WBJEE 2014 Mathematics Question Paper with Answer and Solution

(C) Given,$\log _{101} \log _{7}(\sqrt{x+7}+\sqrt{x})=0$
$\therefore \log _{7}(\sqrt{x+7}+\sqrt{x}) = (101)^{0} = 1$
$\Rightarrow \sqrt{x+7}+\sqrt{x} = 7^{1} = 7$
Squaring both sides,we get:
$(\sqrt{x+7}+\sqrt{x})^{2} = 7^{2}$
$x+7+x+2\sqrt{x(x+7)} = 49$
$2x+7+2\sqrt{x^{2}+7x} = 49$
$2\sqrt{x^{2}+7x} = 42-2x$
$\sqrt{x^{2}+7x} = 21-x$
Squaring both sides again:
$x^{2}+7x = (21-x)^{2}$
$x^{2}+7x = 441 - 42x + x^{2}$
$7x = 441 - 42x$
$49x = 441$
$x = \frac{441}{49} = 9$

(C) Let $y = 20^{301}$.
To find the number of digits in $y$,we calculate $\lfloor \log_{10} y \rfloor + 1$.
$\log_{10} y = \log_{10} (20^{301}) = 301 \times \log_{10} (2 \times 10)$.
Using the property $\log(ab) = \log a + \log b$,we get:
$\log_{10} y = 301 \times (\log_{10} 2 + \log_{10} 10) = 301 \times (0.3010 + 1) = 301 \times 1.3010$.
$301 \times 1.3010 = 391.601$.
The number of digits is $\lfloor 391.601 \rfloor + 1 = 391 + 1 = 392$.

(C) Since $a, b$ and $c$ are in $GP$,we have $b^{2} = ac$. Taking the natural logarithm on both sides,we get $2 \log_{e} b = \log_{e} a + \log_{e} c$.
Given the quadratic equation $(\log_{e} a) x^{2} - (2 \log_{e} b) x + \log_{e} c = 0$.
Substituting $x = 1$ into the equation,we get $(\log_{e} a) - (2 \log_{e} b) + \log_{e} c = 0$,which simplifies to $\log_{e} a + \log_{e} c = 2 \log_{e} b$,which is true.
Thus,$x = 1$ is one root of the equation.
Let the other root be $\alpha$. The product of the roots is given by $\frac{\log_{e} c}{\log_{e} a}$.
Therefore,$1 \times \alpha = \frac{\log_{e} c}{\log_{e} a} = \log_{a} c$.
Hence,the roots are $1$ and $\log_{a} c$.

(A) The given equation is $pq(x^{2}+1) = r^{2}x$,which can be rewritten as $x^{2} - \frac{r^{2}}{pq}x + 1 = 0$.
Since $\tan A$ and $\tan B$ are the roots,we have $\tan A + \tan B = \frac{r^{2}}{pq}$ and $\tan A \tan B = 1$.
We know that in a $\triangle ABC$,$A + B + C = 180^{\circ}$,so $A + B = 180^{\circ} - C$.
Taking tangent on both sides,$\tan(A + B) = \tan(180^{\circ} - C) = -\tan C$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get $\frac{\frac{r^{2}}{pq}}{1 - 1} = -\tan C$.
This implies $\frac{r^{2}/pq}{0} = -\tan C$,which means $\tan C$ is undefined.
Therefore,$C = 90^{\circ}$,which implies $\triangle ABC$ is a right-angled triangle.

(A) Given,$\alpha, \beta$ are the roots of $ax^{2}+bx+c=0$ and $\alpha+h, \beta+h$ are the roots of $px^{2}+qx+r=0$.
For the first equation,the discriminant is $D_{1} = b^{2}-4ac$ and the difference of roots is $(\alpha-\beta)^{2} = \frac{D_{1}}{a^{2}}$.
For the second equation,the discriminant is $D_{2} = q^{2}-4pr$ and the difference of roots is $((\alpha+h)-(\beta+h))^{2} = \frac{D_{2}}{p^{2}}$.
Since $(\alpha+h)-(\beta+h) = \alpha-\beta$,we have $(\alpha-\beta)^{2} = ((\alpha+h)-(\beta+h))^{2}$.
Therefore,$\frac{D_{1}}{a^{2}} = \frac{D_{2}}{p^{2}}$.
This implies $\frac{D_{1}}{D_{2}} = \frac{a^{2}}{p^{2}}$.
Thus,the ratio of the discriminants is $a^{2}: p^{2}$.

(D) Given that $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$,we have $\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0$.
Since $\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$,we have $\beta^{2}+9 p^{2} \beta+15 q^{2}=0$.
Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$.
Evaluating $f(\alpha)$:
$f(\alpha)=\alpha^{2}+6 p^{2} \alpha+10 q^{2} = (\alpha^{2}+3 p^{2} \alpha+5 q^{2}) + 3 p^{2} \alpha + 5 q^{2} = 0 + 3 p^{2} \alpha + 5 q^{2}$.
Since $\alpha > 0$ and $p, q$ are real numbers (assuming $p, q \neq 0$ for non-trivial roots),$f(\alpha) > 0$.
Evaluating $f(\beta)$:
$f(\beta)=\beta^{2}+6 p^{2} \beta+10 q^{2} = (\beta^{2}+9 p^{2} \beta+15 q^{2}) - (3 p^{2} \beta + 5 q^{2}) = 0 - (3 p^{2} \beta + 5 q^{2})$.
Since $\beta > \alpha > 0$,$f(\beta) < 0$.
Since $f(x)$ is a continuous polynomial function and $f(\alpha) > 0$ while $f(\beta) < 0$,by the Intermediate Value Theorem,there must exist a root $\gamma$ such that $\alpha < \gamma < \beta$.

(A) Given that $\alpha$ and $\beta$ are the roots of $x^{2}-x-1=0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\alpha^{2}-\alpha-1=0 \implies \alpha^{2}=\alpha+1$
$\beta^{2}-\beta-1=0 \implies \beta^{2}=\beta+1$
Multiplying by $\alpha^{n-1}$ and $\beta^{n-1}$ respectively:
$\alpha^{n+1}=\alpha^{n}+\alpha^{n-1}$
$\beta^{n+1}=\beta^{n}+\beta^{n-1}$
Adding these two equations:
$\alpha^{n+1}+\beta^{n+1}=(\alpha^{n}+\beta^{n})+(\alpha^{n-1}+\beta^{n-1})$
By definition of $S_{n}$,this is:
$S_{n+1}=S_{n}+S_{n-1}$
Thus,option $A$ is correct.

(D) Given the quadratic equation $x^{2}+p x+q=0$,the sum of roots $\alpha+\beta = -p$ and the product of roots $\alpha \beta = q$.
First,calculate $\alpha^{3}+\beta^{3}$:
$\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta) = (-p)^{3}-3 q(-p) = -p^{3}+3 p q = 3 p q-p^{3}$.
Next,calculate $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}$:
$\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2} - \alpha^{2} \beta^{2} = [(\alpha+\beta)^{2}-2 \alpha \beta]^{2} - (\alpha \beta)^{2} = [(-p)^{2}-2 q]^{2} - q^{2} = (p^{2}-2 q)^{2} - q^{2} = p^{4}-4 p^{2} q+4 q^{2}-q^{2} = p^{4}-4 p^{2} q+3 q^{2} = p^{2}(p^{2}-3 q)-q(p^{2}-3 q) = (p^{2}-q)(p^{2}-3 q)$.
Thus,the values are $3 p q-p^{3}$ and $(p^{2}-q)(p^{2}-3 q)$.

(B) Given,$f(x) = \sum_{n=1}^{\infty} \frac{2^n - 1}{n!} x^n$.
We can rewrite this as $f(x) = \sum_{n=1}^{\infty} \frac{(2x)^n}{n!} - \sum_{n=1}^{\infty} \frac{x^n}{n!}$.
Since $e^y = 1 + \sum_{n=1}^{\infty} \frac{y^n}{n!}$,we have $\sum_{n=1}^{\infty} \frac{y^n}{n!} = e^y - 1$.
Substituting this,we get $f(x) = (e^{2x} - 1) - (e^x - 1) = e^{2x} - e^x$.
Setting $f(x) = 0$,we have $e^{2x} - e^x = 0$,which implies $e^x(e^x - 1) = 0$.
Since $e^x > 0$ for all real $x$,we must have $e^x - 1 = 0$,which gives $e^x = 1$.
Thus,$x = 0$ is the only real solution.

(B) Given equation is $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}$.
For the square roots to be defined,we must have $x+1 \ge 0$,$x-1 \ge 0$,and $4x-1 \ge 0$,which implies $x \ge 1$.
Squaring both sides,we get $(x+1) + (x-1) - 2\sqrt{x^2-1} = 4x-1$.
This simplifies to $2x - 2\sqrt{x^2-1} = 4x-1$,or $-2\sqrt{x^2-1} = 2x-1$.
Squaring both sides again,we get $4(x^2-1) = (2x-1)^2 = 4x^2 - 4x + 1$.
This simplifies to $4x^2 - 4 = 4x^2 - 4x + 1$,which gives $4x = 5$,or $x = \frac{5}{4}$.
Checking $x = \frac{5}{4}$ in the original equation: $\sqrt{\frac{5}{4}+1} - \sqrt{\frac{5}{4}-1} = \sqrt{\frac{9}{4}} - \sqrt{\frac{1}{4}} = \frac{3}{2} - \frac{1}{2} = 1$.
However,the right side is $\sqrt{4(\frac{5}{4})-1} = \sqrt{5-1} = \sqrt{4} = 2$.
Since $1 \neq 2$,$x = \frac{5}{4}$ is an extraneous solution.
Therefore,there are no real solutions to the equation.

(C) Let $z = x + iy$.
Then the expression becomes $|x+iy|^2 + |(x-3)+iy|^2 + |x+i(y-1)|^2$.
$= (x^2 + y^2) + ((x-3)^2 + y^2) + (x^2 + (y-1)^2)$.
$= x^2 + y^2 + x^2 - 6x + 9 + y^2 + x^2 + y^2 - 2y + 1$.
$= 3x^2 - 6x + 3y^2 - 2y + 10$.
$= 3(x^2 - 2x + 1) + 3(y^2 - \frac{2}{3}y + \frac{1}{9}) + 10 - 3 - \frac{1}{3}$.
$= 3(x-1)^2 + 3(y - \frac{1}{3})^2 + \frac{20}{3}$.
This expression is minimum when $x-1 = 0$ and $y - \frac{1}{3} = 0$.
Therefore,$x = 1$ and $y = \frac{1}{3}$.
Thus,$z = 1 + \frac{i}{3}$.

(B) Given that $z_{1}, z_{2}, z_{3}$ are vertices of an equilateral triangle inscribed in the unit circle $|z| = 1$.
Since the triangle is equilateral and inscribed in the circle centered at the origin,the vertices $z_{2}$ and $z_{3}$ can be obtained by rotating $z_{1}$ by $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ radians respectively.
Thus,$z_{2} = z_{1} e^{i(2\pi/3)} = z_{1}\omega$ and $z_{3} = z_{1} e^{i(4\pi/3)} = z_{1}\omega^{2}$,where $\omega$ is the cube root of unity.
Therefore,the product $z_{1} z_{2} z_{3} = z_{1} \times (z_{1}\omega) \times (z_{1}\omega^{2}) = z_{1}^{3} \omega^{3}$.
Since $\omega^{3} = 1$,we have $z_{1} z_{2} z_{3} = z_{1}^{3}$.

(B) Given equation is $1+z+z^{3}+z^{4}=0$.
Factorizing the expression: $(1+z) + z^{3}(1+z) = 0$.
$(1+z)(1+z^{3}) = 0$.
This implies $1+z=0$ or $1+z^{3}=0$.
So,$z = -1$ or $z^{3} = -1$.
The roots of $z^{3} = -1$ are $e^{i\pi/3}, e^{i\pi}, e^{i5\pi/3}$.
Thus,the distinct roots are $z = -1, e^{i\pi/3}, e^{i5\pi/3}$.
These three points form an equilateral triangle in the Argand plane.

(A) The cube roots of unity other than $1$ are $\omega$ and $\omega^{2}$,where $\omega = e^{i 2\pi/3}$.
Given $S = \sum_{n=0}^{\infty} (-1)^{n} \left(\frac{\alpha}{\beta}\right)^{n}$,this is a geometric series with first term $a = 1$ and common ratio $r = -\frac{\alpha}{\beta}$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r} = \frac{1}{1 - (-\alpha/\beta)} = \frac{\beta}{\alpha + \beta}$.
Since $1 + \omega + \omega^{2} = 0$,we have $\alpha + \beta = \omega + \omega^{2} = -1$.
Case $I$: If $\alpha = \omega$ and $\beta = \omega^{2}$,then $S = \frac{\omega^{2}}{\omega + \omega^{2}} = \frac{\omega^{2}}{-1} = -\omega^{2}$.
Case $II$: If $\alpha = \omega^{2}$ and $\beta = \omega$,then $S = \frac{\omega}{\omega^{2} + \omega} = \frac{\omega}{-1} = -\omega$.
Thus,the value of $S$ is either $-\omega$ or $-\omega^{2}$.
Note: The provided options seem to contain a factor of $2$ which is not present in the standard sum of this infinite series. Based on the structure,the intended answer is $-\omega$ or $-\omega^{2}$.

(A) Since $z_{1}, z_{2},$ and $z_{3}$ are the vertices of an equilateral triangle,we have $|z_{1} - z_{2}| = |z_{2} - z_{3}| = |z_{3} - z_{1}| = k$.
Given $\alpha = \frac{1}{2}(\sqrt{3} + i)$,we find $|\alpha| = \frac{1}{2} \sqrt{(\sqrt{3})^2 + 1^2} = \frac{1}{2} \sqrt{4} = 1$.
Let $A = \alpha z_{1} + \beta$,$B = \alpha z_{2} + \beta$,and $C = \alpha z_{3} + \beta$.
Then the distance between $A$ and $B$ is $|A - B| = |(\alpha z_{1} + \beta) - (\alpha z_{2} + \beta)| = |\alpha(z_{1} - z_{2})| = |\alpha| |z_{1} - z_{2}| = 1 \cdot k = k$.
Similarly,$|B - C| = |\alpha(z_{2} - z_{3})| = k$ and $|C - A| = |\alpha(z_{3} - z_{1})| = k$.
Since the distances between the new points are equal,they form an equilateral triangle.

(A) The equation $|z-z_{1}|+|z-z_{2}|=k$ represents an ellipse if $k > |z_{1}-z_{2}|$.
In the given equation,$k = 2|z_{1}-z_{2}|$.
Since $2|z_{1}-z_{2}| > |z_{1}-z_{2}|$,the condition for an ellipse is satisfied.
Therefore,the locus of $z$ is an ellipse with foci at $z_{1}$ and $z_{2}$.

(C) The number of ways to select $3$ consonants from $7$ is given by ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ ways.
The number of ways to select $2$ vowels from $4$ is given by ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$ ways.
The total number of ways to select the $5$ letters is $35 \times 6 = 210$ ways.
Since these $5$ selected letters can be arranged among themselves in $5!$ ways,the total number of words is $210 \times 5! = 210 \times 120 = 25200$.

(A) We know that by the $AM-GM$ inequality,for positive real numbers $a$ and $b$,$\frac{a+b}{2} \geq \sqrt{ab}$.
Applying this to $2^{\sin x}$ and $2^{\cos x}$,we get:
$\frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x+\cos x}{2}} = 2^{1+\frac{\sin x+\cos x}{2}}$.
We know that the range of $\sin x+\cos x$ is $[-\sqrt{2}, \sqrt{2}]$.
To minimize the expression $2^{1+\frac{\sin x+\cos x}{2}}$,we must minimize the exponent $1+\frac{\sin x+\cos x}{2}$.
The minimum value of $\sin x+\cos x$ is $-\sqrt{2}$.
Substituting this into the expression,the minimum value is $2^{1+\frac{-\sqrt{2}}{2}} = 2^{1-\frac{1}{\sqrt{2}}}$.

(A) Given,$f(x) = x + \frac{1}{2} = \frac{2x+1}{2}$.
$f(2x) = 2x + \frac{1}{2} = \frac{4x+1}{2}$.
$f(4x) = 4x + \frac{1}{2} = \frac{8x+1}{2}$.
Since $f(x), f(2x), f(4x)$ are in $HP$,their reciprocals $\frac{1}{f(x)}, \frac{1}{f(2x)}, \frac{1}{f(4x)}$ are in $AP$.
Therefore,$\frac{2}{f(2x)} = \frac{1}{f(x)} + \frac{1}{f(4x)}$.
Substituting the values: $\frac{2}{(4x+1)/2} = \frac{2}{2x+1} + \frac{2}{8x+1}$.
$\frac{4}{4x+1} = 2 \left( \frac{8x+1 + 2x+1}{(2x+1)(8x+1)} \right)$.
$\frac{2}{4x+1} = \frac{10x+2}{(2x+1)(8x+1)}$.
$2(2x+1)(8x+1) = (4x+1)(10x+2)$.
$2(16x^2 + 10x + 1) = 40x^2 + 18x + 2$.
$32x^2 + 20x + 2 = 40x^2 + 18x + 2$.
$8x^2 - 2x = 0$.
$2x(4x - 1) = 0$.
This gives $x = 0$ or $x = 1/4$.
If $x = 0$,the terms are $1/2, 1/2, 1/2$,which are not unequal.
If $x = 1/4$,the terms are $3/4, 1, 3/2$,which are in $HP$ (since $4/3, 1, 2/3$ are in $AP$ with common difference $-1/3$).
Thus,there is only $1$ real value of $x$.

(C) The general term $T_n$ of the series $1, 8, 21, 40, 65, \ldots$ in the numerator can be found using the method of differences. The first differences are $7, 13, 19, 25, \ldots$,which form an arithmetic progression with common difference $6$.
Thus,$T_n = an^2 + bn + c$.
For $n=1, T_1 = a+b+c = 1$.
For $n=2, T_2 = 4a+2b+c = 8$.
For $n=3, T_3 = 9a+3b+c = 21$.
Solving these,we get $a=3, b=-2, c=0$. So,$T_n = 3n^2 - 2n = n(3n-2)$.
The series is $S = \sum_{n=1}^{\infty} \frac{n(3n-2)}{n!} = \sum_{n=1}^{\infty} \frac{3n^2-2n}{n!}$.
Since the first term is $1$ (for $n=1$),we can write $S = \sum_{n=1}^{\infty} \frac{3n-2}{(n-1)!}$.
Let $k = n-1$,then $n = k+1$.
$S = \sum_{k=0}^{\infty} \frac{3(k+1)-2}{k!} = \sum_{k=0}^{\infty} \frac{3k+1}{k!} = 3\sum_{k=1}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!} = 3\sum_{k=1}^{\infty} \frac{1}{(k-1)!} + e = 3e + e = 4e$.
Since $2 < e < 3$,we have $8 < 4e < 12$.
Therefore,$8 < S < 12$.

(A) The general term in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is $T_{r+1} = {}^{13}C_{r} (a x^{2})^{13-r} (b^{-1} x^{-1})^{r} = {}^{13}C_{r} a^{13-r} b^{-r} x^{26-3r}$.
For the coefficient of $x^{8}$,set $26-3r = 8$,which gives $3r = 18$,so $r = 6$.
The coefficient is ${}^{13}C_{6} a^{7} b^{-6}$.
The general term in $\left(a x-\frac{1}{b x^{2}}\right)^{13}$ is $T'_{r+1} = {}^{13}C_{r} (a x)^{13-r} (-b^{-1} x^{-2})^{r} = {}^{13}C_{r} a^{13-r} (-1)^{r} b^{-r} x^{13-3r}$.
For the coefficient of $x^{-8}$,set $13-3r = -8$,which gives $3r = 21$,so $r = 7$.
The coefficient is ${}^{13}C_{7} a^{6} (-1)^{7} b^{-7} = -{}^{13}C_{7} a^{6} b^{-7}$.
Equating the two coefficients:
${}^{13}C_{6} a^{7} b^{-6} = -{}^{13}C_{7} a^{6} b^{-7}$.
Since ${}^{13}C_{6} = {}^{13}C_{7}$,we have $a^{7} b^{-6} = -a^{6} b^{-7}$.
Dividing by $a^{6} b^{-6}$,we get $a = -b^{-1}$,which implies $ab = -1$,or $ab+1=0$.

(B) We can use partial fractions to decompose the expression: $\frac{2}{(1-x)(2-x)} = \frac{A}{1-x} + \frac{B}{2-x}$.
Solving for $A$ and $B$: $2 = A(2-x) + B(1-x)$.
For $x=1$,$2 = A(1) \implies A=2$.
For $x=2$,$2 = B(-1) \implies B=-2$.
So,$\frac{2}{(1-x)(2-x)} = \frac{2}{1-x} - \frac{2}{2-x} = 2(1-x)^{-1} - (1-\frac{x}{2})^{-1}$.
Expanding using the binomial series $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$:
$2(1 + x + x^2 + x^3 + \dots) - (1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots)$.
The coefficient of $x^3$ is $2(1) - \frac{1}{8} = 2 - \frac{1}{8} = \frac{16-1}{8} = \frac{15}{8}$.

(D) We know that the binomial expansion is given by:
$(1+x)^{n} = { }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}$
and $(x+1)^{n} = { }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}+{ }^{n} C_{2} x^{n-2}+\ldots+{ }^{n} C_{n}$.
Multiplying these two expressions,we get $(1+x)^{2n} = (1+x)^{n}(x+1)^{n}$.
The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is ${ }^{2n} C_{n}$.
By multiplying the two series,the coefficient of $x^{n}$ is the sum of the products of corresponding coefficients:
$({ }^{n} C_{0})^{2} + ({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n}$.
The given sum starts from $({ }^{n} C_{1})^{2}$,so we subtract the first term $({ }^{n} C_{0})^{2} = 1$:
$({ }^{n} C_{1})^{2} + ({ }^{n} C_{2})^{2} + \ldots + ({ }^{n} C_{n})^{2} = { }^{2n} C_{n} - ({ }^{n} C_{0})^{2} = { }^{2n} C_{n} - 1$.

(C) We know that $\sin(k\pi) = 0$ for any integer $k$.
The given series is $E = \sum_{n=1}^{\infty} \sin \left(\frac{n! \pi}{720}\right)$.
Expanding the terms:
$n=1: \sin \left(\frac{1! \pi}{720}\right) = \sin \left(\frac{\pi}{720}\right)$
$n=2: \sin \left(\frac{2! \pi}{720}\right) = \sin \left(\frac{2 \pi}{720}\right) = \sin \left(\frac{\pi}{360}\right)$
$n=3: \sin \left(\frac{3! \pi}{720}\right) = \sin \left(\frac{6 \pi}{720}\right) = \sin \left(\frac{\pi}{120}\right)$
$n=4: \sin \left(\frac{4! \pi}{720}\right) = \sin \left(\frac{24 \pi}{720}\right) = \sin \left(\frac{\pi}{30}\right)$
$n=5: \sin \left(\frac{5! \pi}{720}\right) = \sin \left(\frac{120 \pi}{720}\right) = \sin \left(\frac{\pi}{6}\right)$
$n=6: \sin \left(\frac{6! \pi}{720}\right) = \sin \left(\frac{720 \pi}{720}\right) = \sin(\pi) = 0$
For all $n \ge 6$,$n!$ is a multiple of $720$,so $\frac{n! \pi}{720}$ is a multiple of $\pi$,making $\sin \left(\frac{n! \pi}{720}\right) = 0$.
Thus,the sum is $\sin \left(\frac{\pi}{720}\right) + \sin \left(\frac{\pi}{360}\right) + \sin \left(\frac{\pi}{120}\right) + \sin \left(\frac{\pi}{30}\right) + \sin \left(\frac{\pi}{6}\right)$.

(A) We know that $\cot \theta = \frac{1}{\tan \theta}$. Also,$\cot \frac{4 \pi}{5} = \cot (\pi - \frac{\pi}{5}) = -\cot \frac{\pi}{5}$.
Using the identity $\tan \theta + \cot \theta = \frac{2}{\sin 2 \theta}$,we simplify the expression.
Let $S = \tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} + 4 \cot \frac{4 \pi}{5}$.
Note that $4 \cot \frac{4 \pi}{5} = -4 \cot \frac{\pi}{5}$.
So,$S = \tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} - 4 \cot \frac{\pi}{5}$.
Using $\tan \theta - \cot \theta = -2 \cot 2 \theta$,we have $\tan \frac{\pi}{5} - \cot \frac{\pi}{5} = -2 \cot \frac{2 \pi}{5}$.
Thus,$S = -2 \cot \frac{2 \pi}{5} + 2 \tan \frac{2 \pi}{5} - 3 \cot \frac{\pi}{5}$ (This approach is complex).
Alternatively,using the identity $\tan \theta + 2 \tan 2 \theta + 4 \tan 4 \theta + 8 \cot 8 \theta = \cot \theta$,we can evaluate the expression.
For the given expression,it simplifies to $\cot \frac{\pi}{5}$.

(C) Let $S = \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7}$.
Multiply by $2 \sin \frac{\pi}{7}$:
$2 \sin \frac{\pi}{7} S = 2 \sin \frac{\pi}{7} \cos \frac{2 \pi}{7} + 2 \sin \frac{\pi}{7} \cos \frac{4 \pi}{7} + 2 \sin \frac{\pi}{7} \cos \frac{6 \pi}{7}$.
Using $2 \sin A \cos B = \sin(A+B) - \sin(B-A)$:
$2 \sin \frac{\pi}{7} S = (\sin \frac{3 \pi}{7} - \sin \frac{\pi}{7}) + (\sin \frac{5 \pi}{7} - \sin \frac{3 \pi}{7}) + (\sin \frac{7 \pi}{7} - \sin \frac{5 \pi}{7})$.
All terms cancel out except $\sin \frac{7 \pi}{7} - \sin \frac{\pi}{7}$.
Since $\sin \pi = 0$,we have $2 \sin \frac{\pi}{7} S = -\sin \frac{\pi}{7}$.
Since $\sin \frac{\pi}{7} \neq 0$,we get $S = -\frac{1}{2}$.
Thus,the sum is a negative number.

(A) The given lines are $L_1: x+y=0$,$L_2: 5x+y=4$,and $L_3: x+5y=4$.
Solving $L_1$ and $L_2$: $x+y=0 \implies y=-x$. Substituting into $L_2$: $5x-x=4 \implies 4x=4 \implies x=1, y=-1$. So,$A = (1, -1)$.
Solving $L_1$ and $L_3$: $x+y=0 \implies x=-y$. Substituting into $L_3$: $-y+5y=4 \implies 4y=4 \implies y=1, x=-1$. So,$C = (-1, 1)$.
Solving $L_2$ and $L_3$: $5x+y=4$ and $x+5y=4$. Subtracting: $4x-4y=0 \implies x=y$. Substituting into $L_2$: $5x+x=4 \implies 6x=4 \implies x=2/3, y=2/3$. So,$B = (2/3, 2/3)$.
Calculating side lengths:
$AB = \sqrt{(2/3 - 1)^2 + (2/3 + 1)^2} = \sqrt{(-1/3)^2 + (5/3)^2} = \sqrt{1/9 + 25/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.
$BC = \sqrt{(-1 - 2/3)^2 + (1 - 2/3)^2} = \sqrt{(-5/3)^2 + (1/3)^2} = \sqrt{25/9 + 1/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.
$CA = \sqrt{(-1 - 1)^2 + (1 - (-1))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.
Since $AB = BC$,the triangle is an isosceles triangle.

(D) The condition that a circle $x^{2}+y^{2}+2gx+2fy+c=0$ cuts another circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$ at the extremities of its diameter is equivalent to the condition of orthogonality: $2(gg_{i}+ff_{i})=c+c_{i}$.
For the first circle $x^{2}+y^{2}-5=0$,$g_{1}=0, f_{1}=0, c_{1}=-5$. Applying the condition: $2(g(0)+f(0))=c-5 \Rightarrow c=5$.
For the second circle $x^{2}+y^{2}-8x-6y+10=0$,$g_{2}=-4, f_{2}=-3, c_{2}=10$. Applying the condition: $2(g(-4)+f(-3))=c+10$ $\Rightarrow 2(-4g-3f)=5+10$ $\Rightarrow 4g+3f=-15/2$.
For the third circle $x^{2}+y^{2}-4x+2y-2=0$,$g_{3}=-2, f_{3}=1, c_{3}=-2$. Applying the condition: $2(g(-2)+f(1))=c-2$ $\Rightarrow 2(-2g+f)=5-2$ $\Rightarrow -2g+f=3/2$.
Solving the system of equations $4g+3f=-7.5$ and $-2g+f=1.5$:
From the second equation,$f=2g+1.5$. Substituting into the first: $4g+3(2g+1.5)=-7.5$ $\Rightarrow 10g+4.5=-7.5$ $\Rightarrow 10g=-12$ $\Rightarrow g=-1.2$.
Then $f=2(-1.2)+1.5 = -2.4+1.5 = -0.9$.
Checking option $D$: $4f = 4(-0.9) = -3.6$ and $3g = 3(-1.2) = -3.6$. Thus,$4f=3g$ is correct.

(A) Given,$y=[|\sin x|+|\cos x|]$ and $x^{2}+y^{2}=10$.
We know that $(|\sin x|+|\cos x|) \in [1, \sqrt{2}]$.
Since $[x]$ is the greatest integer function,$y = [|\sin x|+|\cos x|] = 1$.
The point of intersection of the given curves is found by substituting $y=1$ into $x^{2}+y^{2}=10$:
$x^{2}+1^{2}=10$ $\Rightarrow x^{2}=9$ $\Rightarrow x=\pm 3$.
The points of intersection are $(3, 1)$ and $(-3, 1)$.
For the circle $x^{2}+y^{2}=10$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$.
At point $(-3, 1)$,the slope $m_{1} = -(-3)/1 = 3$.
The curve $y=1$ is a horizontal line,so its slope $m_{2} = 0$.
The angle $\theta$ between the curves is given by $\tan \theta = |\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}| = |\frac{3-0}{1+3(0)}| = 3$.
Therefore,$\theta = \tan^{-1}(3)$.

(B) The equation of the parabola is $y^{2}=12x$. Comparing with $y^{2}=4ax$,we get $4a=12$,so $a=3$.
The slope of the given line $y=4x+3$ is $m=4$.
The slope of the normal to the parabola $y^{2}=4ax$ at point $(x_{1}, y_{1})$ is given by $m_{n} = -\frac{y_{1}}{2a}$.
Since the normal is parallel to the line $y=4x+3$,its slope must be $4$. Thus,$-\frac{y_{1}}{2(3)} = 4$,which gives $y_{1} = -24$.
Substituting $y_{1} = -24$ into the parabola equation $y^{2}=12x$,we get $(-24)^{2} = 12x$,so $576 = 12x$,which gives $x_{1} = 48$.
The point on the parabola is $(48, -24)$.
The equation of the normal at $(48, -24)$ is $y - (-24) = 4(x - 48)$,which simplifies to $y + 24 = 4x - 192$,or $4x - y - 216 = 0$.
The given line is $4x - y + 3 = 0$.
The distance between two parallel lines $Ax + By + C_{1} = 0$ and $Ax + By + C_{2} = 0$ is given by $d = \frac{|C_{1} - C_{2}|}{\sqrt{A^{2} + B^{2}}}$.
Here,$d = \frac{|3 - (-216)|}{\sqrt{4^{2} + (-1)^{2}}} = \frac{|3 + 216|}{\sqrt{16 + 1}} = \frac{219}{\sqrt{17}}$.

(A) The given equation of the parabola is $y^{2} = 64x$ $(i)$.
The point on the parabola nearest to the line $4x + 3y + 35 = 0$ is the point where the tangent is parallel to the given line.
The slope of the line $4x + 3y + 35 = 0$ is $m = -\frac{4}{3}$.
Differentiating $(i)$ with respect to $x$,we get $2y \frac{dy}{dx} = 64$,which implies $\frac{dy}{dx} = \frac{32}{y}$.
Since the tangent is parallel to the line,their slopes must be equal:
$\frac{32}{y} = -\frac{4}{3} \Rightarrow y = -24$.
Substituting $y = -24$ into $(i)$:
$(-24)^{2} = 64x$ $\Rightarrow 576 = 64x$ $\Rightarrow x = 9$.
Thus,the required point is $(9, -24)$.

(A) Given that $\alpha$ and $\beta$ are the roots of $f(x) = x^{2} + bx + c = 0$.
By the properties of quadratic equations,the sum of roots is $\alpha + \beta = -b$.
The vertex of the parabola $y = f(x)$ occurs at $x = -\frac{b}{2} = \frac{\alpha + \beta}{2}$.
Now,find the derivative of $f(x)$:
$f'(x) = \frac{dy}{dx} = 2x + b$.
At the point $x = \frac{\alpha + \beta}{2} = -\frac{b}{2}$,the slope of the tangent is:
$f'\left(-\frac{b}{2}\right) = 2\left(-\frac{b}{2}\right) + b = -b + b = 0$.
$A$ slope of $0$ indicates that the tangent line is horizontal,meaning it is parallel to the $x$-axis.
Therefore,the angle between the tangent and the positive direction of the $x$-axis is $0^{\circ}$.

(B) The given equation is $(7x+5)^{2}+(7y+3)^{2}=\lambda^{2}(4x+3y-24)^{2}$.
This is of the form $SP^{2} = \lambda^{2} PM^{2}$,where $S$ is the focus,$P$ is a point on the curve,and $PM$ is the perpendicular distance from $P$ to the directrix.
For the curve to be a parabola,the eccentricity $e$ must be equal to $1$.
Here,the distance from the point $(x, y)$ to the focus $S$ is $\sqrt{(x - x_s)^2 + (y - y_s)^2}$.
Rewriting the equation: $\frac{(7x+5)^{2}+(7y+3)^{2}}{49} = \lambda^{2} \frac{(4x+3y-24)^{2}}{49}$.
$\left(\sqrt{(x + 5/7)^2 + (y + 3/7)^2}\right)^2 = \lambda^{2} \left(\frac{4x+3y-24}{7}\right)^2$.
Comparing this with the definition of a conic $SP = e PM$,we have $e = \lambda \cdot \frac{\sqrt{4^2+3^2}}{7} = \lambda \cdot \frac{5}{7}$.
For a parabola,$e = 1$,so $\lambda \cdot \frac{5}{7} = 1$.
Thus,$\lambda = \pm \frac{7}{5}$.

(C) The given equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$.
Here,$a^{2}=144$ and $b^{2}=25$.
The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}}=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}$.
The foci of the ellipse are $(\pm ae, 0) = (\pm 12 \times \frac{\sqrt{119}}{12}, 0) = (\pm \sqrt{119}, 0)$.
The circle has its center at $(0, \sqrt{2})$ and passes through the foci $(\pm \sqrt{119}, 0)$.
The radius $r$ of the circle is the distance between the center $(0, \sqrt{2})$ and one of the foci $(\sqrt{119}, 0)$.
$r = \sqrt{(\sqrt{119}-0)^{2} + (0-\sqrt{2})^{2}}$
$r = \sqrt{119 + 2} = \sqrt{121} = 11$.

(A) Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
Given,foci are $(\pm 8, 0) = (\pm ae, 0)$,so $ae = 8$.
The length of the latus rectum is $\frac{2b^{2}}{a} = 24$,which implies $b^{2} = 12a$.
We know that $b^{2} = a^{2}(e^{2} - 1) = a^{2}e^{2} - a^{2}$.
Substituting $ae = 8$ and $b^{2} = 12a$,we get $12a = 8^{2} - a^{2}$.
$a^{2} + 12a - 64 = 0$.
$(a + 16)(a - 4) = 0$.
Since $a > 0$,we have $a = 4$.
Then $b^{2} = 12(4) = 48$.
Substituting $a^{2} = 16$ and $b^{2} = 48$ into the standard equation: $\frac{x^{2}}{16} - \frac{y^{2}}{48} = 1$.
Multiplying by $48$,we get $3x^{2} - y^{2} = 48$.

(A) The equation of a tangent to the parabola $y^{2}=8 \sqrt{3} x$ in slope form is $y=mx+c$,where $c=\frac{a}{m}$.
Here,$4a=8 \sqrt{3}$,so $a=2 \sqrt{3}$.
Thus,$c=\frac{2 \sqrt{3}}{m}$.
For the hyperbola $4x^{2}-y^{2}=4$,we have $\frac{x^{2}}{1}-\frac{y^{2}}{4}=1$,where $a^{2}=1$ and $b^{2}=4$.
The condition for the line $y=mx+c$ to be a tangent to the hyperbola is $c^{2}=a^{2}m^{2}-b^{2}$.
Substituting the values,$c^{2}=m^{2}-4$.
Equating the two expressions for $c^{2}$:
$\left(\frac{2 \sqrt{3}}{m}\right)^{2}=m^{2}-4$
$\frac{12}{m^{2}}=m^{2}-4$
$m^{4}-4m^{2}-12=0$
$(m^{2}-6)(m^{2}+2)=0$.
Since $m^{2}=-2$ is not possible,we have $m^{2}=6$,so $m=\sqrt{6}$ (as the slope is positive).
Then $c=\frac{2 \sqrt{3}}{\sqrt{6}}=\sqrt{2}$.
Therefore,the equation of the tangent is $y=\sqrt{6}x+\sqrt{2}$.

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x} = 1$.
Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + \dots$ and $\sin 2x = 2x - \frac{8x^3}{6} + \dots$ and $\tan x = x + \frac{x^3}{3} + \dots$:
$\lim _{x \rightarrow 0} \frac{2 a(x - \frac{x^3}{6}) - (2x - \frac{8x^3}{6})}{x^3} = 1$.
$\lim _{x \rightarrow 0} \frac{(2a - 2)x + (\frac{8}{6} - \frac{2a}{6})x^3}{x^3} = 1$.
For the limit to exist,the coefficient of $x$ must be $0$,so $2a - 2 = 0$,which gives $a = 1$.
Substituting $a = 1$ into the expression: $\lim _{x \rightarrow 0} \frac{(8/6 - 2/6)x^3}{x^3} = \frac{6/6}{1} = 1$.
Thus,the value of $a$ is $1$.

(D) Given,$f^{\prime}(4)=5$.
We need to evaluate the limit $L = \lim _{x \rightarrow 2} \frac{f(4)-f\left(x^{2}\right)}{x-2}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$.
$L = \lim _{x \rightarrow 2} \frac{\frac{d}{dx}[f(4)-f(x^2)]}{\frac{d}{dx}[x-2]}$
$L = \lim _{x \rightarrow 2} \frac{0 - f^{\prime}(x^2) \cdot 2x}{1}$
Substituting $x=2$,we get:
$L = -f^{\prime}(2^2) \cdot 2(2) = -f^{\prime}(4) \cdot 4$
Since $f^{\prime}(4)=5$,we have $L = -(5) \cdot 4 = -20$.

(C) We know that for any real number $x$,$x-1 < [x] \leq x$.
Substituting $x = n\sqrt{2}$,we get $n\sqrt{2}-1 < [n\sqrt{2}] \leq n\sqrt{2}$.
Dividing the entire inequality by $n$ (where $n > 0$),we get $\frac{n\sqrt{2}-1}{n} < \frac{[n\sqrt{2}]}{n} \leq \frac{n\sqrt{2}}{n}$.
This simplifies to $\sqrt{2} - \frac{1}{n} < \frac{[n\sqrt{2}]}{n} \leq \sqrt{2}$.
Taking the limit as $n \rightarrow \infty$,we have $\lim_{n \rightarrow \infty} (\sqrt{2} - \frac{1}{n}) = \sqrt{2}$ and $\lim_{n \rightarrow \infty} \sqrt{2} = \sqrt{2}$.
By the Sandwich Theorem,$\lim_{n \rightarrow \infty} \frac{[n\sqrt{2}]}{n} = \sqrt{2}$.

(B) Let the numerator of the $n^{th}$ term be $a_{n} = 1, 10, 21, 34, 49, \ldots$.
This is a quadratic sequence. The first differences are $9, 11, 13, 15, \ldots$,which form an arithmetic progression.
The general term $a_{n}$ is given by $a_{n} = An^{2} + Bn + C$.
For $n=1, a_{1} = A + B + C = 1$.
For $n=2, a_{2} = 4A + 2B + C = 10$.
For $n=3, a_{3} = 9A + 3B + C = 21$.
Solving these equations: $(a_{2}-a_{1}) = 3A + B = 9$ and $(a_{3}-a_{2}) = 5A + B = 11$.
Subtracting gives $2A = 2 \Rightarrow A = 1$.
Then $3(1) + B = 9 \Rightarrow B = 6$.
Then $1 + 6 + C = 1 \Rightarrow C = -6$.
So,$a_{n} = n^{2} + 6n - 6$.
The $n^{th}$ term of the series is $t_{n} = \frac{n^{2} + 6n - 6}{n!}$.
As $n \rightarrow \infty$,the factorial $n!$ grows much faster than the polynomial $n^{2} + 6n - 6$.
Therefore,$\lim _{n \rightarrow \infty} t_{n} = \lim _{n \rightarrow \infty} \frac{n^{2} + 6n - 6}{n!} = 0$.

(A) Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $a = k \sin A, b = k \sin B, c = k \sin C$.
The expression is $\sum a^{3} \sin (B-C) = k^{3} \sum \sin^{3} A \sin (B-C)$.
Since $A = 180^{\circ} - (B+C)$,$\sin A = \sin (B+C)$.
Thus,the expression becomes $k^{3} \sum \sin^{2} A \sin (B+C) \sin (B-C)$.
Using $\sin (B+C) \sin (B-C) = \sin^{2} B - \sin^{2} C$,we get:
$k^{3} [\sin^{2} A (\sin^{2} B - \sin^{2} C) + \sin^{2} B (\sin^{2} C - \sin^{2} A) + \sin^{2} C (\sin^{2} A - \sin^{2} B)]$.
Expanding this,we get $k^{3} [\sin^{2} A \sin^{2} B - \sin^{2} A \sin^{2} C + \sin^{2} B \sin^{2} C - \sin^{2} B \sin^{2} A + \sin^{2} C \sin^{2} A - \sin^{2} C \sin^{2} B] = 0$.

(A) Let $S$,$D$,and $P$ denote the sets of persons who like singing,dancing,and painting,respectively.
Given: $n(S \cup D \cup P) = 265$,$n(S) = 200$,$n(D) = 110$,$n(P) = 55$,$n(S \cap D) = 60$,$n(S \cap P) = 30$,and $n(S \cap D \cap P) = 10$.
Using the principle of inclusion-exclusion:
$n(S \cup D \cup P) = n(S) + n(D) + n(P) - n(S \cap D) - n(S \cap P) - n(D \cap P) + n(S \cap D \cap P)$
$265 = 200 + 110 + 55 - 60 - 30 - n(D \cap P) + 10$
$265 = 285 - n(D \cap P)$
$n(D \cap P) = 285 - 265 = 20$
The number of persons who like only dancing and painting is given by $n(D \cap P) - n(S \cap D \cap P)$.
Therefore,$20 - 10 = 10$.

(B) Given that the number of elements in set $A$ is $n(A) = p$ and the number of elements in set $B$ is $n(B) = q$.
The total number of elements in the Cartesian product $A \times B$ is $n(A \times B) = n(A) \times n(B) = pq$.
$A$ relation from set $A$ to set $B$ is a subset of $A \times B$.
The total number of subsets of a set with $n$ elements is $2^n$.
Therefore,the total number of relations from set $A$ to set $B$ is $2^{pq}$.

(A) Given,$X_{n} = \{z = x + iy : |z|^{2} \leq \frac{1}{n}\}$.
This represents a disk centered at the origin with radius $\frac{1}{\sqrt{n}}$.
For $n = 1$,$X_{1} = \{x^{2} + y^{2} \leq 1\}$.
For $n = 2$,$X_{2} = \{x^{2} + y^{2} \leq \frac{1}{2}\}$.
As $n \to \infty$,the radius $\frac{1}{\sqrt{n}} \to 0$.
The intersection of all such sets $X_{n}$ for $n \geq 1$ is the set of points that satisfy $x^{2} + y^{2} \leq \frac{1}{n}$ for all $n \geq 1$.
This implies $x^{2} + y^{2} \leq 0$.
Since $x^{2} + y^{2}$ cannot be negative,the only solution is $x^{2} + y^{2} = 0$,which means $x = 0$ and $y = 0$.
Thus,$\bigcap_{n=1}^{\infty} X_{n} = \{0 + 0i\} = \{0\}$.
Therefore,the intersection is a singleton set.

(A) The total number of ways to choose $5$ cards from $52$ is $^{52}C_5 = 2598960$.
To form a full house (a pair and a triple):
$1$. Choose the face value for the triple: $^{13}C_1 = 13$ ways.
$2$. Choose $3$ cards of that face value: $^4C_3 = 4$ ways.
$3$. Choose the face value for the pair from the remaining $12$ values: $^{12}C_1 = 12$ ways.
$4$. Choose $2$ cards of that face value: $^4C_2 = 6$ ways.
Total favorable outcomes = $13 \times 4 \times 12 \times 6 = 3744$.
Probability = $\frac{3744}{2598960} = \frac{6}{4165}$.

(A) Let $S = 1! + 2! + 3! + 4! + \ldots + 11!$.
We know that for any $n \ge 4$,$n!$ contains $4 \times 3 = 12$ as a factor.
Therefore,$4!, 5!, \ldots, 11!$ are all divisible by $12$.
Thus,the remainder of $S$ when divided by $12$ is the same as the remainder of $(1! + 2! + 3!)$ when divided by $12$.
$1! + 2! + 3! = 1 + 2 + 6 = 9$.
Since $9 < 12$,the remainder is $9$.

(C) Given $f(x) = \frac{x-2}{x-3}$. Let $y = \frac{x-2}{x-3}$.
$y(x-3) = x-2 \implies yx - 3y = x - 2 \implies x(y-1) = 3y-2 \implies x = \frac{3y-2}{y-1}$.
Thus,$f^{-1}(x) = \frac{3x-2}{x-1}$.
Given $g(x) = 2x-3$. Let $y = 2x-3$.
$y+3 = 2x \implies x = \frac{y+3}{2}$.
Thus,$g^{-1}(x) = \frac{x+3}{2}$.
We are given $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$.
$\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$.
Multiply by $2(x-1)$: $2(3x-2) + (x+3)(x-1) = 13(x-1)$.
$6x - 4 + x^2 + 2x - 3 = 13x - 13$.
$x^2 + 8x - 7 = 13x - 13$.
$x^2 - 5x + 6 = 0$.
$(x-2)(x-3) = 0$.
The roots are $x = 2$ and $x = 3$. However,$f^{-1}(x)$ is defined for $x \in R - \{1\}$. Since $x=3$ is in the domain,we check the equation: $f^{-1}(3) + g^{-1}(3) = \frac{3(3)-2}{3-1} + \frac{3+3}{2} = \frac{7}{2} + 3 = \frac{13}{2}$. Both values are valid.
The sum of the values of $x$ is $2 + 3 = 5$.

(C) Given,$f(x) = 2x^2 + 5x + 1$.
Also,$f(x) = a(x+1)(x-2) + b(x-2)(x-1) + c(x-1)(x+1)$.
Expanding the right side:
$f(x) = a(x^2 - x - 2) + b(x^2 - 3x + 2) + c(x^2 - 1)$
$f(x) = (a+b+c)x^2 + (-a-3b)x + (-2a+2b-c)$.
On equating the coefficients of $x^2$,$x$,and the constant term,we get a system of linear equations:
$1) \; a + b + c = 2$
$2) \; -a - 3b = 5$
$3) \; -2a + 2b - c = 1$
Since we have $3$ linear equations in $3$ variables $(a, b, c)$ and the determinant of the coefficient matrix is non-zero,there exists a unique solution for $a, b, c$.
Solving these,we get $a = -\frac{35}{4}$,$b = \frac{5}{4}$,and $c = \frac{38}{4}$.
Thus,there is exactly one choice for each of $a, b, c$.

(B) We know that $(1+x)^{n} = {}^{n}C_{0} + x{}^{n}C_{1} + x^{2}{}^{n}C_{2} + \ldots + x^{n}{}^{n}C_{n}$.
Integrating both sides with respect to $x$ from $0$ to $2$,we get:
$\int_{0}^{2} (1+x)^{n} dx = \int_{0}^{2} \left( {}^{n}C_{0} + x{}^{n}C_{1} + x^{2}{}^{n}C_{2} + \ldots + x^{n}{}^{n}C_{n} \right) dx$.
Evaluating the integral on the left side:
$\left[ \frac{(1+x)^{n+1}}{n+1} \right]_{0}^{2} = \frac{(1+2)^{n+1}}{n+1} - \frac{(1+0)^{n+1}}{n+1} = \frac{3^{n+1}-1}{n+1}$.
Evaluating the integral on the right side:
$\left[ x{}^{n}C_{0} + \frac{x^{2}}{2}{}^{n}C_{1} + \frac{x^{3}}{3}{}^{n}C_{2} + \ldots + \frac{x^{n+1}}{n+1}{}^{n}C_{n} \right]_{0}^{2} = \frac{2}{1}{}^{n}C_{0} + \frac{2^{2}}{2}{}^{n}C_{1} + \frac{2^{3}}{3}{}^{n}C_{2} + \ldots + \frac{2^{n+1}}{n+1}{}^{n}C_{n}$.
Thus,$S = \frac{3^{n+1}-1}{n+1}$.

(A) Let $L = \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L$' Hospital's rule.
Using the Leibniz rule for differentiation under the integral sign,the derivative of the numerator is $\cos(x^4) \cdot \frac{d}{dx}(x^2) = 2x \cos(x^4)$.
The derivative of the denominator $x \sin x$ is $\sin x + x \cos x$.
So,$L = \lim _{x \rightarrow 0} \frac{2x \cos(x^4)}{\sin x + x \cos x}$.
Divide numerator and denominator by $x$:
$L = \lim _{x \rightarrow 0} \frac{2 \cos(x^4)}{\frac{\sin x}{x} + \cos x}$.
As $x \rightarrow 0$,$\frac{\sin x}{x} \rightarrow 1$,$\cos(x^4) \rightarrow 1$,and $\cos x \rightarrow 1$.
Therefore,$L = \frac{2(1)}{1 + 1} = \frac{2}{2} = 1$.

(D) Given relation is defined as $\theta R \phi$ such that $\sec^{2} \theta - \tan^{2} \phi = 1$.
$1$. Reflexive: For any $\theta$,$\theta R \theta$ implies $\sec^{2} \theta - \tan^{2} \theta = 1$. Since $1 + \tan^{2} \theta = \sec^{2} \theta$,this is $1 = 1$,which is true. Thus,$R$ is reflexive.
$2$. Symmetric: If $\theta R \phi$,then $\sec^{2} \theta - \tan^{2} \phi = 1$. Using $\sec^{2} \theta = 1 + \tan^{2} \theta$ and $\tan^{2} \phi = \sec^{2} \phi - 1$,we get $(1 + \tan^{2} \theta) - (\sec^{2} \phi - 1) = 1$,which simplifies to $\tan^{2} \theta - \sec^{2} \phi = -1$,or $\sec^{2} \phi - \tan^{2} \theta = 1$. This implies $\phi R \theta$. Thus,$R$ is symmetric.
$3$. Transitive: If $\theta R \phi$ and $\phi R \psi$,then $\sec^{2} \theta - \tan^{2} \phi = 1$ and $\sec^{2} \phi - \tan^{2} \psi = 1$. Adding these,$\sec^{2} \theta - \tan^{2} \phi + \sec^{2} \phi - \tan^{2} \psi = 2$. Since $\sec^{2} \phi - \tan^{2} \phi = 1$,we have $\sec^{2} \theta - \tan^{2} \psi + 1 = 2$,which means $\sec^{2} \theta - \tan^{2} \psi = 1$. This implies $\theta R \psi$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) Let the relation be defined as $R = \{(A, B) : B = P A Q^{-1}\}$ where $P$ and $Q$ are invertible matrices.
For reflexive: Since $A = I A I^{-1}$ where $I$ is the identity matrix (which is invertible),$(A, A) \in R$. Thus,$R$ is reflexive.
For symmetric: Let $(A, B) \in R$. Then $B = P A Q^{-1}$. Multiplying by $P^{-1}$ on the left and $Q$ on the right,we get $P^{-1} B Q = A$. Since $P^{-1}$ and $Q$ are invertible,let $P' = P^{-1}$ and $Q' = Q^{-1}$. Then $A = P' B (Q')^{-1}$,so $(B, A) \in R$. Thus,$R$ is symmetric.
For transitive: Let $(A, B) \in R$ and $(B, C) \in R$. Then $A = P_1 B Q_1^{-1}$ and $B = P_2 C Q_2^{-1}$. Substituting $B$,we get $A = P_1 (P_2 C Q_2^{-1}) Q_1^{-1} = (P_1 P_2) C (Q_1 Q_2)^{-1}$. Since the product of invertible matrices is invertible,$(A, C) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(A) The matrix $A$ represents a rotation matrix in $3D$ space about the $z$-axis by an angle $\theta = \frac{2\pi}{n}$.
By the property of rotation matrices,$A^k = \begin{bmatrix} \cos (k\theta) & \sin (k\theta) & 0 \\ -\sin (k\theta) & \cos (k\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
For $A^n$,we have $k = n$,so $k\theta = n \times \frac{2\pi}{n} = 2\pi$.
Thus,$A^n = \begin{bmatrix} \cos (2\pi) & \sin (2\pi) & 0 \\ -\sin (2\pi) & \cos (2\pi) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
For $A^{n-1}$,the angle is $(n-1) \times \frac{2\pi}{n} = 2\pi - \frac{2\pi}{n}$.
Since $n \geq 2$,$\frac{2\pi}{n}$ is not a multiple of $2\pi$,so $A^{n-1} \neq I$.
Therefore,the correct option is $A$.

(B) The identity matrix $I$ is given by $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Rearranging the columns of $I$ results in permutation matrices.
For a $3 \times 3$ matrix,the number of ways to rearrange the $3$ columns is given by $3! = 3 \times 2 \times 1 = 6$.
Thus,there are $6$ distinct choices for $P$.
Since $P$ is a permutation matrix,its determinant is equal to the sign of the permutation,which is either $1$ or $-1$.
Therefore,$\operatorname{det}(P) = \pm 1$.
Hence,there are six distinct choices for $P$ and $\operatorname{det}(P) = \pm 1$.

(A) Given,$\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} \quad ...(i)$
We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$.
Therefore,$\operatorname{cosec}^{-1}\left(\frac{13}{12}\right) = \sin^{-1}\left(\frac{12}{13}\right)$.
Substituting this into equation $(i)$,we get:
$\sin ^{-1}\left(\frac{x}{13}\right)+\sin ^{-1}\left(\frac{12}{13}\right)=\frac{\pi}{2}$.
We know the identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$.
Also,$\sin^{-1}(\frac{12}{13}) = \cos^{-1}(\sqrt{1 - (\frac{12}{13})^2}) = \cos^{-1}(\sqrt{1 - \frac{144}{169}}) = \cos^{-1}(\sqrt{\frac{25}{169}}) = \cos^{-1}(\frac{5}{13})$.
So,$\sin^{-1}(\frac{x}{13}) + \cos^{-1}(\frac{5}{13}) = \frac{\pi}{2}$.
Comparing this with $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$,we must have $\frac{x}{13} = \frac{5}{13}$.
Thus,$x = 5$.

(B) Given the function $f(x) = 3x^2 + 1$.
We need to find the set $f^{-1}([1, 6])$,which consists of all $x$ such that $f(x) \in [1, 6]$.
So,$1 \le 3x^2 + 1 \le 6$.
Subtracting $1$ from all parts: $0 \le 3x^2 \le 5$.
Dividing by $3$: $0 \le x^2 \le \frac{5}{3}$.
Taking the square root: $-\sqrt{\frac{5}{3}} \le x \le \sqrt{\frac{5}{3}}$.
Thus,the set is $[ -\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}} ]$.

(C) Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$.
For the function to be defined,we must have $\frac{\pi^{2}}{16}-x^{2} \geq 0$,which implies $x^{2} \leq \frac{\pi^{2}}{16}$,so $x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$.
Let $u = \sqrt{\frac{\pi^{2}}{16}-x^{2}}$. As $x$ varies from $0$ to $\frac{\pi}{4}$,$u$ varies from $\frac{\pi}{4}$ to $0$.
Thus,the range of $u$ is $\left[0, \frac{\pi}{4}\right]$.
Now,$y = 3 \sin(u)$. Since $u \in \left[0, \frac{\pi}{4}\right]$,$\sin(u)$ varies from $\sin(0) = 0$ to $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.
Therefore,$y$ varies from $3 \times 0 = 0$ to $3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}}$.
Hence,the range of the function is $\left[0, \frac{3}{\sqrt{2}}\right]$.

(D) The given function is $f(x) = x^{2} + bx + c$.
This is a quadratic function representing a parabola.
For a function to be one-to-one,$f(x_1) = f(x_2)$ must imply $x_1 = x_2$.
Here,$f(x) = (x + \frac{b}{2})^2 + (c - \frac{b^2}{4})$.
Since the square term is always non-negative,$f(x_1) = f(x_2)$ does not necessarily imply $x_1 = x_2$ (e.g.,$f(x) = x^2$ gives $f(1) = f(-1) = 1$),so it is many-to-one.
Furthermore,the range of this function is $[c - \frac{b^2}{4}, \infty)$,which is not equal to the codomain $\mathbb{R}$ (assuming the codomain is $\mathbb{R}$),so it is not onto.
Therefore,the function is neither one-to-one nor onto.

(A) Given,$f(x)=\frac{\tan \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$.
Since $[x-\frac{\pi}{2}]$ is an integer for all $x$,let $[x-\frac{\pi}{2}] = k$,where $k \in \mathbb{Z}$.
Then the numerator becomes $\tan(\pi k)$,which is equal to $0$ for all integers $k$.
Since the denominator $2+[x]^{2}$ is always $\geq 2$ and never zero,the function simplifies to $f(x) = \frac{0}{2+[x]^{2}} = 0$ for all $x \in \mathbb{R}$.
$A$ constant function $f(x) = 0$ is continuous and differentiable for all values of $x$.
Therefore,the function is continuous for all values of $x$.

(C) For $f(x) = a \sin |x| + b e^{|x|}$ to be differentiable at $x = 0$,the left-hand derivative $(LHD)$ and right-hand derivative $(RHD)$ must be equal.
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{a \sin |-h| + b e^{|-h|} - (a \sin 0 + b e^0)}{h} = \lim_{h \to 0} \frac{a \sin(-h) + b e^{-h} - b}{h} = \lim_{h \to 0} \frac{-a \sin h + b(e^{-h} - 1)}{h} = -a - b$.
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{a \sin h + b e^h - b}{h} = \lim_{h \to 0} \frac{a \sin h + b(e^h - 1)}{h} = a + b$.
For differentiability,$LHD = RHD \implies -a - b = a + b \implies 2a + 2b = 0 \implies a + b = 0$.

(C) We are given $g(x) = x f^{\prime}(x)$.
To find $g^{\prime}(0)$,we use the definition of the derivative:
$g^{\prime}(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h}$.
Since $g(0) = 0 \cdot f^{\prime}(0) = 0$,we have:
$g^{\prime}(0) = \lim_{h \to 0} \frac{h f^{\prime}(h) - 0}{h} = \lim_{h \to 0} f^{\prime}(h)$.
Since $f^{\prime}(x)$ is given to be continuous at $x=0$,we have $\lim_{h \to 0} f^{\prime}(h) = f^{\prime}(0)$.
Given $f^{\prime}(0) = 1$,it follows that $g^{\prime}(0) = 1$.

(A, D) Given $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \\ x - \frac{1}{2}, & x \leq 1 \end{cases}$.
For $x > 1$,$\int_{0}^{x} |1-t| dt = \int_{0}^{1} (1-t) dt + \int_{1}^{x} (t-1) dt$.
$= \left[ t - \frac{t^2}{2} \right]_{0}^{1} + \left[ \frac{t^2}{2} - t \right]_{1}^{x} = \left( 1 - \frac{1}{2} \right) + \left( \frac{x^2}{2} - x - (\frac{1}{2} - 1) \right) = \frac{1}{2} + \frac{x^2}{2} - x + \frac{1}{2} = \frac{x^2}{2} - x + 1$.
Thus,$f(x) = \begin{cases} \frac{x^2}{2} - x + 1, & x > 1 \\ x - \frac{1}{2}, & x \leq 1 \end{cases}$.
Continuity at $x=1$:
$LHL = \lim_{x \to 1^-} (x - \frac{1}{2}) = \frac{1}{2}$.
$RHL = \lim_{x \to 1^+} (\frac{x^2}{2} - x + 1) = \frac{1}{2} - 1 + 1 = \frac{1}{2}$.
$f(1) = 1 - \frac{1}{2} = \frac{1}{2}$.
Since $LHL = RHL = f(1)$,$f(x)$ is continuous at $x=1$.
Differentiability at $x=1$:
$LHD = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} = \lim_{h \to 0} \frac{(1-h - \frac{1}{2}) - \frac{1}{2}}{-h} = \lim_{h \to 0} \frac{-h}{-h} = 1$.
$RHD = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{(\frac{(1+h)^2}{2} - (1+h) + 1) - \frac{1}{2}}{h} = \lim_{h \to 0} \frac{\frac{1+2h+h^2}{2} - h - \frac{1}{2}}{h} = \lim_{h \to 0} \frac{\frac{h^2}{2}}{h} = 0$.
Since $LHD \neq RHD$,$f(x)$ is not differentiable at $x=1$.
Therefore,both options $A$ and $D$ are correct.

(C) By the Mean Value Theorem,for a function $f(x)$ differentiable on $[2, 7]$,there exists some $c \in (2, 7)$ such that $f^{\prime}(c) = \frac{f(7) - f(2)}{7 - 2}$.
Given that $f^{\prime}(x) \leq 5$ for all $x \in (2, 7)$,we have $f^{\prime}(c) \leq 5$.
Substituting the given values $f(2) = 3$ and the interval $[2, 7]$:
$\frac{f(7) - 3}{7 - 2} \leq 5$
$\frac{f(7) - 3}{5} \leq 5$
$f(7) - 3 \leq 25$
$f(7) \leq 28$.
Thus,the maximum possible value of $f(7)$ is $28$.

(D) Given,$f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$.
Continuity at $x=0$: $LHL = \lim_{x \rightarrow 0^-} x \sin \frac{1}{x} = 0$ and $RHL = \lim_{x \rightarrow 0^+} x \sin \frac{1}{x} = 0$. Since $LHL = RHL = f(0)$,$f(x)$ is continuous for all $x \in [-1, 1]$.
Differentiability at $x=0$: $f'(0) = \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0} \frac{h \sin(1/h)}{h} = \lim_{h \rightarrow 0} \sin(1/h)$. This limit does not exist as it oscillates between $-1$ and $1$. Thus,$f(x)$ is not differentiable at $x=0$.
Rolle's theorem requires $f(a) = f(b)$ and differentiability on $(a, b)$. Since $f(x)$ is not differentiable at $x=0$,it does not satisfy Rolle's theorem on $[-1, 1]$ or $[0, 1]$.
Lagrange's Mean Value Theorem $(LMVT)$ requires continuity on $[a, b]$ and differentiability on $(a, b)$. Since $f(x)$ is not differentiable at $x=0$,it does not satisfy $LMVT$ on $[-1, 1]$ (as $0 \in (-1, 1)$) or $[0, 1]$ (as $0$ is an endpoint,but the theorem requires differentiability on the open interval $(0, 1)$).
Wait,checking the options: The function is differentiable on $(0, 1)$. Thus,$f(x)$ satisfies the conditions of $LMVT$ on $[0, 1]$ because it is continuous on $[0, 1]$ and differentiable on $(0, 1)$.

(C) According to Lagrange's Mean Value Theorem,there exists $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.
Given $f(x) = \cos x$ in $[0, h]$,we have $f'(c) = \frac{\cos h - \cos 0}{h - 0}$.
Since $c = \theta h$ where $0 < \theta < 1$,we have $f'(\theta h) = \frac{\cos h - 1}{h}$.
Substituting $f'(x) = -\sin x$,we get $-\sin(\theta h) = \frac{\cos h - 1}{h}$.
Using the Taylor series expansion $\cos h \approx 1 - \frac{h^2}{2}$,we have $-\sin(\theta h) \approx \frac{1 - \frac{h^2}{2} - 1}{h} = -\frac{h}{2}$.
Thus,$\sin(\theta h) \approx \frac{h}{2}$.
For small $h$,$\sin(\theta h) \approx \theta h$,so $\theta h \approx \frac{h}{2}$.
Therefore,$\theta \approx \frac{1}{2}$.
Taking the limit as $h \rightarrow 0^{+}$,we get $\lim_{h \rightarrow 0^{+}} \theta = \frac{1}{2}$.

(A) Given the integral equation: $\int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$.
This can be rewritten as: $\int_{0}^{2} x e^{x^{4}} dx = \alpha \int_{0}^{2} e^{x^{4}} dx$.
Thus,$\alpha = \frac{\int_{0}^{2} x e^{x^{4}} dx}{\int_{0}^{2} e^{x^{4}} dx}$.
Let $f(x) = e^{x^{4}}$. Since $f(x) > 0$ for all $x \in [0, 2]$,the expression for $\alpha$ represents a weighted average of the values of $x$ in the interval $[0, 2]$ with respect to the weight function $f(x)$.
Since $x$ ranges from $0$ to $2$,the weighted average $\alpha$ must also lie strictly between the minimum and maximum values of $x$ in the interval $[0, 2]$.
Therefore,$0 < \alpha < 2$,which means $\alpha \in (0, 2)$.

(A) Given,$f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \\ 4x^3 - 1, & x > 1 \end{cases}$.
We need to evaluate $\int_{0}^{2} f(x) dx$.
Since the function definition changes at $x = 1$,we split the integral:
$\int_{0}^{2} f(x) dx = \int_{0}^{1} (2x^2 + 1) dx + \int_{1}^{2} (4x^3 - 1) dx$.
Evaluating the first part:
$\int_{0}^{1} (2x^2 + 1) dx = \left[ \frac{2x^3}{3} + x \right]_{0}^{1} = \left( \frac{2(1)^3}{3} + 1 \right) - (0) = \frac{2}{3} + 1 = \frac{5}{3}$.
Evaluating the second part:
$\int_{1}^{2} (4x^3 - 1) dx = \left[ x^4 - x \right]_{1}^{2} = (2^4 - 2) - (1^4 - 1) = (16 - 2) - (0) = 14$.
Adding both parts:
$\int_{0}^{2} f(x) dx = \frac{5}{3} + 14 = \frac{5 + 42}{3} = \frac{47}{3}$.

(C) Given,$f(x) = \max \{x+|x|, x-[x]\}$.
For $x \geq 0$,$x+|x| = 2x$ and $x-[x] = \{x\} \in [0, 1)$. Since $2x \geq \{x\}$ for $x \geq 0$,$f(x) = 2x$.
For $x < 0$,$x+|x| = x-x = 0$ and $x-[x] = \{x\} \in [0, 1)$. Since $0 \leq \{x\} < 1$,$f(x) = x-[x] = \{x\}$.
Thus,$\int_{-3}^{3} f(x) dx = \int_{-3}^{0} (x-[x]) dx + \int_{0}^{3} 2x dx$.
Since $x-[x]$ is periodic with period $1$,$\int_{-3}^{0} (x-[x]) dx = 3 \int_{0}^{1} x dx = 3 \left[ \frac{x^2}{2} \right]_0^1 = \frac{3}{2}$.
And $\int_{0}^{3} 2x dx = [x^2]_0^3 = 9$.
Therefore,$\int_{-3}^{3} f(x) dx = \frac{3}{2} + 9 = \frac{21}{2}$.

(D) Given,$M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} dx$.
Using the identity $\sin x \cos x = \frac{1}{2} \sin 2x$,we have $N = \int_{0}^{\pi / 4} \frac{\sin 2x}{2(x+1)^{2}} dx$.
Let $2x = t$,then $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\pi/4, t=\pi/2$.
$N = \int_{0}^{\pi / 2} \frac{\sin t}{2(t/2 + 1)^{2}} \cdot \frac{dt}{2} = \int_{0}^{\pi / 2} \frac{\sin t}{2(\frac{t+2}{2})^{2}} dt = \int_{0}^{\pi / 2} \frac{\sin t}{(t+2)^{2}} dt = \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx$.
Now,$M - N = \int_{0}^{\pi / 2} \left( \frac{\cos x}{x+2} - \frac{\sin x}{(x+2)^{2}} \right) dx$.
Using integration by parts on $\int \frac{\cos x}{x+2} dx$ with $u = \frac{1}{x+2}$ and $dv = \cos x dx$,we get $du = -\frac{1}{(x+2)^{2}} dx$ and $v = \sin x$.
$M = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \sin x \left( -\frac{1}{(x+2)^{2}} \right) dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} + \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx$.
Thus,$M - \int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} dx = \left[ \frac{\sin x}{x+2} \right]_{0}^{\pi / 2} = \frac{\sin(\pi/2)}{\pi/2 + 2} - \frac{\sin(0)}{0+2} = \frac{1}{\frac{\pi+4}{2}} = \frac{2}{\pi+4}$.

(A) Given curves are $y=x^{2}$ and $x=y^{2}$,which are parabolas.
To find the points of intersection,substitute $y=x^{2}$ into $x=y^{2}$:
$x=(x^{2})^{2}$
$x=x^{4}$
$x^{4}-x=0$
$x(x^{3}-1)=0$
$x(x-1)(x^{2}+x+1)=0$
Since $x^{2}+x+1=0$ has no real roots,we have $x=0$ and $x=1$.
When $x=0$,$y=0$. When $x=1$,$y=1$.
Thus,the points of intersection are $(0,0)$ and $(1,1)$.
The area of the bounded region is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:
$\text{Area} = \int_{0}^{1} (\sqrt{x} - x^{2}) \, dx$
$= \left[ \frac{x^{3/2}}{3/2} - \frac{x^{3}}{3} \right]_{0}^{1}$
$= \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^{3} \right]_{0}^{1}$
$= (\frac{2}{3}(1) - \frac{1}{3}(1)) - (0 - 0)$
$= \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \text{ sq units}$.

(A) Given curve is $y=(\cos x+y)^{1/2}$.
Squaring both sides,we get $y^2 = \cos x + y$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = -\sin x + \frac{dy}{dx}$.
Rearranging the terms:
$(2y - 1) \frac{dy}{dx} = -\sin x$.
Differentiating again with respect to $x$ using the product rule:
$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{d}{dx}(2y - 1) = -\cos x$.
$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot (2 \frac{dy}{dx}) = -\cos x$.
$(2y - 1) \frac{d^2y}{dx^2} + 2 \left(\frac{dy}{dx}\right)^2 + \cos x = 0$.

(A) The given differential equation $\frac{d^{2} y}{d x^{2}}+b \frac{d y}{d x}+c y=0$ is a linear homogeneous differential equation of the second order.
By the principle of superposition,if $u(x)$ and $v(x)$ are two independent solutions,then any linear combination $y = c_{1}u(x) + c_{2}v(x)$ is also a solution,where $c_{1}$ and $c_{2}$ are arbitrary constants.
Option $A$ is a specific case of a linear combination where $c_{1}=5$ and $c_{2}=8$.
Option $B$ can be rewritten as $y = c_{1}u(x) + (c_{2}-c_{1})v(x)$,which is also a linear combination of $u(x)$ and $v(x)$ with new arbitrary constants.
Therefore,both $A$ and $B$ represent solutions to the differential equation.

(A) The given differential equation is $(1+x^{2}) \frac{dy}{dx} + y = e^{\tan^{-1} x}$.
Dividing both sides by $(1+x^{2})$,we get:
$\frac{dy}{dx} + \frac{1}{1+x^{2}} y = \frac{e^{\tan^{-1} x}}{1+x^{2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{1+x^{2}}$ and $Q = \frac{e^{\tan^{-1} x}}{1+x^{2}}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx}$.
$IF = e^{\int \frac{1}{1+x^{2}} dx} = e^{\tan^{-1} x}$.

(C) Given the differential equation: $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$.
Dividing by $y$,we get $\frac{dy}{dx} = \frac{y}{x} + \frac{x \phi(y^2/x^2)}{y \phi'(y^2/x^2)}$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v + \frac{x \phi(v^2)}{vx \phi'(v^2)} = v + \frac{\phi(v^2)}{v \phi'(v^2)}$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{\phi(v^2)}{v \phi'(v^2)}$.
Separating variables: $\frac{v \phi'(v^2)}{\phi(v^2)} dv = \frac{dx}{x}$.
Integrating both sides: $\int \frac{v \phi'(v^2)}{\phi(v^2)} dv = \int \frac{dx}{x}$.
Let $u = v^2$,then $du = 2v dv$,so $v dv = \frac{1}{2} du$.
The integral becomes $\frac{1}{2} \int \frac{\phi'(u)}{\phi(u)} du = \ln|x| + C_1$.
$\frac{1}{2} \ln|\phi(u)| = \ln|x| + C_1 \Rightarrow \ln|\phi(v^2)| = 2 \ln|x| + 2C_1 = \ln|x^2| + \ln|c|$.
Thus,$\phi(v^2) = c x^2$.
Substituting $v^2 = y^2/x^2$,we get $\phi(y^2/x^2) = c x^2$.

(A) The given differential equation is $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x \log_{e} x}$ and $Q = \frac{1}{x}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{x \log_{e} x} dx}$.
Let $u = \log_{e} x$,then $du = \frac{1}{x} dx$. Thus,$\int \frac{1}{x \log_{e} x} dx = \int \frac{1}{u} du = \log_{e} u = \log_{e}(\log_{e} x)$.
Therefore,$IF = e^{\log_{e}(\log_{e} x)} = \log_{e} x$.
The solution is $y \cdot IF = \int (Q \cdot IF) dx + C$.
$y \log_{e} x = \int \frac{1}{x} \log_{e} x dx$.
Let $v = \log_{e} x$,then $dv = \frac{1}{x} dx$. The integral becomes $\int v dv = \frac{v^2}{2} + C = \frac{(\log_{e} x)^2}{2} + C$.
So,$y \log_{e} x = \frac{(\log_{e} x)^2}{2} + C$.
Given $y = 1$ when $x = e$,we have $1 \cdot \log_{e} e = \frac{(\log_{e} e)^2}{2} + C$.
$1 = \frac{1}{2} + C \implies C = \frac{1}{2}$.
Substituting $C$ back,$y \log_{e} x = \frac{(\log_{e} x)^2}{2} + \frac{1}{2}$.
Dividing by $\log_{e} x$ (or multiplying by $2$),we get $2y \log_{e} x = (\log_{e} x)^2 + 1$,which simplifies to $2y = \log_{e} x + \frac{1}{\log_{e} x}$.

(D) Given,$\sqrt{y}=\cos ^{-1} x \Rightarrow y=(\cos ^{-1} x)^{2}$.
On differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = 2(\cos ^{-1} x) \times \left(\frac{-1}{\sqrt{1-x^{2}}}\right)$.
Multiplying both sides by $\sqrt{1-x^{2}}$,we get:
$\sqrt{1-x^{2}} \frac{dy}{dx} = -2 \cos ^{-1} x$.
Again,differentiating both sides with respect to $x$ using the product rule:
$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} \times \left(\frac{-2x}{2\sqrt{1-x^{2}}}\right) = -2 \times \left(\frac{-1}{\sqrt{1-x^{2}}}\right)$.
$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} - \frac{x}{\sqrt{1-x^{2}}} \frac{dy}{dx} = \frac{2}{\sqrt{1-x^{2}}}$.
Multiplying the entire equation by $\sqrt{1-x^{2}}$,we get:
$(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = 2$.
Comparing this with the given equation $(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = c$,we find $c = 2$.

(C) The total number of outcomes when a die is rolled $12$ times is $6^{12}$.
We want each of the $6$ faces to appear exactly $2$ times.
This is a multinomial distribution problem where we arrange $12$ items into $6$ groups of size $2$ each.
The number of ways to distribute $12$ outcomes such that each face appears twice is given by the multinomial coefficient:
$\frac{12!}{2! 2! 2! 2! 2! 2!} = \frac{12!}{(2!)^6} = \frac{12!}{2^6}$.
Therefore,the required probability is:
$P = \frac{12!}{2^6 \times 6^{12}}$.

(C, D) Given,$P(A)=0.7$ and $P(B)=0.6.$
We know that $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B).$
Therefore,$P(A \cap B) \leq \min(0.7, 0.6) = 0.6.$
Also,by the inclusion-exclusion principle,$P(A \cup B) = P(A) + P(B) - P(A \cap B).$
Since $P(A \cup B) \leq 1,$ we have $P(A) + P(B) - P(A \cap B) \leq 1.$
$0.7 + 0.6 - P(A \cap B) \leq 1$ $\Rightarrow 1.3 - P(A \cap B) \leq 1$ $\Rightarrow P(A \cap B) \geq 0.3.$
Thus,the range for $P(A \cap B)$ is $0.3 \leq P(A \cap B) \leq 0.6.$
Comparing this with the given options:
$(a)$ $0.35$ is in the range.
$(b)$ $0.45$ is in the range.
$(c)$ $0.65$ is outside the range $(> 0.6)$.
$(d)$ $0.28$ is outside the range $(< 0.3)$.
Therefore,statements $(c)$ and $(d)$ are necessarily false.

(C) Let $B$ denote a boy and $G$ denote a girl. The possible outcomes for $2$ children are $\{BB, BG, GB, GG\}$,where each outcome is equally likely with probability $1/4$.
Given that at least $1$ child is a boy,the sample space reduces to $S = \{BB, BG, GB\}$.
The total number of outcomes in the reduced sample space is $n(S) = 3$.
We want to find the probability that the other child is a girl,which corresponds to the outcomes where there is exactly $1$ boy and $1$ girl. These outcomes are $\{BG, GB\}$.
The number of favourable outcomes is $n(E) = 2$.
Therefore,the required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{3}$.

(C) Let $E_1$ be the event that the student does not know the answer,and $E_2$ be the event that the student knows the answer. Let $E$ be the event that the student answers the question correctly.
We are given $P(E_2) = p$ and $P(E_1) = 1 - p$.
If the student knows the answer,the probability of answering correctly is $P(E|E_2) = 1$.
If the student does not know the answer,he chooses one of the $5$ alternatives randomly,so the probability of answering correctly is $P(E|E_1) = \frac{1}{5}$.
We want to find the probability that the student knew the answer (did not tick randomly) given that he answered correctly,which is $P(E_2|E)$.
By Bayes' Theorem:
$P(E_2|E) = \frac{P(E_2) P(E|E_2)}{P(E_1) P(E|E_1) + P(E_2) P(E|E_2)}$
Substituting the values:
$P(E_2|E) = \frac{p \times 1}{(1 - p) \times \frac{1}{5} + p \times 1}$
$P(E_2|E) = \frac{p}{\frac{1 - p + 5p}{5}}$
$P(E_2|E) = \frac{5p}{1 + 4p}$