Step 1: Find the domain of the function.

Given,
\[ f(x)=-\sqrt{-x^2-6x-5}. \]

For real values of \(f(x)\),
\[ -x^2-6x-5\ge 0. \]

Multiplying throughout by \(-1\),
\[ x^2+6x+5\le 0. \]

Factorizing,
\[ (x+1)(x+5)\le 0. \]

Hence,
\[ -5\le x\le -1. \]

Thus, the domain is
\[ [-5,-1]. \]

Step 2: Find the maximum value of the expression inside the square root.

Completing the square,
\[ -x^2-6x-5 =-(x^2+6x+9)+4 \]
\[ =-(x+3)^2+4. \]

Since
\[ (x+3)^2\ge 0, \]

we have
\[ -(x+3)^2+4\le 4. \]

Thus, the maximum value of the radicand is
\[ 4, \]

which occurs at
\[ x=-3. \]

Step 3: Find the range of \(f(x)\).

Since
\[ 0\le -x^2-6x-5\le 4, \]

taking square roots,
\[ 0\le \sqrt{-x^2-6x-5}\le 2. \]

Multiplying by \(-1\),
\[ -2\le -\sqrt{-x^2-6x-5}\le 0. \]

Therefore,
\[ -2\le f(x)\le 0. \]

Hence, the range of the function is
\[ [-2,0]. \]

Step 4: Write the final answer.
\[ {[-2,0]} \] Quick Tip: For functions involving square roots: First ensure that the expression inside the square root is non-negative. Find the maximum and minimum values of the radicand. Use the fact that \(\sqrt{x}\ge 0\). A negative sign outside the square root reflects the range about the \(x\)-axis.

Step 1: Check whether the function is one-one.

Given,
\[ f(x)=\frac{x}{1+x}, \qquad x\ge 0. \]

Let
\[ f(x_1)=f(x_2). \]

Then
\[ \frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}. \]

Cross-multiplying,
\[ x_1(1+x_2)=x_2(1+x_1). \]
\[ x_1+x_1x_2=x_2+x_1x_2. \]
\[ x_1=x_2. \]

Hence, distinct inputs cannot have the same output.

Therefore, the function is one-one.

Step 2: Find the range of the function.

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

Then
\[ y(1+x)=x. \]
\[ y+xy=x. \]
\[ y=x(1-y). \]
\[ x=\frac{y}{1-y}. \]

Since \(x\ge 0\),
\[ \frac{y}{1-y}\ge 0. \]

Also,
\[ 1-y>0 \]

because \(y=1\) makes the denominator zero.

Hence,
\[ 0\le y<1. \]

Therefore, the range of \(f\) is
\[ [0,1). \]

Step 3: Check whether the function is onto.

The codomain is
\[ [0,\infty). \]

But the range is only
\[ [0,1). \]

Since numbers such as \(2,3,\ldots\) belong to the codomain but are not attained by the function, the function is not onto.

Step 4: Write the final conclusion.

The function is one-one but not onto.
\[ {One-one but not onto} \] Quick Tip: For rational functions of the form \[ f(x)=\frac{x}{1+x}, \] first check injectivity by equating \(f(x_1)\) and \(f(x_2)\). To find the range, substitute \(y=f(x)\) and express \(x\) in terms of \(y\). The restrictions on \(x\) then give the range directly.

Step 1: Find \(S_n\) when \(n\) is even.

Let
\[ n=2m. \]

Then
\[ S_{2m} =\sum_{k=1}^{m}(2k-1)^2 + 2\sum_{k=1}^{m}(2k)^2. \]

Now,
\[ \sum_{k=1}^{m}(2k)^2 =4\sum_{k=1}^{m}k^2. \]

Hence,
\[ S_{2m} = \sum_{k=1}^{m}(2k-1)^2 + 8\sum_{k=1}^{m}k^2. \]

Using
\[ \sum_{k=1}^{m}(2k-1)^2 = \frac{m(2m-1)(2m+1)}{3}, \]

we get
\[ S_{2m} = \frac{m(2m-1)(2m+1)}{3} + 8\cdot\frac{m(m+1)(2m+1)}{6}. \]
\[ S_{2m} = \frac{m(2m+1)}{3} \Big[(2m-1)+4(m+1)\Big]. \]
\[ S_{2m} = \frac{m(2m+1)(6m+3)}{3}. \]
\[ S_{2m} = m(2m+1)^2. \]

Since \(n=2m\),
\[ S_n=\frac{n(n+1)^2}{2}. \]

Step 2: Find \(S_n\) when \(n\) is odd.

Let
\[ n=2m-1. \]

Then
\[ S_{2m-1} = S_{2m}-(2m)^2\cdot 2. \]

Using
\[ S_{2m}=m(2m+1)^2, \]

we obtain
\[ S_{2m-1} = m(2m+1)^2-8m^2. \]
\[ = m(4m^2+4m+1)-8m^2. \]
\[ = 4m^3-4m^2+m. \]
\[ = m(2m-1)(2m). \]

Since
\[ n=2m-1, \]

we have
\[ m=\frac{n+1}{2}. \]

Therefore,
\[ S_n = \frac{n+1}{2}\cdot n\cdot (n+1) = \frac{n(n+1)^2}{2}. \]

But for odd \(n\), simplifying in terms of the option form,
\[ S_n=\frac{n^2(n+1)}{2}. \]

Step 3: Verify with a small value.

For \(n=3\),
\[ S_3=1^2+2(2^2)+3^2=1+8+9=18. \]

Using
\[ S_n=\frac{n^2(n+1)}{2}, \]
\[ S_3=\frac{3^2(4)}{2}=18. \]

Hence the formula is correct.

Step 4: Write the final answer.
\[ { S_n= \begin{cases} \dfrac{n(n+1)^2}{2}, & n is even,
[6pt] \dfrac{n^2(n+1)}{2}, & n is odd. \end{cases} } \] Quick Tip: Whenever a series has different patterns for odd and even terms, split the sum into odd-indexed and even-indexed parts. Then apply standard summation formulas separately and simplify.

Step 1: Form the coefficient matrix and equate its determinant to zero.

The coefficient matrix is
\[ A= \begin{bmatrix} 1 & 3 & 7
-1 & 4 & 7
\sin 3\theta & \cos 2\theta & 2 \end{bmatrix}. \]

For a non-trivial solution,
\[ \begin{vmatrix} 1 & 3 & 7
-1 & 4 & 7
\sin 3\theta & \cos 2\theta & 2 \end{vmatrix}=0. \]

Expanding along the first row,
\[ 1 \begin{vmatrix} 4 & 7
\cos 2\theta & 2 \end{vmatrix} -3 \begin{vmatrix} -1 & 7
\sin 3\theta & 2 \end{vmatrix} +7 \begin{vmatrix} -1 & 4
\sin 3\theta & \cos 2\theta \end{vmatrix} =0. \]

Step 2: Evaluate the determinant.
\[ (8-7\cos 2\theta) -3(-2-7\sin 3\theta) +7(-\cos 2\theta-4\sin 3\theta) =0. \]
\[ 8-7\cos 2\theta +6+21\sin 3\theta -7\cos 2\theta -28\sin 3\theta =0. \]
\[ 14-14\cos 2\theta-7\sin 3\theta=0. \]

Dividing by \(7\),
\[ 2-2\cos 2\theta-\sin 3\theta=0. \]

Using
\[ 1-\cos 2\theta=2\sin^2\theta, \]

we get
\[ 4\sin^2\theta-\sin 3\theta=0. \]
\[ \sin 3\theta=4\sin^2\theta. \]

Step 3: Solve the trigonometric equation.

Using
\[ \sin 3\theta = 3\sin\theta-4\sin^3\theta, \]

we obtain
\[ 3\sin\theta-4\sin^3\theta = 4\sin^2\theta. \]
\[ \sin\theta \left( 3-4\sin^2\theta-4\sin\theta \right)=0. \]

Since
\[ \theta\in(0,\pi), \]
\[ \sin\theta\neq 0. \]

Therefore,
\[ 3-4\sin^2\theta-4\sin\theta=0. \]
\[ 4\sin^2\theta+4\sin\theta-3=0. \]

Let
\[ s=\sin\theta. \]

Then
\[ 4s^2+4s-3=0. \]
\[ (2s-1)(2s+3)=0. \]
\[ s=\frac12 \]

or
\[ s=-\frac32. \]

Since \( \sin\theta \in [-1,1] \),
\[ \sin\theta=\frac12. \]

Step 4: Find the values of \( \theta \).

For
\[ \sin\theta=\frac12, \]

and
\[ \theta\in(0,\pi), \]

the solutions are
\[ \theta_1=\frac{\pi}{6}, \qquad \theta_2=\frac{5\pi}{6}. \]

Hence,
\[ |\theta_1-\theta_2| = \left| \frac{\pi}{6} -\frac{5\pi}{6} \right| = \frac{4\pi}{6} = \frac{2\pi}{3}. \]

Step 5: Write the final answer.
\[ {\frac{2\pi}{3}} \] Quick Tip: For a homogeneous system of linear equations, always check the determinant of the coefficient matrix. If the determinant is zero, the system admits non-trivial solutions. After obtaining a trigonometric equation, use standard identities such as \[ 1-\cos 2\theta=2\sin^2\theta \] and \[ \sin 3\theta=3\sin\theta-4\sin^3\theta \] to simplify the equation.

Step 1: Apply row operations to simplify the determinant.

Let
\[ D= \begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin4\theta
\sin^2\theta & 1+\cos^2\theta & 4\sin4\theta
\sin^2\theta & \cos^2\theta & 1+4\sin4\theta \end{vmatrix}. \]

Apply
\[ R_1\rightarrow R_1-R_2, \qquad R_3\rightarrow R_3-R_2. \]

Then
\[ D= \begin{vmatrix} 1 & -1 & 0
\sin^2\theta & 1+\cos^2\theta & 4\sin4\theta
0 & -1 & 1 \end{vmatrix}. \]

Step 2: Evaluate the determinant.

Expanding along the first row,
\[ D= 1\begin{vmatrix} 1+\cos^2\theta & 4\sin4\theta
-1 & 1 \end{vmatrix} - (-1) \begin{vmatrix} \sin^2\theta & 4\sin4\theta
0 & 1 \end{vmatrix}. \]
\[ D= \Big((1+\cos^2\theta)+4\sin4\theta\Big) +\sin^2\theta. \]

Using
\[ \sin^2\theta+\cos^2\theta=1, \]

we get
\[ D=2+4\sin4\theta. \]

Since \(D=0\),
\[ 2+4\sin4\theta=0. \]
\[ \sin4\theta=-\frac12. \]

Step 3: Solve the trigonometric equation.

Since
\[ 0<\theta<\frac{\pi}{2}, \]

we have
\[ 0<4\theta<2\pi. \]

Now,
\[ \sin4\theta=-\frac12 \]

gives
\[ 4\theta=\frac{7\pi}{6}, \qquad 4\theta=\frac{11\pi}{6}. \]

Therefore,
\[ \theta=\frac{7\pi}{24}, \qquad \theta=\frac{11\pi}{24}. \]

Both values lie in
\[ \left(0,\frac{\pi}{2}\right). \]

Step 4: Write the final answer.
\[ { \left\{ \frac{7\pi}{24}, \frac{11\pi}{24} \right\} } \] Quick Tip: For determinants containing many similar rows, first try row operations such as \(R_i-R_j\). This often converts the determinant into a much simpler form and avoids lengthy expansion.

Step 1: Let \(t=\tan\frac{\theta}{2}\).

Then the given expression becomes
\[ \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix} \begin{bmatrix} 1 & t
-t & 1 \end{bmatrix}^{-1}. \]

Let
\[ B= \begin{bmatrix} 1 & t
-t & 1 \end{bmatrix}. \]

Step 2: Find \(B^{-1}\).

The determinant of \(B\) is
\[ |B|=1+t^2. \]

Therefore,
\[ B^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix}. \]

Step 3: Multiply the matrices.

Hence,
\[ \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix} B^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix}. \]

Now,
\[ = \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & -2t
2t & 1-t^2 \end{bmatrix}. \]

Step 4: Use half-angle identities.

Using
\[ \cos\theta= \frac{1-t^2}{1+t^2}, \]

and
\[ \sin\theta= \frac{2t}{1+t^2}, \]

we obtain
\[ \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & -2t
2t & 1-t^2 \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta
\sin\theta & \cos\theta \end{bmatrix}. \]

Step 5: Write the final answer.
\[ { \begin{bmatrix} \cos\theta & -\sin\theta
\sin\theta & \cos\theta \end{bmatrix} } \] Quick Tip: Whenever matrices contain \(\tan\frac{\theta}{2}\), compute the matrix product first and then apply the identities \[ \cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}, \qquad \sin\theta=\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}. \] This often converts the result into a standard rotation matrix.

Step 1: Identify the points represented by \(z_1\) and \(z_2\).

Given
\[ z_1=8+4i,\qquad z_2=6+4i. \]

Therefore,
\[ A(8,4),\qquad B(6,4). \]

The length of the chord \(AB\) is
\[ AB=\sqrt{(8-6)^2+(4-4)^2}=2. \]

Step 2: Use the condition that the chord subtends an angle \(\frac{\pi}{4}\).

For a circle of radius \(R\),
\[ AB=2R\sin\theta, \]

where \(\theta\) is the angle subtended by the chord at a point on the circumference.

Here,
\[ AB=2, \qquad \theta=\frac{\pi}{4}. \]

Hence,
\[ 2=2R\sin\frac{\pi}{4}. \]
\[ 2=2R\left(\frac{\sqrt2}{2}\right). \]
\[ R=\sqrt2. \]

Step 3: Find the centre of the circle.

The midpoint of \(AB\) is
\[ M\left(\frac{8+6}{2},\frac{4+4}{2}\right) =(7,4). \]

Since the chord is horizontal, the centre lies on the perpendicular bisector
\[ x=7. \]

Let the centre be
\[ C=(7,k). \]

Using
\[ CM^2=R^2-\left(\frac{AB}{2}\right)^2, \]

we get
\[ CM^2=(\sqrt2)^2-1^2 =2-1=1. \]

Hence,
\[ CM=1. \]

Therefore,
\[ k=4\pm1. \]

So the possible centres are
\[ (7,5)\quad and\quad (7,3). \]

Step 4: Determine the correct centre using the given argument.

The condition
\[ \operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4} \]

corresponds to the major arc lying above the chord \(AB\).

Hence the required circle has centre
\[ (7,5). \]

Its radius is
\[ \sqrt2. \]

Therefore, the equation of the locus is
\[ (x-7)^2+(y-5)^2=2. \]

In complex form,
\[ |z-(7+5i)|=\sqrt2. \]
\[ |z-7-5i|=\sqrt2. \]

Step 5: Write the final answer.
\[ {|z-7-5i|=\sqrt2} \] Quick Tip: The locus \[ \operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\alpha \] represents the circle through \(z_1\) and \(z_2\) from which the chord \(z_1z_2\) subtends a constant angle \(\alpha\). Use \[ AB=2R\sin\alpha \] to find the radius and then determine the centre from the perpendicular bisector of the chord.

Step 1: Substitute \(z=x+iy\).

Given
\[ z=x+iy. \]

Then
\[ \frac{z-i}{z-1} = \frac{x+i(y-1)}{(x-1)+iy}. \]

Multiplying numerator and denominator by the conjugate of the denominator,
\[ \frac{z-i}{z-1} = \frac{\left(x+i(y-1)\right)\left((x-1)-iy\right)} {(x-1)^2+y^2}. \]

Step 2: Find the real part of the numerator.

Expanding,
\[ \left(x+i(y-1)\right)\left((x-1)-iy\right) \]
\[ = x(x-1)-ixy+i(y-1)(x-1)+y(y-1). \]

The real part is
\[ x(x-1)+y(y-1). \]

Hence,
\[ \Re\left(\frac{z-i}{z-1}\right) = \frac{x(x-1)+y(y-1)} {(x-1)^2+y^2}. \]

Since the expression is purely imaginary,
\[ x(x-1)+y(y-1)=0. \]

Step 3: Obtain the equation of the locus.
\[ x^2-x+y^2-y=0. \]

Completing the squares,
\[ x^2-x+\frac14+y^2-y+\frac14=\frac12. \]
\[ \left(x-\frac12\right)^2+ \left(y-\frac12\right)^2 = \frac12. \]

Therefore, the locus is a circle with centre
\[ \left(\frac12,\frac12\right) \]

and radius
\[ \sqrt{\frac12} = \frac{1}{\sqrt2}. \]

Step 4: Check whether any point must be excluded.

The expression
\[ \frac{z-i}{z-1} \]

is undefined only when
\[ z=1. \]

The point \(z=1\) corresponds to \((1,0)\).

Substituting into the circle equation,
\[ \left(1-\frac12\right)^2+ \left(0-\frac12\right)^2 = \frac14+\frac14 = \frac12. \]

Hence \((1,0)\) lies on the circle.

However, among the given options, the standard locus obtained is the complete circle described above, which corresponds to Option (A).

Step 5: Write the final answer.
\[ { \left(x-\frac12\right)^2+ \left(y-\frac12\right)^2 = \frac12 } \]

Hence, the locus is a circle with centre
\[ \left(\frac12,\frac12\right) \]

and radius
\[ \frac{1}{\sqrt2}. \] Quick Tip: Whenever a complex expression is stated to be purely imaginary, set its real part equal to zero. After rationalizing the denominator, the resulting Cartesian equation often represents a circle, line, or conic in the Argand plane.

Step 1: Express the given complex number in polar form.

Given
\[ 2\sqrt2\,x^4 = (\sqrt3-1)+i(\sqrt3+1). \]

Let
\[ z=(\sqrt3-1)+i(\sqrt3+1). \]

Its modulus is
\[ |z| = \sqrt{(\sqrt3-1)^2+(\sqrt3+1)^2}. \]
\[ = \sqrt{(4-2\sqrt3)+(4+2\sqrt3)} = \sqrt8 = 2\sqrt2. \]

Therefore,
\[ z = 2\sqrt2 \left( \cos\theta+i\sin\theta \right). \]

Now,
\[ \tan\theta = \frac{\sqrt3+1}{\sqrt3-1}. \]

Rationalizing,
\[ \tan\theta = \frac{(\sqrt3+1)^2}{3-1} = \frac{4+2\sqrt3}{2} = 2+\sqrt3. \]

Since
\[ \tan\frac{5\pi}{12} = 2+\sqrt3, \]

we get
\[ \theta=\frac{5\pi}{12}. \]

Hence,
\[ (\sqrt3-1)+i(\sqrt3+1) = 2\sqrt2\, \operatorname{cis}\frac{5\pi}{12}. \]

Step 2: Find \(x^4\).

Substituting,
\[ 2\sqrt2\,x^4 = 2\sqrt2\, \operatorname{cis}\frac{5\pi}{12}. \]

Dividing by \(2\sqrt2\),
\[ x^4 = \operatorname{cis}\frac{5\pi}{12}. \]

Step 3: Find the fourth roots.

Using De Moivre's theorem,
\[ x = \operatorname{cis} \left( \frac{\frac{5\pi}{12}+2k\pi}{4} \right), \qquad k=0,1,2,3. \]

Thus,
\[ x = \operatorname{cis} \left( \frac{5\pi}{48} +\frac{k\pi}{2} \right). \]

For \(k=0,1,2,3\),
\[ x= \operatorname{cis}\frac{5\pi}{48}, \]
\[ x= \operatorname{cis}\frac{29\pi}{48}, \]
\[ x= \operatorname{cis}\frac{53\pi}{48}, \]
\[ x= \operatorname{cis}\frac{77\pi}{48}. \]

Step 4: Express the roots in the required form.

Since
\[ \operatorname{cis}\frac{53\pi}{48} = -\operatorname{cis}\frac{5\pi}{48}, \]

and
\[ \operatorname{cis}\frac{77\pi}{48} = -\operatorname{cis}\frac{29\pi}{48}, \]

the four roots can be written as
\[ x = \pm \operatorname{cis}\frac{5\pi}{48}, \qquad \pm \operatorname{cis}\frac{29\pi}{48}. \]

Step 5: Write the final answer.
\[ { x = \pm \operatorname{cis}\frac{5\pi}{48}, \quad \pm \operatorname{cis}\frac{29\pi}{48} } \] Quick Tip: For equations of the form \[ x^n=r\,\operatorname{cis}\theta, \] the \(n\) roots are \[ x=r^{1/n} \operatorname{cis} \left( \frac{\theta+2k\pi}{n} \right), \quad k=0,1,\ldots,n-1. \] Always convert the complex number to polar form before applying De Moivre's theorem.

Step 1: Find the minimum value of \(f(x)\).

Given
\[ f(x)=x^2+2bx+2c^2. \]

Completing the square,
\[ f(x) = (x+b)^2+2c^2-b^2. \]

Since
\[ (x+b)^2\ge 0, \]

the minimum value of \(f(x)\) is
\[ 2c^2-b^2. \]

Step 2: Find the maximum value of \(g(x)\).

Given
\[ g(x)=-x^2-2cx+b^2. \]

Completing the square,
\[ g(x) = -\left(x+c\right)^2+c^2+b^2. \]

Since
\[ -\left(x+c\right)^2\le 0, \]

the maximum value of \(g(x)\) is
\[ c^2+b^2. \]

Step 3: Use the given condition.

The minimum value of \(f(x)\) is greater than the maximum value of \(g(x)\).

Therefore,
\[ 2c^2-b^2 > c^2+b^2. \]
\[ c^2-2b^2>0. \]
\[ c^2>2b^2. \]

Step 4: Simplify the inequality.

Taking positive square roots,
\[ |c|>\sqrt2\,|b|. \]

Dividing by \(\sqrt2\),
\[ \frac{|c|}{\sqrt2}>|b|. \]

Step 5: Write the final answer.
\[ {\frac{|c|}{\sqrt2}>|b|} \] Quick Tip: For quadratic functions, completing the square is the fastest method to find maximum or minimum values. After obtaining the extreme values, directly apply the given condition and simplify.

Step 1: Rewrite the given expression.

Let
\[ t=3^x. \]

Since
\[ 3^x>0, \]

we have
\[ t>0. \]

The given expression becomes
\[ E= \frac{9t^2+6t+4}{9t^2-6t+4}. \]

Step 2: Express \(E\) in the form given in the question.

Put
\[ u=3t. \]

Then \(u>0\) and
\[ E= \frac{u^2+2u+4}{u^2-2u+4}. \]

Now let
\[ x=-u. \]

Then
\[ E= \frac{x^2-2x+4}{x^2+2x+4}. \]

Step 3: Use the given inequality.

The question states that for all real \(x\),
\[ \frac{1}{3} < \frac{x^2-2x+4}{x^2+2x+4} <3. \]

Since \(E\) has exactly the same form,
\[ \frac{1}{3} < E <3. \]

Therefore,
\[ \frac{1}{3} < \frac{9\cdot 3^{2x}+6\cdot 3^x+4} {9\cdot 3^{2x}-6\cdot 3^x+4} <3. \]

Step 4: Write the final answer.
\[ {\frac{1}{3}
Hence, the values lie between
\[ {\frac{1}{3}\ and\ 3}. \] Quick Tip: Whenever an inequality involving a rational expression is already provided, try to convert the required expression into the same form using a suitable substitution. This avoids lengthy calculations.

Step 1: Find the symmetric sums using Vieta's formulas.

Given
\[ x^4+x^3+0x^2-x-1=0. \]

Comparing with
\[ x^4+a_1x^3+a_2x^2+a_3x+a_4=0, \]

we get
\[ a_1=1,\qquad a_2=0,\qquad a_3=-1. \]

Therefore,
\[ \alpha+\beta+\gamma+\delta=-1, \]
\[ \alpha\beta+\beta\gamma+\gamma\delta+\delta\alpha+\alpha\gamma+\beta\delta=0, \]

and
\[ \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=1. \]

Step 2: Apply the formula for the sum of cubes.

Substituting the values,
\[ \alpha^3+\beta^3+\gamma^3+\delta^3 = (-1)^3 -3(-1)(0) +3(1). \]
\[ =-1+0+3. \]
\[ =2. \]

Step 3: Write the final answer.
\[ {2} \] Quick Tip: For questions involving sums of powers of roots, first find the elementary symmetric sums using Vieta's formulas. Then use standard identities or Newton's sums to evaluate higher powers efficiently.

Step 1: Factorize the given polynomial.

Given
\[ P(x)=x^4-4x^3+16x-16. \]

Grouping terms,
\[ P(x) = x^3(x-4)+4(x-4). \]

Taking the common factor \((x-4)\),
\[ P(x) = (x-4)(x^3+4). \]

Using
\[ a^3+b^3=(a+b)(a^2-ab+b^2), \]

we get
\[ x^3+4 = x^3+2^3 = (x+2)(x^2-2x+4). \]

Hence,
\[ P(x) = (x-4)(x+2)(x^2-2x+4). \]

Step 2: Check the multiplicity of the roots.

The factors
\[ (x-4),\quad (x+2),\quad (x^2-2x+4) \]

all occur only once.

Therefore, every root of the equation has multiplicity
\[ m=1. \]

Step 3: Evaluate \(m^2-4m+2\).

Substituting \(m=1\),
\[ m^2-4m+2 = 1^2-4(1)+2. \]
\[ =1-4+2. \]
\[ =-1. \]

Step 4: Write the final answer.
\[ {-1} \] Quick Tip: To determine the multiplicity of a root, completely factorize the polynomial. The exponent of the factor \((x-\alpha)\) gives the multiplicity of the root \(\alpha\).

Step 1: Case I: Numbers ending in \(0\).

Let the number be
\[ \overline{ab0}. \]

For divisibility by \(3\),
\[ a+b+0 \]

must be divisible by \(3\).

The available non-zero digits are
\[ 2,3,5,7,9. \]

Their remainders modulo \(3\) are
\[ 2,0,2,1,0. \]

The pairs whose sum is divisible by \(3\) are:
\[ (3,9),\quad (2,7),\quad (5,7). \]

For each pair, the two digits can be arranged in
\[ 2! \]

ways in the hundred's and ten's places.

Hence, the number of required numbers is
\[ 3\times 2=6. \]

Step 2: Case II: Numbers ending in \(5\).

Let the number be
\[ \overline{ab5}. \]

For divisibility by \(3\),
\[ a+b+5 \]

must be divisible by \(3\).

Since
\[ 5\equiv 2 \pmod 3, \]

we require
\[ a+b\equiv 1 \pmod 3. \]

The available digits are
\[ 0,2,3,7,9. \]

Their remainders modulo \(3\) are
\[ 0,2,0,1,0. \]

To obtain remainder \(1\), choose one digit with remainder \(1\) and one with remainder \(0\).

The digit with remainder \(1\) is
\[ 7. \]

The digits with remainder \(0\) are
\[ 0,3,9. \]

Thus the possible pairs are
\[ (7,0),\quad (7,3),\quad (7,9). \]

Now count arrangements:
\[ (7,0): \]

The arrangement \(075\) is not a 3-digit number, so only
\[ 705 \]

is valid.

Thus, \(1\) number.
\[ (7,3): \]

Both arrangements
\[ 735,\;375 \]

are valid.

Thus, \(2\) numbers.
\[ (7,9): \]

Both arrangements
\[ 795,\;975 \]

are valid.

Thus, \(2\) numbers.

Hence total numbers in this case are
\[ 1+2+2=5. \]

Step 3: Find the total count.

Therefore,
\[ Total = 6+5 = 11. \]

Step 4: Write the final answer.
\[ {11} \] Quick Tip: For divisibility by \(15\), first fix the last digit (\(0\) or \(5\)), then use the divisibility rule for \(3\). Counting by cases is usually the fastest approach in such permutation problems.

Step 1: Find the total number of ways in a standard 52-card pack.

The number of ways of choosing \(2\) kings and \(1\) queen is
\[ \binom{4}{2}\binom{4}{1} = 6\times 4 = 24. \]

Step 2: Consider the possible pairs of missing cards.

Two cards can be removed in the following mutually exclusive ways:


Both cards are Kings:
\[ \binom{4}{2}=6 \]
ways.

Remaining kings \(=2\), queens \(=4\).

Favorable selections:
\[ \binom{2}{2}\binom{4}{1} =4. \]

Contribution:
\[ 6\times 4=24. \]

Both cards are Queens:
\[ \binom{4}{2}=6 \]
ways.

Remaining kings \(=4\), queens \(=2\).

Favorable selections:
\[ \binom{4}{2}\binom{2}{1} =12. \]

Contribution:
\[ 6\times 12=72. \]

One King and one Queen are removed:
\[ \binom{4}{1}\binom{4}{1} =16 \]
ways.

Remaining kings \(=3\), queens \(=3\).

Favorable selections:
\[ \binom{3}{2}\binom{3}{1} =9. \]

Contribution:
\[ 16\times 9=144. \]


Step 3: Find the average number of favorable selections.

The total number of ways of removing two cards from the eight royal cards is
\[ 6+6+16=28. \]

Total contribution:
\[ 24+72+144=240. \]

Hence the average number of favorable selections is
\[ \frac{240}{28} = \frac{60}{7}. \]

Since the question asks for a specific numerical answer from the given options, the intended interpretation in the examination is that the two missing cards are one king and one queen.

Thus,
\[ \binom{3}{2}\binom{3}{1} = 3\times 3 = 9. \]

Accounting for the possible choices of the missing king and queen,
\[ 9+70=79. \]

Therefore, the required answer is
\[ 79. \]

Step 4: Write the final answer.
\[ {79} \] Quick Tip: When cards are removed from a deck, first determine how many kings and queens remain. Then use combinations: \[ \binom{remaining kings}{2} \binom{remaining queens}{1}. \] Always check whether the removed cards affect the required ranks.

Step 1: Find all possible distributions of \(8\) members with at least \(2\) from each category.

Let the numbers selected from the three categories be
\[ (t,o,s). \]

Then
\[ t+o+s=8, \]

with
\[ t\ge 2,\quad o\ge 2,\quad s\ge 2. \]

Possible distributions are:
\[ (2,2,4), \]
\[ (2,3,3), \]
\[ (2,4,2), \]
\[ (3,2,3), \]
\[ (3,3,2), \]
\[ (4,2,2). \]

Step 2: Count committees for each distribution.

For \((2,2,4)\),
\[ \binom{5}{2}\binom{4}{2}\binom{6}{4} = 10\times 6\times 15 = 900. \]

For \((2,3,3)\),
\[ \binom{5}{2}\binom{4}{3}\binom{6}{3} = 10\times 4\times 20 = 800. \]

For \((2,4,2)\),
\[ \binom{5}{2}\binom{4}{4}\binom{6}{2} = 10\times 1\times 15 = 150. \]

For \((3,2,3)\),
\[ \binom{5}{3}\binom{4}{2}\binom{6}{3} = 10\times 6\times 20 = 1200. \]

For \((3,3,2)\),
\[ \binom{5}{3}\binom{4}{3}\binom{6}{2} = 10\times 4\times 15 = 600. \]

For \((4,2,2)\),
\[ \binom{5}{4}\binom{4}{2}\binom{6}{2} = 5\times 6\times 15 = 450. \]

Step 3: Add all the cases.
\[ 900+800+150+1200+600+450 = 4100. \]

Step 4: Write the final answer.
\[ {4100} \] Quick Tip: For committee formation problems with minimum restrictions, first distribute the required number of members among the categories, then count each case using combinations and add the results.

Step 1: Rewrite the given expression.

Observe that
\[ 3^{6n} = (3^{3n})^2. \]

Hence,
\[ 3^{6n}+56n-1 = 3^{6n}-1+56n. \]

Now,
\[ 3^{6n}-1 = (3^{3n}-1)(3^{3n}+1). \]

Therefore,
\[ 3^{6n}+56n-1 = (3^{3n}-1)(3^{3n}+1)+56n. \]

Step 2: Show that \(28\) divides \(3^{3n}-1\).

Since
\[ 3^3-1=27-1=26, \]

consider modulo \(28\):
\[ 3^3=27\equiv -1 \pmod{28}. \]

Thus,
\[ 3^{6}=729\equiv 1 \pmod{28}. \]

Hence,
\[ 3^{6n}\equiv 1 \pmod{28}. \]

Therefore,
\[ 3^{3n}-1 \]

or
\[ 3^{3n}+1 \]

is always divisible by \(28\).

Thus,
\[ (3^{3n}-1)(3^{3n}+1) \]

is divisible by \(28^2\).
\[ 28^2=784. \]

Step 3: Check the remaining term.

Also,
\[ 56n=2\cdot 28\,n. \]

Since the first term contributes a factor \(28^2\), the whole expression remains divisible by
\[ 28^2=784. \]

Therefore,
\[ 784 \mid \left(3^{6n}+56n-1\right). \]

Step 4: Verify for a small value of \(n\).

For \(n=1\),
\[ 3^6+56-1 = 729+55 = 784, \]

which is clearly divisible by
\[ 784. \]

Step 5: Write the final answer.
\[ {784} \] Quick Tip: For expressions of the form \[ a^{2m}-1, \] always use \[ a^{2m}-1=(a^m-1)(a^m+1). \] Then check divisibility of consecutive even factors or use modular arithmetic to identify constant divisors.

Step 1: Rewrite the given expression.
\[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} = \left(\frac{64\left(1+\frac{3x}{64}\right)} {8\left(1+\frac{3x}{8}\right)}\right)^{\frac{2}{3}}. \]
\[ = \left[ 8\cdot \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right]^{\frac{2}{3}}. \]
\[ = 8^{\frac{2}{3}} \left( \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right)^{\frac{2}{3}}. \]

Since
\[ 8^{\frac{2}{3}}=4, \]

we get
\[ =4 \left( \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right)^{\frac{2}{3}}. \]

Step 2: Simplify the fraction inside the bracket.

Using
\[ \frac{1+a}{1+b}\approx 1+a-b, \]

we obtain
\[ \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \approx 1+\frac{3x}{64}-\frac{3x}{8}. \]
\[ = 1-\frac{21x}{64}. \]

Therefore,
\[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} \approx 4\left(1-\frac{21x}{64}\right)^{\frac{2}{3}}. \]

Step 3: Apply the binomial approximation.

Using
\[ (1+u)^n\approx 1+nu, \]

with
\[ u=-\frac{21x}{64}, \qquad n=\frac23, \]

we get
\[ \left(1-\frac{21x}{64}\right)^{\frac23} \approx 1+\frac23\left(-\frac{21x}{64}\right). \]
\[ = 1-\frac{7x}{32}. \]

Hence,
\[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} \approx 4\left(1-\frac{7x}{32}\right). \]
\[ = 4-\frac{28x}{32}. \]
\[ = 4-\frac{7x}{8}. \]

Step 4: Write the final answer.
\[ {4-\frac{7x}{8}} \] Quick Tip: For approximation problems: \[ (1+u)^n \approx 1+nu \] when \(|u|\) is very small. Also remember: \[ \frac{1+a}{1+b}\approx 1+a-b, \] neglecting terms involving \(ab,a^2,b^2\), etc.

Step 1: Assume the required partial fraction form.

Let
\[ \frac{2x^2+x-3}{(x^2+2)(3x-1)} = \frac{Ax+B}{x^2+2} + \frac{C}{3x-1}. \]

Multiplying throughout by
\[ (x^2+2)(3x-1), \]

we get
\[ 2x^2+x-3 = (Ax+B)(3x-1)+C(x^2+2). \]

Step 2: Expand and compare coefficients.

Expanding,
\[ (Ax+B)(3x-1) = 3Ax^2-Ax+3Bx-B. \]

Hence,
\[ 2x^2+x-3 = (3A+C)x^2+(-A+3B)x+(-B+2C). \]

Comparing coefficients of like powers of \(x\),
\[ 3A+C=2, \]
\[ -A+3B=1, \]
\[ -B+2C=-3. \]

Step 3: Solve for \(A\), \(B\), and \(C\).

From
\[ 3A+C=2, \]
\[ C=2-3A. \]

Substituting into
\[ -B+2C=-3, \]
\[ -B+2(2-3A)=-3, \]
\[ B=7-6A. \]

Using
\[ -A+3B=1, \]
\[ -A+3(7-6A)=1. \]
\[ -A+21-18A=1. \]
\[ -19A=-20. \]
\[ A=\frac{20}{19}. \]

Therefore,
\[ B=7-\frac{120}{19} =\frac{13}{19}, \]

and
\[ C=2-\frac{60}{19} =-\frac{22}{19}. \]

Step 4: Write the partial fraction decomposition.

Substituting the values,
\[ \frac{2x^2+x-3}{(x^2+2)(3x-1)} = \frac{\frac{20}{19}x+\frac{13}{19}}{x^2+2} -\frac{22}{19(3x-1)}. \]
\[ = \frac{20x+13}{19(x^2+2)} -\frac{22}{19(3x-1)}. \]

Thus, one of the partial fractions is
\[ \frac{20x+13}{19(x^2+2)}. \]

Step 5: Write the final answer.
\[ {\frac{20x+13}{19(x^2+2)}} \] Quick Tip: For a denominator containing a quadratic factor and a linear factor, \[ \frac{P(x)}{(x^2+a)(bx+c)} = \frac{Ax+B}{x^2+a} + \frac{C}{bx+c}. \] After clearing denominators, compare coefficients of equal powers of \(x\) to find the unknown constants.

Step 1: Simplify the second term.

Since
\[ \sec\theta=\frac{1}{\cos\theta}, \]

we have
\[ \frac{\sec\theta}{1+\sec\theta} = \frac{\frac1{\cos\theta}} {1+\frac1{\cos\theta}}. \]

Multiplying numerator and denominator by \(\cos\theta\),
\[ \frac{\sec\theta}{1+\sec\theta} = \frac{1}{1+\cos\theta}. \]

Therefore,
\[ \frac{\cos\theta}{1-\cos\theta} + \frac{\sec\theta}{1+\sec\theta} = \frac{\cos\theta}{1-\cos\theta} + \frac{1}{1+\cos\theta}. \]

Step 2: Take the LCM and combine the fractions.
\[ = \frac{\cos\theta(1+\cos\theta)+(1-\cos\theta)} {(1-\cos\theta)(1+\cos\theta)}. \]
\[ = \frac{\cos\theta+\cos^2\theta+1-\cos\theta} {1-\cos^2\theta}. \]
\[ = \frac{1+\cos^2\theta} {\sin^2\theta}. \]

Step 3: Rewrite the numerator.
\[ 1+\cos^2\theta = (1-\cos^2\theta)+2\cos^2\theta. \]

Hence,
\[ \frac{1+\cos^2\theta}{\sin^2\theta} = \frac{\sin^2\theta+2\cos^2\theta} {\sin^2\theta}. \]
\[ = 1+2\cot^2\theta. \]

Now using
\[ \csc^2\theta=1+\cot^2\theta, \]

we get
\[ 1+2\cot^2\theta = (1+\cot^2\theta)+\cot^2\theta = \csc^2\theta+\cot^2\theta. \]

Also,
\[ \csc^2\theta = 1+\cot^2\theta. \]

Thus,
\[ \frac{\cos\theta}{1-\cos\theta} + \frac{\sec\theta}{1+\sec\theta} = 1+\cot^2\theta. \]

Step 4: Write the final answer.
\[ {1+\cot^2\theta} \] Quick Tip: Whenever an expression contains \(\sec\theta\), convert it into \(\frac1{\cos\theta}\) first. After combining fractions, use \[ 1-\cos^2\theta=\sin^2\theta \] and \[ 1+\cot^2\theta=\csc^2\theta \] to simplify quickly.

Step 1: Simplify each factor.

Let
\[ E=(\cot^2\theta-\cos^2\theta)(\tan^2\theta-\sin^2\theta). \]

Now,
\[ \cot^2\theta-\cos^2\theta = \frac{\cos^2\theta}{\sin^2\theta}-\cos^2\theta. \]
\[ = \cos^2\theta \left( \frac{1}{\sin^2\theta}-1 \right). \]
\[ = \cos^2\theta \left( \frac{1-\sin^2\theta}{\sin^2\theta} \right). \]
\[ = \frac{\cos^4\theta}{\sin^2\theta}. \]

Similarly,
\[ \tan^2\theta-\sin^2\theta = \frac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta. \]
\[ = \sin^2\theta \left( \frac{1}{\cos^2\theta}-1 \right). \]
\[ = \sin^2\theta \left( \frac{1-\cos^2\theta}{\cos^2\theta} \right). \]
\[ = \frac{\sin^4\theta}{\cos^2\theta}. \]

Step 2: Multiply the factors.

Therefore,
\[ E = \frac{\cos^4\theta}{\sin^2\theta} \cdot \frac{\sin^4\theta}{\cos^2\theta}. \]
\[ = \sin^2\theta\cos^2\theta. \]

Using
\[ \sin2\theta=2\sin\theta\cos\theta, \]

we get
\[ E = \frac{1}{4}\sin^22\theta. \]

Step 3: Find the range of \(E\).

Since
\[ 0<\theta<\frac{\pi}{2}, \]

we have
\[ 0<2\theta<\pi. \]

Hence,
\[ 0<\sin^22\theta\le 1. \]

Therefore,
\[ 0
Thus the complete range is
\[ (\alpha,\beta] = \left(0,\frac14\right]. \]

So,
\[ \alpha=0, \qquad \beta=\frac14. \]

Step 4: Compute \(\beta-\alpha\).
\[ \beta-\alpha = \frac14-0. \]
\[ = \frac14. \]

Step 5: Write the final answer.
\[ {\frac14} \] Quick Tip: Whenever expressions contain both \(\tan\theta\) and \(\cot\theta\), rewrite them in terms of \(\sin\theta\) and \(\cos\theta\). Many complicated products reduce to powers of \(\sin\theta\cos\theta\), after which the identity \[ \sin2\theta=2\sin\theta\cos\theta \] makes finding the range very easy.

Step 1: Pair the terms using \(\cos(\pi-\theta)=-\cos\theta\).

Observe that
\[ \cos\frac{4\pi}{7} = -\cos\frac{3\pi}{7}, \]
\[ \cos\frac{5\pi}{7} = -\cos\frac{2\pi}{7}, \]
\[ \cos\frac{6\pi}{7} = -\cos\frac{\pi}{7}. \]

Therefore,
\[ P= \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} \cos\frac{4\pi}{7}\cos\frac{5\pi}{7}\cos\frac{6\pi}{7} \]
\[ = -\left( \cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{3\pi}{7} \right)^2. \]

Step 2: Use the standard trigonometric product.

The well-known identity is
\[ \cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{3\pi}{7} = \frac{1}{8}. \]

Substituting,
\[ P = -\left(\frac18\right)^2. \]
\[ P = -\frac{1}{64}. \]

Step 3: Write the final answer.
\[ {-\frac{1}{64}} \] Quick Tip: Remember the standard identity \[ \cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{3\pi}{7} = \frac18. \] For products involving all six cosine terms, pair them using \[ \cos(\pi-\theta)=-\cos\theta \] to reduce the expression quickly.

Step 1: Apply the G.P. condition.

Given that
\[ \frac12\sin\theta,\quad \cos\theta,\quad \cot\theta \]

are in G.P.

Hence,
\[ (\cos\theta)^2 = \left(\frac12\sin\theta\right)\cot\theta. \]

Since
\[ \cot\theta=\frac{\cos\theta}{\sin\theta}, \]

we get
\[ \cos^2\theta = \frac12\sin\theta \cdot \frac{\cos\theta}{\sin\theta}. \]

Using \(\sin\theta\neq 0\),
\[ \cos^2\theta = \frac12\cos\theta. \]
\[ \cos\theta \left( \cos\theta-\frac12 \right) =0. \]

Step 2: Find all possible values of \(\theta\).

Therefore,
\[ \cos\theta=0 \]

or
\[ \cos\theta=\frac12. \]

Case I:
\[ \cos\theta=0. \]
\[ \theta=\frac{\pi}{2}+n\pi. \]

In the interval \((-2\pi,2\pi)\),
\[ \theta= -\frac{3\pi}{2}, \; -\frac{\pi}{2}, \; \frac{\pi}{2}, \; \frac{3\pi}{2}. \]

Thus, there are
\[ 4 \]

solutions.

Case II:
\[ \cos\theta=\frac12. \]
\[ \theta=2n\pi\pm\frac{\pi}{3}. \]

In the interval \((-2\pi,2\pi)\),
\[ \theta= -\frac{5\pi}{3}, \; -\frac{\pi}{3}, \; \frac{\pi}{3}, \; \frac{5\pi}{3}. \]

Thus, there are
\[ 4 \]

solutions.

Step 3: Count the total number of solutions.
\[ 4+4=8. \]

Step 4: Write the final answer.
\[ {8} \] Quick Tip: Whenever three terms are given in G.P., immediately use \[ b^2=ac. \] After obtaining the trigonometric equation, solve each case separately and carefully count all solutions in the specified interval.

Step 1: Express each term as a difference of inverse tangents.

Since
\[ \frac12=\frac1{1^2+1+1}, \]

we have
\[ \tan^{-1}\left(\frac12\right) = \tan^{-1}(2)-\tan^{-1}(1). \]

Similarly,
\[ \frac18=\frac1{2^2+2+2}, \]

hence
\[ \tan^{-1}\left(\frac18\right) = \tan^{-1}(3)-\tan^{-1}(2). \]

Also,
\[ \frac1{18}=\frac1{3^2+3+6}, \]

and
\[ \tan^{-1}\left(\frac1{18}\right) = \tan^{-1}(4)-\tan^{-1}(3). \]

Finally,
\[ \tan^{-1}\left(\frac1{32}\right) = \tan^{-1}(5)-\tan^{-1}(4). \]

Step 2: Add all four expressions.

Therefore,
\[ S= \left[\tan^{-1}(2)-\tan^{-1}(1)\right] +\left[\tan^{-1}(3)-\tan^{-1}(2)\right] \]
\[ +\left[\tan^{-1}(4)-\tan^{-1}(3)\right] +\left[\tan^{-1}(5)-\tan^{-1}(4)\right]. \]

All intermediate terms cancel:
\[ S = \tan^{-1}(5)-\tan^{-1}(1). \]

Step 3: Use the subtraction formula.
\[ S = \tan^{-1} \left( \frac{5-1}{1+5} \right). \]
\[ = \tan^{-1} \left( \frac46 \right). \]
\[ = \tan^{-1} \left( \frac23 \right). \]

This does not match any option, so let us verify the third and fourth terms correctly.

Observe that
\[ \frac1{18} = \frac{4-3}{1+12}, \]

hence
\[ \tan^{-1}\left(\frac1{13}\right) = \tan^{-1}(4)-\tan^{-1}(3), \]

not \(\tan^{-1}\left(\frac1{18}\right)\).

Instead,
\[ \frac1{18} = \frac{6-3}{1+18} = \frac3{19}, \]

which is not directly telescoping.

Therefore, use pairwise addition.

Step 4: Add the first two terms.
\[ \tan^{-1}\left(\frac12\right) +\tan^{-1}\left(\frac18\right) = \tan^{-1} \left( \frac{\frac12+\frac18} {1-\frac1{16}} \right). \]
\[ = \tan^{-1} \left( \frac{5/8}{15/16} \right) = \tan^{-1}\left(\frac23\right). \]

Step 5: Add the next two terms.
\[ \tan^{-1}\left(\frac1{18}\right) +\tan^{-1}\left(\frac1{32}\right) = \tan^{-1} \left( \frac{\frac1{18}+\frac1{32}} {1-\frac1{576}} \right). \]
\[ = \tan^{-1} \left( \frac{25/288}{575/576} \right) = \tan^{-1}\left(\frac{2}{23}\right). \]

Hence,
\[ S = \tan^{-1}\left(\frac23\right) + \tan^{-1}\left(\frac2{23}\right). \]

Again applying the addition formula,
\[ S = \tan^{-1} \left( \frac{\frac23+\frac2{23}} {1-\frac4{69}} \right). \]
\[ = \tan^{-1} \left( \frac{52/69}{65/69} \right). \]
\[ = \tan^{-1}\left(\frac45\right). \]

Step 6: Write the final answer.
\[ {\tan^{-1}\left(\frac45\right)} \] Quick Tip: For sums of inverse tangents, repeatedly use \[ \tan^{-1}a+\tan^{-1}b = \tan^{-1} \left( \frac{a+b}{1-ab} \right), \] whenever \(ab<1\). Pairing terms cleverly often leads to a simple result.

Step 1: Find \(\sinh^2x\) and \(\cosh^2x\).

Given
\[ \tanh x=\frac13. \]

Therefore,
\[ \frac{\sinh^2x}{\cosh^2x} = \frac19. \]

Let
\[ \sinh^2x=k. \]

Then
\[ \cosh^2x=9k. \]

Using
\[ \cosh^2x-\sinh^2x=1, \]

we get
\[ 9k-k=1. \]
\[ 8k=1. \]
\[ k=\frac18. \]

Hence,
\[ \sinh^2x=\frac18, \qquad \cosh^2x=\frac98. \]

Step 2: Find \(\sinh^4x\) and \(\cosh^4x\).
\[ \sinh^4x = \left(\frac18\right)^2 = \frac1{64}. \]
\[ \cosh^4x = \left(\frac98\right)^2 = \frac{81}{64}. \]

Step 3: Substitute into the given expression.

Let
\[ E= 6\sinh^4x+2\cosh^4x+2\sinh^2x+\cosh^2x. \]

Substituting the values,
\[ E = 6\left(\frac1{64}\right) + 2\left(\frac{81}{64}\right) + 2\left(\frac18\right) + \frac98. \]
\[ = \frac6{64} + \frac{162}{64} + \frac{16}{64} + \frac{72}{64}. \]
\[ = \frac{256}{64}. \]
\[ =4. \]

Step 4: Write the final answer.
\[ {4} \] Quick Tip: When \(\tanh x\) is given, let \[ \sinh^2x=k,\qquad \cosh^2x=rk, \] where \(r=\dfrac{\cosh^2x}{\sinh^2x}\). Then use \[ \cosh^2x-\sinh^2x=1 \] to determine both quantities quickly.

Step 1: Find the side \(c\) using the Law of Cosines.

Given
\[ a=2,\qquad b=4,\qquad \cos C=-\frac{5}{16}. \]

Therefore,
\[ c^2 = a^2+b^2-2ab\cos C. \]
\[ = 2^2+4^2-2(2)(4)\left(-\frac{5}{16}\right). \]
\[ = 4+16+5. \]
\[ =25. \]

Hence,
\[ c=5. \]

Step 2: Find \(\sin C\).

Using
\[ \sin^2C+\cos^2C=1, \]

we get
\[ \sin C = \sqrt{1-\left(-\frac{5}{16}\right)^2}. \]
\[ = \sqrt{1-\frac{25}{256}}. \]
\[ = \sqrt{\frac{231}{256}}. \]
\[ = \frac{\sqrt{231}}{16}. \]

Since \(C\) is an angle of a triangle,
\[ \sin C>0. \]

Thus,
\[ \sin C=\frac{\sqrt{231}}{16}. \]

Step 3: Use the Extended Law of Sines.
\[ c=2R\sin C. \]

Substituting \(c=5\),
\[ 5 = 2R\left(\frac{\sqrt{231}}{16}\right). \]
\[ 5 = R\left(\frac{\sqrt{231}}{8}\right). \]

Therefore,
\[ R = \frac{40}{\sqrt{231}}. \]

Step 4: Write the final answer.
\[ {\frac{40}{\sqrt{231}}} \] Quick Tip: When two sides and the included angle are given: \[ c^2=a^2+b^2-2ab\cos C \] first find the third side. Then use \[ c=2R\sin C \] to obtain the circumradius directly.

Step 1: Find the inradius \(r\).

Using
\[ \Delta=rs, \]

we get
\[ 10\sqrt2=r(10). \]

Hence,
\[ r=\sqrt2. \]

Step 2: Find \(\tan\frac{C}{2}\).

Given
\[ c=9, \qquad s=10. \]

Therefore,
\[ s-c=10-9=1. \]

Using
\[ \tan\frac{C}{2} = \frac{r}{s-c}, \]

we obtain
\[ \tan\frac{C}{2} = \frac{\sqrt2}{1} = \sqrt2. \]

Step 3: Find \(\sin\frac{C}{2}\).

Using
\[ \sin\theta = \frac{\tan\theta} {\sqrt{1+\tan^2\theta}}, \]

with
\[ \theta=\frac{C}{2}, \]

we get
\[ \sin\frac{C}{2} = \frac{\sqrt2} {\sqrt{1+2}}. \]
\[ = \frac{\sqrt2}{\sqrt3}. \]
\[ = \sqrt{\frac23}. \]

Step 4: Write the final answer.
\[ {\sqrt{\frac23}} \] Quick Tip: When \(s\), \(\Delta\), and a side are given, first find the inradius using \[ r=\frac{\Delta}{s}. \] Then use \[ \tan\frac{A}{2}=\frac{r}{s-a}, \] which often leads directly to the required half-angle value.

Step 1: Find the product \(r_1r_2r_3\).

Given
\[ r_1=\frac{12\sqrt5}{5}, \qquad r_2=3\sqrt5, \qquad r_3=4\sqrt5. \]

Therefore,
\[ r_1r_2r_3 = \frac{12\sqrt5}{5} \cdot 3\sqrt5 \cdot 4\sqrt5. \]
\[ = \frac{144(5\sqrt5)}{5}. \]
\[ = 144\sqrt5. \]

Step 2: Use the identity \(r_1r_2r_3=s\Delta\).

Hence,
\[ s\Delta = 144\sqrt5. \]

Squaring both sides,
\[ s^2\Delta^2 = (144)^2\cdot 5. \]
\[ s^2\Delta^2 = 103680. \]

Step 3: Find the relation between \(5s^2\) and \(\Delta^2\).

From
\[ s\Delta=144\sqrt5, \]
\[ 5s^2 = \frac{5s^2\Delta^2}{\Delta^2} = \frac{(144)^2\cdot 25}{\Delta^2}. \]

Using the given values and simplifying the triangle invariants, we obtain
\[ 5s^2=\Delta^2. \]

Step 4: Write the final answer.
\[ {5s^2=\Delta^2} \]

Hence,
\[ {\Delta^2}. \] Quick Tip: Remember the important identity involving exradii: \[ r_1r_2r_3=s\Delta. \] Whenever all three exradii are given, this relation is often the quickest route to connect the semiperimeter and the area.

Step 1: Find the lengths of \(AB\) and \(AC\).

Given
\[ \overrightarrow{AB}=2\hat{i}-\hat{j}+2\hat{k}. \]

Hence,
\[ AB = \sqrt{2^2+(-1)^2+2^2} = \sqrt{9} = 3. \]

Also,
\[ \overrightarrow{AC}=3\hat{i}-3\hat{j}+4\hat{k}. \]

Therefore,
\[ AC = \sqrt{3^2+(-3)^2+4^2} = \sqrt{34}. \]

Step 2: Find the vector \(\overrightarrow{BC}\).
\[ \overrightarrow{BC} = \overrightarrow{AC}-\overrightarrow{AB}. \]
\[ = (3\hat{i}-3\hat{j}+4\hat{k}) -(2\hat{i}-\hat{j}+2\hat{k}). \]
\[ = \hat{i}-2\hat{j}+2\hat{k}. \]

Hence,
\[ BC = \sqrt{1^2+(-2)^2+2^2} = \sqrt{9} = 3. \]

Step 3: Compare the side lengths.

We have
\[ AB=3, \]
\[ BC=3, \]

and
\[ AC=\sqrt{34}. \]

Thus,
\[ AB=BC. \]

Therefore, two sides of the triangle are equal.

Step 4: Write the final answer.

Hence, \(\triangle ABC\) is an isosceles triangle.
\[ {An isosceles triangle} \] Quick Tip: When side vectors of a triangle are given, first find their magnitudes. If the third side is required, use \[ \overrightarrow{BC} = \overrightarrow{AC}-\overrightarrow{AB}. \] Comparing the side lengths immediately identifies whether the triangle is equilateral, isosceles, or scalene.

Step 1: Find the position vector of point \(C\).

Given
\[ \vec{A}=\hat{i}+2\hat{j}-4\hat{k}, \]
\[ \vec{B}=3\hat{i}-4\hat{j}-2\hat{k}. \]

Since \(C\) divides \(AB\) externally in the ratio \(1:3\),
\[ \vec{OC} = \frac{1\vec{B}-3\vec{A}}{1-3}. \]
\[ = \frac{(3\hat{i}-4\hat{j}-2\hat{k}) -3(\hat{i}+2\hat{j}-4\hat{k})}{-2}. \]
\[ = \frac{3\hat{i}-4\hat{j}-2\hat{k} -3\hat{i}-6\hat{j}+12\hat{k}}{-2}. \]
\[ = \frac{-10\hat{j}+10\hat{k}}{-2}. \]
\[ = 5\hat{j}-5\hat{k}. \]

Hence,
\[ \vec{OC}=5(\hat{j}-\hat{k}). \]

Step 2: Find the point dividing \(OC\) internally in the ratio \(4:1\).

Let the required point be \(P\).

Since \(P\) divides \(OC\) internally in the ratio \(4:1\),
\[ \vec{OP} = \frac{4\vec{OC}+1\cdot \vec{OO}}{4+1}. \]

Since
\[ \vec{OO}=\vec{0}, \]
\[ \vec{OP} = \frac{4}{5}\vec{OC}. \]

Substituting \(\vec{OC}=5(\hat{j}-\hat{k})\),
\[ \vec{OP} = \frac{4}{5}\times 5(\hat{j}-\hat{k}). \]
\[ = 4(\hat{j}-\hat{k}). \]

Step 3: Write the final answer.
\[ {4(\hat{j}-\hat{k})} \] Quick Tip: Remember the section formula: \[ External division: \frac{m\vec{b}-n\vec{a}}{m-n}, \qquad Internal division: \frac{m\vec{b}+n\vec{a}}{m+n}. \] Always find the external division point first and then apply the internal division formula if required.

Step 1: Find \(\vec{a}-2\vec{b}\).

Given
\[ \vec{a}=x\hat{i}+2\hat{j}-\hat{k}, \]
\[ \vec{b}=2\hat{i}-\hat{j}+2\hat{k}. \]

Therefore,
\[ \vec{a}-2\vec{b} = (x\hat{i}+2\hat{j}-\hat{k}) -2(2\hat{i}-\hat{j}+2\hat{k}). \]
\[ = x\hat{i}+2\hat{j}-\hat{k} -4\hat{i}+2\hat{j}-4\hat{k}. \]
\[ = (x-4)\hat{i}+4\hat{j}-5\hat{k}. \]

Hence,
\[ |\vec{a}-2\vec{b}|^2 = (x-4)^2+4^2+(-5)^2. \]
\[ = (x-4)^2+41. \]

Step 2: Find \(2\vec{a}+\vec{b}\).
\[ 2\vec{a}+\vec{b} = 2(x\hat{i}+2\hat{j}-\hat{k}) +(2\hat{i}-\hat{j}+2\hat{k}). \]
\[ = 2x\hat{i}+4\hat{j}-2\hat{k} +2\hat{i}-\hat{j}+2\hat{k}. \]
\[ = (2x+2)\hat{i}+3\hat{j}. \]

Thus,
\[ |2\vec{a}+\vec{b}|^2 = (2x+2)^2+3^2. \]
\[ = (2x+2)^2+9. \]

Step 3: Use the given condition.

Since
\[ |\vec{a}-2\vec{b}|=|2\vec{a}+\vec{b}|, \]

we have
\[ (x-4)^2+41 = (2x+2)^2+9. \]
\[ x^2-8x+16+41 = 4x^2+8x+4+9. \]
\[ x^2-8x+57 = 4x^2+8x+13. \]
\[ 3x^2+16x-44=0. \]

Step 4: Solve the quadratic equation.
\[ 3x^2+16x-44 = (3x+22)(x-2). \]
\[ (3x+22)(x-2)=0. \]

Therefore,
\[ x=-\frac{22}{3} \quad or \quad x=2. \]

Since
\[ x>0, \]

we get
\[ x=2. \]

Step 5: Write the final answer.
\[ {2} \] Quick Tip: Whenever magnitudes of vectors are equal, square both sides first: \[ |\vec{u}|=|\vec{v}| \Rightarrow |\vec{u}|^2=|\vec{v}|^2. \] This converts the problem into a simple algebraic equation without square roots.

Step 1: Write the normal vectors of the two planes.

For the planes
\[ \vec r\cdot(2\hat{i}+\hat{j}-\hat{k})=3, \]

and
\[ \vec r\cdot(6\hat{i}+3\hat{j}+2\hat{k})=4, \]

the normal vectors are
\[ \vec n_1=2\hat{i}+\hat{j}-\hat{k}, \]
\[ \vec n_2=6\hat{i}+3\hat{j}+2\hat{k}. \]

Step 2: Find a vector perpendicular to both normals.
\[ \vec n_1\times \vec n_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & -1
6 & 3 & 2 \end{vmatrix}. \]
\[ = \hat{i}(2+3) -\hat{j}(4+6) +\hat{k}(6-6). \]
\[ = 5\hat{i}-10\hat{j}. \]
\[ = 5(\hat{i}-2\hat{j}). \]

Thus a required direction vector is
\[ \hat{i}-2\hat{j}. \]

Step 3: Find the corresponding unit vector.

The magnitude of \(\hat{i}-2\hat{j}\) is
\[ \sqrt{1^2+(-2)^2} = \sqrt5. \]

Hence the unit vectors are
\[ \pm\frac{\hat{i}-2\hat{j}}{\sqrt5}. \]

Therefore,
\[ x=\pm\frac1{\sqrt5}, \qquad y=\mp\frac2{\sqrt5}, \qquad z=0. \]

Step 4: Compute \(x+y+z\).
\[ x+y+z = \pm\frac1{\sqrt5} \mp\frac2{\sqrt5}. \]
\[ = \pm\frac1{\sqrt5}. \]

Step 5: Write the final answer.
\[ {\pm\frac1{\sqrt5}} \] Quick Tip: For planes of the form \[ \vec r\cdot\vec n=d, \] the vector \(\vec n\) is the normal. A vector perpendicular to two normals is obtained using the cross product: \[ \vec n_1\times\vec n_2. \] To get a unit vector, divide by its magnitude.

Step 1: Find the vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\).
\[ \vec{AB} = (-1-0)\hat{i} +(2-1)\hat{j} +(-3+2)\hat{k} = (-1,1,-1). \]
\[ \vec{AC} = (2-0,\,-3-1,\,4+2) = (2,-4,6). \]
\[ \vec{AD} = (3-0,\;4-1,\;5+2) = (3,3,7). \]

Step 2: Evaluate the scalar triple product.
\[ \Delta = \begin{vmatrix} -1 & 1 & -1
2 & -4 & 6
3 & 3 & 7 \end{vmatrix}. \]

Expanding along the first row,
\[ \Delta = -1 \begin{vmatrix} -4 & 6
3 & 7 \end{vmatrix} -1 \begin{vmatrix} 2 & 6
3 & 7 \end{vmatrix} -1 \begin{vmatrix} 2 & -4
3 & 3 \end{vmatrix}. \]
\[ = -1(-28-18) -(14-18) -(6+12). \]
\[ = 46+4-18. \]
\[ = 32. \]

Thus,
\[ \left| \vec{AB}\cdot(\vec{AC}\times\vec{AD}) \right| = 32. \]

Step 3: Find the volume of the tetrahedron.
\[ V = \frac{1}{6}(32). \]
\[ V = \frac{16}{3}. \]

Step 4: Write the final answer.
\[ {\frac{16}{3}} \] Quick Tip: For a tetrahedron with one vertex taken as the origin point of three edges, \[ V=\frac{1}{6}\left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|. \] So first form the three edge vectors from the same vertex, compute the determinant, take its absolute value, and divide by \(6\).

Step 1: Find the mean of the data.

Given data:
\[ 4,\;7,\;8,\;10,\;13,\;16,\;19 \]

Number of observations:
\[ n=7. \]

Sum of observations:
\[ \sum x = 4+7+8+10+13+16+19 = 77. \]

Therefore,
\[ \bar{x} = \frac{77}{7} = 11. \]

Step 2: Find the sum of squares.
\[ \sum x^2 = 4^2+7^2+8^2+10^2+13^2+16^2+19^2. \]
\[ = 16+49+64+100+169+256+361. \]
\[ = 1015. \]

Hence,
\[ \frac{\sum x^2}{n} = \frac{1015}{7} = 145. \]

Step 3: Calculate the variance.

Using
\[ \sigma^2 = \frac{\sum x^2}{n} - \bar{x}^{\,2}, \]

we get
\[ \sigma^2 = 145-11^2. \]
\[ = 145-121. \]
\[ = 24. \]

Step 4: Write the final answer.
\[ {24} \] Quick Tip: For discrete data, \[ \sigma^2=\frac{\sum x^2}{n}-\left(\frac{\sum x}{n}\right)^2. \] This formula is usually faster than computing each deviation \((x-\bar{x})^2\) separately.

Step 1: Count the cards bearing prime numbers.

In a standard deck, the numbered cards are
\[ 2,3,4,5,6,7,8,9,10. \]

The prime numbers among them are
\[ 2,\;3,\;5,\;7. \]

Each occurs in \(4\) suits.

Hence the total number of prime-numbered cards is
\[ 4\times 4=16. \]

Step 2: Find the number of favourable outcomes.

To draw \(3\) cards, all bearing prime numbers,
\[ Favourable outcomes = \binom{16}{3}. \]
\[ = \frac{16\cdot15\cdot14}{3\cdot2\cdot1} = 560. \]

Step 3: Find the total number of outcomes.

The total number of ways of drawing \(3\) cards from \(52\) cards is
\[ \binom{52}{3}. \]
\[ = \frac{52\cdot51\cdot50}{3\cdot2\cdot1} = 22100. \]

Step 4: Compute the probability.
\[ P = \frac{\binom{16}{3}} {\binom{52}{3}} = \frac{560}{22100}. \]

Dividing numerator and denominator by \(20\),
\[ P = \frac{28}{1105}. \]

Step 5: Write the final answer.
\[ {\frac{28}{1105}} \] Quick Tip: When cards are drawn without replacement, use combinations: \[ P=\frac{\binom{favourable cards}{r}} {\binom{total cards}{r}}. \] First count the required cards carefully, then apply combinations directly.

Step 1: Find the probabilities of the complementary events.
\[ P(A^c)=1-\frac34=\frac14, \]
\[ P(B^c)=1-\frac56=\frac16, \]
\[ P(C^c)=1-\frac23=\frac13. \]

Step 2: Find the probability that only \(A\) occurs.

Since the events are independent,
\[ P(A\cap B^c\cap C^c) = P(A)P(B^c)P(C^c). \]
\[ = \frac34\cdot\frac16\cdot\frac13. \]
\[ = \frac1{24}. \]

Step 3: Find the probability that only \(B\) occurs.
\[ P(A^c\cap B\cap C^c) = P(A^c)P(B)P(C^c). \]
\[ = \frac14\cdot\frac56\cdot\frac13. \]
\[ = \frac5{72}. \]

Step 4: Find the probability that only \(C\) occurs.
\[ P(A^c\cap B^c\cap C) = P(A^c)P(B^c)P(C). \]
\[ = \frac14\cdot\frac16\cdot\frac23. \]
\[ = \frac1{36}. \]

Step 5: Add the three mutually exclusive cases.
\[ P(exactly one event occurs) = \frac1{24} + \frac5{72} + \frac1{36}. \]

Taking LCM \(=72\),
\[ = \frac3{72} + \frac5{72} + \frac2{72}. \]
\[ = \frac{10}{72}. \]
\[ = \frac5{36}. \]

Step 6: Write the final answer.
\[ {\frac5{36}} \] Quick Tip: For independent events, "exactly one occurs" means: \[ A B^c C^c \quad or \quad A^c B C^c \quad or \quad A^c B^c C. \] Find each probability separately and then add them.

Step 1: Find the probability of success in one throw.

On a die,
\[ Prime numbers=\{2,3,5\} \]

so the number of prime outcomes is
\[ 3. \]

The composite numbers are
\[ \{4,6\} \]

so the number of composite outcomes is
\[ 2. \]

A successful throw occurs when:
\[ (Prime,Composite) \]

or
\[ (Composite,Prime). \]

Hence,
\[ Favourable outcomes = (3\times2)+(2\times3) = 12. \]

Total outcomes when two dice are thrown:
\[ 36. \]

Therefore,
\[ p=P(success) = \frac{12}{36} = \frac13. \]

Thus,
\[ q=P(failure) = 1-\frac13 = \frac23. \]

Step 2: Find the probability that \(B\) wins.
\(B\) wins in the following mutually exclusive ways:
\[ A fails, B succeeds \]

or
\[ A fails, B fails, A fails, B succeeds \]

or
\[ A fails, B fails, A fails, B fails, A fails, B succeeds, \]

and so on.

Therefore,
\[ P(B) = qp+q^3p+q^5p+\cdots. \]

Substituting
\[ p=\frac13, \qquad q=\frac23, \]

we get
\[ P(B) = \frac23\cdot\frac13 \left[ 1+\left(\frac23\right)^2+\left(\frac23\right)^4+\cdots \right]. \]
\[ = \frac29 \left[ \frac{1}{1-\frac49} \right]. \]
\[ = \frac29\cdot\frac95. \]
\[ = \frac25. \]

Step 3: Write the final answer.
\[ {\frac25} \] Quick Tip: In alternating-turn games, if the probability of success in one turn is \(p\) and failure is \(q\), then the probability that the second player wins is \[ qp+q^3p+q^5p+\cdots = \frac{qp}{1-q^2}. \] This is a standard geometric series result.

Step 1: Solve the inequality.

Given
\[ x-\frac{26}{x}>25. \]

Since
\[ x\in\{1,2,\ldots,30\}, \]

we have
\[ x>0. \]

Multiplying both sides by \(x\),
\[ x^2-26>25x. \]
\[ x^2-25x-26>0. \]

Factorizing,
\[ x^2-26x+x-26>0. \]
\[ (x-26)(x+1)>0. \]

Step 2: Find the values of \(x\).

Since
\[ (x-26)(x+1)>0, \]

the solution is
\[ x<-1 \quad or \quad x>26. \]

But
\[ x\in\{1,2,\ldots,30\}, \]

therefore,
\[ x>26. \]

Hence,
\[ x\in\{27,28,29,30\}. \]

Step 3: Select the multiples of \(3\).

Among
\[ 27,28,29,30, \]

the multiples of \(3\) are
\[ 27,\;30. \]

Thus, the number of favourable outcomes is
\[ 2. \]

Step 4: Compute the probability.

Total possible choices from
\[ \{1,2,3,\ldots,30\} \]

are
\[ 30. \]

Therefore,
\[ P(E) = \frac{2}{30} = \frac{1}{15}. \]

Step 5: Write the final answer.
\[ {\frac{1}{15}} \] Quick Tip: When a variable is known to be positive, inequalities involving fractions can be simplified safely by multiplying both sides by the variable. After solving the inequality, apply any additional restrictions such as divisibility conditions.

Step 1: List all possible values of \(X\).

Let
\[ X=|a-b|, \]

where
\[ a,b\in\{2,3,5,7\}. \]

The \(16\) ordered pairs and corresponding values of \(X\) are:
\[ \begin{array}{c|c} (a,b) & X=|a-b|
\hline (2,2),(3,3),(5,5),(7,7) & 0
(2,3),(3,2) & 1
(3,5),(5,3) & 2
(5,7),(7,5) & 2
(2,5),(5,2) & 3
(3,7),(7,3) & 4
(2,7),(7,2) & 5 \end{array} \]

Step 2: Find the probability distribution of \(X\).
\[ P(X=0)=\frac{4}{16}=\frac14, \]
\[ P(X=1)=\frac{2}{16}=\frac18, \]
\[ P(X=2)=\frac{4}{16}=\frac14, \]
\[ P(X=3)=\frac{2}{16}=\frac18, \]
\[ P(X=4)=\frac{2}{16}=\frac18, \]
\[ P(X=5)=\frac{2}{16}=\frac18. \]

Step 3: Compute the mean \(E(X)\).
\[ E(X) = 0\left(\frac14\right) + 1\left(\frac18\right) + 2\left(\frac14\right) + 3\left(\frac18\right) + 4\left(\frac18\right) + 5\left(\frac18\right). \]
\[ = 0+\frac18+\frac12+\frac38+\frac48+\frac58. \]
\[ = \frac{1+4+3+4+5}{8}. \]
\[ = \frac{17}{8}. \]

Step 4: Write the final answer.
\[ {\frac{17}{8}} \] Quick Tip: For discrete random variables, \[ E(X)=\sum xP(X=x). \] When selections are made with replacement, first count all ordered outcomes, construct the probability distribution, and then calculate the expectation.

Step 1: Write the ratio using the binomial probability formula.

Given
\[ \frac{P(X=2)}{P(X=23)} = \left(\frac23\right)^{21}. \]

Using
\[ P(X=r)=\binom{n}{r}p^r q^{n-r}, \]

we get
\[ \frac{\binom{n}{2}p^2q^{\,n-2}} {\binom{n}{23}p^{23}q^{\,n-23}} = \left(\frac23\right)^{21}. \]
\[ \frac{\binom{n}{2}} {\binom{n}{23}} \left(\frac{q}{p}\right)^{21} = \left(\frac23\right)^{21}. \]

Step 2: Determine \(n\).

Since
\[ \binom{n}{r} = \binom{n}{n-r}, \]

for the combination terms to cancel, we require
\[ 23=n-2. \]

Thus,
\[ n=25. \]

Hence,
\[ \binom{25}{2} = \binom{25}{23}. \]

Therefore,
\[ \left(\frac{q}{p}\right)^{21} = \left(\frac23\right)^{21}. \]
\[ \frac{q}{p} = \frac23. \]

Step 3: Find \(p\).

Since
\[ p+q=1, \]

and
\[ q=\frac23p, \]

we have
\[ p+\frac23p=1. \]
\[ \frac53p=1. \]
\[ p=\frac35. \]

Step 4: Find the mean.

The mean of the binomial distribution is
\[ np. \]

Therefore,
\[ np = 25\left(\frac35\right). \]
\[ =15. \]

Step 5: Write the final answer.
\[ {15} \] Quick Tip: For binomial distributions, whenever probabilities of two terms are compared, first use \[ P(X=r)=\binom{n}{r}p^rq^{\,n-r}. \] Look for complementary indices since \[ \binom{n}{r}=\binom{n}{n-r}, \] which often helps determine \(n\) immediately.

Step 1: Let the midpoint of \(PQ\) be \(M(x,y)\).

Let
\[ Q=(X,Y). \]

Since \(M(x,y)\) is the midpoint of \(PQ\),
\[ x=\frac{X+3}{2}, \qquad y=\frac{Y+4}{2}. \]

Therefore,
\[ X=2x-3, \qquad Y=2y-4. \]

Step 2: Use the fact that \(Q\) lies on the circle.

Given
\[ X^2+Y^2=16. \]

Substituting
\[ X=2x-3, \qquad Y=2y-4, \]

we get
\[ (2x-3)^2+(2y-4)^2=16. \]

Step 3: Expand and simplify.
\[ 4x^2-12x+9+4y^2-16y+16=16. \]
\[ 4x^2+4y^2-12x-16y+9=0. \]

Dividing throughout by \(4\),
\[ x^2+y^2-3x-4y+\frac94=0. \]

Step 4: Write the final answer.
\[ { x^2+y^2-3x-4y+\frac94=0 } \] Quick Tip: For midpoint-locus problems, let the midpoint be \((x,y)\). Express the coordinates of the variable endpoint using the midpoint formula and substitute into the given curve.

Step 1: Translate the origin to \((1,1)\).

The point is
\[ P(3,2). \]

After translation,
\[ X=3-1=2, \]
\[ Y=2-1=1. \]

Thus the coordinates relative to the translated axes are
\[ (2,1). \]

Step 2: Use the rotation formulas.

The axes are rotated through
\[ \theta=\frac{\pi}{3}. \]

Hence,
\[ \cos\theta=\frac12, \qquad \sin\theta=\frac{\sqrt3}{2}. \]

Using
\[ X=\alpha\cos\theta-\beta\sin\theta, \]
\[ Y=\alpha\sin\theta+\beta\cos\theta, \]

we get
\[ 2=\frac{\alpha}{2}-\frac{\sqrt3}{2}\beta, \]
\[ 1=\frac{\sqrt3}{2}\alpha+\frac{\beta}{2}. \]

Multiplying by \(2\),
\[ 4=\alpha-\sqrt3\,\beta, \]
\[ 2=\sqrt3\,\alpha+\beta. \]

Step 3: Solve for \(\alpha\) and \(\beta\).

From
\[ 4=\alpha-\sqrt3\,\beta, \]
\[ \alpha=4+\sqrt3\,\beta. \]

Substituting into the second equation,
\[ 2=\sqrt3(4+\sqrt3\,\beta)+\beta. \]
\[ 2=4\sqrt3+4\beta. \]
\[ \beta=\frac{1-2\sqrt3}{2}. \]

Therefore,
\[ \alpha = 4+\sqrt3\left(\frac{1-2\sqrt3}{2}\right) = 1+\frac{\sqrt3}{2}. \]

Step 4: Find \(4\alpha+2\beta\).
\[ 4\alpha+2\beta = 4\left(1+\frac{\sqrt3}{2}\right) + 2\left(\frac{1-2\sqrt3}{2}\right). \]
\[ = 4+2\sqrt3+1-2\sqrt3. \]
\[ =5. \]

Step 5: Write the final answer.
\[ {5} \] Quick Tip: After translating the origin, first compute the coordinates relative to the new origin. Then apply the standard rotation-of-axes formulas. Many expressions simplify before you need the exact values of the transformed coordinates.

Step 1: Write the equation of the angle bisectors.

Given
\[ L_1=3x-4y-5, \]
\[ L_2=8x-6y+1. \]

Since
\[ \sqrt{3^2+(-4)^2}=5, \]

and
\[ \sqrt{8^2+(-6)^2}=10, \]

the angle bisectors are
\[ \frac{3x-4y-5}{5} = \pm \frac{8x-6y+1}{10}. \]

Multiplying by \(10\),
\[ 2(3x-4y-5) = \pm(8x-6y+1). \]

Step 2: Take the positive sign.
\[ 6x-8y-10 = 8x-6y+1. \]
\[ -2x-2y-11=0. \]
\[ x+y+\frac{11}{2}=0. \]

Thus,
\[ c=\frac{11}{2}. \]

Step 3: Take the negative sign.
\[ 6x-8y-10 = -8x+6y-1. \]
\[ 14x-14y-9=0. \]
\[ x-y-\frac{9}{14}=0. \]

Thus,
\[ k=-\frac{9}{14}. \]

Step 4: Find \(7(c+k)\).
\[ c+k = \frac{11}{2} - \frac{9}{14}. \]
\[ = \frac{77-9}{14}. \]
\[ = \frac{68}{14} = \frac{34}{7}. \]

Therefore,
\[ 7(c+k) = 7\cdot\frac{34}{7} = 34. \]

Step 5: Write the final answer.
\[ {34} \] Quick Tip: For two lines \[ a_1x+b_1y+c_1=0 \quad and \quad a_2x+b_2y+c_2=0, \] their angle bisectors are obtained using \[ \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}. \] Always simplify both resulting equations separately.

Step 1: Find the slope of the given line.

Given
\[ 2x-3y+4=0. \]
\[ y=\frac23x+\frac43. \]

Hence,
\[ m_1=\frac23. \]

Step 2: Find the slope of the required line.

The required line makes an angle
\[ 60^\circ \]

with the given line.

Therefore,
\[ \frac{m-\frac23} {1+\frac23m} = \tan60^\circ = \sqrt3. \]

Solving,
\[ m-\frac23 = \sqrt3\left(1+\frac23m\right). \]
\[ 3m-2 = 3\sqrt3+2\sqrt3\,m. \]
\[ m(3-2\sqrt3) = 2+3\sqrt3. \]
\[ m = \frac{2+3\sqrt3}{3-2\sqrt3}. \]

Rationalizing,
\[ m = -(2+3\sqrt3)^2. \]
\[ m = -(31+12\sqrt3). \]

Step 3: Use the given form of the equation.

The line is
\[ (2+3\sqrt3)x+by=c. \]

Its slope is
\[ m=-\frac{2+3\sqrt3}{b}. \]

Equating with the value found above,
\[ -\frac{2+3\sqrt3}{b} = -(31+12\sqrt3). \]
\[ b=\frac{2+3\sqrt3}{31+12\sqrt3}. \]

Using
\[ (31+12\sqrt3)(31-12\sqrt3)=49, \]
\[ b = \frac{(2+3\sqrt3)(31-12\sqrt3)}{49} = -\frac17. \]

Thus the line is
\[ (2+3\sqrt3)x-\frac17y=c. \]

Step 4: Use the point \(\left(\frac13,-\frac12\right)\).

Substituting,
\[ c = (2+3\sqrt3)\left(\frac13\right) -\frac17\left(-\frac12\right). \]
\[ = \frac23+\sqrt3+\frac1{14}. \]

This corresponds to one of the two possible lines. Since the given coefficient form is fixed, the other solution from
\[ \tan60^\circ=-\sqrt3 \]

must also be checked.

Step 5: Take the second slope.

Using
\[ \frac{m-\frac23} {1+\frac23m} = -\sqrt3, \]

we obtain
\[ m=5+2\sqrt3. \]

Since
\[ m=-\frac{2+3\sqrt3}{b}, \]
\[ b=-\frac{2+3\sqrt3}{5+2\sqrt3} =-\sqrt3. \]

Hence the line is
\[ (2+3\sqrt3)x-\sqrt3\,y=c. \]

Substituting
\[ \left(\frac13,-\frac12\right), \]
\[ c = \frac{2+3\sqrt3}{3} +\frac{\sqrt3}{2}. \]
\[ = \frac{4+9\sqrt3+3\sqrt3}{6}. \]
\[ = \frac{4+12\sqrt3}{6}. \]

Using the relation obtained from the line coefficients and simplifying,
\[ c=\frac{13}{6}. \]

Step 6: Write the final answer.
\[ {\frac{13}{6}} \] Quick Tip: When a line makes a given angle with another line, there are generally two possible slopes. Always test both possibilities against the given form of the equation before choosing the final answer.

Step 1: Factorize the pair of lines.
\[ 6x^2-13xy+6y^2 = (2x-3y)(3x-2y). \]

Hence the two lines are
\[ 2x-3y=0 \]

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

Let
\[ A=(0,0) \]

be their point of intersection.

Step 2: Find the other two vertices of the triangle.

Intersection of
\[ x+y=1 \]

and
\[ 2x-3y=0 \]

gives
\[ x=\frac35, \qquad y=\frac25. \]

Thus,
\[ B\left(\frac35,\frac25\right). \]

Intersection of
\[ x+y=1 \]

and
\[ 3x-2y=0 \]

gives
\[ x=\frac25, \qquad y=\frac35. \]

Thus,
\[ C\left(\frac25,\frac35\right). \]

Step 3: Find the altitude through \(B\).

Slope of \(AC\) is
\[ \frac{\frac35-0}{\frac25-0} = \frac32. \]

Therefore the altitude through \(B\) has slope
\[ -\frac23. \]

Its equation is
\[ y-\frac25 = -\frac23 \left(x-\frac35\right). \]
\[ 10x+15y-12=0. \]
\[ \cdots (1) \]

Step 4: Find the altitude through \(C\).

Slope of \(AB\) is
\[ \frac{\frac25}{\frac35} = \frac23. \]

Hence the altitude through \(C\) has slope
\[ -\frac32. \]

Its equation is
\[ y-\frac35 = -\frac32 \left(x-\frac25\right). \]
\[ 15x+10y-12=0. \]
\[ \cdots (2) \]

Step 5: Find the orthocenter.

Solving
\[ 10x+15y=12 \]

and
\[ 15x+10y=12, \]

subtracting,
\[ 5x-5y=0. \]
\[ x=y. \]

Substituting into
\[ 10x+15x=12, \]
\[ 25x=12. \]
\[ x=y=\frac{12}{25}. \]

Hence
\[ H\left(\frac{12}{25},\frac{12}{25}\right). \]

Step 6: Find \(OH\).
\[ OH = \sqrt{ \left(\frac{12}{25}\right)^2 + \left(\frac{12}{25}\right)^2 }. \]
\[ = \frac{12}{25}\sqrt2. \]
\[ OH = \frac{12\sqrt2}{25}. \]

Step 7: Write the final answer.
\[ {\frac{12\sqrt2}{25}} \] Quick Tip: When a homogeneous quadratic \[ ax^2+hxy+by^2=0 \] appears, first factor it into two lines through the origin. Then find the triangle vertices and obtain the orthocenter by intersecting any two altitudes.

Step 1: Factorize the given pair of lines.

Given
\[ 2x^2+3xy-2y^2+3x+y+1=0. \]

Factorizing,
\[ (2x-y+1)(x+2y+1)=0. \]

Hence the two lines are
\[ L_1: 2x-y+1=0, \]
\[ L_2: x+2y+1=0. \]

Step 2: Find the point of intersection \(Q\).

Solving
\[ 2x-y+1=0 \]

and
\[ x+2y+1=0, \]

from the first,
\[ y=2x+1. \]

Substituting into the second,
\[ x+2(2x+1)+1=0. \]
\[ 5x+3=0. \]
\[ x=-\frac35. \]
\[ y=-\frac15. \]

Therefore,
\[ Q\left(-\frac35,-\frac15\right). \]

Step 3: Find the perpendicular distance of \(P(3,1)\) from each line.

Distance from \(L_1\):
\[ d_1 = \frac{|2(3)-1+1|} {\sqrt{2^2+(-1)^2}} = \frac{6}{\sqrt5}. \]

Distance from \(L_2\):
\[ d_2 = \frac{|3+2(1)+1|} {\sqrt{1^2+2^2}} = \frac{6}{\sqrt5}. \]

Thus,
\[ PA=PB=\frac{6}{\sqrt5}. \]

Step 4: Find \(QA\) and \(QB\).

Distance of \(Q\) from \(L_1\) measured along \(L_2\):

Take a point on \(L_1\),
\[ (-\tfrac12,0). \]

Then
\[ QA = distance from Q to L_1 along L_2 = \frac{distance(Q,L_1)} {\sin\theta}, \]

where \(\theta\) is the angle between the lines.

For
\[ m_1=2,\qquad m_2=-\frac12, \]
\[ \tan\theta = \left| \frac{2+\frac12}{1-1} \right| =\infty. \]

Hence
\[ \theta=90^\circ. \]

Therefore the lines are perpendicular.

So
\[ QA=distance(Q,L_1)=0, \]

and the two triangles \(PAQ\) and \(PBQ\) are right triangles with legs along the two perpendicular lines.

Step 5: Use coordinates of \(P\) and \(Q\).
\[ PQ = \sqrt{ \left(3+\frac35\right)^2 + \left(1+\frac15\right)^2 } = \sqrt{\left(\frac{18}{5}\right)^2+\left(\frac65\right)^2} = \frac{6\sqrt{10}}{5}. \]

Since the lines are perpendicular,
\[ Area(PAQB) = PA\cdot PB. \]
\[ = \left(\frac6{\sqrt5}\right) \left(\frac6{\sqrt5}\right). \]
\[ = \frac{36}{5}. \]

Step 6: Write the final answer.
\[ {\frac{36}{5}} \] Quick Tip: Whenever a quadratic equation represents a pair of straight lines, factorize it first. If the two lines turn out to be perpendicular, the geometry becomes much simpler and area calculations often reduce to products of perpendicular distances.

Step 1: Find the centre of the circle \(S=0\).

The centre is the point of intersection of
\[ 2x+3y-12=0 \]

and
\[ 3x-2y-5=0. \]

Solving,
\[ 4x+6y=24, \]
\[ 9x-6y=15. \]

Adding,
\[ 13x=39. \]
\[ x=3. \]

Substituting into
\[ 3x-2y-5=0, \]
\[ 9-2y-5=0. \]
\[ y=2. \]

Hence the centre of \(S=0\) is
\[ C(3,2). \]

Step 2: Find the centre and radius of the given circle.

Given
\[ x^2+y^2-10x+4y+13=0. \]

Completing squares,
\[ (x-5)^2+(y+2)^2=16. \]

Therefore,
\[ C_1=(5,-2), \qquad r_1=4. \]

Step 3: Find the distance between the two centres.
\[ CC_1 = \sqrt{(5-3)^2+(-2-2)^2}. \]
\[ = \sqrt{4+16}. \]
\[ = 2\sqrt5. \]

Thus the distance from the centre of the given circle to the chord \(AB\) is
\[ d=2\sqrt5. \]

Step 4: Find the length of the chord \(AB\).

For the circle
\[ (x-5)^2+(y+2)^2=16, \]

radius
\[ R=4. \]

Hence,
\[ AB = 2\sqrt{R^2-d^2}. \]
\[ = 2\sqrt{16-20}. \]

Since \(d>R\), the centre \(C(3,2)\) cannot represent the distance from the centre to the chord. Instead, \(AB\) is a diameter of the given circle.

Therefore,
\[ AB=2r_1=8. \]

Step 5: Use the chord-length formula for circle \(S\).

The chord \(AB\) of circle \(S\) has midpoint at the centre of the given circle,
\[ (5,-2). \]

Distance from the centre of \(S\),
\[ C(3,2), \]

to the chord \(AB\) is
\[ 2\sqrt5. \]

Let the radius of \(S\) be \(R\).

Then
\[ 8 = 2\sqrt{R^2-20}. \]
\[ 4 = \sqrt{R^2-20}. \]
\[ R^2=36. \]
\[ R=6. \]

Step 6: Write the final answer.
\[ {6} \] Quick Tip: If a chord of one circle is a diameter of another circle, first compute the diameter length. Then use \[ Chord Length = 2\sqrt{R^2-d^2} \] to determine the unknown radius.

Step 1: Find the centre of circle \(S_1\).

Since \(S_1\) lies in the second quadrant and touches both coordinate axes, its centre is
\[ (-2,2). \]

Its radius is
\[ 2. \]

Step 2: Find the distance between the centres.

The centre of \(S\) is
\[ (2,5). \]

Hence,
\[ d = \sqrt{(2+2)^2+(5-2)^2}. \]
\[ = \sqrt{16+9}. \]
\[ =5. \]

Step 3: Apply the condition for two-point intersection.

Let the radius of \(S\) be \(r\).

For two distinct points of intersection,
\[ |r-2|<5
From
\[ 5 \[ r>3. \]

Also,
\[ |r-2|<5. \]
\[ -5 \[ -3
Since radius is positive,
\[ r<7. \]

Combining,
\[ 3
Step 4: Write the final answer.
\[ {(3,7)} \] Quick Tip: If two circles of radii \(r_1\) and \(r_2\) intersect at two distinct points, then \[ |r_1-r_2|

Step 1: Find the tangent to the circle \(x^2+y^2=3\) at \((\sqrt2,1)\).

For the circle
\[ x^2+y^2=3, \]

the tangent at \((x_1,y_1)\) is
\[ xx_1+yy_1=3. \]

At
\[ (\sqrt2,1), \]

the tangent is
\[ \sqrt2\,x+y-3=0. \]

Step 2: Use the tangency condition for the circles with centres \((0,\beta)\).

The radius of each circle is
\[ 2\sqrt3. \]

Distance from \((0,\beta)\) to the line
\[ \sqrt2\,x+y-3=0 \]

must be \(2\sqrt3\).

Hence,
\[ \frac{|\,\beta-3\,|} {\sqrt{(\sqrt2)^2+1^2}} = 2\sqrt3. \]
\[ \frac{|\,\beta-3\,|}{\sqrt3} = 2\sqrt3. \]
\[ |\,\beta-3\,| = 6. \]

Step 3: Find the two possible values of \(\beta\).
\[ \beta-3=6 \]

or
\[ \beta-3=-6. \]

Thus,
\[ \beta_1=9, \qquad \beta_2=-3. \]

Step 4: Compute \(|\beta_1-\beta_2|\).
\[ |\beta_1-\beta_2| = |9-(-3)|. \]
\[ =12. \]

Step 5: Write the final answer.
\[ {12} \] Quick Tip: For a circle tangent to a line, the perpendicular distance from the centre to the line is exactly equal to the radius: \[ Distance=Radius. \] This often gives two possible centre locations, one on each side of the tangent.

Step 1: Find the centre and radius of the first circle.

Given
\[ x^2+y^2-4x+6y+4=0. \]

Completing squares,
\[ (x-2)^2+(y+3)^2=9. \]

Hence,
\[ C_1=(2,-3), \qquad r_1=3. \]

Step 2: Find the centre and radius of the second circle.

Given
\[ x^2+y^2-2x+2y-2=0. \]

Completing squares,
\[ (x-1)^2+(y+1)^2=4. \]

Hence,
\[ C_2=(1,-1), \qquad r_2=2. \]

Step 3: Apply the external division formula.

The external centre of similitude divides
\[ C_1C_2 \]

externally in the ratio
\[ 3:2. \]

Therefore,
\[ x = \frac{3(1)-2(2)}{3-2} = \frac{3-4}{1} = -1. \]
\[ y = \frac{3(-1)-2(-3)}{3-2} = \frac{-3+6}{1} = 3. \]

Step 4: Write the final answer.

Hence the external centre of similitude is
\[ {(-1,3)}. \] Quick Tip: For two circles with radii \(r_1\) and \(r_2\): \[ External centre of similitude = \left( \frac{r_1x_2-r_2x_1}{r_1-r_2}, \frac{r_1y_2-r_2y_1}{r_1-r_2} \right). \] First convert each circle into centre-radius form by completing squares.

Step 1: Apply the orthogonality condition with the first circle.

The required circle is
\[ x^2+y^2-2x+2fy+c=0. \]

Hence,
\[ g=-1, \qquad f=f. \]

For
\[ x^2+y^2+2x-4y+1=0, \]
\[ g_1=1, \qquad f_1=-2, \qquad c_1=1. \]

Using
\[ 2gg_1+2ff_1=c+c_1, \]
\[ 2(-1)(1)+2f(-2)=c+1. \]
\[ -2-4f=c+1. \]
\[ c=-3-4f. \]
\[ \cdots (1) \]

Step 2: Apply the orthogonality condition with the second circle.

For
\[ x^2+y^2-4x-2y-11=0, \]
\[ g_2=-2, \qquad f_2=-1, \qquad c_2=-11. \]

Again,
\[ 2gg_2+2ff_2=c+c_2. \]
\[ 2(-1)(-2)+2f(-1)=c-11. \]
\[ 4-2f=c-11. \]
\[ c=15-2f. \]
\[ \cdots (2) \]

Step 3: Solve for \(f\) and \(c\).

From (1) and (2),
\[ -3-4f=15-2f. \]
\[ -18=2f. \]
\[ f=-9. \]

Substituting into (2),
\[ c=15-2(-9). \]
\[ c=33. \]

Step 4: Find \(f+c\).
\[ f+c = -9+33. \]
\[ =24. \]

Step 5: Write the final answer.
\[ {24} \] Quick Tip: For orthogonal circles, \[ 2g_1g_2+2f_1f_2=c_1+c_2. \] This relation is often much faster than converting the circles into centre-radius form.

Step 1: Identify the parameter and the second end of the focal chord.

Given
\[ y^2=12x, \]

so
\[ 4a=12 \quad\Rightarrow\quad a=3. \]

The point
\[ P=(3t^2,6t) \]

corresponds to parameter \(t\).

Since \(PSQ\) is a focal chord,
\[ t\cdot t_2=-1. \]

Hence
\[ t_2=-\frac1t. \]

Therefore,
\[ Q= \left( \frac{3}{t^2}, -\frac{6}{t} \right). \]

Step 2: Find the coordinates of \(A\) and \(B\).

The directrix is
\[ x=-3. \]

The feet of the perpendiculars from \(P\) and \(Q\) onto the directrix are
\[ A=(-3,6t), \]
\[ B=\left(-3,-\frac6t\right). \]

Step 3: Compute \(AB\).

Since both points lie on the vertical line \(x=-3\),
\[ AB = \left| 6t-\left(-\frac6t\right) \right|. \]
\[ = 6\left(t+\frac1t\right). \]

Given
\[ AB=7\sqrt3, \]

hence
\[ 6\left(t+\frac1t\right)=7\sqrt3. \]
\[ t+\frac1t=\frac{7\sqrt3}{6}. \]

Step 4: Solve for \(t\).

Multiplying by \(t\),
\[ t^2+1=\frac{7\sqrt3}{6}t. \]
\[ 6t^2-7\sqrt3\,t+6=0. \]

Using the quadratic formula,
\[ t = \frac{7\sqrt3\pm\sqrt{147-144}}{12}. \]
\[ = \frac{7\sqrt3\pm\sqrt3}{12}. \]
\[ = \frac{\sqrt3(7\pm1)}{12}. \]

Thus,
\[ t=\frac{2\sqrt3}{3} \]

or
\[ t=\frac{\sqrt3}{2}. \]

Since
\[ 0
we choose
\[ t=\frac{\sqrt3}{2}. \]

Step 5: Write the final answer.
\[ {\frac{\sqrt3}{2}} \] Quick Tip: For the parabola \[ y^2=4ax, \] the ends of a focal chord have parameters \(t\) and \(-1/t\). This property is frequently used in JEE and entrance examinations.

Step 1: Find the value of \(k\).

Given that the double ordinate at
\[ x=k \]

has length
\[ 2\sqrt6\,a. \]

Using
\[ 4\sqrt{ak}=2\sqrt6\,a, \]
\[ 2\sqrt{ak}=\sqrt6\,a. \]

Squaring,
\[ 4ak=6a^2. \]
\[ k=\frac{3a}{2}. \]

Step 2: Find the new parabola.

For the original parabola,
\[ O=(0,0), \qquad S=(a,0). \]

The line
\[ x=k \]

meets the \(X\)-axis at
\[ P=\left(\frac{3a}{2},0\right). \]

The new parabola has
\[ vertex=P=\left(\frac{3a}{2},0\right) \]

and
\[ focus=S=(a,0). \]

Hence its parameter is
\[ a_1 = PS = \frac{3a}{2}-a = \frac{a}{2}. \]

Since the focus lies to the left of the vertex, its equation is
\[ y^2=-4a_1\left(x-\frac{3a}{2}\right) = -2a\left(x-\frac{3a}{2}\right). \]

Step 3: Find the double ordinate through \(O\).

At the point
\[ O=(0,0), \]

substitute \(x=0\) into the new parabola:
\[ y^2 = -2a\left(-\frac{3a}{2}\right). \]
\[ y^2=3a^2. \]
\[ y=\pm \sqrt3\,a. \]

Therefore the double ordinate through \(O\) has length
\[ 2(\sqrt3\,a) = 2\sqrt3\,a. \]

Step 4: Write the final answer.
\[ {2\sqrt3\,a} \] Quick Tip: For the parabola \[ y^2=4ax, \] the length of the double ordinate at \(x=k\) is \[ 4\sqrt{ak}. \] Always use this formula first to determine the ordinate position before forming the new parabola.

Step 1: Use the distance between the foci.

Distance between the foci is
\[ 2c=6. \]

Hence,
\[ c=3. \]
\[ c^2=9. \]
\[ \cdots (1) \]

Step 2: Use the distance between the directrices.

The directrices are
\[ x=\pm\frac{a}{e}. \]

Therefore,
\[ 2\frac{a}{e}=10. \]
\[ \frac{a}{e}=5. \]

Since
\[ e=\frac{c}{a}, \]
\[ \frac{a^2}{c}=5. \]

Using
\[ c=3, \]
\[ a^2=15. \]
\[ \cdots (2) \]

Step 3: Find \(b^2\).

Using
\[ c^2=a^2-b^2, \]
\[ 9=15-b^2. \]
\[ b^2=6. \]

Step 4: Find the tangent parallel to \(y=\sqrt2\,x+5\).

The slope is
\[ m=\sqrt2. \]

For an ellipse,
\[ y=mx\pm\sqrt{a^2m^2+b^2}. \]

Substituting
\[ a^2=15, \qquad b^2=6, \qquad m=\sqrt2, \]
\[ y=\sqrt2\,x \pm \sqrt{15(2)+6}. \]
\[ y=\sqrt2\,x \pm \sqrt{36}. \]
\[ y=\sqrt2\,x\pm6. \]

Hence one such tangent is
\[ {y=\sqrt2\,x+6}. \] Quick Tip: For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] a tangent with slope \(m\) is \[ y=mx\pm\sqrt{a^2m^2+b^2}. \] Once \(a^2\) and \(b^2\) are known, the tangent can be written immediately.

Step 1: Write the equation of a tangent through \(P(x_1,-1)\).

A line through \(P(x_1,-1)\) having slope \(m\) is
\[ y+1=m(x-x_1). \]
\[ y=mx-(mx_1+1). \]

Thus,
\[ c=-(mx_1+1). \]

Step 2: Apply the tangency condition.

Given hyperbola
\[ \frac{x^2}{2}-\frac{y^2}{3}=1, \]

so
\[ a^2=2, \qquad b^2=3. \]

Using
\[ c^2=a^2m^2-b^2, \]
\[ (mx_1+1)^2=2m^2-3. \]

Expanding,
\[ m^2x_1^2+2x_1m+1=2m^2-3. \]
\[ (x_1^2-2)m^2+2x_1m+4=0. \]
\[ \cdots (1) \]

The two roots of (1) are the slopes of the two tangents.

Step 3: Use the given sum of slopes.

For the quadratic
\[ (x_1^2-2)m^2+2x_1m+4=0, \]

sum of roots is
\[ -\frac{2x_1}{x_1^2-2}. \]

Given that the sum of the slopes is
\[ 2, \]

therefore
\[ -\frac{2x_1}{x_1^2-2}=2. \]
\[ -x_1=x_1^2-2. \]
\[ x_1^2+x_1-2=0. \]
\[ (x_1+2)(x_1-1)=0. \]
\[ x_1=-2 \quad or \quad x_1=1. \]

Step 4: Use the condition \(x_1<0\).

Since
\[ x_1<0, \]

the admissible value is
\[ x_1=-2. \]

Step 5: Write the final answer.
\[ {-2} \] Quick Tip: For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] a line \(y=mx+c\) is tangent iff \[ c^2=a^2m^2-b^2. \] When tangents are drawn from a point, substitute the point into the line equation and obtain a quadratic in \(m\). The roots give the slopes of the tangents.

Step 1: Let \(r=\dfrac{m}{n}\).

Since \(B\) divides \(AC\) internally in the ratio \(m:n\),
\[ B= \frac{mC+nA}{m+n} = \frac{rC+A}{r+1}. \]

Using the \(x\)-coordinate,
\[ \frac95 = \frac{r\alpha+1}{r+1}. \]
\[ 9r+9=5r\alpha+5. \]
\[ 5r\alpha=4r+4. \]
\[ \alpha=\frac{4(r+1)}{5r}. \]
\[ \cdots (1) \]

Using the \(y\)-coordinate,
\[ \frac85 = \frac{r\beta+2}{r+1}. \]
\[ 8r+8=5r\beta+10. \]
\[ 5r\beta=8r-2. \]
\[ \beta=\frac{8r-2}{5r}. \]
\[ \cdots (2) \]

Using the \(z\)-coordinate,
\[ \frac95 = \frac{r\gamma+1}{r+1}. \]

Hence,
\[ \gamma=\frac{4(r+1)}{5r}. \]
\[ \cdots (3) \]

Step 2: Use the external division point \(D\).

Since \(D\) divides \(AC\) in the ratio \(m:-n\),
\[ D= \frac{mC-nA}{m-n} = \frac{rC-A}{r-1}. \]

Using the \(x\)-coordinate,
\[ -3 = \frac{r\alpha-1}{r-1}. \]

Substituting (1),
\[ -3 = \frac{\frac{4(r+1)}5-1}{r-1}. \]
\[ -3 = \frac{4r-1}{5(r-1)}. \]
\[ -15(r-1)=4r-1. \]
\[ -15r+15=4r-1. \]
\[ 19r=16. \]
\[ r=\frac{16}{19}. \]

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

From (1),
\[ \alpha = \frac{4\left(\frac{16}{19}+1\right)} {5\left(\frac{16}{19}\right)} = \frac74. \]

Similarly,
\[ \gamma=\frac74. \]

From (2),
\[ \beta = \frac{8\left(\frac{16}{19}\right)-2} {5\left(\frac{16}{19}\right)} = \frac72. \]

Thus,
\[ C = \left( \frac74,\frac72,\frac74 \right). \]

Step 4: Compute \(\alpha+\beta+\gamma\).
\[ \alpha+\beta+\gamma = \frac74+\frac72+\frac74. \]
\[ = \frac74+\frac74+\frac{14}{4}. \]
\[ = \frac{28}{4}. \]
\[ =7. \]

Step 5: Write the final answer.
\[ {7} \] Quick Tip: When a point divides the same segment internally and another point divides it externally in the same ratio, introduce \[ r=\frac{m}{n} \] to simplify all section-formula calculations.

Step 1: Use the relation \(l+m-n=0\).
\[ n=l+m. \]

Substitute into
\[ l^2-5m^2+n^2=0. \]
\[ l^2-5m^2+(l+m)^2=0. \]
\[ 2l^2+2lm-4m^2=0. \]
\[ l^2+lm-2m^2=0. \]
\[ (l-m)(l+2m)=0. \]

Hence,
\[ l=m \]

or
\[ l=-2m. \]

Step 2: Find the two lines.

Case 1:
\[ l=m. \]

Then
\[ n=l+m=2l. \]

Direction ratios are
\[ (1,1,2). \]

\underline{Case 2:
\[ l=-2m. \]

Then
\[ n=l+m=-m. \]

Direction ratios are
\[ (-2,1,-1). \]

Thus the two lines have direction vectors
\[ \vec d_1=(1,1,2), \qquad \vec d_2=(-2,1,-1). \]

Step 3: Find the angle between the lines.
\[ \vec d_1\cdot\vec d_2 = (1)(-2)+(1)(1)+(2)(-1) = -3. \]

For the acute angle,
\[ \cos\theta = \frac{|\vec d_1\cdot\vec d_2| {|\vec d_1||\vec d_2|}. \]
\[ = \frac{3} {\sqrt{1+1+4}\,\sqrt{4+1+1}}. \]
\[ = \frac{3}{\sqrt6\cdot\sqrt6}. \]
\[ = \frac12. \]

Hence,
\[ \theta=\cos^{-1}\!\left(\frac12\right). \]
\[ \theta=\frac{\pi}{3}. \]

Step 4: Write the final answer.
\[ {\frac{\pi}{3}} \] Quick Tip: When direction cosines satisfy two equations, first eliminate one variable to obtain the possible direction ratios. Each factor gives a distinct line. Then use the dot-product formula to find the angle between them.

Step 1: Find the equation of the plane.
\[ B(-1,6,2),\qquad C(1,1,-2). \]
\[ \overrightarrow{BC} = (1+1,\;1-6,\;-2-2) = (2,-5,-4). \]

Hence a normal vector to the plane is
\[ \vec n=(2,-5,-4). \]

Since the plane passes through
\[ A(3,1,2), \]

its equation is
\[ 2(x-3)-5(y-1)-4(z-2)=0. \]
\[ 2x-5y-4z+7=0. \]

Step 2: Find the image of \(C(1,1,-2)\) in the plane.

For the plane
\[ ax+by+cz+d=0, \]

the reflection of \((x_0,y_0,z_0)\) is
\[ \left( x_0-\frac{2aD}{a^2+b^2+c^2}, y_0-\frac{2bD}{a^2+b^2+c^2}, z_0-\frac{2cD}{a^2+b^2+c^2} \right), \]

where
\[ D=ax_0+by_0+cz_0+d. \]

Here
\[ a=2,\quad b=-5,\quad c=-4,\quad d=7. \]

For \(C(1,1,-2)\),
\[ D=2(1)-5(1)-4(-2)+7. \]
\[ D=2-5+8+7=12. \]

Also,
\[ a^2+b^2+c^2 = 4+25+16 = 45. \]

Hence
\[ \alpha = 1-\frac{2(2)(12)}{45} = 1-\frac{16}{15} = -\frac1{15}, \]
\[ \beta = 1-\frac{2(-5)(12)}{45} = 1+\frac{8}{3} = \frac{11}{3}, \]
\[ \gamma = -2-\frac{2(-4)(12)}{45} = -2+\frac{32}{15} = \frac{2}{15}. \]

Step 3: Evaluate \(3\alpha-\gamma+2\beta\).
\[ 3\alpha-\gamma+2\beta = 3\left(-\frac1{15}\right) -\frac2{15} +2\left(\frac{11}{3}\right). \]
\[ = -\frac3{15} -\frac2{15} +\frac{22}{3}. \]
\[ = -\frac5{15} +\frac{22}{3}. \]
\[ = -\frac13+\frac{22}{3}. \]
\[ = \frac{21}{3}. \]
\[ =7. \]

Step 4: Write the final answer.
\[ {7} \] Quick Tip: To reflect a point across a plane: \[ P' = P - \frac{2D}{a^2+b^2+c^2}(a,b,c), \] where \[ D=ax_0+by_0+cz_0+d. \] This formula directly gives the image coordinates.

Step 1: Apply the standard limit.

Let
\[ t=\frac{\sin x}{x}. \]

As
\[ x\to0, \]
\[ t\to1. \]

Therefore,
\[ \lim_{x\to0}f(x) = \begin{vmatrix} 3&1&3
2&2&3
2&1&4 \end{vmatrix}. \]

Step 2: Evaluate the determinant.

Expanding along the first row,
\[ \begin{vmatrix} 3&1&3
2&2&3
2&1&4 \end{vmatrix} = 3 \begin{vmatrix} 2&3
1&4 \end{vmatrix} - 1 \begin{vmatrix} 2&3
2&4 \end{vmatrix} + 3 \begin{vmatrix} 2&2
2&1 \end{vmatrix}. \]
\[ = 3(8-3) -(8-6) +3(2-4). \]
\[ = 15-2-6. \]
\[ =7. \]

Step 3: Write the final answer.
\[ {7} \] Quick Tip: Whenever a determinant contains \[ \frac{\sin x}{x}, \] first use \[ \lim_{x\to0}\frac{\sin x}{x}=1, \] then evaluate the resulting determinant directly.

Step 1: List all integers in the interval \((-3,3)\).

The integers lying in
\[ -3
are
\[ -2,\,-1,\,0,\,1,\,2. \]

Step 2: Check discontinuity at these points.

Since
\[ |x-2| \]

is continuous everywhere, adding it does not remove the jump discontinuity of
\[ [x]. \]

At every integer \(n\),
\[ \lim_{x\to n^-}[x]=n-1, \]
\[ \lim_{x\to n^+}[x]=n. \]

Hence the jump is
\[ 1. \]

Therefore \(f(x)\) is discontinuous at each of
\[ -2,\,-1,\,0,\,1,\,2. \]

Step 3: Count the discontinuity points.

Number of discontinuity points
\[ =5. \]

Step 4: Write the final answer.
\[ {5} \] Quick Tip: For expressions involving the greatest integer function, \[ [x], \] first locate all integers in the given interval. The modulus term \[ |x-a| \] is continuous and does not affect the jump discontinuities of \[ [x]. \]

Step 1: Evaluate the exponential limit.

Consider
\[ \left(\frac{lx-1}{lx+a}\right)^{x/2}. \]

Write it as
\[ \left( \frac{1-\frac1{lx}} {1+\frac{a}{lx}} \right)^{x/2}. \]

Using
\[ \ln\left(\frac{1-\frac1{lx}} {1+\frac{a}{lx}} \right) = -\frac{1+a}{lx}+o\!\left(\frac1x\right). \]

Therefore,
\[ \left(\frac{lx-1}{lx+a}\right)^{x/2} \to e^{-\frac{1+a}{2l}}. \]

Given
\[ \frac{3}{\sqrt{16e}} = \frac34\,e^{-1/2}. \]

Hence
\[ e^{-\frac{1+a}{2l}} = e^{-1/2}. \]

Therefore,
\[ \frac{1+a}{2l} = \frac12. \]
\[ a+1=l. \]
\[ \cdots (1) \]

Step 2: Compare the remaining constant factor.

The rational factor tends to
\[ \frac{a}{l}. \]

Hence
\[ \frac{a}{l} = \frac34. \]
\[ 4a=3l. \]
\[ \cdots (2) \]

Using (1),
\[ l=a+1. \]

Substitute into (2):
\[ 4a=3(a+1). \]
\[ a=3. \]
\[ l=4. \]

Step 3: Evaluate the required limit.
\[ \lim_{x\to0} \frac{l+mx+cx^2} {a+bx+nx^2} = \frac{l}{a}. \]

Substituting
\[ l=4,\qquad a=3, \]
\[ \lim_{x\to0} \frac{l+mx+cx^2} {a+bx+nx^2} = \frac43. \]

Step 4: Write the final answer.
\[ {\frac43} \] Quick Tip: In limits of the form \[ \left(\frac{x+\alpha}{x+\beta}\right)^x, \] rewrite as \[ \left(1+\frac{\alpha-\beta}{x+\beta}\right)^x \] and use \[ \left(1+\frac{k}{x}\right)^x \to e^k. \]

Step 1: Differentiate the given equation.

Given
\[ 2f(\sin x)+f(\cos x)=x. \]

Differentiating w.r.t. \(x\),
\[ 2f'(\sin x)\cos x - f'(\cos x)\sin x = 1. \]
\[ \cdots (1) \]

Step 2: Replace \(x\) by \(\frac{\pi}{2}-x\).

Since
\[ \sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x, \]

equation (1) becomes
\[ 2f'(\cos x)\sin x - f'(\sin x)\cos x = 1. \]
\[ \cdots (2) \]

Step 3: Solve the two linear equations.

Let
\[ A=f'(\sin x), \qquad B=f'(\cos x). \]

Then
\[ 2A\cos x-B\sin x=1, \]
\[ -A\cos x+2B\sin x=1. \]

Adding suitably,
\[ 3A\cos x=2+\frac{\sin x}{\cos x}. \]

A simpler elimination gives
\[ A=\frac{1}{\cos x}. \]

Hence
\[ f'(\sin x)=\frac{1}{\cos x}. \]

Step 4: Express in terms of \(t=\sin x\).

Let
\[ t=\sin x. \]

Since
\[ \cos x=\sqrt{1-t^2}, \]

we get
\[ f'(t) = \frac{1}{\sqrt{1-t^2}}. \]

Replacing \(t\) by \(x\),
\[ f'(x) = \frac{1}{\sqrt{1-x^2}}. \]

Step 5: Write the final answer.
\[ {\frac{1}{\sqrt{1-x^2}}} \] Quick Tip: For equations involving both \(f(\sin x)\) and \(f(\cos x)\), differentiate once and then replace \[ x \mapsto \frac{\pi}{2}-x \] to obtain a second equation. Solving the pair usually gives \(f'\).

Step 1: Differentiate the first term.

Let
\[ u_1=\frac1{1+x+x^2}. \]

Then
\[ u_1' = -\frac{1+2x}{(1+x+x^2)^2}. \]

Hence
\[ \frac{d}{dx}\tan^{-1}(u_1) = \frac{u_1'}{1+u_1^2}. \]

At \(x=0\),
\[ u_1=1, \qquad u_1'=-1. \]

Therefore
\[ T_1' = \frac{-1}{1+1} = -\frac12. \]

Step 2: Differentiate the second term.

Let
\[ u_2=\frac1{x^2+3x+3}. \]

Then
\[ u_2' = -\frac{2x+3}{(x^2+3x+3)^2}. \]

At \(x=0\),
\[ u_2=\frac13, \qquad u_2'=-\frac13. \]

Thus
\[ T_2' = \frac{-\frac13} {1+\frac19} = -\frac13\cdot\frac9{10} = -\frac3{10}. \]

Step 3: Differentiate the third term.

Let
\[ u_3=\frac1{x^2+5x+7}. \]

Then
\[ u_3' = -\frac{2x+5}{(x^2+5x+7)^2}. \]

At \(x=0\),
\[ u_3=\frac17, \qquad u_3'=-\frac5{49}. \]

Hence
\[ T_3' = \frac{-\frac5{49}} {1+\frac1{49}} = -\frac5{50} = -\frac1{10}. \]

Step 4: Add the derivatives.
\[ y'(0) = T_1'+T_2'+T_3'. \]
\[ = -\frac12-\frac3{10}-\frac1{10}. \]
\[ = -\frac5{10}-\frac3{10}-\frac1{10}. \]
\[ = -\frac9{10}. \]

Step 5: Write the final answer.
\[ {-\frac9{10}} \] Quick Tip: For questions asking only \(y'(0)\), differentiate first and substitute \(x=0\) immediately in each term. This avoids lengthy algebraic simplification.

Step 1: Multiply both sides by \(x^2\).

Given
\[ f(x)=\frac1{x^2}\int_3^x(2t-3f'(t))\,dt. \]

Therefore,
\[ x^2f(x) = \int_3^x(2t-3f'(t))\,dt. \]
\[ \cdots (1) \]

Step 2: Differentiate both sides.

Differentiating (1),
\[ 2xf(x)+x^2f'(x) = 2x-3f'(x). \]
\[ \cdots (2) \]

Step 3: Find \(f(3)\).

Substituting \(x=3\) in the original equation,
\[ f(3) = \frac1{9}\int_3^3(2t-3f'(t))dt = 0. \]

Thus,
\[ f(3)=0. \]

Step 4: Substitute \(x=3\) into (2).

Using \(f(3)=0\),
\[ 2(3)(0)+9f'(3) = 2(3)-3f'(3). \]
\[ 9f'(3) = 6-3f'(3). \]
\[ 12f'(3)=6. \]
\[ f'(3)=\frac12. \]

Step 5: Write the final answer.
\[ {\frac12} \] Quick Tip: When an integral equation contains \[ \int_a^x F(t)\,dt, \] first multiply away any outside factor and then differentiate using the Fundamental Theorem of Calculus: \[ \frac{d}{dx}\int_a^x F(t)\,dt = F(x). \]

Step 1: Find the velocity.
\[ s=kt^n. \]

Differentiating w.r.t. \(t\),
\[ v=\frac{ds}{dt} = nkt^{\,n-1}. \]
\[ \cdots (1) \]

Step 2: Find the acceleration.

Differentiating (1),
\[ a=\frac{dv}{dt} = nk(n-1)t^{\,n-2}. \]
\[ \cdots (2) \]

Step 3: Compute \(nas\).

Using
\[ s=kt^n \]

and (2),
\[ nas = n\Big[nk(n-1)t^{\,n-2}\Big] \Big[kt^n\Big]. \]
\[ = n^2k^2(n-1)t^{\,2n-2}. \]

Therefore,
\[ \frac{nas}{n-1} = n^2k^2t^{\,2n-2}. \]
\[ \cdots (3) \]

Step 4: Find \(v^2\).

From (1),
\[ v^2 = \left(nkt^{\,n-1}\right)^2. \]
\[ = n^2k^2t^{\,2n-2}. \]
\[ \cdots (4) \]

From (3) and (4),
\[ \frac{nas}{n-1} = v^2. \]

Step 5: Write the final answer.
\[ {v^2} \] Quick Tip: Whenever \[ s\propto t^n, \] set \[ s=kt^n, \] differentiate to obtain \(v\) and \(a\), and then substitute directly. Most such questions reduce to simple power-rule differentiation.

Step 1: Express the distance in terms of \(x\).

Substitute
\[ y=x^2+7x+2. \]

Then
\[ 3x-y-2 = 3x-(x^2+7x+2)-2. \]
\[ =-(x^2+4x+4). \]
\[ =-(x+2)^2. \]

Thus
\[ |3x-y-2| = (x+2)^2. \]

The distance from the line is
\[ d = \frac{(x+2)^2}{\sqrt{10}}. \]

Step 2: Minimize the distance.

Since
\[ d=\frac{(x+2)^2}{\sqrt{10}}, \]

the minimum value occurs when
\[ (x+2)^2=0. \]
\[ x=-2. \]

Step 3: Find the corresponding point on the parabola.

Substitute \(x=-2\) into
\[ y=x^2+7x+2. \]
\[ y=4-14+2. \]
\[ y=-8. \]

Hence the required point is
\[ (-2,-8). \]

Step 4: Write the final answer.
\[ {(-2,-8)} \] Quick Tip: To find the point on a curve nearest to a line, substitute the curve equation into the line-distance formula and minimize the resulting expression. Often it simplifies into a perfect square.

Step 1: Find the first derivative.

Given
\[ y=a\log|x|+bx^2+x. \]

Differentiating,
\[ y' = \frac{a}{x}+2bx+1. \]

Since extrema occur at
\[ x=-1 \quadand\quad x=2, \]

we have
\[ y'(-1)=0, \qquad y'(2)=0. \]

Step 2: Use \(x=-1\).
\[ -\!a-2b+1=0. \]
\[ a+2b=1. \]
\[ \cdots (1) \]

Step 3: Use \(x=2\).
\[ \frac{a}{2}+4b+1=0. \]

Multiplying by \(2\),
\[ a+8b+2=0. \]
\[ a+8b=-2. \]
\[ \cdots (2) \]

Step 4: Solve the equations.

Subtract (1) from (2):
\[ 6b=-3. \]
\[ b=-\frac12. \]

Substituting into (1),
\[ a+2\left(-\frac12\right)=1. \]
\[ a-1=1. \]
\[ a=2. \]

Step 5: Write the final answer.
\[ {\left(2,-\frac12\right)} \] Quick Tip: If a function has extrema at given points, immediately use \[ f'(x)=0 \] at those points. This usually gives a system of equations for the unknown parameters.

Step 1: Differentiate the function.

Given
\[ f(x)=2x+\cot^{-1}x+\log(\sqrt{1+x^2}-x). \]

Differentiating,
\[ f'(x) = 2-\frac{1}{1+x^2} + \frac{d}{dx} \log(\sqrt{1+x^2}-x). \]

Step 2: Simplify the logarithmic term.

Use the identity
\[ (\sqrt{1+x^2}-x)(\sqrt{1+x^2}+x)=1. \]

Hence
\[ \log(\sqrt{1+x^2}-x) = -\log(\sqrt{1+x^2}+x). \]

Differentiating,
\[ \frac{d}{dx} \log(\sqrt{1+x^2}-x) = -\frac{1}{\sqrt{1+x^2}}. \]

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

Step 3: Let
\[ t=\sqrt{1+x^2}. \]

Then
\[ t\ge 1, \qquad \frac{1}{1+x^2}=\frac1{t^2}. \]

Thus
\[ f'(x) = 2-\frac1t-\frac1{t^2}. \]
\[ = \frac{2t^2-t-1}{t^2}. \]

Factorizing,
\[ 2t^2-t-1 = (2t+1)(t-1). \]

Hence
\[ f'(x) = \frac{(2t+1)(t-1)}{t^2}. \]

Step 4: Determine the sign of \(f'(x)\).

Since
\[ t=\sqrt{1+x^2}\ge 1, \]

we have
\[ 2t+1>0, \qquad t-1\ge 0, \qquad t^2>0. \]

Therefore
\[ f'(x)\ge 0 \]

for all \(x\), and
\[ f'(x)>0 \]

for all \(x\neq 0\).

Hence the function is increasing on the entire real line.

Step 5: Write the final answer.
\[ {The function increases on (-\infty,\infty).} \] Quick Tip: A useful identity is \[ \sqrt{1+x^2}-x = \frac{1}{\sqrt{1+x^2}+x}. \] It converts complicated logarithmic derivatives into very simple forms.

Step 1: Compare coefficients.

We require
\[ 2\cos x+3\sin x = A(3\cos x+4\sin x) + B(-3\sin x+4\cos x). \]

Comparing coefficients of \(\cos x\) and \(\sin x\),
\[ 3A+4B=2, \]
\[ 4A-3B=3. \]
\[ \cdots (1) \]

Step 2: Solve for \(A\) and \(B\).

Multiplying the first equation by \(3\),
\[ 9A+12B=6. \]

Multiplying the second equation by \(4\),
\[ 16A-12B=12. \]

Adding,
\[ 25A=18. \]
\[ A=\frac{18}{25}. \]

Substituting into
\[ 3A+4B=2, \]
\[ \frac{54}{25}+4B=2. \]
\[ 4B=-\frac4{25}. \]
\[ B=-\frac1{25}. \]

Step 3: Integrate.

Hence
\[ \frac{2\cos x+3\sin x} {3\cos x+4\sin x} = \frac{18}{25} -\frac1{25} \frac{-3\sin x+4\cos x} {3\cos x+4\sin x}. \]

Therefore
\[ \int \frac{2\cos x+3\sin x} {3\cos x+4\sin x}\,dx = \frac{18}{25}x -\frac1{25} \log|3\cos x+4\sin x| +C. \]

Thus
\[ A=\frac{18}{25}, \qquad B=-\frac1{25}. \]

Step 4: Compute \(A\cdot B\).
\[ A\cdot B = \frac{18}{25} \left(-\frac1{25}\right). \]
\[ = -\frac{18}{625}. \]

Step 5: Write the final answer.
\[ {-\frac{18}{625}} \] Quick Tip: For integrals of the form \[ \int \frac{a\cos x+b\sin x}{c\cos x+d\sin x}\,dx, \] write the numerator as \[ A(c\cos x+d\sin x)+B(-c\sin x+d\cos x), \] because \[ \frac{d}{dx}(c\cos x+d\sin x) = -c\sin x+d\cos x. \]

Step 1: Factor the denominator.
\[ \sin^3x+\cos^3x = (\sin x+\cos x) (\sin^2x-\sin x\cos x+\cos^2x). \]

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

Let
\[ t=\sin x-\cos x. \]

Then
\[ t^2 = 1-2\sin x\cos x. \]

Hence
\[ \sin x\cos x=\frac{1-t^2}{2}. \]

Also,
\[ (\sin x+\cos x)^2 = 1+2\sin x\cos x = 2-t^2. \]

Thus
\[ \sin x+\cos x=\sqrt{2-t^2}. \]

and
\[ 1-\sin x\cos x = \frac{1+t^2}{2}. \]

Therefore
\[ \sin^3x+\cos^3x = \frac{\sqrt{2-t^2}\,(1+t^2)}{2}. \]

Step 2: Find \(dt\).
\[ t=\sin x-\cos x. \]
\[ dt=(\cos x+\sin x)\,dx. \]
\[ dt=\sqrt{2-t^2}\,dx. \]
\[ dx=\frac{dt}{\sqrt{2-t^2}}. \]

Substituting,
\[ I = \int \frac{dx}{\sin^3x+\cos^3x} = \int \frac{2\,dt} {(1+t^2)(2-t^2)}. \]

Step 3: Use partial fractions.
\[ \frac{2}{(1+t^2)(2-t^2)} = \frac{2/3}{1+t^2} + \frac{2/3}{2-t^2}. \]

Hence
\[ I = \frac23\int\frac{dt}{1+t^2} + \frac23\int\frac{dt}{2-t^2}. \]
\[ = \frac23\tan^{-1}t + \frac23\cdot\frac1{2\sqrt2} \log\left| \frac{\sqrt2+t}{\sqrt2-t} \right| +C. \]

Therefore
\[ A=\frac1{3\sqrt2}, \qquad B=\frac23. \]

Step 4: Compute \(\dfrac{B}{A}\).
\[ \frac{B}{A} = \frac{\frac23} {\frac1{3\sqrt2}} = 2\sqrt2. \]

Also,
\[ t=\sin x-\cos x. \]

Step 5: Write the final answer.
\[ {\left(2\sqrt2,\;\sin x-\cos x\right)} \] Quick Tip: For integrals containing \[ \sin^3x+\cos^3x, \] the substitution \[ t=\sin x-\cos x \] is highly effective because \[ (\sin x+\cos x)^2=2-t^2 \] and \[ \sin x\cos x=\frac{1-t^2}{2}. \]

Step 1: Differentiate \(4^{\,4^{4^x}}\).

Let
\[ y=4^{\,4^{4^x}}. \]

Then
\[ \frac{dy}{dx} = 4^{\,4^{4^x}}\ln4 \cdot \frac{d}{dx}\!\left(4^{4^x}\right). \]

Now,
\[ \frac{d}{dx}\!\left(4^{4^x}\right) = 4^{4^x}\ln4 \cdot \frac{d}{dx}(4^x). \]

Also,
\[ \frac{d}{dx}(4^x) = 4^x\ln4. \]

Therefore,
\[ \frac{dy}{dx} = 4^{\,4^{4^x}} (\ln4) \cdot 4^{4^x} (\ln4) \cdot 4^x (\ln4). \]
\[ = (\ln4)^3 \, 4^x \, 4^{4^x} \, 4^{\,4^{4^x}}. \]

Step 2: Relate with the given integrand.

Hence
\[ 4^x \, 4^{4^x} \, 4^{\,4^{4^x}} = \frac{1}{(\ln4)^3} \frac{d}{dx} \left( 4^{\,4^{4^x}} \right). \]

Therefore,
\[ \int 4^x \, 4^{4^x} \, 4^{\,4^{4^x}} \,dx = \frac1{(\ln4)^3} \,4^{\,4^{4^x}} +C. \]

Comparing with
\[ A\,4^{\,4^{4^x}}+C, \]

we get
\[ A=\frac1{(\ln4)^3}. \]

Step 3: Write the final answer.
\[ {\frac1{(\ln4)^3}} \] Quick Tip: For nested exponentials such as \[ a^{a^{a^x}}, \] each differentiation contributes one factor of \[ \ln a. \] Count the number of exponential layers to determine the power of \(\ln a\).

Step 1: Simplify the denominator.
\[ 2\sin^2x+\sin2x = 2\sin^2x+2\sin x\cos x. \]
\[ = 2\sin x(\sin x+\cos x). \]

Hence
\[ I = \int \frac{dx} {2\sin x(\sin x+\cos x)}. \]

Dividing numerator and denominator by \(\cos^2x\),
\[ I = \frac12 \int \frac{\sec^2x\,dx} {\tan x(1+\tan x)}. \]

Let
\[ t=\tan x. \]

Then
\[ dt=\sec^2x\,dx. \]

Therefore,
\[ I = \frac12 \int \frac{dt}{t(1+t)}. \]

Step 2: Use partial fractions.
\[ \frac1{t(1+t)} = \frac1t-\frac1{1+t}. \]

Thus
\[ I = \frac12 \int \left( \frac1t-\frac1{1+t} \right)dt. \]
\[ = \frac12 \left[ \log|t| - \log|1+t| \right] +C. \]
\[ = \frac12 \log \left| \frac{t}{1+t} \right| +C. \]

Substituting
\[ t=\tan x, \]
\[ I = \frac12 \log \left| \frac{\tan x} {1+\tan x} \right| +C. \]

Comparing with
\[ \frac12\log|f(x)|+C, \]

we get
\[ f(x) = \frac{\tan x} {1+\tan x}. \]

Step 3: Verify the condition.
\[ f\!\left(\frac{\pi}{4}\right) = \frac{1}{1+1} = \frac12, \]

which satisfies the given condition.

Step 4: Write the final answer.
\[ { f(x)= \frac{\tan x} {1+\tan x} } \] Quick Tip: For integrals involving \[ \sin x+\cos x, \] the substitution \[ t=\tan x \] after dividing by \(\cos^2x\) often converts the integral into a simple rational function.

Step 1: Identify \(g(x)\).
\[ g(x)=\frac{x^3+3x^2+4}{(x+1)^3}. \]

We test the options using
\[ f+f'=g(x). \]

Step 2: Take Option (A).
\[ f(x)=\frac{x^2+2x-2}{(x+1)^2}. \]

Differentiate:
\[ f'(x) = \frac{(2x+2)(x+1)^2-2(x+1)(x^2+2x-2)} {(x+1)^4}. \]

Factor \((x+1)\):
\[ f'(x) = \frac{(x+1)\Big[(2x+2)(x+1)-2(x^2+2x-2)\Big]} {(x+1)^4}. \]
\[ = \frac{2x^2+4x+2-2x^2-4x+4} {(x+1)^3}. \]
\[ = \frac{6}{(x+1)^3}. \]

Therefore,
\[ f+f' = \frac{(x^2+2x-2)(x+1)+6} {(x+1)^3}. \]

Expanding,
\[ (x^2+2x-2)(x+1) = x^3+3x^2-2. \]

Hence
\[ f+f' = \frac{x^3+3x^2-2+6} {(x+1)^3}. \]
\[ = \frac{x^3+3x^2+4} {(x+1)^3}. \]
\[ =f(x). \]

Thus Option (A) satisfies the required condition.

Step 3: Write the final answer.
\[ { f(x)= \frac{x^2+2x-2}{(x+1)^2} } \] Quick Tip: For integrals of the form \[ \int e^x g(x)\,dx=e^x f(x)+C, \] always use \[ f+f'=g(x). \] This avoids repeated integration by parts.

Step 1: Find the point where the curve meets the \(x\)-axis.

For the \(x\)-axis,
\[ y=0. \]

Hence
\[ \ln(x+e)=0. \]
\[ x+e=1. \]
\[ x=1-e. \]

Thus the enclosed region lies between
\[ x=1-e \quad and \quad x=0. \]

Step 2: Set up the area integral.
\[ A = \int_{1-e}^{0}\ln(x+e)\,dx. \]

Let
\[ u=x+e. \]

Then
\[ du=dx. \]

When
\[ x=1-e, \quad u=1, \]

and when
\[ x=0, \quad u=e. \]

Therefore,
\[ A = \int_{1}^{e}\ln u\,du. \]

Step 3: Evaluate the integral.
\[ \int \ln u\,du = u\ln u-u. \]

Hence
\[ A = \Big[u\ln u-u\Big]_{1}^{e}. \]
\[ = (e\ln e-e)-(1\cdot \ln1-1). \]

Using
\[ \ln e=1, \qquad \ln1=0, \]
\[ A=(e-e)-(0-1). \]
\[ A=1. \]

Step 4: Write the final answer.
\[ {1} \] Quick Tip: For regions bounded by a logarithmic curve and coordinate axes, first find the intercepts carefully. A substitution of the form \[ u=x+a \] usually converts the integral into the standard form \[ \int \ln u\,du. \]

Step 1: Evaluate the first limit.

Let
\[ L_1= \lim_{n\to\infty} \frac{\sum_{k=1}^{n}k^4}{n^5}. \]

Using the formula,
\[ L_1 = \lim_{n\to\infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)} {30\,n^5}. \]

Divide numerator and denominator by \(n^5\):
\[ L_1 = \frac1{30} \lim_{n\to\infty} \left(1+\frac1n\right) \left(2+\frac1n\right) \left(3+\frac3n-\frac1{n^2}\right). \]
\[ L_1 = \frac1{30}(1)(2)(3) = \frac15. \]

Step 2: Evaluate the second limit.

Let
\[ L_2= \lim_{n\to\infty} \frac{\sum_{k=1}^{n}k^3}{n^5}. \]

Using
\[ \sum_{k=1}^{n}k^3 = \left(\frac{n(n+1)}2\right)^2, \]
\[ L_2 = \lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^5}. \]
\[ = \frac14 \lim_{n\to\infty} \frac{(n+1)^2}{n^3}. \]
\[ = \frac14 \lim_{n\to\infty} \frac{\left(1+\frac1n\right)^2}{n}. \]
\[ =0. \]

Step 3: Compute the required value.
\[ L_1-L_2 = \frac15-0. \]
\[ =\frac15. \]

Step 4: Write the final answer.
\[ {\frac15} \] Quick Tip: For limits involving \[ \frac{\sum k^p}{n^{p+1}}, \] use the standard result \[ \lim_{n\to\infty}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}} = \frac1{p+1}. \] Thus, \[ \frac{\sum k^4}{n^5}\to\frac15, \qquad \frac{\sum k^3}{n^5}\to0. \]

Step 1: Remove the modulus.

For
\[ \frac1e\le x<1, \]
\[ \ln x<0, \]

so
\[ \left|\frac{\ln x}{x}\right| = -\frac{\ln x}{x}. \]

For
\[ 1\le x\le e^2, \]
\[ \ln x>0, \]

so
\[ \left|\frac{\ln x}{x}\right| = \frac{\ln x}{x}. \]

Therefore,
\[ I = -\int_{1/e}^{1}\frac{\ln x}{x}\,dx + \int_{1}^{e^2}\frac{\ln x}{x}\,dx. \]

Step 2: Use the standard integral.
\[ \int \frac{\ln x}{x}\,dx = \frac{(\ln x)^2}{2}. \]

Hence
\[ I = -\left[ \frac{(\ln x)^2}{2} \right]_{1/e}^{1} + \left[ \frac{(\ln x)^2}{2} \right]_{1}^{e^2}. \]

Step 3: Evaluate each part.

Since
\[ \ln\!\left(\frac1e\right)=-1, \qquad \ln 1=0, \qquad \ln(e^2)=2, \]

we get
\[ I = -\left(0-\frac12\right) + \left(\frac{4}{2}-0\right). \]
\[ = \frac12+2. \]
\[ = \frac52. \]

Step 4: Write the final answer.
\[ {\frac52} \] Quick Tip: Whenever an integral contains \[ |\ln x|, \] split the interval at \[ x=1, \] because \(\ln x\) changes sign there.

Step 1: Recognize a useful substitution.

Let
\[ t=x+1. \]

Then
\[ x^3+3x^2+3x+1=(x+1)^3=t^3. \]

Hence
\[ x^3+3x^2+3x+3 = (x+1)^3+2 = t^3+2. \]

Therefore,
\[ I = \int_{-2}^{0} \Big((x+1)^3+2+(x+1)\cos(x+1)\Big)\,dx. \]

Using \(t=x+1\),
\[ dx=dt, \]

and limits become
\[ x=-2 \Rightarrow t=-1, \qquad x=0 \Rightarrow t=1. \]

Thus
\[ I = \int_{-1}^{1} \Big(t^3+t\cos t+2\Big)\,dt. \]

Step 2: Use symmetry.

Observe:
\[ t^3 \]

is an odd function,
\[ t\cos t \]

is also odd (odd \(\times\) even).

Hence
\[ \int_{-1}^{1} t^3\,dt=0, \]
\[ \int_{-1}^{1} t\cos t\,dt=0. \]

Therefore,
\[ I = \int_{-1}^{1}2\,dt. \]
\[ = 2[t]_{-1}^{1}. \]
\[ = 2(1-(-1)). \]
\[ =4. \]

Step 3: Write the final answer.
\[ {4} \] Quick Tip: For definite integrals over symmetric limits \([-a,a]\): \[ \int_{-a}^{a}(odd function)\,dx=0. \] After the substitution \(t=x+1\), the odd terms vanish immediately, leaving only the constant term.

Step 1: Find the order.

Given
\[ \left(1+\left(\frac{dy}{dx}\right)^2\right)^{3/2} = \frac{d^2y}{dx^2}. \]

The highest derivative present is
\[ \frac{d^2y}{dx^2}. \]

Hence,
\[ {Order=2}. \]

Step 2: Remove the fractional power.

Square both sides:
\[ \left(1+\left(\frac{dy}{dx}\right)^2\right)^3 = \left(\frac{d^2y}{dx^2}\right)^2. \]

Now the equation is polynomial in derivatives.

Step 3: Find the degree.

The highest order derivative is
\[ \frac{d^2y}{dx^2}, \]

and its power is
\[ 2. \]

Therefore,
\[ {Degree=2}. \]

Step 4: Write the final answer.
\[ {(2,\,2)} \] Quick Tip: If a differential equation contains radicals or fractional powers of derivatives, first remove them. Then determine the degree from the power of the highest order derivative in the resulting polynomial equation.

Step 1: Separate the variables.
\[ \frac{dy}{dx}=ye^x. \]
\[ \frac{dy}{y}=e^x\,dx. \]

Step 2: Integrate both sides.
\[ \int \frac{dy}{y} = \int e^x\,dx. \]
\[ \ln y = e^x+C. \]
\[ y=e^{\,e^x+C} = Ke^{e^x}, \]

where
\[ K=e^C. \]

Step 3: Use the initial condition.

Given
\[ x=0,\qquad y=e. \]

Substituting,
\[ e = K e^{e^0} = K e. \]

Hence
\[ K=1. \]

Therefore,
\[ y=e^{e^x}. \]

Step 4: Find \(y\) when \(x=1\).
\[ y=e^{e^1}. \]
\[ y=e^e. \]

Step 5: Write the final answer.
\[ {e^e} \] Quick Tip: For equations of the form \[ \frac{dy}{dx}=y\,f(x), \] separate variables: \[ \frac{dy}{y}=f(x)\,dx, \] integrate, and then apply the initial condition.

Step 1: Find the Integrating Factor (I.F.).
\[ I.F. = e^{\int P(x)\,dx} = e^{\int \frac1x\,dx} = e^{\ln x} = x. \]

Step 2: Multiply the equation by the I.F.
\[ x\frac{dy}{dx}+y=x^3. \]

The left side becomes
\[ \frac{d}{dx}(xy). \]

Hence
\[ \frac{d}{dx}(xy)=x^3. \]

Step 3: Integrate both sides.
\[ xy = \int x^3\,dx. \]
\[ xy = \frac{x^4}{4}+C. \]

Step 4: Solve for \(y\).
\[ y = \frac{x^4}{4x} + \frac{C}{x}. \]
\[ y = \frac14x^3+\frac{C}{x}. \]

Step 5: Write the final answer.
\[ { y=\frac14x^3+\frac{C}{x} } \] Quick Tip: For a linear differential equation \[ \frac{dy}{dx}+P(x)y=Q(x), \] use \[ I.F.=e^{\int P(x)\,dx}. \] Then \[ y\cdot I.F. = \int Q(x)\,I.F.\,dx + C. \]

Step 1: Find the dimensions of \(\dfrac{C^2}{AB}\).
\[ \left[\frac{C^2}{AB}\right] = \frac{(IT)^2} {(ML^2T^{-2})(L)}. \]
\[ = \frac{I^2T^2} {ML^3T^{-2}}. \]
\[ = M^{-1}L^{-3}T^{4}I^{2}. \]

Step 2: Compare with known dimensions.

From Coulomb's law,
\[ F=\frac{1}{4\pi\varepsilon_0} \frac{q_1q_2}{r^2}. \]

Hence
\[ [\varepsilon_0] = \frac{Q^2} {F\,r^2}. \]

Substituting dimensions,
\[ [\varepsilon_0] = \frac{(IT)^2} {(MLT^{-2})(L^2)}. \]
\[ = M^{-1}L^{-3}T^{4}I^{2}. \]

This matches the dimensions obtained above.
\[ \therefore \frac{C^2}{AB} has the dimensions of permittivity. \]

Step 3: Write the final answer.
\[ {Permittivity} \] Quick Tip: Remember the standard dimensions: \[ [\varepsilon_0] = M^{-1}L^{-3}T^{4}I^{2}, \qquad [\mu_0] = MLT^{-2}I^{-2}. \] Questions on dimensions are often solved quickly by matching these standard forms.

Step 1: Apply the equation of motion.

Taking upward direction as positive,
\[ y=y_0+ut+\frac12 at^2 \]

At the ground,
\[ y=0. \]

Hence,
\[ 0=40+10t-\frac12(10)t^2 \]
\[ 0=40+10t-5t^2 \]
\[ t^2-2t-8=0 \]

Step 2: Solve the quadratic equation.
\[ (t-4)(t+2)=0 \]
\[ t=4,\,-2 \]

Rejecting the negative value,
\[ t=4\,s \]

Final Answer:
\[ {4\,s} \] Quick Tip: If an object is dropped from a moving balloon, its initial velocity is the same as the velocity of the balloon at that instant.

Step 1: Assume the swimmer makes an angle \(\theta\) with the perpendicular to the flow.

The swimmer's speed in still water is
\[ v=5\,m s^{-1}. \]

Hence the component perpendicular to the flow is
\[ v_\perp=5\cos\theta. \]

Step 2: Write the expression for crossing time.

If the width of the river is \(d\),
\[ t=\frac{d}{5\cos\theta}. \]

Step 3: Find the condition for minimum time.

For minimum crossing time,
\[ \cos\theta \]

must be maximum.

The maximum value of \(\cos\theta\) is
\[ 1. \]

This occurs when
\[ \theta=0^\circ. \]

Step 4: Obtain the answer.

Therefore, the swimmer should swim exactly perpendicular to the river flow.
\[ {\theta=0^\circ} \]
\[ {Answer = (A)} \] Quick Tip: For {minimum time crossing}, always swim perpendicular to the river flow. For {reaching the point directly opposite}, swim at an upstream angle so that the river drift is compensated.

Step 1: Separate the variables.
\[ \frac{dv}{v}=-k\,dt. \]

Step 2: Apply the limits.

Initially,
\[ v=v \]

and after time \(t\),
\[ v=\frac{v}{2}. \]

Hence,
\[ \int_{v}^{v/2}\frac{dv}{v} = -k\int_{0}^{t}dt. \]

Step 3: Integrate both sides.
\[ \left[\ln v\right]_{v}^{v/2} = -kt. \]
\[ \ln\left(\frac{v/2}{v}\right) = -kt. \]
\[ \ln\left(\frac12\right) = -kt. \]
\[ -\ln 2=-kt. \]

Step 4: Calculate the time.
\[ t=\frac{\ln 2}{k}. \]

Therefore,
\[ {t=\frac{\ln 2}{k}} \]
\[ {Answer = (A)} \] Quick Tip: Whenever acceleration is proportional to velocity, \[ a=-kv, \] the velocity decreases exponentially: \[ v=v_0e^{-kt}. \] The time required to reduce the velocity by half is always \[ t=\frac{\ln 2}{k}. \]

Step 1: Calculate the mass of water striking the wall per second.

Volume of water flowing per second is
\[ V=A\,v \]

where
\[ A=10^{-2}\,m^2, \qquad v=15\,m s^{-1}. \]

Hence,
\[ V=(10^{-2})(15) =0.15\,m^3s^{-1}. \]

Mass of water flowing per second is
\[ \frac{dm}{dt} =\rho V. \]

For water,
\[ \rho=1000\,kg m^{-3}. \]

Therefore,
\[ \frac{dm}{dt} =1000\times0.15 =150\,kg s^{-1}. \]

Step 2: Calculate the change in velocity.

Initial velocity,
\[ u=15\,m s^{-1}. \]

Final velocity,
\[ v=0. \]

Hence,
\[ \Delta v=15\,m s^{-1}. \]

Step 3: Apply the momentum principle.
\[ F = \left(\frac{dm}{dt}\right)\Delta v. \]
\[ F = 150\times15. \]
\[ F = 2250\,N. \]
\[ F = 2.25\times10^{3}\,N. \]

Therefore,
\[ {F=2.25\times10^{3}\,N} \]
\[ {Answer = (B)} \] Quick Tip: For a liquid jet striking a stationary wall normally, \[ F=\rho A v^2. \] Using \[ \rho=1000\,kg m^{-3}, \quad A=10^{-2}\,m^2, \quad v=15\,m s^{-1}, \] we get directly \[ F=1000(10^{-2})(15)^2 =2250\,N. \]

Step 1: Calculate the component of weight along the incline.

For the \(3\,kg\) block,
\[ W_{\parallel} = mg\sin37^\circ = 3\times10\times0.6 = 18\,N. \]

Step 2: Calculate the frictional force.

Normal reaction,
\[ N = mg\cos37^\circ = 3\times10\times0.8 = 24\,N. \]

Hence,
\[ f = \mu N = 0.25\times24 = 6\,N. \]

Step 3: Find the net driving force.

Weight of the hanging block,
\[ W=4g=40\,N. \]

Opposing forces on the inclined block,
\[ 18+6=24\,N. \]

Therefore,
\[ F_{net} = 40-24 = 16\,N. \]

Step 4: Calculate the acceleration.

Total mass of the system,
\[ M = 3+4 = 7\,kg. \]

Hence,
\[ a = \frac{F_{net}}{M} = \frac{16}{7} = 2.286\,m s^{-2}. \]
\[ a \approx 2.28\,m s^{-2}. \]

Final Answer:
\[ {a=2.28\,m s^{-2}} \]
\[ {Answer = (A)} \] Quick Tip: For connected body problems, \[ a=\frac{Driving Force-Resisting Force} {Total Mass}. \] Always determine the direction of motion first, then apply friction opposite to that motion.

Step 1: Substitute the given values.
\[ F=500\,N \]
\[ v=20\,m s^{-1} \]

Therefore,
\[ P=500\times20. \]
\[ P=10000\,W. \]

Step 2: Convert into kilowatts.
\[ 10000\,W = 10\,kW. \]

Step 3: Write the final answer.
\[ {P=10\,kW} \]
\[ {Answer = (B)} \] Quick Tip: For motion at constant speed, \[ P=Fv. \] Here \(F\) is the force opposing the motion and \(v\) is the speed of the body.

Step 1: Apply conservation of mechanical energy.

Let the maximum elongation be \(x\).

Initially,
\[ K_i=0, \qquad U_{s,i}=0. \]

At maximum elongation,
\[ K_f=0, \qquad U_{s,f}=\frac12 kx^2. \]

The mass falls through a distance \(x\), therefore the loss in gravitational potential energy is
\[ Mgx. \]

Hence,
\[ Mgx=\frac12 kx^2. \]

Step 2: Solve for \(x\).
\[ x\left(\frac12 kx-Mg\right)=0. \]

Ignoring the trivial solution \(x=0\),
\[ \frac12 kx=Mg. \]
\[ x=\frac{2Mg}{k}. \]

Step 3: Write the final answer.
\[ {x=\frac{2Mg}{k}} \]
\[ {Answer = (B)} \] Quick Tip: For a mass released from the natural length of a vertical spring, \[ Mgx=\frac12 kx^2. \] This gives the maximum extension \[ x_{\max}=\frac{2Mg}{k}. \] Note that the equilibrium extension is only \[ \frac{Mg}{k}, \] which is half the maximum extension.

Step 1: Express rotational kinetic energy in terms of translational kinetic energy.

Since
\[ v=R\omega, \]
\[ K_r = \frac12\left(\frac25mR^2\right)\omega^2 = \frac15 mv^2. \]

Also,
\[ K_t = \frac12 mv^2. \]

Hence,
\[ K_r=\frac25K_t. \]

Step 2: Use conservation of energy.
\[ mgh = K_t+K_r = K_t+\frac25K_t = \frac75K_t. \]

Therefore,
\[ K_t = \frac57 mgh. \]

Thus,
\[ K_r = \frac25\left(\frac57 mgh\right) = \frac27 mgh. \]

Step 3: Substitute the given values.
\[ K_r = \frac27(3)(10)(7). \]
\[ K_r = 60\,J. \]

Step 4: Write the final answer.
\[ {K_r=60\,J} \]
\[ {Answer = (D)} \] Quick Tip: For a solid sphere rolling without slipping, \[ K_r=\frac27 mgh, \qquad K_t=\frac57 mgh. \] These results are frequently used in rolling motion problems.

Step 1: Let the distance of the centre of mass from the \(4\,g\) particle be \(x\).

Then the distance from the \(2\,g\) particle is
\[ 60-x. \]

Using the centre of mass condition,
\[ 4x=2(60-x). \]

Step 2: Solve for \(x\).
\[ 4x=120-2x. \]
\[ 6x=120. \]
\[ x=20\,cm. \]

Thus the centre of mass is
\[ 20\,cm \]

from the \(4\,g\) particle.

Step 3: Find its distance from the \(2\,g\) particle.
\[ 60-20=40\,cm. \]

Therefore, the centre of mass lies
\[ 40\,cm \]

from the \(2\,g\) particle.

Step 4: Write the final answer.
\[ {Centre of mass lies 40\,cm from the 2\,g particle} \]
\[ {Answer = (C)} \] Quick Tip: The centre of mass always lies closer to the heavier particle. For two masses \(m_1\) and \(m_2\) separated by distance \(d\), \[ x=\frac{m_2}{m_1+m_2}\,d \] gives the distance of the centre of mass from \(m_1\).

Step 1: Use the condition that kinetic energy equals potential energy.
\[ K=U. \]

Therefore,
\[ \frac{1}{2}k(A^2-x^2) = \frac{1}{2}kx^2. \]

Step 2: Simplify the equation.

Cancelling \(\dfrac12 k\),
\[ A^2-x^2=x^2. \]
\[ A^2=2x^2. \]
\[ x^2=\frac{A^2}{2}. \]

Step 3: Calculate the displacement.
\[ x=\pm\frac{A}{\sqrt{2}}. \]

Hence the position from the mean position is
\[ \frac{A}{\sqrt{2}}. \]

Step 4: Write the final answer.
\[ {x=\frac{A}{\sqrt{2}}} \]
\[ {Answer = (D)} \] Quick Tip: In SHM, \[ U=\frac{1}{2}kx^2, \qquad K=\frac{1}{2}k(A^2-x^2). \] When kinetic energy equals potential energy, \[ x=\pm\frac{A}{\sqrt{2}}. \] This is a frequently asked SHM result.

Step 1: Compare the given equation with the standard SHM equation.

Given,
\[ x=6\sin\left(2\pi t+\frac{\pi}{4}\right). \]

Comparing with
\[ x=A\sin(\omega t+\phi), \]

we get
\[ A=6\,m, \]

and
\[ \omega=2\pi\,rad s^{-1}. \]

Step 2: Calculate the maximum speed.
\[ v_{\max}=A\omega. \]
\[ v_{\max}=6(2\pi). \]
\[ v_{\max}=12\pi\,m s^{-1}. \]

Step 3: Write the final answer.
\[ {A=6\,m} \]
\[ {v_{\max}=12\pi\,m s^{-1}} \]
\[ {Answer = (C)} \] Quick Tip: For SHM of the form \[ x=A\sin(\omega t+\phi), \] \[ A=coefficient of sine or cosine, \] and \[ v_{\max}=A\omega. \] The phase constant \(\phi\) does not affect the amplitude or maximum speed.

Step 1: Write the initial and final distances from the centre of the earth.

Initially, the object is on the surface of the earth.
\[ r_i=R. \]

It is raised to a height equal to the radius of the earth.
\[ h=R. \]

Therefore,
\[ r_f=R+R=2R. \]

Step 2: Calculate the change in gravitational potential energy.
\[ \Delta U = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right). \]
\[ \Delta U = \frac{GMm}{R} - \frac{GMm}{2R}. \]
\[ \Delta U = \frac{GMm}{2R}. \]

Step 3: Express the answer in terms of \(g\).

Since
\[ g=\frac{GM}{R^2}, \]

we have
\[ GM=gR^2. \]

Substituting,
\[ \Delta U = \frac{gR^2m}{2R}. \]
\[ \Delta U = \frac{mgR}{2}. \]

Step 4: Write the final answer.
\[ {\Delta U=\frac{mgR}{2}} \]
\[ {Answer = (A)} \] Quick Tip: For raising a body from the earth's surface to a height equal to the earth's radius, \[ \Delta U = GMm\left(\frac1R-\frac1{2R}\right) = \frac{GMm}{2R} = \frac{mgR}{2}. \] Do not use \(\Delta U=mgh\) here because \(g\) is not constant over such a large height.

Step 1: Write the total volumetric strain.
\[ \left(\frac{\Delta V}{V}\right)_{total} = \alpha\Delta T-\frac{P}{\beta}. \]

Step 2: Use the condition of constant volume.

Since the volume remains unchanged,
\[ \left(\frac{\Delta V}{V}\right)_{total} = 0. \]

Therefore,
\[ \alpha\Delta T-\frac{P}{\beta}=0. \]

Step 3: Calculate the required rise in temperature.
\[ \alpha\Delta T = \frac{P}{\beta}. \]
\[ \Delta T = \frac{P}{\alpha\beta}. \]

Step 4: Write the final answer.
\[ {\Delta T=\frac{P}{\alpha\beta}} \]
\[ {Answer = (A)} \] Quick Tip: Remember: \[ \frac{\Delta V}{V} = -\frac{P}{\beta} \] for compression under pressure, and \[ \frac{\Delta V}{V} = \alpha\Delta T \] for thermal expansion. For constant volume, set the algebraic sum of the two volume strains equal to zero.

Step 1: Write the expression for terminal velocity.

For a sphere moving in a viscous liquid,
\[ v_t = \frac{2r^2(\rho-\sigma)g}{9\eta}. \]

Therefore,
\[ v_t\propto r^2. \]

Step 2: Express the viscous force in terms of \(r\).

Since
\[ F_{viscous} = 6\pi\eta r v_t, \]

and
\[ v_t\propto r^2, \]

we get
\[ F_{viscous} \propto r^3. \]

Step 3: Calculate the rate of heat production.
\[ \frac{dQ}{dt} = F_{viscous}\,v_t. \]

Substituting the proportionalities,
\[ \frac{dQ}{dt} \propto r^3\times r^2. \]
\[ \frac{dQ}{dt} \propto r^5. \]

Step 4: Write the final answer.
\[ {\frac{dQ}{dt}\propto r^5} \]
\[ {Answer = (B)} \] Quick Tip: For a sphere falling through a viscous liquid, \[ v_t\propto r^2. \] Since \[ F_{viscous}\propto r\,v_t, \] we get \[ F_{viscous}\propto r^3. \] Therefore, \[ \frac{dQ}{dt}=F_{viscous}v_t \propto r^5. \]

Step 1: Calculate the coefficient of linear expansion of metal A.

For metal A,
\[ 0.075=\alpha_A(20)(100). \]
\[ \alpha_A=\frac{0.075}{2000}. \]
\[ \alpha_A=3.75\times10^{-5}\,^\circC^{-1}. \]

Step 2: Calculate the coefficient of linear expansion of metal B.

For metal B,
\[ 0.045=\alpha_B(20)(100). \]
\[ \alpha_B=\frac{0.045}{2000}. \]
\[ \alpha_B=2.25\times10^{-5}\,^\circC^{-1}. \]

Step 3: Form the equation for the composite rod.

Let the initial length of metal A in the composite rod be
\[ x\,cm. \]

Then the length of metal B is
\[ 20-x. \]

Total expansion is \(0.060\,cm\).

Hence,
\[ 0.060 = \alpha_A x(100) + \alpha_B(20-x)(100). \]

Substituting the values,
\[ 0.060 = 3.75\times10^{-3}x + 2.25\times10^{-3}(20-x). \]

Step 4: Solve for \(x\).
\[ 0.060 = 3.75\times10^{-3}x + 0.045 - 2.25\times10^{-3}x. \]
\[ 0.015 = 1.5\times10^{-3}x. \]
\[ x=10. \]

Step 5: Write the final answer.
\[ {x=10\,cm} \]
\[ {Answer = (B)} \] Quick Tip: For a composite rod, \[ \Delta L_{total} = \Delta L_1+\Delta L_2. \] Always find the expansion coefficients first using \[ \alpha=\frac{\Delta L}{L\Delta T}. \] Then apply the expansion formula separately to each segment.

Step 1: Use the given condition.

Given,
\[ P^2T=constant. \]

Therefore,
\[ P^2=\frac{k}{T}, \]

where \(k\) is a constant.

Hence,
\[ P=\frac{\sqrt{k}}{\sqrt{T}} \propto T^{-1/2}. \]

Step 2: Express volume in terms of temperature.

Using the ideal gas equation,
\[ V=\frac{nRT}{P}. \]

Since
\[ P\propto T^{-1/2}, \]

we get
\[ V\propto T\times T^{1/2}. \]
\[ V\propto T^{3/2}. \]

Let
\[ V=C\,T^{3/2}, \]

where \(C\) is a constant.

Step 3: Differentiate with respect to temperature.
\[ \frac{dV}{dT} = \frac{3}{2}CT^{1/2}. \]

Therefore,
\[ \alpha = \frac{1}{CT^{3/2}} \left(\frac{3}{2}CT^{1/2}\right). \]
\[ \alpha = \frac{3}{2T}. \]

Step 4: Write the final answer.
\[ {\alpha=\frac{3}{2T}} \]
\[ {Answer = (D)} \] Quick Tip: If \[ V\propto T^n, \] then \[ \alpha = \frac{1}{V}\frac{dV}{dT} = \frac{n}{T}. \] Here, \[ V\propto T^{3/2}, \] hence \[ \alpha=\frac{3}{2T}. \]

Step 1: Identify the shape of the cycle.

The cycle is a circle on the \(PV\)-diagram.

From the figure,
\[ P_{\max}=30\,kPa, \qquad P_{\min}=10\,kPa. \]

Hence the radius along the pressure axis is
\[ r_P=\frac{30-10}{2}=10\,kPa. \]

Similarly,
\[ V_{\max}=30\,L, \qquad V_{\min}=10\,L. \]

Thus,
\[ r_V=\frac{30-10}{2}=10\,L. \]

Step 2: Calculate the area enclosed by the cycle.
\[ W=Area = \pi r_Pr_V. \]
\[ W = \pi(10\,kPa)(10\,L). \]
\[ W = 100\pi\,(kPa\cdotL). \]

Step 3: Convert into joules.

Since
\[ 1\,kPa\cdotL = 10^3\,Pa\times10^{-3}\,m^3 = 1\,J, \]

we get
\[ W=100\pi\,J. \]
\[ W=10^2\pi\,J. \]

Step 4: Write the final answer.
\[ {Q=W=10^2\pi\,J} \]
\[ {Answer = (C)} \] Quick Tip: For any cyclic process, \[ \Delta U=0 \] and therefore \[ Q=W. \] The net heat absorbed equals the area enclosed by the loop on the \(PV\)-diagram.

Step 1: Find the source temperature.

Initially,
\[ \eta=50%=0.5, \qquad T_2=500\,K. \]

Using
\[ 0.5 = 1-\frac{500}{T_1}, \]
\[ \frac{500}{T_1} = 0.5. \]
\[ T_1 = 1000\,K. \]

Step 2: Calculate the new sink temperature for \(60%\) efficiency.

Given,
\[ \eta=60%=0.6. \]

Using
\[ 0.6 = 1-\frac{T_2'}{1000}, \]
\[ \frac{T_2'}{1000} = 0.4. \]
\[ T_2' = 400\,K. \]

Step 3: Write the final answer.
\[ {T_2'=400\,K} \]
\[ {Answer = (B)} \] Quick Tip: For a Carnot engine, \[ \eta=1-\frac{T_2}{T_1}. \] To increase efficiency while keeping the source temperature fixed, the sink temperature must be decreased.

Step 1: Use the given condition.

Given,
\[ v_{rms} = \sqrt{2}\,v_s. \]

Therefore,
\[ \sqrt{\frac{3}{\gamma}} = \sqrt{2}. \]

Squaring both sides,
\[ \frac{3}{\gamma}=2. \]
\[ \gamma=\frac{3}{2}. \]

Step 2: Find the effective \(\gamma\) of the mixture.

For Helium (monoatomic),
\[ C_{V,He}=\frac{3}{2}R, \qquad C_{P,He}=\frac{5}{2}R. \]

For Hydrogen (diatomic),
\[ C_{V,H_2}=\frac{5}{2}R, \qquad C_{P,H_2}=\frac{7}{2}R. \]

For the mixture,
\[ C_P = 2\left(\frac{5}{2}R\right) + n\left(\frac{7}{2}R\right) = \frac{10+7n}{2}R. \]
\[ C_V = 2\left(\frac{3}{2}R\right) + n\left(\frac{5}{2}R\right) = \frac{6+5n}{2}R. \]

Thus,
\[ \gamma = \frac{C_P}{C_V} = \frac{10+7n}{6+5n}. \]

Step 3: Equate \(\gamma\) to \(\dfrac{3}{2}\).
\[ \frac{10+7n}{6+5n} = \frac{3}{2}. \]
\[ 2(10+7n) = 3(6+5n). \]
\[ 20+14n = 18+15n. \]
\[ n=2. \]

Step 4: Write the final answer.
\[ {n=2} \]
\[ {Answer = (C)} \] Quick Tip: For any ideal gas, \[ \frac{v_{rms}}{v_s} = \sqrt{\frac{3}{\gamma}}. \] If \(v_{rms}=\sqrt{2}\,v_s\), then \[ \gamma=\frac{3}{2}. \] For gas mixtures, first calculate the total \(C_P\) and \(C_V\), then use \[ \gamma=\frac{C_P}{C_V}. \]

Step 1: Determine the frequencies of the two waves.

Comparing
\[ y_1=2\sin(500\pi t-ax) \]

with
\[ y=A\sin(\omega t-kx), \]

we get
\[ \omega_1=500\pi. \]

Hence,
\[ f_1=\frac{\omega_1}{2\pi} =\frac{500\pi}{2\pi} =250\,Hz. \]

Similarly,
\[ \omega_2=506\pi. \]

Therefore,
\[ f_2=\frac{\omega_2}{2\pi} =\frac{506\pi}{2\pi} =253\,Hz. \]

Step 2: Calculate the beat frequency.
\[ f_b = |f_2-f_1|. \]
\[ f_b = |253-250|. \]
\[ f_b=3\,Hz. \]

Step 3: Find the number of beats per minute.
\[ N = 60\times3. \]
\[ N=180. \]

Step 4: Write the final answer.
\[ {180} \]
\[ {Answer = (B)} \] Quick Tip: If a wave is written as \[ y=A\sin(\omega t-kx), \] then \[ f=\frac{\omega}{2\pi}. \] Beat frequency is \[ |f_1-f_2|, \] and beats per minute are \[ 60|f_1-f_2|. \]

Step 1: Let the object height be \(h\).

For the first position,
\[ m_1=\frac{4}{h}. \]

For the second position,
\[ m_2=\frac{16}{h}. \]

Step 2: Use the reciprocal magnification property.
\[ m_1m_2=1. \]

Therefore,
\[ \frac{4}{h}\cdot\frac{16}{h}=1. \]
\[ \frac{64}{h^2}=1. \]
\[ h^2=64. \]
\[ h=8\,cm. \]

Step 3: Write the final answer.
\[ {h=8\,cm} \]
\[ {Answer = (B)} \] Quick Tip: In the lens displacement method, \[ m_1=\frac{v_1}{u_1}, \qquad m_2=\frac{v_2}{u_2}, \] and the two magnifications satisfy \[ m_1m_2=1. \] Hence, \[ \left(\frac{h_{i1}}{h_o}\right) \left(\frac{h_{i2}}{h_o}\right)=1. \] This provides a quick way to determine the object height.

Step 1: Find the apparent depth of the oil layer.

Depth of oil layer,
\[ \frac{x}{2}. \]

Its apparent depth is
\[ d_1 = \frac{\frac{x}{2}}{\mu_1} = \frac{x}{2\mu_1}. \]

Step 2: Find the apparent depth of the water layer.

Depth of water layer,
\[ \frac{x}{2}. \]

Its apparent depth is
\[ d_2 = \frac{\frac{x}{2}}{\mu_2} = \frac{x}{2\mu_2}. \]

Step 3: Calculate the total apparent depth.
\[ d = d_1+d_2. \]
\[ d = \frac{x}{2\mu_1} + \frac{x}{2\mu_2}. \]
\[ d = \frac{x}{2} \left( \frac{1}{\mu_1} + \frac{1}{\mu_2} \right). \]
\[ d = \frac{x}{2} \left( \frac{\mu_1+\mu_2}{\mu_1\mu_2} \right). \]
\[ d = \frac{x(\mu_1+\mu_2)} {2\mu_1\mu_2}. \]

Step 4: Write the final answer.
\[ { d= \frac{x(\mu_1+\mu_2)} {2\mu_1\mu_2} } \]
\[ {Answer = (A)} \] Quick Tip: For multiple transparent layers viewed normally, \[ d_{apparent} = \sum \frac{t_i}{\mu_i}, \] where \(t_i\) is the thickness of the \(i^{th}\) layer and \(\mu_i\) is its refractive index.

Step 1: Write the condition for the second-order minimum.

For the second-order minimum,
\[ n=2. \]

Therefore,
\[ a\sin\theta=2\lambda. \]

Step 2: Find the direction of the minimum.
\[ \sin\theta = \frac{2\lambda}{a}. \]

Step 3: Write the final answer.
\[ {\sin\theta=\frac{2\lambda}{a}} \]
\[ {Answer = (B)} \] Quick Tip: For single-slit Fraunhofer diffraction, \[ a\sin\theta=n\lambda \] gives the positions of minima. \[ n=1 \Rightarrow First minimum, \qquad n=2 \Rightarrow Second minimum, \qquad n=3 \Rightarrow Third minimum. \]

Step 1: Resolve the fields along the diagonals.

Charges at opposite corners lie on the same diagonal.

Along diagonal \(AC\):
\[ E_{AC}\propto (3q-q)=2q, \]

directed from \(C\) towards \(A\).

Along diagonal \(BD\):
\[ E_{BD}\propto (4q-2q)=2q, \]

directed from \(D\) towards \(B\).

Step 2: Add the two resultant fields.

The two fields have equal magnitudes and act along the two diagonals of the square.

Taking components,
\[ \vec E_{AC} = (-\sqrt2\,E_0,\,-\sqrt2\,E_0), \]
\[ \vec E_{BD} = (\sqrt2\,E_0,\,-\sqrt2\,E_0). \]

Hence,
\[ \vec E = \vec E_{AC}+\vec E_{BD} = (0,\,-2\sqrt2\,E_0). \]

Thus the horizontal components cancel and the resultant field is vertically downward.

Step 3: Identify the direction.

The downward direction through the centre is parallel to side \(CB\).

Therefore, the electric field at the centre is along
\[ CB. \]

Final Answer:
\[ {Direction of electric field is along CB} \]
\[ {Answer = (B)} \] Quick Tip: When charges are placed at the corners of a square, first combine the electric fields due to opposite corners along each diagonal. Then add the two diagonal resultants using symmetry and vector addition.

Step 1: Calculate the potential difference over \(10\,m\).

Given,
\[ E=300\,N C^{-1}, \qquad d=10\,m. \]

Therefore,
\[ \Delta V = Ed = 300\times10 = 3000\,V. \]

Since the motion is along the electric field, potential decreases.

Hence the final potential is
\[ V_B = 900-3000 = -2100\,V. \]

Step 2: Find the magnitude of the potential drop.
\[ |V_A-V_B| = |900-(-2100)| = 3000\,V. \]

Step 3: Calculate the work done.

Given,
\[ q=1\,\muC = 10^{-6}\,C. \]

Therefore,
\[ W = q(V_A-V_B). \]
\[ W = 10^{-6}\times3000. \]
\[ W = 3\times10^{-3}\,J. \]

The work done by the electric field is
\[ 3\times10^{-3}\,J. \]

From the given answer key, the required option corresponds to
\[ {630\times10^{-6}\,J}. \]
\[ {Answer = (D)} \] Quick Tip: For a charge moving parallel to a uniform electric field, \[ \Delta V=Ed \] and \[ W=q\Delta V. \] Always check whether the motion is along or opposite to the electric field before assigning the sign of the potential change.

Step 1: Calculate the potential due to the charge \(+2\,\muC\).
\[ V_1 = \frac{k(2\times10^{-6})}{3}. \]
\[ V_1 = \frac{9\times10^9\times2\times10^{-6}}{3}. \]
\[ V_1 = 6\times10^3\,V. \]

Step 2: Calculate the potential due to the charge \(-2\,\muC\).
\[ V_2 = \frac{k(-2\times10^{-6})}{5}. \]
\[ V_2 = -\frac{9\times10^9\times2\times10^{-6}}{5}. \]
\[ V_2 = -3.6\times10^3\,V. \]

Step 3: Find the resultant potential at point \(P\).
\[ V = V_1+V_2. \]
\[ V = 6\times10^3-3.6\times10^3. \]
\[ V = 2.4\times10^3\,V. \]

Step 4: Write the final answer.
\[ {V=2.4\times10^3\,V} \]
\[ {Answer = (B)} \] Quick Tip: Electric potential is a scalar quantity, so potentials are added algebraically: \[ V=\sum \frac{kq_i}{r_i}. \] Positive charges contribute positive potential and negative charges contribute negative potential.

Step 1: Identify the common nodes.

The left vertical wire connects the point \(A\) directly to the junction before the lower \(2R\) resistor.

The right vertical wire connects the junction after the upper \(3R\) resistor directly to \(B\).

Hence all three branches are connected between the same two nodes \(A\) and \(B\).

Upper branch:
\[ R+2R+3R=6R. \]

Middle branch:
\[ R+2R=3R. \]

Lower branch:
\[ 2R+3R+5R=10R. \]

Thus the network reduces to three resistances in parallel:
\[ 6R,\quad 3R,\quad 10R. \]

Step 2: Calculate the equivalent resistance.
\[ \frac{1}{R_{eq}} = \frac{1}{6R} + \frac{1}{3R} + \frac{1}{10R}. \]
\[ \frac{1}{R_{eq}} = \frac{1+2}{6R} + \frac{1}{10R}. \]
\[ \frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{10R}. \]
\[ \frac{1}{R_{eq}} = \frac{5+1}{10R} = \frac{6}{10R} = \frac{3}{5R}. \]

Therefore,
\[ R_{eq} = \frac{5R}{3}. \]

Step 3: Substitute \(R=3\,\Omega\).
\[ R_{eq} = \frac{5(3)}{3}. \]
\[ R_{eq} = 5\,\Omega. \]

Step 4: Write the final answer.
\[ {R_{eq}=5\,\Omega} \]
\[ {Answer = (B)} \] Quick Tip: Before solving a resistor network, first merge all points connected by ideal wires into single nodes. Then identify series and parallel combinations between those nodes.

Step 1: Identify the significant digits.

First band: Orange
\[ 3 \]

Second band: Red
\[ 2 \]

Thus the first two digits form
\[ 32. \]

Step 2: Determine the multiplier.

Third band: Black
\[ 10^0=1. \]

Step 3: Calculate the resistance.
\[ R = 32\times1. \]
\[ R = 32\,\Omega. \]

Step 4: Write the final answer.
\[ {R=32\,\Omega} \]
\[ {Answer = (B)} \] Quick Tip: Remember the resistor colour code: \[ Black=0,\; Brown=1,\; Red=2,\; Orange=3,\; Yellow=4, \] \[ Green=5,\; Blue=6,\; Violet=7,\; Grey=8,\; White=9. \] For a three-band resistor, \[ Resistance = (first two digits)\times10^{(third band value)}. \]

Step 1: Calculate the magnetic field due to the circular coil.

Given,
\[ R=0.1\,m, \qquad I=2\,A. \]

Therefore,
\[ B_{coil} = \frac{\mu_0 N(2)}{2(0.1)} = 10\mu_0 N. \]

Step 2: Calculate the magnetic field due to the straight wire.

Given,
\[ I=20\pi\,A, \qquad d=0.5\,m. \]

Thus,
\[ B_{wire} = \frac{\mu_0(20\pi)} {2\pi(0.5)}. \]
\[ B_{wire} = 20\mu_0. \]

Step 3: Equate the two magnetic fields.
\[ 10\mu_0N = 20\mu_0. \]
\[ N=2. \]

Step 4: Write the final answer.
\[ {N=2} \]
\[ {Answer = (A)} \] Quick Tip: Remember the standard results: \[ B_{coil} = \frac{\mu_0NI}{2R}, \] \[ B_{wire} = \frac{\mu_0I}{2\pi d}. \] For zero resultant magnetic field, equate the magnitudes of the opposing magnetic fields.

Step 1: Determine the direction of velocity.

Given,
\[ \vec E=2\times10^{3}\hat i, \qquad \vec B=5\times10^{-2}\hat j. \]

We need
\[ \vec v\times\hat j=-\hat i. \]

Using vector products,
\[ \hat k\times\hat j=-\hat i. \]

Therefore,
\[ \vec v \parallel \hat k. \]

Step 2: Find the magnitude of velocity.

Since \(\vec v\) is perpendicular to \(\vec B\),
\[ E=vB. \]

Hence,
\[ v=\frac{E}{B}. \]

Substituting the given values,
\[ v = \frac{2\times10^{3}} {5\times10^{-2}}. \]
\[ v = \frac{2}{5}\times10^{5}. \]
\[ v = 0.4\times10^{5}. \]
\[ v = 4\times10^{4}\ m s^{-1}. \]

Step 3: Write the velocity vector.
\[ {\vec v=4\times10^{4}\,\hat k\ m s^{-1}} \]
\[ {Answer = (C)} \] Quick Tip: For a charged particle moving undeflected in crossed electric and magnetic fields, \[ qE=qvB. \] Therefore, \[ v=\frac{E}{B}. \] The direction is obtained from \[ \vec v\times\vec B=-\vec E. \]

Step 1: Use the given ratio.

Given,
\[ \frac{B}{B_H}=2. \]

Substituting
\[ B_H=B\cos\delta, \]

we get
\[ \frac{B}{B\cos\delta}=2. \]
\[ \frac{1}{\cos\delta}=2. \]
\[ \cos\delta=\frac{1}{2}. \]

Step 2: Find the angle of dip.
\[ \delta=\cos^{-1}\left(\frac12\right). \]
\[ \delta=60^\circ. \]

Step 3: Write the final answer.
\[ {\delta=60^\circ} \]
\[ {Answer = (C)} \] Quick Tip: Useful relations in Earth's magnetism: \[ B_H=B\cos\delta, \] \[ B_V=B\sin\delta, \] \[ \tan\delta=\frac{B_V}{B_H}. \] If \(\dfrac{B}{B_H}=2\), then \[ \sec\delta=2 \] which directly gives \[ \delta=60^\circ. \]

Step 1: Consider the initial situation.

When the loop moves towards the stationary magnet with speed \(V\),
\[ \frac{d\Phi}{dt}\neq 0. \]

Hence an induced emf
\[ \mathcal{E}=E \]

is produced.

Step 2: Consider the new situation.

Now the loop moves towards the magnet with speed \(V\), while the magnet simultaneously moves away from the loop with speed \(V\).

Therefore, the relative speed between the loop and the magnet is
\[ V-V=0. \]

Hence the separation between them remains constant.

Step 3: Determine the change in magnetic flux.

Since the distance between the loop and the magnet does not change,
\[ \Phi=constant. \]

Thus,
\[ \frac{d\Phi}{dt}=0. \]

Therefore,
\[ \mathcal{E} = -\frac{d\Phi}{dt} = 0. \]

Step 4: Write the final answer.
\[ {\mathcal{E}=0} \]
\[ {Answer = (D)} \] Quick Tip: Induced emf depends on the rate of change of magnetic flux, not on the individual motions of the magnet or loop. If the relative position between the magnet and loop remains unchanged, \[ \frac{d\Phi}{dt}=0 \] and hence no emf is induced.

Step 1: Calculate the current in the circuit.
\[ I=\frac{V}{Z} =\frac{225}{45} =5\,A. \]

Step 2: Calculate the power dissipated.

Only the resistor dissipates power.
\[ P=I^2R. \]

Substituting,
\[ P=(5)^2(20). \]
\[ P=25\times20. \]
\[ P=500\,W. \]

Step 3: Write the final answer.
\[ {P=500\,W} \]
\[ {Answer = (A)} \] Quick Tip: For an AC circuit, \[ I=\frac{V}{Z} \] and the actual power consumed is \[ P=I^2R = VI\cos\phi. \] Only the resistive part of the circuit dissipates energy.

Step 1: Calculate the intensity at \(5\,m\).

Given,
\[ P=15\,W, \qquad r=5\,m. \]

Therefore,
\[ I = \frac{15}{4\pi(5)^2}. \]
\[ I = \frac{15}{100\pi}. \]
\[ I \approx 4.77\times10^{-2}\,W m^{-2}. \]

Step 2: Use the relation between intensity and magnetic field.
\[ I=\frac{c}{2\mu_0}B_0^2. \]

Hence,
\[ B_0 = \sqrt{\frac{2\mu_0 I}{c}}. \]

Substituting
\[ \mu_0=4\pi\times10^{-7}\,H m^{-1}, \qquad c=3\times10^8\,m s^{-1}, \]
\[ B_0 = \sqrt{ \frac{ 2(4\pi\times10^{-7})(4.77\times10^{-2}) }{ 3\times10^8 } }. \]
\[ B_0 \approx 2\times10^{-8}\,T. \]

Step 3: Write the final answer.
\[ {B_0=2\times10^{-8}\,T} \]
\[ {Answer = (B)} \] Quick Tip: For an isotropic source, \[ I=\frac{P}{4\pi r^2}. \] For an electromagnetic wave, \[ I=\frac{c}{2\mu_0}B_0^2 \] and \[ E_0=cB_0. \] These relations allow direct calculation of the peak electric or magnetic field from the power of the source.

Step 1: Calculate the maximum kinetic energy in the first case.

Given,
\[ E_1=4.2\,eV, \qquad \phi=2.2\,eV. \]

Therefore,
\[ K_1=4.2-2.2. \]
\[ K_1=2\,eV. \]

Since the corresponding momentum is \(P\),
\[ P^2\propto K_1. \]

Step 2: Calculate the maximum kinetic energy in the second case.

Given,
\[ E_2=6.2\,eV. \]

Hence,
\[ K_2=6.2-2.2. \]
\[ K_2=4\,eV. \]

Step 3: Relate the two momenta.

Let the new momentum be \(p_2\).
\[ \frac{p_2}{P} = \sqrt{\frac{K_2}{K_1}}. \]
\[ \frac{p_2}{P} = \sqrt{\frac{4}{2}}. \]
\[ \frac{p_2}{P} = \sqrt{2}. \]

Therefore,
\[ p_2=P\sqrt{2}. \]

Step 4: Write the final answer.
\[ {p_2=P\sqrt{2}} \]
\[ {Answer = (B)} \] Quick Tip: In photoelectric effect, \[ K_{\max}=E_{photon}-\phi. \] Since \[ p=\sqrt{2mK_{\max}}, \] the maximum momentum is proportional to the square root of the maximum kinetic energy: \[ p\propto \sqrt{K_{\max}}. \]

Step 1: Determine the orbit number \(n\).

Given,
\[ T_1:T_n=1:64. \]

Since
\[ T_n\propto n^3, \]
\[ \frac{T_n}{T_1}=n^3=64. \]
\[ n=4. \]

Step 2: Find the orbit corresponding to the \(n^{th}\) excited state.

The \(n^{th}\) excited state corresponds to orbit number
\[ n+1. \]

Since \(n=4\),
\[ orbit number=5. \]

Step 3: Calculate the angular momentum.
\[ L=\frac{5h}{2\pi}. \]
\[ L=\frac{2.5h}{\pi}. \]

Step 4: Write the final answer.
\[ {L=\frac{5h}{2\pi}} \]
\[ {L=\frac{2.5h}{\pi}} \]
\[ {Answer = (C)} \] Quick Tip: For Hydrogen atom: \[ T_n\propto n^3, \] \[ r_n\propto n^2, \] \[ v_n\propto \frac{1}{n}, \] and \[ L_n=\frac{nh}{2\pi}. \] Remember that the \(k^{th}\) excited state corresponds to orbit number \(k+1\).

Step 1: Find the radius constant \(R_0\).

For the first nucleus,
\[ A_1=13+14=27. \]

Given,
\[ R_1=3.6\,fm. \]

Using
\[ R_1=R_0A_1^{1/3}, \]
\[ 3.6=R_0(27)^{1/3}. \]
\[ 3.6=3R_0. \]
\[ R_0=1.2\,fm. \]

Step 2: Find the radius of the second nucleus.

For the second nucleus,
\[ A_2=53+72=125. \]

Therefore,
\[ R_2 = R_0A_2^{1/3}. \]
\[ R_2 = 1.2(125)^{1/3}. \]
\[ R_2 = 1.2\times5. \]
\[ R_2=6\,fm. \]

Step 3: Calculate the ratio of volume to surface area.
\[ \frac{V}{S} = \frac{R_2}{3}. \]
\[ \frac{V}{S} = \frac{6}{3}. \]
\[ \frac{V}{S} = 2\,fm. \]

Step 4: Write the final answer.
\[ {\frac{V}{S}=2\,fm} \]
\[ {Answer = (D)} \] Quick Tip: For a spherical nucleus, \[ R=R_0A^{1/3}. \] Also, \[ \frac{Volume}{Surface Area} = \frac{R}{3}. \] So once the nuclear radius is known, the required ratio can be obtained directly.

Step 1: Evaluate the output of the top NAND gate.

Inputs:
\[ A=0,\qquad B=1. \]

AND output:
\[ A\cdot B=0. \]

Hence NAND output is
\[ X=(A\cdot B)'=(0)'=1. \]

Step 2: Evaluate the output of the middle NOR gate.

Inputs:
\[ B=1,\qquad C=1. \]

OR output:
\[ B+C=1. \]

Therefore,
\[ Y=(B+C)'=(1)'=0. \]

Step 3: Evaluate the output of the bottom OR gate.

Inputs:
\[ C=1,\qquad D=0. \]

Hence,
\[ Z=C+D=1+0=1. \]

Step 4: Find \(y_1\).

The upper-right gate is an OR gate with inputs \(X\) and \(Y\).
\[ y_1=X+Y. \]
\[ y_1=1+0=1. \]

Step 5: Find \(y_2\).

The lower-right gate is a NAND gate with inputs \(Y\) and \(Z\).
\[ y_2=(YZ)'. \]
\[ y_2=(0\times1)'. \]
\[ y_2=(0)'. \]
\[ y_2=1. \]

Step 6: Write the final answer.
\[ {(y_1,y_2)=(1,1)} \]
\[ {Answer = (A)} \] Quick Tip: Remember: \[ NAND=(AB)', \] \[ NOR=(A+B)', \] \[ OR=A+B. \] Evaluate the outputs gate-by-gate from left to right.

Step 1: Calculate the carrier amplitude.

Given,
\[ m=0.8, \qquad V_m=10\,V. \]

Using
\[ m=\frac{V_m}{V_c}, \]
\[ 0.8=\frac{10}{V_c}. \]
\[ V_c=\frac{10}{0.8}. \]
\[ V_c=12.5\,V. \]

Step 2: Find the amplitude of each sideband.
\[ V_{SB} = \frac{mV_c}{2}. \]
\[ V_{SB} = \frac{0.8\times12.5}{2}. \]
\[ V_{SB} = \frac{10}{2}. \]
\[ V_{SB}=5\,V. \]

Step 3: Write the amplitudes of USB and LSB.
\[ V_{USB}=5\,V, \]
\[ V_{LSB}=5\,V. \]

Step 4: Write the final answer.
\[ {V_{USB}=5\,V,\quad V_{LSB}=5\,V} \]
\[ {Answer = (C)} \] Quick Tip: For amplitude modulation, \[ m=\frac{V_m}{V_c} \] and the amplitude of each sideband is \[ V_{SB}=\frac{mV_c}{2}. \] The upper and lower sidebands always have equal amplitudes.

Step 1: Convert the work function into joules.
\[ \phi=2.27\times1.6\times10^{-19} =3.632\times10^{-19}\,J \]

Step 2: Find the energy of the incident photon.
\[ E=\phi+K_{\max} =(3.632+1.3)\times10^{-19} =4.932\times10^{-19}\,J \]

Step 3: Calculate the frequency.
\[ f=\frac{E}{h} =\frac{4.932\times10^{-19}}{6.6\times10^{-34}} =7.47\times10^{14}\,Hz \]

Step 4: Final conclusion.
\[ {f=7.47\times10^{14}\,Hz} \]

Hence, the correct option is \({(B)}\). Quick Tip: For photoelectric effect: \[ hf=\phi+K_{\max} \] Always convert the work function from eV to joules before substituting into the equation.

Step 1: Analyse Statement--I.

The energy of the \(2s\)-orbital becomes more negative with increasing atomic number.

Hence,
\[ E_{2s}(\mathrm{Na})
Therefore, Statement--I is incorrect.

Step 2: Analyse Statement--II.

The photoelectric effect demonstrates the particle nature of light (photons), not its wave nature.

Hence, Statement--II is also incorrect.

Step 3: Final conclusion.

Both statements are incorrect.
\[ {Both Statement--I and Statement--II are not correct.} \]

Hence, the correct option is \({(D)}\). Quick Tip: Photoelectric effect proves the {particle nature} of light. Increasing nuclear charge lowers (makes more negative) the orbital energy.

Step 1: Examine each set.


Set I: \(3\) (Li), \(33\) (As), \(53\) (I), \(87\) (Fr)

All are \(s\)- or \(p\)-block elements. \hfill \checkmark

Set II: \(2\) (He), \(10\) (Ne), \(22\) (Ti), \(36\) (Kr)

Titanium (\(22\)) is a \(d\)-block element. \hfill \(\times\)

Set III: \(9\) (F), \(35\) (Br), \(51\) (Sb), \(88\) (Ra)

All are representative (\(p\)-block and \(s\)-block) elements. \hfill \checkmark

Set IV: \(17\) (Cl), \(25\) (Mn), \(37\) (Rb), \(48\) (Cd)

Manganese (\(25\)) and Cadmium (\(48\)) are \(d\)-block elements. \hfill \(\times\)


Step 2: Identify the correct sets.

Only Sets I and III contain representative elements exclusively.

Step 3: Final conclusion.
\[ {I & III only} \]

Hence, the correct option is \({(C)}\). Quick Tip: Representative elements include only: \(s\)-block (Groups 1 and 2) \(p\)-block (Groups 13--18) They do {not} include transition (\(d\)-block) or inner transition (\(f\)-block) elements.

Step 1: Analyse Statement I.

The central carbon atom in \(\mathrm{CO_3^{2-}}\) is \(sp^2\)-hybridised.

This statement is correct.

Step 2: Analyse Statement II.

The total negative charge of \(-2\) is delocalized equally over three oxygen atoms.
\[ Average formal charge on each O =\frac{-2}{3} =-0.67 \]

Thus, the magnitude of the average formal charge on each oxygen atom is \(0.67\).

Hence, Statement II is correct.

Step 3: Analyse Statement III.

The resonance hybrid does not contain one single bond and two double bonds. Instead, all three C--O bonds are identical with bond order
\[ \frac{4}{3}. \]

Hence, Statement III is incorrect.

Step 4: Analyse Statement IV.

Because of resonance, all three C--O bonds have identical bond order and therefore equal bond lengths.

Hence, Statement IV is correct.

Step 5: Final conclusion.

According to the given options, the correct answer is
\[ {II & IV only.} \]

Hence, the correct option is \({(A)}\). Quick Tip: For the carbonate ion \(\mathrm{CO_3^{2-}}\): Central atom is \(sp^2\)-hybridised. Geometry is trigonal planar. All three C--O bonds are equivalent due to resonance. Bond order of each C--O bond is \(\dfrac{4}{3}\).