WBJEE 2012 Mathematics Question Paper with Answer and Solution

(A) Given the equations of the lines are $3x + 4y = 9$ and $y = mx + 1$.
Substituting $y = mx + 1$ into the first equation:
$3x + 4(mx + 1) = 9$
$3x + 4mx + 4 = 9$
$x(3 + 4m) = 5$
$x = \frac{5}{3 + 4m}$
For $x$ to be an integer,$(3 + 4m)$ must be a divisor of $5$.
The divisors of $5$ are $\{1, -1, 5, -5\}$.
Case $1$: $3 + 4m = 1 \implies 4m = -2 \implies m = -0.5$ (Not an integer).
Case $2$: $3 + 4m = -1 \implies 4m = -4 \implies m = -1$ (Integer).
Case $3$: $3 + 4m = 5 \implies 4m = 2 \implies m = 0.5$ (Not an integer).
Case $4$: $3 + 4m = -5 \implies 4m = -8 \implies m = -2$ (Integer).
Thus,there are $2$ integer values of $m$,which are $\{-1, -2\}$.

(C) Given equation: $\log _{2}\left(x^{2}+2 x-1\right)=1$
By the definition of logarithm,$\log _{b}(a) = c \implies a = b^{c}$.
So,$x^{2}+2 x-1 = 2^{1} = 2$.
Rearranging the terms: $x^{2}+2 x-3 = 0$.
Factoring the quadratic equation: $(x+3)(x-1) = 0$.
This gives $x = -3$ or $x = 1$.
Check the domain: For $\log _{2}(f(x))$ to be defined,$f(x) > 0$.
If $x = 1$,$x^{2}+2 x-1 = 1+2-1 = 2 > 0$ (Valid).
If $x = -3$,$x^{2}+2 x-1 = 9-6-1 = 2 > 0$ (Valid).
Both solutions are valid,so there are $2$ solutions.

(A) For the logarithmic inequality $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$ to be defined,the arguments must be positive:
$1) \ x^{2}-16 > 0$ $\Rightarrow (x-4)(x+4) > 0$ $\Rightarrow x \in (-\infty, -4) \cup (4, \infty)$
$2) \ 4x-11 > 0 \Rightarrow x > \frac{11}{4} = 2.75$
Combining these,the domain is $x > 4$.
Now,solving the inequality:
$x^{2}-16 \leq 4x-11$
$x^{2}-4x-5 \leq 0$
$(x-5)(x+1) \leq 0$
This holds for $x \in [-1, 5]$.
Taking the intersection of the domain $(x > 4)$ and the solution $(x \in [-1, 5])$,we get $4 < x \leq 5$.

(A) Given that $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are roots of $x^{2}+px+q=0$.
Sum of roots: $(\alpha+\sqrt{\beta}) + (\alpha-\sqrt{\beta}) = -p \Rightarrow 2\alpha = -p \Rightarrow \alpha = -\frac{p}{2}$.
Product of roots: $(\alpha+\sqrt{\beta})(\alpha-\sqrt{\beta}) = q \Rightarrow \alpha^{2}-\beta = q \Rightarrow \beta = \alpha^{2}-q = \frac{p^{2}}{4}-q$.
Thus,$p^{2}-4q = 4\beta$.
Substitute into the equation $(p^{2}-4q)(p^{2}x^{2}+4px)-16q=0$:
$4\beta(p^{2}x^{2}+4px) - 16q = 0$.
Since $p = -2\alpha$,$p^{2} = 4\alpha^{2}$ and $q = \alpha^{2}-\beta$:
$4\beta(4\alpha^{2}x^{2}-8\alpha x) - 16(\alpha^{2}-\beta) = 0$.
Divide by $4$:
$\beta(\alpha^{2}x^{2}-2\alpha x) - (\alpha^{2}-\beta) = 0$.
$\alpha^{2}\beta x^{2} - 2\alpha\beta x - \alpha^{2} + \beta = 0$.
Rearranging terms: $\alpha^{2}(\beta x^{2}-1) - \beta(2\alpha x - 1) = 0$ is not straightforward,so let's simplify $4\beta(p^{2}x^{2}+4px) = 16q$:
$4\beta(4\alpha^{2}x^{2}-8\alpha x) = 16(\alpha^{2}-\beta) \Rightarrow \beta(\alpha^{2}x^{2}-2\alpha x) = \alpha^{2}-\beta$.
$\alpha^{2}\beta x^{2} - 2\alpha\beta x + \beta = \alpha^{2} \Rightarrow \beta(\alpha x - 1)^{2} = \alpha^{2}$.
$(\alpha x - 1)^{2} = \frac{\alpha^{2}}{\beta} \Rightarrow \alpha x - 1 = \pm \frac{\alpha}{\sqrt{\beta}}$.
$\alpha x = 1 \pm \frac{\alpha}{\sqrt{\beta}} \Rightarrow x = \frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}$.

(A) Since $a, b$ and $c$ are in $AP$,we have $2b = a + c$.
Given the quadratic equation $a x^{2} - 2b x + c = 0$.
Substituting $2b = a + c$ into the equation,we get:
$a x^{2} - (a + c) x + c = 0$
$a x^{2} - a x - c x + c = 0$
$a x(x - 1) - c(x - 1) = 0$
$(x - 1)(a x - c) = 0$
Thus,the roots are $x = 1$ and $x = \frac{c}{a}$.

(B) Let $\alpha$ be the common root.
Then,$\alpha^{2}+\alpha+a=0$ ... $(i)$
And $\alpha^{2}+a\alpha+1=0$ ... (ii)
Subtracting (ii) from $(i)$,we get:
$(\alpha^{2}+\alpha+a) - (\alpha^{2}+a\alpha+1) = 0$
$\alpha(1-a) + (a-1) = 0$
$\alpha(1-a) - (1-a) = 0$
$(1-a)(\alpha-1) = 0$
This implies $a=1$ or $\alpha=1$.
Case $1$: If $a=1$,the equations become $x^{2}+x+1=0$,which has no real roots.
Case $2$: If $\alpha=1$,substituting into $(i)$ gives $1^{2}+1+a=0$,so $a=-2$.
For $a=-2$,the equations are $x^{2}+x-2=0$ and $x^{2}-2x+1=0$.
The roots of $x^{2}+x-2=0$ are $x=1, -2$.
The roots of $x^{2}-2x+1=0$ are $x=1, 1$.
The common root is $x=1$,which is real.
Thus,there is exactly one value of $a$ $(a=-2)$ for which the equations have a common real root.

(B) For a quadratic equation $Ax^{2} + Bx + C = 0$ to have roots of opposite signs,the product of the roots must be less than zero.
Product of the roots = $\frac{C}{A} < 0$.
Here,$A = 2$ and $C = a^{2} - 4a$.
So,$\frac{a^{2} - 4a}{2} < 0$.
$a^{2} - 4a < 0$.
$a(a - 4) < 0$.
This inequality holds when $0 < a < 4$.

(D) Let $z = \frac{3}{2} + i \frac{\sqrt{3}}{2} = \sqrt{3} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) = \sqrt{3} e^{i \frac{\pi}{6}}$.
Then,$\left( \frac{3}{2} + i \frac{\sqrt{3}}{2} \right)^{50} = (\sqrt{3})^{50} e^{i \frac{50\pi}{6}} = 3^{25} e^{i \frac{25\pi}{3}}$.
Since $\frac{25\pi}{3} = 8\pi + \frac{\pi}{3}$,we have $e^{i \frac{25\pi}{3}} = e^{i \frac{\pi}{3}} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.
Thus,$3^{25} (x + iy) = 3^{25} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)$.
Comparing real and imaginary parts,we get $x = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$.

(C) Given the condition $\left|z+\frac{2}{z}\right|=2$.
Using the triangle inequality,$|z| = \left|z+\frac{2}{z}-\frac{2}{z}\right| \leq \left|z+\frac{2}{z}\right| + \left|-\frac{2}{z}\right|$.
Substituting the given value,$|z| \leq 2 + \frac{2}{|z|}$.
Multiplying by $|z|$ (since $|z| > 0$),we get $|z|^2 \leq 2|z| + 2$.
Rearranging the terms,$|z|^2 - 2|z| \leq 2$.
Completing the square,$|z|^2 - 2|z| + 1 \leq 2 + 1$,which gives $(|z|-1)^2 \leq 3$.
Taking the square root,$-\sqrt{3} \leq |z|-1 \leq \sqrt{3}$.
Adding $1$ to all sides,$1-\sqrt{3} \leq |z| \leq 1+\sqrt{3}$.
Since $|z| \geq 0$,the maximum value of $|z|$ is $1+\sqrt{3}$.

(B) Let $\omega = \frac{z-1}{z+1}$ be a purely imaginary number.
$A$ complex number $\omega$ is purely imaginary if and only if $\omega + \overline{\omega} = 0$ (where $\omega \neq 0$).
So,$\frac{z-1}{z+1} + \overline{\left(\frac{z-1}{z+1}\right)} = 0$
$\frac{z-1}{z+1} + \frac{\overline{z}-1}{\overline{z}+1} = 0$
$(z-1)(\overline{z}+1) + (\overline{z}-1)(z+1) = 0$
$(z\overline{z} + z - \overline{z} - 1) + (z\overline{z} - z + \overline{z} - 1) = 0$
$2z\overline{z} - 2 = 0$
$z\overline{z} = 1$
Since $|z|^2 = z\overline{z}$,we have $|z|^2 = 1$,which implies $|z| = 1$.

(A) Let $z = x + iy$. The given equation is $\text{arg}\left(\frac{z-2}{z+2}\right) = \frac{\pi}{3}$.
This is the locus of a point $z$ such that the angle subtended by the segment joining $2$ and $-2$ at $z$ is $\frac{\pi}{3}$.
Using the property $\text{arg}\left(\frac{z-z_1}{z-z_2}\right) = \theta$,the locus is an arc of a circle.
Specifically,$\frac{z-2}{z+2} = k(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$ for some $k > 0$.
This simplifies to $\frac{z-2}{z+2} = k(\frac{1}{2} + i\frac{\sqrt{3}}{2})$.
Solving for $z$ leads to the equation of a circle.

(D) The total number of ways to choose $2$ letters from the English alphabet (assuming repetition is allowed as it is not specified otherwise) is $26 \times 26 = 26^{2}$.
The total number of ways to choose $4$ digits such that the first digit is not zero is $9 \times 10 \times 10 \times 10 = 9 \times 10^{3}$.
Therefore,the total number of distinct registration numbers is $26^{2} \times 9 \times 10^{3}$.

(A) The word "$IRRATIONAL$" contains $10$ letters in total.
Counting the frequency of each letter:
$I$ appears $2$ times.
$R$ appears $2$ times.
$A$ appears $2$ times.
$T, O, N, L$ appear $1$ time each.
Using the formula for permutations of a multiset,the total number of arrangements is given by $\frac{n!}{n_1! n_2! n_3! \dots n_k!}$.
Here,$n = 10$,$n_1 = 2$ (for $I$),$n_2 = 2$ (for $R$),and $n_3 = 2$ (for $A$).
Therefore,the total number of words $= \frac{10!}{2! 2! 2!} = \frac{10!}{(2!)^3}$.

(D) The total number of ways to arrange $4$ speakers is $4! = 4 \times 3 \times 2 \times 1 = 24$.
In any arrangement,there are only two possibilities for the relative order of speakers $P$ and $Q$: either $P$ speaks before $Q$ or $Q$ speaks before $P$.
Since these two cases are equally likely,exactly half of the total arrangements will have $Q$ speaking before $P$.
Therefore,the required number of ways is $\frac{24}{2} = 12$.

(B) The number of diagonals in a regular polygon of $n$ sides is given by the formula $\frac{n(n-3)}{2}$ or ${ }^{n} C_{2}-n$.
For a polygon with $n = 100$ sides:
Number of diagonals $= { }^{100} C_{2} - 100$
$= \frac{100 \times 99}{2} - 100$
$= 50 \times 99 - 100$
$= 4950 - 100$
$= 4850$

(B) The coefficients of the $2nd$,$3rd$,and $4th$ terms in the expansion of $(1+x)^{n}$ are ${}^{n}C_{1}$,${}^{n}C_{2}$,and ${}^{n}C_{3}$ respectively.
Given that these are in arithmetic progression,we have $2({}^{n}C_{2}) = {}^{n}C_{1} + {}^{n}C_{3}$.
Substituting the values,$2 \times \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$.
Dividing by $n$ (since $n \neq 0$),we get $n-1 = 1 + \frac{(n-1)(n-2)}{6}$.
$6(n-2) = (n-1)(n-2)$.
Since $n \neq 2$,we divide by $(n-2)$ to get $6 = n-1$,which implies $n = 7$.
The sum of the coefficients of odd powers of $x$ in the expansion of $(1+x)^{n}$ is given by $2^{n-1}$.
Substituting $n=7$,the sum is $2^{7-1} = 2^{6} = 64$.

(D) Let the six terms of the $AP$ be $a-5d, a-3d, a-d, a+d, a+3d, a+5d$ with common difference $2d$.
Sum of the terms $= (a-5d) + (a-3d) + (a-d) + (a+d) + (a+3d) + (a+5d) = 6a$.
Given $6a = 3$,so $a = \frac{1}{2}$.
Given $T_1 = 4T_3$,where $T_1 = a-5d$ and $T_3 = a-d$.
$a-5d = 4(a-d)$ $\Rightarrow a-5d = 4a-4d$ $\Rightarrow -3a = d$.
Substituting $a = \frac{1}{2}$,we get $d = -3 \times \frac{1}{2} = -\frac{3}{2}$.
The fifth term is $T_5 = a+3d$.
$T_5 = \frac{1}{2} + 3(-\frac{3}{2}) = \frac{1}{2} - \frac{9}{2} = -\frac{8}{2} = -4$.

(B) The given series is $S = 1 + \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \dots$
This is a binomial series of the form $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$
We can rewrite the terms as $1 + \frac{1}{2}(\frac{2}{3}) + \frac{\frac{1}{2}(\frac{1}{2}+1)}{2!}(\frac{2}{3})^2 + \frac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)}{3!}(\frac{2}{3})^3 + \dots$
Comparing this with the binomial expansion $(1-x)^{-n}$,we have $n = \frac{1}{2}$ and $x = \frac{2}{3}$.
Thus,the sum is $(1 - \frac{2}{3})^{-1/2} = (\frac{1}{3})^{-1/2} = 3^{1/2} = \sqrt{3}$.

(C) Let $a$ be the first term and $r$ be the common ratio of a $GP$. The $P^{\text{th}}$,$Q^{\text{th}}$,and $R^{\text{th}}$ terms are $ar^{P-1}$,$ar^{Q-1}$,and $ar^{R-1}$ respectively.
According to the question:
$ar^{P-1} = 64$ $(i)$
$ar^{Q-1} = 27$ (ii)
$ar^{R-1} = 36$ (iii)
Dividing $(i)$ by (ii):
$r^{P-Q} = \frac{64}{27} = \left(\frac{4}{3}\right)^3$ (iv)
Dividing (ii) by (iii):
$r^{Q-R} = \frac{27}{36} = \frac{3}{4}$
Raising to the power of $3$:
$r^{3(Q-R)} = \left(\frac{3}{4}\right)^3 = \left(\frac{4}{3}\right)^{-3}$ $(v)$
Multiplying (iv) and $(v)$:
$r^{P-Q} \times r^{3Q-3R} = \left(\frac{4}{3}\right)^3 \times \left(\frac{4}{3}\right)^{-3} = 1$
$r^{P+2Q-3R} = r^0$
Therefore,$P+2Q-3R = 0$,which implies $P+2Q = 3R$.

(C) Let $a^{p} = b^{q} = c^{r} = k$.
Since $a, b, c$ are positive real numbers,we can write $a = k^{1/p}$,$b = k^{1/q}$,and $c = k^{1/r}$.
Given that $a, b, c$ are in $GP$,we have $\frac{b}{a} = \frac{c}{b}$.
Substituting the values of $a, b, c$,we get $\frac{k^{1/q}}{k^{1/p}} = \frac{k^{1/r}}{k^{1/q}}$.
This implies $k^{(1/q - 1/p)} = k^{(1/r - 1/q)}$.
Equating the exponents,we get $\frac{1}{q} - \frac{1}{p} = \frac{1}{r} - \frac{1}{q}$,which simplifies to $\frac{2}{q} = \frac{1}{p} + \frac{1}{r}$.
This shows that $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in $A.P.$
Therefore,$p, q, r$ are in $H.P.$

(D) The sum of an infinite $GP$ with first term $a=k$ and common ratio $r=\frac{k}{k+1}$ is given by $S_{k} = \frac{a}{1-r} = \frac{k}{1-\frac{k}{k+1}} = \frac{k}{\frac{1}{k+1}} = k(k+1)$.
We need to calculate $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}} = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k(k+1)}$.
Using partial fractions,$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.
So,the sum is $\sum_{k=1}^{\infty} (-1)^{k} \left( \frac{1}{k} - \frac{1}{k+1} \right) = -(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) - (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) - \dots$
$= -1 + \frac{1}{2} + \frac{1}{2} - \frac{1}{3} - \frac{1}{3} + \frac{1}{4} + \frac{1}{4} - \dots$
$= -1 + 2 \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots \right)$.
Recall the expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$. For $x=1$,$\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$,so $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.
Substituting this back,the sum is $-1 + 2(1 - \log_{e} 2) = -1 + 2 - 2 \log_{e} 2 = 1 - 2 \log_{e} 2 = 1 - \log_{e} 4$.

(B) The general term of the series is $T_n = n \times n!$.
We can rewrite $T_n$ as $T_n = ((n + 1) - 1) \times n! = (n + 1)! - n!$.
The sum $S_{50} = \sum_{n=1}^{50} T_n = \sum_{n=1}^{50} ((n + 1)! - n!)$.
This is a telescoping series:
$S_{50} = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + (51! - 50!)$.
All intermediate terms cancel out,leaving $S_{50} = 51! - 1!$.
Since $1! = 1$,the sum is $51! - 1$.

(C) The given series is a geometric progression with $a=1$,common ratio $r=(1+x)$,and $n=21$ terms.
The sum of the series is $S = \frac{1((1+x)^{21}-1)}{(1+x)-1} = \frac{(1+x)^{21}-1}{x}$.
We need to find the coefficient of $x^{10}$ in $S = \frac{(1+x)^{21}-1}{x}$.
This is equivalent to finding the coefficient of $x^{11}$ in the expansion of $(1+x)^{21}-1$.
The general term in the expansion of $(1+x)^{21}$ is given by $T_{r+1} = {}^{21}C_{r} x^{r}$.
For $r=11$,the term is ${}^{21}C_{11} x^{11}$.
Thus,the coefficient of $x^{11}$ in $(1+x)^{21}$ is ${}^{21}C_{11}$.
Therefore,the coefficient of $x^{10}$ in the given series is ${}^{21}C_{11}$.

(A) Let the sum be $S = \sum_{k=0}^{n} \frac{1}{k+1} {}^{n}C_{k}$.
Using the identity $\frac{1}{k+1} {}^{n}C_{k} = \frac{1}{n+1} {}^{n+1}C_{k+1}$,we have:
$S = \sum_{k=0}^{n} \frac{1}{n+1} {}^{n+1}C_{k+1}$
$S = \frac{1}{n+1} \sum_{k=0}^{n} {}^{n+1}C_{k+1}$
$S = \frac{1}{n+1} [{}^{n+1}C_{1} + {}^{n+1}C_{2} + \dots + {}^{n+1}C_{n+1}]$
Since $\sum_{r=0}^{m} {}^{m}C_{r} = 2^{m}$,we know that $\sum_{r=1}^{m} {}^{m}C_{r} = 2^{m} - {}^{m}C_{0} = 2^{m} - 1$.
Here $m = n+1$,so the sum is $2^{n+1} - 1$.
Therefore,$S = \frac{2^{n+1}-1}{n+1}$.

(C) The sum of the first $(r-1)$ natural numbers is given by $\frac{(r-1)r}{2}$.
Substituting this into the summation,we get $\sum_{r=2}^{\infty} \frac{(r-1)r}{2 \cdot r!}$.
Since $r! = r(r-1)(r-2)!$,the expression simplifies to $\sum_{r=2}^{\infty} \frac{1}{2(r-2)!}$.
Factoring out the constant $\frac{1}{2}$,we have $\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-2)!}$.
Let $k = r-2$. As $r$ goes from $2$ to $\infty$,$k$ goes from $0$ to $\infty$.
Thus,the expression becomes $\frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{2} e$.

(B) Given $P = \sum_{r=0}^{5} c_{2r} = c_{0} + c_{2} + c_{4} + c_{6} + c_{8} + c_{10}$.
Since $c_{r} = {}^{10}C_{r}$,we have $P = {}^{10}C_{0} + {}^{10}C_{2} + {}^{10}C_{4} + {}^{10}C_{6} + {}^{10}C_{8} + {}^{10}C_{10} = 2^{10-1} = 2^{9}$.
Given $Q = \sum_{r=0}^{3} d_{2r+1} = d_{1} + d_{3} + d_{5} + d_{7}$.
Since $d_{r} = {}^{7}C_{r}$,we have $Q = {}^{7}C_{1} + {}^{7}C_{3} + {}^{7}C_{5} + {}^{7}C_{7} = 2^{7-1} = 2^{6}$.
Therefore,$\frac{P}{Q} = \frac{2^{9}}{2^{6}} = 2^{9-6} = 2^{3} = 8$.

(C) Given equation: $\tan x + \sec x = 2 \cos x$
Using $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$,we have:
$\frac{\sin x + 1}{\cos x} = 2 \cos x$
$\sin x + 1 = 2 \cos^2 x$
Since $\cos^2 x = 1 - \sin^2 x$,the equation becomes:
$\sin x + 1 = 2(1 - \sin^2 x)$
$\sin x + 1 = 2 - 2 \sin^2 x$
$2 \sin^2 x + \sin x - 1 = 0$
Factoring the quadratic equation:
$2 \sin^2 x + 2 \sin x - \sin x - 1 = 0$
$2 \sin x(\sin x + 1) - 1(\sin x + 1) = 0$
$(2 \sin x - 1)(\sin x + 1) = 0$
This gives $\sin x = \frac{1}{2}$ or $\sin x = -1$.
For $x \in [0, \pi]$,$\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
If $\sin x = -1$,then $x = \frac{3\pi}{2}$,which is outside the interval $[0, \pi]$.
Also,at $x = \frac{\pi}{2}$,$\tan x$ and $\sec x$ are undefined,so $x = \frac{\pi}{2}$ is not a solution.
Thus,the solutions are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
The number of solutions is $2$.

(A) Given $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.
Since $M(x, y)$ divides $AB$ externally in the ratio $b : a$,the coordinates of $M$ are given by the section formula for external division:
$x = \frac{b(a \cos \beta) - a(b \cos \alpha)}{b - a} = \frac{ab(\cos \beta - \cos \alpha)}{b - a}$
$y = \frac{b(a \sin \beta) - a(b \sin \alpha)}{b - a} = \frac{ab(\sin \beta - \sin \alpha)}{b - a}$
Now,consider the expression $E = x \cos \frac{\alpha+\beta}{2} + y \sin \frac{\alpha+\beta}{2}$.
Substituting $x$ and $y$:
$E = \frac{ab}{b-a} [(\cos \beta - \cos \alpha) \cos \frac{\alpha+\beta}{2} + (\sin \beta - \sin \alpha) \sin \frac{\alpha+\beta}{2}]$
Using the identity $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$E = \frac{ab}{b-a} [(-2 \sin \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \cos \frac{\alpha+\beta}{2} + (2 \cos \frac{\beta+\alpha}{2} \sin \frac{\beta-\alpha}{2}) \sin \frac{\alpha+\beta}{2}]$
$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [-\sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2} + \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha+\beta}{2}]$
$E = \frac{2ab \sin \frac{\beta-\alpha}{2}}{b-a} [0] = 0$.

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.
The coordinates of the points where the line intersects the axes are $A(0, b)$ and $B(a, 0)$.
Since the point $(\alpha, \beta)$ is the midpoint of the segment $AB$,we have:
$\alpha = \frac{0 + a}{2} \Rightarrow a = 2\alpha$
$\beta = \frac{b + 0}{2} \Rightarrow b = 2\beta$
Substituting these values into the intercept form equation:
$\frac{x}{2\alpha} + \frac{y}{2\beta} = 1$
Multiplying both sides by $2$:
$\frac{x}{\alpha} + \frac{y}{\beta} = 2$

(A) Let the coordinates of vertex $R$ be $(h, k)$.
The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
Substituting the given vertices $P(2, -3), Q(-2, 1),$ and $R(h, k)$,the centroid is $\left(\frac{2 - 2 + h}{3}, \frac{-3 + 1 + k}{3}\right) = \left(\frac{h}{3}, \frac{k - 2}{3}\right)$.
Since the centroid lies on the line $2x + 3y = 1$,the coordinates must satisfy the equation:
$2\left(\frac{h}{3}\right) + 3\left(\frac{k - 2}{3}\right) = 1$
Multiplying by $3$,we get:
$2h + 3(k - 2) = 3$
$2h + 3k - 6 = 3$
$2h + 3k = 9$
Replacing $(h, k)$ with $(x, y)$,the locus of $R$ is $2x + 3y = 9$.

(B) Given lines are $x+2y=4$ $(i)$ and $2x+y=4$ (ii).
Solving equations $(i)$ and (ii),we get the point of intersection as $(\frac{4}{3}, \frac{4}{3})$.
Let the equation of the line passing through this point be $\frac{x}{a} + \frac{y}{b} = 1$.
Since it passes through $(\frac{4}{3}, \frac{4}{3})$,we have $\frac{4}{3a} + \frac{4}{3b} = 1$,which simplifies to $\frac{1}{a} + \frac{1}{b} = \frac{3}{4}$ (iii).
Let the mid-point of $AB$ be $(h, k)$. Then $h = \frac{a}{2}$ and $k = \frac{b}{2}$,so $a = 2h$ and $b = 2k$.
Substituting these into equation (iii),we get $\frac{1}{2h} + \frac{1}{2k} = \frac{3}{4}$.
Multiplying by $2hk$,we get $k + h = \frac{3}{2}hk$,or $2(h+k) = 3hk$.
Replacing $(h, k)$ with $(x, y)$,the locus is $2(x+y) = 3xy$.

(C) The points $(2,0)$,$(0,3)$,and $(0,0)$ form a right-angled triangle at the origin $(0,0)$ because the angle between the $x$-axis and $y$-axis is $90^{\circ}$.
Since the angle subtended by the line segment joining $(2,0)$ and $(0,3)$ at the point $(0,0)$ is $90^{\circ}$,this line segment must be the diameter of the circle.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the points $(2,0)$ and $(0,3)$,we get:
$(x-2)(x-0) + (y-0)(y-3) = 0$
$x^2 - 2x + y^2 - 3y = 0$
$x^2 + y^2 - 2x - 3y = 0$
Since the point $(2k, 3k)$ lies on this circle,it must satisfy the equation:
$(2k)^2 + (3k)^2 - 2(2k) - 3(3k) = 0$
$4k^2 + 9k^2 - 4k - 9k = 0$
$13k^2 - 13k = 0$
$13k(k - 1) = 0$
This gives $k = 0$ or $k = 1$.
If $k = 0$,the point $(2k, 3k)$ becomes $(0,0)$,which is not a distinct point from the given point $(0,0)$.
Therefore,for the four points to be distinct,we must have $k = 1$.

(A) Two circles $x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ intersect orthogonally if and only if $2(g_{1}g_{2}+f_{1}f_{2})=c_{1}+c_{2}$.
For the given circles:
Circle $1$: $g_{1}=1, f_{1}=k, c_{1}=6$
Circle $2$: $g_{2}=0, f_{2}=k, c_{2}=k$
Substituting these values into the condition:
$2((1)(0) + (k)(k)) = 6 + k$
$2k^{2} = 6 + k$
$2k^{2} - k - 6 = 0$
Factoring the quadratic equation:
$2k^{2} - 4k + 3k - 6 = 0$
$2k(k-2) + 3(k-2) = 0$
$(k-2)(2k+3) = 0$
Thus,$k = 2$ or $k = -\frac{3}{2}$.

(C) The given equations of the circles are:
$C_{1}: x^{2}+y^{2}=4$ (centre $C_{1} = (0,0)$,radius $r_{1} = 2$)
$C_{2}: (x-2)^{2}+y^{2}=1$ (centre $C_{2} = (2,0)$,radius $r_{2} = 1$)
Let $N$ be the point of intersection of the common chord $PQ$ with the line joining the centres $C_{1}C_{2}$.
Subtracting the two equations:
$(x^{2}+y^{2}) - ((x-2)^{2}+y^{2}) = 4 - 1$
$x^{2} - (x^{2}-4x+4) = 3$
$4x - 4 = 3 \implies 4x = 7 \implies x = \frac{7}{4}$
Since $N$ lies on the line $C_{1}C_{2}$ (the $x$-axis),the coordinate of $N$ is $(\frac{7}{4}, 0)$.
The distance $C_{1}N = \frac{7}{4}$.
The distance $C_{2}N = |C_{1}C_{2} - C_{1}N| = |2 - \frac{7}{4}| = \frac{1}{4}$.
Both triangles $\Delta C_{1}PQ$ and $\Delta C_{2}PQ$ share the same base $PQ$.
The ratio of their areas is the ratio of their heights from the base $PQ$ to the vertices $C_{1}$ and $C_{2}$ respectively,which is the ratio of the distances $C_{1}N$ and $C_{2}N$:
$\frac{\text{Area}(\Delta C_{1}PQ)}{\text{Area}(\Delta C_{2}PQ)} = \frac{\frac{1}{2} \times PQ \times C_{1}N}{\frac{1}{2} \times PQ \times C_{2}N} = \frac{C_{1}N}{C_{2}N} = \frac{7/4}{1/4} = 7: 1$.
Thus,the correct option is $C$.

(D) Given equation: $y^{2}+4x+4y+k=0$
Complete the square for $y$:
$y^{2}+4y+4-4+4x+k=0$
$(y+2)^{2} = -4x+4-k$
$(y+2)^{2} = -4(x - \frac{4-k}{4})$
Comparing this with the standard form of a parabola $(y-k')^{2} = -4a(x-h')$,we get $4a = 4$.
Therefore,the length of the latus rectum is $4$ units.

(B) The vertex of the parabola $y^{2}=8x$ is $M(0,0)$.
Let the other point on the parabola be $N(2t^{2}, 4t)$,where $t$ is a parameter.
Let $P(x, y)$ be the mid-point of the chord $MN$.
Using the mid-point formula,we have:
$x = \frac{0 + 2t^{2}}{2} = t^{2}$
$y = \frac{0 + 4t}{2} = 2t$
From the second equation,$t = \frac{y}{2}$.
Substituting this into the first equation:
$x = (\frac{y}{2})^{2}$
$x = \frac{y^{2}}{4}$
$y^{2} = 4x$
Thus,the locus of $P$ is $y^{2}=4x$.

(C) Let the coordinates of points $P$ and $Q$ on the parabola $y^{2}=4x$ be $(t^{2}, 2t)$ and $(m^{2}, 2m)$ respectively. The vertex of the parabola is $X(0, 0)$.
Since $PQ$ subtends a right angle at the vertex $X$,the product of the slopes of $XP$ and $XQ$ is $-1$.
Slope of $XP = \frac{2t-0}{t^{2}-0} = \frac{2}{t}$.
Slope of $XQ = \frac{2m-0}{m^{2}-0} = \frac{2}{m}$.
Given,$(\frac{2}{t}) \times (\frac{2}{m}) = -1 \Rightarrow tm = -4$.
The equation of the line $PQ$ passing through $(t^{2}, 2t)$ and $(m^{2}, 2m)$ is:
$y - 2t = \frac{2m-2t}{m^{2}-t^{2}}(x - t^{2})$
$y - 2t = \frac{2(m-t)}{(m-t)(m+t)}(x - t^{2})$
$y - 2t = \frac{2}{m+t}(x - t^{2})$.
Since $PQ$ intersects the axis of the parabola ($x$-axis) at $R(\alpha, 0)$,we substitute $y=0$ and $x=\alpha$:
$0 - 2t = \frac{2}{m+t}(\alpha - t^{2})$
$-t(m+t) = \alpha - t^{2}$
$-tm - t^{2} = \alpha - t^{2}$
$\alpha = -tm$.
Since $tm = -4$,we have $\alpha = -(-4) = 4$.
Thus,the distance of the vertex $X(0, 0)$ from $R(4, 0)$ is $4$.

(B) Let $P(\sqrt{10} \cos \theta, \sqrt{8} \sin \theta)$ be the point on the ellipse $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$.
Given that the distance of $P$ from the centre $(0,0)$ is $3$ units.
So,$OP^2 = 3^2 = 9$.
$10 \cos^2 \theta + 8 \sin^2 \theta = 9$.
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we can write $9 = 9(\sin^2 \theta + \cos^2 \theta)$.
$10 \cos^2 \theta + 8 \sin^2 \theta = 9 \sin^2 \theta + 9 \cos^2 \theta$.
Rearranging the terms,we get $\cos^2 \theta = \sin^2 \theta$.
$\tan^2 \theta = 1$.
Since the point is in the first quadrant,$\theta = \frac{\pi}{4}$.

(D) Given equations are $x=2y$ $(i)$ and $\frac{x^{2}}{4}+y^{2}=1$ (ii).
Substituting $x=2y$ into (ii),we get:
$\frac{(2y)^{2}}{4}+y^{2}=1$
$\frac{4y^{2}}{4}+y^{2}=1$
$y^{2}+y^{2}=1$
$2y^{2}=1$ $\Rightarrow y^{2}=\frac{1}{2}$ $\Rightarrow y=\pm \frac{1}{\sqrt{2}}$.
From $(i)$,$x=2y$,so $x=\pm 2(\frac{1}{\sqrt{2}}) = \pm \sqrt{2}$.
Thus,the points of intersection are $P(\sqrt{2}, \frac{1}{\sqrt{2}})$ and $Q(-\sqrt{2}, -\frac{1}{\sqrt{2}})$.
The equation of a circle with diameter $PQ$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
$(x-\sqrt{2})(x+\sqrt{2}) + (y-\frac{1}{\sqrt{2}})(y+\frac{1}{\sqrt{2}}) = 0$
$(x^{2}-2) + (y^{2}-\frac{1}{2}) = 0$
$x^{2}+y^{2} = 2 + \frac{1}{2}$
$x^{2}+y^{2} = \frac{5}{2}$.

(D) Given the ellipse equation is $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$.
Comparing with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we have $a^{2}=9$ and $b^{2}=1$.
The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$.
The foci are $(\pm ae, 0) = (\pm 3 \times \frac{2\sqrt{2}}{3}, 0) = (\pm 2\sqrt{2}, 0)$.
Let $P(h, k)$ be a point such that the foci $F_{1}(2\sqrt{2}, 0)$ and $F_{2}(-2\sqrt{2}, 0)$ subtend a right angle at $P$.
Thus,the product of the slopes of $PF_{1}$ and $PF_{2}$ is $-1$.
$\frac{k-0}{h-2\sqrt{2}} \times \frac{k-0}{h+2\sqrt{2}} = -1$.
$\frac{k^{2}}{h^{2}-(2\sqrt{2})^{2}} = -1$.
$k^{2} = -(h^{2}-8)$.
$h^{2}+k^{2} = 8$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2}=8$.

(D) Let $e$ be the eccentricity of the hyperbola. The centre is $(0, 0)$,the vertex is $(a, 0)$,and the focus is $(ae, 0)$.
Since the vertex $(a, 0)$ bisects the line segment joining the centre $(0, 0)$ and the focus $(ae, 0)$,we have:
$a = \frac{ae + 0}{2}$ $\Rightarrow 2a = ae$ $\Rightarrow e = 2$.
For a hyperbola,$b^{2} = a^{2}(e^{2} - 1)$.
Substituting $e = 2$,we get $b^{2} = a^{2}(2^{2} - 1) = 3a^{2}$.
The equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.
Substituting $b^{2} = 3a^{2}$,we get $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{3a^{2}} = 1$.
Multiplying by $3a^{2}$,we obtain $3x^{2} - y^{2} = 3a^{2}$.

(D) The definition of a hyperbola states that it is the locus of a point such that the absolute difference of its distances from two fixed points (foci) is constant.
Given the fixed points are $(8,0)$ and $(-8,0)$,these are the foci of the hyperbola.
The constant difference is given as $2a = 4$.
Since the absolute difference of distances from two fixed points is a constant value less than the distance between the two fixed points (which is $16$),this locus represents a hyperbola.

(B) Given limit is $\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$.
This is a $\frac{0}{0}$ indeterminate form.
Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx}(\pi^{x}-1)}{\frac{d}{dx}(\sqrt{1+x}-1)}$
$= \lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$
$= \lim _{x \rightarrow 0} 2 \sqrt{1+x} \cdot \pi^{x} \log _{e} \pi$
Substituting $x = 0$:
$= 2 \sqrt{1+0} \cdot \pi^{0} \log _{e} \pi$
$= 2 \cdot 1 \cdot 1 \cdot \log _{e} \pi$
$= 2 \log _{e} \pi = \log _{e} \pi^{2}$.

(D) Let $L = \lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n} = \lim _{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{1/n}$.
Taking the natural logarithm on both sides:
$\ln L = \lim _{n \rightarrow \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \ln \left( \frac{r}{n} \right)$.
This is a Riemann sum for the integral $\int_{0}^{1} \ln(x) \, dx$.
$\ln L = \int_{0}^{1} \ln(x) \, dx = [x \ln x - x]_{0}^{1}$.
Evaluating the limit as $x \rightarrow 0^+$,we get $\lim_{x \rightarrow 0^+} x \ln x = 0$.
So,$\ln L = (1 \ln 1 - 1) - (0) = -1$.
Therefore,$L = e^{-1} = \frac{1}{e}$.

(D) Using the Sine Rule in $\Delta PQR$,we have $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c}$,where $R_{c}$ is the circumradius of the triangle.
Thus,$\sin P = \frac{p}{2R_{c}}$,$\sin Q = \frac{q}{2R_{c}}$,and $\sin R = \frac{r}{2R_{c}}$.
Given the equation $r^{2} \sin P \sin Q = pq$.
Substituting the values of $\sin P$ and $\sin Q$:
$r^{2} \left( \frac{p}{2R_{c}} \right) \left( \frac{q}{2R_{c}} \right) = pq$
$r^{2} \frac{pq}{4R_{c}^{2}} = pq$
Since $p, q \neq 0$,we can divide both sides by $pq$:
$\frac{r^{2}}{4R_{c}^{2}} = 1$
$r^{2} = 4R_{c}^{2}$
$r = 2R_{c}$
Since $r = 2R_{c} \sin R$,we have $2R_{c} \sin R = 2R_{c}$.
$\sin R = 1$
$R = 90^{\circ}$.
Therefore,the triangle is a right-angled triangle.

(B) In $\Delta PQR$,the sum of angles is $P+Q+R = 180^{\circ}$.
Since $P+R = 180^{\circ}-Q$,we substitute this into the expression:
$2pr \sin \left(\frac{P+R-Q}{2}\right) = 2pr \sin \left(\frac{180^{\circ}-Q-Q}{2}\right)$
$= 2pr \sin \left(\frac{180^{\circ}-2Q}{2}\right)$
$= 2pr \sin (90^{\circ}-Q)$
$= 2pr \cos Q$
Using the Law of Cosines,$\cos Q = \frac{p^{2}+r^{2}-q^{2}}{2pr}$.
Substituting this value:
$= 2pr \left(\frac{p^{2}+r^{2}-q^{2}}{2pr}\right)$
$= p^{2}+r^{2}-q^{2}$.

(B) Let the sides of the triangle be $a, b, c$. The area $S$ of the triangle is given by:
$S = \frac{1}{2} a p = \frac{1}{2} b q = \frac{1}{2} c r$
From this,we can express the reciprocals of the altitudes as:
$\frac{1}{p} = \frac{a}{2S}, \frac{1}{q} = \frac{b}{2S}, \frac{1}{r} = \frac{c}{2S}$
Adding these expressions,we get:
$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{a+b+c}{2S}$
Since the perimeter is $2t = a+b+c$,we substitute this into the equation:
$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{2t}{2S} = \frac{t}{S}$

(B) In an equilateral triangle,the incentre,centroid,circumcentre,and orthocentre coincide. Let the incentre be $G(1, 1)$.
The distance $r$ from the incentre $G(1, 1)$ to the side $3x + 4y + 3 = 0$ is the inradius:
$r = \frac{|3(1) + 4(1) + 3|}{\sqrt{3^{2} + 4^{2}}} = \frac{|3 + 4 + 3|}{\sqrt{9 + 16}} = \frac{10}{5} = 2$.
In an equilateral triangle,the circumradius $R$ is twice the inradius $r$. Therefore,$R = 2r = 2(2) = 4$.
The circumcircle has its centre at the incentre $(1, 1)$ and radius $R = 4$.
The equation of the circle is $(x - 1)^{2} + (y - 1)^{2} = 4^{2}$.
$x^{2} - 2x + 1 + y^{2} - 2y + 1 = 16$.
$x^{2} + y^{2} - 2x - 2y + 2 = 16$.
$x^{2} + y^{2} - 2x - 2y - 14 = 0$.

(C) Let $M, P, B$ be the sets of students who failed in Mathematics,Physics,and Biology respectively. Let the regions in the Venn diagram be represented by $a, b, c, d, e, f, g$ as shown,and $h$ be the number of students who passed in all subjects.
Total students = $100$,so $a+b+c+d+e+f+g+h = 100$. Given $h = 1$,so $a+b+c+d+e+f+g = 99$.
Given:
$1) \text{ Failed in Mathematics: } a+b+d+e = 50$
$2) \text{ Failed in Physics: } b+c+d+f = 45$
$3) \text{ Failed in Biology: } d+e+f+g = 40$
$4) \text{ Failed in exactly two subjects: } b+e+f = 32$
Summing equations $(1), (2), (3)$:
$(a+b+d+e) + (b+c+d+f) + (d+e+f+g) = 50 + 45 + 40 = 135$
$(a+b+c+d+e+f+g) + (b+d+e+f) + d = 135$
$99 + (b+e+f) + 2d = 135$
$99 + 32 + 2d = 135$
$131 + 2d = 135$
$2d = 4 \implies d = 2$.
Thus,the number of students failing in all three subjects is $2$.

(D) Let $A = \{a_{1}, a_{2}, a_{3}, a_{4}\}$ and $B = \{b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\}$.
Here,the number of elements in set $A$ is $n(A) = 4$ and the number of elements in set $B$ is $n(B) = 7$.
An injection (one-one mapping) from set $A$ to set $B$ exists if we choose $4$ distinct elements from $B$ and assign them to the $4$ elements of $A$.
The total number of such injections is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$,where $n = 7$ and $r = 4$.
Total injections $= {}^{7}P_{4} = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.

(A) Given,$P = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \end{bmatrix}$.
We need to find $Q = P P^{T}$.
$P^{T} = \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$.
$Q = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$.
$Q = \begin{bmatrix} (1 \times 1 + 2 \times 2 + 1 \times 1) & (1 \times 1 + 2 \times 3 + 1 \times 1) \\ (1 \times 1 + 3 \times 2 + 1 \times 1) & (1 \times 1 + 3 \times 3 + 1 \times 1) \end{bmatrix}$.
$Q = \begin{bmatrix} (1 + 4 + 1) & (1 + 6 + 1) \\ (1 + 6 + 1) & (1 + 9 + 1) \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 8 & 11 \end{bmatrix}$.
Now,the determinant of $Q$ is $|Q| = (6 \times 11) - (8 \times 8)$.
$|Q| = 66 - 64 = 2$.

(B) Let $\Delta = \left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$.
Expanding the determinant along the first row:
$\Delta = -1(1 - \cos^2 P) - \cos R(-\cos R - \cos Q \cos P) + \cos Q(\cos R \cos P + \cos Q)$
$\Delta = -(1 - \cos^2 P) + \cos^2 R + \cos R \cos Q \cos P + \cos Q \cos R \cos P + \cos^2 Q$
$\Delta = -1 + \cos^2 P + \cos^2 R + \cos^2 Q + 2 \cos P \cos Q \cos R$.
Since $P, Q, R$ are angles of a triangle,$P+Q+R = \pi$,so $R = \pi - (P+Q)$.
Using the identity for the sum of squares of cosines in a triangle: $\cos^2 P + \cos^2 Q + \cos^2 R = 1 - 2 \cos P \cos Q \cos R$.
Substituting this into the expression for $\Delta$:
$\Delta = -1 + (1 - 2 \cos P \cos Q \cos R) + 2 \cos P \cos Q \cos R = 0$.

(A) The given system of equations can be rewritten as:
$(1-\alpha)x + 3y + 5z = 0$
$5x + (1-\alpha)y + 3z = 0$
$3x + 5y + (1-\alpha)z = 0$
For the system to have infinite solutions,the determinant of the coefficient matrix must be zero:
$\left|\begin{array}{ccc} 1-\alpha & 3 & 5 \\ 5 & 1-\alpha & 3 \\ 3 & 5 & 1-\alpha \end{array}\right| = 0$
Applying $C_1 \rightarrow C_1 + C_2 + C_3$,we get:
$\left|\begin{array}{ccc} 9-\alpha & 3 & 5 \\ 9-\alpha & 1-\alpha & 3 \\ 9-\alpha & 5 & 1-\alpha \end{array}\right| = 0$
Taking $(9-\alpha)$ common from the first column:
$(9-\alpha) \left|\begin{array}{ccc} 1 & 3 & 5 \\ 1 & 1-\alpha & 3 \\ 1 & 5 & 1-\alpha \end{array}\right| = 0$
Performing $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$(9-\alpha) \left|\begin{array}{ccc} 1 & 3 & 5 \\ 0 & -\alpha-2 & -2 \\ 0 & 2 & -\alpha-4 \end{array}\right| = 0$
$(9-\alpha) [(-\alpha-2)(-\alpha-4) - (-4)] = 0$
$(9-\alpha) [\alpha^2 + 6\alpha + 8 + 4] = 0$
$(9-\alpha) (\alpha^2 + 6\alpha + 12) = 0$
For $\alpha^2 + 6\alpha + 12 = 0$,the discriminant $D = 6^2 - 4(1)(12) = 36 - 48 = -12 < 0$. Thus,there are no real roots for this quadratic part.
Therefore,the only real value is $\alpha = 9$. The number of real values is $1$.

(B) The augmented matrix $[A|B]$ is given by $\begin{bmatrix} \lambda & 1 & 1 & 3 \\ 1 & -1 & -2 & 6 \\ -1 & 1 & 1 & \mu \end{bmatrix}$.
Performing row operations:
$R_1 \leftrightarrow R_2$ gives $\begin{bmatrix} 1 & -1 & -2 & 6 \\ \lambda & 1 & 1 & 3 \\ -1 & 1 & 1 & \mu \end{bmatrix}$.
$R_3 \rightarrow R_3 + R_1$ gives $\begin{bmatrix} 1 & -1 & -2 & 6 \\ \lambda & 1 & 1 & 3 \\ 0 & 0 & -1 & \mu + 6 \end{bmatrix}$.
$R_2 \rightarrow R_2 - \lambda R_1$ gives $\begin{bmatrix} 1 & -1 & -2 & 6 \\ 0 & 1+\lambda & 1+2\lambda & 3-6\lambda \\ 0 & 0 & -1 & \mu + 6 \end{bmatrix}$.
For infinite solutions,the rank of the matrix must be less than the number of variables $(3)$.
If $\lambda = -1$,the second row becomes $[0, 0, -1, 9]$.
The system becomes consistent with infinite solutions if the equations are dependent,which occurs when $\mu = 3$.

(C) We know that the range of $\sin ^{-1} \theta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given that $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$.
Since the maximum value of each term $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $\frac{\pi}{2}$,the sum can be $\frac{3 \pi}{2}$ only if $\sin ^{-1} x = \frac{\pi}{2}$,$\sin ^{-1} y = \frac{\pi}{2}$,and $\sin ^{-1} z = \frac{\pi}{2}$.
This implies $x = \sin(\frac{\pi}{2}) = 1$,$y = \sin(\frac{\pi}{2}) = 1$,and $z = \sin(\frac{\pi}{2}) = 1$.
Substituting these values into the expression:
$x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} = (1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}}$
$= 1+1+1-\frac{1}{1 \times 1 \times 1}$
$= 3-1 = 2$.

(A) Given $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$.
We need to find $x$ such that $f(g(x)) = g(f(x))$.
First,calculate $f(g(x))$:
$f(g(x)) = f(x + 1) = (x + 1)^{2} + 2(x + 1) - 3 = (x^{2} + 2x + 1) + 2x + 2 - 3 = x^{2} + 4x$.
Next,calculate $g(f(x))$:
$g(f(x)) = g(x^{2} + 2x - 3) = (x^{2} + 2x - 3) + 1 = x^{2} + 2x - 2$.
Equating both expressions:
$x^{2} + 4x = x^{2} + 2x - 2$.
Subtracting $x^{2}$ from both sides:
$4x = 2x - 2$.
$4x - 2x = -2$.
$2x = -2$.
$x = -1$.

(B) Let $y = \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right)$.
Using the trigonometric identities $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2})$:
$\frac{\cos x}{1 + \sin x} = \frac{\sin(\frac{\pi}{2} - x)}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \frac{2 \sin(\frac{\pi}{4} - \frac{x}{2}) \cos(\frac{\pi}{4} - \frac{x}{2})}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \tan(\frac{\pi}{4} - \frac{x}{2})$.
Thus,$y = \tan^{-1} \left( \tan(\frac{\pi}{4} - \frac{x}{2}) \right) = \frac{\pi}{4} - \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \frac{x}{2} \right) = -\frac{1}{2}$.

(A) Given,$y = \left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x + \log_{e}(1+x)$.
Applying the product rule to the first term and the chain rule to the second term:
$\frac{dy}{dx} = \left(\frac{3^{x}-1}{3^{x}+1}\right) \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right) + \frac{1}{1+x}$.
Using the quotient rule for $\frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right)$:
$\frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right) = \frac{(3^{x}+1)(3^{x} \ln 3) - (3^{x}-1)(3^{x} \ln 3)}{(3^{x}+1)^{2}} = \frac{3^{x} \ln 3 (3^{x} + 1 - 3^{x} + 1)}{(3^{x}+1)^{2}} = \frac{2 \cdot 3^{x} \ln 3}{(3^{x}+1)^{2}}$.
Substituting this back into the derivative expression:
$\frac{dy}{dx} = \left(\frac{3^{x}-1}{3^{x}+1}\right) \cos x + \sin x \left(\frac{2 \cdot 3^{x} \ln 3}{(3^{x}+1)^{2}}\right) + \frac{1}{1+x}$.
Evaluating at $x = 0$:
$\left(\frac{dy}{dx}\right)_{x=0} = \left(\frac{3^{0}-1}{3^{0}+1}\right) \cos(0) + \sin(0) \left(\frac{2 \cdot 3^{0} \ln 3}{(3^{0}+1)^{2}}\right) + \frac{1}{1+0}$.
$= \left(\frac{1-1}{1+1}\right) \cdot 1 + 0 \cdot \left(\frac{2 \ln 3}{4}\right) + 1 = 0 + 0 + 1 = 1$.

(D) Given,$f(x) f^{\prime}(x) < 0$. This implies that $f(x)$ and $f^{\prime}(x)$ have opposite signs for all $x \in R$.
Consider the function $g(x) = |f(x)|$.
Then $g(x)^2 = |f(x)|^2 = f(x)^2$.
Differentiating both sides with respect to $x$,we get $2g(x) g^{\prime}(x) = 2f(x) f^{\prime}(x)$.
Thus,$g^{\prime}(x) = \frac{f(x) f^{\prime}(x)}{g(x)} = \frac{f(x) f^{\prime}(x)}{|f(x)|}$.
Since $f(x) f^{\prime}(x) < 0$ and $|f(x)| > 0$ for all $x$ where $f(x) \neq 0$,it follows that $g^{\prime}(x) < 0$.
Therefore,$|f(x)|$ is a decreasing function.

(D) Given function is $f(x) = \frac{x}{8} + \frac{2}{x}$.
First,find the derivative $f'(x) = \frac{1}{8} - \frac{2}{x^2} = \frac{x^2 - 16}{8x^2}$.
To find critical points,set $f'(x) = 0$,which gives $x^2 - 16 = 0$,so $x = 4$ or $x = -4$.
Since the interval is $[1, 6]$,we only consider $x = 4$.
Now,evaluate $f(x)$ at the critical point and the endpoints of the interval $[1, 6]$:
$f(1) = \frac{1}{8} + \frac{2}{1} = \frac{1}{8} + 2 = \frac{17}{8} = 2.125$.
$f(4) = \frac{4}{8} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$.
$f(6) = \frac{6}{8} + \frac{2}{6} = \frac{3}{4} + \frac{1}{3} = \frac{9+4}{12} = \frac{13}{12} \approx 1.083$.
Comparing the values $f(1) = \frac{17}{8}$,$f(4) = 1$,and $f(6) = \frac{13}{12}$,the maximum value is $\frac{17}{8}$.

(B) For Rolle's theorem to be applicable to a function $f(x)$ on the interval $[a, b]$,the following conditions must be met:
$(i)$ $f(x)$ must be continuous on $[a, b]$.
(ii) $f(x)$ must be differentiable on $(a, b)$.
(iii) $f(a) = f(b)$.
Let us check the options for the interval $[-2, 2]$:
$(A)$ $f(x) = x^{3} \Rightarrow f(-2) = -8, f(2) = 8$. Since $f(-2) \neq f(2)$,Rolle's theorem is not applicable.
$(B)$ $f(x) = 4x^{4} \Rightarrow f(-2) = 4(-2)^{4} = 64$ and $f(2) = 4(2)^{4} = 64$. Since $f(-2) = f(2)$,and the function is a polynomial (continuous and differentiable everywhere),Rolle's theorem is applicable.
$(C)$ $f(x) = 2x^{3} + 3 \Rightarrow f(-2) = 2(-8) + 3 = -13, f(2) = 2(8) + 3 = 19$. Since $f(-2) \neq f(2)$,Rolle's theorem is not applicable.
$(D)$ $f(x) = \pi|x|$ is not differentiable at $x = 0$,which lies in $(-2, 2)$. Thus,Rolle's theorem is not applicable.
Therefore,the correct option is $B$.

(B) Given that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$.
Integrating both sides with respect to $x$,we get $f^{\prime}(x)=g^{\prime}(x)+C_1$.
Substituting $x=1$,we have $f^{\prime}(1)=g^{\prime}(1)+C_1$.
Given $f^{\prime}(1)=4$ and $g^{\prime}(1)=6$,so $4=6+C_1$,which implies $C_1=-2$.
Thus,$f^{\prime}(x)=g^{\prime}(x)-2$.
Integrating again with respect to $x$,we get $f(x)=g(x)-2x+C_2$.
Substituting $x=2$,we have $f(2)=g(2)-2(2)+C_2$.
Given $f(2)=3$ and $g(2)=9$,so $3=9-4+C_2$,which implies $3=5+C_2$,so $C_2=-2$.
Thus,$f(x)=g(x)-2x-2$.
Rearranging the terms,we get $f(x)-g(x)=-2x-2$.
For $x=1$,$f(1)-g(1)=-2(1)-2=-4$.

(A) Let $I = \int_{-1}^{1}(|x|-2[x]) \, dx$.
We split the integral at $x=0$:
$I = \int_{-1}^{0}(|x|-2[x]) \, dx + \int_{0}^{1}(|x|-2[x]) \, dx$.
For $x \in [-1, 0)$,$|x| = -x$ and $[x] = -1$.
For $x \in [0, 1)$,$|x| = x$ and $[x] = 0$.
Substituting these values:
$I = \int_{-1}^{0}(-x - 2(-1)) \, dx + \int_{0}^{1}(x - 2(0)) \, dx$.
$I = \int_{-1}^{0}(-x + 2) \, dx + \int_{0}^{1} x \, dx$.
Evaluating the integrals:
$I = \left[ -\frac{x^2}{2} + 2x \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{1}$.
$I = (0 - 0) - \left( -\frac{(-1)^2}{2} + 2(-1) \right) + \left( \frac{1^2}{2} - 0 \right)$.
$I = - \left( -\frac{1}{2} - 2 \right) + \frac{1}{2}$.
$I = \frac{1}{2} + 2 + \frac{1}{2} = 1 + 2 = 3$.

(D) Let $I = \int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} \right) dx$
Using the identities $\sin 2x = 2\sin x \cos x$ and $\cos 2x = 2\cos^2 x - 1$:
$I = \int_{\pi / 6}^{\pi / 2} \left( \frac{1 + 2\sin x \cos x + 2\cos^2 x - 1}{\sin x + \cos x} \right) dx$
$I = \int_{\pi / 6}^{\pi / 2} \left( \frac{2\cos x(\sin x + \cos x)}{\sin x + \cos x} \right) dx$
$I = \int_{\pi / 6}^{\pi / 2} 2\cos x dx$
$I = 2[\sin x]_{\pi / 6}^{\pi / 2}$
$I = 2\left( \sin \frac{\pi}{2} - \sin \frac{\pi}{6} \right)$
$I = 2\left( 1 - \frac{1}{2} \right) = 2 \times \frac{1}{2} = 1$

(D) Let $I = \int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\pi / 2} \frac{1}{1+(\tan(\frac{\pi}{2}-x))^{-101}} dx$
$I = \int_{0}^{\pi / 2} \frac{1}{1+(\cot x)^{-101}} dx$
$I = \int_{0}^{\pi / 2} \frac{1}{1+(\frac{1}{\tan x})^{-101}} dx = \int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{101}} dx$.
Adding the two expressions for $I$:
$2I = \int_{0}^{\pi / 2} (\frac{1}{1+(\tan x)^{-101}} + \frac{1}{1+(\tan x)^{101}}) dx$
$2I = \int_{0}^{\pi / 2} (\frac{(\tan x)^{101}}{(\tan x)^{101}+1} + \frac{1}{1+(\tan x)^{101}}) dx$
$2I = \int_{0}^{\pi / 2} \frac{1+(\tan x)^{101}}{1+(\tan x)^{101}} dx = \int_{0}^{\pi / 2} 1 dx = \frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{4}$.

(D) Let $I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$.
Since $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,we can rewrite the denominator as $3 + 1 - (\sin x - \cos x)^2 = 4 - (\sin x - \cos x)^2$.
Thus,$I = \int_{0}^{\pi / 4} \frac{\sin x + \cos x}{4 - (\sin x - \cos x)^2} dx$.
Let $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi / 4$,$t = \sin(\pi / 4) - \cos(\pi / 4) = 0$.
Substituting these into the integral,we get $I = \int_{-1}^{0} \frac{dt}{4 - t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log |\frac{a+x}{a-x}|$,we have:
$I = \frac{1}{2(2)} [\log |\frac{2+t}{2-t}|]_{-1}^{0} = \frac{1}{4} [\log |\frac{2+0}{2-0}| - \log |\frac{2-1}{2+1}|]$.
$I = \frac{1}{4} [\log 1 - \log (1/3)] = \frac{1}{4} [0 - (-\log 3)] = \frac{1}{4} \log 3$.

(C) Let $I = \int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$.
We can split the integral as:
$I = \int_{-2}^{2} e^{|x|} d x + 2 \int_{-2}^{2} \sin x e^{|x|} d x$.
Consider the first part: $f(x) = e^{|x|}$. Since $f(-x) = e^{|-x|} = e^{|x|} = f(x)$,$f(x)$ is an even function.
Thus,$\int_{-2}^{2} e^{|x|} d x = 2 \int_{0}^{2} e^{x} d x = 2[e^{x}]_{0}^{2} = 2(e^{2}-1)$.
Consider the second part: $g(x) = \sin x e^{|x|}$. Since $g(-x) = \sin(-x) e^{|-x|} = -\sin x e^{|x|} = -g(x)$,$g(x)$ is an odd function.
Thus,$\int_{-2}^{2} \sin x e^{|x|} d x = 0$.
Therefore,$I = 2(e^{2}-1) + 2(0) = 2(e^{2}-1)$.

(C) Let $I = \int_{1}^{5} [|x-3| + |1-x|] dx$.
Since $x \in [1, 5]$,$|1-x| = x-1$.
Thus,$I = \int_{1}^{5} |x-3| dx + \int_{1}^{5} (x-1) dx$.
For the first part,$\int_{1}^{5} |x-3| dx = \int_{1}^{3} -(x-3) dx + \int_{3}^{5} (x-3) dx$.
$= \int_{1}^{3} (3-x) dx + \int_{3}^{5} (x-3) dx = [3x - \frac{x^2}{2}]_{1}^{3} + [\frac{x^2}{2} - 3x]_{3}^{5}$.
$= (9 - 4.5) - (3 - 0.5) + (12.5 - 15) - (4.5 - 9) = 4.5 - 2.5 - 2.5 + 4.5 = 4$.
For the second part,$\int_{1}^{5} (x-1) dx = [\frac{x^2}{2} - x]_{1}^{5} = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$.
Therefore,$I = 4 + 8 = 12$.

(B) The curves $y=x^{3}$ and $y=\frac{1}{x}$ intersect at $x^{3} = \frac{1}{x}$,which implies $x^{4} = 1$. Since we are in the first quadrant,$x=1$. At $x=1$,$y=1$.
The region is bounded by $y=x^{3}$ from $x=0$ to $x=1$ and by $y=\frac{1}{x}$ from $x=1$ to $x=2$.
Required Area $= \int_{0}^{1} x^{3} dx + \int_{1}^{2} \frac{1}{x} dx$
$= \left[ \frac{x^{4}}{4} \right]_{0}^{1} + \left[ \log_{e} x \right]_{1}^{2}$
$= (\frac{1}{4} - 0) + (\log_{e} 2 - \log_{e} 1)$
$= \frac{1}{4} + \log_{e} 2$.

(C) The function $y = \sin^{-1} x$ is defined for $-1 \leq x \leq 1$. In the first quadrant,we have $0 \leq x \leq 1$.
Since $x(1-x) \geq 0$ for $x \in [0, 1]$,the curve $y_1 = \sin^{-1} x + x(1-x)$ lies above the curve $y_2 = \sin^{-1} x - x(1-x)$.
The intersection points are found by setting $y_1 = y_2$,which gives $2x(1-x) = 0$,implying $x = 0$ or $x = 1$.
The required area is given by the integral:
$A = \int_{0}^{1} (y_1 - y_2) dx = \int_{0}^{1} [(\sin^{-1} x + x - x^2) - (\sin^{-1} x - x + x^2)] dx$
$A = \int_{0}^{1} (2x - 2x^2) dx = 2 \int_{0}^{1} (x - x^2) dx$
$A = 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 2 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3}$.

(D) Given differential equation: $\frac{d y}{d x}=\frac{x+y+1}{2 x+2 y+1}$
Let $x+y=v$. Then $1+\frac{d y}{d x}=\frac{d v}{d x}$,so $\frac{d y}{d x}=\frac{d v}{d x}-1$.
Substituting this into the equation: $\frac{d v}{d x}-1=\frac{v+1}{2 v+1}$
$\frac{d v}{d x}=\frac{v+1}{2 v+1}+1 = \frac{v+1+2 v+1}{2 v+1} = \frac{3 v+2}{2 v+1}$
Separating variables: $\frac{2 v+1}{3 v+2} d v=d x$
Rewrite the numerator: $\frac{\frac{2}{3}(3 v+2)-\frac{1}{3}}{3 v+2} d v=d x$
$\left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v=d x$
Integrating both sides: $\int \left(\frac{2}{3}-\frac{1}{3(3 v+2)}\right) d v = \int d x + C'$
$\frac{2}{3} v - \frac{1}{9} \log |3 v+2| = x + C'$
Substitute $v=x+y$: $\frac{2}{3}(x+y) - \frac{1}{9} \log |3 x+3 y+2| = x + C'$
Multiply by $9$: $6(x+y) - \log |3 x+3 y+2| = 9 x + 9 C'$
$6 x + 6 y - 9 x - \log |3 x+3 y+2| = C$
$-3 x + 6 y - \log |3 x+3 y+2| = C$
Multiplying by $-1$: $3 x - 6 y + \log |3 x+3 y+2| = C$
Thus,the correct option is $D$.

(C) The given differential equation is $25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$.
The auxiliary equation is $25 m^{2}-10 m+1=0$.
This can be written as $(5 m-1)^{2}=0$,which gives $m=\frac{1}{5}, \frac{1}{5}$.
Since the roots are real and equal,the general solution is $y=(c_{1}+c_{2} x) e^{x / 5} \quad \dots(i)$.
Given $y(0)=1$,substituting $x=0$ in $(i)$ gives $1=(c_{1}+0) e^{0} \Rightarrow c_{1}=1$.
Given $y(1)=2 e^{1 / 5}$,substituting $x=1$ and $c_{1}=1$ in $(i)$ gives $2 e^{1 / 5}=(1+c_{2}) e^{1 / 5}$.
Dividing by $e^{1 / 5}$,we get $2=1+c_{2} \Rightarrow c_{2}=1$.
Substituting $c_{1}=1$ and $c_{2}=1$ in $(i)$,the particular solution is $y=(1+x) e^{x / 5}$.

(D) Given the differential equation: $3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x$.
Dividing both sides by $3 x \log _{e} x$,we get: $\frac{d y}{d x} + \frac{1}{3 x \log _{e} x} y = \frac{2}{3 x}$.
This is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$,where $P = \frac{1}{3 x \log _{e} x}$.
The integrating factor $(IF)$ is given by $e^{\int P d x}$.
$IF = e^{\int \frac{1}{3 x \log _{e} x} d x}$.
Let $t = \log _{e} x$,then $d t = \frac{1}{x} d x$.
$IF = e^{\frac{1}{3} \int \frac{1}{t} d t} = e^{\frac{1}{3} \log _{e} t} = e^{\log _{e} (t^{1/3})} = t^{1/3}$.
Substituting back $t = \log _{e} x$,we get $IF = (\log _{e} x)^{1/3}$.

(B) Given the differential equation: $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$.
Rearranging the terms,we get: $\frac{1 - y \log x}{y^2} dy = \frac{1}{x} dx$.
Integrating both sides: $\int (y^{-2} - \frac{\log x}{y}) dy = \int \frac{1}{x} dx$ is not direct. Let us check the options.
If $y = x^y$,then taking $\log$ on both sides: $\log y = y \log x$.
Differentiating w.r.t. $x$: $\frac{1}{y} \frac{dy}{dx} = y \cdot \frac{1}{x} + \log x \cdot \frac{dy}{dx}$.
Rearranging for $\frac{dy}{dx}$: $\frac{dy}{dx} (\frac{1}{y} - \log x) = \frac{y}{x}$.
$\frac{dy}{dx} (\frac{1 - y \log x}{y}) = \frac{y}{x}$.
$x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$.
This matches the given differential equation. Also,for $x=1$,$y=1^y=1$,which satisfies the condition $y(1)=1$.

(D) There are $52$ distinct types of cards in a standard deck,and each type appears twice in two decks combined (total $104$ cards).
To get $26$ distinct cards,we must first choose $26$ types out of $52$ types,which can be done in ${ }^{52} C_{26}$ ways.
For each of these $26$ chosen types,we can pick either of the $2$ available cards,which gives $2^{26}$ ways.
The total number of ways to choose $26$ cards from $104$ is ${ }^{104} C_{26}$.
Therefore,the probability is $\frac{{ }^{52} C_{26} \times 2^{26}}{{ }^{104} C_{26}}$.

(A) Given: $P(A^{C}) = 0.3 \implies P(A) = 1 - 0.3 = 0.7$.
$P(B) = 0.4 \implies P(B^{C}) = 1 - 0.4 = 0.6$.
$P(A \cap B^{C}) = 0.5$.
We know $P(A) = P(A \cap B) + P(A \cap B^{C})$,so $P(A \cap B) = P(A) - P(A \cap B^{C}) = 0.7 - 0.5 = 0.2$.
We need to find $P(B \mid A \cup B^{C}) = \frac{P(B \cap (A \cup B^{C}))}{P(A \cup B^{C})}$.
Numerator: $P(B \cap (A \cup B^{C})) = P((B \cap A) \cup (B \cap B^{C})) = P(A \cap B) = 0.2$.
Denominator: $P(A \cup B^{C}) = P(A) + P(B^{C}) - P(A \cap B^{C}) = 0.7 + 0.6 - 0.5 = 0.8$.
Thus,$P(B \mid A \cup B^{C}) = \frac{0.2}{0.8} = \frac{1}{4}$.

(B) Let $F$ denote the fair coin,$T$ denote the two-headed coin,and $H$ denote the event that the outcome is a head.
Given probabilities are $P(F) = \frac{3}{4}$ and $P(T) = 1 - \frac{3}{4} = \frac{1}{4}$.
The probability of getting a head given the fair coin is $P(H|F) = \frac{1}{2}$.
The probability of getting a head given the two-headed coin is $P(H|T) = 1$.
We need to find the probability that the two-headed coin was chosen given that the outcome is a head,$P(T|H)$.
By Bayes' theorem:
$P(T|H) = \frac{P(H|T) \cdot P(T)}{P(H|T) \cdot P(T) + P(H|F) \cdot P(F)}$
$P(T|H) = \frac{1 \cdot \frac{1}{4}}{1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}}$
$P(T|H) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{3}{8}} = \frac{\frac{1}{4}}{\frac{2+3}{8}} = \frac{\frac{1}{4}}{\frac{5}{8}} = \frac{1}{4} \times \frac{8}{5} = \frac{2}{5}$.