WBJEE 2016 Mathematics Question Paper with Answer and Solution

(C) To find the roots of the equation $\tan x - x = 0$,we look for the intersection points of the graphs $y = \tan x$ and $y = x$.
At $x = 0$,both functions are zero,but we are looking for the smallest positive root.
For $x \in (0, \frac{\pi}{2})$,$\tan x > x$,so there is no root in this interval.
For $x \in (\frac{\pi}{2}, \pi)$,$\tan x$ is negative while $x$ is positive,so there is no root.
For $x \in (\pi, \frac{3\pi}{2})$,the graph of $y = \tan x$ starts from $-\infty$ and increases to $+\infty$,while $y = x$ is a line with a positive slope. They intersect at a point in the interval $(\pi, \frac{3\pi}{2})$.
Thus,the smallest positive root lies in the interval $(\pi, \frac{3\pi}{2})$.

(A) Let $f(n) = n^{3} + 2n$.
We can rewrite this as $f(n) = n^{3} - n + 3n = n(n^{2} - 1) + 3n = (n-1)n(n+1) + 3n$.
Here,$(n-1)n(n+1)$ is the product of three consecutive integers,which is always divisible by $3! = 6$.
However,the expression $n^{3} + 2n$ is specifically divisible by $3$ because $n^{3} \equiv n \pmod{3}$ by Fermat's Little Theorem,so $n^{3} + 2n \equiv n + 2n = 3n \equiv 0 \pmod{3}$.
Testing values:
For $n=1$,$1^{3} + 2(1) = 3$ (divisible by $3$).
For $n=2$,$2^{3} + 2(2) = 8 + 4 = 12$ (divisible by $3$).
For $n=3$,$3^{3} + 2(3) = 27 + 6 = 33$ (divisible by $3$).
Thus,it is always divisible by $3$.

(A) Given that $p$ and $q$ are the roots of the quadratic equation $x^{2}+px+q=0$.
According to the relation between roots and coefficients:
Sum of roots: $p+q = -p \Rightarrow 2p+q=0$ (Equation $1$)
Product of roots: $pq = q$ (Equation $2$)
From Equation $2$,$pq - q = 0 \Rightarrow q(p-1) = 0$.
This implies either $q=0$ or $p=1$.
Case $I$: If $q=0$,then from Equation $1$,$2p+0=0 \Rightarrow p=0$. However,if $p=0$ and $q=0$,the equation becomes $x^2=0$,which has roots $0, 0$. This is a valid solution,but let us check the options.
Case $II$: If $p=1$,then from Equation $1$,$2(1)+q=0 \Rightarrow q=-2$.
Thus,the pair $(p, q) = (1, -2)$ satisfies the given condition.

(A) Given that $\alpha$ and $\beta$ are roots of $ax^{2}+bx+c=0$.
Sum of roots: $\alpha+\beta = -\frac{b}{a}$.
Product of roots: $\alpha\beta = \frac{c}{a}$.
For the new equation with roots $\alpha^{2}$ and $\beta^{2}$:
Sum of roots: $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta = (-\frac{b}{a})^{2}-2(\frac{c}{a}) = \frac{b^{2}}{a^{2}}-\frac{2c}{a} = \frac{b^{2}-2ac}{a^{2}}$.
Product of roots: $\alpha^{2}\beta^{2} = (\alpha\beta)^{2} = (\frac{c}{a})^{2} = \frac{c^{2}}{a^{2}}$.
The required quadratic equation is $x^{2}-(\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^{2}-(\frac{b^{2}-2ac}{a^{2}})x + \frac{c^{2}}{a^{2}} = 0$.
Multiplying by $a^{2}$,we get $a^{2}x^{2}-(b^{2}-2ac)x+c^{2}=0$.

(C) Given equation: $x^{2} - 10x + y^{2} + 21 = 0$.
For $x$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
Treating the equation as a quadratic in $x$: $x^{2} - 10x + (y^{2} + 21) = 0$.
$D = (-10)^{2} - 4(1)(y^{2} + 21) \geq 0$.
$100 - 4y^{2} - 84 \geq 0$ $\Rightarrow 16 - 4y^{2} \geq 0$ $\Rightarrow y^{2} \leq 4$.
Thus,$-2 \leq y \leq 2$.
Similarly,for $y$ to have real roots,treat the equation as a quadratic in $y$: $y^{2} + (x^{2} - 10x + 21) = 0$.
Since $y^{2} = -x^{2} + 10x - 21$,for $y$ to be real,$y^{2} \geq 0$.
$-x^{2} + 10x - 21 \geq 0 \Rightarrow x^{2} - 10x + 21 \leq 0$.
$(x-7)(x-3) \leq 0 \Rightarrow 3 \leq x \leq 7$.

(D) Let $f(x) = x^{2}-3x+k$.
For the roots to lie in the interval $(0,1)$,the vertex of the parabola must lie within $(0,1)$.
The $x$-coordinate of the vertex is given by $x = -\frac{b}{2a} = -\frac{-3}{2(1)} = \frac{3}{2} = 1.5$.
Since $1.5 \notin (0,1)$,it is impossible for both roots to lie in the interval $(0,1)$ simultaneously.
Therefore,there is no value of $k$ that satisfies the given condition.

(B) We have $\sum_{n=1}^{13}(i^{n}+i^{n+1}) = \sum_{n=1}^{13} i^{n} + \sum_{n=1}^{13} i^{n+1}$.
Since $i^{n}$ is a geometric progression with first term $a=i$ and common ratio $r=i$,the sum of $13$ terms is $S_{13} = i \frac{1-i^{13}}{1-i}$.
Note that $i^{13} = (i^{4})^{3} \times i = 1^{3} \times i = i$.
Thus,$\sum_{n=1}^{13} i^{n} = i \frac{1-i}{1-i} = i$.
Similarly,$\sum_{n=1}^{13} i^{n+1} = i^{2} \frac{1-i^{13}}{1-i} = -1 \frac{1-i}{1-i} = -1$.
Therefore,the total sum is $i + (-1) = i-1$.

(A) Given $|z_{1}|=|z_{2}|=|z_{3}|=1$.
Since $|z|=1$ $\Rightarrow z\bar{z}=1$ $\Rightarrow \bar{z}=\frac{1}{z}$.
Thus,$\frac{1}{z_{1}}=\bar{z}_{1}$,$\frac{1}{z_{2}}=\bar{z}_{2}$,and $\frac{1}{z_{3}}=\bar{z}_{3}$.
We are given $|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}|=1$.
Substituting the conjugates,we get $|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}|=1$.
Using the property $|\bar{z}|=|z|$,we have $|\overline{z_{1}+z_{2}+z_{3}}|=1$.
Therefore,$|z_{1}+z_{2}+z_{3}|=1$.

(A) Given,$z = \sin \theta - i \cos \theta$
$z = \cos \left(\theta - \frac{\pi}{2}\right) + i \sin \left(\theta - \frac{\pi}{2}\right) = e^{i(\theta - \frac{\pi}{2})}$
Using De Moivre's Theorem,$z^{n} = e^{i(n\theta - \frac{n\pi}{2})} = \cos \left(n \theta - \frac{n \pi}{2}\right) + i \sin \left(n \theta - \frac{n \pi}{2}\right)$
Then,$\frac{1}{z^{n}} = z^{-n} = e^{-i(n\theta - \frac{n\pi}{2})} = \cos \left(n \theta - \frac{n \pi}{2}\right) - i \sin \left(n \theta - \frac{n \pi}{2}\right)$
Adding these,$z^{n} + \frac{1}{z^{n}} = 2 \cos \left(n \theta - \frac{n \pi}{2}\right) = 2 \cos \left(\frac{n \pi}{2} - n \theta\right)$
Thus,option $A$ is correct.

(A) The given expression is the real part of the binomial expansion of $(1+e^{i\theta})^n$.
Let $S = 1+{ }^{n} C_{1} \cos \theta+{ }^{n} C_{2} \cos 2 \theta+\ldots+{ }^{n} C_{n} \cos n \theta$.
Then $S = \operatorname{Re}\left(\sum_{k=0}^{n} { }^{n} C_{k} e^{ik\theta}\right) = \operatorname{Re}((1+e^{i\theta})^n)$.
Using $1+e^{i\theta} = 1+\cos \theta + i \sin \theta = 2 \cos^2 \frac{\theta}{2} + i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$.
$1+e^{i\theta} = 2 \cos \frac{\theta}{2} \left(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right) = 2 \cos \frac{\theta}{2} e^{i\theta/2}$.
Thus,$(1+e^{i\theta})^n = (2 \cos \frac{\theta}{2})^n e^{in\theta/2} = (2 \cos \frac{\theta}{2})^n \left(\cos \frac{n\theta}{2} + i \sin \frac{n\theta}{2}\right)$.
Taking the real part,we get $S = (2 \cos \frac{\theta}{2})^n \cos \frac{n\theta}{2}$.

(A) The letters of the word $COCHIN$ are $C, C, H, I, N, O$. Arranging them in alphabetical order,we get $C, C, H, I, N, O$.
To find the number of words appearing before $COCHIN$,we consider words starting with letters that come before the letters in $COCHIN$ at each position.
$1$. Words starting with $C$ (at the first position): The remaining letters are $C, H, I, N, O$. Since there are two $C$s,the number of arrangements is $\frac{5!}{2!} = \frac{120}{2} = 60$.
$2$. However,we need words before $COCHIN$. Let's fix the first letter as $C$:
- Words starting with $CC$: Remaining letters are $H, I, N, O$. Number of arrangements $= 4! = 24$.
- Words starting with $CH$: Remaining letters are $C, I, N, O$. Number of arrangements $= 4! = 24$.
- Words starting with $CI$: Remaining letters are $C, H, N, O$. Number of arrangements $= 4! = 24$.
- Words starting with $CN$: Remaining letters are $C, H, I, O$. Number of arrangements $= 4! = 24$.
- The next words start with $CO$. The first word starting with $CO$ is $COCHIN$.
Total words before $COCHIN = 24 + 24 + 24 + 24 = 96$.

(C) The word is $ARRANGE$. It contains $7$ letters: $A(2), R(2), N(1), G(1), E(1)$.
To ensure the two $R$'s occur together,we treat the block $(RR)$ as a single unit.
Now,the letters to be arranged are $(RR), A, A, N, G, E$. This gives us $6$ units in total.
Among these $6$ units,the letter $A$ repeats $2$ times.
The number of ways to arrange these $6$ units is $\frac{6!}{2!}$.
Within the block $(RR)$,the two $R$'s can be arranged in $\frac{2!}{2!} = 1$ way.
Therefore,the total number of permutations is $\frac{6!}{2!} \times 1 = \frac{6!}{2!}$.

(C) Given that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $AP$.
Therefore,$2 \log _{b} x = \log _{a} x + \log _{c} x$.
Using the change of base formula $\log _{m} n = \frac{1}{\log _{n} m}$,we get:
$\frac{2}{\log _{x} b} = \frac{1}{\log _{x} a} + \frac{1}{\log _{x} c}$.
$\frac{2}{\log _{x} b} = \frac{\log _{x} c + \log _{x} a}{\log _{x} a \cdot \log _{x} c} = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
Rearranging,we have $\frac{2}{\log _{x} b} = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
Since $\frac{1}{\log _{x} a} = \log _{a} x$,we can write $\frac{\log _{x} a}{\log _{x} b} = \log _{b} a$,so $\frac{1}{\log _{x} b} = \log _{b} x$.
Actually,$\frac{2}{\log _{x} b} = 2 \log _{b} x$. Thus,$2 \log _{b} x = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
Using $\log _{x} a \cdot \log _{a} b = \log _{x} b$,we get $\log _{x} b = \log _{x} a \cdot \log _{a} b$.
Substituting this,$2 \log _{b} x = \frac{\log _{x} (ac)}{\log _{x} a \cdot \log _{x} c}$.
This simplifies to $\log _{x} c^2 = \log _{x} (ac)^{\log _{a} b}$.
Thus,$c^2 = (ac)^{\log _{a} b}$.

(C) The given series is $S = \sum_{n=1}^{\infty} (1+a+a^2+\dots+a^{n-1})x^{n-1}$.
Using the sum of a geometric progression,$1+a+a^2+\dots+a^{n-1} = \frac{1-a^n}{1-a}$.
So,$S = \sum_{n=1}^{\infty} \frac{1-a^n}{1-a} x^{n-1} = \frac{1}{1-a} \sum_{n=1}^{\infty} (x^{n-1} - a^n x^{n-1})$.
$S = \frac{1}{1-a} \left( \sum_{n=1}^{\infty} x^{n-1} - a \sum_{n=1}^{\infty} (ax)^{n-1} \right)$.
Using the sum of an infinite geometric series $\sum_{k=0}^{\infty} r^k = \frac{1}{1-r}$ for $|r| < 1$:
$S = \frac{1}{1-a} \left( \frac{1}{1-x} - \frac{a}{1-ax} \right)$.
$S = \frac{1}{1-a} \left( \frac{(1-ax) - a(1-x)}{(1-x)(1-ax)} \right) = \frac{1}{1-a} \left( \frac{1-ax-a+ax}{(1-x)(1-ax)} \right)$.
$S = \frac{1}{1-a} \left( \frac{1-a}{(1-x)(1-ax)} \right) = \frac{1}{(1-x)(1-ax)}$.

(A) The $n^{th}$ term of the series is $T_{n} = (2n-1)^{3}$.
Expanding this,we get $T_{n} = 8n^{3} - 12n^{2} + 6n - 1$.
The sum of $n$ terms is $S_{n} = \sum_{k=1}^{n} T_{k} = \sum_{k=1}^{n} (8k^{3} - 12k^{2} + 6k - 1)$.
Using standard summation formulas:
$S_{n} = 8 \left[ \frac{n(n+1)}{2} \right]^{2} - 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] - n$.
$S_{n} = 2n^{2}(n+1)^{2} - 2n(n+1)(2n+1) + 3n(n+1) - n$.
$S_{n} = 2n^{2}(n^{2}+2n+1) - 2n(2n^{2}+3n+1) + 3n^{2} + 3n - n$.
$S_{n} = 2n^{4} + 4n^{3} + 2n^{2} - 4n^{3} - 6n^{2} - 2n + 3n^{2} + 2n$.
$S_{n} = 2n^{4} - n^{2} = n^{2}(2n^{2}-1)$.

(A) Let the general term be $T_k = (k-1)(k-\omega)(k-\omega^2)$ for $k=2$ to $n$.
Since $\omega$ is an imaginary cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,which implies $\omega + \omega^2 = -1$.
Expanding the term: $T_k = (k-1)(k^2 - k(\omega + \omega^2) + \omega^3) = (k-1)(k^2 + k + 1) = k^3 - 1$.
The sum is $S = \sum_{k=2}^{n} (k^3 - 1)$.
This can be written as $\sum_{k=1}^{n} (k^3 - 1) - (1^3 - 1) = \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} 1$.
Using the standard summation formulas $\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$ and $\sum_{k=1}^{n} 1 = n$,we get:
$S = \frac{n^2(n+1)^2}{4} - n$.

(D) Let the first term be $x$ and the $(2n-1)$-th term be $y$.
Since $a, b, c$ are the $n$-th terms of $AP, GP, HP$ respectively,they are the Arithmetic Mean,Geometric Mean,and Harmonic Mean of $x$ and $y$.
We know that for any two positive numbers $x$ and $y$,the relationship between their means is $AM \geq GM \geq HM$.
Therefore,$a \geq b \geq c$.
Also,the relationship between $AM, GM,$ and $HM$ is $AM \cdot HM = GM^2$,which implies $a \cdot c = b^2$ or $ac - b^2 = 0$.
Since the question asks for the relationship that is 'always' true,both $a \geq b \geq c$ and $ac = b^2$ are correct. However,in standard competitive mathematics,$ac = b^2$ is the specific identity derived from the definition of these means.

(B) Given,$\frac{1}{{ }^{5}C_{r}} + \frac{1}{{ }^{6}C_{r}} = \frac{1}{{ }^{4}C_{r}}$
Using the formula ${ }^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:
$\frac{r!(5-r)!}{5!} + \frac{r!(6-r)!}{6!} = \frac{r!(4-r)!}{4!}$
Dividing by $r!$ and multiplying by $4!$:
$\frac{(5-r)!}{5} + \frac{(6-r)(5-r)!}{6 \times 5} = (4-r)!$
Dividing by $(4-r)!$:
$\frac{(5-r)}{5} + \frac{(6-r)(5-r)}{30} = 1$
Multiply by $30$:
$6(5-r) + (6-r)(5-r) = 30$
$30 - 6r + 30 - 11r + r^{2} = 30$
$r^{2} - 17r + 30 = 0$
$(r-2)(r-15) = 0$
Since $r \leq 4$ (as ${ }^{4}C_{r}$ is defined),we have $r = 2$.

(C) We know the property of binomial coefficients: $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$.
Using this,we have:
$1) \frac{{}^n C_r}{{}^n C_{r-1}} = \frac{84}{36} = \frac{7}{3}$ $\Rightarrow \frac{n-r+1}{r} = \frac{7}{3}$ $\Rightarrow 3n - 3r + 3 = 7r$ $\Rightarrow 3n - 10r = -3$ (Eq. $1$)
$2) \frac{{}^n C_{r+1}}{{}^n C_r} = \frac{126}{84} = \frac{3}{2}$ $\Rightarrow \frac{n-(r+1)+1}{r+1} = \frac{3}{2}$ $\Rightarrow \frac{n-r}{r+1} = \frac{3}{2}$ $\Rightarrow 2n - 2r = 3r + 3$ $\Rightarrow 2n - 5r = 3$ (Eq. $2$)
Multiplying Eq. $2$ by $2$,we get $4n - 10r = 6$ (Eq. $3$).
Subtracting Eq. $1$ from Eq. $3$: $(4n - 10r) - (3n - 10r) = 6 - (-3) \Rightarrow n = 9$.
Substituting $n=9$ into Eq. $2$: $2(9) - 5r = 3$ $\Rightarrow 18 - 5r = 3$ $\Rightarrow 5r = 15$ $\Rightarrow r = 3$.
Thus,${}^n C_8 = {}^9 C_8 = {}^9 C_{9-8} = {}^9 C_1 = 9$.

(B) The expression is given by the product $P(x) = (x-1)(x-2) \ldots (x-18)$.
This is a polynomial of degree $18$.
The coefficient of $x^{n-1}$ in the expansion of $(x-a_1)(x-a_2) \ldots (x-a_n)$ is given by $-(a_1 + a_2 + \ldots + a_n)$.
Here,$n = 18$ and the terms are $a_1=1, a_2=2, \ldots, a_{18}=18$.
The coefficient of $x^{17}$ is $-(1 + 2 + 3 + \ldots + 18)$.
Using the sum formula for the first $n$ natural numbers,$S_n = \frac{n(n+1)}{2}$.
$S_{18} = \frac{18 \times 19}{2} = 9 \times 19 = 171$.
Therefore,the coefficient is $-171$.

(B) We are given the expression: $\cos 15^{\circ} \cos 7.5^{\circ} \sin 7.5^{\circ}$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we can write $\sin 7.5^{\circ} \cos 7.5^{\circ} = \frac{1}{2} \sin(2 \times 7.5^{\circ}) = \frac{1}{2} \sin 15^{\circ}$.
Substituting this into the original expression:
$\cos 15^{\circ} \times (\frac{1}{2} \sin 15^{\circ}) = \frac{1}{2} \sin 15^{\circ} \cos 15^{\circ}$.
Again,using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin 15^{\circ} \cos 15^{\circ} = \frac{1}{2} \sin(2 \times 15^{\circ}) = \frac{1}{2} \sin 30^{\circ}$.
Thus,the expression becomes $\frac{1}{2} \times (\frac{1}{2} \sin 30^{\circ}) = \frac{1}{4} \sin 30^{\circ}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,the final value is $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.

(B) Given the three sides of a triangle are:
$x+8y-22=0$ $(i)$
$5x+2y-34=0$ $(ii)$
$2x-3y+13=0$ $(iii)$
Solving $(i)$ and $(ii)$:
$x=6, y=2$. Vertex $A = (6, 2)$.
Solving $(ii)$ and $(iii)$:
$x=4, y=7$. Vertex $B = (4, 7)$.
Solving $(i)$ and $(iii)$:
$x=-2, y=3$. Vertex $C = (-2, 3)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
$\text{Area} = \frac{1}{2} |6(7-3) + 4(3-2) + (-2)(2-7)|$
$\text{Area} = \frac{1}{2} |6(4) + 4(1) - 2(-5)|$
$\text{Area} = \frac{1}{2} |24 + 4 + 10|$
$\text{Area} = \frac{1}{2} |38| = 19$ sq units.

(B) The sum of the distances of a point $P(x, y)$ from $A(8, 0)$ and $B(0, 12)$ is minimized when $P$ lies on the line segment $AB$.
The equation of the line passing through $(8, 0)$ and $(0, 12)$ is given by $\frac{x}{8} + \frac{y}{12} = 1$.
Multiplying by $24$,we get $3x + 2y = 24$.
Since $x$ and $y$ must be natural numbers $(x, y \in \mathbb{N})$,we check for integer solutions on the line segment $0 < x < 8$ and $0 < y < 12$.
If $x = 2$,$3(2) + 2y = 24$ $\Rightarrow 2y = 18$ $\Rightarrow y = 9$.
If $x = 4$,$3(4) + 2y = 24$ $\Rightarrow 2y = 12$ $\Rightarrow y = 6$.
If $x = 6$,$3(6) + 2y = 24$ $\Rightarrow 2y = 6$ $\Rightarrow y = 3$.
These points $(2, 9), (4, 6), (6, 3)$ are all in $S$.
Thus,there are $3$ such points.

(C) Let the given points be $A(a, b)$ and $B(-a, -b)$.
The slope $m$ of the line passing through $A$ and $B$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-b - b}{-a - a} = \frac{-2b}{-2a} = \frac{b}{a}$.
The equation of the line passing through $(a, b)$ with slope $\frac{b}{a}$ is:
$y - b = \frac{b}{a}(x - a)$
$ay - ab = bx - ab$
$bx = ay$
Now,we check the given options to see which point satisfies the equation $bx = ay$:
For option $(C)$,substituting $x = a^2$ and $y = ab$:
$b(a^2) = a(ab)$
$a^2b = a^2b$
Since the equation is satisfied,the line passes through $(a^2, ab)$.

(A) Let the four points be $A(-a,-b)$,$B(a, b)$,$C(0,0)$,and $D(a^{2}, ab)$.
To check if $A, B$,and $C$ are collinear,we calculate the determinant:
$\left|\begin{array}{ccc}-a & -b & 1 \\ a & b & 1 \\ 0 & 0 & 1\end{array}\right| = -a(b-0) + b(a-0) + 1(0) = -ab + ab = 0$.
Since the determinant is $0$,points $A, B$,and $C$ are collinear.
Now,check if $B, C$,and $D$ are collinear:
$\left|\begin{array}{ccc}a & b & 1 \\ 0 & 0 & 1 \\ a^{2} & ab & 1\end{array}\right| = a(0-ab) - b(0-a^{2}) + 1(0) = -a^{2}b + a^{2}b = 0$.
Since the determinant is $0$,points $B, C$,and $D$ are collinear.
Since $A, B, C$ are collinear and $B, C, D$ are collinear,all four points $A, B, C$,and $D$ must lie on the same line.

(B, D) Let $(h, k)$ be a point on the line $x+y+1=0.$
Since the point lies on the line,we have $h+k+1=0,$ which implies $h = -k-1.$
The perpendicular distance from $(h, k)$ to the line $3x+4y+2=0$ is given by $\frac{|3h+4k+2|}{\sqrt{3^2+4^2}} = \frac{1}{5}.$
Substituting $h = -k-1$ into the distance formula:
$\frac{|3(-k-1)+4k+2|}{5} = \frac{1}{5}$
$|-3k-3+4k+2| = 1$
$|k-1| = 1$
This gives two cases:
$k-1 = 1 \implies k=2,$ then $h = -2-1 = -3.$ Point is $(-3, 2).$
$k-1 = -1 \implies k=0,$ then $h = 0-1 = -1.$ Point is $(-1, 0).$
Thus,the required points are $(-3, 2)$ and $(-1, 0).$ Both options $B$ and $D$ are correct.

(B) The equation of the line $AB$ with intercepts $2a$ on both axes is given by the intercept form: $\frac{x}{2a} + \frac{y}{2a} = 1$,which simplifies to $x + y = 2a$.
Let the coordinates of any point $P$ on the line $AB$ be $(2h, 2k)$.
Since $P$ lies on the line $x + y = 2a$,we have $2h + 2k = 2a$,which simplifies to $h + k = a$.
Perpendiculars $PR$ and $PS$ are drawn to the axes,so $R$ is $(2h, 0)$ and $S$ is $(0, 2k)$.
The mid-point of $RS$ is given by $(\frac{2h+0}{2}, \frac{0+2k}{2}) = (h, k)$.
Let the coordinates of the mid-point be $(x, y)$,so $x = h$ and $y = k$.
Substituting these into the relation $h + k = a$,we get $x + y = a$.
Thus,the locus of the mid-point of $RS$ is $x + y = a$.

(D) Given equations of straight lines are:
$\frac{x}{a} + \frac{y}{b} = K$ $(1)$
$\frac{x}{a} - \frac{y}{b} = \frac{1}{K}$ $(2)$
Let the point of intersection be $(x, y)$.
Multiplying equation $(1)$ and $(2)$,we get:
$(\frac{x}{a} + \frac{y}{b})(\frac{x}{a} - \frac{y}{b}) = K \times \frac{1}{K}$
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
This is the equation of a hyperbola. Therefore,the locus is a hyperbola.

(B) $x^{3}-y x^{2}+x-y=0$
Factorizing by grouping:
$x^{2}(x-y)+1(x-y)=0$
$(x^{2}+1)(x-y)=0$
Since $x^{2}+1=0$ has no real solutions for $x$,the only real locus is given by:
$x-y=0$
$x=y$
Thus,the equation represents a straight line.

(A) The equation of the circle is $x^{2} + y^{2} = 9$,which has center $(0, 0)$ and radius $r = 3$.
Any line parallel to $3x + 4y = 0$ is of the form $3x + 4y = k$.
The perpendicular distance from the center $(0, 0)$ to the line $3x + 4y - k = 0$ must be equal to the radius $r = 3$.
Using the formula for distance from a point to a line: $\frac{|3(0) + 4(0) - k|}{\sqrt{3^{2} + 4^{2}}} = 3$.
$\frac{|-k|}{\sqrt{9 + 16}} = 3$ $\Rightarrow \frac{|k|}{5} = 3$ $\Rightarrow |k| = 15$.
So,$k = 15$ or $k = -15$.
The lines are $3x + 4y = 15$ and $3x + 4y = -15$.
For the line to touch the circle in the first quadrant,the intercept form $\frac{x}{5} + \frac{y}{3.75} = 1$ (for $k=15$) shows positive intercepts,which lies in the first quadrant.
Thus,the required equation is $3x + 4y = 15$.

(B) Let $(h, k)$ be the coordinates of the mid-point of a chord of the circle $x^{2}+y^{2}=1$. The equation of the chord with mid-point $(h, k)$ is given by $T=S_1$,which is $hx+ky = h^{2}+k^{2}$.
The equation of the pair of lines joining the origin to the points of intersection of the circle and the chord is obtained by homogenizing the circle equation using the chord equation:
$x^{2}+y^{2} = 1 \cdot \left(\frac{hx+ky}{h^{2}+k^{2}}\right)^{2}$
$(h^{2}+k^{2})^{2}(x^{2}+y^{2}) = (hx+ky)^{2}$
$(h^{2}+k^{2})^{2}(x^{2}+y^{2}) = h^{2}x^{2} + k^{2}y^{2} + 2hkxy$
Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero:
$(h^{2}+k^{2})^{2} - h^{2} + (h^{2}+k^{2})^{2} - k^{2} = 0$
$2(h^{2}+k^{2})^{2} - (h^{2}+k^{2}) = 0$
Since $h^{2}+k^{2} \neq 0$,we have $2(h^{2}+k^{2}) = 1$,or $h^{2}+k^{2} = \frac{1}{2}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $x^{2}+y^{2} = \frac{1}{2}$.

(B) Given equations are $x+y=4$ $(i)$ and $x-y=2$ $(ii)$.
Solving $(i)$ and $(ii)$,we get $x=3$ and $y=1$.
The line passing through $(3, 1)$ with slope $m = \tan(\tan^{-1}(3/4)) = 3/4$ is given by:
$(y-1) = \frac{3}{4}(x-3) \Rightarrow y = \frac{3x-5}{4}$.
Substituting this into the parabola equation $y^{2}=4(x-3)$:
$\left(\frac{3x-5}{4}\right)^{2} = 4(x-3)$
$\frac{9x^{2}-30x+25}{16} = 4x-12$
$9x^{2}-30x+25 = 64x-192$
$9x^{2}-94x+217 = 0$.
For this quadratic equation,$x_{1}+x_{2} = \frac{94}{9}$ and $x_{1}x_{2} = \frac{217}{9}$.
Then $|x_{1}-x_{2}| = \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}$
$= \sqrt{\left(\frac{94}{9}\right)^{2} - 4\left(\frac{217}{9}\right)}$
$= \sqrt{\frac{8836}{81} - \frac{868}{9}} = \sqrt{\frac{8836-7812}{81}} = \sqrt{\frac{1024}{81}} = \frac{32}{9}$.

(A, B) Given,the equation of the parabola is $x^{2}=ay$,which implies $y=\frac{x^{2}}{a}$.
The equation of the line is $y=2x+1$.
Substituting $y$ in the parabola equation: $\frac{x^{2}}{a}=2x+1 \Rightarrow x^{2}-2ax-a=0$.
Let the roots be $x_{1}$ and $x_{2}$. Then $x_{1}+x_{2}=2a$ and $x_{1}x_{2}=-a$.
The difference $|x_{1}-x_{2}| = \sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = \sqrt{4a^{2}+4a} = 2\sqrt{a^{2}+a}$.
Since the points lie on $y=2x+1$,the distance $d$ between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.
Since $y_{2}-y_{1} = 2(x_{2}-x_{1})$,we have $d = \sqrt{(x_{2}-x_{1})^{2}+4(x_{2}-x_{1})^{2}} = |x_{2}-x_{1}|\sqrt{5}$.
Given $d=\sqrt{40}$,so $\sqrt{40} = 2\sqrt{a^{2}+a} \cdot \sqrt{5} \Rightarrow \sqrt{40} = \sqrt{20(a^{2}+a)}$.
Squaring both sides: $40 = 20(a^{2}+a) \Rightarrow a^{2}+a-2=0$.
$(a+2)(a-1)=0$,so $a=1$ or $a=-2$.

(B) The given equation is $y^{2}-4y=4x-4a$.
Completing the square for $y$,we get $(y-2)^{2}-4=4x-4a$.
This simplifies to $(y-2)^{2}=4x-4a+4$,or $(y-2)^{2}=4(x-(a-1))$.
Comparing this with the standard form $(y-k)^{2}=4A(x-h)$,the vertex is $(h, k) = (a-1, 2)$.
The vertex lies between the lines $L_1: x+y-3=0$ and $L_2: 2x+2y-1=0$.
For a point $(x_0, y_0)$ to lie between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$,the expressions $(ax_0+by_0+c_1)$ and $(ax_0+by_0+c_2)$ must have opposite signs,i.e.,$(ax_0+by_0+c_1)(ax_0+by_0+c_2) < 0$.
Substituting the vertex $(a-1, 2)$ into the lines:
$L_1(a-1, 2) = (a-1)+2-3 = a-2$.
$L_2(a-1, 2) = 2(a-1)+2(2)-1 = 2a-2+4-1 = 2a+1$.
Thus,$(a-2)(2a+1) < 0$.
Solving this inequality,we find $a \in \left(-\frac{1}{2}, 2\right)$.

(D) Let the chord pass through the vertex $V(0, 0)$ and intersect the parabola $y^{2}=4ax$ at $P(at^{2}, 2at)$.
Let $(h, k)$ be the mid-point of the chord $VP$.
Then,$h = \frac{at^{2}+0}{2} = \frac{at^{2}}{2}$ and $k = \frac{2at+0}{2} = at$.
From $k = at$,we have $t = \frac{k}{a}$.
Substituting $t$ in the expression for $h$: $h = \frac{a}{2} \left(\frac{k}{a}\right)^{2} = \frac{k^{2}}{2a}$.
Thus,$k^{2} = 2ah$.
The locus of the mid-point $(h, k)$ is $y^{2} = 2ax$.
This can be rewritten as $(y-0)^{2} = 4\left(\frac{a}{2}\right)(x-0)$.
For a parabola $Y^{2} = 4AX$,the directrix is $X = -A$.
Here,$A = \frac{a}{2}$,so the directrix is $x = -\frac{a}{2}$.

(A) The given equation of the ellipse is $16 x^{2}+25 y^{2}+32 x-100 y=284$.
Rearranging the terms,we get $16(x^{2}+2 x)+25(y^{2}-4 y)=284$.
Completing the square,$16(x^{2}+2 x+1)+25(y^{2}-4 y+4)=284+16+100$.
$16(x+1)^{2}+25(y-2)^{2}=400$.
Dividing by $400$,we get $\frac{(x+1)^{2}}{25}+\frac{(y-2)^{2}}{16}=1$.
Here,$a^{2}=25$. The auxiliary circle of an ellipse $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$ is $(x-h)^{2}+(y-k)^{2}=a^{2}$.
Thus,the equation of the auxiliary circle is $(x+1)^{2}+(y-2)^{2}=25$.
Expanding this,$x^{2}+2 x+1+y^{2}-4 y+4=25$.
Therefore,$x^{2}+y^{2}+2 x-4 y-20=0$.

(B, D) The given equation of the ellipse is $4x^{2} + 9y^{2} = 1$.
Differentiating with respect to $x$,we get $8x + 18yy' = 0$,which implies $y' = -\frac{8x}{18y} = -\frac{4x}{9y}$.
The slope of the line $8x = 9y$ is $y = \frac{8}{9}x$,so the slope $m = \frac{8}{9}$.
Since the tangents are parallel to the line,their slopes must be equal: $-\frac{4x}{9y} = \frac{8}{9}$.
This simplifies to $-4x = 8y$,or $x = -2y$.
Substituting $x = -2y$ into the ellipse equation: $4(-2y)^{2} + 9y^{2} = 1$.
$4(4y^{2}) + 9y^{2} = 1$ $\Rightarrow 16y^{2} + 9y^{2} = 1$ $\Rightarrow 25y^{2} = 1$.
Thus,$y^{2} = \frac{1}{25}$,which gives $y = \pm \frac{1}{5}$.
If $y = \frac{1}{5}$,then $x = -2(\frac{1}{5}) = -\frac{2}{5}$.
If $y = -\frac{1}{5}$,then $x = -2(-\frac{1}{5}) = \frac{2}{5}$.
The required points are $\left(-\frac{2}{5}, \frac{1}{5}\right)$ and $\left(\frac{2}{5}, -\frac{1}{5}\right)$.

(C) The equation of the line is $y=x+\lambda$.
On comparing it with $y=mx+c$,we get $m=1$ and $c=\lambda$.
The equation of the ellipse is $2x^{2}+3y^{2}=1$,which can be written as $\frac{x^{2}}{1/2} + \frac{y^{2}}{1/3} = 1$.
On comparing it with $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we get $a^{2}=\frac{1}{2}$ and $b^{2}=\frac{1}{3}$.
If the line touches the ellipse,the condition for tangency is $c^{2}=a^{2}m^{2}+b^{2}$.
Substituting the values,we get $\lambda^{2} = \frac{1}{2}(1)^{2} + \frac{1}{3} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.
Therefore,$\lambda = \pm \sqrt{\frac{5}{6}}$.
Given the options,the correct value is $\lambda = \sqrt{\frac{5}{6}}$.

(D) Given the equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $\Delta OPQ$ is an equilateral triangle. $PQ$ is a double ordinate of the hyperbola.
Let the coordinates of $P$ and $Q$ be $(a \sec \theta, b \tan \theta)$ and $(a \sec \theta, -b \tan \theta)$ respectively.
In $\Delta OPQ$,$OP = PQ$.
Since $\Delta OPQ$ is equilateral,$OP^2 = PQ^2$.
$OP^2 = (a \sec \theta)^2 + (b \tan \theta)^2 = a^2 \sec^2 \theta + b^2 \tan^2 \theta$.
$PQ^2 = (2b \tan \theta)^2 = 4b^2 \tan^2 \theta$.
Equating them: $a^2 \sec^2 \theta + b^2 \tan^2 \theta = 4b^2 \tan^2 \theta$.
$a^2 \sec^2 \theta = 3b^2 \tan^2 \theta$.
$a^2 (1 + \tan^2 \theta) = 3b^2 \tan^2 \theta$.
$a^2 = (3b^2 - a^2) \tan^2 \theta$.
$\tan^2 \theta = \frac{a^2}{3b^2 - a^2}$.
Since $\tan^2 \theta > 0$,we must have $3b^2 - a^2 > 0$,so $3b^2 > a^2$,or $\frac{b^2}{a^2} > \frac{1}{3}$.
We know $e^2 = 1 + \frac{b^2}{a^2}$.
Since $\frac{b^2}{a^2} > \frac{1}{3}$,$e^2 > 1 + \frac{1}{3} = \frac{4}{3}$.
Therefore,$e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

(C) We have,$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{1-x}}$
$= \lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{(1+\sqrt{x})(1-\sqrt{x})}}$
$= \lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1}{1+\sqrt{x}}}$
$= \left(\frac{1+1}{2+1}\right)^{\frac{1}{1+1}} = \left(\frac{2}{3}\right)^{\frac{1}{2}} = \sqrt{\frac{2}{3}}$

(C) Let the observations be $x_{i} = a_{i}$ for $i = 1, 2, \ldots, n$. The standard deviation $\sigma$ is given by $\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_{i} - \bar{x})^2}$.
Let the new observations be $y_{i} = \lambda a_{i} = \lambda x_{i}$.
The mean of the new observations is $\bar{y} = \frac{1}{n} \sum_{i=1}^{n} \lambda x_{i} = \lambda \left( \frac{1}{n} \sum_{i=1}^{n} x_{i} \right) = \lambda \bar{x}$.
The standard deviation of the new observations $\sigma_{y}$ is:
$\sigma_{y} = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (y_{i} - \bar{y})^2}$
$= \sqrt{\frac{1}{n} \sum_{i=1}^{n} (\lambda x_{i} - \lambda \bar{x})^2}$
$= \sqrt{\frac{1}{n} \sum_{i=1}^{n} \lambda^2 (x_{i} - \bar{x})^2}$
$= \sqrt{\lambda^2 \cdot \frac{1}{n} \sum_{i=1}^{n} (x_{i} - \bar{x})^2}$
$= \sqrt{\lambda^2} \cdot \sigma$
$= |\lambda| \sigma$.

(A) Let the circumradius of $\triangle DEF$ be $R'$.
In $\triangle ABC$,the pedal triangle $\triangle DEF$ has angles $\angle FDE = 180^{\circ} - 2A$,$\angle DEF = 180^{\circ} - 2B$,and $\angle EFD = 180^{\circ} - 2C$.
The side length $EF$ of the pedal triangle is given by $EF = R \sin 2A$.
Using the sine rule in $\triangle DEF$,we have $2R' = \frac{EF}{\sin(\angle FDE)}$.
Substituting the values,$2R' = \frac{R \sin 2A}{\sin(180^{\circ} - 2A)} = \frac{R \sin 2A}{\sin 2A} = R$.
Therefore,$R' = \frac{R}{2}$.

(C) We have $A = 5^{n} - 4n - 1 = (1 + 4)^{n} - 4n - 1$.
Using the binomial expansion,$(1 + 4)^{n} = {}^{n}C_{0} + {}^{n}C_{1}(4) + {}^{n}C_{2}(4^{2}) + \dots + {}^{n}C_{n}(4^{n})$.
So,$A = (1 + 4n + 16({}^{n}C_{2} + {}^{n}C_{3}(4) + \dots + {}^{n}C_{n}(4^{n-2}))) - 4n - 1$.
$A = 16({}^{n}C_{2} + {}^{n}C_{3}(4) + \dots + {}^{n}C_{n}(4^{n-2}))$.
This shows that every element of $A$ is a multiple of $16$.
For $n=1$,$5^{1}-4(1)-1 = 0$.
For $n=2$,$5^{2}-4(2)-1 = 16$.
For $n=3$,$5^{3}-4(3)-1 = 112 = 16 \times 7$.
Thus,$A = \{0, 16, 112, \dots\}$.
$B = \{16(n-1) : n \in N\} = \{0, 16, 32, 48, \dots\}$.
Since every element of $A$ is a multiple of $16$,$A \subseteq B$.

(A) Given,$\log _{0.3}(x-1) < \log _{0.09}(x-1)$.
For the logarithm to be defined,we must have $x-1 > 0$,which implies $x > 1$.
We can rewrite the inequality as $\log _{0.3}(x-1) < \log _{0.3^{2}}(x-1)$.
Using the property $\log _{a^n} b = \frac{1}{n} \log_a b$,we get $\log _{0.3}(x-1) < \frac{1}{2} \log _{0.3}(x-1)$.
Multiplying by $2$,we get $2 \log _{0.3}(x-1) < \log _{0.3}(x-1)$.
Subtracting $\log _{0.3}(x-1)$ from both sides,we get $\log _{0.3}(x-1) < 0$.
Since the base $0.3 < 1$,the inequality reverses when we remove the logarithm: $x-1 > (0.3)^0$.
$x-1 > 1$.
$x > 2$.
Thus,$x$ lies in the interval $(2, \infty)$.

(A) Given,$P(A \cap B) = \frac{1}{6}$,$P(A \cup B) = \frac{31}{45}$,and $P(\bar{B}) = \frac{7}{10}$.
Since $P(B) = 1 - P(\bar{B})$,we have $P(B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we substitute the values:
$\frac{31}{45} = P(A) + \frac{3}{10} - \frac{1}{6}$.
$P(A) = \frac{31}{45} + \frac{1}{6} - \frac{3}{10} = \frac{62 + 15 - 27}{90} = \frac{50}{90} = \frac{5}{9}$.
Now,check for independence: $P(A) \times P(B) = \frac{5}{9} \times \frac{3}{10} = \frac{15}{90} = \frac{1}{6}$.
Since $P(A \cap B) = P(A) \times P(B)$,the events $A$ and $B$ are independent.

(C) Reflexivity: For $x \in Z$,we check if $(x, x) \in R$.
$x + 2x = 3x$,which is clearly divisible by $3$.
Thus,$xRx$ holds for all $x \in Z$,so $R$ is reflexive.
Symmetry: Let $(x, y) \in R$,which means $x + 2y = 3\lambda$ for some integer $\lambda$.
Then $x = 3\lambda - 2y$.
We check $y + 2x$:
$y + 2x = y + 2(3\lambda - 2y) = y + 6\lambda - 4y = 6\lambda - 3y = 3(2\lambda - y)$.
Since $3(2\lambda - y)$ is divisible by $3$,$(y, x) \in R$.
Thus,$R$ is symmetric.
Transitivity: Let $(x, y) \in R$ and $(y, z) \in R$.
Then $x + 2y = 3\lambda$ and $y + 2z = 3\mu$ for some integers $\lambda, \mu$.
Adding these: $(x + 2y) + (y + 2z) = 3\lambda + 3\mu \Rightarrow x + 3y + 2z = 3(\lambda + \mu)$.
$x + 2z = 3(\lambda + \mu) - 3y = 3(\lambda + \mu - y)$.
Since $x + 2z$ is divisible by $3$,$(x, z) \in R$.
Thus,$R$ is transitive.
Conclusion: Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(C) Given $Q = \begin{bmatrix} \cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix}$ and $x = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$.
We know that the matrix $Q(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ represents a rotation matrix.
By the property of rotation matrices,$Q^{n}(\theta) = Q(n\theta)$.
Therefore,$Q^{3} = Q\left(3 \times \frac{\pi}{4}\right) = Q\left(\frac{3\pi}{4}\right) = \begin{bmatrix} \cos \frac{3\pi}{4} & -\sin \frac{3\pi}{4} \\ \sin \frac{3\pi}{4} & \cos \frac{3\pi}{4} \end{bmatrix}$.
Evaluating the trigonometric values: $\cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}}$ and $\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.
So,$Q^{3} = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$.
Now,calculate $Q^{3}x = \begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}$.
$Q^{3}x = \begin{bmatrix} (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) \\ (\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}) \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - \frac{1}{2} \\ \frac{1}{2} - \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$.

(D) Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2 \end{bmatrix}$.
We can write $A = 2I + B$,where $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix}$.
Note that $B^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$.
Since $2I$ and $B$ commute,we use the Binomial Theorem:
$A^n = (2I + B)^n = \sum_{k=0}^n \binom{n}{k} (2I)^{n-k} B^k = \binom{n}{0} (2I)^n + \binom{n}{1} (2I)^{n-1} B + 0 + ...$
$A^n = 2^n I + n(2^{n-1}) B = 2^n \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + n 2^{n-1} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix}$
$A^n = \begin{bmatrix} 2^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 2^n \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ n 2^n & 0 & 0 \end{bmatrix} = \begin{bmatrix} 2^n & 0 & 0 \\ 0 & 2^n & 0 \\ n 2^n & 0 & 2^n \end{bmatrix}$.
Comparing this with the given form,we get $a = 2^n$ and $b = n 2^n$.

(C) We know that for a square matrix $A$ of order $n$,the determinant of its adjoint matrix is given by $|\operatorname{adj} A| = |A|^{n-1}$.
Given that $A$ is a $3 \times 3$ matrix,so $n = 3$.
Therefore,$|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.
Given $|B| = |\operatorname{adj} A| = 64$.
So,$|A|^2 = 64$.
Taking the square root on both sides,we get $|A| = \pm \sqrt{64} = \pm 8$.

(C) Let $\Delta = \left|\begin{array}{ccc}1 & \log _{x}y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$.
Using the change of base formula $\log_{a}b = \frac{\log b}{\log a}$,we can rewrite the determinant as:
$\Delta = \left|\begin{array}{ccc}\frac{\log x}{\log x} & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & \frac{\log y}{\log y} & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & \frac{\log z}{\log z}\end{array}\right|$.
Now,take $\frac{1}{\log x}$ common from $R_{1}$,$\frac{1}{\log y}$ common from $R_{2}$,and $\frac{1}{\log z}$ common from $R_{3}$:
$\Delta = \frac{1}{\log x \cdot \log y \cdot \log z} \left|\begin{array}{ccc}\log x & \log y & \log z \\ \log x & \log y & \log z \\ \log x & \log y & \log z\end{array}\right|$.
Since all three rows are identical,the value of the determinant is $0$.

(C) Given that $f(x)$ is an odd differentiable function.
By definition of an odd function,$f(-x) = -f(x)$.
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[-f(x)]$
$-f^{\prime}(-x) = -f^{\prime}(x)$
$f^{\prime}(-x) = f^{\prime}(x)$
This shows that the derivative of an odd function is an even function.
Now,substitute $x = 3$ into the equation $f^{\prime}(-x) = f^{\prime}(x)$:
$f^{\prime}(-3) = f^{\prime}(3)$
Since it is given that $f^{\prime}(3) = 2$,we have:
$f^{\prime}(-3) = 2$.

(C) Given,$f(x) = (x^{2} + 1)^{35}$ for all $x \in R$.
For one-one: Check if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
$f(1) = (1^{2} + 1)^{35} = 2^{35}$ and $f(-1) = ((-1)^{2} + 1)^{35} = 2^{35}$.
Since $f(1) = f(-1)$ but $1 \neq -1$,the function is not one-one.
For onto: The range of $f(x)$ must be equal to the codomain $R$.
Since $x^{2} \geq 0$,we have $x^{2} + 1 \geq 1$,which implies $(x^{2} + 1)^{35} \geq 1^{35} = 1$.
Thus,the range is $[1, \infty)$,which is not equal to the codomain $R$.
Therefore,the function is neither one-one nor onto.

(D) Given that $f(f(x)) = x$ for all $x \in X$.
To check for one-to-one (injective):
Let $f(x_1) = f(x_2)$.
Applying $f$ on both sides,we get $f(f(x_1)) = f(f(x_2))$.
Since $f(f(x)) = x$,this implies $x_1 = x_2$.
Thus,$f$ is one-to-one.
To check for onto (surjective):
For any $y \in X$,let $x = f(y)$.
Then $f(x) = f(f(y)) = y$.
Since for every $y \in X$,there exists an $x \in X$ such that $f(x) = y$,$f$ is onto.
Therefore,$f$ is both one-to-one and onto (bijective).

(C) Given,$y=(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{n}})$.
Taking the natural logarithm on both sides:
$\log y = \log(1+x) + \log(1+x^{2}) + \log(1+x^{4}) + \ldots + \log(1+x^{2^{n}})$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}}$.
Thus,$\frac{d y}{d x} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}} \right]$.
At $x=0$,$y = (1+0)(1+0) \ldots (1+0) = 1$.
Substituting $x=0$ in the derivative expression:
$\left(\frac{d y}{d x}\right)_{x=0} = 1 \left[ \frac{1}{1+0} + 0 + 0 + \ldots + 0 \right] = 1 \times 1 = 1$.

(C) We are given the function $f(x) = \max \{a-x, a+x, b\}$.
To find the points of non-differentiability,we analyze the intersections of the functions $y_1 = a-x$,$y_2 = a+x$,and $y_3 = b$.
$1$. Intersection of $y_1$ and $y_3$: $a-x = b \implies x = a-b$.
$2$. Intersection of $y_2$ and $y_3$: $a+x = b \implies x = b-a$.
$3$. Intersection of $y_1$ and $y_2$: $a-x = a+x \implies 2x = 0 \implies x = 0$.
Since $0 < a < b$,we have $a-b < 0$ and $b-a > 0$.
At $x = a-b$,$f(x)$ transitions from $a-x$ to $b$,creating a sharp turn.
At $x = b-a$,$f(x)$ transitions from $b$ to $a+x$,creating a sharp turn.
At $x = 0$,the value of $f(0) = \max \{a, a, b\} = b$ (since $b > a$). The function is $b$ in the interval $[a-b, b-a]$,so it is differentiable at $x=0$.
Thus,there are exactly $2$ points of non-differentiability: $x = a-b$ and $x = b-a$.

(D) Given,$f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}\right)}{\log \left(e x^{2}\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$
Using the properties of logarithms,$\log(e/x^2) = \log e - 2 \log x = 1 - 2 \log x$ and $\log(ex^2) = \log e + 2 \log x = 1 + 2 \log x$.
So,$f(x) = \tan^{-1}\left[\frac{1 - 2 \log x}{1 + 2 \log x}\right] + \tan^{-1}\left[\frac{3 + 2 \log x}{1 - 3(2 \log x)}\right]$.
Let $2 \log x = u$. Then $f(x) = \tan^{-1}\left[\frac{1 - u}{1 + u}\right] + \tan^{-1}\left[\frac{3 + u}{1 - 3u}\right]$.
Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ and $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$:
$f(x) = (\tan^{-1} 1 - \tan^{-1} u) + (\tan^{-1} 3 + \tan^{-1} u)$.
$f(x) = \frac{\pi}{4} + \tan^{-1} 3$.
Since $f(x)$ is a constant,its derivative $f'(x) = 0$ and the second derivative $f''(x) = 0$.

(A) Given that the ordinate $y$ decreases at the same rate at which the abscissa $x$ increases,we have $\frac{dy}{dt} = -\frac{dx}{dt}$.
The equation of the ellipse is $16x^{2} + 9y^{2} = 400$.
Differentiating both sides with respect to $t$,we get:
$16(2x) \frac{dx}{dt} + 9(2y) \frac{dy}{dt} = 0$
$32x \frac{dx}{dt} + 18y \frac{dy}{dt} = 0$
$16x \frac{dx}{dt} + 9y \frac{dy}{dt} = 0$.
Substituting $\frac{dy}{dt} = -\frac{dx}{dt}$ into the equation:
$16x \frac{dx}{dt} + 9y \left(-\frac{dx}{dt}\right) = 0$
$(16x - 9y) \frac{dx}{dt} = 0$.
Since $\frac{dx}{dt} \neq 0$ for the points to be moving,we have $16x - 9y = 0$,which implies $y = \frac{16}{9}x$.
Substituting $y = \frac{16}{9}x$ into the ellipse equation:
$16x^{2} + 9\left(\frac{16}{9}x\right)^{2} = 400$
$16x^{2} + 9 \cdot \frac{256}{81}x^{2} = 400$
$16x^{2} + \frac{256}{9}x^{2} = 400$
$\frac{144x^{2} + 256x^{2}}{9} = 400$
$\frac{400x^{2}}{9} = 400$
$x^{2} = 9 \Rightarrow x = \pm 3$.
If $x = 3$,then $y = \frac{16}{9}(3) = \frac{16}{3}$.
If $x = -3$,then $y = \frac{16}{9}(-3) = -\frac{16}{3}$.
Thus,the points are $\left(3, \frac{16}{3}\right)$ and $\left(-3, -\frac{16}{3}\right)$.

(B) Given,$T = 2 \pi \sqrt{\frac{l}{g}}$.
Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.
Differentiating both sides,we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.
Given that the length is increased by $2 \%$,so $\frac{dl}{l} = 0.02$.
Substituting this value,we get $\frac{dT}{T} = \frac{1}{2} \times 0.02 = 0.01$.
Therefore,the percentage change in the time period is $\frac{dT}{T} \times 100 = 0.01 \times 100 = 1 \%$.
Hence,the approximate change in the time period is $1 \%$.

(C) Given $f^{\prime}(x)=(x-1)^{2}(4-x).$
To find the critical points,we set $f^{\prime}(x)=0.$
$(x-1)^{2}(4-x)=0 \implies x=1, 4.$
We analyze the sign of $f^{\prime}(x)$ in the intervals $(-\infty, 1), (1, 4),$ and $(4, \infty).$
For $x \in (-\infty, 1),$ $f^{\prime}(x) > 0.$
For $x \in (1, 4),$ $f^{\prime}(x) > 0.$
For $x \in (4, \infty),$ $f^{\prime}(x) < 0.$
Since $f^{\prime}(x) > 0$ for $x \in (-\infty, 4),$ the function is increasing in $(-\infty, 4).$
Since $f^{\prime}(x) < 0$ for $x \in (4, \infty),$ the function is decreasing in $(4, \infty).$
At $x=4,$ $f^{\prime}(x)=0,$ so $x=4$ is a critical point.
Thus,option $C$ is correct.

(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} d x$.
Substitute $z = \log \sqrt{x} = \frac{1}{2} \log x$.
Then,$dz = \frac{1}{2} \cdot \frac{1}{x} dx$,which implies $\frac{dx}{x} = 2 dz$.
Substituting these into the integral:
$I = \int \frac{z}{3} (2 dz) = \frac{2}{3} \int z dz$.
Integrating $z$ with respect to $z$:
$I = \frac{2}{3} \cdot \frac{z^2}{2} + C = \frac{1}{3} z^2 + C$.
Substituting back $z = \log \sqrt{x}$:
$I = \frac{1}{3} (\log \sqrt{x})^2 + C$.

(C) Let $I = \int 2^{x} [f^{\prime}(x) + f(x) \log 2] \, dx$.
We know that the derivative of the product of two functions is given by the product rule: $\frac{d}{dx} [u(x)v(x)] = u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$.
Consider the function $g(x) = 2^{x} f(x)$.
Differentiating $g(x)$ with respect to $x$ using the product rule:
$g^{\prime}(x) = \frac{d}{dx}(2^{x}) \cdot f(x) + 2^{x} \cdot \frac{d}{dx}(f(x))$
$g^{\prime}(x) = 2^{x} \log 2 \cdot f(x) + 2^{x} f^{\prime}(x)$
$g^{\prime}(x) = 2^{x} [f^{\prime}(x) + f(x) \log 2]$.
Since the integrand is exactly the derivative of $g(x)$,we have:
$I = \int g^{\prime}(x) \, dx = g(x) + C = 2^{x} f(x) + C$.

(B) Let $I = \int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$.
Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we get:
$I = \int_{0}^{1} \log \left(\frac{1-(1-x)}{1-x}\right) d x$
$I = \int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x$
$I = \int_{0}^{1} \log \left(\left(\frac{1-x}{x}\right)^{-1}\right) d x$
$I = -\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$
$I = -I$
$2I = 0 \implies I = 0$.

(B) Let $I = \int_{0}^{2} x^{2}[x] d x$.
Since $[x]$ is the greatest integer function,we split the integral at the integer points in the interval $[0, 2]$.
$I = \int_{0}^{1} x^{2}[x] d x + \int_{1}^{2} x^{2}[x] d x$.
For $0 \le x < 1$,$[x] = 0$.
For $1 \le x < 2$,$[x] = 1$.
Substituting these values,we get:
$I = \int_{0}^{1} x^{2}(0) d x + \int_{1}^{2} x^{2}(1) d x$.
$I = 0 + \int_{1}^{2} x^{2} d x$.
$I = \left[ \frac{x^{3}}{3} \right]_{1}^{2}$.
$I = \frac{2^{3}}{3} - \frac{1^{3}}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$.

(A) Given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^{3/2}} \sum_{r=1}^{n-1} \sqrt{n+r}$
Dividing the numerator and denominator by $n$,we get:
$= \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \sqrt{\frac{n+r}{n}} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} \sqrt{1+\frac{r}{n}}$
This is a Riemann sum of the form $\int_{0}^{1} f(x) dx$ where $f(x) = \sqrt{1+x}$
$= \int_{0}^{1} (1+x)^{1/2} dx = \left[ \frac{(1+x)^{3/2}}{3/2} \right]_{0}^{1} = \frac{2}{3} \left[ (1+x)^{3/2} \right]_{0}^{1}$
$= \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2\sqrt{2} - 1)$

(B) We are given $\phi(t) = 1$ for $0 \leq t < 1$ and $0$ otherwise. This means $\phi(t-k) = 1$ for $k \leq t < k+1$ and $0$ otherwise.
The integral is $I = \int_{-3000}^{3000} \sum_{r'=2014}^{2016} \phi(t-r') \phi(t-2016) dt$.
Expanding the summation: $I = \int_{-3000}^{3000} [\phi(t-2014)\phi(t-2016) + \phi(t-2015)\phi(t-2016) + \phi(t-2016)\phi(t-2016)] dt$.
Note that $\phi(t-2014)\phi(t-2016) = 0$ because the intervals $[2014, 2015)$ and $[2016, 2017)$ are disjoint.
Similarly,$\phi(t-2015)\phi(t-2016) = 0$ because the intervals $[2015, 2016)$ and $[2016, 2017)$ are disjoint.
Thus,the expression simplifies to $\int_{-3000}^{3000} \phi(t-2016)^2 dt$.
Since $\phi(t-2016) = 1$ for $2016 \leq t < 2017$ and $0$ otherwise,$\phi(t-2016)^2 = \phi(t-2016)$.
Therefore,$I = \int_{2016}^{2017} 1 dt = [t]_{2016}^{2017} = 2017 - 2016 = 1$.

(C) The curves are $y=\sqrt{5-x^2}$ (a semicircle of radius $\sqrt{5}$) and $y=|x-1|$.
To find the intersection points,set $\sqrt{5-x^2} = |x-1|$.
Squaring both sides: $5-x^2 = x^2-2x+1 \implies 2x^2-2x-4=0 \implies x^2-x-2=0 \implies (x-2)(x+1)=0$.
So,the intersection points are $x=-1$ and $x=2$.
The area $A$ is given by $\int_{-1}^{2} (\sqrt{5-x^2} - |x-1|) dx$.
$A = \int_{-1}^{2} \sqrt{5-x^2} dx - \int_{-1}^{2} |x-1| dx$.
For the first integral,$\int_{-1}^{2} \sqrt{5-x^2} dx = \left[ \frac{x}{2}\sqrt{5-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt{5}}\right) \right]_{-1}^{2} = (1 + \frac{5}{2}\sin^{-1}\frac{2}{\sqrt{5}}) - (-1 + \frac{5}{2}\sin^{-1}\frac{-1}{\sqrt{5}}) = 2 + \frac{5}{2}(\sin^{-1}\frac{2}{\sqrt{5}} + \sin^{-1}\frac{1}{\sqrt{5}})$.
Since $\sin^{-1}\frac{2}{\sqrt{5}} + \sin^{-1}\frac{1}{\sqrt{5}} = \sin^{-1}(\frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}} + \frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}}) = \sin^{-1}(1) = \frac{\pi}{2}$,the first integral is $2 + \frac{5\pi}{4}$.
For the second integral,$\int_{-1}^{2} |x-1| dx = \int_{-1}^{1} (1-x) dx + \int_{1}^{2} (x-1) dx = [x - \frac{x^2}{2}]_{-1}^{1} + [\frac{x^2}{2} - x]_{1}^{2} = (1 - \frac{1}{2} - (-1 - \frac{1}{2})) + (2 - 2 - (\frac{1}{2} - 1)) = (1.5) + (0.5) = 2$.
Thus,$A = 2 + \frac{5\pi}{4} - 2 = \frac{5\pi}{4}$ sq units.
Wait,re-evaluating the integral $\int_{-1}^{2} |x-1| dx$: $\int_{-1}^{1} (1-x) dx = [x - x^2/2]_{-1}^{1} = (1-0.5) - (-1-0.5) = 0.5 + 1.5 = 2$. $\int_{1}^{2} (x-1) dx = [x^2/2 - x]_1^2 = (2-2) - (0.5-1) = 0.5$. Total is $2.5 = 5/2$.
$A = 2 + \frac{5\pi}{4} - 2.5 = \frac{5\pi}{4} - 0.5$ sq units.

(A) The general equation of a parabola whose axis of symmetry is along the $X$-axis is given by $y^2 = 4a(x - h)$,which can be rewritten as $y^2 = Ax + B$,where $A$ and $B$ are arbitrary constants.
Since there are $2$ arbitrary constants,we need to differentiate the equation twice to eliminate them.
First derivative: $2y \frac{dy}{dx} = A$.
Second derivative: $2 \left( \left( \frac{dy}{dx} \right)^2 + y \frac{d^2y}{dx^2} \right) = 0$.
Since the highest order derivative present in the differential equation is the second derivative,the order of the differential equation is $2$.

(B) Given the differential equation: $y \frac{dy}{dx} + by^2 = a \cos x$.
Let $y^2 = z$. Then,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = \frac{dz}{dx}$,which implies $y \frac{dy}{dx} = \frac{1}{2} \frac{dz}{dx}$.
Substituting this into the original equation: $\frac{1}{2} \frac{dz}{dx} + bz = a \cos x$.
Multiplying by $2$: $\frac{dz}{dx} + 2bz = 2a \cos x$.
This is a linear differential equation of the form $\frac{dz}{dx} + Pz = Q$,where $P = 2b$ and $Q = 2a \cos x$.
The integrating factor $(IF)$ is $e^{\int 2b \, dx} = e^{2bx}$.
The solution is given by $z \cdot IF = \int Q \cdot IF \, dx + c$.
$z e^{2bx} = \int 2a \cos x \cdot e^{2bx} \, dx + c$.
Using the standard integral $\int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)$,we get:
$z e^{2bx} = 2a \left[ \frac{e^{2bx}}{(2b)^2 + 1^2} (2b \cos x + \sin x) \right] + c$.
$y^2 e^{2bx} = \frac{2a}{4b^2 + 1} e^{2bx} (2b \cos x + \sin x) + c$.
Multiplying by $e^{-2bx}$: $y^2 = \frac{2a}{4b^2 + 1} (2b \cos x + \sin x) + c e^{-2bx}$.
Rearranging gives: $(4b^2 + 1) y^2 = 2a(\sin x + 2b \cos x) + c e^{-2bx}$.

(B) Given the differential equation: $x \frac{dy}{dx} + y = x e^x$.
Dividing by $x$ (assuming $x \neq 0$),we get: $\frac{dy}{dx} + \frac{1}{x} y = e^x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = e^x$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.
The solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + C$.
$xy = \int x e^x dx + C$.
Using integration by parts for $\int x e^x dx$:
$\int x e^x dx = x e^x - \int 1 \cdot e^x dx = x e^x - e^x = e^x(x-1)$.
Thus,$xy = e^x(x-1) + C$.
Comparing this with the given form $xy = e^x \phi(x) + C$,we find $\phi(x) = x-1$.

(D) Given,$|a+b| < |a-b|$.
Squaring both sides,we get $|a+b|^2 < |a-b|^2$.
Using the property $|x|^2 = x \cdot x$,we have $(a+b) \cdot (a+b) < (a-b) \cdot (a-b)$.
Expanding the dot product,$|a|^2 + |b|^2 + 2(a \cdot b) < |a|^2 + |b|^2 - 2(a \cdot b)$.
Subtracting $|a|^2 + |b|^2$ from both sides,$2(a \cdot b) < -2(a \cdot b)$.
This simplifies to $4(a \cdot b) < 0$,which implies $a \cdot b < 0$.
Since $a \cdot b = |a||b| \cos \theta < 0$ and $|a|, |b| > 0$,we must have $\cos \theta < 0$.
Therefore,the angle $\theta$ between $a$ and $b$ is an obtuse angle.

(A) Let the vertices of a cube be $(0,0,0)$ and $(a,a,a)$. The four diagonals of the cube can be represented by the vectors connecting opposite vertices: $\vec{d_1} = (a,a,a)$,$\vec{d_2} = (-a,a,a)$,$\vec{d_3} = (a,-a,a)$,and $\vec{d_4} = (a,a,-a)$.
Consider two diagonals with direction ratios $(1,1,1)$ and $(-1,1,1)$.
The cosine of the angle $\theta$ between two vectors $\vec{u} = (a_1, b_1, c_1)$ and $\vec{v} = (a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values: $\cos \theta = \frac{|(1)(-1) + (1)(1) + (1)(1)|}{\sqrt{1^2 + 1^2 + 1^2} \sqrt{(-1)^2 + 1^2 + 1^2}} = \frac{|-1 + 1 + 1|}{\sqrt{3} \sqrt{3}} = \frac{1}{3}$.
Thus,the cosine of the angle between any two diagonals of a cube is $1/3$.

(C) The equations of the given planes are $x+y+2z-6=0$ and $2x-y+z-9=0$.
Comparing these with the general form $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$,we get the normal vectors $\vec{n_1} = (1, 1, 2)$ and $\vec{n_2} = (2, -1, 1)$.
The angle $\theta$ between the two planes is given by the formula:
$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$.
Substituting the values:
$\cos \theta = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+(-1)^2+1^2}}$.
$\cos \theta = \frac{|2 - 1 + 2|}{\sqrt{1+1+4} \sqrt{4+1+1}} = \frac{3}{\sqrt{6} \times \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(B) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
For the points $(1, 1, 1)$ and $(0, 0, 0)$,the equation is $\frac{x-0}{1-0} = \frac{y-0}{1-0} = \frac{z-0}{1-0} = \lambda$.
This implies $x = \lambda, y = \lambda, z = \lambda$.
Any point on this line is of the form $(\lambda, \lambda, \lambda)$.
Since this point lies on the plane $2x + 2y + z = 10$,we substitute the coordinates into the plane equation:
$2(\lambda) + 2(\lambda) + \lambda = 10$.
$5\lambda = 10$.
$\lambda = 2$.
Substituting $\lambda = 2$ back into the point coordinates,we get $(2, 2, 2)$.

(D) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Therefore,$P(A) + P(B) = P(A \cup B) + P(A \cap B)$.
Given that $\frac{3}{4} \leq P(A \cup B) \leq 1$ and $\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}$.
Adding these inequalities,we get:
$\frac{3}{4} + \frac{1}{8} \leq P(A \cup B) + P(A \cap B) \leq 1 + \frac{3}{8}$.
$\frac{6+1}{8} \leq P(A) + P(B) \leq \frac{8+3}{8}$.
$\frac{7}{8} \leq P(A) + P(B) \leq \frac{11}{8}$.
Thus,both $P(A) + P(B) \geq \frac{7}{8}$ and $P(A) + P(B) \leq \frac{11}{8}$ are correct.

(B) Let $F$ be the event that the selected person is a female.
Let $A$ be the event that the selected person is aged above $40 \text{ yr}$.
We are given:
Total number of females $n(F) = 6$.
Number of females aged above $40 \text{ yr}$ is $n(A \cap F) = 3$.
We need to find the conditional probability $P(A|F)$,which is the probability that the person is aged above $40 \text{ yr}$ given that the person is a female.
The formula for conditional probability is $P(A|F) = \frac{n(A \cap F)}{n(F)}$.
Substituting the values:
$P(A|F) = \frac{3}{6} = \frac{1}{2}$.