Step 1: Understanding the Concept:

The given expression is a functional equation involving \( f(x) \) and \( f(1/x) \).

To find the explicit form of the function \( f(x) \), we can substitute \( 1/x \) for \( x \) to generate a second equation and solve the resulting system.


Step 2: Key Formula or Approach:

Given: \[ 3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10 \quad \dots (1) \]
Replace \( x \) with \( \frac{1}{x} \): \[ 3f\left(\frac{1}{x}\right) + 2f(x) = x - 10 \quad \dots (2) \]

Step 3: Detailed Explanation:

Multiply equation (1) by 3 and equation (2) by 2:
\[ 9f(x) + 6f\left(\frac{1}{x}\right) = \frac{3}{x} - 30 \] \[ 4f(x) + 6f\left(\frac{1}{x}\right) = 2x - 20 \]
Subtracting the second from the first: \[ 5f(x) = \frac{3}{x} - 2x - 10 \implies f(x) = \frac{1}{5} \left( \frac{3}{x} - 2x - 10 \right) \]
Calculate \( f(3) \): \[ f(3) = \frac{1}{5} \left( \frac{3}{3} - 2(3) - 10 \right) = \frac{1}{5} (1 - 6 - 10) = \frac{-15}{5} = -3 \]
Now find the derivative \( f'(x) \): \[ f'(x) = \frac{d}{dx} \left[ \frac{1}{5} \left( \frac{3}{x} - 2x - 10 \right) \right] = \frac{1}{5} \left( -\frac{3}{x^2} - 2 \right) \]
Calculate \( f'\left(\frac{1}{4}\right) \): \[ f'\left(\frac{1}{4}\right) = \frac{1}{5} \left( -\frac{3}{(1/4)^2} - 2 \right) = \frac{1}{5} (-3(16) - 2) = \frac{-50}{5} = -10 \]
The required value is: \[ \left| f(3) + f'\left(\frac{1}{4}\right) \right| = | -3 + (-10) | = | -13 | = 13 \]

Step 4: Final Answer:

The value of the expression is 13.
Quick Tip: In functional equations involving \( f(x) \) and \( f(1/x) \), simple substitution and elimination of the unwanted term is the fastest way to find the function.

Step 1: Understanding the Concept:

The number of roots of \( f(x) = -a \) depends on the monotonicity of the function \( f(x) = x|x-1| + |x+2| \).

If the function is strictly monotonic on its entire domain, it will cross any horizontal line \( y = -a \) exactly once.


Step 2: Detailed Explanation:

Let \( f(x) = x|x-1| + |x+2| \). We analyze this in intervals based on critical points \( x = 1 \) and \( x = -2 \):

Case 1: \( x < -2 \)
\( f(x) = x(-(x-1)) - (x+2) = -x^2 + x - x - 2 = -x^2 - 2 \).
\( f'(x) = -2x \). For \( x < -2 \), \( f'(x) > 4 > 0 \). (Increasing)

Case 2: \( -2 \le x < 1 \)
\( f(x) = x(-(x-1)) + (x+2) = -x^2 + x + x + 2 = -x^2 + 2x + 2 \).
\( f'(x) = -2x + 2 \). For \( x \in (-2, 1) \), \( f'(x) > 0 \). (Increasing)

Case 3: \( x \ge 1 \)
\( f(x) = x(x-1) + x + 2 = x^2 - x + x + 2 = x^2 + 2 \).
\( f'(x) = 2x \). For \( x \ge 1 \), \( f'(x) \ge 2 > 0 \). (Increasing)

Since \( f(x) \) is continuous and strictly increasing from \( -\infty \) to \( \infty \), its range is \( \mathbb{R} \).

Thus, the equation \( f(x) = -a \) has exactly one real root for every \( a \in \mathbb{R} \).


Step 3: Final Answer:

The set of all such \( a \) is \( (-\infty, \infty) \).
Quick Tip: Whenever you need to find the range of \( a \) for a specific number of roots, sketching the graph or checking the monotonicity of the function is the most reliable approach.

Step 1: Understanding the Concept:

We use the matrix property \( |adj(A)| = |A|^{n-1} \). For a \( 3 \times 3 \) matrix, \( |B| = |A|^2 \).


Step 2: Key Formula or Approach:

Given \( |A| = 2 \), then \( |B| = 2^2 = 4 \).

Compute \( |B| \): \[ |B| = 1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha) = 8 - 3\alpha - 12 + 9\alpha - \alpha^2 \] \[ |B| = -\alpha^2 + 6\alpha - 4 \]

Step 3: Detailed Explanation:

Set \( -\alpha^2 + 6\alpha - 4 = 4 \implies \alpha^2 - 6\alpha + 8 = 0 \).

Factoring gives \( (\alpha-2)(\alpha-4) = 0 \). Since \( \alpha > 2 \), we have \( \alpha = 4 \).

The required calculation is \( V^T B V \) where \( V = \begin{bmatrix} 4
-8
4 \end{bmatrix} = 4 \begin{bmatrix} 1
-2
1 \end{bmatrix} \).

Let \( X = \begin{bmatrix} 1
-2
1 \end{bmatrix} \). Then the expression is \( (4X)^T B (4X) = 16 (X^T B X) \).

Calculate \( BX \): \[ \begin{bmatrix} 1 & 3 & 4
1 & 2 & 3
4 & 4 & 4 \end{bmatrix} \begin{bmatrix} 1
-2
1 \end{bmatrix} = \begin{bmatrix} 1 - 6 + 4
1 - 4 + 3
4 - 8 + 4 \end{bmatrix} = \begin{bmatrix} -1
0
0 \end{bmatrix} \]
Now calculate \( X^T (BX) \): \[ \begin{bmatrix} 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} -1
0
0 \end{bmatrix} = -1 \]
Total value = \( 16 \times (-1) = -16 \).


Step 4: Final Answer:

The resulting value is -16.
Quick Tip: Identifying common factors in vectors (\( \alpha \) in this case) before performing matrix multiplication significantly simplifies the arithmetic.

Step 1: Understanding the Concept:

For a system of linear equations, a unique solution exists if the determinant of the coefficient matrix \( \Delta \neq 0 \).

Infinitely many solutions or no solution occurs when \( \Delta = 0 \).


Step 2: Key Formula or Approach:

Calculate \( \Delta \): \[ \Delta = \begin{vmatrix} 2 & 4 & 2a
1 & 2 & 3
2 & -5 & 2 \end{vmatrix} = 2(4+15) - 4(2-6) + 2a(-5-4) \] \[ \Delta = 38 + 16 - 18a = 54 - 18a = 18(3-a) \]
Unique solution exists if \( a \neq 3 \).


Step 3: Detailed Explanation:

Check \( a = 3 \): In this case \( \Delta = 0 \).

Check \( \Delta_z \): \[ \Delta_z = \begin{vmatrix} 2 & 4 & b
1 & 2 & 4
2 & -5 & 8 \end{vmatrix} = 2(16+20) - 4(8-8) + b(-5-4) = 72 - 9b = 9(8-b) \]
For infinitely many solutions, \( \Delta_z \) must be 0, which implies \( b = 8 \).

Statement (C) is correct (\( a=3, b=8 \)).

Statement (D) claims infinitely many solutions for \( a=3, b=6 \). However, for \( b=6 \), \( \Delta_z = 18 \neq 0 \), meaning the system is inconsistent (No Solution).

Statements (A) and (B) are correct as \( a \neq 3 \) guarantees a unique solution.


Step 4: Final Answer:

The incorrect statement is (D).
Quick Tip: Always check the determinant of the coefficient matrix first. If it's non-zero, the system is guaranteed to have a unique solution regardless of the constants on the right side.

Step 1: Understanding the Concept:

A symmetric matrix \( A \) is defined by \( a_{ij} = a_{ji} \).

For an \( n \times n \) matrix, the number of independent entries is \( \frac{n(n+1)}{2} \).


Step 2: Detailed Explanation:

For order \( n = 3 \):
Number of independent entries = \( \frac{3 \times 4}{2} = 6 \).

These 6 entries consist of the 3 diagonal elements and 3 elements either above or below the diagonal.

The set \( \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \) has 10 elements.

Each of the 6 independent positions can be filled in 10 ways.

Total symmetric matrices = \( 10^6 \).


Step 3: Final Answer:

The number of such matrices is \( 10^6 \).
Quick Tip: For a symmetric matrix, you only need to choose the diagonal and one half of the off-diagonal elements. For a skew-symmetric matrix, the diagonal must be zero.

Step 1: Understanding the Concept:

We find a general expression for \( s_i \), the sum of 12 terms of the \( i \)-th A.P., and then sum these values.


Step 2: Key Formula or Approach:

For the \( i \)-th A.P.:
First term \( a_i = i \)

Common difference \( d_i = 2i - 1 \)

Sum formula \( S_n = \frac{n}{2}[2a + (n-1)d] \)


Step 3: Detailed Explanation:
\[ s_i = \frac{12}{2} [2(i) + (12-1)(2i - 1)] \] \[ s_i = 6 [2i + 11(2i - 1)] = 6 [2i + 22i - 11] = 6(24i - 11) = 144i - 66 \]
Now calculate \( \sum_{i=1}^{10} s_i \): \[ \sum_{i=1}^{10} (144i - 66) = 144 \sum_{i=1}^{10} i - \sum_{i=1}^{10} 66 \] \[ = 144 \left( \frac{10 \times 11}{2} \right) - 66 \times 10 \] \[ = 144(55) - 660 = 7920 - 660 = 7260 \]

Step 4: Final Answer:

The sum is 7260.
Quick Tip: Expressing the general term \( s_i \) in terms of \( i \) and then applying linearity of summation makes complex arithmetic manageable.

Step 1: Understanding the Concept:

These limits can be evaluated using summation formulas or by treating them as Riemann sums for definite integrals.


Step 2: Detailed Explanation:

For (S1): \[ Sum = 2 + 4 + \dots + 2n = 2(1+2+\dots+n) = 2 \frac{n(n+1)}{2} = n^2 + n \] \[ \lim_{n \to \infty} \frac{n^2 + n}{n^2} = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) = 1 \]
Thus, (S1) is true.

For (S2):
The sum is \( \sum_{k=1}^n k^{15} \). The limit can be expressed as: \[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left( \frac{k}{n} \right)^{15} \]
This is the Riemann sum for the integral \( \int_0^1 x^{15} dx \): \[ \int_0^1 x^{15} dx = \left[ \frac{x^{16}}{16} \right]_0^1 = \frac{1}{16} \]
Thus, (S2) is true.


Step 3: Final Answer:

Both statements (S1) and (S2) are true.
Quick Tip: For large \( n \), the sum \( \sum k^p \) is approximately \( \frac{n^{p+1}}{p+1} \). This shortcut allows for immediate evaluation of such limits.

Step 1: Understanding the Concept:

The family of planes passing through the intersection of \( P_1 = 0 \) and \( P_2 = 0 \) is \( P_1 + \lambda P_2 = 0 \).


Step 2: Key Formula or Approach:
\[ (x + 2y + az - 2) + \lambda(x - y + z - 3) = 0 \] \[ (1+\lambda)x + (2-\lambda)y + (a+\lambda)z = 2 + 3\lambda \]

Step 3: Detailed Explanation:

Comparing with \( 5x - 11y + bz = 6a - 1 \): \[ \frac{1+\lambda}{5} = \frac{2-\lambda}{-11} \implies -11 - 11\lambda = 10 - 5\lambda \implies 6\lambda = -21 \implies \lambda = -\frac{7}{2} \]
Using the ratio \( \frac{1+\lambda}{5} = \frac{1 - 3.5}{5} = -0.5 \): \[ \frac{2 + 3\lambda}{6a - 1} = -0.5 \implies \frac{2 - 10.5}{6a - 1} = -0.5 \implies 17 = 6a - 1 \implies a = 3 \] \[ \frac{a + \lambda}{b} = -0.5 \implies \frac{3 - 3.5}{b} = -0.5 \implies b = 1 \]
The plane is \( 5x - 11y + z - 17 = 0 \). Distance from \( (3, -c, c) \): \[ \frac{|15 + 11c + c - 17|}{\sqrt{25 + 121 + 1}} = \frac{2}{\sqrt{3}} \implies \frac{|12c - 2|}{\sqrt{147}} = \frac{2}{\sqrt{3}} \] \[ \frac{|12c - 2|}{7\sqrt{3}} = \frac{2}{\sqrt{3}} \implies |12c - 2| = 14 \]
Since \( c \in \mathbb{Z} \), \( 12c - 2 = -14 \implies c = -1 \). (Other case gives \( c = 4/3 \)).

Result: \( \frac{a+b}{c} = \frac{3+1}{-1} = -4 \).


Step 4: Final Answer:

The value of the expression is -4.
Quick Tip: For problems involving the family of planes, always compare the ratio of coefficients of \( x \), \( y \), \( z \), and the constant term to solve for unknowns.

Step 1: Understanding the Concept:

The equation \( \vec{d} \times \vec{b} = \vec{c} \times \vec{b} \) can be written as \( (\vec{d} - \vec{c}) \times \vec{b} = 0 \).

This implies that \( \vec{d} - \vec{c} \) is parallel to \( \vec{b} \).


Step 2: Key Formula or Approach:

Let \( \vec{d} = \vec{c} + \lambda \vec{b} \) for some scalar \( \lambda \).


Step 3: Detailed Explanation:

Given \( \vec{d} \cdot \vec{a} = 24 \): \[ (\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24 \implies \vec{c} \cdot \vec{a} + \lambda(\vec{b} \cdot \vec{a}) = 24 \]
Calculate dot products: \[ \vec{c} \cdot \vec{a} = (2)(1) + (-1)(4) + (4)(2) = 2 - 4 + 8 = 6 \] \[ \vec{b} \cdot \vec{a} = (3)(1) + (-2)(4) + (7)(2) = 3 - 8 + 14 = 9 \]
Substituting: \( 6 + 9\lambda = 24 \implies 9\lambda = 18 \implies \lambda = 2 \).

Now, \( \vec{d} = \vec{c} + 2\vec{b} = (2, -1, 4) + 2(3, -2, 7) = (8, -5, 18) \).

Calculate \( |\vec{d}|^2 \): \[ |\vec{d}|^2 = 8^2 + (-5)^2 + 18^2 = 64 + 25 + 324 = 413 \]

Step 4: Final Answer:

The value of \( |\vec{d}|^2 \) is 413.
Quick Tip: The relation \( \vec{A} \times \vec{B} = \vec{C} \times \vec{B} \) immediately tells you that \( \vec{A} \) lies in the form \( \vec{C} + \lambda \vec{B} \). This reduces the vector equation to a scalar one.

Step 1: Understanding the Concept:

To find the maximum of \( f(x) \) on a closed interval \( [0, \pi] \), we compare the values of the function at critical points and boundaries.


Step 2: Key Formula or Approach:

Let \( f(x) = x - \sin 2x + \frac{1}{3}\sin 3x \).

Differentiate \( f(x) \): \[ f'(x) = 1 - 2\cos 2x + \cos 3x \]

Step 3: Detailed Explanation:

Set \( f'(x) = 0 \): \[ 1 - 2(2\cos^2 x - 1) + (4\cos^3 x - 3\cos x) = 0 \] \[ 4\cos^3 x - 4\cos^2 x - 3\cos x + 3 = 0 \] \[ 4\cos^2 x(\cos x - 1) - 3(\cos x - 1) = 0 \implies (4\cos^2 x - 3)(\cos x - 1) = 0 \]
Critical points: \( \cos x = 1 \implies x = 0 \) and \( \cos x = \pm \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6} \).

Evaluate \( f(x) \): \( f(0) = 0 \)
\( f(\pi/6) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{\pi - 3\sqrt{3} + 2}{6} \)
\( f(5\pi/6) = \frac{5\pi}{6} - \sin(5\pi/3) + \frac{1}{3}\sin(5\pi/2) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 3\sqrt{3} + 2}{6} \)
\( f(\pi) = \pi \)

Comparing \( f(5\pi/6) \) and \( f(\pi) \): Since \( \frac{2\pi}{6} + \frac{3\sqrt{3}}{6} + \frac{2}{6} > 0 \), \( f(5\pi/6) \) is the absolute maximum.


Step 4: Final Answer:

The maximum value is \( \frac{5\pi + 2 + 3\sqrt{3}}{6} \).
Quick Tip: Factorization of the derivative in terms of \( \cos x \) is the most efficient way to find critical points for trigonometric polynomial functions.

Step 1: Understanding the Concept:

The given integral involves an exponential expression in the denominator.

To solve it, we can use a substitution \( u = e^x \) to transform it into an integral of a rational function, followed by partial fraction decomposition.


Step 2: Key Formula or Approach:

Substitute \( u = e^x \implies du = e^x dx \implies dx = \frac{du}{u} \).

Change the limits: As \( x \to 0, u \to 1 \). As \( x \to \infty, u \to \infty \).

The denominator factors as \( e^{3x} + 6e^{2x} + 11e^x + 6 = (e^x + 1)(e^x + 2)(e^x + 3) \).


Step 3: Detailed Explanation:

The integral becomes: \[ I = \int_1^{\infty} \frac{6}{u(u+1)(u+2)(u+3)} du \]
Using partial fraction decomposition: \[ \frac{6}{u(u+1)(u+2)(u+3)} = \frac{A}{u} + \frac{B}{u+1} + \frac{C}{u+2} + \frac{D}{u+3} \]
Solving for constants: \[ A = \frac{6}{(1)(2)(3)} = 1 \] \[ B = \frac{6}{(-1)(1)(2)} = -3 \] \[ C = \frac{6}{(-2)(-1)(1)} = 3 \] \[ D = \frac{6}{(-3)(-2)(-1)} = -1 \]
So, \( I = \int_1^{\infty} \left( \frac{1}{u} - \frac{3}{u+1} + \frac{3}{u+2} - \frac{1}{u+3} \right) du \)

Integrating each term: \[ I = \left[ \ln|u| - 3\ln|u+1| + 3\ln|u+2| - \ln|u+3| \right]_1^{\infty} \] \[ I = \left[ \ln \left| \frac{u(u+2)^3}{(u+1)^3(u+3)} \right| \right]_1^{\infty} \]
As \( u \to \infty \), the argument of the logarithm \( \frac{u^4 + \dots}{u^4 + \dots} \to 1 \), so \( \ln(1) = 0 \).

At \( u = 1 \): \[ \ln \left| \frac{1(3)^3}{(2)^3(4)} \right| = \ln \left( \frac{27}{32} \right) \]
Thus, \( I = 0 - \ln\left(\frac{27}{32}\right) = \ln\left(\frac{32}{27}\right) \).


Step 4: Final Answer:

The value of the integral is \( \log_e \left( \frac{32}{27} \right) \).
Quick Tip: For expressions like \( \frac{1}{u(u+1)(u+2)(u+3)} \), notice the symmetry in coefficients \( (+1, -3, +3, -1) \). These are the binomial coefficients of \( (x-1)^3 \) which often appear in such partial fraction problems.

Step 1: Understanding the Concept:

The function \( f(x) = \max\{\sin x, \cos x\} \) switches between \( \sin x \) and \( \cos x \) at their intersection points.

Intersection points in \( [-\pi, \pi] \) are \( x = \frac{\pi}{4} \) and \( x = -\frac{3\pi}{4} \).


Step 2: Detailed Explanation:

Analyze the function in intervals:

1. For \( x \in [-\pi, -\frac{3\pi}{4}] \): \( \sin x \ge \cos x \), so \( f(x) = \sin x \).

2. For \( x \in [-\frac{3\pi}{4}, \frac{\pi}{4}] \): \( \cos x \ge \sin x \), so \( f(x) = \cos x \).

3. For \( x \in [\frac{\pi}{4}, \pi] \): \( \sin x \ge \cos x \), so \( f(x) = \sin x \).

The area is given by \( \int_{-\pi}^{\pi} |f(x)| dx \).

Splitting the integral based on sign changes relative to the x-axis:

- In \( [-\pi, -\frac{3\pi}{4}] \), \( \sin x \) is negative. Area \( A_1 = \int_{-\pi}^{-3\pi/4} -\sin x dx = [\cos x]_{-\pi}^{-3\pi/4} = -\frac{1}{\sqrt{2}} - (-1) = 1 - \frac{1}{\sqrt{2}} \).

- In \( [-\frac{3\pi}{4}, -\frac{\pi}{2}] \), \( \cos x \) is negative. Area \( A_2 = \int_{-3\pi/4}^{-\pi/2} -\cos x dx = [-\sin x]_{-3\pi/4}^{-\pi/2} = 1 - \frac{1}{\sqrt{2}} \).

- In \( [-\frac{\pi}{2}, \frac{\pi}{4}] \), \( \cos x \) is positive. Area \( A_3 = \int_{-\pi/2}^{\pi/4} \cos x dx = [\sin x]_{-\pi/2}^{\pi/4} = \frac{1}{\sqrt{2}} - (-1) = 1 + \frac{1}{\sqrt{2}} \).

- In \( [\frac{\pi}{4}, \pi] \), \( \sin x \) is positive. Area \( A_4 = \int_{\pi/4}^{\pi} \sin x dx = [-\cos x]_{\pi/4}^{\pi} = 1 + \frac{1}{\sqrt{2}} \).

Total Area \( = (1 - \frac{1}{\sqrt{2}}) + (1 - \frac{1}{\sqrt{2}}) + (1 + \frac{1}{\sqrt{2}}) + (1 + \frac{1}{\sqrt{2}}) = 4 \).


Step 3: Final Answer:

The total area enclosed is 4 square units.
Quick Tip: Sketching the graphs of \( \sin x \) and \( \cos x \) quickly identifies which function is higher. The area under the "upper shell" of two trigonometric functions over a full period is often a simple integer.

Step 1: Understanding the Concept:

The fractional part of \( \frac{n}{d} \) is \( \frac{n \pmod d}{d} \).

We need to find the remainder when \( 4^{2022} \) is divided by 15.


Step 2: Key Formula or Approach:

Use modular arithmetic properties.
\( 4^1 \equiv 4 \pmod{15} \)
\( 4^2 \equiv 16 \equiv 1 \pmod{15} \)


Step 3: Detailed Explanation:

Since \( 4^2 \equiv 1 \pmod{15} \), any even power of 4 will leave a remainder of 1.
\[ 4^{2022} = (4^2)^{1011} \equiv 1^{1011} \pmod{15} \] \[ 4^{2022} \equiv 1 \pmod{15} \]
This means \( 4^{2022} = 15k + 1 \) for some integer \( k \).

The number is \( \frac{15k + 1}{15} = k + \frac{1}{15} \).

The fractional part is \( \{ k + \frac{1}{15} \} = \frac{1}{15} \).


Step 4: Final Answer:

The fractional part is \( \frac{1}{15} \).
Quick Tip: For finding the fractional part of \( \frac{a^n}{b} \), look for the smallest power \( a^k \) such that \( a^k \equiv 1 \pmod b \) or \( a^k \equiv -1 \pmod b \).

Step 1: Understanding the Concept:

We solve the linear differential equation and use the initial conditions to find the specific curves.


Step 2: Key Formula or Approach:

The equation \( \frac{dy}{dx} - y = 7 \) is a first-order linear differential equation.

Integrating factor (I.F.) \( = e^{\int -1 dx} = e^{-x} \).


Step 3: Detailed Explanation:

The general solution is: \[ y e^{-x} = \int 7 e^{-x} dx = -7 e^{-x} + C \] \[ y = -7 + C e^x \]
For \( y_1(x) \): \( y_1(0) = 0 \implies 0 = -7 + C_1 \implies C_1 = 7 \).
So, \( y_1(x) = 7 e^x - 7 \).

For \( y_2(x) \): \( y_2(0) = 1 \implies 1 = -7 + C_2 \implies C_2 = 8 \).
So, \( y_2(x) = 8 e^x - 7 \).

To find intersection: \( 7 e^x - 7 = 8 e^x - 7 \).

This simplifies to \( 7 e^x = 8 e^x \implies e^x = 0 \).

Since \( e^x \) is never zero for any real \( x \), there is no solution.


Step 4: Final Answer:

The curves do not intersect at any point.
Quick Tip: According to the Uniqueness Theorem for differential equations, two different solution curves of a first-order ODE (with a continuous derivative) can never intersect.

Step 1: Understanding the Concept:

Find the equations of the tangent and normal to determine points \( A \) and \( B \). Then use circle properties and the concept of "pole and polar".


Step 2: Key Formula or Approach:

Tangent at \( (x_1, y_1) \): \( \frac{xx_1}{36} + \frac{yy_1}{4} = 1 \).

Normal at \( (x_1, y_1) \): \( \frac{x - x_1}{x_1/36} = \frac{y - y_1}{y_1/4} \).


Step 3: Detailed Explanation:

At \( (3\sqrt{3}, 1) \):
Tangent: \( \frac{3\sqrt{3}x}{36} + \frac{y}{4} = 1 \implies \frac{\sqrt{3}x}{12} + \frac{y}{4} = 1 \).
Meet y-axis (\( x=0 \)): \( A = (0, 4) \).

Normal: Slope \( m_T = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}} \), so \( m_N = \sqrt{3} \).
Eq: \( y - 1 = \sqrt{3}(x - 3\sqrt{3}) \).
Meet y-axis (\( x=0 \)): \( y - 1 = -9 \implies B = (0, -8) \).

Circle with diameter \( AB \): Center \( (0, -2) \), radius \( R = \frac{4 - (-8)}{2} = 6 \).
Eq: \( x^2 + (y+2)^2 = 36 \).

Line \( x = 2\sqrt{5} \) is the polar of the point \( (\alpha, \beta) \).
Equation of polar of \( (\alpha, \beta) \) w.r.t circle: \( x\alpha + (y+2)(\beta+2) = 36 \).
Compare with \( x = 2\sqrt{5} \):
Coefficient of \( y \) must be 0: \( \beta + 2 = 0 \implies \beta = -2 \).
Comparing \( x \) terms: \( \frac{\alpha}{1} = \frac{36}{2\sqrt{5}} \implies \alpha = \frac{18}{\sqrt{5}} \).

Calculate \( \alpha^2 - \beta^2 \): \[ \alpha^2 - \beta^2 = \left( \frac{18}{\sqrt{5}} \right)^2 - (-2)^2 = \frac{324}{5} - 4 = \frac{324 - 20}{5} = \frac{304}{5} \]

Step 4: Final Answer:

The value of \( \alpha^2 - \beta^2 \) is \( \frac{304}{5} \).
Quick Tip: If tangents at \( P \) and \( Q \) on a circle meet at \( T \), then the line \( PQ \) is the "polar" of point \( T \) with respect to the circle. The equation of the polar for \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is \( xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0 \).

Step 1: Understanding the Concept:

Determine the coordinates of \( P \) and \( Q \) using focal chord length properties. Find \( M \) using the section formula, then find the equation of the line perpendicular to \( PQ \).


Step 2: Key Formula or Approach:

Length of focal chord for \( y^2 = 4ax \) is \( L = a(t + \frac{1}{t})^2 \).

Coordinates: \( P(at^2, 2at) \) and \( Q(\frac{a}{t^2}, -\frac{2a}{t}) \).


Step 3: Detailed Explanation:

For \( y^2 = 36x \), \( a = 9 \). \( 9(t + \frac{1}{t})^2 = 100 \implies t + \frac{1}{t} = \frac{10}{3} \). \( 3t^2 - 10t + 3 = 0 \implies (3t-1)(t-3) = 0 \implies t = 3 \) (since ordinate \( P \) is positive).
\( P = (9(3^2), 18(3)) = (81, 54) \).
\( Q = (9(1/3)^2, -18(3)) = (1, -6) \).

Section formula for \( M \) (\( PM:MQ = 3:1 \)): \( M = \frac{1(P) + 3(Q)}{4} = \frac{(81, 54) + (3, -18)}{4} = \frac{(84, 36)}{4} = (21, 9) \).

Slope of \( PQ \): \( m_{PQ} = \frac{54 - (-6)}{81 - 1} = \frac{60}{80} = \frac{3}{4} \).

Slope of perpendicular line: \( m_{\perp} = -\frac{4}{3} \).

Equation of line through \( M(21, 9) \): \( y - 9 = -\frac{4}{3}(x - 21) \implies 3y - 27 = -4x + 84 \implies 4x + 3y = 111 \).

Testing points:

(A) \( 4(3) + 3(33) = 12 + 99 = 111 \) (Lies on line)

(B) \( 4(-6) + 3(45) = -24 + 135 = 111 \) (Lies on line)

(C) \( 4(-3) + 3(43) = -12 + 129 = 117 \neq 111 \) (Does NOT lie on line)

(D) \( 4(6) + 3(29) = 24 + 87 = 111 \) (Lies on line)


Step 4: Final Answer:

Point \( (-3, 43) \) does not lie on the line.
Quick Tip: The length of a focal chord can also be written as \( 4a \csc^2 \theta \), where \( \theta \) is the angle it makes with the x-axis. Here, \( 36 \csc^2 \theta = 100 \implies \sin \theta = \frac{6}{10} = \frac{3}{5} \), which gives the slope \( \tan \theta = \frac{3}{4} \).

Step 1: Understanding the Concept:

We need to find the distance along a specific direction. That direction is given by the line of shortest distance (S.D.) between two skew lines.


Step 2: Key Formula or Approach:

Direction of S.D. line between lines with directions \( \vec{b_1} \) and \( \vec{b_2} \) is \( \vec{v} = \vec{b_1} \times \vec{b_2} \).


Step 3: Detailed Explanation:

Directions: \( \vec{b_1} = (2, 0, 1) \) and \( \vec{b_2} = (1, -1, 1) \).
\( \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 0 & 1
1 & -1 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(2-1) + \hat{k}(-2) = \hat{i} - \hat{j} - 2\hat{k} \).

Equation of line through \( P(-1, 2, 3) \) parallel to \( \vec{v} \):
\( \vec{r} = (-1, 2, 3) + k(1, -1, -2) \implies (x, y, z) = (k-1, -k+2, -2k+3) \).

Substitute into the plane eq \( x - 2y + 3z = 10 \): \[ (k-1) - 2(-k+2) + 3(-2k+3) = 10 \] \[ k - 1 + 2k - 4 - 6k + 9 = 10 \implies -3k + 4 = 10 \implies k = -2 \]
Intersection point \( Q = (-2-1, -(-2)+2, -2(-2)+3) = (-3, 4, 7) \).

Distance \( PQ = \sqrt{(-3 - (-1))^2 + (4 - 2)^2 + (7 - 3)^2} \) \[ PQ = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt{6} \]

Step 4: Final Answer:

The distance is \( 2\sqrt{6} \).
Quick Tip: "Distance measured parallel to a line" means you should parameterize a line through the point in that direction and find its intersection with the plane. The scalar parameter \( k \) multiplied by the magnitude of the direction vector gives the distance.

Step 1: Understanding the Concept:

Identify the probability distribution of the random variable \( X \).

Given: \( P(H) = 3P(T) \) and \( P(H) + P(T) = 1 \implies P(H) = \frac{3}{4}, P(T) = \frac{1}{4} \).


Step 2: Detailed Explanation:

Possible values of \( X \) and their paths:

- \( X = 1 \): First toss is H. \( P(X=1) = P(H) = \frac{3}{4} \).

- \( X = 2 \): First toss is T, second is H. \( P(X=2) = P(T) \cdot P(H) = \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16} \).

- \( X = 3 \): First two are T, third is H or third is T (experiment stops anyway). \( P(X=3) = P(T) \cdot P(T) \cdot P(H) + P(T) \cdot P(T) \cdot P(T) \) \( P(X=3) = (\frac{1}{4})^2 \cdot (\frac{3}{4}) + (\frac{1}{4})^3 = \frac{3}{64} + \frac{1}{64} = \frac{4}{64} = \frac{1}{16} \).

Mean \( E[X] = \sum x_i P(X=x_i) \): \[ E[X] = 1 \left(\frac{3}{4}\right) + 2 \left(\frac{3}{16}\right) + 3 \left(\frac{1}{16}\right) \] \[ E[X] = \frac{12}{16} + \frac{6}{16} + \frac{3}{16} = \frac{21}{16} \]

Step 3: Final Answer:

The mean of \( X \) is \( \frac{21}{16} \).
Quick Tip: For stopping criteria like "three tails", remember that the last outcome could be anything to fulfill the condition. Always check if probabilities sum to 1: \( \frac{12}{16} + \frac{3}{16} + \frac{1}{16} = 1 \).

Step 1: Understanding the Concept:

The problem involves composite functions. We need to express \( f(x) \) explicitly by substituting for \( g(x) \).


Step 2: Detailed Explanation:

Given \( g(x) = \sqrt{x} + 1 \). Let \( g(x) = t \).

Then \( t = \sqrt{x} + 1 \implies \sqrt{x} = t - 1 \).

Squaring gives \( x = (t - 1)^2 \).

Substitute these into the expression for \( f(g(x)) \): \[ f(t) = (t - 1)^2 + 3 - (t - 1) \] \[ f(t) = (t^2 - 2t + 1) + 3 - t + 1 \] \[ f(t) = t^2 - 3t + 5 \]
To find \( f(0) \), substitute \( t = 0 \): \[ f(0) = 0^2 - 3(0) + 5 = 5 \]

Step 3: Final Answer:

The value of \( f(0) \) is 5.
Quick Tip: Instead of finding the full quadratic \( f(t) \), you can directly set \( g(x) = 0 \). Here \( \sqrt{x} + 1 = 0 \implies \sqrt{x} = -1 \). Though \( \sqrt{x} \) isn't real, the algebraic substitution \( \sqrt{x} = -1 \) and \( x = (-1)^2 = 1 \) in the final expression yields \( 1 + 3 - (-1) = 5 \) correctly.

Step 1: Understanding the Concept:

We use logical equivalences to simplify the given compound statement before negating it.


Step 2: Detailed Explanation:

Let \( S_1 = (A \wedge (B \vee C)) \implies (A \vee B) \).

By definition of \( \implies \), \( P \implies Q \equiv \sim P \vee Q \).

Note that \( A \wedge (B \vee C) \) implies \( A \).

Also \( A \) implies \( A \vee B \).

Therefore, \( A \wedge (B \vee C) \implies A \vee B \) is a Tautology (\( T \)).

The whole statement is \( T \implies A \).

By truth table or logic laws, \( T \implies A \equiv \sim T \vee A \equiv F \vee A \equiv A \).

The original statement is simply equivalent to \( A \).

The negation of the statement is \( \sim A \).


Step 3: Final Answer:

The negation is equivalent to \( \sim A \).
Quick Tip: Always look for patterns like "If \( P \subset Q \), then \( P \implies Q \) is always True." In this case, the intersection of \( A \) with anything is a subset of the union of \( A \) with anything.

Step 1: Understanding the Concept:

We find the equation of the circle from the geometric conditions and equate it to the real part of the complex expression. Then, we find the line corresponding to the imaginary part and calculate the chord length.


Step 2: Key Formula or Approach:

For a circle touching the \( y \)-axis at \( (0, y_0) \) with radius \( r \), the center is \( (r, y_0) \).

The real part of \( w \) represents the circle equation, and the imaginary part represents a line.


Step 3: Detailed Explanation:

Let \( z = x + iy \). Then \( z\bar{z} = x^2 + y^2 \).
\( w = (x^2 + y^2) + k_1(x + iy) + k_2 i(x + iy) + \lambda + i\lambda \)
\( w = (x^2 + y^2 + k_1 x - k_2 y + \lambda) + i(k_1 y + k_2 x + \lambda) \)

The circle \( C \) is \(Re(w) = x^2 + y^2 + k_1 x - k_2 y + \lambda = 0 \).

Circle \( C \) has radius 1 and touches the \( y \)-axis. Since it's in the first quadrant, its center is \( (1, y_c) \).

It also touches \( y = 1 \). The distance from center \( (1, y_c) \) to \( y = 1 \) is \( |y_c - 1| = 1 \).

Since it's in the first quadrant, \( y_c = 2 \). Center is \( (1, 2) \).

Circle equation: \( (x-1)^2 + (y-2)^2 = 1 \implies x^2 + y^2 - 2x - 4y + 4 = 0 \).

Comparing coefficients: \( k_1 = -2, k_2 = 4, \lambda = 4 \).

Now, \(Im(w) = 0 \implies k_1 y + k_2 x + \lambda = 0 \):
\( -2y + 4x + 4 = 0 \implies y - 2x - 2 = 0 \).

Distance from center \( (1, 2) \) to the line \( 2x - y + 2 = 0 \):
\[ p = \frac{|2(1) - 2 + 2|}{\sqrt{2^2 + (-1)^2}} = \frac{2}{\sqrt{5}} \]
Chord length \( AB = 2\sqrt{r^2 - p^2} \):
\[ AB = 2\sqrt{1^2 - \left(\frac{2}{\sqrt{5}}\right)^2} = 2\sqrt{1 - \frac{4}{5}} = 2\sqrt{\frac{1}{5}} \]
Then \( (AB)^2 = \frac{4}{5} \).
\( 30(AB)^2 = 30 \times \frac{4}{5} = 24 \).


Step 4: Final Answer:

The value of \( 30(AB)^2 \) is 24.
Quick Tip: Always simplify complex expressions into \( x + iy \) form to separate real and imaginary geometric curves. Chord length is best calculated using perpendicular distance from the center.

Step 1: Understanding the Concept:

This is a problem of finding the number of integer solutions to an equation with constraints, followed by counting permutations.


Step 2: Key Formula or Approach:

Let the digits be \( x_1, x_2, \dots, x_7 \). We need \( \sum_{i=1}^7 x_i = 12 \) with \( 1 \le x_i \le 4 \).

Let \( y_i = x_i - 1 \), then \( \sum_{i=1}^7 y_i = 5 \) with \( 0 \le y_i \le 3 \).


Step 3: Detailed Explanation:

The number of solutions is the coefficient of \( x^5 \) in the expansion of \( (1 + x + x^2 + x^3)^7 \).
\[ (1 + x + x^2 + x^3)^7 = \left( \frac{1-x^4}{1-x} \right)^7 = (1 - x^4)^7 (1 - x)^{-7} \]
We need the coefficient of \( x^5 \) in \( (1 - 7x^4 + \dots) \times \sum_{r=0}^{\infty} \binom{7+r-1}{r} x^r \).

Coefficient of \( x^5 = \binom{7+5-1}{5} - 7 \binom{7+1-1}{1} \)
\[ = \binom{11}{5} - 7 \binom{7}{1} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} - 7(7) \] \[ = 462 - 49 = 413 \]

Step 4: Final Answer:

The number of such integers is 413.
Quick Tip: Using multinomial theorem or generating functions is often faster than listing manual cases when dealing with sums of digits and specific constraints.

Step 1: Understanding the Concept:

We use the general term of a binomial expansion to find conditions for the constant term and the coefficient of \( x^{-n} \).


Step 2: Detailed Explanation:

General term \( T_{r+1} = \binom{n}{r} (\sqrt{x})^{n-r} \left(-\frac{6}{x^{3/2}}\right)^r = \binom{n}{r} (-6)^r x^{\frac{n-r}{2} - \frac{3r}{2}} = \binom{n}{r} (-6)^r x^{\frac{n-4r}{2}} \).

For constant term \(\alpha\), \( n - 4r = 0 \implies n = 4r \).

Since \( n \le 15 \), \( n \) can be 4, 8, or 12.

Sum of all coefficients (put \( x=1 \)) is \( (1 - 6)^n = (-5)^n \).

Sum of remaining coefficients \( = (-5)^n - \alpha = 649 \).

Case 1: \( n=4 \). Then \( r=1 \). \(\alpha = \binom{4}{1}(-6)^1 = -24 \).
\( (-5)^4 - (-24) = 625 + 24 = 649 \). This matches. So \( n=4 \).

We need coefficient of \( x^{-n} = x^{-4} \).
\( \frac{4-4r}{2} = -4 \implies 2-2r = -4 \implies r = 3 \).

Coefficient \( = \binom{4}{3}(-6)^3 = 4 \times (-216) = -864 \).

Given: \( -864 = \lambda \alpha \implies -864 = \lambda (-24) \).
\( \lambda = \frac{864}{24} = 36 \).


Step 3: Final Answer:

The value of \( \lambda \) is 36.
Quick Tip: To find the sum of coefficients in any polynomial expansion \( (f(x))^n \), simply evaluate \( f(1)^n \). The constant term corresponds to the term independent of \( x \).

Step 1: Understanding the Concept:

The series has terms that alternate in structure. We can group the terms into pairs to find a general expression for the sum.


Step 2: Detailed Explanation:

The given series is \( \sum_{k=1}^{10} \left[ 2(2k)^2 - (2k+1)^2 \right] \).

The general term \( T_k = 2(4k^2) - (4k^2 + 4k + 1) = 4k^2 - 4k - 1 \).

Sum \( S = \sum_{k=1}^{10} (4k^2 - 4k - 1) = 4 \sum_{k=1}^{10} k^2 - 4 \sum_{k=1}^{10} k - \sum_{k=1}^{10} 1 \)

Using sum formulas:
\[ S = 4 \left( \frac{10 \times 11 \times 21}{6} \right) - 4 \left( \frac{10 \times 11}{2} \right) - 10 \] \[ S = 4(385) - 4(55) - 10 = 1540 - 220 - 10 = 1310 \]

Step 3: Final Answer:

The sum of the series is 1310.
Quick Tip: Grouping terms into pairs helps in identifying a simplified polynomial form for the general term of alternating or piecewise series.

Step 1: Understanding the Concept:

This problem involves calculating a sequence of functions and constants defined recursively by integration.


Step 2: Detailed Explanation:
\( k=1 \): \( C_1 = 1 - \int_0^1 x dx = 1 - \frac{1}{2} = \frac{1}{2} \).
\( S_1(x) = \frac{1}{2}x + 1 \int_0^x t dt = \frac{x}{2} + \frac{x^2}{2} \).
\( k=2 \): \( C_2 = 1 - \int_0^1 (\frac{x}{2} + \frac{x^2}{2}) dx = 1 - (\frac{1}{4} + \frac{1}{6}) = 1 - \frac{5}{12} = \frac{7}{12} \).
\( S_2(x) = \frac{7}{12}x + 2 \int_0^x (\frac{t}{2} + \frac{t^2}{2}) dt = \frac{7}{12}x + \frac{x^2}{2} + \frac{x^3}{3} \).
\( k=3 \): \( C_3 = 1 - \int_0^1 (\frac{7}{12}x + \frac{x^2}{2} + \frac{x^3}{3}) dx = 1 - (\frac{7}{24} + \frac{1}{6} + \frac{1}{12}) = 1 - \frac{13}{24} = \frac{11}{24} \).

Now, \( S_2(3) = \frac{7}{12}(3) + \frac{3^2}{2} + \frac{3^3}{3} = \frac{7}{4} + \frac{9}{2} + 9 = \frac{7 + 18 + 36}{4} = \frac{61}{4} \).
\( 6C_3 = 6 \left(\frac{11}{24}\right) = \frac{11}{4} \).

Sum \( = \frac{61}{4} + \frac{11}{4} = \frac{72}{4} = 18 \).


Step 3: Final Answer:

The result is 18.
Quick Tip: Be careful with the recursive integration steps. Polynomial integration \( \int x^n = \frac{x^{n+1}}{n+1} \) is the core method here.

Step 1: Understanding the Concept:

We find the slopes of tangents from a point to a hyperbola using the condition of tangency. Then we find point \( Q \) and the \( x \)-axis intercepts.


Step 2: Detailed Explanation:
Hyperbola: \( \frac{y^2}{25} - \frac{x^2}{16} = 1 \). Condition for tangency \( y = mx \pm \sqrt{25 - 16m^2} \).

Since it passes through \( (4, 1) \): \( 1 - 4m = \pm \sqrt{25 - 16m^2} \).

Squaring: \( 1 + 16m^2 - 8m = 25 - 16m^2 \implies 32m^2 - 8m - 24 = 0 \).
\( 4m^2 - m - 3 = 0 \implies (4m+3)(m-1) = 0 \).

Slopes are \( m_1 = 1 \) and \( m_2 = -3/4 \).

Given slopes from \( Q \) are \( |m_1| = 1 \) and \( |m_2| = 3/4 \).

Intercepts on \( x \)-axis are positive. Tangent lines from \( Q \):
\( y = 1(x - \alpha) \implies y = x - \alpha \) and \( y = \frac{3}{4}(x - \beta) \implies y = \frac{3}{4}x - \frac{3}{4}\beta \).

Using tangency condition \( c^2 = b^2 - a^2 m^2 \):
\( \alpha^2 = 25 - 16(1) = 9 \implies \alpha = 3 \) (since intercept is positive).
\( \left(\frac{3}{4}\beta\right)^2 = 25 - 16\left(\frac{3}{4}\right)^2 = 25 - 9 = 16 \implies \frac{3}{4}\beta = 4 \implies \beta = \frac{16}{3} \).

Lines are \( y = x - 3 \) and \( y = \frac{3}{4}x - 4 \).

Intersection \( Q \): \( x - 3 = \frac{3}{4}x - 4 \implies \frac{1}{4}x = -1 \implies x = -4, y = -7 \).
\( P(4, 1) \), \( Q(-4, -7) \).
\( (PQ)^2 = (4 - (-4))^2 + (1 - (-7))^2 = 64 + 64 = 128 \).
\( \alpha \beta = 3 \times \frac{16}{3} = 16 \).

Result \( = \frac{128}{16} = 8 \).


Step 3: Final Answer:

The ratio is 8.
Quick Tip: For a conjugate hyperbola \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the tangency condition is \( c^2 = b^2 - a^2 m^2 \).

Step 1: Understanding the Concept:

First, find the coordinates of the image point \( P \) using the standard formula. Then apply the distance formula between \( P \) and \( Q \).


Step 2: Detailed Explanation:

Image \( P(x_1, y_1, z_1) \) of \( (x_0, y_0, z_0) \) in plane \( ax + by + cz + d = 0 \):
\[ \frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = \frac{z_1 - z_0}{c} = -2 \frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2} \]
For \( (5/3, 5/3, 8/3) \) and \( x - 2y + z - 2 = 0 \):
\[ \frac{x_1 - 5/3}{1} = \frac{y_1 - 5/3}{-2} = \frac{z_1 - 8/3}{1} = -2 \frac{5/3 - 10/3 + 8/3 - 2}{1 + 4 + 1} = -2 \frac{1/3}{6} = \frac{1}{3} \]
So, \( x_1 = 2, y_1 = 1, z_1 = 3 \). \( P = (2, 1, 3) \).

Distance \( PQ = 13 \):
\[ (6 - 2)^2 + (-2 - 1)^2 + (\alpha - 3)^2 = 13^2 \] \[ 16 + 9 + (\alpha - 3)^2 = 169 \implies (\alpha - 3)^2 = 144 \]
Since \( \alpha > 0 \), \( \alpha - 3 = 12 \implies \alpha = 15 \).


Step 3: Final Answer:

The value of \( \alpha \) is 15.
Quick Tip: Double check the sign in the image formula. For an image, the factor is \(-2\). For the foot of the perpendicular, the factor is \(-1\).

Step 1: Understanding the Concept:

We use the magnitude of the cross product and dot product properties to find the term in the final expression.


Step 2: Detailed Explanation:
\( |\vec{a}|^2 = |\vec{b} \times \vec{c}|^2 = |\vec{b}|^2 |\vec{c}|^2 - (\vec{b} \cdot \vec{c})^2 \).
\( |\vec{a}|^2 = 9 + 1 + 1 = 11 \).
\( |\vec{c}|^2 = 4 + 9 + 9 = 22 \).
\( 11 = 50 \times 22 - (\vec{b} \cdot \vec{c})^2 \implies (\vec{b} \cdot \vec{c})^2 = 1100 - 11 = 1089 \).
\( \vec{b} \cdot \vec{c} = \pm 33 \).

Now, \( |\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c} = 50 + 22 \pm 66 \).

If \( \vec{b} \cdot \vec{c} = 33 \), then \( |\vec{b} + \vec{c}|^2 = 138 \).

If \( \vec{b} \cdot \vec{c} = -33 \), then \( |\vec{b} + \vec{c}|^2 = 6 \).

In both cases, \( |72 - 138| = 66 \) or \( |72 - 6| = 66 \).


Step 3: Final Answer:

The result is 66.
Quick Tip: The relation \( |\vec{u} \times \vec{v}|^2 + (\vec{u} \cdot \vec{v})^2 = |\vec{u}|^2 |\vec{v}|^2 \) is known as Lagrange's Identity and is very useful in vector magnitude problems.

Step 1: Understanding the Concept:

First, use the mean to find the missing frequency \( \alpha \). Then calculate the mean deviation and variance using standard definitions.


Step 2: Detailed Explanation:

Mean \( = \frac{1(4) + 3(24) + 5(28) + 7\alpha + 9(8)}{64 + \alpha} = 5 \).
\( 4 + 72 + 140 + 7\alpha + 72 = 320 + 5\alpha \implies 288 + 7\alpha = 320 + 5\alpha \implies \alpha = 16 \).

Total frequency \( N = 64 + 16 = 80 \).

Mean deviation \( m = \frac{1}{80} \sum f_i |x_i - \bar{x}| = \frac{1}{80} [4(4) + 24(2) + 0 + 16(2) + 8(4)] = \frac{128}{80} = 1.6 \).

Variance \( \sigma^2 = \frac{1}{80} \sum f_i (x_i - \bar{x})^2 = \frac{1}{80} [4(16) + 24(4) + 0 + 16(4) + 8(16)] = \frac{352}{80} = 4.4 \).

Required value \( = \frac{3(16)}{1.6 + 4.4} = \frac{48}{6} = 8 \).


Step 3: Final Answer:

The result is 8.
Quick Tip: For mean deviation and variance, calculating \( x_i - \bar{x} \) (deviations) separately helps in reducing calculation errors.

Step 1: Understanding the Concept:

The terms in the arcsine functions represent components of right triangles. Converting them to arctangent form simplifies the equation.


Step 2: Detailed Explanation:

Let \( \sin^{-1} \left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) = \tan^{-1}(x+1) \) and \( \sin^{-1} \left(\frac{x}{\sqrt{x^2+1}}\right) = \tan^{-1}(x) \).

Equation becomes \( \tan^{-1}(x+1) - \tan^{-1}(x) = \frac{\pi}{4} \).
\( \tan^{-1} \left( \frac{(x+1) - x}{1 + x(x+1)} \right) = \frac{\pi}{4} \implies \frac{1}{x^2+x+1} = 1 \).
\( x^2 + x = 0 \implies x = 0, -1 \). \( S = \{0, -1\} \).

For \( x=0 \) and \( x=-1 \), \( x^2+x+5 = 5 \).

The summation is \( \sum_{x \in \{0, -1\}} \left( \sin \left( \frac{5\pi}{2} \right) - \cos(5\pi) \right) \).
\( \sin(5\pi/2) = 1 \) and \( \cos(5\pi) = -1 \).

Term for each \( x = 1 - (-1) = 2 \).

Sum \( = 2 + 2 = 4 \).


Step 3: Final Answer:

The sum is 4.
Quick Tip: Replacing \( \sin^{-1}\left(\frac{a}{\sqrt{a^2+1}}\right) \) with \( \tan^{-1}(a) \) is a common and effective trick for solving inverse trigonometric equations.