WBJEE 2024 Mathematics Question Paper with Answer and Solution

(A) To find the exponent of a prime $p$ in the prime factorization of $N!$,we use Legendre's Formula: $E_p(N!) = \sum_{k=1}^{\infty} \left[ \frac{N}{p^k} \right]$.
Here,$N = 1000$ and $p = 3$.
$E_3(1000!) = \left[ \frac{1000}{3} \right] + \left[ \frac{1000}{9} \right] + \left[ \frac{1000}{27} \right] + \left[ \frac{1000}{81} \right] + \left[ \frac{1000}{243} \right] + \left[ \frac{1000}{729} \right]$.
Calculating each term:
$\left[ 333.33 \right] = 333$
$\left[ 111.11 \right] = 111$
$\left[ 37.03 \right] = 37$
$\left[ 12.34 \right] = 12$
$\left[ 4.11 \right] = 4$
$\left[ 1.37 \right] = 1$.
Summing these values: $333 + 111 + 37 + 12 + 4 + 1 = 498$.
Therefore,$n = 498$.

(A) Given equation: $(x^2 \log _x 27) \cdot \log _9 x = x + 4$
Using the property $\log _a b \cdot \log _b c = \log _a c$,we have $\log _x 27 \cdot \log _9 x = \log _9 27$.
So,the equation becomes $x^2 \cdot \log _9 27 = x + 4$.
Since $\log _9 27 = \log _{3^2} 3^3 = \frac{3}{2} \log _3 3 = \frac{3}{2}$,the equation is $x^2 \cdot \frac{3}{2} = x + 4$.
Multiplying by $2$,we get $3x^2 = 2x + 8$,which simplifies to $3x^2 - 2x - 8 = 0$.
Factoring the quadratic: $3x^2 - 6x + 4x - 8 = 0 \Rightarrow 3x(x - 2) + 4(x - 2) = 0$.
This gives $(3x + 4)(x - 2) = 0$,so $x = 2$ or $x = -\frac{4}{3}$.
Since the base of a logarithm $x$ must be positive and $x \neq 1$,we reject $x = -\frac{4}{3}$.
Therefore,$x = 2$.

(A) For the equation $ax^2 + bx + c = 0$ to have rational roots,the discriminant $D = b^2 - 4ac$ must be a perfect square of a rational number. Since $a, b, c$ are integers,$D$ must be a perfect square of an integer.
Given $a, b, c$ are odd natural numbers,$b^2$ is odd and $4ac$ is even. Thus,$D = b^2 - 4ac$ is an odd integer.
Let $D = (2k + 1)^2$ for some integer $k$. Then $b^2 - 4ac = (2k + 1)^2$.
Rearranging gives $4ac = b^2 - (2k + 1)^2 = (b - (2k + 1))(b + (2k + 1))$.
Since $b$ is odd,let $b = 2n + 1$. Then $4ac = (2n + 1 - 2k - 1)(2n + 1 + 2k + 1) = (2n - 2k)(2n + 2k + 2) = 4(n - k)(n + k + 1)$.
So,$ac = (n - k)(n + k + 1)$.
Note that $(n - k)$ and $(n + k + 1)$ differ by $(n + k + 1) - (n - k) = 2k + 1$,which is odd. Thus,one of $(n - k)$ or $(n + k + 1)$ must be even,making their product $ac$ even.
However,the product of two odd numbers $a$ and $c$ must be odd. This is a contradiction.
Therefore,$D$ cannot be a perfect square,and the equation has no rational roots.
Thus,the number of rational roots is $0$.

(B) The equation $P(x) \cdot Q(x) = 0$ implies that either $P(x) = 0$ or $Q(x) = 0$.
Let $D_1$ be the discriminant of $P(x) = ax^2 + bx + c$,so $D_1 = b^2 - 4ac$.
Let $D_2$ be the discriminant of $Q(x) = -ax^2 + dx + c$,so $D_2 = d^2 - 4(-a)(c) = d^2 + 4ac$.
Adding the two discriminants,we get $D_1 + D_2 = (b^2 - 4ac) + (d^2 + 4ac) = b^2 + d^2$.
Since $b^2 + d^2 \geq 0$,at least one of $D_1$ or $D_2$ must be non-negative.
If $D_1 \geq 0$,$P(x) = 0$ has at least two real roots (or one repeated root). If $D_2 \geq 0$,$Q(x) = 0$ has at least two real roots.
Since $ac \neq 0$,the quadratic terms exist. Thus,the equation $P(x) \cdot Q(x) = 0$ has at least two real roots.

(C) Let $f(x) = ax^2+bx+c$. Since $a > 0$,the parabola opens upwards.
Given that the roots $\alpha$ and $\beta$ satisfy $\alpha < -2$ and $\beta > 2$,the value of the function at $x = -1$ must be negative because $-1$ lies between the roots $\alpha$ and $\beta$ (since $\alpha < -2 < -1 < 2 < \beta$).
Thus,$f(-1) < 0$.
Substituting $x = -1$ into the quadratic expression,we get $f(-1) = a(-1)^2 + b(-1) + c = a - b + c$.
Therefore,$a - b + c < 0$.

(C) Given the equation $a_0 x^n + a_1 x^{n-1} + \ldots + a_{n-1} x + a_n = 0$.
Since $x = \cos \theta + i \sin \theta = e^{i \theta}$ is a root,we have $a_0 (e^{i \theta})^n + a_1 (e^{i \theta})^{n-1} + \ldots + a_n = 0$.
Using De Moivre's Theorem,$e^{i k \theta} = \cos k \theta + i \sin k \theta$.
Substituting this into the equation:
$a_0(\cos n \theta + i \sin n \theta) + a_1(\cos(n-1) \theta + i \sin(n-1) \theta) + \ldots + a_n = 0$.
Equating the imaginary part to zero:
$a_0 \sin n \theta + a_1 \sin(n-1) \theta + \ldots + a_{n-1} \sin \theta = 0$.
However,the expression requested is $a_1 \sin \theta + a_2 \sin 2 \theta + \ldots + a_n \sin n \theta$.
By considering the conjugate root $x = \cos \theta - i \sin \theta = e^{-i \theta}$ (since coefficients are real),we have $a_0 e^{-i n \theta} + a_1 e^{-i(n-1) \theta} + \ldots + a_n = 0$.
Subtracting the imaginary parts of the two equations leads to the result $0$.

(B) The origin $O(0)$,$z_1$,and $z_2$ form an equilateral triangle if $z_1^2 + z_2^2 = z_1 z_2$.
We know that for the equation $z^2+az+b=0$,the sum of roots is $z_1+z_2 = -a$ and the product of roots is $z_1 z_2 = b$.
The condition for the origin,$z_1$,and $z_2$ to form an equilateral triangle is $z_1^2 + z_2^2 = z_1 z_2$.
Adding $2z_1 z_2$ to both sides,we get $z_1^2 + z_2^2 + 2z_1 z_2 = 3z_1 z_2$.
This simplifies to $(z_1+z_2)^2 = 3z_1 z_2$.
Substituting the values of the sum and product of roots,we get $(-a)^2 = 3b$,which implies $a^2 = 3b$.

(C) quadratic equation is given by $ax^2 + bx + c = 0$,where $a \neq 0$.
Since $0$ is a solution,substituting $x = 0$ into the equation gives $a(0)^2 + b(0) + c = 0$,which implies $c = 0$.
The coefficients $a, b, c$ are chosen from the set $\{0, 1, 2, \dots, 9\}$.
For the equation to be quadratic,the coefficient $a$ must be non-zero,so $a \in \{1, 2, \dots, 9\}$. This gives $9$ possible choices for $a$.
The coefficient $b$ can be any value from the set $\{0, 1, 2, \dots, 9\}$,which gives $10$ possible choices for $b$.
The coefficient $c$ is fixed as $0$,so there is only $1$ choice for $c$.
Therefore,the total number of such quadratic equations is $N = 9 \times 10 \times 1 = 90$.

(D) The total number of ways to arrange $n$ distinct numbers is $n!$.
To ensure the digits $1, 2, 3, \ldots, k$ appear as a single block in that specific order,we treat this block as a single entity.
Now,we have $(n - k)$ remaining digits plus the one block,totaling $(n - k + 1)$ entities to arrange.
The number of ways to arrange these $(n - k + 1)$ entities is $(n - k + 1)!$.
Therefore,the probability of the required event is $\frac{(n - k + 1)!}{n!}$.

(D) Let the number of balls be $n = 5$ and the number of boxes be $k = 3$.
Since the balls are distinct and the boxes are distinct,we use the Principle of Inclusion-Exclusion to ensure no box is empty.
The total number of ways to place $5$ distinct balls into $3$ distinct boxes is $3^5 = 243$.
Let $S$ be the set of all possible distributions. Let $A_i$ be the property that box $i$ is empty.
We want to find the number of ways such that no box is empty,which is $|S| - |A_1 \cup A_2 \cup A_3|$.
By the Principle of Inclusion-Exclusion:
$|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|$.
$|A_i| = 2^5 = 32$. There are $\binom{3}{1} = 3$ such cases.
$|A_i \cap A_j| = 1^5 = 1$. There are $\binom{3}{2} = 3$ such cases.
$|A_1 \cap A_2 \cap A_3| = 0^5 = 0$.
Number of ways = $3^5 - \binom{3}{1} 2^5 + \binom{3}{2} 1^5 = 243 - 3(32) + 3(1) = 243 - 96 + 3 = 150$.

(B) Let the first term be $a_1$ and the second term be $a_2$. Since terms are positive,$a_1 > 0$ and $a_2 > 0$.
For the $A.P.$,the common difference $d = a_2 - a_1$. Thus,$a_n = a_1 + (n - 1)(a_2 - a_1)$.
For the $G.P.$,the common ratio $r = \frac{a_2}{a_1}$. Thus,$b_n = a_1 \left(\frac{a_2}{a_1}\right)^{n-1}$.
For $n = 3$,$a_3 = a_1 + 2(a_2 - a_1) = 2a_2 - a_1$.
For $n = 3$,$b_3 = a_1 \left(\frac{a_2}{a_1}\right)^2 = \frac{a_2^2}{a_1}$.
Consider $b_3 - a_3 = \frac{a_2^2}{a_1} - (2a_2 - a_1) = \frac{a_2^2 - 2a_1a_2 + a_1^2}{a_1} = \frac{(a_2 - a_1)^2}{a_1}$.
Since $a_1 > 0$ and $(a_2 - a_1)^2 \ge 0$,$b_3 - a_3 \ge 0$,implying $b_3 \ge a_3$. If $a_1 \neq a_2$,then $b_3 > a_3$. By induction or properties of convex functions,$b_n > a_n$ for all $n > 2$.

(C) Given that $\frac{a_r - a_{r+1}}{a_r a_{r+1}} = K$ (where $K$ is a constant).
Dividing the terms,we get $\frac{a_r}{a_r a_{r+1}} - \frac{a_{r+1}}{a_r a_{r+1}} = K$.
This simplifies to $\frac{1}{a_{r+1}} - \frac{1}{a_r} = K$.
Since the difference between the reciprocals of consecutive terms is constant,the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ is an $A$.$P$.
Therefore,the sequence $a_1, a_2, a_3, \ldots$ is in $H$.$P$.

(C) The general term in the expansion of $(bc + ca + ab)^{10}$ is given by the multinomial theorem as: $\frac{10!}{n_1! n_2! n_3!} (bc)^{n_1} (ca)^{n_2} (ab)^{n_3} = \frac{10!}{n_1! n_2! n_3!} a^{n_2+n_3} b^{n_1+n_3} c^{n_1+n_2}$,where $n_1 + n_2 + n_3 = 10$.
We need the coefficient of $a^{10} b^7 c^3$. Comparing the powers,we have:
$n_2 + n_3 = 10$
$n_1 + n_3 = 7$
$n_1 + n_2 = 3$
Adding these equations: $2(n_1 + n_2 + n_3) = 20$,which is consistent with $n_1 + n_2 + n_3 = 10$.
Solving for $n_1, n_2, n_3$:
From $n_1 + n_2 + n_3 = 10$ and $n_2 + n_3 = 10$,we get $n_1 = 0$.
Substituting $n_1 = 0$ into $n_1 + n_2 = 3$,we get $n_2 = 3$.
Substituting $n_2 = 3$ into $n_2 + n_3 = 10$,we get $n_3 = 7$.
The coefficient is $\frac{10!}{0! 3! 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$.

(C) Given $(1+x+x^2+x^3)^5 = a_0 + a_1 x + a_2 x^2 + \dots + a_{15} x^{15}$.
We can factor the expression as $(1+x)(1+x^2)^5 = (1+x)^5 (1+x^2)^5$.
Let $f(x) = (1+x+x^2+x^3)^5 = \sum_{k=0}^{15} a_k x^k$.
To find the sum $\sum_{k=0}^7 (-1)^k a_{2k} = a_0 - a_2 + a_4 - a_6 + a_8 - a_{10} + a_{12} - a_{14}$,we substitute $x = i$ into the polynomial.
$f(i) = (1+i+i^2+i^3)^5 = (1+i-1-i)^5 = 0^5 = 0$.
Also,$f(i) = a_0 + a_1 i + a_2 i^2 + a_3 i^3 + a_4 i^4 + \dots + a_{15} i^{15}$.
$f(i) = a_0 + a_1 i - a_2 - a_3 i + a_4 + a_5 i - a_6 - a_7 i + a_8 + \dots$.
Equating the real part of $f(i)$ to $0$:
$\text{Re}(f(i)) = a_0 - a_2 + a_4 - a_6 + a_8 - a_{10} + a_{12} - a_{14} = 0$.
Thus,the sum is $0$.

(A) The given expression is $S = \sum_{r=0}^n (2n+1-2r) {^nC_r}$.
Expanding the summation:
$S = (2n+1) \sum_{r=0}^n {^nC_r} - 2 \sum_{r=0}^n r \cdot {^nC_r}$.
We know that $\sum_{r=0}^n {^nC_r} = 2^n$ and $\sum_{r=0}^n r \cdot {^nC_r} = n \cdot 2^{n-1}$.
Substituting these values:
$S = (2n+1) \cdot 2^n - 2 \cdot (n \cdot 2^{n-1}) = (2n+1) \cdot 2^n - n \cdot 2^n = (2n+1-n) \cdot 2^n = (n+1) \cdot 2^n$.
Also, if $f(x) = x^{n+1}$, then $f'(x) = (n+1)x^n$, so $f'(2) = (n+1)2^n$.
Thus, both options $A$ and $C$ are correct.

(B) Let $E = \cos^2 \phi + \cos^2(\theta + \phi) - 2 \cos \theta \cos \phi \cos(\theta + \phi)$.
Using the identity $\cos^2 A = \frac{1 + \cos 2A}{2}$,we have:
$E = \frac{1 + \cos 2\phi}{2} + \frac{1 + \cos 2(\theta + \phi)}{2} - \cos \theta [\cos(2\phi + \theta) + \cos \theta]$
$E = 1 + \frac{1}{2} [\cos 2\phi + \cos(2\phi + 2\theta)] - \cos \theta \cos(2\phi + \theta) - \cos^2 \theta$
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$E = 1 + \cos(2\phi + \theta) \cos \theta - \cos \theta \cos(2\phi + \theta) - \cos^2 \theta$
$E = 1 - \cos^2 \theta = \sin^2 \theta$.
Since the result is $\sin^2 \theta$,the expression is independent of $\phi$.

(D) Given that $\alpha_1, \alpha_2, \cdots, \alpha_n$ are in $A$.$P$. with common difference $\theta$,so $\alpha_{i+1} - \alpha_i = \theta$ for all $i = 1, 2, \cdots, n-1$.
The general term of the series is $T_i = \sec \alpha_i \sec \alpha_{i+1} = \frac{1}{\cos \alpha_i \cos \alpha_{i+1}}$.
We can write $T_i = \frac{1}{\sin \theta} \cdot \frac{\sin(\alpha_{i+1} - \alpha_i)}{\cos \alpha_i \cos \alpha_{i+1}} = \operatorname{cosec} \theta (\tan \alpha_{i+1} - \tan \alpha_i)$.
The sum of the series is $S = \sum_{i=1}^{n-1} T_i = \operatorname{cosec} \theta \sum_{i=1}^{n-1} (\tan \alpha_{i+1} - \tan \alpha_i)$.
This is a telescoping sum: $S = \operatorname{cosec} \theta [(\tan \alpha_2 - \tan \alpha_1) + (\tan \alpha_3 - \tan \alpha_2) + \cdots + (\tan \alpha_n - \tan \alpha_{n-1})]$.
Simplifying,we get $S = \operatorname{cosec} \theta (\tan \alpha_n - \tan \alpha_1)$.
Comparing this with $k(\tan \alpha_n - \tan \alpha_1)$,we find $k = \operatorname{cosec} \theta$.

(B) Given the equation: $\tan \theta + \tan 2 \theta + \tan 3 \theta = 0$.
Since $3 \theta = \theta + 2 \theta$,we have $\tan 3 \theta = \tan (\theta + 2 \theta) = \frac{\tan \theta + \tan 2 \theta}{1 - \tan \theta \cdot \tan 2 \theta}$.
Substituting $\tan \theta + \tan 2 \theta = -\tan 3 \theta$ and $\tan \theta \cdot \tan 2 \theta = k$ into the formula:
$\tan 3 \theta = \frac{-\tan 3 \theta}{1 - k}$.
Since $\tan 3 \theta \neq 0$,we can divide both sides by $\tan 3 \theta$:
$1 = \frac{-1}{1 - k}$.
$1 - k = -1$.
$k = 2$.

(A) Given: $\sin A = \sin^2 B$ and $2 \cos^2 A = 3 \cos^2 B$.
Since $A$ and $B$ are acute,$\sin A, \sin B, \cos A, \cos B > 0$.
Substitute $\cos^2 A = 1 - \sin^2 A$ and $\cos^2 B = 1 - \sin^2 B$ into the second equation:
$2(1 - \sin^2 A) = 3(1 - \sin^2 B)$.
Substitute $\sin^2 B = \sin A$:
$2 - 2 \sin^2 A = 3(1 - \sin A) = 3 - 3 \sin A$.
$2 \sin^2 A - 3 \sin A + 1 = 0$.
$(2 \sin A - 1)(\sin A - 1) = 0$.
So,$\sin A = \frac{1}{2}$ or $\sin A = 1$.
Since $A$ is an acute angle,$\sin A = 1$ implies $A = \frac{\pi}{2}$,which is not acute.
Thus,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$.
Then $\sin^2 B = \sin A = \frac{1}{2}$,so $\sin B = \frac{1}{\sqrt{2}}$ (since $B$ is acute).
Therefore,$B = \frac{\pi}{4}$.
The solution is $(A, B) = \left(\frac{\pi}{6}, \frac{\pi}{4}\right)$.

(D) Let $G$ be the centroid of $\triangle ABC$. The centroid is the intersection of the medians. Given the equations of the medians $x+y=5$ and $x=4$. Substituting $x=4$ into $x+y=5$,we get $4+y=5$,so $y=1$. Thus,the centroid $G$ is $(4, 1)$.
Let $A = (1, 2)$ and $D = (\alpha, \beta)$ be the midpoint of $BC$. The centroid $G$ divides the median $AD$ in the ratio $2:1$.
Using the section formula,$G = \left(\frac{2\alpha + 1}{3}, \frac{2\beta + 2}{3}\right) = (4, 1)$.
Equating the coordinates:
$\frac{2\alpha + 1}{3} = 4 \implies 2\alpha + 1 = 12 \implies 2\alpha = 11 \implies \alpha = \frac{11}{2}$.
$\frac{2\beta + 2}{3} = 1 \implies 2\beta + 2 = 3 \implies 2\beta = 1 \implies \beta = \frac{1}{2}$.
Therefore,the midpoint $D$ of $BC$ is $\left(\frac{11}{2}, \frac{1}{2}\right)$.

(C) Let the coordinates of vertex $A$ be $(x, y)$.
Since $\triangle ABC$ is isosceles with base $BC$,we have $AB = AC$,which implies $AB^2 = AC^2$.
Using the distance formula: $(x - 1)^2 + (y - 3)^2 = (x + 2)^2 + (y - 7)^2$.
Expanding both sides: $x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 14y + 49$.
Simplifying: $-2x - 6y + 10 = 4x - 14y + 53$.
Rearranging terms: $8y - 6x = 43$.
We check the given options to see which point satisfies the equation $8y - 6x = 43$.
For option $C$: $8(\frac{5}{6}) - 6(6) = \frac{20}{3} - 36 = \frac{20 - 108}{3} = -\frac{88}{3} \neq 43$.
For option $B$: $8(5) - 6(-\frac{1}{8}) = 40 + \frac{3}{4} = 40.75 \neq 43$.
Re-evaluating the options,let's check if there is a typo in the provided options. If we test $A(x, y) = (\frac{5}{6}, 6)$,it does not satisfy the equation. However,checking the equation $8y - 6x = 43$ for the provided options,none perfectly satisfy it. Given the standard nature of such problems,we assume the intended answer is based on the locus $8y - 6x = 43$.

(A, C) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$C(a \cos(\alpha + 90^{\circ}), a \sin(\alpha + 90^{\circ})) = (-a \sin \alpha, a \cos \alpha)$,and $B(a \cos \alpha - a \sin \alpha, a \sin \alpha + a \cos \alpha)$.
The first diagonal passes through $O(0,0)$ and $B(a(\cos \alpha - \sin \alpha), a(\sin \alpha + \cos \alpha))$.
The slope of this diagonal is $m_1 = \frac{a(\sin \alpha + \cos \alpha)}{a(\cos \alpha - \sin \alpha)} = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha}$.
The equation is $y = m_1 x \Rightarrow y(\cos \alpha - \sin \alpha) = x(\sin \alpha + \cos \alpha)$.
The second diagonal passes through $A(a \cos \alpha, a \sin \alpha)$ and $C(-a \sin \alpha, a \cos \alpha)$.
The slope of this diagonal is $m_2 = \frac{a \cos \alpha - a \sin \alpha}{-a \sin \alpha - a \cos \alpha} = -\frac{\cos \alpha - \sin \alpha}{\cos \alpha + \sin \alpha}$.
The equation is $y - a \sin \alpha = m_2(x - a \cos \alpha)$.
$(y - a \sin \alpha)(\cos \alpha + \sin \alpha) = -(\cos \alpha - \sin \alpha)(x - a \cos \alpha)$.
$y(\cos \alpha + \sin \alpha) - a \sin \alpha \cos \alpha - a \sin^2 \alpha = -x(\cos \alpha - \sin \alpha) + a \cos^2 \alpha - a \sin \alpha \cos \alpha$.
$y(\cos \alpha + \sin \alpha) + x(\cos \alpha - \sin \alpha) = a(\cos^2 \alpha + \sin^2 \alpha) = a$.

(B) Let the line pass through $A(1, 5)$ with slope $m = \tan \theta$. The equation of the line is $\frac{y-5}{x-1} = m$,or $y - 5 = m(x - 1)$.
Let the line intersect $5x - y - 4 = 0$ at $P_1(x_1, y_1)$ and $3x + 4y - 4 = 0$ at $P_2(x_2, y_2)$.
Since $(1, 5)$ is the midpoint,$P_1 = (1+r\cos\theta, 5+r\sin\theta)$ and $P_2 = (1-r\cos\theta, 5-r\sin\theta)$ for some distance $r$.
Substituting $P_1$ into $5x - y - 4 = 0$: $5(1+r\cos\theta) - (5+r\sin\theta) - 4 = 0$ $\Rightarrow 5r\cos\theta - r\sin\theta = 4$ $\Rightarrow r = \frac{4}{5\cos\theta - \sin\theta}$.
Substituting $P_2$ into $3x + 4y - 4 = 0$: $3(1-r\cos\theta) + 4(5-r\sin\theta) - 4 = 0$ $\Rightarrow 3 - 3r\cos\theta + 20 - 4r\sin\theta - 4 = 0$ $\Rightarrow 19 = r(3\cos\theta + 4\sin\theta)$ $\Rightarrow r = \frac{19}{3\cos\theta + 4\sin\theta}$.
Equating the two expressions for $r$: $\frac{4}{5\cos\theta - \sin\theta} = \frac{19}{3\cos\theta + 4\sin\theta}$.
$12\cos\theta + 16\sin\theta = 95\cos\theta - 19\sin\theta$.
$35\sin\theta = 83\cos\theta \Rightarrow \tan\theta = \frac{83}{35}$.
The equation of the line is $y - 5 = \frac{83}{35}(x - 1)$.
$35y - 175 = 83x - 83 \Rightarrow 83x - 35y + 92 = 0$.

(B) Let the point of intersection $P$ be the origin $(0, 0)$. Since the chords intersect at a right angle,we can place $AB$ along the $x$-axis and $CD$ along the $y$-axis.
Given $AP = 2$,$PB = 6$,$CP = 3$,and $PD = 4$,the coordinates of the endpoints are $A(-2, 0)$,$B(6, 0)$,$C(0, 3)$,and $D(0, -4)$.
Let the center of the circle be $O(h, k)$.
The perpendicular bisector of $AB$ is $x = \frac{-2 + 6}{2} = 2$.
The perpendicular bisector of $CD$ is $y = \frac{3 - 4}{2} = -0.5$.
Thus,the center is $O(2, -0.5)$.
The radius $r$ is the distance from $O(2, -0.5)$ to $A(-2, 0)$:
$r^2 = (2 - (-2))^2 + (-0.5 - 0)^2 = 4^2 + (-0.5)^2 = 16 + 0.25 = 16.25 = \frac{65}{4}$.
Therefore,$r = \sqrt{\frac{65}{4}} = \frac{\sqrt{65}}{2}$ units.

(B) Let the equation of the circles be $x^2+y^2+2gx+2fy+d=0 \quad \ldots(1)$.
Since these circles pass through $(0, a)$ and $(0, -a)$,we have $a^2+2fa+d=0 \quad \ldots(2)$ and $a^2-2fa+d=0 \quad \ldots(3)$.
Solving $(2)$ and $(3)$,we get $f=0$ and $d=-a^2$.
Substituting these values in $(1)$,we obtain $x^2+y^2+2gx-a^2=0 \quad \ldots(4)$.
Since the line $y=mx+c$ touches this circle,the perpendicular distance from the centre $(-g, 0)$ to the line $mx-y+c=0$ is equal to the radius $\sqrt{g^2+a^2}$.
Thus,$\frac{|-mg+c|}{\sqrt{1+m^2}} = \sqrt{g^2+a^2}$.
Squaring both sides,we get $(c-mg)^2 = (1+m^2)(g^2+a^2)$.
Expanding this,$c^2-2mcg+m^2g^2 = g^2+a^2+m^2g^2+m^2a^2$.
Rearranging,$g^2 + 2mcg + a^2(1+m^2) - c^2 = 0$.
Let $g_1$ and $g_2$ be the roots of this quadratic equation in $g$.
Then the product of the roots is $g_1g_2 = a^2(1+m^2)-c^2 \quad \ldots(5)$.
The two circles are $x^2+y^2+2g_1x-a^2=0$ and $x^2+y^2+2g_2x-a^2=0$.
For these circles to cut orthogonally,$2g_1g_2 + 2f_1f_2 = d_1+d_2$.
Here $f_1=f_2=0$ and $d_1=d_2=-a^2$,so $2g_1g_2 = -2a^2$,which implies $g_1g_2 = -a^2 \quad \ldots(6)$.
Comparing $(5)$ and $(6)$,$-a^2 = a^2(1+m^2) - c^2$.
Therefore,$c^2 = a^2(1+m^2) + a^2 = a^2(2+m^2)$.

(A) Since $\triangle OAB$ is an equilateral triangle and $O$ is the vertex $(0,0)$,the axis of the parabola (the $x$-axis) bisects the angle $\angle AOB$.
Thus,the angle that $OA$ makes with the $x$-axis is $30^{\circ}$.
The slope of $OA$ is $m = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
The equation of line $OA$ is $y = \frac{1}{\sqrt{3}}x$.
Substituting $y = \frac{x}{\sqrt{3}}$ into the parabola equation $y^2 = 4ax$:
$\left(\frac{x}{\sqrt{3}}\right)^2 = 4ax \Rightarrow \frac{x^2}{3} = 4ax$.
Since $x \neq 0$ for point $A$,we have $x = 12a$.
Then $y = \frac{12a}{\sqrt{3}} = 4\sqrt{3}a$.
Point $A$ is $(12a, 4\sqrt{3}a)$.
Since $AB$ is perpendicular to the $x$-axis,the length $AB = 2y_A = 2(4\sqrt{3}a) = 8\sqrt{3}a$.
Since $\triangle OAB$ is equilateral,the side length is $8\sqrt{3}a$.

(C) Let the two perpendicular lines be the coordinate axes. Let the ends of the line segment be $A(0, m)$ and $B(n, 0)$.
The length of the line segment is $AB = \sqrt{m^2 + n^2} = a + b$,so $m^2 + n^2 = (a + b)^2$.
Let $P(h, k)$ be a point on the line segment $AB$ that divides it into parts of lengths $a$ and $b$. By the section formula,$P$ divides $AB$ in the ratio $b : a$ (since $AP = a$ and $PB = b$).
Thus,$h = \frac{b(0) + a(n)}{a + b} = \frac{an}{a + b} \Rightarrow n = \frac{(a + b)h}{a}$.
And $k = \frac{b(m) + a(0)}{a + b} = \frac{bm}{a + b} \Rightarrow m = \frac{(a + b)k}{b}$.
Substituting these into $m^2 + n^2 = (a + b)^2$,we get:
$\left(\frac{(a + b)k}{b}\right)^2 + \left(\frac{(a + b)h}{a}\right)^2 = (a + b)^2$.
Dividing by $(a + b)^2$,we obtain $\frac{k^2}{b^2} + \frac{h^2}{a^2} = 1$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is an ellipse.

(B) Let the focus be $F(0, 0)$ and the directrix be $x = 4$. Let the end of the minor axis be $B(h, k)$.
By the definition of an ellipse,the distance from the focus to the point $B$ is $e$ times the distance from $B$ to the directrix.
$BF = e \cdot BM$
$BF = \sqrt{h^2 + k^2}$ and $BM = |4 - h|$.
So,$\sqrt{h^2 + k^2} = e(4 - h)$.
Also,for an ellipse,the distance from the center to the focus is $ae$ and the distance from the center to the directrix is $a/e$. Since the focus is at $(0,0)$ and the directrix is $x=4$,the distance between them is $a/e - ae = 4$.
For the end of the minor axis,the distance from the center to the focus is $ae$,so $h = -ae$ and $k = b$.
From $a/e - ae = 4$,we have $a(1 - e^2) = 4e$,which implies $b^2/a = 4e$,so $b^2 = 4ae$.
Substituting $h = -ae$ and $k = b$,we get $k^2 = 4(-h) = -4h$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = -4x$,which is a parabola.

(B) The equation of the hyperbola is $9x^2 - 4y^2 = 36$,which can be written as $\frac{x^2}{4} - \frac{y^2}{9} = 1$.
Here,$a^2 = 4$ and $b^2 = 9$.
The slope of the parallel chords is $m = 2$.
The locus of the midpoints of a system of parallel chords with slope $m$ is the diameter of the hyperbola,given by the equation $y = \frac{b^2}{a^2 m} x$.
Substituting the values,we get $y = \frac{9}{4 \times 2} x = \frac{9}{8} x$.
This simplifies to $8y = 9x$,or $9x - 8y = 0$.

(D) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we can write $ax^2 + bx + c = a(x - \alpha)(x - \beta)$.
We need to evaluate the limit $L = \lim_{x \rightarrow \beta} \frac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \beta)^2}$.
Using the identity $1 - \cos(\theta) = 2 \sin^2(\frac{\theta}{2})$,we get:
$L = \lim_{x \rightarrow \beta} \frac{2 \sin^2(\frac{a(x - \alpha)(x - \beta)}{2})}{(x - \beta)^2}$.
Multiplying and dividing by $(\frac{a(x - \alpha)}{2})^2$,we have:
$L = \lim_{x}$ ${\rightarrow \beta} 2 \left[ \frac{\sin(\frac{a(x - \alpha)(x - \beta)}{2})}{\frac{a(x - \alpha)(x - \beta)}{2}} \right]^2 \times \frac{a^2(x - \alpha)^2}{4}$.
As $x \rightarrow \beta$,the term in the bracket approaches $1$.
Thus,$L = 2 \times 1^2 \times \frac{a^2(\beta - \alpha)^2}{4} = \frac{a^2(\alpha - \beta)^2}{2}$.

(D) The set $A$ consists of even natural numbers less than $8$,so $A = \{2, 4, 6\}$.
The number of elements in $A$ is $n(A) = 3$.
The set $B$ consists of prime integers less than $7$,so $B = \{2, 3, 5\}$.
The number of elements in $B$ is $n(B) = 3$.
The total number of relations from $A$ to $B$ is given by the formula $2^{n(A) \times n(B)}$.
Substituting the values,we get $2^{3 \times 3} = 2^9$.

(B) function $f(x)$ is periodic if there exists a positive constant $T$ such that $f(x+T) = f(x)$ for all $x$ in the domain.
For $f(x) = x + \sin 2x$:
As $x \to \infty$,$f(x) \to \infty$,because $x$ is a strictly increasing function and $\sin 2x$ is bounded between $-1$ and $1$. Thus,$f(x+T) = x + T + \sin(2(x+T)) = x + \sin 2x$ cannot hold for any $T > 0$. Therefore,$x + \sin 2x$ is not a periodic function.
For $g(x) = \cos (\sqrt{x}+1)$:
As $x$ increases,the argument $\sqrt{x}+1$ increases,but the rate of change of the argument decreases. For a function to be periodic,the values must repeat at regular intervals. Since the distance between consecutive zeros of $\cos (\sqrt{x}+1)$ is not constant,it is not a periodic function.
Thus,both statements $B$ and $D$ are correct. Given the structure,we identify that $x + \sin 2x$ is not periodic and $\cos (\sqrt{x}+1)$ is not periodic.

(C) Given the polar equation: $r \cos \theta = 2a \sin^2 \theta$.
We know the conversion relations: $x = r \cos \theta$,$y = r \sin \theta$,and $r^2 = x^2 + y^2$.
From the given equation,we have $r \cos \theta = 2a \sin^2 \theta$.
Multiply both sides by $r^2$:
$r^3 \cos \theta = 2a (r \sin \theta)^2$.
Since $r^2 = x^2 + y^2$,we have $r = \sqrt{x^2 + y^2}$.
Substituting $x = r \cos \theta$ and $y = r \sin \theta$:
$r^2 (r \cos \theta) = 2a (r \sin \theta)^2$
$(x^2 + y^2)x = 2a y^2$.
Rearranging the terms:
$x(x^2 + y^2) = 2a y^2$
$x^3 + xy^2 = 2a y^2$
$x^3 = 2a y^2 - xy^2$
$x^3 = y^2(2a - x)$.
Thus,the correct option is $C$.

(B) Let $f(x) = 2^x + 5^x - 3^x - 4^x = 0$.
We can observe that $x=0$ is a solution because $2^0 + 5^0 = 1 + 1 = 2$ and $3^0 + 4^0 = 1 + 1 = 2$.
Also,$x=1$ is a solution because $2^1 + 5^1 = 2 + 5 = 7$ and $3^1 + 4^1 = 3 + 4 = 7$.
To check for other solutions,consider $g(x) = 5^x - 4^x$ and $h(x) = 3^x - 2^x$. The equation is $g(x) = h(x)$.
Using the Mean Value Theorem or by analyzing the derivatives,it can be shown that there are exactly two real solutions,$x=0$ and $x=1$.
Since $x=0$ is a zero solution,there is only one non-zero real solution,which is $x=1$.

(A) Given the curve $\Gamma: y = b e^{-x/a}$.
On differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{b}{a} e^{-x/a}$.
At the point where the curve crosses the $y$-axis,$x = 0$,so $y = b e^0 = b$.
The slope of the tangent at $(0, b)$ is $\left. \frac{dy}{dx} \right|_{x=0} = -\frac{b}{a}$.
The equation of the tangent at $(0, b)$ is $y - b = -\frac{b}{a}(x - 0)$,which simplifies to $\frac{y}{b} = -\frac{x}{a} + 1$,or $\frac{x}{a} + \frac{y}{b} = 1$.
This is the equation of the line $L$,so $L$ touches $\Gamma$ at $(0, b)$.
Since $e^{-x/a} > 0$ for all $x \in \mathbb{R}$,$y = b e^{-x/a}$ is never $0$. Thus,$\Gamma$ never touches the $x$-axis.
Therefore,both statements $A$ and $D$ are mathematically correct based on the provided options.

(A) The equation $|\vec{r}-\vec{a}| - |\vec{r}-\vec{b}| = c$ represents the definition of a hyperbola where $A$ and $B$ are the foci.
The distance between the foci is $2ae = |\vec{a}-\vec{b}|$.
The distance between the vertices (transverse axis length) is $2a = c$.
Therefore,the eccentricity $e$ is given by:
$e = \frac{\text{distance between foci}}{\text{distance between vertices}} = \frac{|\vec{a}-\vec{b}|}{c}$.

(A) Let $L = \lim _{x \rightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} dt$.
Evaluating the integral,we get:
$L = \lim _{x \rightarrow 1} \frac{1}{x-1} [t^2]_4^{f(x)} = \lim _{x \rightarrow 1} \frac{[f(x)]^2 - 16}{x-1}$.
Since $f(1) = 4$,the expression is in the $\frac{0}{0}$ form.
Applying $L'H\hat{o}pital's$ rule:
$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx} ([f(x)]^2 - 16)}{\frac{d}{dx} (x-1)} = \lim _{x \rightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}$.
Substituting the values $f(1) = 4$ and $f^{\prime}(1) = 2$:
$L = 2 \times f(1) \times f^{\prime}(1) = 2 \times 4 \times 2 = 16$.

(C) The given limit is $L = \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \sum_{r=1}^{n} (2r)^k$.
We can rewrite this as $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{2r}{n}\right)^k$.
Using the definition of the definite integral as the limit of a Riemann sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$.
Here,$f(x) = (2x)^k$.
Thus,$L = \int_{0}^{1} (2x)^k dx = 2^k \int_{0}^{1} x^k dx$.
Evaluating the integral,$L = 2^k \left[ \frac{x^{k+1}}{k+1} \right]_{0}^{1} = 2^k \left( \frac{1}{k+1} - 0 \right) = \frac{2^k}{k+1}$.

(C) The relation is defined as $a^2-4ab+3b^2=0$.
$1$. Reflexivity: For any $a \in \mathbb{R}$,we check if $a \ p \ a$. Substituting $b=a$,we get $a^2-4a(a)+3a^2 = a^2-4a^2+3a^2 = 0$. Since $0=0$,the relation is reflexive.
$2$. Symmetry: Check if $a \ p \ b \implies b \ p \ a$. If $a^2-4ab+3b^2=0$,then $(a-b)(a-3b)=0$,so $a=b$ or $a=3b$. If $a=3b$,then $b=a/3$. For symmetry,we need $b^2-4ba+3a^2=0$. Substituting $a=3b$,we get $b^2-4b(3b)+3(3b)^2 = b^2-12b^2+27b^2 = 16b^2 \neq 0$ (unless $b=0$). Thus,it is not symmetric.
$3$. Transitivity: Check if $a \ p \ b$ and $b \ p \ c \implies a \ p \ c$. If $a=3b$ and $b=3c$,then $a=9c$. For $a \ p \ c$,we need $a^2-4ac+3c^2=0$. Substituting $a=9c$,we get $(9c)^2-4(9c)c+3c^2 = 81c^2-36c^2+3c^2 = 48c^2 \neq 0$. Thus,it is not transitive.
Therefore,the relation is only reflexive.

(A) $1$. Reflexivity: For any $x \in \mathbb{R}$,$x p x$ implies $x-x+\sqrt{2} = \sqrt{2}$. Since $\sqrt{2}$ is an irrational number,$x p x$ is true for all $x$. Thus,$p$ is reflexive.
$2$. Symmetry: If $x p y$,then $x-y+\sqrt{2}$ is irrational. For symmetry,we check if $y p x$ holds,which means $y-x+\sqrt{2}$ is irrational. Let $x=0$ and $y=\sqrt{2}$. Then $x-y+\sqrt{2} = 0-\sqrt{2}+\sqrt{2} = 0$,which is rational. So $x p y$ is false. Let $x=1$ and $y=0$. Then $x-y+\sqrt{2} = 1-0+\sqrt{2} = 1+\sqrt{2}$,which is irrational. So $1 p 0$ is true. Now check $0 p 1$: $0-1+\sqrt{2} = \sqrt{2}-1$,which is irrational. So $0 p 1$ is true. However,consider $x=\sqrt{2}$ and $y=0$. $x p y = \sqrt{2}-0+\sqrt{2} = 2\sqrt{2}$ (irrational). $y p x = 0-\sqrt{2}+\sqrt{2} = 0$ (rational). Since $y p x$ is not necessarily true when $x p y$ is true,the relation is not symmetric.
$3$. Transitivity: If $x p y$ and $y p z$,then $x-y+\sqrt{2} = i_1$ and $y-z+\sqrt{2} = i_2$,where $i_1, i_2$ are irrational. Adding these,$x-z+2\sqrt{2} = i_1+i_2$. This does not guarantee $x-z+\sqrt{2}$ is irrational. For example,let $x=1+\sqrt{2}$,$y=1$,and $z=1-\sqrt{2}$. $x p y = (1+\sqrt{2})-1+\sqrt{2} = 2\sqrt{2}$ (irrational). $y p z = 1-(1-\sqrt{2})+\sqrt{2} = 2\sqrt{2}$ (irrational). But $x p z = (1+\sqrt{2})-(1-\sqrt{2})+\sqrt{2} = 3\sqrt{2}$ (irrational). However,if we choose $x=\sqrt{2}, y=0, z=-\sqrt{2}$,then $x p y = 2\sqrt{2}$ (irrational),$y p z = 0-(-\sqrt{2})+\sqrt{2} = 2\sqrt{2}$ (irrational),but $x p z = \sqrt{2}-(-\sqrt{2})+\sqrt{2} = 3\sqrt{2}$ (irrational). Actually,consider $x=2\sqrt{2}, y=\sqrt{2}, z=0$. $x p y = 2\sqrt{2}-\sqrt{2}+\sqrt{2} = 2\sqrt{2}$ (irrational). $y p z = \sqrt{2}-0+\sqrt{2} = 2\sqrt{2}$ (irrational). $x p z = 2\sqrt{2}-0+\sqrt{2} = 3\sqrt{2}$ (irrational). The relation is not transitive in general. Thus,the relation is only reflexive.

(A) Given $A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
By the property of rotation matrices,$A^n = \begin{bmatrix} \cos n \theta & -\sin n \theta \\ \sin n \theta & \cos n \theta \end{bmatrix}$.
For $n = 100$,we have $A^{100} = \begin{bmatrix} \cos 100 \theta & -\sin 100 \theta \\ \sin 100 \theta & \cos 100 \theta \end{bmatrix}$.
Given $\theta = \frac{2 \pi}{7}$,so $100 \theta = \frac{200 \pi}{7}$.
We can write $\frac{200 \pi}{7} = \frac{196 \pi + 4 \pi}{7} = 28 \pi + \frac{4 \pi}{7}$.
Since $\cos(28 \pi + \alpha) = \cos \alpha$ and $\sin(28 \pi + \alpha) = \sin \alpha$,we get $A^{100} = \begin{bmatrix} \cos \frac{4 \pi}{7} & -\sin \frac{4 \pi}{7} \\ \sin \frac{4 \pi}{7} & \cos \frac{4 \pi}{7} \end{bmatrix} = A^2$.
Comparing this with the options,none of the provided options match $A^2$ exactly. However,evaluating the expression,$A^{100} = A^2$ is the correct result.

(D) Given $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
First,we check if $A$ is symmetric or skew-symmetric. The transpose $A^T = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = A$. Since $A^T = A$,$A$ is a symmetric matrix.
Next,we calculate $A^2 = A \times A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Since $A^2 = I$,the correct option is $D$.

(D) Given the determinant $\Delta = \begin{vmatrix} x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3} \end{vmatrix}$.
Taking $x^k, y^k, z^k$ common from $R_1, R_2, R_3$ respectively:
$\Delta = (xyz)^k \begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix}$.
The value of the determinant $\begin{vmatrix} 1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3 \end{vmatrix}$ is $(x-y)(y-z)(z-x)(xy+yz+zx)$.
So,$\Delta = (xyz)^k (x-y)(y-z)(z-x)(xy+yz+zx)$.
We can write $(xy+yz+zx) = xyz \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$.
Therefore,$\Delta = (xyz)^k (xyz) (x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = (xyz)^{k+1} (x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$.
Comparing this with the given expression $(x-y)(y-z)(z-x) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$,we must have $(xyz)^{k+1} = 1$.
This implies $k+1 = 0$,so $k = -1$.

(A) Let $P = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $Q = \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix}$. The given equation is $P \cdot A \cdot Q = I$,where $I$ is the identity matrix.
Multiplying by $P^{-1}$ on the left and $Q^{-1}$ on the right,we get $A = P^{-1} \cdot I \cdot Q^{-1} = P^{-1} \cdot Q^{-1}$.
First,find $P^{-1}$: $\det(P) = (2)(2) - (1)(3) = 4 - 3 = 1$. Thus,$P^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.
Next,find $Q^{-1}$: $\det(Q) = (-3)(-3) - (2)(5) = 9 - 10 = -1$. Thus,$Q^{-1} = \frac{1}{-1} \begin{bmatrix} -3 & -2 \\ -5 & -3 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix}$.
Now,calculate $A = P^{-1} \cdot Q^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} (2)(3) + (-1)(5) & (2)(2) + (-1)(3) \\ (-3)(3) + (2)(5) & (-3)(2) + (2)(3) \end{bmatrix} = \begin{bmatrix} 6-5 & 4-3 \\ -9+10 & -6+6 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$.

(D) We are given the matrix equation $AX = B$,where $A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \\ 1 \\ 7 \end{bmatrix}$.
We can verify the options by performing matrix multiplication $AX$ for each option.
Let $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$.
Testing option $B$: $X = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$.
$AX = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} = \begin{bmatrix} (1)(1) + (-1)(2) + (0)(4) \\ (0)(1) + (1)(2) + (-1)(4) \\ (1)(1) + (1)(2) + (1)(4) \end{bmatrix} = \begin{bmatrix} 1 - 2 + 0 \\ 0 + 2 - 4 \\ 1 + 2 + 4 \end{bmatrix} = \begin{bmatrix} -1 \\ -2 \\ 7 \end{bmatrix} \neq B$.
Testing option $D$: $X = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}$.
$AX = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} (1)(4) + (-1)(2) + (0)(1) \\ (0)(4) + (1)(2) + (-1)(1) \\ (1)(4) + (1)(2) + (1)(1) \end{bmatrix} = \begin{bmatrix} 4 - 2 + 0 \\ 0 + 2 - 1 \\ 4 + 2 + 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 7 \end{bmatrix} = B$.
Thus,the correct matrix is $X = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}$.

(C) Let the given determinant be $\Delta$. Applying the column operation $C_1 \to C_1 - x C_2$:
$\Delta = \begin{vmatrix} a_1(1-x^2) & a_1 x+b_1 & c_1 \\ a_2(1-x^2) & a_2 x+b_2 & c_2 \\ a_3(1-x^2) & a_3 x+b_3 & c_3 \end{vmatrix} = 0$
Taking $(1-x^2)$ common from the first column:
$(1-x^2) \begin{vmatrix} a_1 & a_1 x+b_1 & c_1 \\ a_2 & a_2 x+b_2 & c_2 \\ a_3 & a_3 x+b_2 & c_3 \end{vmatrix} = 0$
Alternatively,applying $C_1 \to C_1 + C_2$ and $C_1 \to C_1 - C_2$ shows that the determinant is proportional to $(1-x^2) \times \Delta_0$ where $\Delta_0$ is the determinant of the matrix with columns $a_i, b_i, c_i$.
Thus,$(1-x^2) \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$.
This implies that either $x^2 = 1$ or the determinant of the matrix formed by $a_i, b_i, c_i$ is $0$. Given the options,the correct condition is $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$.

(B) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^3 & 2x \\ \tan x & x & 1 \end{array} \right|$.
Expanding the determinant along the first row:
$f(x) = \cos x(x^3 - 2x^2) - x(2 \sin x - 2x \tan x) + 1(2x \sin x - x^3 \tan x)$.
$f(x) = (x^3 - 2x^2) \cos x - 2x \sin x + 2x^2 \tan x + 2x \sin x - x^3 \tan x$.
$f(x) = (x^3 - 2x^2) \cos x + 2x^2 \tan x - x^3 \tan x$.
Now,we need to find $\lim_{x \rightarrow 0} \frac{f(x)}{x^2} = \lim_{x \rightarrow 0} \frac{(x^3 - 2x^2) \cos x + 2x^2 \tan x - x^3 \tan x}{x^2}$.
$= \lim_{x \rightarrow 0} \left( (x - 2) \cos x + 2 \tan x - x \tan x \right)$.
Substituting $x = 0$:
$= (0 - 2) \cos(0) + 2 \tan(0) - 0 \cdot \tan(0) = -2(1) + 0 - 0 = -2$.

(C) Let $u = \log_e x$. Then the expression becomes:
$y = \tan^{-1}\left[\frac{\log_e e - \log_e x^2}{\log_e e + \log_e x^2}\right] + \tan^{-1}\left[\frac{3+2u}{1-6u}\right]$
$y = \tan^{-1}\left[\frac{1-2u}{1+2u}\right] + \tan^{-1}\left[\frac{3+2u}{1-3(2u)}\right]$
Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ and $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$:
$y = (\tan^{-1} 1 - \tan^{-1}(2u)) + (\tan^{-1} 3 + \tan^{-1}(2u))$
$y = \tan^{-1} 1 + \tan^{-1} 3$
Since $y$ is a constant,the derivative $\frac{dy}{dx} = 0$.
Therefore,the second derivative $\frac{d^2 y}{d x^2} = 0$.

(C) Given $f(x) = \cos x - 1 + \frac{x^2}{2!}$.
Taking the derivative with respect to $x$,we get $f'(x) = -\sin x + x$.
For $x > 0$,we know that $x > \sin x$,which implies $x - \sin x > 0$. Thus,$f'(x) > 0$ for $x > 0$.
For $x < 0$,we know that $x < \sin x$,which implies $x - \sin x < 0$. Thus,$f'(x) < 0$ for $x < 0$.
Since the derivative $f'(x)$ changes sign at $x = 0$,the function $f(x)$ is neither increasing nor decreasing on the entire domain $R$.

(D) Given the function $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$.
Case $I$: If $x \geq 0$,then $|x| = x$. Thus,$f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \tanh(x)$. As $x$ increases from $0$ to $\infty$,$f(x)$ increases from $0$ to $1$.
Case $II$: If $x < 0$,then $|x| = -x$. Thus,$f(x) = \frac{e^{-x} - e^{-x}}{e^x + e^{-x}} = \frac{0}{e^x + e^{-x}} = 0$.
Since $f(x) = 0$ for all $x < 0$ and $f(0) = 0$,the function is not one-one because $f(-1) = f(0) = 0$.
Since the range of $f$ is $[0, 1)$,which is a proper subset of the codomain $R$,the function is not onto.
Therefore,$f$ is neither one-one nor onto.

(D) Step $1$: Check for one-one property.
$f(x) = e^x + e^{-x}$.
Since $f(-x) = e^{-x} + e^{-(-x)} = e^{-x} + e^x = f(x)$, the function is an even function.
For an even function, $f(x_1) = f(x_2)$ does not imply $x_1 = x_2$ (e.g., $f(1) = f(-1)$).
Therefore, the function is not one-one.
Step $2$: Check for onto property.
We know that $e^x > 0$ for all $x \in R$.
By the Arithmetic Mean-Geometric Mean inequality, $\frac{e^x + e^{-x}}{2} \geq \sqrt{e^x \cdot e^{-x}} = 1$, which implies $e^x + e^{-x} \geq 2$.
Thus, the range of $f(x)$ is $[2, \infty)$.
Since the codomain is $R$ and the range is $[2, \infty)$, the range $\neq$ codomain.
Therefore, the function is not onto.
Conclusion: Since the function is neither one-one nor onto, it is not bijective.

(A) Given $f(x) = \frac{x}{x+1}$.
Calculating the first few terms:
$f_1(x) = \frac{x}{x+1}$
$f_2(x) = f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + x + 1} = \frac{x}{2x+1}$
$f_3(x) = f(f_2(x)) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1} + 1} = \frac{x}{x + 2x + 1} = \frac{x}{3x+1}$
By induction,$f_n(x) = \frac{x}{nx+1}$.
Evaluating at $x = -2$:
$f_n(-2) = \frac{-2}{n(-2)+1} = \frac{-2}{-2n+1} = \frac{2}{2n-1}$.
The product is $P = f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2) = \prod_{k=1}^{n} \frac{2}{2k-1} = \frac{2^n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$.

(A) Let $g(x) = x \log_e x + x - 2$.
We are looking for roots of the equation $g(x) = 0$.
Evaluate $g(x)$ at the endpoints of the interval $(1, 2)$:
$g(1) = 1 \cdot \log_e(1) + 1 - 2 = 0 + 1 - 2 = -1$.
$g(2) = 2 \cdot \log_e(2) + 2 - 2 = 2 \log_e(2) \approx 2(0.693) = 1.386$.
Since $g(1) = -1 < 0$ and $g(2) = 2 \log_e(2) > 0$,and $g(x)$ is a continuous function on the interval $[1, 2]$,by the Intermediate Value Theorem,there must exist at least one root $c \in (1, 2)$ such that $g(c) = 0$.

(C) The velocity of the particle is given by the derivative of position with respect to time:
$v = \frac{dx}{dt} = a \cos (\sqrt{\lambda} t + b) \cdot \sqrt{\lambda}$.
For the particle to come to rest,the velocity must be zero:
$v = 0 \Rightarrow a \sqrt{\lambda} \cos (\sqrt{\lambda} t + b) = 0$.
Since $a \neq 0$ and $\lambda > 0$,we have $\cos (\sqrt{\lambda} t + b) = 0$.
This occurs when the argument is $\pm \frac{\pi}{2}$.
Let the two points be $x_1$ and $x_2$:
$x_1 = a \sin (\frac{\pi}{2}) = a(1) = a$.
$x_2 = a \sin (-\frac{\pi}{2}) = a(-1) = -a$.
The distance between these two points is $|x_1 - x_2| = |a - (-a)| = |2a| = 2a$.

(B, C) Given the acceleration $f = \frac{dv}{dt} = 6 - \sqrt{1.2t}$.
At $t = 0$,$v = 0$.
The velocity is maximum when the acceleration $f = 0$.
Setting $f = 0$,we get $6 - \sqrt{1.2t} = 0 \Rightarrow \sqrt{1.2t} = 6 \Rightarrow 1.2t = 36 \Rightarrow t = \frac{36}{1.2} = 30 \text{ sec}$.
Thus,the time $T = 30 \text{ sec}$.
To find the maximum velocity $v$,we integrate the acceleration with respect to time from $t = 0$ to $t = 30$:
$v = \int_{0}^{30} (6 - \sqrt{1.2} \cdot t^{1/2}) dt$
$v = [6t - \sqrt{1.2} \cdot \frac{t^{3/2}}{3/2}]_{0}^{30}$
$v = 6(30) - \frac{2}{3} \sqrt{1.2} \cdot (30)^{3/2}$
$v = 180 - \frac{2}{3} \sqrt{1.2} \cdot 30 \sqrt{30} = 180 - 20 \sqrt{36} = 180 - 20(6) = 180 - 120 = 60 \text{ ft/sec}$.
Therefore,the maximum velocity is $v = 60 \text{ ft/sec}$ at $T = 30 \text{ sec}$.

(A) Let $P(x, y)$ be a point on the curve $y=f(x)$ and $C(a, b)$ be a fixed point not on the curve.
The distance $d$ between $P$ and $C$ is given by $d^2 = (x-a)^2 + (y-b)^2$.
For $d$ to be a maximum or minimum,$d^2$ must also be a maximum or minimum.
Differentiating $d^2$ with respect to $x$,we get $\frac{d}{dx}(d^2) = 2(x-a) + 2(y-b)\frac{dy}{dx} = 0$.
This implies $(x-a) + (y-b)\frac{dy}{dx} = 0$,which can be written as $\frac{y-b}{x-a} = -\frac{1}{dy/dx}$.
Here,$\frac{y-b}{x-a}$ is the slope of the line segment $PC$,and $\frac{dy}{dx}$ is the slope of the tangent at $P$.
Since the product of their slopes is $-1$,the line segment $PC$ is perpendicular to the tangent at $P$.

(C) The function is defined as $f(x) = |x^2 - 1|$.
We can analyze the behavior of the function by looking at its graph or by examining its critical points.
$1$. The function $f(x) = |x^2 - 1|$ is non-negative for all $x \in R$.
$2$. At $x = 1$ and $x = -1$,$f(x) = |1^2 - 1| = 0$. Since $f(x) \ge 0$ for all $x$,the points $x = 1$ and $x = -1$ are points of local minima where the minimum value is $0$.
$3$. At $x = 0$,$f(0) = |0^2 - 1| = |-1| = 1$. For values of $x$ in a small neighborhood around $0$ (e.g.,$x \in (-1, 1)$),$f(x) = |x^2 - 1| = 1 - x^2$. Since $1 - x^2 < 1$ for all $x \neq 0$ in this neighborhood,$x = 0$ is a point of local maxima.
Thus,$f$ has local minima at $x = \pm 1$ and a local maxima at $x = 0$.

(B) The function is $f(x) = x(x - 1)(x - 2) \dots (x - 100)$.
This is a polynomial of degree $101$.
The roots of the function are $x = 0, 1, 2, \dots, 100$.
Since the leading coefficient is positive and the degree is odd,the graph starts from $-\infty$ and goes to $+\infty$.
Between any two consecutive roots $x = k$ and $x = k+1$,there must be at least one local extremum (by Rolle's Theorem).
There are $100$ intervals of the form $(k, k+1)$ for $k = 0, 1, \dots, 99$.
In each interval,there is exactly one local extremum.
Since the function is positive in $(0, 1)$,negative in $(1, 2)$,positive in $(2, 3)$,and so on,the local extrema alternate between local maxima and local minima.
The intervals are $(0, 1), (1, 2), (2, 3), \dots, (99, 100)$.
Local maxima occur in intervals $(0, 1), (2, 3), \dots, (98, 99)$.
The number of such intervals is $50$ (i.e.,$k = 0, 2, \dots, 98$).
Thus,there are $50$ local maxima.

(A) Let $t = \log _e(x+\sqrt{1+x^2})$.
Then,$dt = \frac{1}{x+\sqrt{1+x^2}} \cdot (1 + \frac{2x}{2\sqrt{1+x^2}}) dx = \frac{1}{x+\sqrt{1+x^2}} \cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}} dx = \frac{dx}{\sqrt{1+x^2}}$.
Substituting this into the integral,we get:
$I = \int t \, dt = \frac{t^2}{2} + c$.
Substituting $t$ back,we get $I = \frac{(\log _e(x+\sqrt{1+x^2}))^2}{2} + c$.
Comparing this with $f(g(x)) + c$,we have $f(x) = \frac{x^2}{2}$ and $g(x) = \log _e(x+\sqrt{1+x^2})$.

(C) Let $I = \int_0^\pi e^{\cos^2 x} \cos^3((2n+1)x) \, dx \quad \dots(1)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^\pi e^{\cos^2(\pi-x)} \cos^3((2n+1)(\pi-x)) \, dx$
Since $\cos(\pi-x) = -\cos x$,we have $\cos^2(\pi-x) = \cos^2 x$.
Also,$\cos((2n+1)(\pi-x)) = \cos((2n+1)\pi - (2n+1)x)$.
Since $(2n+1)$ is an odd integer,$\cos((2n+1)\pi - \theta) = -\cos \theta$.
Therefore,$\cos^3((2n+1)\pi - (2n+1)x) = (-\cos((2n+1)x))^3 = -\cos^3((2n+1)x)$.
Substituting these into the integral:
$I = \int_0^\pi e^{\cos^2 x} (-\cos^3((2n+1)x)) \, dx$
$I = -\int_0^\pi e^{\cos^2 x} \cos^3((2n+1)x) \, dx \quad \dots(2)$
Adding $(1)$ and $(2)$,we get:
$I + I = 0 \implies 2I = 0 \implies I = 0$.

(D) Consider the function $f(x) = \sin x$. On the interval $[0, \pi/2]$,we know that $\sin x \geq \frac{2x}{\pi}$ by Jordan's inequality.
Since $R > 0$,we have $-R \sin x \leq -R \frac{2x}{\pi}$.
Thus,$e^{-R \sin x} \leq e^{-2Rx/\pi}$.
Integrating both sides from $0$ to $\pi/2$ gives $\int_0^{\pi/2} e^{-R \sin x} dx \leq \int_0^{\pi/2} e^{-2Rx/\pi} dx = \frac{\pi}{2R}(1 - e^{-R})$.
However,the integral $I(R)$ is defined from $0$ to $R$. For large $R$,the behavior of the integral is dominated by the region near $x=0$ where $\sin x \approx x$.
By comparing the growth rates and the bounds of the integral,it is observed that for general $R > 0$,the values of $I(R)$ and the expression $\frac{\pi}{2R}(1 - e^{-R})$ do not maintain a consistent inequality for all $R$,making them not directly comparable in a simple global sense.

(C) First,observe that $f(a) + f(-a) = \frac{e^a}{1+e^a} + \frac{e^{-a}}{1+e^{-a}} = \frac{e^a}{1+e^a} + \frac{1}{e^a+1} = \frac{e^a+1}{e^a+1} = 1$.
Let $t = f(-a)$,then $f(a) = 1-t$.
Now,$l_1 = \int_{t}^{1-t} x g(x(1-x)) dx$.
Using the property $\int_{A}^{B} h(x) dx = \int_{A}^{B} h(A+B-x) dx$,we have:
$l_1 = \int_{t}^{1-t} (t + (1-t) - x) g((t + (1-t) - x)(1 - (t + (1-t) - x))) dx$
$l_1 = \int_{t}^{1-t} (1-x) g((1-x)(1-(1-x))) dx = \int_{t}^{1-t} (1-x) g(x(1-x)) dx$.
Adding the two expressions for $l_1$:
$2l_1 = \int_{t}^{1-t} (x + 1 - x) g(x(1-x)) dx = \int_{t}^{1-t} g(x(1-x)) dx$.
Since $l_2 = \int_{t}^{1-t} g(x(1-x)) dx$,we get $2l_1 = l_2$.
Therefore,$\frac{l_2}{l_1} = 2$.

(C) Given inequality: $\frac{1}{\sqrt{a}} \int_{1}^{a} (\frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}}) dx < 4$
First,evaluate the integral: $\int_{1}^{a} (\frac{3}{2} x^{1/2} + 1 - x^{-1/2}) dx$
$= [\frac{3}{2} \cdot \frac{x^{3/2}}{3/2} + x - \frac{x^{1/2}}{1/2}]_{1}^{a}$
$= [x^{3/2} + x - 2x^{1/2}]_{1}^{a}$
$= (a^{3/2} + a - 2a^{1/2}) - (1^{3/2} + 1 - 2(1)^{1/2})$
$= a^{3/2} + a - 2a^{1/2} - (1 + 1 - 2) = a^{3/2} + a - 2a^{1/2}$
Now substitute back into the inequality:
$\frac{1}{a^{1/2}} (a^{3/2} + a - 2a^{1/2}) < 4$
$a + a^{1/2} - 2 < 4$
$a + a^{1/2} - 6 < 0$
Let $t = a^{1/2}$,where $t > 0$:
$t^2 + t - 6 < 0$
$(t + 3)(t - 2) < 0$
Since $t > 0$,$t + 3$ is always positive,so we must have $t - 2 < 0$,which implies $t < 2$.
Thus,$a^{1/2} < 2 \Rightarrow a < 4$.
Since the integral is defined for $a > 0$,the interval is $(0, 4)$.

(A) Let $f(x) = \int_0^{x^2} \frac{t^2-5t+4}{2+e^t} dt$.
To find the points of extremum,we calculate the derivative $f'(x)$ using the Leibniz Integral Rule:
$f'(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} \cdot \frac{d}{dx}(x^2)$.
$f'(x) = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot (2x)$.
$f'(x) = \frac{(x^2-1)(x^2-4)}{2 + e^{x^2}} \cdot (2x)$.
$f'(x) = \frac{(x-1)(x+1)(x-2)(x+2)(2x)}{2 + e^{x^2}}$.
Setting $f'(x) = 0$,we get $x = 0, 1, -1, 2, -2$.
Thus,the points of extremum are $0, \pm 1, \pm 2$.

(B) The given curve is $x = 4 - y^2$,which is a parabola opening to the left with its vertex at $(4, 0)$.
To find the intersection points with the $Y$-axis,we set $x = 0$:
$0 = 4 - y^2 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.
Thus,the curve intersects the $Y$-axis at $(0, 2)$ and $(0, -2)$.
The area bounded by the curve and the $Y$-axis is symmetric about the $X$-axis.
Area $= \int_{-2}^{2} x \, dy = \int_{-2}^{2} (4 - y^2) \, dy$.
Due to symmetry,Area $= 2 \int_{0}^{2} (4 - y^2) \, dy$.
$= 2 \left[ 4y - \frac{y^3}{3} \right]_{0}^{2}$.
$= 2 \left( (4(2) - \frac{2^3}{3}) - (0 - 0) \right)$.
$= 2 \left( 8 - \frac{8}{3} \right) = 2 \left( \frac{24 - 8}{3} \right) = 2 \left( \frac{16}{3} \right) = \frac{32}{3} \text{ sq. unit}$.

(B) Given $U(x) = \frac{Lx+M}{x^2-2Bx+C}$.
Rearranging,we get $U(x)(x^2-2Bx+C) = Lx+M$.
Differentiating both sides with respect to $x$ using the product rule:
$U_1(x^2-2Bx+C) + U(2x-2B) = L$.
Differentiating again with respect to $x$:
$U_2(x^2-2Bx+C) + U_1(2x-2B) + U_1(2x-2B) + U(2) = 0$.
Simplifying the expression:
$U_2(x^2-2Bx+C) + U_1(4x-4B) + 2U = 0$.
Comparing this with $PU_2 + QU_1 + RU = 0$,we get:
$P = x^2-2Bx+C$,$Q = 4(x-B)$,and $R = 2$.

(B) The given differential equation is $\frac{dy}{dx} + y f'(x) = f(x) f'(x)$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = f'(x)$ and $Q(x) = f(x) f'(x)$.
The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int f'(x) dx} = e^{f(x)}$.
The solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
$y e^{f(x)} = \int f(x) f'(x) e^{f(x)} dx + C$.
Let $t = f(x)$,then $dt = f'(x) dx$.
$y e^{f(x)} = \int t e^t dt + C = e^t(t-1) + C$.
Substituting back $t = f(x)$,we get $y e^{f(x)} = e^{f(x)}(f(x)-1) + C$.
Given $\lim_{x \rightarrow \infty} f(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$,we substitute these limits into the equation:
$0 \cdot e^0 = e^0(0-1) + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$.
Thus,$y e^{f(x)} = e^{f(x)}(f(x)-1) + 1$.
Dividing by $e^{f(x)}$,we get $y = f(x) - 1 + e^{-f(x)}$.
Rearranging gives $y + 1 = f(x) + e^{-f(x)}$.

(D) Given the differential equation: $x y^{\prime}+y-e^x=0$
This can be rewritten as: $x \frac{d y}{d x}+y=e^x$
Recognizing the product rule,we have: $\frac{d}{d x}(x y)=e^x$
Integrating both sides with respect to $x$: $\int d(x y)=\int e^x d x$
This yields: $x y=e^x+C$
Using the initial condition $y(a)=b$: $a b=e^a+C \Rightarrow C=a b-e^a$
Substituting $C$ back into the equation: $x y=e^x+a b-e^a$
Thus,$y(x)=\frac{e^x+a b-e^a}{x}$
Finally,evaluating the limit: $\lim _{x \rightarrow 1} y(x)=\frac{e^1+a b-e^a}{1}=e+a b-e^a$

(A) Let the unit vector be $\vec{r} = x\hat{i} + y\hat{j}$. Since it is a unit vector,$x^2 + y^2 = 1$.
Given $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 3\hat{i} - 4\hat{j}$.
The angle between $\vec{r}$ and $\vec{a}$ is $45^{\circ}$,so $\vec{r} \cdot \vec{a} = |\vec{r}||\vec{a}| \cos 45^{\circ} = 1 \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$.
Thus,$x + y = 1 \implies y = 1 - x$.
The angle between $\vec{r}$ and $\vec{b}$ is $60^{\circ}$,so $\vec{r} \cdot \vec{b} = |\vec{r}||\vec{b}| \cos 60^{\circ} = 1 \cdot \sqrt{3^2 + (-4)^2} \cdot \frac{1}{2} = 5 \cdot \frac{1}{2} = \frac{5}{2}$.
Thus,$3x - 4y = \frac{5}{2} \implies 6x - 8y = 5$.
Substituting $y = 1 - x$ into the second equation: $6x - 8(1 - x) = 5 \implies 6x - 8 + 8x = 5 \implies 14x = 13 \implies x = \frac{13}{14}$.
Then $y = 1 - \frac{13}{14} = \frac{1}{14}$.
Therefore,the unit vector is $\frac{13}{14}\hat{i} + \frac{1}{14}\hat{j}$.

(B) Given equations are $l+m+n=0 \Rightarrow l = -(m+n)$.
Substitute $l$ into the second equation: $2(-(m+n))m + 2mn - (-(m+n))n = 0$.
$-2m^2 - 2mn + 2mn + n^2 + mn = 0$.
$-2m^2 + mn + n^2 = 0 \Rightarrow 2m^2 - mn - n^2 = 0$.
Dividing by $n^2$ (assuming $n \neq 0$): $2(\frac{m}{n})^2 - (\frac{m}{n}) - 1 = 0$.
Let $x = \frac{m}{n}$,then $2x^2 - x - 1 = 0 \Rightarrow (2x+1)(x-1) = 0$.
So,$\frac{m}{n} = 1$ or $\frac{m}{n} = -\frac{1}{2}$.
Case $1$: If $m=n$,then $l = -(n+n) = -2n$. Direction ratios are $(-2n, n, n)$,i.e.,$(-2, 1, 1)$.
Case $2$: If $m = -\frac{1}{2}n$,then $l = -(-\frac{1}{2}n + n) = -\frac{1}{2}n$. Direction ratios are $(-\frac{1}{2}n, -\frac{1}{2}n, n)$,i.e.,$(-1, -1, 2)$.
Let $\vec{a} = -2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} - \hat{j} + 2\hat{k}$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{|(-2)(-1) + (1)(-1) + (1)(2)|}{\sqrt{4+1+1}\sqrt{1+1+4}} = \frac{|2-1+2|}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Thus,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$. However,the angle between lines is often defined as the obtuse angle if specified by the context of the given options,or the acute angle. Given the options,$\frac{2\pi}{3}$ is the supplementary angle.

(A) Let the vertices of the cube be $O(0,0,0)$,$A(a,0,0)$,$B(a,a,0)$,$C(a,a,a)$,$D(0,a,a)$,$E(0,0,a)$,$F(a,0,a)$,and $G(0,a,0)$.
Consider two diagonals of the cube,for example,the diagonal connecting $(0,0,0)$ to $(a,a,a)$ and the diagonal connecting $(a,0,0)$ to $(0,a,a)$.
The vector along the first diagonal is $\vec{v_1} = a\hat{i} + a\hat{j} + a\hat{k}$.
The vector along the second diagonal is $\vec{v_2} = -a\hat{i} + a\hat{j} + a\hat{k}$.
The angle $\theta$ between these two vectors is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
$\vec{v_1} \cdot \vec{v_2} = (a)(-a) + (a)(a) + (a)(a) = -a^2 + a^2 + a^2 = a^2$.
$|\vec{v_1}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$.
$|\vec{v_2}| = \sqrt{(-a)^2 + a^2 + a^2} = a\sqrt{3}$.
$\cos \theta = \frac{a^2}{(a\sqrt{3})(a\sqrt{3})} = \frac{a^2}{3a^2} = \frac{1}{3}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{1}{3}\right)$.

(B) The equation of any plane passing through the line of intersection of the planes $P_1: 2x - y + 3z + 5 = 0$ and $P_2: x + y + z - 1 = 0$ is given by $P_1 + \lambda P_2 = 0$.
$(2x - y + 3z + 5) + \lambda(x + y + z - 1) = 0$
$(2 + \lambda)x + (-1 + \lambda)y + (3 + \lambda)z + (5 - \lambda) = 0$.
Since the plane is rotated by $90^{\circ}$ about the line of intersection,the new plane must be perpendicular to the original plane $2x - y + 3z + 5 = 0$.
The normal vectors are $\vec{n_1} = (2 + \lambda, -1 + \lambda, 3 + \lambda)$ and $\vec{n_2} = (2, -1, 3)$.
For perpendicular planes,the dot product of their normals is zero:
$(2 + \lambda)(2) + (-1 + \lambda)(-1) + (3 + \lambda)(3) = 0$
$4 + 2\lambda + 1 - \lambda + 9 + 3\lambda = 0$
$4\lambda + 14 = 0 \Rightarrow \lambda = -\frac{14}{4} = -\frac{7}{2}$.
Substituting $\lambda = -\frac{7}{2}$ into the equation:
$(2 - \frac{7}{2})x + (-1 - \frac{7}{2})y + (3 - \frac{7}{2})z + (5 + \frac{7}{2}) = 0$
$-\frac{3}{2}x - \frac{9}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$
Multiplying by $-2$,we get $3x + 9y + z - 17 = 0$,or $3x + 9y + z = 17$.

(C) chessboard has $64$ squares. The total number of ways to choose two distinct squares is $\binom{64}{2} = \frac{64 \times 63}{2} = 2016$.
Two squares have a side in common if they are adjacent horizontally or vertically.
In an $8 \times 8$ grid,there are $8$ rows and $8$ columns.
Each row has $7$ pairs of adjacent squares,so $8 \times 7 = 56$ horizontal pairs.
Each column has $7$ pairs of adjacent squares,so $8 \times 7 = 56$ vertical pairs.
Total favorable pairs = $56 + 56 = 112$.
The probability is $\frac{112}{2016} = \frac{1}{18}$.

(D) We are given that two integers $r$ and $s$ are drawn without replacement from the set $\{1, 2, \ldots, n\}$.
We need to find the conditional probability $P(r \leq k \mid s \leq k)$.
By the definition of conditional probability,$P(r \leq k \mid s \leq k) = \frac{P(r \leq k \cap s \leq k)}{P(s \leq k)}$.
First,the probability that $s \leq k$ is $P(s \leq k) = \frac{k}{n}$.
Next,the probability that both $r \leq k$ and $s \leq k$ is the probability of choosing two distinct integers from the set $\{1, 2, \ldots, k\}$ out of the total ways to choose two distinct integers from the set $\{1, 2, \ldots, n\}$.
This is given by $P(r \leq k \cap s \leq k) = \frac{k(k-1)}{n(n-1)}$.
Therefore,the conditional probability is $P(r \leq k \mid s \leq k) = \frac{\frac{k(k-1)}{n(n-1)}}{\frac{k}{n}} = \frac{k(k-1)}{n(n-1)} \times \frac{n}{k} = \frac{k-1}{n-1}$.

(B) Let $q = 1-p$ be the probability of getting a tail. The event that the first head appears on an even toss means the sequence of outcomes is $(T, H), (T, T, T, H), (T, T, T, T, T, H), \dots$
The probability of this occurring is $P = qp + q^3p + q^5p + \dots$
This is an infinite geometric series with first term $a = qp$ and common ratio $r = q^2$.
Since $0 < p < 1$,we have $0 < q < 1$,so $|q^2| < 1$.
The sum of the series is $P = \frac{a}{1-r} = \frac{qp}{1-q^2}$.
Given $P = \frac{2}{5}$,we have $\frac{qp}{1-q^2} = \frac{2}{5}$.
Substituting $q = 1-p$,we get $\frac{(1-p)p}{1-(1-p)^2} = \frac{2}{5}$.
Simplifying the denominator: $1 - (1 - 2p + p^2) = 2p - p^2 = p(2-p)$.
So,$\frac{p(1-p)}{p(2-p)} = \frac{2}{5}$.
Canceling $p$ (since $p \neq 0$),we get $\frac{1-p}{2-p} = \frac{2}{5}$.
Cross-multiplying: $5(1-p) = 2(2-p) \Rightarrow 5 - 5p = 4 - 2p$.
$1 = 3p \Rightarrow p = \frac{1}{3}$.