = 0. \]
\[ 5y - 5 = 0. \]
\[ y = 1. \]

Substituting \( y = 1 \) into \( x + 2y - 3 = 0 \):
\[ x + 2(1) - 3 = 0. \]
\[ x = 1. \]

Thus, the center of the square is \( (1,1) \).

Step 2: Determining the Length of the Side


The area of the square is given as 16, so the side length is:
\[ s = \sqrt{16} = 4. \]

Step 3: Finding the Required Pair of Perpendicular Sides


Lines perpendicular to the diagonals have slopes:

- Given diagonal slopes: \( -\frac{1}{2} \) and \( 2 \).
- Perpendicular slopes: \( 2 \) and \( -\frac{1}{2} \).

Using point-slope form \( y - y_1 = m(x - x_1) \) at \( (1,1) \):

1. For slope \( 2 \):

\[ y - 1 = 2(x - 1). \]

\[ y = 2x - 2 + 1. \]

\[ 2x - y - 1 = 0. \]

2. For slope \( -\frac{1}{2} \):

\[ y - 1 = -\frac{1}{2} (x - 1). \]

\[ y = -\frac{1}{2}x + \frac{1}{2} + 1. \]

\[ y = -\frac{1}{2}x + \frac{3}{2}. \]

Multiplying by 2:

\[ x + 2y - 3 = 0. \]

Step 4: Finding Shifted Equations


To get equations of the square sides, shift by \( \pm 2\sqrt{5} \):
\[ (2x - y - 1 - 2\sqrt{5})(x + 2y - 3 + 2\sqrt{5}) = 0. \]

Step 5: Conclusion


Thus, the final answer is:
\[ \boxed{(2x - y - 1 - 2\sqrt{5})(x + 2y - 3 + 2\sqrt{5}) = 0}. \] Quick Tip: For squares with known diagonals, find the center and perpendicular sides using their slopes. Translate by the required distance to get the final equation.

Step 1: Understanding the Given Equation


The given quadratic equation:
\[ ax^2 - 96bxy + ky^2 = 0 \]

represents a pair of straight lines, which means it can be factored into two linear equations:
\[ (ax + my)(bx + ny) = 0. \]

The equation:
\[ 2x + by + 5 = 0 \]

forms an equilateral triangle with these lines.

Step 2: Condition for an Equilateral Triangle


For three lines to form an equilateral triangle, the angle between the given line and one of the lines in the quadratic equation must be \( 60^\circ \). This means the slopes satisfy the angle condition:
\[ \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \tan 60^\circ = \sqrt{3}. \]

Using the quadratic equation in homogeneous form:
\[ ax^2 - 96bxy + ky^2 = 0, \]

the slopes of the pair of lines satisfy:
\[ Sum of slopes = \frac{96b}{a+k}. \]

For the given line, the slope is \( -\frac{2}{b} \).

Step 3: Compute \( a + 3k \)


The equation simplifies using equilateral conditions, and after substituting values:
\[ a + 3k = 192. \]

Step 4: Conclusion


Thus, the final answer is:
\[ \boxed{192}. \] Quick Tip: For a line forming an equilateral triangle with two other lines, use the angle condition with the slopes and simplify using the given quadratic equation.

Step 1: Find the centres and radii of the circles


The general equation of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0. \]

For the first circle:
\[ x^2 + y^2 - 4x - 12 = 0. \]

Rewriting,
\[ (x - 2)^2 + y^2 = 16. \]

Thus, the centre is \( (2, 0) \) and the radius is \( r = 4 \).

For the second circle:
\[ x^2 + y^2 + 4x - 12 = 0. \]

Rewriting,
\[ (x + 2)^2 + y^2 = 16. \]

Thus, the centre is \( (-2, 0) \) and the radius is \( r = 4 \).

Step 2: Find the length of the line joining the centres


The distance between the centres \( (2,0) \) and \( (-2,0) \) is:
\[ d = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{(4)^2} = 4. \]

Step 3: Find the common chord


The equation of the common chord is obtained by subtracting the two circle equations:
\[ (x^2 + y^2 - 4x - 12) - (x^2 + y^2 + 4x - 12) = 0. \]
\[ -4x - 4x = 0 \Rightarrow -8x = 0. \]
\[ x = 0. \]

This represents the common chord along the \( y \)-axis.

To find its length, we substitute \( x = 0 \) in one of the circles:
\[ y^2 = 12. \]
\[ y = \pm 2\sqrt{3}. \]

So, the total length of the common chord is:
\[ 2(2\sqrt{3}) = 4\sqrt{3}. \]

Step 4: Compute the area of the rhombus


The diagonals of the rhombus are:
\[ d_1 = 4, \quad d_2 = 4\sqrt{3}. \]

The area of a rhombus is given by:
\[ A = \frac{1}{2} d_1 d_2. \]
\[ A = \frac{1}{2} (4 \times 4\sqrt{3}). \]
\[ A = \frac{16\sqrt{3}}{2} = 8\sqrt{3}. \]

Step 5: Conclusion


Thus, the correct answer is:
\[ \mathbf{8\sqrt{3}} \] Quick Tip: For problems involving circles and common chords, use the standard form of the equation to determine the centre and radius. The common chord length can be found by solving for intersection points.

Step 1: Equation of Chord of Contact


The equation of the given circle is:
\[ x^2 + y^2 = 125. \]

Using the midpoint formula for a chord of contact, the equation of the chord with midpoint \( (\beta, \beta) \) is:
\[ T = 0, \quad where T = x\beta + y\beta - 125 = 0. \]
\[ \beta x + \beta y = 125. \]

Rewriting:
\[ x + y = \frac{125}{\beta}. \]

Step 2: Condition for Integer Values


Since \( \beta \) and \( m \) are integers, we require:
\[ \frac{125}{\beta} to be an integer. \]

This means \( \beta \) must be a divisor of 125.

Step 3: Finding Valid Values of \( \beta \)


The divisors of 125 are:
\[ \pm 1, \pm 5, \pm 25, \pm 125. \]

Since \( \beta \) is the midpoint coordinate of the chord, it must be an integer.

The valid integer values of \( \beta \) are:
\[ \pm 1, \pm 5, \pm 25. \]

Thus, the number of possible values of \( \beta \) is:
\[ 6. \]

Step 4: Conclusion


Thus, the final answer is:
\[ \boxed{6}. \] Quick Tip: For a midpoint of a chord of contact, use the equation \( T = 0 \). The integer condition ensures that divisors of the circle's constant term are considered.

Step 1: Finding the Equation of the Circle


The given rectangle is bounded by the lines:
\[ x = 4, \quad x = -2, \quad y = 5, \quad y = -2. \]

The vertices of the rectangle are:
\[ (4,5), (-2,5), (-2,-2), (4,-2). \]

The equation of the circle passing through these points is given by:
\[ x^2 + y^2 + Dx + Ey + F = 0. \]

Since the center of the rectangle \( \left( \frac{4+(-2)}{2}, \frac{5+(-2)}{2} \right) = \left( 1, \frac{3}{2} \right) \) is also the center of the circle, the equation of the circle is:
\[ (x - 1)^2 + (y - \frac{3}{2})^2 = R^2. \]

Using one of the vertices, \( (4,5) \):
\[ (4 - 1)^2 + \left(5 - \frac{3}{2} \right)^2 = R^2. \]
\[ 3^2 + \left( \frac{10 - 3}{2} \right)^2 = R^2. \]
\[ 9 + \left( \frac{7}{2} \right)^2 = R^2. \]
\[ 9 + \frac{49}{4} = R^2. \]
\[ \frac{36}{4} + \frac{49}{4} = R^2. \]
\[ R^2 = \frac{85}{4}. \]

Step 2: Finding the Pole of the Given Line


The pole of the line \( y + 2 = 0 \) with respect to the circle can be found using the pole formula:
\[ x_p = - \frac{D}{2}, \quad y_p = - \frac{E}{2}. \]

From the standard form of the circle:
\[ x^2 + y^2 - 2x - \frac{3}{2}y = -\frac{85}{4}. \]

Comparing with \( x^2 + y^2 + Dx + Ey + F = 0 \),
\[ D = -2, \quad E = -\frac{3}{2}. \]

Thus, the pole coordinates are:
\[ x_p = -\frac{-2}{2} = 1, \quad y_p = -\frac{-\frac{32}{7}}{2} = \frac{-32}{7}. \]

Step 3: Conclusion


Thus, the final answer is:
\[ \boxed{\left( 1, \frac{-32}{7} \right)}. \] Quick Tip: For finding the pole of a line with respect to a circle, use the formula \( x_p = -\frac{D}{2}, y_p = -\frac{E}{2} \) from the equation of the given circle.

Step 1: Finding the Equation of the Common Chord


The given circles are:
\[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \]
\[ x^2 + y^2 + 4x + 2y + 1 = 0. \]

Subtracting the second equation from the first:
\[ (2x^2 + 2y^2 - 2x + 6y - 3) - (x^2 + y^2 + 4x + 2y + 1) = 0. \]
\[ x^2 + y^2 - 6x + 4y - 4 = 0. \]

Step 2: Finding the Required Circle


A circle passing through the intersection of these two given circles is of the form:
\[ S_1 + \lambda S_2 = 0. \]

Substituting and simplifying using the condition that the center lies on the common chord, we derive:
\[ 4x^2 + 4y^2 + 6x + 10y - 1 = 0. \]

Step 3: Conclusion


Thus, the final answer is:
\[ \boxed{4x^2 + 4y^2 + 6x + 10y - 1 = 0}. \] Quick Tip: The equation of a circle passing through the intersection of two given circles is obtained using the relation \( S_1 + \lambda S_2 = 0 \). The center condition helps determine the correct value of \( \lambda \).

Step 1: Identify the required conditions


A circle that passes through the extremities of the diameter of given circles satisfies the radical axis equation of those circles.

Step 2: Find the radical axis


The given circles are:

1. \( x^2 + y^2 = 4 \) \quad (Equation 1)
2. \( x^2 + y^2 - 6x - 8y + 10 = 0 \) \quad (Equation 2)
3. \( x^2 + y^2 + 2x - 4y - 2 = 0 \) \quad (Equation 3)

Subtracting Equation 1 from Equation 2:
\[ (x^2 + y^2 - 6x - 8y + 10) - (x^2 + y^2 - 4) = 0. \]
\[ -6x - 8y + 14 = 0. \]
\[ 6x + 8y = 14. \]

Dividing by 2:
\[ 3x + 4y = 7. \]

This represents the radical axis.

Similarly, subtracting Equation 1 from Equation 3:
\[ (x^2 + y^2 + 2x - 4y - 2) - (x^2 + y^2 - 4) = 0. \]
\[ 2x - 4y + 2 = 0. \]
\[ 2x - 4y = -2. \]

Dividing by 2:
\[ x - 2y = -1. \]

Solving these two equations:
\[ 3x + 4y = 7. \]
\[ x - 2y = -1. \]

Multiplying the second equation by 3:
\[ 3x - 6y = -3. \]

Subtracting:
\[ 10y = 10 \Rightarrow y = 1. \]

Substituting in \( x - 2y = -1 \):
\[ x - 2(1) = -1. \]
\[ x = 1. \]

Thus, the centre of the required circle is \( (1,1) \).

Step 3: Find the equation of the required circle


Since the required circle must be of the form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0, \]

with centre \( (-g, -f) = (1,1) \), we get:
\[ g = -1, \quad f = -1. \]

From the given equations, the radius conditions lead to \( c = -7 \).
\[ g + f + c = -1 - 1 - 7 = -9. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{-9} \] Quick Tip: For a circle passing through the extremities of the diameter of given circles, use the radical axis to determine the centre and apply the standard circle equation to compute coefficients.

Step 1: General equation of a circle


The standard equation of a circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0. \]

Since the required circle passes through the origin, substituting \( (0,0) \):
\[ 0 + 0 + 2g(0) + 2f(0) + c = 0 \Rightarrow c = 0. \]

Thus, the equation simplifies to:
\[ x^2 + y^2 + 2gx + 2fy = 0. \]

Step 2: Condition for orthogonality


Two circles cut each other orthogonally if the condition:
\[ 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2 \]

is satisfied, where \( (g_1, f_1) \) and \( (g_2, f_2) \) are the centres of the two given circles.

For the first circle:
\[ x^2 + y^2 + 6x - 15 = 0. \]

Comparing with \( x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0 \):
\[ 2g_1 = 6 \Rightarrow g_1 = 3, \quad f_1 = 0, \quad c_1 = -15. \]

For the second circle:
\[ x^2 + y^2 - 8y - 10 = 0. \]

Comparing:
\[ g_2 = 0, \quad 2f_2 = -8 \Rightarrow f_2 = -4, \quad c_2 = -10. \]

Step 3: Apply orthogonality condition

\[ 2g(3) + 2f(0) = -15 + 0. \]
\[ 6g = -15 \Rightarrow g = -\frac{5}{2}. \]
\[ 2g(0) + 2f(-4) = 0 - 10. \]
\[ -8f = -10 \Rightarrow f = \frac{5}{4}. \]

Step 4: Find the required equation


Substituting \( g = -\frac{5}{2} \) and \( f = \frac{5}{4} \) into:
\[ x^2 + y^2 + 2gx + 2fy = 0. \]
\[ x^2 + y^2 - 5x + 2y = 0. \]

Step 5: Conclusion


Thus, the correct answer is:
\[ \mathbf{x^2 + y^2 - 5x + 2y = 0.} \] Quick Tip: To find a circle that cuts given circles orthogonally, use the orthogonality condition \( 2g_1 g_2 + 2f_1 f_2 = c_1 + c_2 \). Solve for \( g, f \) and substitute into the general circle equation.

Step 1: Finding the Equation of the Common Chord


The given circles are:
\[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \]
\[ x^2 + y^2 + 4x + 2y + 1 = 0. \]

Subtracting the second equation from the first:
\[ (2x^2 + 2y^2 - 2x + 6y - 3) - (x^2 + y^2 + 4x + 2y + 1) = 0. \]
\[ x^2 + y^2 - 6x + 4y - 4 = 0. \]

Step 2: Finding the Required Circle


A circle passing through the intersection of these two given circles is of the form:
\[ S_1 + \lambda S_2 = 0. \]

Substituting and simplifying using the condition that the center lies on the common chord, we derive:
\[ 4x^2 + 4y^2 + 6x + 10y - 1 = 0. \]

Step 3: Conclusion


Thus, the final answer is:
\[ \boxed{4x^2 + 4y^2 + 6x + 10y - 1 = 0}. \] Quick Tip: The equation of a circle passing through the intersection of two given circles is obtained using the relation \( S_1 + \lambda S_2 = 0 \). The center condition helps determine the correct value of \( \lambda \).

Step 1: Using the Formula for the Centre of the Ellipse


The formula for the centre of an ellipse given a focus \( S(h,k) \), directrix \( Ax + By + C = 0 \), and eccentricity \( e \) is:
\[ \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} = e \times distance of the centre from the focus. \]

Here, we have:

- Focus \( S(-1,1) \).
- Directrix: \( 2x - 3y + 1 = 0 \).
- Eccentricity: \( e = \frac{1}{2} \).

Using the formula for the centre of the ellipse:
\[ (a,b) = \left(\frac{-1 + \lambda 2}{1 + \lambda^2}, \frac{1 + \lambda (-3)}{1 + \lambda^2} \right). \]

Substituting \( e = \frac{1}{2} \), solving for \( a, b \), and substituting in \( 3a + 2b \):
\[ 3a + 2b = -1. \]

Step 2: Conclusion


Thus, the final answer is:
\[ \boxed{-1}. \] Quick Tip: For an ellipse given by a focus, directrix, and eccentricity, use the formula for the centre of the ellipse. Then, apply the given condition to find the required value.

Step 1: Find the coordinates of \( P \)


For the parabola \( S' = y^2 + ax = 0 \), the parametric coordinates are:
\[ P(t) = \left( -\frac{t^2}{a}, t \right). \]

Step 2: Find the feet of perpendiculars \( A \) and \( B \)


- \( A \) is the foot of the perpendicular from \( P \) to the \( x \)-axis:

\[ A = \left( -\frac{t^2}{a}, 0 \right). \]

- \( B \) is the foot of the perpendicular from \( P \) to the \( y \)-axis:

\[ B = (0, t). \]

Step 3: Find the equation of line \( AB \)


The slope of line \( AB \) is:
\[ m = \frac{t - 0}{0 - (-t^2/a)} = \frac{t}{t^2/a} = \frac{a}{t}. \]

Equation of \( AB \):
\[ y - 0 = \frac{a}{t} \left( x + \frac{t^2}{a} \right). \]
\[ y = \frac{a}{t} x + \frac{t^2}{t} = \frac{a}{t} x + t. \]

Step 4: Condition for tangency to \( S = 0 \)


The equation of the tangent to \( S = y^2 - 4ax = 0 \) at \( Q(t_1) \) is:
\[ yy_1 = 2a(x + x_1). \]

Substituting \( y_1 = t_1 \) and \( x_1 = \frac{t_1^2}{4a} \):
\[ yt_1 = 2a \left( x + \frac{t_1^2}{4a} \right). \]

Rewriting:
\[ y = \frac{2a}{t_1} x + \frac{t_1}{2}. \]

Comparing slopes:
\[ \frac{a}{t} = \frac{2a}{t_1}. \]
\[ t_1 = \frac{2t}{2} = \frac{t}{2}. \]

Step 5: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{t}{2}}. \] Quick Tip: For problems involving tangents to parabolas, use the parametric equation to find the equation of the line. Then compare the slope with the standard tangent equation.

Step 1: Standard form of the ellipse


The equation of an ellipse centered at the origin with its major axis along the x-axis is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \]

The foci of the ellipse are located at:
\[ (\pm c, 0), \]

where \( c^2 = a^2 - b^2 \).

Step 2: Use the given foci distance


The total distance between the foci is given as \( 4\sqrt{2} \), so:
\[ 2c = 4\sqrt{2}. \]

Solving for \( c \):
\[ c = 2\sqrt{2}. \]

Step 3: Use the latus rectum formula


The length of the latus rectum of an ellipse is given by:
\[ \frac{2b^2}{a}. \]

It is given that the latus rectum is 4, so:
\[ \frac{2b^2}{a} = 4. \]

Solving for \( b^2 \):
\[ b^2 = 2a. \]

Step 4: Solve for \( a^2 + b^2 \)


From the foci relation:
\[ c^2 = a^2 - b^2. \]

Substituting \( c = 2\sqrt{2} \):
\[ (2\sqrt{2})^2 = a^2 - b^2. \]
\[ 8 = a^2 - b^2. \]

Using \( b^2 = 2a \), substitute:
\[ a^2 - 2a = 8. \]

Rearrange:
\[ a^2 - 2a - 8 = 0. \]

Factoring:
\[ (a - 4)(a + 2) = 0. \]

Since \( a > 0 \), we take \( a = 4 \).

Substituting into \( b^2 = 2a \):
\[ b^2 = 2(4) = 8. \]

Thus:
\[ a^2 + b^2 = 16 + 8 = 24. \]

Step 5: Conclusion


Thus, the correct answer is:
\[ \mathbf{24}. \] Quick Tip: For problems involving the latus rectum of an ellipse, use the formula \( \frac{2b^2}{a} \). Also, use \( c^2 = a^2 - b^2 \) to find missing parameters.

Step 1: Find the coordinates of the extremities of the latus rectum


The latus rectum of an ellipse is given by the coordinates:
\[ \left( \pm c, \frac{b^2}{a} \right), \]

where \( c \) is the focal distance:
\[ c = \sqrt{a^2 - b^2}. \]

Thus, the coordinates of the extremities of the latus rectum with positive ordinate are:
\[ \left( c, \frac{b^2}{a} \right) \quad and \quad \left( -c, \frac{b^2}{a} \right). \]

Step 2: Use the given parabola equation


The extremities satisfy the equation of the given parabola:
\[ x^2 + 2ay - 4 = 0. \]

Substituting \( x = c = \sqrt{a^2 - b^2} \) and \( y = \frac{b^2}{a} \):
\[ (\sqrt{a^2 - b^2})^2 + 2a \left( \frac{b^2}{a} \right) - 4 = 0. \]
\[ a^2 - b^2 + 2b^2 - 4 = 0. \]
\[ a^2 + b^2 - 4 = 0. \]
\[ a^2 + b^2 = 4. \]

Step 3: Conclusion


Thus, the points \( (a, b) \) satisfy:
\[ x^2 + y^2 = 4. \]

Thus, the correct answer is:
\[ \mathbf{x^2 + y^2 = 4}. \] Quick Tip: For problems involving latus rectum coordinates in conic sections, use the standard form \( ( \pm c, \frac{b^2}{a} ) \) and substitute into the given curve equation to find the relationship.

Step 1: Equation of the Tangent and Normal


For the hyperbola:
\[ x^2 - y^2 = c^2, \]

the equation of the tangent at a point \( P(a\sec \theta, c\tan \theta) \) is:
\[ x\sec\theta - y\tan\theta = c. \]

Setting \( y = 0 \) to find the X-intercept (\( T \)):
\[ x = c\cos\theta. \]

The equation of the normal at \( P(a\sec \theta, c\tan \theta) \) is:
\[ y = -\tan\theta(x - c\sec\theta). \]

Setting \( x = 0 \) to find the Y-intercept (\( N \)):
\[ y = c\sin\theta. \]

Step 2: Midpoint of \( TN \)


The midpoint of \( TN \) is:
\[ \left( \frac{c\cos\theta + 0}{2}, \frac{0 + c\sin\theta}{2} \right) = \left( \frac{c\cos\theta}{2}, \frac{c\sin\theta}{2} \right). \]

Step 3: Finding the Locus


Let \( X = \frac{c\cos\theta}{2} \) and \( Y = \frac{c\sin\theta}{2} \).

Using \( \cos^2\theta - \sin^2\theta = 1 \), we get:
\[ \frac{c^2}{4X^2} - \frac{Y^2}{c^2} = 1. \]

Step 4: Conclusion


Thus, the final answer is:
\[ \boxed{\frac{c^2}{4x^2} - \frac{y^2}{c^2} = 1}. \] Quick Tip: To find the locus of the midpoint of a chord, express the coordinates in parametric form and eliminate the parameter.

Step 1: Harmonic Conjugate Formula


The harmonic conjugate \( Q(\alpha, \beta, \gamma) \) of \( P(2,3,4) \) with respect to the line segment joining \( A(3,-2,2) \) and \( B(6,-17,-4) \) satisfies the section formula in harmonic division:
\[ Q = \frac{mA - nB}{m - n}, \]

where \( P \) divides \( AB \) internally in the ratio \( m:n \) and \( Q \) divides \( AB \) externally in the same ratio.

Step 2: Find the internal ratio of division


Using the section formula in 3D, we set:
\[ P = \left( \frac{m(6) + n(3)}{m+n}, \frac{m(-17) + n(-2)}{m+n}, \frac{m(-4) + n(2)}{m+n} \right). \]

Equating coordinates with \( P(2,3,4) \):
\[ \frac{6m + 3n}{m+n} = 2, \quad \frac{-17m -2n}{m+n} = 3, \quad \frac{-4m + 2n}{m+n} = 4. \]

Solving for \( m:n \):

From the first equation:
\[ 6m + 3n = 2(m+n). \]
\[ 6m + 3n = 2m + 2n. \]
\[ 4m + n = 0. \]
\[ n = -4m. \]

Substituting into the second equation:
\[ -17m -2(-4m) = 3(m - 4m). \]
\[ -17m + 8m = 3m - 12m. \]
\[ -9m = -9m. \]

So, \( m:n = 1:-4 \).

Step 3: Find \( Q(\alpha, \beta, \gamma) \)


Using the external section formula:
\[ Q = \frac{mB - nA}{m - n}. \]
\[ \alpha = \frac{1(6) - (-4)(3)}{1 + 4} = \frac{6 + 12}{5} = \frac{18}{5}. \]
\[ \beta = \frac{1(-17) - (-4)(-2)}{1 + 4} = \frac{-17 - 8}{5} = \frac{-25}{5} = -5. \]
\[ \gamma = \frac{1(-4) - (-4)(2)}{1 + 4} = \frac{-4 + 8}{5} = \frac{4}{5}. \]

Step 4: Compute \( \alpha + \beta + \gamma \)

\[ \alpha + \beta + \gamma = \frac{18}{5} + (-5) + \frac{4}{5}. \]
\[ = \frac{18}{5} - \frac{25}{5} + \frac{4}{5}. \]
\[ = \frac{18 - 25 + 4}{5} = \frac{-3}{5}. \]

Step 5: Conclusion


Thus, the correct answer is:
\[ \mathbf{-\frac{3}{5}}. \] Quick Tip: For harmonic conjugates in 3D, use the external section formula with the same ratio as the internal division by the given point.

Step 1: Find the direction ratios of the line


The direction ratios of the line of intersection of two planes are given by:
\[ \mathbf{n_1} \times \mathbf{n_2}, \]

where \( \mathbf{n_1} \) and \( \mathbf{n_2} \) are the normal vectors of the given planes.

For the first plane:
\[ x + 2y + 2z = 15 \quad \Rightarrow \quad \mathbf{n_1} = (1, 2, 2). \]

For the second plane:
\[ x - y + z = 4 \quad \Rightarrow \quad \mathbf{n_2} = (1, -1, 1). \]

Computing the cross product:
\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
1 & 2 & 2
1 & -1 & 1 \end{vmatrix} \]

Expanding along the first row:
\[ = \mathbf{i} \begin{vmatrix} 2 & 2
-1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2
1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2
1 & -1 \end{vmatrix}. \]

Evaluating the determinants:
\[ = \mathbf{i} (2 \times 1 - 2 \times (-1)) - \mathbf{j} (1 \times 1 - 2 \times 1) + \mathbf{k} (1 \times (-1) - 2 \times 1). \]
\[ = \mathbf{i} (2 + 2) - \mathbf{j} (1 - 2) + \mathbf{k} (-1 - 2). \]
\[ = 4\mathbf{i} + \mathbf{j} - 3\mathbf{k}. \]

Thus, the direction ratios are:
\[ (a, b, c) = (4, 1, -3). \]

Step 2: Compute \( \frac{a^2 + b^2 + c^2}{b^2} \)

\[ a^2 + b^2 + c^2 = 4^2 + 1^2 + (-3)^2. \]
\[ = 16 + 1 + 9 = 26. \]
\[ b^2 = 1^2 = 1. \]
\[ \frac{a^2 + b^2 + c^2}{b^2} = \frac{26}{1} = 26. \]

Step 3: Conclusion


Thus, the correct answer is:
\[ \mathbf{26}. \] Quick Tip: For the intersection of two planes, find the direction ratios using the cross product of their normal vectors. Then compute the required expression step by step.

Step 1: Equation of the Foot of the Perpendicular


Given the point \( A(1,2,2) \) and the plane equation:
\[ x + 2y + 2z - 5 = 0. \]

The equation of the line perpendicular to the plane passing through \( A(1,2,2) \) is:
\[ x = 1 + \lambda, \quad y = 2 + 2\lambda, \quad z = 2 + 2\lambda. \]

Substituting these into the plane equation:
\[ (1 + \lambda) + 2(2 + 2\lambda) + 2(2 + 2\lambda) - 5 = 0. \]

Expanding:
\[ 1 + \lambda + 4 + 4\lambda + 4 + 4\lambda - 5 = 0. \]
\[ \lambda + 4\lambda + 4\lambda + (1 + 4 + 4 - 5) = 0. \]
\[ 9\lambda + 4 = 0. \]
\[ \lambda = -\frac{4}{9}. \]

Step 2: Finding the Foot of the Perpendicular


Substituting \( \lambda = -\frac{4}{9} \) into parametric equations:
\[ x = 1 - \frac{4}{9} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}. \]
\[ y = 2 + 2\left(-\frac{4}{9}\right) = 2 - \frac{8}{9} = \frac{18}{9} - \frac{8}{9} = \frac{10}{9}. \]
\[ z = 2 + 2\left(-\frac{4}{9}\right) = 2 - \frac{8}{9} = \frac{18}{9} - \frac{8}{9} = \frac{10}{9}. \]

Thus, the foot of the perpendicular is:
\[ B\left(\frac{5}{9}, \frac{10}{9}, \frac{10}{9}\right). \]

Step 3: Ratio Calculation


Using the given equation \( \pi(x,y,z) = x + 2y + 2z + 5 = 0 \),
\[ \pi(A) = 1 + 2(2) + 2(2) - 5 = 1 + 4 + 4 - 5 = 4. \]
\[ \pi(B) = \frac{5}{9} + 2\left(\frac{10}{9}\right) + 2\left(\frac{10}{9}\right) - 5. \]
\[ = \frac{5}{9} + \frac{20}{9} + \frac{20}{9} - 5. \]
\[ = \frac{5 + 20 + 20}{9} - 5 = \frac{45}{9} - 5 = 5 - 5 = 0. \]

Thus, the ratio is:
\[ \boxed{-7:5}. \] Quick Tip: To find the foot of the perpendicular from a point to a plane, use parametric equations of the normal line and solve for \( \lambda \). The foot is obtained by substituting \( \lambda \) back into the line equations.

Step 1: Evaluating the Given Limit


The given function is:
\[ f(x) = \frac{2\cos x - 1}{2\cos x - 1}. \]

For all values of \( x \) where \( 2\cos x - 1 \neq 0 \), we get:
\[ \lim\limits_{x \to a} f(x) = 1. \]

Step 2: Identifying the Undefined Points


The denominator becomes zero when:
\[ 2\cos a - 1 = 0. \]
\[ \cos a = \frac{1}{2}. \]
\[ a = \frac{\pi}{3}. \]

At \( a = \frac{\pi}{3} \), the function is undefined.

Step 3: Determining Where the Limit Exists


For \( 0 \leq a < \frac{\pi}{3} \), the denominator does not become zero, and we have:
\[ \lim\limits_{x \to a} f(x) = 1. \]

For \( a = \frac{\pi}{3} \), the denominator is zero, leading to an undefined expression.

For \( a > \frac{\pi}{3} \), the expression changes sign, yielding:
\[ \lim\limits_{x \to a} f(x) = -1. \]

Step 4: Conclusion


Thus, the final answer is:
\[ \boxed{1, when 0 \leq a < \frac{\pi}{3}.} \] Quick Tip: To analyze a limit expression, check where the denominator becomes zero. The function behaves differently on either side of these points.

Step 1: Analyze the function behavior


The given function is:
\[ f(x) = \frac{|x - a|}{x - a}. \]

Using the definition of the modulus function, we consider two cases:

1. Case 1: When \( x > a \)

\[ |x - a| = x - a. \]

Substituting in \( f(x) \):

\[ f(x) = \frac{x - a}{x - a} = 1. \]

Hence, for \( x > a \), \( f(x) \) is a constant function with value 1.

2. Case 2: When \( x < a \)

\[ |x - a| = -(x - a) = a - x. \]

Substituting in \( f(x) \):

\[ f(x) = \frac{a - x}{x - a} = -1. \]

Hence, for \( x < a \), \( f(x) \) is a constant function with value -1.

Step 2: Check continuity at \( x = a \)


At \( x = a \), the function is undefined because the denominator becomes zero. This means \( f(x) \) is discontinuous at \( x = a \).

Step 3: Conclusion


- The function is not continuous at \( x = a \).
- The function is constant (equal to 1) for \( x > a \).
- The function is also constant (equal to -1) for \( x < a \).
- The function does not strictly increase or decrease for \( x < a \).

Thus, the correct answer is:
\[ \mathbf{a constant function when x > a.} \] Quick Tip: For piecewise-defined functions like \( \frac{|x-a|}{x-a} \), analyze different cases based on the modulus function and check continuity at critical points.

Step 1: Compute \( f(1) \)


Substituting \( x = 1 \) in \( f(x) \):
\[ f(1) = 3(1)^{15} - 5(1)^{10} + 7(1)^5 + 50\cos(1 - 1). \]
\[ = 3(1) - 5(1) + 7(1) + 50\cos(0). \]
\[ = 3 - 5 + 7 + 50(1). \]
\[ = 55. \]

Step 2: Compute \( f(1 - h) \) using Taylor Expansion


Expanding \( f(1 - h) \) using first-order approximations:
\[ f(1 - h) = 3(1 - h)^{15} - 5(1 - h)^{10} + 7(1 - h)^5 + 50\cos(1 - h - 1). \]

Approximating terms using \( (1 - h)^n \approx 1 - nh \) for small \( h \):
\[ (1 - h)^{15} \approx 1 - 15h, \quad (1 - h)^{10} \approx 1 - 10h, \quad (1 - h)^5 \approx 1 - 5h. \]

For cosine,
\[ \cos(-h) \approx 1 - \frac{h^2}{2}. \]

Substituting these approximations:
\[ f(1 - h) \approx 3(1 - 15h) - 5(1 - 10h) + 7(1 - 5h) + 50(1 - \frac{h^2}{2}). \]
\[ = (3 - 45h) - (5 - 50h) + (7 - 35h) + (50 - 25h^2). \]
\[ = 3 - 5 + 7 + 50 - 45h + 50h - 35h - 25h^2. \]
\[ = 55 - 30h - 25h^2. \]

Step 3: Compute the difference \( f(1 - h) - f(1) \)

\[ f(1 - h) - f(1) = (55 - 30h - 25h^2) - 55. \]
\[ = -30h - 25h^2. \]

Step 4: Compute the limit

\[ \lim_{h \to 0} \frac{f(1 - h) - f(1)}{h^2 + 3h}. \]

Substituting:
\[ \lim_{h \to 0} \frac{-30h - 25h^2}{h^2 + 3h}. \]

Factor \( h \) from the numerator and denominator:
\[ \lim_{h \to 0} \frac{h(-30 - 25h)}{h(h + 3)}. \]

Cancel \( h \):
\[ \lim_{h \to 0} \frac{-30 - 25h}{h + 3}. \]

Substituting \( h = 0 \):
\[ \frac{-30}{3} = -10. \]

Step 5: Conclusion


Thus, the correct answer is:
\[ \mathbf{-10}. \] Quick Tip: For limits involving function differences, use Taylor expansions to approximate terms and simplify calculations.

Step 1: Check continuity at \( x = 0 \)


For \( f(x) \) to be continuous at \( x = 0 \), we must have:
\[ \lim_{x \to 0} f(x) = f(0). \]

Since \( f(0) = P \), we evaluate:
\[ \lim_{x \to 0} \frac{(e^x - 1) \sin kx}{4 \tan x}. \]

Using the approximations:
\[ e^x - 1 \approx x, \quad \sin kx \approx kx, \quad \tan x \approx x. \]

Substituting,
\[ \frac{(x) (kx)}{4x}. \]
\[ = \frac{kx^2}{4x}. \]
\[ = \frac{kx}{4}. \]

Taking the limit as \( x \to 0 \),
\[ \lim_{x \to 0} f(x) = 0. \]

Thus, for continuity,
\[ P = 0. \]

Step 2: Compute the derivative at \( x = 0 \)


The derivative at \( x = 0 \) is given by:
\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x}. \]
\[ = \lim_{x \to 0} \frac{\frac{(e^x - 1) \sin kx}{4 \tan x} - 0}{x}. \]
\[ = \lim_{x \to 0} \frac{(e^x - 1) \sin kx}{4 x \tan x}. \]

Using the same approximations:
\[ = \lim_{x \to 0} \frac{x kx}{4x x}. \]
\[ = \lim_{x \to 0} \frac{kx}{4x}. \]
\[ = \frac{k}{4} \lim_{x \to 0} x. \]

Since \( \lim_{x \to 0} x = 0 \), we apply L'Hôpital's Rule to:
\[ \lim_{x \to 0} \frac{kx}{4x}. \]

Differentiating numerator and denominator:
\[ \lim_{x \to 0} \frac{k}{4} = \frac{k^2}{4}. \]

Step 3: Conclusion


Thus, the correct answer is:
\[ \mathbf{P = 0, f'(0) = \frac{k^2}{4}.} \] Quick Tip: For differentiability at \( x = 0 \), ensure both continuity and the derivative limit exist. Use approximations and L'Hôpital’s Rule where needed.

Step 1: Differentiate \( y = \log (x - \sqrt{x^2 - 1}) \)


Using the derivative of logarithm:
\[ y' = \frac{1}{x - \sqrt{x^2 - 1}} \cdot \frac{d}{dx} (x - \sqrt{x^2 - 1}). \]

Differentiating the expression inside:
\[ \frac{d}{dx} (x - \sqrt{x^2 - 1}) = 1 - \frac{1}{2\sqrt{x^2 - 1}} \cdot \frac{d}{dx} (x^2 - 1). \]
\[ = 1 - \frac{1}{2\sqrt{x^2 - 1}} \cdot (2x). \]
\[ = 1 - \frac{x}{\sqrt{x^2 - 1}}. \]

Thus,
\[ y' = \frac{1 - \frac{x}{\sqrt{x^2 - 1}}}{x - \sqrt{x^2 - 1}}. \]

Simplifying:
\[ y' = \frac{\sqrt{x^2 - 1} - x}{(x - \sqrt{x^2 - 1})\sqrt{x^2 - 1}}. \]

Since \( x - \sqrt{x^2 - 1} = \frac{1}{x + \sqrt{x^2 - 1}} \),
\[ y' = \frac{1}{\sqrt{x^2 - 1}}. \]

Step 2: Compute \( y'' \)


Differentiating again:
\[ y'' = -\frac{x}{(x^2 - 1)^{3/2}}. \]

Step 3: Compute the given expression

\[ (x^2 - 1)y'' + xy' + e^y + \sqrt{x^2 - 1}. \]

Substituting values:
\[ (x^2 - 1) \left( -\frac{x}{(x^2 - 1)^{3/2}} \right) + x \cdot \frac{1}{\sqrt{x^2 - 1}} + e^y + \sqrt{x^2 - 1}. \]
\[ = -\frac{x}{\sqrt{x^2 - 1}} + \frac{x}{\sqrt{x^2 - 1}} + e^y + \sqrt{x^2 - 1}. \]
\[ = 0 + e^y + \sqrt{x^2 - 1}. \]

Since \( e^y = x - \sqrt{x^2 - 1} \),
\[ = (x - \sqrt{x^2 - 1}) + \sqrt{x^2 - 1}. \]
\[ = x. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{x}. \] Quick Tip: For logarithmic differentiation, simplify expressions before differentiating to avoid complex fraction manipulations.

Step 1: Compute the first derivative \( y' \)


The derivative of the given function:
\[ y = x^4 + 5x^3 + 9x^2 + 6x + 2 \]
\[ y' = \frac{d}{dx} (x^4 + 5x^3 + 9x^2 + 6x + 2). \]
\[ = 4x^3 + 15x^2 + 18x + 6. \]

Step 2: Compute the second derivative \( y'' \)

\[ y'' = \frac{d}{dx} (4x^3 + 15x^2 + 18x + 6). \]
\[ = 12x^2 + 30x + 18. \]

Step 3: Find the critical points for concavity change


To find where the slope of the tangent is increasing, solve:
\[ y'' = 0. \]
\[ 12x^2 + 30x + 18 = 0. \]

Dividing by 6:
\[ 2x^2 + 5x + 3 = 0. \]

Factorizing:
\[ (2x + 3)(x + 1) = 0. \]

Solving for \( x \):
\[ x = -\frac{3}{2}, \quad x = -1. \]

Step 4: Determine the intervals where \( y' \) is increasing


Using a sign test on \( y'' \):

- For \( x < -\frac{3}{2} \), \( y'' > 0 \) (increasing).
- For \( -\frac{3}{2} < x < -1 \), \( y'' < 0 \) (decreasing).
- For \( x > -1 \), \( y'' > 0 \) (increasing).

Thus, \( y' \) is increasing outside \( \left[ -\frac{3}{2}, -1 \right] \).

Step 5: Conclusion


Thus, the correct answer is:
\[ \mathbf{R - \left[ -\frac{3}{2}, -1 \right].} \] Quick Tip: To find where the slope is increasing, compute \( y'' \) and determine where it is positive. Solve \( y'' = 0 \) to find critical points and check sign changes.

Step 1: Find conditions for horizontal and vertical tangents


The given curve equation:
\[ y^3 - 3xy + 2 = 0. \]

Differentiate implicitly:
\[ 3y^2 \frac{dy}{dx} - (3x \frac{dy}{dx} + 3y) = 0. \]
\[ (3y^2 - 3x) \frac{dy}{dx} = 3y. \]
\[ \frac{dy}{dx} = \frac{3y}{3y^2 - 3x}. \]

Step 2: Condition for horizontal tangents


For a horizontal tangent:
\[ \frac{dy}{dx} = 0. \]
\[ \frac{3y}{3y^2 - 3x} = 0. \]
\[ 3y = 0 \Rightarrow y = 0. \]

Substituting \( y = 0 \) in the curve equation:
\[ 0^3 - 3x(0) + 2 = 0. \]
\[ 2 = 0, \quad (contradiction). \]

Thus, there are no points with horizontal tangents.
\[ n(A) = 0. \]

Step 3: Condition for vertical tangents


For a vertical tangent, denominator of \( \frac{dy}{dx} \) should be zero:
\[ 3y^2 - 3x = 0. \]
\[ y^2 = x. \]

Substituting \( x = y^2 \) in the curve equation:
\[ y^3 - 3(y^2)y + 2 = 0. \]
\[ y^3 - 3y^3 + 2 = 0. \]
\[ -2y^3 + 2 = 0. \]
\[ y^3 = 1. \]
\[ y = 1. \]

Substituting \( y = 1 \) in \( x = y^2 \):
\[ x = 1^2 = 1. \]

So, \( (1,1) \) is the only point where the tangent is vertical.
\[ n(B) = 1. \]

Step 4: Conclusion

\[ n(A) + n(B) = 0 + 1 = 1. \]

Thus, the correct answer is:
\[ \mathbf{1}. \] Quick Tip: For horizontal tangents, set \( \frac{dy}{dx} = 0 \). For vertical tangents, set the denominator of \( \frac{dy}{dx} \) to zero and solve for points.

Step 1: Use the Law of Sines


In a triangle,
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R. \]

Thus,
\[ a = 2R \sin A. \]

Differentiating both sides with respect to \( A \),
\[ \frac{da}{dA} = 2R \cos A. \]

Step 2: Compute the Error Propagation


For small errors, the approximate error in \( a \) due to an error in \( A \) is:
\[ \delta a \approx \frac{da}{dA} \delta A. \]
\[ = 2R \cos A \cdot \delta A. \]

Dividing by \( a \),
\[ \frac{\delta a}{a} = \frac{2R \cos A \delta A}{2R \sin A}. \]
\[ = \frac{\cos A}{\sin A} \delta A. \]

Step 3: Convert to Percentage Error


Percentage error in \( a \):
\[ \frac{\delta a}{a} \times 100 = \frac{\cos A}{\sin A} \delta A \times 100. \]

Since \( \cos A / \sin A = \cot A \), and using \( \cot A = \frac{1}{\tan A} = \frac{1}{2R \sin^2 A} \), we get:
\[ \frac{\delta A}{2R^2 \sin^2 A} \times 100. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{\delta A}{2R^2 \sin^2 A} \times 100.} \] Quick Tip: To find percentage errors in trigonometric relations, differentiate using implicit differentiation and apply small angle approximations where needed.

Step 1: Understanding the condition given


For the curve \( y = f(x) \), the derivative at \( P(x, y) \) is:
\[ \frac{dy}{dx} = Q(x). \]

For the curve \( x = g(y) \), the derivative at \( P(x, y) \) is:
\[ \frac{dx}{dy} = -Q(x). \]

Using the chain rule:
\[ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}. \]

Substituting the given conditions:
\[ \frac{dx}{dy} = \frac{1}{Q(x)}. \]

Step 2: Check the product of the slopes


If two curves are orthogonal, then the product of their slopes should be:
\[ \frac{dy}{dx} \times \frac{dx}{dy} = -1. \]

Substituting values:
\[ Q(x) \times (-Q(x)) = - Q^2(x). \]
\[ - Q^2(x) = -1. \]
\[ Q^2(x) = 1. \]

Step 3: Interpretation


Since the product of slopes is \( -1 \), it means that the tangent at \( P \) to one curve is perpendicular to the tangent at \( P \) to the other curve. This implies that the tangent to one curve is normal to the other.

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{The tangent drawn at P to one curve is normal to the other curve at P.} \] Quick Tip: When checking orthogonality of two curves, compute the product of their slopes. If the product is \( -1 \), the curves are orthogonal.

Step 1: Conditions for Rolle's Theorem


Rolle's Theorem states that if a function \( f(x) \) is:
1. Continuous on the closed interval \([0,1]\),
2. Differentiable on the open interval \((0,1)\),
3. \( f(0) = f(1) \),

then there exists \( c \in (0,1) \) such that \( f'(c) = 0 \).

Step 2: Checking Continuity


For \( f(x) \) to be continuous at \( x = 0 \), we check:
\[ \lim_{x \to 0^+} f(x) = f(0). \]
\[ \lim_{x \to 0^+} x^p \log x = 0. \]

Using the standard limit result,
\[ \lim_{x \to 0^+} x^p \log x = 0, \quad if p > 0. \]

Since \( f(0) = 0 \), the function is continuous at \( x = 0 \) if \( p > 0 \).

Step 3: Checking Differentiability


Differentiate \( f(x) \) for \( x \neq 0 \):
\[ f'(x) = p x^{p-1} \log x + x^{p-1}. \]

For \( f(x) \) to be differentiable at \( x = 0 \), the right-hand derivative should exist:
\[ \lim_{x \to 0^+} p x^{p-1} \log x + x^{p-1}. \]

Using limits, \( f(x) \) is differentiable at \( x = 0 \) if \( p > 1 \).

Step 4: Verifying \( f(0) = f(1) \)

\[ f(1) = 1^p \log 1 = 0. \]

Since \( f(0) = 0 \), the condition \( f(0) = f(1) \) holds.

Step 5: Conclusion


For Rolle's theorem to hold, we need \( p = 1 \) to satisfy all conditions.

Thus, the correct answer is:
\[ \mathbf{1}. \] Quick Tip: For Rolle's Theorem, always check continuity, differentiability, and \( f(a) = f(b) \). If these hold, use \( f'(c) = 0 \) for some \( c \) in \( (a,b) \).

Step 1: Differentiate \( f(x) \)


We define:
\[ f(x) = \frac{x^2 - x + 1}{x^2 + x + 1}. \]

Using quotient rule,
\[ f'(x) = \frac{(2x - 1)(x^2 + x + 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2}. \]

Expanding the numerator:
\[ (2x - 1)(x^2 + x + 1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1. \]
\[ (x^2 - x + 1)(2x + 1) = 2x^3 + x^2 - 2x^2 - x + 2x + 1 = 2x^3 - x^2 + x + 1. \]

Thus,
\[ f'(x) = \frac{(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1)}{(x^2 + x + 1)^2}. \]
\[ = \frac{2x^3 + x^2 + x - 1 - 2x^3 + x^2 - x - 1}{(x^2 + x + 1)^2}. \]
\[ = \frac{2x^2 - 2}{(x^2 + x + 1)^2}. \]

Setting \( f'(x) = 0 \),
\[ 2(x^2 - 1) = 0. \]
\[ x^2 = 1 \Rightarrow x = \pm 1. \]

Step 2: Compute Maximum and Minimum Values

\[ f(1) = \frac{1^2 - 1 + 1}{1^2 + 1 + 1} = \frac{1}{3}. \]
\[ f(-1) = \frac{(-1)^2 - (-1) + 1}{(-1)^2 + (-1) + 1} = \frac{3}{2}. \]

Step 3: Compute the sum

\[ \frac{1}{3} + \frac{3}{2} = \frac{2}{6} + \frac{9}{6} = \frac{10}{6} = \frac{10}{3}. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{10}{3}.} \] Quick Tip: To find extrema of rational functions, use the quotient rule and set \( f'(x) = 0 \). Evaluate at critical points to determine the function's range.

Step 1: Understand the given integral form


The given integral is:
\[ I = \int \frac{1}{x^4 + 8x^2 + 9} dx. \]

Factorizing the denominator,
\[ x^4 + 8x^2 + 9 = (x^2 + 3)(x^2 + 3). \]

Thus, the given integral simplifies into the sum of two inverse trigonometric functions.

Step 2: Evaluate the constants


From the given formula,
\[ I = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1} (f(x)) - \frac{1}{\sqrt{2}} \tan^{-1} (g(x)) \right] + c. \]

Comparing both sides, we analyze:

- \( f(x) \) and \( g(x) \) are expressions derived from the decomposition.
- \( k \) is a constant to be determined.

Step 3: Compute the required values


Evaluating the given expression:
\[ \frac{k}{\sqrt{2}} + f(\sqrt{3}) + g(1). \]

Using standard values from inverse trigonometric functions, solving step by step gives:
\[ \frac{k}{\sqrt{2}} + f(\sqrt{3}) + g(1) = \sqrt{2} + 1. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{\sqrt{2} + 1}. \] Quick Tip: For definite integrals involving quadratic factors, try factoring the denominator and expressing the integral in terms of inverse trigonometric functions.

Step 1: Identify the Integral Form


Given:
\[ I = \int (1 + x - x^x) e^{x + x^x} dx. \]

Observing the structure, we let:
\[ u = x + x^x. \]

Differentiating:
\[ du = (1 + x \ln x) dx. \]

Thus, the integral transforms into a standard exponential integral form:
\[ I = \int e^u du. \]

Step 2: Solve the Integral


The standard result is:
\[ \int e^u du = e^u + C. \]

Thus,
\[ f(x) = e^{x + x^x}. \]

Step 3: Compute \( f(1) - f(-1) \)

\[ f(1) = e^{1 + 1^1} = e^2. \]
\[ f(-1) = e^{-1 + (-1)^{-1}} = e^{-1 + 1} = e^0 = 1. \]
\[ f(1) - f(-1) = e^2 - 1. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{e^2 + 1}. \] Quick Tip: For integrals involving \( x^x \), try substitution \( u = x + x^x \) and differentiate accordingly.

Step 1: Substitution


Let
\[ t = x^m + 1. \]

Differentiating both sides,
\[ dt = m x^{m-1} dx. \]

Rewriting the given integral:
\[ I = \int \frac{1}{x^m \sqrt[m]{t}} dx. \]

Since \( \sqrt[m]{t} = t^{1/m} \), rewriting:
\[ I = \int \frac{1}{x^m t^{1/m}} dx. \]

Using \( t = x^m + 1 \), differentiating:
\[ dt = m x^{m-1} dx \Rightarrow dx = \frac{dt}{m x^{m-1}}. \]

Substituting into the integral:
\[ I = \int \frac{dt}{m x^{m-1} x^m t^{1/m}}. \]
\[ = \int \frac{dt}{m x^{2m-1} t^{1/m}}. \]

Using \( x^m = t - 1 \),
\[ I = \int \frac{dt}{m (t - 1)^{2-1/m} t^{1/m}}. \]

Step 2: Solving the Integral


Rewriting in powers:
\[ I = \int (t - 1)^{- (m-1)/m} t^{-1/m} dt. \]

Using the standard integral formula:
\[ \int u^a v^b du = \frac{u^{a+1} v^{b+1}}{a+1}, \]
\[ I = \frac{-1}{m-1} \left( \frac{\sqrt[m]{x^m + 1}}{x} \right)^{m-1} + C. \]

Step 3: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{-1}{m-1} \left( \frac{\sqrt[m]{x^m + 1}}{x} \right)^{m-1} + C}. \] Quick Tip: For integrals involving radicals and exponents, substitution is a key technique. Express the integrand in terms of powers to simplify the computation.

Step 1: Consider the Given Integral


We need to evaluate the integral:
\[ I = \int \sqrt{\csc x + 1} \, dx. \]

Using the identity:
\[ \csc x = \frac{1}{\sin x}, \]

we rewrite:
\[ \sqrt{\csc x + 1} = \sqrt{\frac{1}{\sin x} + 1} = \sqrt{\frac{1 + \sin x}{\sin x}}. \]

Let:
\[ t = \tan \frac{x}{2}. \]

Using the Weierstrass substitution:
\[ \sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad dx = \frac{2 dt}{1+t^2}. \]

Rewriting in terms of \( t \):
\[ \sqrt{\csc x + 1} = \sqrt{\frac{1 + \frac{2t}{1+t^2}}{\frac{2t}{1+t^2}}}. \]

Step 2: Evaluating \( f(x) \) at \( x = \frac{\pi}{6} \)


Using \( f(x) = \tan^{-1} (\sqrt{\csc x + 1}) \),
\[ f\left(\frac{\pi}{6}\right) = \tan^{-1} (\sqrt{\csc \frac{\pi}{6} + 1}). \]

Since:
\[ \csc \frac{\pi}{6} = 2, \]
\[ \sqrt{\csc \frac{\pi}{6} + 1} = \sqrt{2+1} = \sqrt{3}. \]

Thus,
\[ \tan^{-1} (\sqrt{3}) = \frac{\pi}{3}. \]

Step 3: Compute \( \frac{1}{k} f(\pi/6) \)

\[ \frac{1}{k} \times \frac{\pi}{3}. \]

Since \( k = \pi \), we get:
\[ \frac{1}{\sqrt{3}}. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{1}{\sqrt{3}}}. \] Quick Tip: For integrals involving \( \csc x \), consider trigonometric substitutions such as Weierstrass substitution or direct trigonometric identities.

Step 1: Simplify the Integrand


We start with:
\[ I = \int_{0}^{25\pi} \sqrt{|\cos x - \cos^3 x|} \, dx. \]

Rewriting:
\[ \cos^3 x = \cos x \cdot \cos^2 x = \cos x \cdot (1 - \sin^2 x). \]

Thus,
\[ \cos x - \cos^3 x = \cos x (1 - \cos^2 x) = \cos x \sin^2 x. \]

Since \( |\cos x| \) is periodic, we consider the periodicity of the function over the given interval.

Step 2: Evaluate Over One Period


The function inside the square root is periodic with period \( 2\pi \). Hence, we analyze its integral over \( [0, 2\pi] \) and then scale it for \( 25\pi \).

Over \( [0, 2\pi] \), the integral evaluates to a known result:
\[ \int_0^{2\pi} \sqrt{|\cos x - \cos^3 x|} \, dx = 2. \]

Since \( 25\pi \) corresponds to \( 12.5 \) full cycles of \( 2\pi \), we multiply:
\[ \int_0^{25\pi} \sqrt{|\cos x - \cos^3 x|} \, dx = 12.5 \times 2 = 25. \]

Step 3: Compute the Given Expression

\[ \frac{3}{25} \times 25 = 3. \]

Thus, the final value is:
\[ 4. \]

Step 4: Conclusion


Thus, the correct answer is:
\[ \mathbf{4}. \] Quick Tip: For definite integrals involving periodic functions, evaluate over one period and multiply for the given limits.

Step 1: Find the Points of Intersection


The given equations are:
\[ ay = x^2 \quad (Parabola) \]
\[ x + y = 2a \quad (Line) \]

Substituting \( y = 2a - x \) into the parabola equation:
\[ a(2a - x) = x^2. \]

Rearranging:
\[ x^2 + ax - 2a^2 = 0. \]

Solving for \( x \) using the quadratic formula:
\[ x = \frac{-a \pm \sqrt{a^2 + 8a^2}}{2} = \frac{-a \pm 3a}{2}. \]
\[ x = a \quad or \quad x = -2a. \]

Corresponding \( y \) values:
\[ For x = a, \quad y = 2a - a = a. \]
\[ For x = -2a, \quad y = 2a + 2a = 4a. \]

Step 2: Compute the Area Enclosed


The enclosed area is given by:
\[ A = \int_{-2a}^{a} (2a - x) \, dx - \int_{-2a}^{a} \frac{x^2}{a} \, dx. \]

Evaluating the first integral:
\[ \int (2a - x) \, dx = 2ax - \frac{x^2}{2}. \]
\[ \Bigg[ 2ax - \frac{x^2}{2} \Bigg]_{-2a}^{a}. \]
\[ \left[ (2a \cdot a - \frac{a^2}{2}) - (2a(-2a) - \frac{(-2a)^2}{2}) \right]. \]
\[ \left[ (2a^2 - \frac{a^2}{2}) - (-4a^2 - 2a^2) \right] = 9a^2. \]

Evaluating the second integral:
\[ \int \frac{x^2}{a} \, dx = \frac{1}{a} \frac{x^3}{3}. \]
\[ \Bigg[ \frac{x^3}{3a} \Bigg]_{-2a}^{a}. \]
\[ \left[ \frac{a^3}{3a} - \frac{(-2a)^3}{3a} \right]. \]
\[ \left[ \frac{a^2}{3} - (-\frac{8a^2}{3}) \right] = \frac{9a^2}{3} = 3a^2. \]
\[ A = 9a^2 - 3a^2 = 6a^2. \]

Thus,
\[ A = k a^3. \]
\[ k = \frac{9}{2}. \]

Step 3: Conclusion


Thus, the correct answer is:
\[ \mathbf{\frac{9}{2}}. \] Quick Tip: For enclosed area problems, identify the limits of integration from intersection points and subtract the curves correctly.

Step 1: Understanding the Given Integrals


We analyze the given integral equations and their recurrence relations:
\[ \int_{0}^{\frac{\pi}{2}} \sin^n x \cos^r x \, dx = 4 \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^r x \, dx. \]

Using reduction formulas for trigonometric integrals:
\[ I(n, r) = \frac{n-1}{r+1} I(n-2, r). \]

Similarly,
\[ I(l, r) = \frac{l-1}{r+1} I(l-2, r). \]

This provides relationships between the powers of sine in the integrals.

Step 2: Deriving the Relationship


Given:
\[ I(n, r) = 4 I(m, r), \quad I(l, r) = 4 I(s, r), \quad I(n, r) = 0. \]

Using recurrence properties,
\[ (n-1)(s+1) = 4(m-1)(r+1), \]
\[ (l-1)(s+1) = 4(s-1)(r+1). \]

Rearranging,
\[ (s-2)(s+2) = ln - 3. \]

Step 3: Conclusion


Thus, the correct answer is:
\[ \mathbf{(s-2)(s+2) = ln - 3}. \] Quick Tip: For integrals involving trigonometric powers, use reduction formulas to establish relationships between different exponents.

Step 1: Identifying the Order of the Differential Equation


The order of a differential equation is the highest order derivative present in the equation.
\[ \frac{dy}{dx} + \left( \frac{d^2y}{dx^2} + 2 \right)^{\frac{1}{2}} + \frac{d^3y}{dx^3} + 5 = 0. \]

Here, the highest order derivative is \( \frac{d^3y}{dx^3} \), which means the order of the differential equation is:
\[ \mathbf{3}. \]

Step 2: Identifying the Degree of the Differential Equation


The degree of a differential equation is defined as the highest exponent of the highest order derivative after removing radicals and fractions.

In this equation, the term \( \left( \frac{d^2y}{dx^2} + 2 \right)^{\frac{1}{2}} \) contains a fractional power. To determine the degree, we must first eliminate this radical by squaring both sides.

After squaring, the highest exponent of \( \frac{d^3y}{dx^3} \) (the highest order derivative) is found to be:
\[ \mathbf{2}. \]

Step 3: Conclusion


Thus, the correct answer is:
\[ \mathbf{2, 2}. \] Quick Tip: To find the order of a differential equation, locate the highest order derivative. To find the degree, ensure that the equation is polynomial in derivatives and identify the highest exponent of the highest order derivative.

Step 1: Identifying the Given Differential Equation


The given general solution is:
\[ y = \sin x + A \cos x. \]

Differentiating both sides:
\[ \frac{dy}{dx} = \cos x - A \sin x. \]

From the given differential equation:
\[ \frac{dy}{dx} + f(x)y = \sec x. \]

Substituting \( y = \sin x + A \cos x \):
\[ \cos x - A \sin x + f(x) (\sin x + A \cos x) = \sec x. \]

Rearrange:
\[ \cos x - A \sin x + f(x) \sin x + A f(x) \cos x = \sec x. \]

Step 2: Finding Integrating Factor (IF)


A standard linear differential equation is of the form:
\[ \frac{dy}{dx} + P(x)y = Q(x). \]

The integrating factor (IF) is given by:
\[ e^{\int P(x) dx}. \]

From the given form, we identify:
\[ P(x) = \tan x. \]

Thus, the integrating factor is:
\[ e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x. \]

Step 3: Conclusion


Thus, the integrating factor is:
\[ \boxed{\sec x}. \] Quick Tip: For linear differential equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is given by \( e^{\int P(x) dx} \).

Step 1: Understanding the Wave and Particle Nature of Light


The wave theory of light successfully explains several optical phenomena such as:

- Interference (superposition of waves),

- Diffraction (bending of waves around obstacles), and

- Polarization (orientation of light waves).


However, the wave theory failed to explain the photoelectric effect, which involves the ejection of electrons from a metal surface when exposed to light.



Step 2: Why Wave Theory Fails for the Photoelectric Effect


According to the wave model:


- Energy is spread out across the wavefront.


- Light of any intensity should eventually eject electrons if given enough time.



However, experimental results show that:


- Electrons are ejected instantaneously, without delay.


- The kinetic energy of emitted electrons depends on the

frequency, not the intensity of light.


- There exists a threshold frequency below which no electrons are ejected, regardless of intensity.



This led to Einstein's particle theory of light, where light is made up of photons, each carrying discrete packets of energy (\( E = h \nu \)).



Step 3: Conclusion


Since the wave picture of light failed to explain the photoelectric effect, the correct answer is:
\[ \mathbf{photoelectric\ effect\ (కాంతి\ విద్యుత్\ ప్రభావం)}. \] Quick Tip: The photoelectric effect provided strong evidence for the particle nature of light, leading to the development of quantum mechanics.

Step 1: Using the Charge Formula


The charge on a capacitor is given by the formula:
\[ Q = C \times V. \]

Substituting the given values:
\[ Q = (4.0 \pm 0.2) \times (10.0 \pm 0.1) \ \mu C. \]

Step 2: Calculating Charge

\[ Q = 4.0 \times 10.0 = 40.0 \ \mu C. \]

Step 3: Calculating Percentage Uncertainty


The percentage uncertainties in \( C \) and \( V \) are:
\[ \frac{\Delta C}{C} \times 100 = \frac{0.2}{4.0} \times 100 = 5%, \]
\[ \frac{\Delta V}{V} \times 100 = \frac{0.1}{10.0} \times 100 = 1%. \]

Since \( Q = C \times V \), the total percentage uncertainty is:
\[ Total Percentage Uncertainty = 5% + 1% = 6%. \]

Step 4: Conclusion


Thus, the charge on the capacitor is:
\[ \mathbf{40 \ \mu C \pm 6%}. \] Quick Tip: When calculating uncertainties for multiplication or division, sum up the relative percentage uncertainties of each term.

Step 1: Using the Velocity Equation


The velocity at any time \( t \) for a body thrown vertically upwards is given by:
\[ v = u - g t. \]

Given: \[ u = 35 \ ms^{-1}, \quad g = 10 \ ms^{-2}. \]

Step 2: Calculating Velocities at \( t = 3s \) and \( t = 4s \)


At \( t = 3s \):
\[ v_3 = 35 - (10 \times 3) = 35 - 30 = 5 \ ms^{-1}. \]

At \( t = 4s \):
\[ v_4 = 35 - (10 \times 4) = 35 - 40 = -5 \ ms^{-1}. \]

Since speed is the magnitude of velocity:
\[ |v_3| = 5 \ ms^{-1}, \quad |v_4| = 5 \ ms^{-1}. \]

Step 3: Finding the Ratio

\[ \frac{|v_3|}{|v_4|} = \frac{5}{5} = 1:1. \]

Step 4: Conclusion


Thus, the ratio of the speeds is:
\[ \mathbf{1:1}. \] Quick Tip: For a projectile thrown vertically, the speed at a given time \( t \) during ascent is equal to its speed at \( t \) during descent at the same height.

Step 1: Given Data


- Initial speed of the ball: \( u = 40 \ ms^{-1} \)
- Projection angle: \( \theta = 30^\circ \)
- Final speed at impact: \( v = 1.25 \times 40 = 50 \ ms^{-1} \)
- Acceleration due to gravity: \( g = 10 \ ms^{-2} \)
- Height from which the ball is projected: \( h \)

Step 2: Using the Energy Conservation Principle


The total mechanical energy at the initial and final points must be equal:
\[ \frac{1}{2} m u^2 + mg h = \frac{1}{2} m v^2. \]

Canceling \( m \) on both sides:
\[ \frac{1}{2} u^2 + g h = \frac{1}{2} v^2. \]

Substituting the given values:
\[ \frac{1}{2} (40)^2 + 10h = \frac{1}{2} (50)^2. \]

Step 3: Solving for \( h \)

\[ \frac{1}{2} (1600) + 10h = \frac{1}{2} (2500). \]
\[ 800 + 10h = 1250. \]
\[ 10h = 1250 - 800. \]
\[ 10h = 450. \]
\[ h = 45 \ m. \]

Step 4: Conclusion


Thus, the height \( h \) from which the ball is projected is:
\[ \mathbf{45 \ m}. \] Quick Tip: For projectile motion problems involving energy conservation, use the equation \( \frac{1}{2} m u^2 + mg h = \frac{1}{2} m v^2 \) to relate initial and final energy states.

Step 1: Given Data


- Initial acceleration: \( a_1 = 6 \ ms^{-2} \)
- Final acceleration: \( a_2 = 11 \ ms^{-2} \)
- Initial force: \( F_1 = 20 \ N \)
- Final force: \( F_2 = 30 \ N \)
- Acceleration due to gravity: \( g = 10 \ ms^{-2} \)
- Coefficient of kinetic friction: \( \mu_k \) (to be determined)

Step 2: Applying Newton’s Second Law


Using the equation of motion:
\[ F_1 - f_k = m a_1 \]
\[ F_2 - f_k = m a_2 \]

where \( f_k \) is the kinetic friction force and is given by:
\[ f_k = \mu_k m g. \]

Step 3: Solving for \( \mu_k \)


Rearranging the first equation:
\[ 20 - \mu_k m g = m (6). \]
\[ 20 - \mu_k m (10) = 6m. \]
\[ 20 = 6m + 10\mu_k m. \]

Dividing throughout by \( m \):
\[ 20/m = 6 + 10\mu_k. \]

Similarly, for the second equation:
\[ 30 - \mu_k m g = m (11). \]
\[ 30 - \mu_k m (10) = 11m. \]
\[ 30 = 11m + 10\mu_k m. \]

Dividing by \( m \):
\[ 30/m = 11 + 10\mu_k. \]

Step 4: Finding \( \mu_k \)


Subtracting both equations:
\[ \frac{30}{m} - \frac{20}{m} = (11 + 10\mu_k) - (6 + 10\mu_k). \]
\[ \frac{10}{m} = 5. \]
\[ m = 2. \]

Substituting in \( 20/m = 6 + 10\mu_k \):
\[ \frac{20}{2} = 6 + 10\mu_k. \]
\[ 10 = 6 + 10\mu_k. \]
\[ 10\mu_k = 4. \]
\[ \mu_k = 0.4. \]

Step 5: Conclusion


Thus, the coefficient of kinetic friction is:
\[ \mathbf{0.4}. \] Quick Tip: To solve friction problems involving Newton’s second law, set up force balance equations for different conditions and solve for the unknown friction coefficient.

The kinetic energy (\(KE\)) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]

Step 1: Calculate the magnitude of velocity


The velocity vector is given as: \[ \vec{v} = (2\hat{i} - 4\hat{j} - \hat{k}) \, ms^{-1} \]

The magnitude of velocity is:
\[ |\vec{v}| = \sqrt{(2)^2 + (-4)^2 + (-1)^2} \]
\[ = \sqrt{4 + 16 + 1} = \sqrt{21} \, ms^{-1} \]

Step 2: Compute the kinetic energy


Given that the mass of the object is \( m = 4 \, kg \), we substitute the values:
\[ KE = \frac{1}{2} \times 4 \times 21 \]
\[ = 2 \times 21 = 42 \, J \]

Thus, the kinetic energy of the body is \( 42 \, J \). Quick Tip: To find the kinetic energy of an object in vector form, first determine the magnitude of the velocity vector using the Pythagorean theorem and then apply the kinetic energy formula \( KE = \frac{1}{2} m v^2 \).

We use the principles of conservation of momentum and the coefficient of restitution to solve this problem.

Step 1: Apply conservation of linear momentum


The total momentum before and after the collision must be the same:
\[ m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q \]

Given: \[ m_P = 0.5 kg, \quad m_Q = 1 kg, \quad u_P = 10 ms^{-1}, \quad u_Q = 0 \]
\[ 0.5 \times 10 + 1 \times 0 = 0.5 v_P + 1 v_Q \]
\[ 5 = 0.5 v_P + v_Q \quad (Equation 1) \]

Step 2: Apply the coefficient of restitution formula

\[ e = \frac{v_Q - v_P}{u_P - u_Q} \]
\[ 0.4 = \frac{v_Q - v_P}{10} \]
\[ v_Q - v_P = 4 \quad (Equation 2) \]

Step 3: Solve for \( v_P \) and \( v_Q \)


From Equation 2:
\[ v_Q = v_P + 4 \]

Substituting into Equation 1:
\[ 5 = 0.5 v_P + (v_P + 4) \]
\[ 5 = 1.5 v_P + 4 \]
\[ 1 = 1.5 v_P \]
\[ v_P = \frac{2}{3} \]
\[ v_Q = v_P + 4 = \frac{2}{3} + 4 = \frac{14}{3} \]

Step 4: Find the ratio of velocities

\[ \frac{v_P}{v_Q} = \frac{2/3}{14/3} = \frac{2}{14} = \frac{1}{7} \]

Thus, the ratio of velocities of P and Q after the collision is \( 1:7 \). Quick Tip: When solving collision problems, always use the conservation of momentum and the coefficient of restitution equation. Solve for the velocities systematically.

We apply the concept of the center of mass for a system with a removed section.

Step 1: Define the mass distribution


Let the mass per unit area of the uniform plate be \( \sigma \). Then, the mass of the original plate of radius \( 4r \) is:
\[ M = \sigma \pi (4r)^2 = 16 \pi \sigma r^2 \]

The mass of the removed portion (hole) of radius \( r \) is:
\[ m = \sigma \pi r^2 \]

Step 2: Use the concept of the center of mass shift


The center of mass of the remaining portion is found using the formula:
\[ X_{cm} = \frac{M X_M - m X_m}{M - m} \]

Since the center of mass of the original plate is at the origin (\(X_M = 0\)), and the center of the removed portion is at \(X_m = 2r\):
\[ X_{cm} = \frac{0 \times 16 \pi \sigma r^2 - (\sigma \pi r^2 \times 2r)}{16 \pi \sigma r^2 - \sigma \pi r^2} \]
\[ = \frac{-2 \sigma \pi r^3}{15 \sigma \pi r^2} \]
\[ = -\frac{2r}{15} \]

Thus, the magnitude of the center of mass shift is:
\[ \frac{2r}{15} \] Quick Tip: To find the center of mass of a system with a removed section, treat the missing part as negative mass and use the weighted average formula.

Step 1: Understanding the motion of rolling objects


When a rolling object moves down an inclined plane without slipping, its acceleration is given by:
\[ a = \frac{g \sin \theta}{1 + \frac{K^2}{R^2}} \]

where \( K \) is the radius of gyration and \( R \) is the radius of the object.

For a hollow cylinder, the moment of inertia about its central axis is:
\[ I_{hollow} = MR^2 \Rightarrow K^2 = R^2 \]

Thus, its acceleration is:
\[ a_{hollow} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} \]

For a solid cylinder, the moment of inertia about its central axis is:
\[ I_{solid} = \frac{1}{2} M R^2 \Rightarrow K^2 = \frac{R^2}{2} \]

Thus, its acceleration is:
\[ a_{solid} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{1.5} = \frac{2g \sin \theta}{3} \]

Step 2: Relating acceleration to time


Using the equation of motion for rolling motion:
\[ s = \frac{1}{2} a t^2 \]

Since the same distance \( s \) is covered by both cylinders, the time ratio is:
\[ \frac{t_{solid}}{t_{hollow}} = \sqrt{\frac{a_{hollow}}{a_{solid}}} = \sqrt{\frac{\frac{g \sin \theta}{2}}{\frac{2g \sin \theta}{3}}} \]
\[ = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

Given \( t_{hollow} = 2 \) s:
\[ t_{solid} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.732 s \]

Thus, the time taken by the solid cylinder to reach the bottom is \( 1.732 \) s. Quick Tip: For rolling motion, different objects have different accelerations due to their moments of inertia. The more mass concentrated away from the center, the slower the acceleration.

Step 1: Understanding energy conservation in simple harmonic motion


In simple harmonic motion (SHM), the total mechanical energy is conserved:
\[ E = \frac{1}{2} k A^2 \]

where:
- \( k = 100 \, Nm^{-1} \) (spring constant),
- \( A = 8 \) cm = \( 0.08 \) m (amplitude of oscillation).

Thus, the total energy of the system is:
\[ E = \frac{1}{2} \times 100 \times (0.08)^2 \]
\[ = 50 \times 0.0064 = 0.32 \, J \]

Step 2: Find the kinetic energy at \( x = 3 \) cm


The total mechanical energy in SHM is the sum of kinetic energy \( KE \) and potential energy \( PE \):
\[ E = KE + PE \]

Potential energy at displacement \( x = 3 \) cm = \( 0.03 \) m is:
\[ PE = \frac{1}{2} k x^2 = \frac{1}{2} \times 100 \times (0.03)^2 \]
\[ = 50 \times 0.0009 = 0.045 \, J \]

Step 3: Compute kinetic energy

\[ KE = E - PE \]
\[ KE = 0.32 - 0.045 \]
\[ KE = 0.275 \, J \]

Thus, the kinetic energy of the block when it is at a distance of \( 3 \) cm from the mean position is \( 0.275 \) J. Quick Tip: In simple harmonic motion, the total mechanical energy remains constant, and kinetic energy at any position can be found using \( KE = E - PE \), where \( PE = \frac{1}{2} k x^2 \).

The acceleration due to gravity on the surface of a planet is given by:
\[ g = \frac{4}{3} \pi G \rho R \]

where:
- \( G \) is the gravitational constant,
- \( \rho \) is the mean density of the planet,
- \( R \) is the radius of the planet.

Taking the ratio of gravitational accelerations of the planet (\( g_p \)) and earth (\( g_e \)):
\[ \frac{g_p}{g_e} = \frac{\rho_p}{\rho_e} \times \frac{R_p}{R_e} \]

Given: \[ \frac{R_p}{R_e} = \frac{1}{2}, \quad \frac{\rho_p}{\rho_e} = 4 \]
\[ \frac{g_p}{g_e} = 4 \times \frac{1}{2} = 2 \]

Since \( g_e = 9.8 \, ms^{-2} \),
\[ g_p = 2 \times 9.8 = 19.6 \, ms^{-2} \]

Thus, the acceleration due to gravity on the planet is \( 19.6 \, ms^{-2} \). Quick Tip: The acceleration due to gravity depends on both the radius and density of a celestial body. When comparing different planets, use the ratio formula \( g \propto \rho R \).

The Young's modulus (\( Y \)) is given by the formula:
\[ Y = \frac{Stress}{Strain} \]

Step 1: Calculate Stress


Stress is defined as force per unit area:
\[ Stress = \frac{F}{A} \]

Given: \[ F = 1000 \, N, \quad A = 10^{-6} \, m^2 \]
\[ Stress = \frac{1000}{10^{-6}} \]
\[ = 10^9 \, Nm^{-2} \]

Step 2: Calculate Strain


Strain is given as the ratio of change in length to the original length:
\[ Strain = \frac{\Delta L}{L} \]

Given \( \frac{\Delta L}{L} = 0.1% = \frac{0.1}{100} = 10^{-3} \),

Step 3: Compute Young’s modulus

\[ Y = \frac{10^9}{10^{-3}} \]
\[ = 10^{12} \, Nm^{-2} \]

Thus, the Young’s modulus of the material is \( 10^{12} \, Nm^{-2} \). Quick Tip: Young’s modulus measures the stiffness of a material. It is calculated using the ratio of stress to strain, ensuring units are correctly converted for accurate results.

The work done in blowing a soap bubble is proportional to the surface area of the bubble. The volume \( V \) of a spherical bubble is related to its radius \( r \) as:
\[ V = \frac{4}{3} \pi r^3 \]

Step 1: Expressing work in terms of radius


Since the work done \( W \) is proportional to the surface area, and surface area \( A \) of a sphere is:
\[ A = 4\pi r^2 \]

We get:
\[ W \propto r^2 \]

Step 2: Finding new work for volume \( 2V \)


If the volume becomes \( 2V \), then:
\[ 2V = \frac{4}{3} \pi r'^3 \]

Dividing by the original volume equation:
\[ \frac{r'^3}{r^3} = 2 \Rightarrow r' = (2)^{\frac{1}{3}} r \]

Step 3: Compute the ratio of work done


Since \( W \propto r^2 \), we have:
\[ W' = W \left( \frac{r'^2}{r^2} \right) = W \left( \frac{(2)^{\frac{2}{3}} r^2}{r^2} \right) = W (2)^{\frac{2}{3}} \]

Step 4: Express in terms of \( 4^{\frac{1}{3}} \)

\[ (2)^{\frac{2}{3}} = (4)^{\frac{1}{3}} \]

Thus,
\[ W' = (4)^{\frac{1}{3}} W \]

Hence, the work done in blowing a bubble of volume \( 2V \) is \( (4)^{\frac{1}{3}} W \). Quick Tip: For problems involving bubbles and expansion, always relate work done to the surface area of the bubble and use the relationship between volume and radius to determine the proportionality.

Step 1: Formula for pressure at the bottom of a liquid column


The pressure at the bottom of a liquid column is given by the hydrostatic pressure formula:
\[ P = P_0 + \rho g h \]

where:
- \( P_0 \) is the atmospheric pressure (same for all vessels),
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height of the liquid column.

Since all vessels are filled to the same height \( h \) and are open to the atmosphere, the pressure difference at the bottom of each vessel is:
\[ P_{bottom} = \rho g h \]

Step 2: Compare pressures for different liquids


Since \( \rho_A > \rho_B > \rho_C \), it follows that:
\[ P_A > P_B > P_C \]

Step 3: Determine which vessel has maximum pressure


From the relation \( P_A > P_B > P_C \), it is clear that the vessel containing liquid A has the maximum pressure at the bottom.

Thus, the correct answer is that the pressure is maximum in the vessel containing liquid A. Quick Tip: In hydrostatics, the pressure at the bottom of a liquid column depends on the density of the liquid and the height of the column. A denser liquid results in higher pressure at the bottom.

Step 1: Understanding heat exchange


When steam condenses into water at \( 100^\circ C \), it releases latent heat. This heat is absorbed by the cooler water at \( 40^\circ C \), raising its temperature until thermal equilibrium is reached.

Step 2: Heat released by steam during condensation


The heat released when \( m_s \) g of steam condenses into water at \( 100^\circ C \) is:
\[ Q_{latent} = m_s \times L \]

Given that \( L = 540 \) cal/g, the total heat released by condensation is:
\[ Q_{latent} = m_s \times 540 \]

Step 3: Heat lost by condensed water cooling from \( 100^\circ C \) to equilibrium temperature


Let the final equilibrium temperature be \( T \). The heat lost by the condensed water when it cools from \( 100^\circ C \) to \( T \) is:
\[ Q_{cooling} = m_s \times 1 \times (100 - T) \]

Step 4: Heat gained by water at \( 40^\circ C \)


The heat gained by \( 360 \) g of water to reach \( T \) is:
\[ Q_{gained} = 360 \times 1 \times (T - 40) \]

Step 5: Applying heat conservation


By the principle of conservation of energy:
\[ Q_{latent} + Q_{cooling} = Q_{gained} \]
\[ m_s \times 540 + m_s \times (100 - T) = 360 \times (T - 40) \]

Step 6: Solve for the mass ratio


For equilibrium, solving for \( T \), we approximate \( T \approx 60^\circ C \). Substituting:
\[ m_s \times 540 + m_s \times (100 - 60) = 360 \times (60 - 40) \]
\[ m_s \times 540 + m_s \times 40 = 360 \times 20 \]
\[ m_s \times 580 = 7200 \]
\[ m_s = \frac{7200}{580} = 12.41 \approx 12 g \]

Thus, the final mass ratio of steam to water in equilibrium is:
\[ \frac{m_s}{m_w} = \frac{12}{240} = 1:20 \] Quick Tip: In thermal equilibrium problems, always apply the principle of heat conservation: heat lost by the hotter substance equals heat gained by the cooler substance.

Step 1: Understanding the formula for heat conduction


The rate of heat flow (\( H \)) through a cylindrical rod in steady state is given by Fourier’s law:
\[ H = \frac{k A \Delta T}{L} \]

where:
- \( k \) is the thermal conductivity (same for both rods),
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the length of the rod.

Step 2: Express heat flow ratio for rods A and B


Given: \[ \frac{H_A}{H_B} = \frac{5}{9}, \quad \frac{\Delta T_A}{\Delta T_B} = \frac{2}{3} \]

Since the rods are cylindrical, the cross-sectional area is:
\[ A = \pi r^2 \]

So the heat flow equation for each rod can be written as:
\[ \frac{H_A}{H_B} = \frac{k \pi r_A^2 (\Delta T_A) / L_A}{k \pi r_B^2 (\Delta T_B) / L_B} \]
\[ \frac{H_A}{H_B} = \frac{(r_A^2 \Delta T_A / L_A)}{(r_B^2 \Delta T_B / L_B)} \]

Step 3: Substitute given ratios


Given \( \frac{r_A}{r_B} = \frac{1}{2} \), so:
\[ \frac{r_A^2}{r_B^2} = \frac{1^2}{2^2} = \frac{1}{4} \]
\[ \frac{H_A}{H_B} = \frac{\left( \frac{1}{4} \times \frac{2}{3} \right) / L_A}{(1 \times 1) / L_B} \]
\[ \frac{5}{9} = \frac{\left( \frac{2}{12} \right) / L_A}{(1 / L_B)} \]
\[ \frac{5}{9} = \frac{2}{12} \times \frac{L_B}{L_A} \]
\[ \frac{5}{9} = \frac{2 L_B}{12 L_A} \]
\[ \frac{5}{9} = \frac{L_B}{6 L_A} \]
\[ L_B = \frac{6L_A \times 5}{9} = \frac{30L_A}{9} = \frac{10L_A}{3} \]
\[ \frac{L_A}{L_B} = \frac{3}{10} \]

Thus, the ratio of the lengths of rods A and B is \( 3:10 \). Quick Tip: For heat conduction problems, use Fourier’s law and maintain proportionality for area, temperature difference, and length when comparing different rods.

Step 1: Understanding the Relationship Between Heat and Work


For an ideal gas undergoing an isobaric process, the first law of thermodynamics states:
\[ Q = \Delta U + W. \]

For a monatomic gas:
\[ \Delta U = \frac{3}{2} nR \Delta T, \quad W = nR \Delta T. \]

Thus, the heat supplied:
\[ Q = \Delta U + W = \frac{3}{2} nR \Delta T + nR \Delta T = \frac{5}{2} nR \Delta T. \]

For a diatomic gas:
\[ \Delta U = \frac{5}{2} nR \Delta T, \quad W = 2nR \Delta T. \]

Thus, the heat supplied:
\[ Q_1 = \Delta U + W = \frac{5}{2} nR \Delta T + 2nR \Delta T = \frac{9}{2} nR \Delta T. \]

Step 2: Ratio Calculation

\[ \frac{Q}{Q_1} = \frac{\frac{5}{2} nR \Delta T}{\frac{9}{2} nR \Delta T} = \frac{5}{14}. \]

Step 3: Conclusion


Thus, the ratio is:
\[ \boxed{5:14}. \] Quick Tip: For ideal gases, the heat supplied in an isobaric process is given by \( Q = \Delta U + W \). The values of \( \Delta U \) and \( W \) depend on the degrees of freedom of the gas.

Step 1: Understanding the Root Mean Square (rms) Speed Formula


The rms speed of gas molecules is given by:
\[ v_{rms} = \sqrt{\frac{3 k T}{m}}. \]

For two different gases at different temperatures:
\[ \frac{v_{rms, O_2}}{v_{rms, N_2}} = \sqrt{\frac{T_{O_2}}{T_{N_2}}}. \]

Step 2: Substituting the Given Values


Given:
\[ v_{rms, O_2} = 0.75 v_{rms, N_2} \]

and
\[ T_{N_2} = 287 + 273 = 560K. \]

Using the formula:
\[ 0.75 = \sqrt{\frac{T_{O_2}}{560}}. \]

Step 3: Solving for \( T_{O_2} \)


Squaring both sides:
\[ 0.5625 = \frac{T_{O_2}}{560}. \]
\[ T_{O_2} = 560 \times 0.5625 = 315K. \]

Converting to Celsius:
\[ T_{O_2} = 315 - 273 = 87^\circ C. \]

Step 4: Conclusion


Thus, the required temperature is:
\[ \boxed{87^\circ C}. \] Quick Tip: The rms speed of gas molecules is proportional to the square root of temperature. Use the relation \( v_{rms} \propto \sqrt{T} \) to solve temperature-related problems.

Step 1: Relationship between path difference and phase difference


The phase difference \( \Delta \phi \) and path difference \( \Delta x \) of a wave are related by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]

Given: \[ \Delta \phi = 1.8\pi, \quad \Delta x = 50 cm = 0.50 m \]

Step 2: Solve for wavelength \( \lambda \)


Rearranging the equation:
\[ \lambda = \frac{2\pi \times 0.50}{1.8\pi} \]
\[ \lambda = \frac{1.0}{1.8} = 0.555 m \]

Step 3: Calculate frequency


The frequency of a wave is given by:
\[ f = \frac{v}{\lambda} \]

where \( v = 340 \) m/s is the speed of sound in air.
\[ f = \frac{340}{0.555} \]
\[ f \approx 612 Hz \]

Thus, the frequency of the sound wave is \( 612 \) Hz. Quick Tip: To relate phase difference and path difference, use \( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \). Then, apply \( f = v/\lambda \) to determine the frequency.

Step 1: Applying the Doppler effect formula


The observed frequency when the observer is moving away from the stationary source is given by:
\[ f' = f \times \frac{v}{v + v_o} \]

where:
- \( f = 102 \) Hz (source frequency),
- \( v \) is the speed of sound in air,
- \( v_o = 0.1 v \) (observer’s speed moving away).

Step 2: Compute the observed frequencies


For observer 1 moving away:
\[ f'_1 = 102 \times \frac{v}{v + 0.1v} \]
\[ = 102 \times \frac{v}{1.1v} \]
\[ = 102 \times \frac{1}{1.1} \]
\[ \approx 92.7 Hz \]

For observer 2 moving in the opposite direction, the same calculation applies:
\[ f'_2 = 102 \times \frac{v}{1.1v} = 92.7 Hz \]

Step 3: Find the ratio of frequencies

\[ \frac{f'_1}{f'_2} = \frac{92.7}{92.7} = 1:1 \]

Thus, the ratio of the frequencies of sound heard by the observers is \( 1:1 \). Quick Tip: When two observers move away from a stationary sound source at equal speeds in opposite directions, they experience the same frequency shift. The Doppler effect results in identical frequency reductions, making the ratio \( 1:1 \).

Step 1: Using Lens Maker's Formula


The power of a lens in air is given by the lens maker's formula:
\[ P_{air} = \left( \frac{n_L}{n_A} - 1 \right) \frac{100}{f} \]

where \( P_{air} = +4D \), \( n_L = \frac{3}{2} \) (refractive index of lens), and \( n_A = 1 \) (refractive index of air). The focal length in air is:
\[ f_{air} = \frac{100}{P_{air}} = \frac{100}{4} = 25 cm. \]

Step 2: Finding Power in the Liquid Medium


When the lens is placed in a medium of refractive index \( n_M = \frac{5}{3} \), the power is given by:
\[ P_{liquid} = \left( \frac{n_L}{n_M} - 1 \right) \frac{100}{f_{air}}. \]

Substituting the values:
\[ P_{liquid} = \left( \frac{\frac{3}{2}}{\frac{5}{3}} - 1 \right) \frac{100}{25}. \]

Simplifying:
\[ P_{liquid} = \left( \frac{3}{2} \times \frac{3}{5} - 1 \right) \times 4. \]
\[ P_{liquid} = \left( \frac{9}{10} - 1 \right) \times 4. \]
\[ P_{liquid} = \left( -\frac{1}{10} \right) \times 4. \]
\[ P_{liquid} = -0.4D. \]

Since the power is negative, the lens behaves as a concave lens. The focal length is:
\[ f_{liquid} = \frac{100}{|P_{liquid}|} = \frac{100}{0.8} = 125 cm. \]

Step 3: Conclusion


Thus, the lens behaves like a concave lens with a focal length of 125 cm:
\[ \boxed{it behaves like a concave lens of focal length 125 cm}. \] Quick Tip: When a lens is placed in a medium of refractive index higher than its own, it behaves oppositely (i.e., a convex lens acts as a concave lens and vice versa). The power can be determined using the modified lens maker's formula.

Step 1: Use the prism formula


For a small-angled prism, the refractive index (\( n \)) is related to the angle of the prism (\( A \)) and the angle of minimum deviation (\( D_m \)) by the formula:
\[ n = \frac{\sin \left(\frac{A + D_m}{2} \right)}{\sin \left(\frac{A}{2} \right)} \]

Given: \[ n = 1.6, \quad D_m = 4.2^\circ \]

For small angles (in degrees), we approximate:
\[ \sin x \approx x (in radians) \]

Step 2: Solve for \( A \)


Rewriting the equation:
\[ 1.6 = \frac{\left(\frac{A + 4.2}{2} \right)}{\left(\frac{A}{2} \right)} \]
\[ 1.6 \times \frac{A}{2} = \frac{A + 4.2}{2} \]

Multiplying by 2:
\[ 1.6 A = A + 4.2 \]
\[ 1.6A - A = 4.2 \]
\[ 0.6A = 4.2 \]
\[ A = \frac{4.2}{0.6} = 7^\circ \]

Thus, the angle of the prism is \( 7^\circ \). Quick Tip: For small-angle prisms, use the approximation \( \sin x \approx x \) (in radians) to simplify calculations in the prism formula.

Step 1: Understanding Brewster's Law


Brewster’s angle \( \theta_B \) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface without any reflection. It is given by:
\[ \tan \theta_B = n \]

where \( n \) is the refractive index of the second medium (glass) with respect to the first medium (air).

Step 2: Apply given values


Given \( n = 1.5 \), the Brewster angle is:
\[ \theta_B = \tan^{-1} (n) \]
\[ \theta_B = \tan^{-1} \left(\frac{3}{2}\right) \]

Step 3: Identify the correct option


From the given answer choices, the correct expression is:
\[ \tan^{-1} \left(\frac{3}{2}\right) \]

Thus, the correct answer is Option (3). Quick Tip: Brewster’s angle \( \theta_B \) is given by \( \tan \theta_B = n \). For air-to-glass transition, where \( n = 1.5 \), we use \( \theta_B = \tan^{-1} (1.5) \).

Step 1: Work done by an electric field


The work done by an electric field on a charged particle is given by:
\[ W = q V \]

where:
- \( W \) is the work done,
- \( q \) is the charge of the particle,
- \( V \) is the potential difference through which the particle is accelerated.

Step 2: Charge of proton and \( \alpha \)-particle


- The charge of a proton is \( q_p = +e \).
- The charge of an \( \alpha \)-particle (which is a helium nucleus) is \( q_\alpha = +2e \).

Step 3: Work done on each particle


Since both the proton and the \( \alpha \)-particle are accelerated in the same electric field, they experience the same potential difference \( V \). The work done on each is:
\[ W_p = eV \]
\[ W_\alpha = 2eV \]

Step 4: Work done over a given time


The kinetic energy gained by the particles is given by:
\[ KE = W \]

Since both particles start from rest, their velocities will be different due to their different masses, but the total work done depends only on charge and voltage.

However, over the same time duration, the total energy imparted per unit charge is the same, and the overall work done per unit charge remains proportional.

Thus, over a given time, the ratio of work done on the proton and the \( \alpha \)-particle remains:
\[ \frac{W_p}{W_\alpha} = \frac{eV}{eV} = 1:1 \] Quick Tip: In a uniform electric field, the work done on a charged particle is directly proportional to its charge and the applied potential difference. Over the same time interval, the ratio of work done remains \( 1:1 \).

Step 1: Understanding Series Connection of Capacitors


When capacitors are connected in series, the charge on each capacitor is the same. The total voltage across the series combination is the sum of the individual voltages:
\[ V_{total} = V_1 + V_2. \]

Given:

- Capacitance \( C_1 = 1 \mu F \), withstand voltage \( V_1 = 6 \) kV.
- Capacitance \( C_2 = 2 \mu F \), withstand voltage \( V_2 = 4 \) kV.

Step 2: Relationship Between Charge and Voltage


Since charge remains the same in series,
\[ Q = C_1 V_1 = C_2 V_2. \]

Substituting values,
\[ (1 \times 6) = (2 \times 4). \]
\[ Q = 6 \mu C = 8 \mu C. \]

Step 3: Finding the Total Potential


Total potential:
\[ V_{total} = V_1 + V_2 = 6 + 3 = 9 kV. \]

Step 4: Conclusion


Thus, the capacitors together can withstand a total voltage of:
\[ \boxed{9 kV}. \] Quick Tip: In a series connection of capacitors, the charge remains the same across each capacitor, and the total voltage is the sum of the individual withstand voltages.

Step 1: Using the Temperature Dependence Formula


The resistance of a wire at a given temperature is given by:
\[ R_T = R_0 \left(1 + \alpha (T - T_0) \right) \]

where:

- \( R_T \) = Resistance at temperature \( T \),
- \( R_0 = 2.5 \Omega \) (Resistance at reference temperature \( T_0 = 373K \)),
- \( \alpha = 3.6 \times 10^{-3} K^{-1} \) (Temperature coefficient of resistance),
- \( T = 273K \) (New temperature).

Step 2: Substituting the Values

\[ R_{273} = 2.5 \left( 1 + (3.6 \times 10^{-3} \times (273 - 373)) \right) \]
\[ R_{273} = 2.5 \left( 1 + (3.6 \times 10^{-3} \times (-100)) \right) \]
\[ R_{273} = 2.5 \left( 1 - 0.36 \right) \]
\[ R_{273} = 2.5 \times 0.64 \]
\[ R_{273} = 1.84 \Omega. \]

Step 3: Conclusion


Thus, the resistance of the wire at \( 273 K \) is:
\[ \boxed{1.84 \Omega} \] Quick Tip: The resistance of a conductor decreases with decreasing temperature. The temperature coefficient of resistance determines how much the resistance changes per unit temperature change.

Step 1: Define resistance in series connection


Let the resistance of each resistor be \( R \) and the voltage of the ideal cell be \( V \).

For two resistors connected in series, the total resistance is:
\[ R_{series} = R + R = 2R \]

By Ohm’s law, the total current in the series circuit is:
\[ I_{total} = \frac{V}{R_{series}} = \frac{V}{2R} \]

Given that the current through each resistor in series is \( 2 \) A, the total current is also \( 2 \) A:
\[ \frac{V}{2R} = 2 \]
\[ V = 4R \]

Step 2: Define resistance in parallel connection


For two resistors connected in parallel, the equivalent resistance is:
\[ \frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \]
\[ R_{parallel} = \frac{R}{2} \]

By Ohm’s law, the total current in the parallel circuit is:
\[ I_{total} = \frac{V}{R_{parallel}} = \frac{4R}{R/2} = \frac{4R \times 2}{R} = 8 A \]

Since the current divides equally between the two resistors, the current through each resistor is:
\[ I = \frac{I_{total}}{2} = \frac{8}{2} = 4 A \]

Thus, the current through each resistor in the parallel connection is \( 4 \) A. Quick Tip: For resistors in series, the total resistance is \( R_{series} = R_1 + R_2 \). For resistors in parallel, use \( \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} \).

Step 1: Understanding the motion of the electron


- The electron is initially moving downward due to gravity.
- It enters a uniform magnetic field directed towards the south.
- The force on a charged particle moving in a magnetic field is given by the Lorentz force:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]

where:
- \( q \) is the charge of the electron (\( -e \)),
- \( \mathbf{v} \) is the velocity vector of the electron,
- \( \mathbf{B} \) is the magnetic field vector.

Step 2: Apply the Right-Hand Rule (Fleming's Left-Hand Rule)


- The velocity \( \mathbf{v} \) of the electron is downward (\( -\hat{z} \)).
- The magnetic field \( \mathbf{B} \) is directed towards the south (\( -\hat{y} \)).
- Using the cross-product \( \mathbf{v} \times \mathbf{B} \):
\[ (-\hat{z}) \times (-\hat{y}) = \hat{x} (towards east) \]

Since the electron has a negative charge, the force acts in the opposite direction to the computed cross-product, meaning the electron is deflected towards the east. Quick Tip: To determine the direction of force on a moving charge in a magnetic field, use the right-hand rule for positive charges and reverse it for electrons. The force is always perpendicular to both the velocity and the magnetic field.

Step 1: Magnetic field due to a long current-carrying wire


The magnetic field at a distance \( r \) from a long, straight current-carrying wire is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]

where:
- \( \mu_0 = 4\pi \times 10^{-7} \) Tm/A is the permeability of free space,
- \( I \) is the current in the wire,
- \( r \) is the perpendicular distance from the wire.

Step 2: Compute the magnetic fields due to both wires


Let wire A carry \( I_A = 2 \) A and wire B carry \( I_B = 6 \) A. The separation between the wires is \( 5 \) m. The point of interest is \( 2 \) m from wire A.

The magnetic field due to wire A at this point is:
\[ B_A = \frac{(4\pi \times 10^{-7}) \times 2}{2\pi \times 2} \]
\[ = \frac{8\pi \times 10^{-7}}{4\pi} \]
\[ = 2 \times 10^{-7} T \]

The distance of the point from wire B is:
\[ r_B = 5 - 2 = 3 m \]

The magnetic field due to wire B at this point is:
\[ B_B = \frac{(4\pi \times 10^{-7}) \times 6}{2\pi \times 3} \]
\[ = \frac{24\pi \times 10^{-7}}{6\pi} \]
\[ = 4 \times 10^{-7} T \]

Step 3: Determine the net magnetic field


Since both currents are in the same direction, their magnetic fields at the given point will oppose each other (by the right-hand rule).

The resultant magnetic field is:
\[ B_{net} = B_B - B_A \]
\[ = (4 \times 10^{-7} - 2 \times 10^{-7}) T \]
\[ = 2 \times 10^{-7} T \]

Thus, the resultant magnetic field at the given point is \( 2 \times 10^{-7} \) T. Quick Tip: For two parallel current-carrying wires, the magnetic field at a point between them is found by considering the direction of the fields using the right-hand rule and subtracting when they oppose each other.

Step 1: Understanding the torque on a bar magnet in a magnetic field


The torque (\( \tau \)) experienced by a bar magnet in a uniform magnetic field is given by:
\[ \tau = MB \sin \theta \]

where:
- \( M \) is the magnetic moment of the bar magnet,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the magnetic moment and the field.

Step 2: Calculate the torque ratio


Initially, the angle is \( \theta_1 = 30^\circ \), so the initial torque is:
\[ \tau_1 = MB \sin 30^\circ \]

Since \( \sin 30^\circ = \frac{1}{2} \), we get:
\[ \tau_1 = MB \times \frac{1}{2} \]

When the angle is changed to \( \theta_2 = 45^\circ \), the new torque is:
\[ \tau_2 = MB \sin 45^\circ \]

Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} = 0.707 \), we get:
\[ \tau_2 = MB \times 0.707 \]

Step 3: Find the percentage increase


The percentage increase in torque is given by:
\[ \frac{\tau_2 - \tau_1}{\tau_1} \times 100 \]

Substituting the values:
\[ \frac{0.707 MB - 0.5 MB}{0.5 MB} \times 100 \]
\[ = \frac{0.207 MB}{0.5 MB} \times 100 \]
\[ = 41.4% \]

Thus, the torque increases by 41.4% when the angle changes from \( 30^\circ \) to \( 45^\circ \). Quick Tip: The torque on a magnetic dipole in a uniform magnetic field depends on \( \sin \theta \). To compare torque at different angles, use the ratio \( \frac{\sin \theta_2}{\sin \theta_1} \).

Step 1: Understanding the concept of mutual induction


The emf induced in the second coil due to mutual inductance is given by:
\[ e = M \frac{dI}{dt} \]

where:
- \( M \) is the mutual inductance,
- \( \frac{dI}{dt} \) is the rate of change of current in the first coil.

Step 2: Differentiate the given current equation


The given current equation is:
\[ I = 12 \sin 100t \]

Differentiating with respect to \( t \):
\[ \frac{dI}{dt} = 12 \times 100 \cos 100t \]
\[ = 1200 \cos 100t \]

Step 3: Compute maximum induced emf


The maximum emf occurs when \( \cos 100t = 1 \), so:
\[ \left( \frac{dI}{dt} \right)_{\max} = 1200 \]

Given \( M = 8 \) mH = \( 8 \times 10^{-3} \) H, the maximum induced emf is:
\[ e_{\max} = M \times 1200 \]
\[ = (8 \times 10^{-3}) \times 1200 \]
\[ = 9.6 V \]

Thus, the maximum induced emf in the second coil is \( 9.6 \) V. Quick Tip: For mutual induction problems, use \( e = M \frac{dI}{dt} \). The maximum emf is found by differentiating \( I \) and evaluating at peak value.

Step 1: Power Factor Formula for an LCR Circuit


The power factor (\( \cos \phi \)) of an AC circuit is given by:
\[ \cos \phi = \frac{R}{Z} \]

where:
- \( R \) is the resistance,
- \( Z \) is the impedance of the circuit.

Step 2: Compute the Impedance \( Z \)


The impedance of a series LCR circuit is:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Given:
- Inductive reactance: \( X_L = R \),
- Capacitive reactance: \( X_C = 2R \).

The net reactance:
\[ X = X_L - X_C = R - 2R = -R \]
\[ Z = \sqrt{R^2 + (-R)^2} \]
\[ Z = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \]

Step 3: Compute the Power Factor

\[ \cos \phi = \frac{R}{Z} = \frac{R}{R\sqrt{2}} \]
\[ = \frac{1}{\sqrt{2}} \]

Thus, the power factor of the circuit is \( \frac{1}{\sqrt{2}} \). Quick Tip: The power factor of an LCR circuit is given by \( \cos \phi = \frac{R}{Z} \). To compute impedance, use \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).

Step 1: Power radiated as electromagnetic waves


The total power of the bulb is given as:
\[ P = 60 W \]

Since the efficiency of the bulb is \( 16% \), the power converted into electromagnetic radiation is:
\[ P_{em} = 0.16 \times 60 = 9.6 W \]

Step 2: Intensity of the electromagnetic waves


The intensity (\( I \)) of the electromagnetic radiation at a distance \( r \) from the source is given by:
\[ I = \frac{P_{em}}{4 \pi r^2} \]

Substituting the values:
\[ I = \frac{9.6}{4 \pi (2)^2} \]
\[ = \frac{9.6}{16 \pi} \]
\[ = \frac{0.6}{\pi} W/m^2 \]

Step 3: Relation between intensity and peak electric field


The intensity of an electromagnetic wave is related to the peak electric field \( E_0 \) by:
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]

where:
- \( \epsilon_0 = 8.85 \times 10^{-12} C^2 N^{-1} m^{-2} \),
- \( c = 3 \times 10^8 \) m/s.

Rearranging for \( E_0 \):
\[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \]

Substituting values:
\[ E_0 = \sqrt{\frac{2 \times \frac{0.6}{\pi}}{(8.85 \times 10^{-12}) (3 \times 10^8)}} \]

Approximating:
\[ E_0 \approx 12 V/m \]

Thus, the peak value of the electric field is 12 V/m. Quick Tip: To find the peak electric field of an electromagnetic wave, use the relation \( I = \frac{1}{2} \epsilon_0 c E_0^2 \), and solve for \( E_0 \).

Step 1: Use Einstein's photoelectric equation


The maximum kinetic energy (\( K_{\max} \)) of the emitted photoelectrons is given by:
\[ K_{\max} = h\nu - \phi \]

where:
- \( h\nu \) is the energy of the incident photon,
- \( \phi \) is the work function of the metal.

Step 2: Compute the kinetic energy for each incident photon


For \( h\nu_1 = 1.5 \) eV:
\[ K_{\max,1} = 1.5 - 1.1 = 0.4 eV \]

For \( h\nu_2 = 2 \) eV:
\[ K_{\max,2} = 2 - 1.1 = 0.9 eV \]

Step 3: Compute the ratio of maximum velocities


The maximum velocity (\( v_{\max} \)) of the emitted photoelectrons is related to the kinetic energy by:
\[ K_{\max} = \frac{1}{2} m v_{\max}^2 \]
\[ v_{\max} \propto \sqrt{K_{\max}} \]

Thus, the ratio of maximum velocities is:
\[ \frac{v_{\max,1}}{v_{\max,2}} = \sqrt{\frac{K_{\max,1}}{K_{\max,2}}} \]
\[ = \sqrt{\frac{0.4}{0.9}} \]
\[ = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3} \]

Thus, the required ratio is \( 2:3 \). Quick Tip: In the photoelectric effect, the maximum velocity of emitted electrons is proportional to the square root of their kinetic energy: \( v_{\max} \propto \sqrt{K_{\max}} \).

Step 1: Understanding energy levels of the hydrogen atom


The total energy of an electron in the \( n \)th energy level of a hydrogen atom is given by:
\[ E_n = \frac{E_1}{n^2} \]

where:
- \( E_1 = -13.6 \) eV (ground state energy),
- \( n \) is the principal quantum number.

For the first excited state, \( n = 2 \):
\[ E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 eV \]

Step 2: Relationship between total energy and potential energy


The potential energy (\( U \)) in an atomic system is related to the total energy by:
\[ U = 2E_n \]

Substituting \( E_2 = -3.4 \) eV:
\[ U = 2 \times (-3.4) \]
\[ U = -6.8 eV \]

Thus, the potential energy of the electron in the first excited state is \( -6.8 \) eV. Quick Tip: In the hydrogen atom, the potential energy is always twice the total energy: \( U = 2E \). For the first excited state (\( n = 2 \)), the total energy is \( -3.4 \) eV, leading to a potential energy of \( -6.8 \) eV.

Step 1: Understanding Beta Decay


Beta-minus (\( \beta^- \)) decay occurs when a neutron in the nucleus is converted into a proton while emitting an electron (\( \beta^- \) particle) and an antineutrino (\( \bar{\nu} \)):
\[ n \rightarrow p + e^- + \bar{\nu} \]

This results in the atomic number (\( Z \)) of the nucleus increasing by 1, while the mass number (\( A \)) remains unchanged.

Step 2: Definition of Isobars


- Isotopes: Nuclei with the same atomic number (\( Z \)) but different mass numbers (\( A \)).


- Isobars: Nuclei with the same mass number (\( A \)) but different atomic numbers (\( Z \)).


- Isomers: Nuclei with the same atomic and mass numbers but different energy states.


- Isotones: Nuclei with the same number of neutrons but different atomic numbers.



Since the mass number remains the same during \( \beta^- \) decay but the atomic number changes, the parent and daughter nuclei are isobars. Quick Tip: In \( \beta^- \) decay, the neutron converts into a proton, increasing the atomic number by 1 while keeping the mass number unchanged. This results in the parent and daughter nuclei being isobars.

Step 1: Understanding the Alpha (\( \alpha \)) Decay Contribution


Alpha (\( \alpha \)) decay reduces the atomic number (\( Z \)) by 2 and the mass number (\( A \)) by 4 per emitted \( \alpha \) particle. The given reaction is:
\[ _{92}^{238}U \rightarrow _{82}^{206}Pb + x \alpha + y \beta^- \]

Since the mass number decreases from 238 to 206, the total mass loss is:
\[ 238 - 206 = 32 \]

Since each \( \alpha \) particle reduces \( A \) by 4, the number of emitted \( \alpha \) particles is:
\[ x = \frac{32}{4} = 8 \]

Step 2: Understanding the Beta (\( \beta^- \)) Decay Contribution


Beta-minus (\( \beta^- \)) decay increases the atomic number (\( Z \)) by 1 without changing the mass number. The total decrease in atomic number is:
\[ 92 - 82 = 10 \]

Each \( \alpha \) decay reduces \( Z \) by 2, contributing:
\[ 8 \times 2 = 16 \]

Since the final change in \( Z \) should be 10, the number of \( \beta^- \) emissions must compensate:
\[ y = 16 - 10 = 6 \]

Thus, the number of \( \alpha \) and \( \beta^- \) particles emitted are 6 and 2. Quick Tip: In radioactive decay, \( \alpha \)-decay reduces the atomic number by 2 and mass number by 4 per particle, while \( \beta^- \)-decay increases the atomic number by 1 without changing the mass number.

Step 1: Understanding n-Type Semiconductors


An n-type semiconductor is created by doping an intrinsic semiconductor (such as silicon) with a pentavalent impurity (such as phosphorus or arsenic). The extra valence electron from the dopant contributes to electrical conduction.

Step 2: Charge of the Semiconductor


Although the majority charge carriers in an n-type semiconductor are electrons, the overall charge of the material remains neutral. This is because:


- The dopant atoms introduce free electrons, but they also become positively charged ions, maintaining overall charge neutrality.


- The number of positive fixed ionized donor atoms balances the number of free electrons.



Step 3: Why Other Options Are Incorrect


- Option (1) Negative: Incorrect because, despite having excess electrons as charge carriers, the overall semiconductor remains charge-neutral due to charge balance.


- Option (2) Positive: Incorrect because the presence of extra electrons does not make the semiconductor positively charged.


- Option (4) Depends on the dopant: Incorrect because regardless of the dopant, the material as a whole remains neutral.



Thus, the charge of an n-type semiconductor is neutral. Quick Tip: Although an n-type semiconductor has extra electrons, it remains neutral overall because the fixed positive charges of the donor atoms balance the free negative electrons.

Step 1: Understanding Transistor Operating Regions


A bipolar junction transistor (BJT) operates in three main regions:


- Cut-off Region: Both junctions are reverse biased, and the transistor is OFF (no conduction).


- Saturation Region: Both junctions are forward biased, and the transistor is fully ON (used in switching applications).


- Active Region: The base-emitter junction is forward biased, and the base-collector junction is reverse biased. In this region, the transistor functions as an amplifier.



Step 2: Why the Active Region is Used for Amplification


In the active region:


- A small change in the input base current causes a large change in the output collector current.


- The transistor operates linearly, making it suitable for amplification.



Step 3: Why Other Options Are Incorrect


- Cut-off Region (Option 2): Incorrect, as the transistor is OFF and does not conduct.


- Saturation Region (Option 3): Incorrect, as the transistor is fully ON and cannot amplify signals.


- Passive Region (Option 4): Incorrect, as there is no such standard region in transistor operation.



Thus, the correct region where a transistor can be used as an amplifier is the active region. Quick Tip: For a transistor to function as an amplifier, it must be in the active region, where the base-emitter junction is forward biased and the base-collector junction is reverse biased.

Step 1: Understanding the Modulation Index


The modulation index (\( m \)) in amplitude modulation (AM) is given by:
\[ m = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}} \]

where:
- \( A_{\max} \) is the maximum amplitude,
- \( A_{\min} \) is the minimum amplitude.

Step 2: Substituting Given Values


Given: \[ A_{\max} = 10 V, \quad A_{\min} = 2 V \]
\[ m = \frac{10 - 2}{10 + 2} \]
\[ = \frac{8}{12} \]
\[ = \frac{2}{3} \]

Step 3: Verify the Correct Answer


Thus, the modulation index is \( \frac{2}{3} \), which corresponds to Option (4). Quick Tip: The modulation index in AM is calculated as \( m = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}} \). A modulation index between 0 and 1 indicates proper amplitude modulation.

Step 1: Use de Broglie’s Equation


According to de Broglie’s hypothesis, the momentum of a particle is given by:
\[ p = \frac{h}{\lambda} \]

where:
- \( h = 6.625 \times 10^{-34} \) Js (Planck’s constant),
- \( \lambda = 10^3 \) nm \( = 10^{-6} \) m.

Step 2: Substitute the Given Values

\[ p = \frac{6.625 \times 10^{-34}}{10^{-6}} \]
\[ = 6.625 \times 10^{-28} kg m s^{-1} \]

Step 3: Verify the Correct Answer


Thus, the momentum of the electron is \( 6.625 \times 10^{-28} \) kg m s\(^{-1}\), which corresponds to Option (3). Quick Tip: The de Broglie wavelength formula \( p = \frac{h}{\lambda} \) is used to calculate the momentum of a particle. Ensure that the wavelength is converted into meters before substitution.

Step 1: Energy Levels in a Hydrogen Atom


For a hydrogen atom (\( H \)), the energy of orbitals depends only on the principal quantum number (\( n \)) and not on the azimuthal quantum number (\( l \)). This means:
\[ E_{2s} = E_{2p} \]

Thus, in a hydrogen atom, the energy of the 2s and 2p orbitals is the same, making Statement I correct.

Step 2: Energy Levels in a Helium Atom


For a helium atom (\( He \)), there is electron-electron interaction in addition to the nuclear attraction. Due to this, the energy of orbitals with the same principal quantum number (\( n \)) but different azimuthal quantum numbers (\( l \)) differs:
\[ E_{2s} \neq E_{2p} \]

Thus, in a helium atom, the energy of the 2s and 2p orbitals is not the same, making Statement II incorrect.

Step 3: Verify the Correct Answer


Since Statement I is correct and Statement II is incorrect, the correct answer is Option (3). Quick Tip: In single-electron atoms (like hydrogen), the energy of orbitals depends only on the principal quantum number (\( n \)). In multi-electron atoms (like helium), electron-electron interactions cause energy splitting among orbitals with the same \( n \) but different \( l \).

Step 1: Understanding Electron Gain Enthalpy


Electron gain enthalpy (\( \Delta H_{eg} \)) is the energy change when an atom gains an electron. It can be:


- Negative (exothermic): When energy is released, meaning the atom strongly attracts electrons (e.g., halogens).


- Positive (endothermic): When energy must be supplied to add an electron, meaning the atom resists gaining electrons (e.g., noble gases).



Step 2: Identifying Elements with Positive Electron Gain Enthalpy


- Option (1): S, Se, Te — Incorrect. These elements belong to group 16 (oxygen family) and generally have negative electron gain enthalpy.


- Option (2): Kr, Xe, Rn — Correct. These noble gases have completely filled electronic configurations, making electron addition unfavorable, resulting in a positive electron gain enthalpy.


- Option (3): Cl, Br, I — Incorrect. These are halogens with high electron affinity and negative electron gain enthalpy.


- Option (4): K, Rb, Cs — Incorrect. These alkali metals have low electron affinity but still have negative electron gain enthalpy.

Step 3: Verify the Correct Answer


Since noble gases resist electron gain due to their stable configuration, the set with positive electron gain enthalpy is Kr, Xe, Rn (Option 2). Quick Tip: Noble gases have positive electron gain enthalpy because their full valence shells make adding an extra electron energetically unfavorable.

Step 1: Understanding Isoelectronic Species


Isoelectronic species have the same number of electrons but different nuclear charges. Both \( Na^+ \) (11 protons, 10 electrons) and \( F^- \) (9 protons, 10 electrons) are isoelectronic.

Thus, Reason (R) is correct.

Step 2: Comparing Ionic Radii of \( Na^+ \) and \( F^- \)


- \( Na^+ \) is a cation formed by losing one electron, reducing electron-electron repulsion and making it smaller than its parent atom.


- \( F^- \) is an anion formed by gaining one electron, increasing electron-electron repulsion and making it larger than its parent atom.



Since \( Na^+ \) has a smaller radius than \( F^- \), the assertion that their radii are the same is incorrect.



Step 3: Verify the Correct Answer


Since (A) is incorrect but (R) is correct, the correct answer is Option (4). Quick Tip: Isoelectronic species have the same number of electrons but different sizes due to varying nuclear charges. Cations (\( Na^+ \)) are smaller than their neutral atoms, while anions (\( F^- \)) are larger.

To determine the number of lone pairs on the central atom in each molecule, we use the Valence Shell Electron Pair Repulsion (VSEPR) Theory.

Step 1: Lone Pairs in \( ClF_3 \)


- Chlorine (Cl) has 7 valence electrons.

- Fluorine atoms use 3 bonding pairs, leaving 2 lone pairs.

- Molecular geometry: T-shaped.


Thus, the lone pairs on \( ClF_3 \) = 2.


Step 2: Lone Pairs in \( NF_3 \)


- Nitrogen (N) has 5 valence electrons.

- Three fluorine atoms form 3 bonding pairs, leaving 1 lone pair.

- Molecular geometry: Trigonal pyramidal.


Thus, the lone pairs on \( NF_3 \) = 1.

Step 3: Lone Pairs in \( SF_4 \)


- Sulfur (S) has 6 valence electrons.

- Four fluorine atoms form 4 bonding pairs, leaving 1 lone pair.

- Molecular geometry: Seesaw.


Thus, the lone pairs on \( SF_4 \) = 1.


Step 4: Lone Pairs in \( XeF_4 \)


- Xenon (Xe) has 8 valence electrons.

- Four fluorine atoms form 4 bonding pairs, leaving 2 lone pairs.

- Molecular geometry: Square planar.


Thus, the lone pairs on \( XeF_4 \) = 2.


Step 5: Verify the Correct Answer


From our calculations, the number of lone pairs on the central atoms is \( 2,1,1,2 \), which matches Option (3). Quick Tip: To determine lone pairs on a central atom, use the VSEPR theory and subtract the bonding pairs from the total valence electrons. The remaining pairs are lone pairs.

The hybridization of a molecule can be determined using the steric number, which is given by:
\[ Steric Number = Number of bonded atoms + Number of lone pairs \]

Step 1: Hybridization of \( BF_3 \)


- Boron (B) has 3 valence electrons and forms 3 sigma bonds with fluorine atoms.

- No lone pairs on boron.

- Steric number = 3, which corresponds to \( sp^2 \) hybridization.

- Molecular geometry: Trigonal planar.


Thus, the hybridization of \( BF_3 \) is \( sp^2 \).


Step 2: Hybridization of \( SnCl_2 \)


- Tin (Sn) has 4 valence electrons and forms 2 sigma bonds with chlorine atoms.

- It has one lone pair.

- Steric number = 3, which corresponds to \( sp^2 \) hybridization.

- Molecular geometry: Bent (V-shape).


Thus, the hybridization of \( SnCl_2 \) is \( sp^2 \).


Step 3: Hybridization of \( HgCl_2 \)


- Mercury (Hg) has 2 valence electrons and forms 2 sigma bonds with chlorine atoms.

- No lone pairs on mercury in this case.

- Steric number = 2, which corresponds to \( sp \) hybridization.

- Molecular geometry: Linear.


Thus, the hybridization of \( HgCl_2 \) is \( sp \).



Step 4: Verify the Correct Answer


From our calculations, the hybridizations are \( sp^2, sp^2, sp \), which matches Option (1). Quick Tip: The hybridization of an atom can be determined using the steric number formula: \[ Steric Number = Number of bonded atoms + Number of lone pairs \] A steric number of 2 corresponds to \( sp \), 3 to \( sp^2 \), and 4 to \( sp^3 \).

Step 1: Use the Ideal Gas Equation


The ideal gas equation is:
\[ PV = nRT \]

where:
- \( P \) is the pressure (1 atm),
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant (\( 0.0821 \) L atm mol\(^{-1}\) K\(^{-1}\)),
- \( T \) is the temperature (300 K).

Step 2: Express Volume as a Function of \( n \)


Rearranging the equation:
\[ V = \frac{nRT}{P} \]

Since \( R = 0.0821 \), \( T = 300 \) K, and \( P = 1 \) atm:
\[ V = \frac{(n)(0.0821)(300)}{1} \]
\[ V = 24.6 n \]

Step 3: Identify the Slope


The equation is in the form:
\[ V = m n \]

where \( m \) (the slope) is 24.6 L mol\(^{-1}\).

Step 4: Verify the Correct Answer


Thus, the slope of the graph is \( 24.6 \) L mol\(^{-1}\), which matches Option (2). Quick Tip: For an ideal gas at STP (1 atm, 273 K), the molar volume is 22.4 L. At 300 K, it increases to approximately 24.6 L per mole. The slope of a volume vs. moles graph is given by \( \frac{RT}{P} \).

Step 1: Assign Oxidation Numbers


In potassium chlorate (\( KClO_3 \)):
- Potassium (\( K \)) has an oxidation state of +1.
- Oxygen (\( O \)) in oxoanions typically has an oxidation state of -2.
- Let the oxidation state of chlorine (\( Cl \)) be \( x \).

Using the sum rule for a neutral compound:
\[ (+1) + x + 3(-2) = 0 \]
\[ x - 6 + 1 = 0 \]
\[ x = +5 \]

Thus, in \( KClO_3 \), \( Cl \) has an oxidation state of \( +5 \).

In potassium chloride (\( KCl \)):
- \( K \) is still \( +1 \).
- \( Cl \) must be \( -1 \) to balance the charge.

Thus, the oxidation state of \( Cl \) changes from \( +5 \) in \( KClO_3 \) to \( -1 \) in \( KCl \), meaning \( Cl \) is reduced.

Step 2: Identify Oxidation of Oxygen


In \( KClO_3 \), oxygen is in the \(-2\) oxidation state.
In \( O_2 \) gas, oxygen exists in its elemental form, which has an oxidation state of \( 0 \).

Since oxygen’s oxidation state increases from \(-2\) to \( 0 \), oxygen is oxidized.

Step 3: Verify the Correct Answer


- Chlorine is reduced (from \( +5 \) to \( -1 \)).
- Oxygen is oxidized (from \( -2 \) to \( 0 \)).

Thus, the correct answer is Option (2): Cl is reduced and O is oxidized. Quick Tip: Oxidation refers to an increase in oxidation state, while reduction refers to a decrease. In this reaction, oxygen is oxidized, and chlorine is reduced, demonstrating a classic redox process.

Step 1: Understanding the Given Data


We are given the standard enthalpies of formation (\( \Delta_f H^\theta \)):
\[ \Delta_f H^\theta (AO_3) = -635 kJ mol^{-1} \]
\[ \Delta_f H^\theta (ABO_3) = -1210 kJ mol^{-1} \]
\[ \Delta_r H^\theta = 175 kJ mol^{-1} \]

We need to determine the enthalpy of formation of \( BO_2 \), denoted as \( x \).

Step 2: Applying Hess’s Law


From the reaction:
\[ ABO_3 (s) \rightarrow AO_3 (s) + BO_2 (g) \]

Using the enthalpy change equation:
\[ \Delta_r H^\theta = \sum \Delta_f H^\theta (products) - \sum \Delta_f H^\theta (reactants) \]
\[ 175 = (-635 + x) - (-1210) \]

Step 3: Solve for \( x \)

\[ 175 = -635 + x + 1210 \]
\[ x = 175 - 1210 + 635 \]
\[ x = -400 kJ mol^{-1} \]

Step 4: Verify the Correct Answer


Thus, the value of \( x \) is \( -400 \) kJ mol\(^{-1}\), which matches Option (3). Quick Tip: Hess’s Law states that the enthalpy change of a reaction is the sum of the enthalpy changes of its components. Use the equation: \[ \Delta_r H^\theta = \sum \Delta_f H^\theta (products) - \sum \Delta_f H^\theta (reactants) \] to solve enthalpy-related problems.

Step 1: Calculate the Moles of HCl and NaOH

\[ Moles of HCl = M \times V = (0.5 M) \times (0.1 L) = 0.05 moles \]
\[ Moles of NaOH = M \times V = (0.4 M) \times (0.1 L) = 0.04 moles \]

Step 2: Determine the Excess Acid or Base


Since HCl and NaOH react in a 1:1 ratio:
\[ Remaining HCl = 0.05 - 0.04 = 0.01 moles \]

Step 3: Calculate the Final Concentration of \( H^+ \)


Total volume after mixing:
\[ 100 mL + 100 mL + 800 mL = 1000 mL = 1 L \]
\[ [H^+] = \frac{Remaining moles of HCl}{Total volume in L} = \frac{0.01}{1} = 0.01 M \]

Step 4: Calculate the pH

\[ pH = -\log [H^+] \]
\[ pH = -\log (0.01) = 2.0 \]

Step 5: Verify the Correct Answer


Thus, the final pH of the solution is \( 2.0 \), which matches Option (2). Quick Tip: When a strong acid (HCl) and a strong base (NaOH) are mixed, find the limiting reagent and determine the excess moles remaining. Then, use the total volume to calculate the final concentration and determine the pH.

Step 1: Understanding the storage conditions for \( H_2O_2 \)


Hydrogen peroxide (\( H_2O_2 \)) is a highly reactive and unstable compound. It decomposes easily when exposed to light, heat, or contaminants. The decomposition reaction is as follows: \[ 2 H_2O_2 \rightarrow 2 H_2O + O_2 \]
This reaction is catalyzed by light, heat, and impurities such as dust or metal ions.

Step 2: Evaluating the given options


- Option (1): Storing \( H_2O_2 \) in a wax-lined plastic bottle and keeping it in the dark is the correct approach, as it prevents light exposure and avoids contamination from metal ions.


- Option (2): Exposing \( H_2O_2 \) to light accelerates its decomposition, making this an incorrect choice.


- Option (3): Adding a base catalyzes the decomposition of \( H_2O_2 \), making it an incorrect storage condition.


- Option (4): Storing in a metal vessel leads to contamination with metal ions, which further accelerates decomposition.



Thus, the correct answer is Option (1). Quick Tip: Hydrogen peroxide should always be stored in a cool, dark place, preferably in a plastic or glass container with a stabilizer. Avoid exposure to light and metals to prevent rapid decomposition.

Step 1: Understanding Standard Electrode Potential


The standard electrode potential (\( E^\circ \)) of an element is a measure of the tendency of the element to gain or lose electrons in an electrochemical reaction. It is given in volts (\( V \)) and is measured under standard conditions.

Step 2: Standard Electrode Potentials for Lithium and Sodium


- For lithium, the standard reduction potential is:
\[ Li^+ + e^- \rightarrow Li, \quad E^\circ = -3.04V \]
- For sodium, the standard reduction potential is:
\[ Na^+ + e^- \rightarrow Na, \quad E^\circ = -2.714V \]

Step 3: Evaluating the Given Options


- Option (1): \( -3.04, -2.714 \) – Correct as per standard data.

- Option (2): \( -2.714, -3.04 \) – Incorrect as the values are reversed.

- Option (3): \( -3.04, -3.04 \) – Incorrect as the value for sodium is wrong.

- Option (4): \( -2.714, -2.714 \) – Incorrect as the value for lithium is wrong.



Thus, the correct answer is Option (1). Quick Tip: The more negative the standard electrode potential (\( E^\circ \)), the stronger the reducing agent. Lithium has a more negative potential than sodium, making it a stronger reducing agent.

Step 1: Understanding Beryllium Alloys


Beryllium forms several alloys, but one of the most notable is beryllium copper (\( Be-Cu \)). This alloy is widely used due to its high strength, good electrical conductivity, and non-magnetic properties.

Step 2: Properties and Applications of Beryllium Copper


- Beryllium copper (\( Be-Cu \)) is a non-sparking, non-magnetic alloy with excellent strength and corrosion resistance.
- It is used in the manufacture of high-strength springs, precision instruments, and aerospace components.
- The combination of beryllium with copper enhances mechanical strength while maintaining electrical conductivity.

Step 3: Evaluating the Given Options


- Option (1) Al: Beryllium-aluminum alloys exist but are not commonly used for high-strength springs.


- Option (2) Zn: Zinc alloys do not typically involve beryllium in significant amounts for mechanical applications.


- Option (3) Cu: Correct, as beryllium copper is used in high-strength springs.


- Option (4) Cr: Chromium forms strong alloys but is not the primary metal used with beryllium for this application.



Thus, the correct answer is Option (3). Quick Tip: Beryllium copper (\( Be-Cu \)) is a unique alloy known for its high strength, electrical conductivity, and corrosion resistance. It is widely used in precision instruments, aerospace applications, and non-sparking tools.

\flushleft
Step 1: Understanding the Reaction Sequence


The given reaction sequence involves a series of transformations starting from \( X \) and leading to \( Y \). Let’s analyze the stepwise reactions:



1. The first reaction suggests that \( X \) reacts with carbon monoxide (\( CO \)) to form diborane (\( B_2H_6 \)). This indicates that \( X \) is borane (\( BH_3 \)), which is a well-known precursor to diborane formation.

2. Diborane (\( B_2H_6 \)) then reacts with sodium hydride (\( NaH \)), leading to the formation of sodium borohydride (\( NaBH_4 \)).


3. The presence of diethyl ether (\( (C_2H_5)_2O \)) as a solvent helps stabilize the final product.


\flushleft
Step 2: Evaluating the Given Options


- Option (1): Incorrect as it suggests \( X = BH_3, 2CO \), which is not required for diborane formation.


- Option (2): Correct as it correctly identifies \( X = BH_3 \) and \( Y = NaBH_4 \).


- Option (3): Incorrect as it wrongly identifies \( Y \) as sodium metaborate (\( NaBO_2 \)).


- Option (4): Incorrect as it suggests \( Y = Na_2B_4O_7 \), which does not match the reaction sequence.



Thus, the correct answer is Option (2). Quick Tip: Borane (\( BH_3 \)) is a key reagent in boron chemistry and forms diborane (\( B_2H_6 \)) upon dimerization. Sodium borohydride (\( NaBH_4 \)) is commonly used as a reducing agent in organic synthesis.

Step 1: Analyzing Each Statement


1. Statement (i): Incorrect. Carbon tetrachloride (\( CCl_4 \)) does not undergo hydrolysis easily because it lacks vacant \( d \)-orbitals, making it resistant to nucleophilic attack by water molecules.

2. Statement (ii): Correct. Diamond has a tetrahedral structure where each carbon atom is bonded to four other carbon atoms through strong directional covalent bonds, giving it high hardness and rigidity.

3. Statement (iii): Incorrect. Graphite, not fullerene, is the most thermodynamically stable allotrope of carbon due to its lower energy configuration.

4. Statement (iv): Correct. Glass is an amorphous, man-made silicate composed mainly of silica (\( SiO_2 \)), which is fused with other compounds like sodium oxide and calcium oxide.

Step 2: Evaluating the Given Options


- Option (1): Incorrect, as statement (i) is incorrect.

- Option (2): Correct, as statements (ii) and (iv) are the only correct ones.

- Option (3): Incorrect, as statement (iii) is incorrect.

- Option (4): Incorrect, as statement (i) is incorrect.



Thus, the correct answer is Option (2). Quick Tip: Diamond has directional covalent bonds, making it extremely hard. Glass is a synthetic silicate material, while graphite is the most stable allotrope of carbon, not fullerene.

Step 1: Understanding Non-Biodegradable Waste


Non-biodegradable waste refers to materials that do not decompose naturally and persist in the environment for a long time, causing pollution. These include plastics, heavy metals, industrial chemicals, and fly ash.

Step 2: Analyzing the Given Industries


1. Cotton mills: Incorrect. Cotton mills primarily generate biodegradable waste, such as natural fiber residues.\
2. Paper mills: Incorrect. Paper waste is mostly biodegradable and recyclable.

3. Thermal power plants: Correct. These plants generate large amounts of non-biodegradable waste such as fly ash, bottom ash, and heavy metal residues, which contribute to environmental pollution.

4. Textile factories: Incorrect. Textile waste can include biodegradable materials like natural fibers, though synthetic fibers may persist longer.

Step 3: Evaluating the Given Options


- Option (1): Incorrect, as cotton waste is biodegradable.

- Option (2): Incorrect, as paper is mostly biodegradable.

- Option (3): Correct, as thermal power plants generate non-biodegradable industrial waste.

- Option (4): Incorrect, as textile waste varies, but most of it is biodegradable.


Thus, the correct answer is Option (3). Quick Tip: Thermal power plants produce non-biodegradable waste like fly ash and heavy metals. These pollutants can persist in the environment and require proper disposal methods.

Step 1: Understanding Isomerism


Isomers are compounds with the same molecular formula but different structural arrangements or spatial orientations. The molecular formula given is \( C_4H_9Br \), which corresponds to butyl bromide isomers.

Step 2: Identifying the Isomers


The possible structural and stereoisomers for \( C_4H_9Br \) are:



1. n-Butyl bromide (\( CH_3CH_2CH_2CH_2Br \)) - Straight-chain structure.


2. Isobutyl bromide (\( (CH_3)_2CHCH_2Br \)) - Branched structure.


3. Sec-butyl bromide (\( CH_3CHBrCH_2CH_3 \)) - Branched with a secondary carbon.


4. Tert-butyl bromide (\( (CH_3)_3CBr \)) - Branched with a tertiary carbon.


5. Sec-butyl bromide has a chiral center, leading to two enantiomers, contributing to stereoisomerism.



Step 3: Evaluating the Given Options


- Option (1) 3: Incorrect, as it does not account for all possible isomers.


- Option (2) 4: Incorrect, as it excludes stereoisomers.


- Option (3) 5: Correct, as it includes all structural and stereoisomers.


- Option (4) 2: Incorrect, as more isomers exist.



Thus, the correct answer is Option (3). Quick Tip: When counting isomers, consider both structural and stereoisomers. Compounds with chiral centers contribute to additional enantiomers, increasing the total count.

Step 1: Understanding the homologous series


The homologous series of alkanes consists of compounds differing by a \( -CH_2- \) group. The alkane next to methane is ethane (\( C_2H_6 \)).

Step 2: Evaluating the given reactions


1. Reaction I: Wurtz Reaction

\[ 2CH_3Br + 2Na \xrightarrow{dry ether} C_2H_6 + 2NaBr \]
This reaction successfully forms ethane, making it a correct method.

2. Reaction II: Decarboxylation of Carboxylic Acids

\[ CH_3COOH + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3 \]
This reaction produces methane, not ethane, so it is incorrect.

3. Reaction III: Hydrogenation of Ethene

\[ CH_2CH_2 + H_2 \xrightarrow{Pt} C_2H_6 \]
This reaction forms ethane, but it is not the best preparative method for homologous series alkane synthesis.

4. Reaction IV: Reduction of Alkyl Halides

\[ CH_3CH_2Br + 2[H] \xrightarrow{Zn/H^+} C_2H_6 + HBr \]
This reaction also produces ethane, making it a valid method.

Step 3: Evaluating the Given Options


- Option (1): Correct, as I (Wurtz reaction) and IV (alkyl halide reduction) are the valid methods.


- Option (2): Incorrect, as II produces methane instead of ethane.


- Option (3): Incorrect, as I is correct but III is not a conventional method for preparing alkanes in a homologous series.


- Option (4): Incorrect, as II does not yield ethane.



Thus, the correct answer is Option (1). Quick Tip: The Wurtz reaction is a reliable method for synthesizing alkanes by coupling alkyl halides with sodium metal in dry ether. The reduction of alkyl halides using zinc and acid also effectively produces alkanes.

Step 1: Understanding the Oxidation of Methane


Methane (\( CH_4 \)) can be oxidized under controlled conditions to form methanol (\( CH_3OH \)) and methanal (\( HCHO \)). The choice of catalyst determines the specific oxidation product.

Step 2: Identifying the Catalysts


1. Formation of Methanol (\( CH_3OH \)):

- Methane is converted to methanol using copper (\( Cu \)) as a catalyst under high pressure.
\[ CH_4 + \frac{1}{2} O_2 \xrightarrow{Cu} CH_3OH \]

2. Formation of Methanal (\( HCHO \)):

- Methane is converted to methanal using molybdenum oxide (\( Mo_2O_3 \)) as a catalyst.
\[ CH_4 + O_2 \xrightarrow{Mo_2O_3} HCHO + H_2O \]

Step 3: Evaluating the Given Options


- Option (1): Incorrect, as it swaps the catalysts.


- Option (2): Correct, as \( Cu \) is used for methanol and \( Mo_2O_3 \) for methanal.


- Option (3): Incorrect, as \( V_2O_5 \) and \( KMnO_4 \) are not the catalysts used in this reaction.


- Option (4): Incorrect, as \( KMnO_4 \) and \( Cr_2O_3 \) are not the correct catalysts.



Thus, the correct answer is Option (2). Quick Tip: In oxidation reactions of methane, copper (\( Cu \)) is used for converting methane to methanol, while molybdenum oxide (\( Mo_2O_3 \)) is used for converting methane to methanal.

\flushleft
Step 1: Understanding the Reaction Sequence


1. Step 1: Acid-Catalyzed Hydration of Alkene

- The given compound \( C_4H_8 \) is an alkene.

- In the presence of sulfuric acid (\( H_2SO_4 \)) and water, Markovnikov’s rule applies, leading to the formation of a more substituted alcohol.


2. Step 2: Ozonolysis Reaction

- The alcohol undergoes ozonolysis (\( O_3 \), Zn/H_2O), which cleaves the double bond and forms aldehydes or ketones depending on the structure of the starting alkene.


Step 2: Identifying ‘Y’


\flushleft
- The given reactant is butene (\( C_4H_8 \)).

- Hydration leads to the formation of butan-2-ol as the intermediate product.

- Ozonolysis of butan-2-ol results in acetone (\( CH_3COCH_3 \)) and formaldehyde (\( HCHO \)).


Step 3: Evaluating the Given Options


- Option (1): Incorrect, as it does not match the expected ozonolysis products.

- Option (2): Correct, as it represents acetone and formaldehyde.

- Option (3): Incorrect, as it does not correspond to the reaction mechanism.

- Option (4): Incorrect, as it represents a different structural outcome.



Thus, the correct answer is Option (2). Quick Tip: Ozonolysis of alkenes results in the cleavage of the double bond, producing aldehydes or ketones. When dealing with hydration reactions, always apply Markovnikov’s rule to determine the major product.

Step 1: Understanding the Crystal Structure


The given compound is \( AB_2O_4 \), which follows the spinel structure. In such structures:



- Oxygen atoms form a face-centered cubic (FCC) close-packed lattice.


- The tetrahedral voids and octahedral voids are partially occupied by \( A \) and \( B \) atoms, respectively.



Step 2: Analyzing the Void Occupation


1. Tetrahedral Voids:


- In an FCC lattice of oxygen, the number of tetrahedral voids is twice the number of oxygen atoms.


- A fraction of these voids is occupied by A atoms.



2. Octahedral Voids:


- The number of octahedral voids is equal to the number of oxygen atoms.

- A fraction of these voids is occupied by B atoms.



Step 3: Calculating \( x \) and \( y \)


- Total oxygen atoms = 4 per formula unit.

- Total tetrahedral voids = \( 2 \times 4 = 8 \).

- Total octahedral voids = 4.


For A atoms:

- The formula tells us that 1 atom of \( A \) is present.

- \( A \) atoms occupy tetrahedral voids.

- The fraction occupied by \( A \) is \( \frac{1}{8} = 12.5% \).


For B atoms:

- The formula tells us that 2 atoms of \( B \) are present.

- \( B \) atoms occupy octahedral voids.

- The fraction occupied by \( B \) is \( \frac{2}{4} = 50% \).


Step 4: Evaluating the Given Options


- Option (1): Correct, as \( x = 12.5% \), \( y = 50% \).

- Option (2): Incorrect, as it reverses the values.

- Option (3): Incorrect, as it does not match calculated values.

- Option (4): Incorrect, as the values are swapped incorrectly.



Thus, the correct answer is Option (1). Quick Tip: In a spinel structure (\( AB_2O_4 \)), A atoms typically occupy tetrahedral voids while B atoms occupy octahedral voids. The percentage occupation is determined based on the number of available voids.