TS EAMCET 2014 Mathematics Question Paper with Answer and Solution

(A) Let $\alpha, \beta, \gamma$ be the roots of $x^3-2x^2+10x-8=0$.
From Vieta's formulas:
$\alpha+\beta+\gamma = 2$,
$\alpha\beta+\beta\gamma+\gamma\alpha = 10$,
$\alpha\beta\gamma = 8$.
We want the equation with roots $\alpha^2, \beta^2, \gamma^2$.
The sum of the new roots is $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (2)^2 - 2(10) = 4 - 20 = -16$.
The sum of the products of the new roots taken two at a time is $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (10)^2 - 2(8)(2) = 100 - 32 = 68$.
The product of the new roots is $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (8)^2 = 64$.
The required cubic equation is $x^3 - (\text{sum of roots})x^2 + (\text{sum of product of roots taken two at a time})x - (\text{product of roots}) = 0$.
Substituting the values: $x^3 - (-16)x^2 + 68x - 64 = 0$,which simplifies to $x^3+16x^2+68x-64=0$.

(D) Given $Z_r = \cos \left(\frac{\pi}{2^r}\right) + i \sin \left(\frac{\pi}{2^r}\right) = e^{i \frac{\pi}{2^r}}$.
The product is $P = Z_1 Z_2 Z_3 \ldots = e^{i \frac{\pi}{2^1}} \cdot e^{i \frac{\pi}{2^2}} \cdot e^{i \frac{\pi}{2^3}} \ldots$
Using the property of exponents,$P = e^{i \left( \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \ldots \right)}$.
The exponent is a geometric series with first term $a = \frac{\pi}{2}$ and common ratio $r = \frac{1}{2}$.
The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{\pi/2}{1 - 1/2} = \frac{\pi/2}{1/2} = \pi$.
Therefore,$P = e^{i \pi} = \cos \pi + i \sin \pi = -1 + i(0) = -1$.

(A) Given,$x=p+q$,$y=p \omega+q \omega^2$ and $z=p \omega^2+q \omega$.
We know that $\omega^3=1$ and $1+\omega+\omega^2=0$,which implies $\omega+\omega^2=-1$.
Now,$xyz = (p+q)(p \omega+q \omega^2)(p \omega^2+q \omega)$
$xyz = (p+q)(p^2 \omega^3 + pq \omega^2 + pq \omega^4 + q^2 \omega^3)$
$xyz = (p+q)(p^2(1) + pq \omega^2 + pq \omega + q^2(1))$
$xyz = (p+q)(p^2 + pq(\omega^2+\omega) + q^2)$
$xyz = (p+q)(p^2 + pq(-1) + q^2)$
$xyz = (p+q)(p^2 - pq + q^2)$
$xyz = p^3+q^3$.

(D) Given,$\cos x = \tan y$,$\cot y = \tan z$ and $\cot z = \tan x$.
From the given equations,we have $\tan y = \cos x$.
Since $\cot y = \tan z$,we have $\frac{1}{\tan y} = \tan z$,which implies $\tan z = \frac{1}{\cos x}$.
Also,$\cot z = \tan x$,so $\frac{1}{\tan z} = \tan x$.
Substituting $\tan z = \frac{1}{\cos x}$ into $\cot z = \tan x$,we get $\cos x = \tan x$.
$\cos x = \frac{\sin x}{\cos x} \implies \cos^2 x = \sin x$.
Using $\cos^2 x = 1 - \sin^2 x$,we get $1 - \sin^2 x = \sin x$,or $\sin^2 x + \sin x - 1 = 0$.
Solving this quadratic equation for $\sin x$ using the quadratic formula: $\sin x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Since $-1 \le \sin x \le 1$,we discard the negative root $\frac{-1-\sqrt{5}}{2} < -1$.
Thus,$\sin x = \frac{\sqrt{5}-1}{2}$.

(D) We have the expression: $\tan 81^{\circ} + \tan 9^{\circ} - (\tan 63^{\circ} + \tan 27^{\circ})$
Using the identity $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B}$:
$\tan 81^{\circ} + \tan 9^{\circ} = \frac{\sin(81^{\circ}+9^{\circ})}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{\sin 90^{\circ}}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} = \frac{2}{\sin 18^{\circ}}$
Similarly,$\tan 63^{\circ} + \tan 27^{\circ} = \frac{\sin(63^{\circ}+27^{\circ})}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{\sin 90^{\circ}}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} = \frac{2}{\sin 54^{\circ}}$
Now,the expression becomes $\frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}} = 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}} \right)$
Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$\sin 54^{\circ} - \sin 18^{\circ} = 2 \cos 36^{\circ} \sin 18^{\circ}$
So,$2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}} \right) = 2 \times 2 = 4$

(C) We use the identity $\cos \theta \cos \left(\frac{\pi}{3}-\theta\right) \cos \left(\frac{\pi}{3}+\theta\right) = \frac{1}{4} \cos 3\theta$.
Given the equation $\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right) = \frac{1}{4}$,we substitute the identity:
$\frac{1}{4} \cos 3x = \frac{1}{4}$
$\cos 3x = 1$
For $x \in (0, 2\pi)$,$3x$ lies in the interval $(0, 6\pi)$.
The solutions for $\cos 3x = 1$ are $3x = 2\pi, 4\pi$.
Thus,$x = \frac{2\pi}{3}, \frac{4\pi}{3}$.
The sum of the solutions is $\frac{2\pi}{3} + \frac{4\pi}{3} = \frac{6\pi}{3} = 2\pi$.

(C) We use the logarithmic forms of inverse hyperbolic functions:
$\sec h^{-1} x = \log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosec} h^{-1} x = \log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right)$.
For $\sec h^{-1}\left(\frac{1}{2}\right)$:
$\sec h^{-1}\left(\frac{1}{2}\right) = \log _e\left(\frac{1+\sqrt{1-(1/2)^2}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3/4}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3}/2}{1/2}\right) = \log _e(2+\sqrt{3})$.
For $\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$:
$\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e\left(\frac{1+\sqrt{1+(3/4)^2}}{3/4}\right) = \log _e\left(\frac{1+\sqrt{25/16}}{3/4}\right) = \log _e\left(\frac{1+5/4}{3/4}\right) = \log _e\left(\frac{9/4}{3/4}\right) = \log _e(3)$.
Subtracting the two values:
$\log _e(2+\sqrt{3}) - \log _e(3) = \log _e\left(\frac{2+\sqrt{3}}{3}\right)$.

(D) Given,$\cos x + \cos y = \frac{3}{2}$
$\Rightarrow 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = \frac{3}{2}$
And,$\sin x + \sin y = \frac{3}{4}$
$\Rightarrow 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right) = \frac{3}{4}$
Dividing the second equation by the first:
$\frac{2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)}{2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)} = \frac{3/4}{3/2}$
$\Rightarrow \tan \left(\frac{x + y}{2}\right) = \frac{1}{2}$
Using the identity $\sin(x + y) = \frac{2 \tan \left(\frac{x + y}{2}\right)}{1 + \tan^2 \left(\frac{x + y}{2}\right)}$:
$\sin(x + y) = \frac{2 \times \frac{1}{2}}{1 + (\frac{1}{2})^2} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$

(B) The given equation is $x^2 + \alpha y^2 + 2 \beta y - a^2 = 0$.
Comparing this with the general equation of a pair of lines $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,we get $A=1, B=\alpha, H=0, G=0, F=\beta, C=-a^2$.
For the lines to be perpendicular,the condition is $A+B=0$.
Thus,$1 + \alpha = 0 \Rightarrow \alpha = -1$.
The condition for the equation to represent a pair of lines is $ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0$.
Substituting the values: $(1)(\alpha)(-a^2) + 2(\beta)(0)(0) - (1)(\beta)^2 - (\alpha)(0)^2 - (-a^2)(0)^2 = 0$.
This simplifies to $-\alpha a^2 - \beta^2 = 0$.
Substituting $\alpha = -1$,we get $-(-1)a^2 - \beta^2 = 0$,which implies $a^2 - \beta^2 = 0$.
Therefore,$\beta^2 = a^2$,which gives $\beta = \pm a$. Given the options,the correct value is $\beta = a$.

(D) The area of the triangle formed by the pair of lines $ax^2+2hxy+by^2=0$ and the line $lx+my+n=0$ is given by the formula $\text{Area} = \frac{n^2\sqrt{h^2-ab}}{|am^2-2hlm+bl^2|}$.
Here,the pair of lines is $x^2-3xy+y^2=0$,so $a=1, h=-\frac{3}{2}, b=1$.
The third line is $x+y+1=0$,so $l=1, m=1, n=1$.
Substituting these values into the formula:
$\text{Area} = \frac{1^2\sqrt{(-\frac{3}{2})^2-(1)(1)}}{|(1)(1)^2-2(-\frac{3}{2})(1)(1)+(1)(1)^2|}$
$\text{Area} = \frac{\sqrt{\frac{9}{4}-1}}{|1+3+1|} = \frac{\sqrt{\frac{5}{4}}}{5} = \frac{\frac{\sqrt{5}}{2}}{5} = \frac{\sqrt{5}}{10} = \frac{1}{2\sqrt{5}}$ sq. units.

(A) Let the coordinates of point $B$ be $(x, y, z)$.
Given that point $P(-2, 3, 5)$ divides the line segment joining $A(1, 3, 4)$ and $B(x, y, z)$ in the ratio $m:n = 1:3$.
Using the section formula for internal division:
$P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$
Substituting the values:
$(-2, 3, 5) = \left( \frac{1(x) + 3(1)}{1+3}, \frac{1(y) + 3(3)}{1+3}, \frac{1(z) + 3(4)}{1+3} \right)$
$(-2, 3, 5) = \left( \frac{x+3}{4}, \frac{y+9}{4}, \frac{z+12}{4} \right)$
Equating the coordinates:
$1) \frac{x+3}{4} = -2 \Rightarrow x+3 = -8 \Rightarrow x = -11$
$2) \frac{y+9}{4} = 3 \Rightarrow y+9 = 12 \Rightarrow y = 3$
$3) \frac{z+12}{4} = 5 \Rightarrow z+12 = 20 \Rightarrow z = 8$
Thus,the coordinates of $B$ are $(-11, 3, 8)$.

(C) Let the roots of the quadratic equation $x^2 - px + q = 0$ be $\alpha$ and $\beta$,where $\alpha, \beta \in \mathbb{Z}^+$.
By the properties of roots,we have:
$\alpha + \beta = p$ (sum of roots)
$\alpha \cdot \beta = q$ (product of roots)
Since $q$ is a prime number,its only factors are $1$ and $q$. Thus,the roots must be $1$ and $q$.
Substituting these into the sum of roots equation: $1 + q = p$.
Since $p$ and $q$ are both prime numbers,we look for two primes that differ by $1$. The only such primes are $2$ and $3$ (where $q=2$ and $p=3$).
Substituting $p=3$ and $q=2$ into the equation: $x^2 - 3x + 2 = 0$.
Factoring the quadratic: $(x - 1)(x - 2) = 0$.
Thus,the roots are $1$ and $2$.

(D) Given,$x_1$ and $x_2$ are the roots of the equation $x^2-kx+c=0$.
From the properties of roots:
$x_1+x_2 = k$
$x_1x_2 = c$
The distance between points $A(x_1, 0)$ and $B(x_2, 0)$ is given by $|x_2-x_1|$.
We know that $(x_2-x_1)^2 = (x_1+x_2)^2 - 4x_1x_2$.
Substituting the values:
$(x_2-x_1)^2 = k^2 - 4c$.
Therefore,the distance $|x_2-x_1| = \sqrt{k^2-4c}$.

(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.
Multiplying both sides by $(x^2+x+1)$,we get $y(x^2+x+1) = x^2-x+1$.
Rearranging the terms as a quadratic in $x$:
$(y-1)x^2 + (y+1)x + (y-1) = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = (y+1)^2 - 4(y-1)^2 \geq 0$.
$(y+1)^2 - [2(y-1)]^2 \geq 0$.
Using $a^2-b^2 = (a-b)(a+b)$:
$[(y+1) - 2(y-1)][(y+1) + 2(y-1)] \geq 0$.
$(-y+3)(3y-1) \geq 0$.
$(y-3)(3y-1) \leq 0$.
This inequality holds for $\frac{1}{3} \leq y \leq 3$.
Thus,the minimum value of $y$ is $\frac{1}{3}$.

(C) Given,$(1+i)^n=(1-i)^n$
$\Rightarrow \frac{(1+i)^n}{(1-i)^n}=1$
$\Rightarrow \left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n=1$
$\Rightarrow \left[\frac{1+i^2+2i}{1-i^2}\right]^n=1$
$\Rightarrow \left[\frac{1-1+2i}{1+1}\right]^n=1$
$\Rightarrow \left(\frac{2i}{2}\right)^n=1$
$\Rightarrow i^n=1$
Since the smallest positive integer $n$ for which $i^n=1$ is $4$,therefore $n=4$.

(A) Given,$z^3+\bar{z}=0$. Let $z=x+iy$.
Substituting $z$ in the equation: $(x+iy)^3 + (x-iy) = 0$.
Expanding: $x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 + x - iy = 0$.
$x^3 + 3x^2yi - 3xy^2 - iy^3 + x - iy = 0$.
Grouping real and imaginary parts: $(x^3 - 3xy^2 + x) + i(3x^2y - y^3 - y) = 0$.
Equating real and imaginary parts to zero:
$1) x(x^2 - 3y^2 + 1) = 0$
$2) y(3x^2 - y^2 - 1) = 0$
Case $1$: If $x=0$,then $-y(y^2+1)=0 \Rightarrow y=0$. Solution: $(0,0)$.
Case $2$: If $y=0$,then $x(x^2+1)=0 \Rightarrow x=0$. Solution: $(0,0)$.
Case $3$: If $x \neq 0$ and $y \neq 0$,then $x^2 - 3y^2 + 1 = 0$ and $3x^2 - y^2 - 1 = 0$.
Adding the two equations: $4x^2 - 4y^2 = 0 \Rightarrow x^2 = y^2$.
Substituting $x^2 = y^2$ into $x^2 - 3y^2 + 1 = 0$: $y^2 - 3y^2 + 1 = 0$ $\Rightarrow 2y^2 = 1$ $\Rightarrow y = \pm \frac{1}{\sqrt{2}}$.
Since $x^2 = y^2$,$x = \pm \frac{1}{\sqrt{2}}$.
Possible pairs $(x,y)$ are $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.
Total solutions: $(0,0)$ and the $4$ pairs above,giving $5$ solutions.

(B) We know that the binomial coefficient ${}^{11}C_n = \frac{11!}{n!(11-n)!}$ is maximum when $n$ is the middle value of the range $[0, 11]$.
Since $11$ is odd,the maximum value of ${}^{11}C_n$ occurs at $n = \frac{11-1}{2} = 5$ and $n = \frac{11+1}{2} = 6$.
Since ${}^{11}C_n$ is inversely proportional to $n!(11-n)!$,the expression $n!(11-n)!$ is minimized when ${}^{11}C_n$ is maximized.
Therefore,$n!(11-n)!$ is minimum when $n = 5$ or $n = 6$.

(C) The total number of points in a plane is $n = 30$.
Since $8$ points are collinear,they do not form distinct lines when joined in pairs,except for the single line they all lie on.
The formula for the number of lines formed by $n$ points where $m$ points are collinear is given by:
$\text{Number of lines} = {}^{n}C_{2} - {}^{m}C_{2} + 1$
Substituting the values $n = 30$ and $m = 8$:
$\text{Number of lines} = {}^{30}C_{2} - {}^{8}C_{2} + 1$
$= \frac{30 \times 29}{2} - \frac{8 \times 7}{2} + 1$
$= 435 - 28 + 1$
$= 408$

(B) The given sum is $S = 1^2 - 2^2 + 3^2 - 4^2 + \dots - (2n)^2 + (2n+1)^2$.
We can group the terms as:
$S = (1^2 - 2^2) + (3^2 - 4^2) + \dots + ((2n-1)^2 - (2n)^2) + (2n+1)^2$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,each pair becomes:
$(1-2)(1+2) + (3-4)(3+4) + \dots + ((2n-1)-2n)((2n-1)+2n) + (2n+1)^2$.
$S = -1(3) - 1(7) - 1(11) - \dots - 1(4n-1) + (2n+1)^2$.
$S = -(3 + 7 + 11 + \dots + (4n-1)) + (2n+1)^2$.
The sum inside the bracket is an arithmetic progression with $n$ terms,first term $a=3$,and last term $l=4n-1$.
Sum $= \frac{n}{2}(3 + 4n - 1) = \frac{n}{2}(4n+2) = n(2n+1)$.
Thus,$S = -n(2n+1) + (2n+1)^2$.
$S = (2n+1)(-n + 2n + 1) = (2n+1)(n+1)$.

(A) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is given by:
$T_{r+1} = { }^{18} C_r (\sqrt{x})^{18-r} \left(-\frac{2}{\sqrt{x}}\right)^r$
$T_{r+1} = { }^{18} C_r (x)^{\frac{18-r}{2}} (-2)^r (x)^{-\frac{r}{2}}$
$T_{r+1} = { }^{18} C_r (-2)^r x^{\frac{18-r-r}{2}}$
$T_{r+1} = { }^{18} C_r (-2)^r x^{9-r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$9 - r = 0 \Rightarrow r = 9$
Substituting $r = 9$ into the general term:
$T_{9+1} = { }^{18} C_9 (-2)^9$
$T_{10} = -{ }^{18} C_9 2^9$

(D) Given the expansion: $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$
We can write $(a+bx)^{-3} = a^{-3}(1 + \frac{bx}{a})^{-3}$.
Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we get:
$a^{-3}(1 + (-3)(\frac{bx}{a}) + \dots) = \frac{1}{a^3} - \frac{3bx}{a^4} + \dots$
Comparing the constant terms:
$\frac{1}{a^3} = \frac{1}{27} \implies a^3 = 27 \implies a = 3$.
Comparing the coefficients of $x$:
$-\frac{3b}{a^4} = \frac{1}{3}$.
Substituting $a=3$:
$-\frac{3b}{3^4} = \frac{1}{3} \implies -\frac{b}{3^3} = \frac{1}{3} \implies -\frac{b}{27} = \frac{1}{3} \implies b = -\frac{27}{3} = -9$.
Thus,the ordered pair $(a, b) = (3, -9)$.

(C) We use the identity $\cos \theta \cos \left(\frac{\pi}{3}-\theta\right) \cos \left(\frac{\pi}{3}+\theta\right) = \frac{1}{4} \cos 3\theta$.
Given the equation $\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right) = \frac{1}{4}$,we substitute the identity:
$\frac{1}{4} \cos 3x = \frac{1}{4}$
$\cos 3x = 1$
For $x \in (0, 2\pi)$,we have $3x \in (0, 6\pi)$.
The solutions for $\cos 3x = 1$ are $3x = 2\pi, 4\pi$.
Thus,$x = \frac{2\pi}{3}, \frac{4\pi}{3}$.
The sum of the solutions is $\frac{2\pi}{3} + \frac{4\pi}{3} = \frac{6\pi}{3} = 2\pi$.

(D) We have the expression: $\tan 81^{\circ} + \tan 9^{\circ} - (\tan 63^{\circ} + \tan 27^{\circ})$.
Using the identity $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B}$,we get:
$\frac{\sin(81^{\circ}+9^{\circ})}{\cos 81^{\circ} \cos 9^{\circ}} - \frac{\sin(63^{\circ}+27^{\circ})}{\cos 63^{\circ} \cos 27^{\circ}}$
$= \frac{\sin 90^{\circ}}{\cos 81^{\circ} \cos 9^{\circ}} - \frac{\sin 90^{\circ}}{\cos 63^{\circ} \cos 27^{\circ}}$
$= \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} - \frac{1}{\sin 27^{\circ} \cos 27^{\circ}}$
$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$
$= 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} \right)$
Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= 2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 54^{\circ} \sin 18^{\circ}} \right)$
$= 4 \cos 36^{\circ} / \sin 54^{\circ}$
Since $\cos 36^{\circ} = \sin 54^{\circ}$,the expression simplifies to $4 \times 1 = 4$.

(C) Let the vertices of the triangle be $A(a \cos \theta, a \sin \theta)$,$B(b \sin \theta, -b \cos \theta)$,and $C(1, 0)$.
Let the centroid be $(x, y)$.
By the centroid formula,$x = \frac{a \cos \theta + b \sin \theta + 1}{3}$ and $y = \frac{a \sin \theta - b \cos \theta}{3}$.
Rearranging,we get $a \cos \theta + b \sin \theta = 3x - 1$ and $a \sin \theta - b \cos \theta = 3y$.
Squaring both equations and adding them:
$(a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 = (3x - 1)^2 + (3y)^2$.
Expanding the left side:
$a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta = (3x - 1)^2 + 9y^2$.
$a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) = (3x - 1)^2 + 9y^2$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get $a^2 + b^2 = (3x - 1)^2 + 9y^2$.

(A) $1$. Reflection of $P(1,3)$ about $y=x$ gives $Q(3,1)$.
$2$. Translation of $Q(3,1)$ by $3$ units along the positive $X$-axis gives $R(3+3, 1) = R(6,1)$.
$3$. Rotation of $R(6,1)$ by $\theta = -\frac{\pi}{6}$ (clockwise) about the origin:
$x' = x \cos(\theta) - y \sin(\theta) = 6 \cos(-\frac{\pi}{6}) - 1 \sin(-\frac{\pi}{6}) = 6(\frac{\sqrt{3}}{2}) - 1(-\frac{1}{2}) = \frac{6\sqrt{3}+1}{2}$
$y' = x \sin(\theta) + y \cos(\theta) = 6 \sin(-\frac{\pi}{6}) + 1 \cos(-\frac{\pi}{6}) = 6(-\frac{1}{2}) + 1(\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}-6}{2}$
The final position is $\left(\frac{6 \sqrt{3}+1}{2}, \frac{\sqrt{3}-6}{2}\right)$.

(D) Let the point $P = \left(-\frac{7}{5}, -\frac{6}{5}\right)$ and its image $Q = (1, 2)$. The line is the perpendicular bisector of $PQ$.
The midpoint $M$ of $PQ$ is $\left(\frac{-\frac{7}{5} + 1}{2}, \frac{-\frac{6}{5} + 2}{2}\right) = \left(-\frac{1}{5}, \frac{2}{5}\right)$.
The slope of $PQ$ is $m_{PQ} = \frac{2 - (-6/5)}{1 - (-7/5)} = \frac{16/5}{12/5} = \frac{4}{3}$.
The slope of the required line is $m = -\frac{1}{m_{PQ}} = -\frac{3}{4}$.
Using the point-slope form $y - y_1 = m(x - x_1)$ at point $M$:
$y - \frac{2}{5} = -\frac{3}{4}\left(x + \frac{1}{5}\right)$
$\frac{5y - 2}{5} = -\frac{3}{4} \cdot \frac{5x + 1}{5}$
$4(5y - 2) = -3(5x + 1)$
$20y - 8 = -15x - 3$
$15x + 20y = 5$
Dividing by $5$,we get $3x + 4y = 1$.

(B) The given equation of the line is $3x - 4y = 6$.
Any line perpendicular to this line is of the form $4x + 3y = k$.
The intercepts of this line on the coordinate axes are $x = \frac{k}{4}$ and $y = \frac{k}{3}$.
The area of the triangle formed by the line with the coordinate axes is given by $\text{Area} = \frac{1}{2} \times |\text{base}| \times |\text{height}|$.
$\frac{1}{2} \times |\frac{k}{4}| \times |\frac{k}{3}| = 6$.
$|\frac{k^2}{24}| = 6$.
$k^2 = 144$.
$k = \pm 12$.
Thus,the required equations of the lines are $4x + 3y = 12$ or $4x + 3y = -12$.

(A) Since the points $A(k, 2k), B(3k, 3k)$,and $C(3, 1)$ are collinear,the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{3k - 2k}{3k - k} = \frac{k}{2k} = \frac{1}{2}$.
Slope of $BC = \frac{1 - 3k}{3 - 3k}$.
Equating the slopes: $\frac{1}{2} = \frac{1 - 3k}{3 - 3k}$.
Cross-multiplying: $3 - 3k = 2(1 - 3k)$ $\Rightarrow 3 - 3k = 2 - 6k$ $\Rightarrow 3k = -1$ $\Rightarrow k = -\frac{1}{3}$.
The equation of the line passing through $B(-1, -1)$ and $C(3, 1)$ is $y - 1 = \frac{1 - (-1)}{3 - (-1)}(x - 3)$.
$y - 1 = \frac{2}{4}(x - 3)$ $\Rightarrow y - 1 = \frac{1}{2}(x - 3)$ $\Rightarrow 2y - 2 = x - 3$ $\Rightarrow x - 2y - 1 = 0$.
The distance from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|-1|}{\sqrt{1^2 + (-2)^2}} = \frac{1}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}$.

(A) Let $r$ be the radius of the circle. The center of the circle is $C(2,4)$.
The perpendicular distance $d$ from the center $C(2,4)$ to the line $x+y+2=0$ is given by:
$d = \frac{|2+4+2|}{\sqrt{1^2+1^2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$.
The chord length is $6$,so the half-chord length is $AB = \frac{6}{2} = 3$.
In the right-angled triangle $\triangle CAB$,by the Pythagorean theorem:
$r^2 = d^2 + (AB)^2$
$r^2 = (4\sqrt{2})^2 + 3^2$
$r^2 = 32 + 9 = 41$
$r = \sqrt{41}$.

(C) The equation of the circle is $x^2+y^2=2^2$,so the radius $r=2$.
Any tangent to the circle with slope $m$ is given by $y=mx \pm r\sqrt{1+m^2}$,which is $y=mx \pm 2\sqrt{1+m^2}$.
The parabola is $y^2=32x$,which is of the form $y^2=4ax$ with $4a=32$,so $a=8$.
The focus of the parabola is $(a, 0) = (8, 0)$.
Since the tangent is a focal chord,it must pass through the focus $(8, 0)$.
Substituting $(8, 0)$ into the tangent equation: $0 = m(8) \pm 2\sqrt{1+m^2}$.
$8m = \mp 2\sqrt{1+m^2} \Rightarrow 4m = \mp \sqrt{1+m^2}$.
Squaring both sides: $16m^2 = 1+m^2$.
$15m^2 = 1 \Rightarrow m^2 = \frac{1}{15}$.
Thus,$m = \pm \frac{1}{\sqrt{15}}$.

(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$,where the centre is $(-g, -f)$.
The centre of the given circle $x^2+y^2-20x+4=0$ is $(10, 0)$ and its radius squared is $g_1^2+f_1^2-c_1 = 10^2+0^2-4 = 96$.
For two circles to cut orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1+c_2$.
Here,$g_1 = -10, f_1 = 0, c_1 = 4$ and $g_2 = g, f_2 = f, c_2 = c$.
So,$2(-10)(g) + 2(0)(f) = 4+c$,which gives $c = -20g-4$.
The circle touches the line $x=2$,so the distance from the centre $(-g, -f)$ to $x=2$ is equal to the radius $r = \sqrt{g^2+f^2-c}$.
$| -g - 2 | = \sqrt{g^2+f^2-c} \Rightarrow (g+2)^2 = g^2+f^2-c$.
$g^2+4g+4 = g^2+f^2-c \Rightarrow f^2-c-4g-4 = 0$.
Substituting $c = -20g-4$ into the equation:
$f^2 - (-20g-4) - 4g - 4 = 0$ $\Rightarrow f^2 + 20g + 4 - 4g - 4 = 0$ $\Rightarrow f^2 + 16g = 0$.
Replacing $(-g, -f)$ with $(x, y)$,we have $g = -x$ and $f = -y$.
$(-y)^2 + 16(-x) = 0$ $\Rightarrow y^2 - 16x = 0$ $\Rightarrow y^2 = 16x$.

(C) The given equations of the circles are $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,the centers and radii are:
For the first circle: $C_1 = (2, 2)$ and $r_1 = \sqrt{2^2+2^2-7} = \sqrt{8-7} = 1$.
For the second circle: $C_2 = (6, 5)$ and $r_2 = \sqrt{6^2+5^2-45} = \sqrt{36+25-45} = \sqrt{16} = 4$.
The distance between centers $C_1C_2 = \sqrt{(6-2)^2+(5-2)^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = 5$.
Since $C_1C_2 = r_1+r_2 = 1+4 = 5$,the circles touch each other externally.
The point of contact $P$ divides the line segment $C_1C_2$ internally in the ratio $r_1:r_2 = 1:4$.
Using the section formula,$P = \left(\frac{1(6)+4(2)}{1+4}, \frac{1(5)+4(2)}{1+4}\right) = \left(\frac{6+8}{5}, \frac{5+8}{5}\right) = \left(\frac{14}{5}, \frac{13}{5}\right)$.

(D) The given equations of the circles are:
$x^2+y^2-4y=0$ $(1)$
$x^2+y^2-8x-4y+11=0$ $(2)$
Subtracting equation $(1)$ from $(2)$,we get the equation of the common chord:
$(x^2+y^2-8x-4y+11) - (x^2+y^2-4y) = 0$
$-8x+11=0$ $\Rightarrow 8x=11$ $\Rightarrow x=\frac{11}{8}$
For the first circle $x^2+y^2-4y=0$,the center is $C(0, 2)$ and the radius $r = \sqrt{0^2 + 2^2 - 0} = 2$.
The perpendicular distance $d$ from the center $(0, 2)$ to the line $x = \frac{11}{8}$ is:
$d = |0 - \frac{11}{8}| = \frac{11}{8}$
Let $L$ be the length of the common chord. The formula for the length of the chord is $L = 2\sqrt{r^2 - d^2}$.
$L = 2\sqrt{2^2 - (\frac{11}{8})^2} = 2\sqrt{4 - \frac{121}{64}}$
$L = 2\sqrt{\frac{256 - 121}{64}} = 2\sqrt{\frac{135}{64}} = 2 \times \frac{\sqrt{135}}{8} = \frac{\sqrt{135}}{4} \text{ units}$.

(A) Two lines $lx + my + n = 0$ and $l_1x + m_1y + n_1 = 0$ are conjugate with respect to the circle $x^2 + y^2 = r^2$ if the pole of the first line lies on the second line.
The pole of the line $lx + my + n = 0$ with respect to the circle $x^2 + y^2 = r^2$ is given by $(x_1, y_1) = (-\frac{lr^2}{n}, -\frac{mr^2}{n})$.
Since this point lies on the line $l_1x + m_1y + n_1 = 0$,we have:
$l_1(-\frac{lr^2}{n}) + m_1(-\frac{mr^2}{n}) + n_1 = 0$
$-l_1lr^2 - m_1mr^2 + n_1n = 0$
$nn_1 = r^2(ll_1 + mm_1)$.

(C) The equation of the normal to the parabola $y^2=4ax$ at point $P(at^2, 2at)$ is $y + tx = 2at + at^3$.
Let this normal meet the parabola again at point $Q$. The combined equation of the lines $OP$ and $OQ$ joining the vertex $O(0,0)$ to the points $P$ and $Q$ is obtained by homogenizing the equation of the parabola $y^2=4ax$ using the equation of the normal:
$y^2 = 4ax \left( \frac{y+tx}{2at+at^3} \right)$
$y^2(2at + at^3) = 4ax(y + tx)$
$y^2(2at + at^3) = 4axy + 4atx^2$
$4atx^2 + 4axy - (2at + at^3)y^2 = 0$
Since $OP$ and $OQ$ are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$4at - (2at + at^3) = 0$
$4at - 2at - at^3 = 0$
$2at - at^3 = 0$
$at(2 - t^2) = 0$
Since $t \neq 0$ for a normal chord,we have $t^2 = 2$.

(B) The $y$-coordinate of the foci is $0$,so the major axis is on the $X$-axis. \\
Given $ae = 4$. \\
Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. \\
Since $b^2 = a^2(1 - e^2) = a^2 - (ae)^2 = a^2 - 16$,we substitute the point $(4\sqrt{2}, 2\sqrt{6})$: \\
$\frac{(4\sqrt{2})^2}{a^2} + \frac{(2\sqrt{6})^2}{a^2 - 16} = 1$ \\
$\frac{32}{a^2} + \frac{24}{a^2 - 16} = 1$ \\
$32(a^2 - 16) + 24a^2 = a^2(a^2 - 16)$ \\
$32a^2 - 512 + 24a^2 = a^4 - 16a^2$ \\
$a^4 - 72a^2 + 512 = 0$ \\
$(a^2 - 64)(a^2 - 8) = 0$ \\
Since $a > ae$,$a^2$ must be greater than $16$,so $a^2 = 64$ and $a = 8$. \\
Using $ae = 4$,we get $8e = 4$,which implies $e = \frac{1}{2}$.

(D) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$,where $a^2 = 16$ and $b^2 = 9$.
The director circle of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $x^2 + y^2 = a^2 + b^2$.
Here,$a^2 + b^2 = 16 + 9 = 25$.
Thus,the given circle $x^2 + y^2 = 25$ is the director circle of the ellipse.
By definition,the locus of the point of intersection of perpendicular tangents to an ellipse is its director circle.
Therefore,the angle between the tangents drawn from any point on the director circle to the ellipse is $\frac{\pi}{2}$.

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
The given equation of the ellipse is $\frac{x^2}{13^2} + \frac{y^2}{5^2} = 1$.
For the ellipse,$a_e = 13$ and $b_e = 5$.
The eccentricity of the ellipse $e$ is given by $e = \sqrt{1 - \frac{b_e^2}{a_e^2}} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
The foci of the ellipse are $(\pm a_e e, 0) = (\pm 13 \times \frac{12}{13}, 0) = (\pm 12, 0)$.
Since the hyperbola passes through $(\pm 12, 0)$,we have $\frac{12^2}{a^2} - \frac{0}{b^2} = 1$,which gives $a^2 = 144$.
The eccentricity of the hyperbola $e'$ is given by $e' = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{b^2}{144}}$.
Given that the product of eccentricities is $1$,we have $e \times e' = 1$.
$\frac{12}{13} \times \sqrt{1 + \frac{b^2}{144}} = 1$.
$\sqrt{1 + \frac{b^2}{144}} = \frac{13}{12}$.
Squaring both sides,$1 + \frac{b^2}{144} = \frac{169}{144}$.
$\frac{b^2}{144} = \frac{169}{144} - 1 = \frac{25}{144}$.
Thus,$b^2 = 25$.
The equation of the hyperbola is $\frac{x^2}{144} - \frac{y^2}{25} = 1$.

(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$.
Rationalizing the numerator,we multiply by $\frac{\sqrt{1+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x^2}+\sqrt{1-x+x^2}}$:
$= \lim _{x \rightarrow 0} \frac{(1+x^2)-(1-x+x^2)}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}$
$= \lim _{x \rightarrow 0} \frac{x}{(3^x-1)(\sqrt{1+x^2}+\sqrt{1-x+x^2})}$
$= \lim _{x}$ ${\rightarrow 0} \left( \frac{1}{\frac{3^x-1}{x}} \right) \times \left( \frac{1}{\sqrt{1+x^2}+\sqrt{1-x+x^2}} \right)$
Using the standard limit $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log _e a$,we get:
$= \frac{1}{\log _e 3} \times \frac{1}{\sqrt{1+0}+\sqrt{1-0+0}}$
$= \frac{1}{\log _e 3} \times \frac{1}{1+1}$
$= \frac{1}{2 \log _e 3} = \frac{1}{\log _e 3^2} = \frac{1}{\log _e 9}$.

(D) We are given $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 80$.
We know that the variance of a set of observations is always non-negative,i.e.,$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \geq 0$.
Substituting the given values:
$\frac{400}{n} - \left(\frac{80}{n}\right)^2 \geq 0$
$\frac{400}{n} - \frac{6400}{n^2} \geq 0$
Multiplying by $n^2$ (since $n > 0$):
$400n - 6400 \geq 0$
$400n \geq 6400$
$n \geq \frac{6400}{400}$
$n \geq 16$.
Thus,the least value of $n$ is $16$.

(C) Let the four observations be $x_1, x_2, x_3,$ and $x_4$.
Given,the mean $(\bar{x}) = 3$ and the sum of squares $\Sigma x_i^2 = 48$.
The number of observations $n = 4$.
The formula for standard deviation $(SD)$ is:
$SD = \sqrt{\frac{\Sigma x_i^2}{n} - (\bar{x})^2}$
Substituting the given values:
$SD = \sqrt{\frac{48}{4} - (3)^2}$
$SD = \sqrt{12 - 9}$
$SD = \sqrt{3}$

(D) Given,the ratio of angles of a triangle is $1: 1: 4$. Let the angles be $A, B$,and $C$.
$\therefore A: B: C = 1: 1: 4$
Let $A = x, B = x$,and $C = 4x$.
Since $A + B + C = 180^{\circ}$,we have $x + x + 4x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$.
Thus,$A = 30^{\circ}, B = 30^{\circ}$,and $C = 120^{\circ}$.
The largest angle is $120^{\circ}$,so the largest side is $c$.
The ratio of the perimeter to the largest side is $(a + b + c) : c$.
Using the sine rule,$a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.
Ratio $= (2R \sin 30^{\circ} + 2R \sin 30^{\circ} + 2R \sin 120^{\circ}) : 2R \sin 120^{\circ}$
$= (\sin 30^{\circ} + \sin 30^{\circ} + \sin 120^{\circ}) : \sin 120^{\circ}$
$= (\frac{1}{2} + \frac{1}{2} + \frac{\sqrt{3}}{2}) : \frac{\sqrt{3}}{2}$
$= (1 + \frac{\sqrt{3}}{2}) : \frac{\sqrt{3}}{2} = (2 + \sqrt{3}) : \sqrt{3}$.

(C) Let $2s = a+b+c$. Then $b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$.
Substituting these into the expression:
$\frac{2s(2s-2a)(2s-2b)(2s-2c)}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2} = 4 \frac{s(s-a)}{bc} \cdot \frac{(s-b)(s-c)}{bc}$.
Using the half-angle formulas $\cos^2(\frac{A}{2}) = \frac{s(s-a)}{bc}$ and $\sin^2(\frac{A}{2}) = \frac{(s-b)(s-c)}{bc}$,we get:
$4 \cos^2(\frac{A}{2}) \sin^2(\frac{A}{2}) = (2 \sin(\frac{A}{2}) \cos(\frac{A}{2}))^2 = \sin^2 A$.

(D) Given,$r_1=2, r_2=3$ and $r_3=6$.
We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$,where $r$ is the inradius.
$\frac{1}{r} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$,so $r=1$.
Also,$\Delta = \sqrt{r r_1 r_2 r_3} = \sqrt{1 \times 2 \times 3 \times 6} = \sqrt{36} = 6$.
Since $r_1 = \frac{\Delta}{s-a}$,we have $2 = \frac{6}{s-a}$,which implies $s-a = 3$.
Also,$s = \frac{\Delta}{r} = \frac{6}{1} = 6$.
Substituting $s=6$ into $s-a=3$,we get $6-a=3$,so $a=3$.

(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of $\triangle ABC$. Then $a+b+c = 2s$,$b+c-a = 2(s-a)$,$c+a-b = 2(s-b)$,and $a+b-c = 2(s-c)$.
Substituting these into the expression:
$\frac{(2s)(2(s-a))(2(s-b))(2(s-c))}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2}$.
Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$,so the expression becomes $\frac{16\Delta^2}{4b^2c^2} = \frac{4\Delta^2}{b^2c^2}$.
Since the area $\Delta = \frac{1}{2}bc \sin A$,we have $\sin A = \frac{2\Delta}{bc}$.
Thus,$\frac{4\Delta^2}{b^2c^2} = (\frac{2\Delta}{bc})^2 = \sin^2 A$.

(A) We know that two conics $\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1$ and $\frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=1$ intersect orthogonally if and only if $a_1^2-b_1^2 = a_2^2-b_2^2$,which can be rearranged as $a_1^2-a_2^2 = b_1^2-b_2^2$.
Given the curves are $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$.
Applying the condition for orthogonal intersection:
$a^2-25 = b^2-16$
Rearranging the terms to find $a^2-b^2$:
$a^2-b^2 = 25-16$
$a^2-b^2 = 9$
Thus,the value is $9$.

(C) Given,$\frac{2x^3+x^2-5}{x^4-25}=\frac{Ax+B}{x^2-5}+\frac{Cx+1}{x^2+5}$
Since $x^4-25 = (x^2-5)(x^2+5)$,we have:
$2x^3+x^2-5 = (Ax+B)(x^2+5) + (Cx+1)(x^2-5)$
$2x^3+x^2-5 = Ax^3 + 5Ax + Bx^2 + 5B + Cx^3 - 5Cx + x^2 - 5$
$2x^3+x^2-5 = x^3(A+C) + x^2(B+1) + x(5A-5C) + (5B-5)$
Equating the coefficients of $x^3, x^2, x$ and the constant term:
$1) A+C = 2$
$2) B+1 = 1 \Rightarrow B = 0$
$3) 5A-5C = 0 \Rightarrow A = C$
$4) 5B-5 = -5 \Rightarrow 5(0)-5 = -5$ (Consistent)
Substituting $A=C$ into $A+C=2$,we get $2C=2$,so $C=1$ and $A=1$.
Thus,$(A, B, C) = (1, 0, 1)$.

(B) Let $S$ be the sample space of outcomes where the sum of the two dice is $7$. The possible outcomes are:
$S = \{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)\}$
Total number of outcomes $n(S) = 6$.
Let $E$ be the event that the number $3$ appears at least once.
The favorable outcomes are:
$E = \{(3, 4), (4, 3)\}$
Number of favorable outcomes $n(E) = 2$.
The required probability is $P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.

(D) Given,$P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B)$.
Since $A, B$ and $C$ are mutually exclusive and exhaustive events,their sum of probabilities is $1$:
$P(A) + P(B) + P(C) = 1$
Substituting the given relations in terms of $P(A)$:
$P(A) + \frac{3}{2} P(A) + \frac{1}{2} \left( \frac{3}{2} P(A) \right) = 1$
$P(A) \left( 1 + \frac{3}{2} + \frac{3}{4} \right) = 1$
$P(A) \left( \frac{4 + 6 + 3}{4} \right) = 1$
$\frac{13}{4} P(A) = 1 \implies P(A) = \frac{4}{13}$
Now,calculate $P(C)$:
$P(C) = \frac{3}{4} P(A) = \frac{3}{4} \times \frac{4}{13} = \frac{3}{13}$
Since $A$ and $C$ are mutually exclusive,$P(A \cup C) = P(A) + P(C)$:
$P(A \cup C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$

(B) Given,$A=\left[\begin{array}{cccc}1 & 2 & 3 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 2 & 1 & 3 \\ 6 & 8 & 7 & \alpha\end{array}\right]$.
Applying row operations $R_2 \rightarrow R_2-2R_1$,$R_3 \rightarrow R_3-3R_1$,and $R_4 \rightarrow R_4-6R_1$:
$A \sim \left[\begin{array}{cccc}1 & 2 & 3 & 0 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -8 & 3 \\ 0 & -4 & -11 & \alpha\end{array}\right]$.
Applying $R_4 \rightarrow R_4-R_3$:
$A \sim \left[\begin{array}{cccc}1 & 2 & 3 & 0 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -8 & 3 \\ 0 & 0 & -3 & \alpha-3\end{array}\right]$.
Applying $R_4 \rightarrow R_4-R_2$:
$A \sim \left[\begin{array}{cccc}1 & 2 & 3 & 0 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -8 & 3 \\ 0 & 0 & 0 & \alpha-5\end{array}\right]$.
Since the rank of matrix $A$ is $3$,the last row must be a zero row.
Therefore,$\alpha-5=0$,which implies $\alpha=5$.

(D) Given,$f(x) = \frac{x}{1+x}$.
We need to find $g(x) = f(f(x))$.
$g(x) = f\left(\frac{x}{1+x}\right) = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$.
Multiplying the numerator and denominator by $(1+x)$,we get:
$g(x) = \frac{x}{1+x+x} = \frac{x}{2x+1}$.
Now,differentiate $g(x)$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
$g^{\prime}(x) = \frac{(1)(2x+1) - (x)(2)}{(2x+1)^2}$.
$g^{\prime}(x) = \frac{2x+1 - 2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}$.

(D) Given,the error in diameter $\Delta D = \pm 0.04 \text{ cm}$.
Since the radius $r = \frac{D}{2}$,the error in radius is $\Delta r = \frac{\Delta D}{2} = \pm \frac{0.04}{2} = \pm 0.02 \text{ cm}$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$,we get $\frac{dV}{dr} = 4 \pi r^2$.
The approximate error in volume is $\Delta V \approx \frac{dV}{dr} \Delta r = 4 \pi r^2 \Delta r$.
The percentage error in volume is given by $\frac{\Delta V}{V} \times 100$.
Substituting the values,we get $\frac{\Delta V}{V} \times 100 = \frac{4 \pi r^2 \Delta r}{\frac{4}{3} \pi r^3} \times 100 = \frac{3 \Delta r}{r} \times 100$.
Given $r = 10 \text{ cm}$ and $\Delta r = \pm 0.02 \text{ cm}$,the percentage error is $\frac{3 \times (\pm 0.02)}{10} \times 100 = \frac{\pm 0.06}{10} \times 100 = \pm 0.6 \%$.

(D) Given the function $f(x) = ax^3 + bx^2 + cx + d$.
To find the extreme values,we first find the derivative of $f(x)$ with respect to $x$:
$f'(x) = 3ax^2 + 2bx + c$.
$A$ function has no extreme values if its derivative $f'(x)$ does not change sign,which means $f'(x) = 0$ has no real roots or has equal roots such that the sign does not change.
For the quadratic equation $3ax^2 + 2bx + c = 0$ to have no real roots,its discriminant $D$ must be less than $0$.
$D = (2b)^2 - 4(3a)(c) < 0$.
$4b^2 - 12ac < 0$.
Dividing by $4$,we get $b^2 - 3ac < 0$,which implies $b^2 < 3ac$.

(B) Given,$\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$.
We can rewrite the integral as:
$\int \frac{d x}{\sqrt{\sin ^4 x \cdot \frac{\cos x}{\sin x}}} = \int \frac{d x}{\sin ^2 x \sqrt{\cot x}}$
This simplifies to:
$\int \operatorname{cosec}^2 x \cdot (\cot x)^{-1/2} d x$
Let $t = \cot x$,then $dt = -\operatorname{cosec}^2 x d x$,which implies $\operatorname{cosec}^2 x d x = -dt$.
Substituting these into the integral:
$\int -t^{-1/2} dt = -\frac{t^{1/2}}{1/2} + c = -2\sqrt{t} + c$
Substituting back $t = \cot x$:
$-2\sqrt{\cot x} + c = -\frac{2}{\sqrt{\tan x}} + c$
Therefore,$g(x) = -\frac{2}{\sqrt{\tan x}}$.

(B) The given limit can be expressed as a Riemann sum:
$S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{n^5( (r/n)^5 + 1 )} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^4}{(r/n)^5 + 1}$.
By the definition of the definite integral,this is equal to $\int_0^1 \frac{x^4}{x^5+1} dx$.
Let $t = x^5 + 1$,then $dt = 5x^4 dx$,or $x^4 dx = \frac{dt}{5}$.
When $x=0, t=1$. When $x=1, t=2$.
Thus,the integral becomes $\int_1^2 \frac{1}{t} \cdot \frac{dt}{5} = \frac{1}{5} [\log |t|]_1^2 = \frac{1}{5} (\log 2 - \log 1) = \frac{1}{5} \log 2$.

(D) The given differential equation is $x \frac{dy}{dx} = y + x e^{y/x}$.
Dividing by $x$,we get $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v + e^v$.
This simplifies to $x \frac{dv}{dx} = e^v$,or $e^{-v} dv = \frac{1}{x} dx$.
Integrating both sides: $\int e^{-v} dv = \int \frac{1}{x} dx$,which gives $-e^{-v} = \log x + c$.
Substituting $v = y/x$,we get $-e^{-y/x} = \log x + c$.
Given the condition $y(1) = 0$,we substitute $x = 1$ and $y = 0$: $-e^0 = \log 1 + c$,which implies $-1 = 0 + c$,so $c = -1$.
Thus,$-e^{-y/x} = \log x - 1$,which rearranges to $e^{-y/x} + \log x = 1$.

(C) Given,$\hat{d} = \frac{1}{x}(\hat{a} + \hat{b} + \hat{c}) \implies x\hat{d} = \hat{a} + \hat{b} + \hat{c} \implies \hat{a} + \hat{b} + \hat{c} - x\hat{d} = 0$.
Also,$\hat{d} = \frac{1}{y}(\hat{b} + \hat{c} + \hat{d}) \implies y\hat{d} = \hat{b} + \hat{c} + \hat{d} \implies \hat{b} + \hat{c} + \hat{d} - y\hat{d} = 0$.
Since $\hat{a}, \hat{b}, \hat{c}$ are non-coplanar,they are linearly independent.
From the given equations,we can deduce that $\hat{a} + \hat{b} + \hat{c} + \hat{d} = 0$.
Therefore,$\frac{1}{xy}(\hat{a} + \hat{b} + \hat{c} + \hat{d}) = \frac{1}{xy}(0) = 0$.

(C) Given vectors are $\vec{a}=x \hat{i}+2 \hat{j}$,$\vec{b}=y \hat{j}+3 \hat{k}$,and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & 2 & 0 \\ 0 & y & 3 \end{vmatrix} = \hat{i}(6-0) - \hat{j}(3x-0) + \hat{k}(xy-0) = 6 \hat{i} - 3x \hat{j} + xy \hat{k}$.
Given $\vec{a} \times \vec{b} = z \hat{i} - 3 \hat{j} + \hat{k}$.
Comparing coefficients,we get $z=6$,$3x=3 \Rightarrow x=1$,and $xy=1 \Rightarrow y=1$.
Now,the scalar triple product is $[\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c}$.
$[\vec{a} \vec{b} \vec{c}] = (6 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (x \hat{i} + y \hat{j} + z \hat{k})$.
Substituting $x=1, y=1, z=6$:
$[\vec{a} \vec{b} \vec{c}] = (6 \hat{i} - 3 \hat{j} + \hat{k}) \cdot (1 \hat{i} + 1 \hat{j} + 6 \hat{k}) = (6)(1) + (-3)(1) + (1)(6) = 6 - 3 + 6 = 9$.

(C) Given points are $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$.
Direction ratios (DRs) of line $DC$ are given by $(x_C - x_D, y_C - y_D, z_C - z_D) = (-1-1, 2-0, 4-5) = (-2, 2, -1)$.
Direction ratios (DRs) of line $AB$ are given by $(x_B - x_A, y_B - y_A, z_B - z_A) = (4-3, 6-4, 3-5) = (1, 2, -2)$.
Let $\theta$ be the angle between lines $AB$ and $DC$. The formula for $\cos \theta$ is:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the values:
$\cos \theta = \frac{|(-2)(1) + (2)(2) + (-1)(-2)|}{\sqrt{(-2)^2 + 2^2 + (-1)^2} \sqrt{1^2 + 2^2 + (-2)^2}}$
$\cos \theta = \frac{|-2 + 4 + 2|}{\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4}}$
$\cos \theta = \frac{4}{\sqrt{9} \sqrt{9}} = \frac{4}{3 \times 3} = \frac{4}{9}$.

(D) Given equations are $2l + m + 2n = 0$ $(1)$ and $3l^2 + 5m^2 - 11n^2 = 0$ $(2)$.
From $(1)$,$m = -2l - 2n$.
Substitute $m$ into $(2)$: $3l^2 + 5(-2l - 2n)^2 - 11n^2 = 0$.
$3l^2 + 5(4l^2 + 8ln + 4n^2) - 11n^2 = 0$.
$3l^2 + 20l^2 + 40ln + 20n^2 - 11n^2 = 0$.
$23l^2 + 40ln + 9n^2 = 0$.
Divide by $n^2$: $23(\frac{l}{n})^2 + 40(\frac{l}{n}) + 9 = 0$.
Let $x = \frac{l}{n}$. Then $23x^2 + 40x + 9 = 0$.
Let the roots be $x_1 = \frac{l_1}{n_1}$ and $x_2 = \frac{l_2}{n_2}$.
Then $x_1 x_2 = \frac{l_1 l_2}{n_1 n_2} = \frac{9}{23}$.
Similarly,substituting $l = -\frac{m+2n}{2}$ into $(2)$ gives $3(\frac{m+2n}{2})^2 + 5m^2 - 11n^2 = 0$,which leads to $3(m^2 + 4mn + 4n^2) + 20m^2 - 44n^2 = 0$,so $23m^2 + 12mn - 32n^2 = 0$.
Dividing by $n^2$,$23(\frac{m}{n})^2 + 12(\frac{m}{n}) - 32 = 0$.
Let $y_1 = \frac{m_1}{n_1}$ and $y_2 = \frac{m_2}{n_2}$. Then $y_1 y_2 = \frac{m_1 m_2}{n_1 n_2} = -\frac{32}{23}$.
For two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Since $l_i^2 + m_i^2 + n_i^2 = 1$,we use the relation $l_1 l_2 + m_1 m_2 + n_1 n_2 = n_1 n_2 (x_1 x_2 + y_1 y_2 + 1) = n_1 n_2 (\frac{9}{23} - \frac{32}{23} + 1) = n_1 n_2 (\frac{-23}{23} + 1) = 0$.
Thus,$\cos \theta = 0$,which means $\theta = \frac{\pi}{2}$.

(C) Given,mean of binomial variable,$np = 8$ and variance of binomial variable,$npq = 4$.
$\therefore q = \frac{npq}{np} = \frac{4}{8} = \frac{1}{2}$.
Since $p = 1 - q$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ in $np = 8$,we get $n \times \frac{1}{2} = 8$,so $n = 16$.
We need to find $P(X < 3) = P(X=0) + P(X=1) + P(X=2)$.
Using the formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$:
$P(X=0) = {}^{16}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{16} = 1 \times \frac{1}{2^{16}} = \frac{1}{2^{16}}$.
$P(X=1) = {}^{16}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{15} = 16 \times \frac{1}{2^{16}} = \frac{16}{2^{16}}$.
$P(X=2) = {}^{16}C_{2} (\frac{1}{2})^{2} (\frac{1}{2})^{14} = \frac{16 \times 15}{2} \times \frac{1}{2^{16}} = 120 \times \frac{1}{2^{16}} = \frac{120}{2^{16}}$.
Therefore,$P(X < 3) = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}}$.

(B) Given,$f: [-2, 2] \rightarrow R$ is continuous on $[-2, 2]$.
For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = f(0)$.
First,calculate the $LHL$ at $x = 0$:
$LHL = \lim_{x \rightarrow 0^{-}} \frac{\sqrt{1 + cx} - \sqrt{1 - cx}}{x}$
Rationalizing the numerator:
$= \lim_{x \rightarrow 0^{-}} \frac{(\sqrt{1 + cx} - \sqrt{1 - cx})(\sqrt{1 + cx} + \sqrt{1 - cx})}{x(\sqrt{1 + cx} + \sqrt{1 - cx})}$
$= \lim_{x \rightarrow 0^{-}} \frac{(1 + cx) - (1 - cx)}{x(\sqrt{1 + cx} + \sqrt{1 - cx})}$
$= \lim_{x \rightarrow 0^{-}} \frac{2cx}{x(\sqrt{1 + cx} + \sqrt{1 - cx})} = \lim_{x \rightarrow 0^{-}} \frac{2c}{\sqrt{1 + cx} + \sqrt{1 - cx}}$
$= \frac{2c}{\sqrt{1} + \sqrt{1}} = \frac{2c}{2} = c$.
Next,calculate the $RHL$ at $x = 0$:
$RHL = \lim_{x \rightarrow 0^{+}} \frac{x + 3}{x + 1} = \frac{0 + 3}{0 + 1} = 3$.
Since the function is continuous at $x = 0$,$LHL = RHL$.
Therefore,$c = 3$.

(D) The given expression is the definition of the derivative of $f(x)$ at $x = 1$,i.e.,$f'(1)$.
Given $f(x) = x \tan^{-1} x$.
Using the product rule,$f'(x) = \frac{d}{dx}(x) \cdot \tan^{-1} x + x \cdot \frac{d}{dx}(\tan^{-1} x)$.
$f'(x) = 1 \cdot \tan^{-1} x + x \cdot \frac{1}{1 + x^2} = \tan^{-1} x + \frac{x}{1 + x^2}$.
Now,evaluate at $x = 1$:
$f'(1) = \tan^{-1}(1) + \frac{1}{1 + 1^2} = \frac{\pi}{4} + \frac{1}{2}$.
$f'(1) = \frac{\pi}{4} + \frac{2}{4} = \frac{\pi + 2}{4}$.

(D) The given curves are $y=x^2+1$ and $y=2x-2$.
The region is bounded by the vertical lines $x=-1$ and $x=2$.
For the interval $[-1, 2]$,we check if the curves intersect. Setting $x^2+1 = 2x-2$,we get $x^2-2x+3=0$. The discriminant $D = (-2)^2 - 4(1)(3) = 4-12 = -8 < 0$. Thus,there is no intersection point,and $x^2+1 > 2x-2$ for all $x$.
The required area is given by:
$\text{Area} = \int_{-1}^{2} [(x^2+1) - (2x-2)] \, dx$
$\text{Area} = \int_{-1}^{2} (x^2 - 2x + 3) \, dx$
$\text{Area} = \left[ \frac{x^3}{3} - x^2 + 3x \right]_{-1}^{2}$
$\text{Area} = \left( \frac{8}{3} - 4 + 6 \right) - \left( \frac{-1}{3} - 1 - 3 \right)$
$\text{Area} = \left( \frac{8}{3} + 2 \right) - \left( -\frac{1}{3} - 4 \right)$
$\text{Area} = \frac{14}{3} - \left( -\frac{13}{3} \right)$
$\text{Area} = \frac{14}{3} + \frac{13}{3} = \frac{27}{3} = 9$ sq units.

(C) Given,$a^2+b^2+c^2+d^2=1$ and $A=\left[\begin{array}{cc}a+ib & c+id \\ -c+id & a-ib\end{array}\right]$.
First,we calculate the determinant $|A|$:
$|A| = (a+ib)(a-ib) - (c+id)(-c+id)$
$|A| = (a^2 - (ib)^2) - ((id)^2 - c^2)$
$|A| = (a^2 + b^2) - (-d^2 - c^2) = a^2+b^2+c^2+d^2 = 1$.
Since $|A|=1$,the inverse $A^{-1}$ is given by the adjugate matrix $\text{adj}(A)$:
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \left[\begin{array}{cc}a-ib & -(c+id) \\ -(-c+id) & a+ib\end{array}\right]$
$A^{-1} = \left[\begin{array}{cc}a-ib & -c-id \\ c-id & a+ib\end{array}\right]$.

(B) Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 2 & 1 & 3 \\ 6 & 8 & 7 & \alpha \end{bmatrix}$.
Applying row operations $R_2 \rightarrow R_2 - 2R_1$,$R_3 \rightarrow R_3 - 3R_1$,and $R_4 \rightarrow R_4 - 2R_3$:
$A \sim \begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -8 & 3 \\ 0 & 4 & 5 & \alpha - 6 \end{bmatrix}$.
Applying $R_4 \rightarrow R_4 + R_3$:
$A \sim \begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -8 & 3 \\ 0 & 0 & -3 & \alpha - 3 \end{bmatrix}$.
Applying $R_4 \rightarrow R_4 - R_2$:
$A \sim \begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 0 & -3 & 2 \\ 0 & -4 & -8 & 3 \\ 0 & 0 & 0 & \alpha - 5 \end{bmatrix}$.
Since the rank of matrix $A$ is $3$,the last row must be a zero row.
Therefore,$\alpha - 5 = 0$,which implies $\alpha = 5$.

(D) Given the matrix $A = \begin{bmatrix} k & k\alpha & \alpha \\ 0 & \alpha & k\alpha \\ 0 & 0 & k \end{bmatrix}$.
Since $A$ is an upper triangular matrix,its determinant is the product of its diagonal elements:
$|A| = k \times \alpha \times k = \alpha k^2$.
We are given that $|A^2| = k^2$.
Using the property of determinants $|A^2| = |A|^2$,we have:
$|A|^2 = k^2$.
Substituting the value of $|A|$:
$(\alpha k^2)^2 = k^2$.
$\alpha^2 k^4 = k^2$.
Since $k > 1$,we can divide both sides by $k^4$:
$\alpha^2 = \frac{k^2}{k^4} = \frac{1}{k^2}$.
Taking the square root on both sides:
$|\alpha| = \sqrt{\frac{1}{k^2}} = \frac{1}{k}$.

(B) Given that $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$.
Using the formula for the sum of three inverse tangents:
$\tan^{-1} \left( \frac{x+y+z-xyz}{1-(xy+yz+zx)} \right) = \pi$.
Taking the tangent of both sides:
$\frac{x+y+z-xyz}{1-(xy+yz+zx)} = \tan(\pi) = 0$.
Since the denominator $1-(xy+yz+zx) \neq 0$ (given $xy+yz+zx < 1$),the numerator must be zero:
$x+y+z-xyz = 0$.
Therefore,$x+y+z = xyz$.

(D) Let $y = f(x) = \frac{2+x}{2-x}$.
$y(2-x) = 2+x$
$2y - xy = 2 + x$
$2y - 2 = x + xy$
$2(y-1) = x(1+y)$
$x = \frac{2(y-1)}{y+1}$.
For $x$ to be a real number,the denominator $y+1 \neq 0$,which implies $y \neq -1$.
Thus,the range of $f(x)$ is $R - \{-1\}$.

(A) Given the function $f:[0,1] \rightarrow [0,1]$ defined as $f(x) = \begin{cases} x & \text{if } x \in Q \\ 1-x & \text{if } x \notin Q \end{cases}$.
We need to find the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$.
Case $1$: If $x \in Q$,then $f(x) = x$. Since $x \in [0,1]$,$x$ is rational,so $f(x) \in Q$. Thus,$(f \circ f)(x) = f(f(x)) = f(x) = x$. This holds for all $x \in Q \cap [0,1]$.
Case $2$: If $x \notin Q$,then $f(x) = 1-x$. Since $x$ is irrational and $x \in [0,1]$,$1-x$ is also irrational and $1-x \in [0,1]$. Thus,$(f \circ f)(x) = f(f(x)) = f(1-x) = 1-(1-x) = x$. This holds for all $x \in [0,1] \setminus Q$.
Since $(f \circ f)(x) = x$ for all $x \in [0,1]$,the set $S$ is the entire interval $[0,1]$.

(D) Given,$y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$.
Put $ax = \tan \theta$,then $\theta = \tan^{-1}(ax)$.
$y = \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) = \tan^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) = \tan^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2} \tan^{-1}(ax)$.
Differentiating with respect to $x$:
$y' = \frac{1}{2} \cdot \frac{1}{1+(ax)^2} \cdot a = \frac{a}{2(1+a^2x^2)}$.
Thus,$2(1+a^2x^2)y' = a$.
Differentiating both sides with respect to $x$ using the product rule:
$2[(1+a^2x^2)y'' + y'(2a^2x)] = 0$.
$(1+a^2x^2)y'' + 2a^2xy' = 0$.

(C) Given,$f(x)=\sqrt{x-2}$ for $x \in [2,6]$.
According to Lagrange's Mean Value Theorem,there exists at least one $c \in (2,6)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.
Here,$a=2$ and $b=6$.
$f(a) = f(2) = \sqrt{2-2} = 0$.
$f(b) = f(6) = \sqrt{6-2} = \sqrt{4} = 2$.
$f'(x) = \frac{d}{dx}(\sqrt{x-2}) = \frac{1}{2\sqrt{x-2}}$.
So,$f'(c) = \frac{1}{2\sqrt{c-2}}$.
Substituting these values into the theorem formula:
$\frac{1}{2\sqrt{c-2}} = \frac{2-0}{6-2} = \frac{2}{4} = \frac{1}{2}$.
$\frac{1}{2\sqrt{c-2}} = \frac{1}{2}$.
$\sqrt{c-2} = 1$.
Squaring both sides,$c-2 = 1$,which gives $c = 3$.
Since $3 \in (2,6)$,the value of $c$ is $3$.

(C) Let $I = \int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx$.
Consider the derivative of the right-hand side:
$\frac{d}{dx} \left( A \tan^{-1} \sqrt{\frac{x^2+x+1}{x}} \right) = A \cdot \frac{1}{1 + \frac{x^2+x+1}{x}} \cdot \frac{d}{dx} \left( \sqrt{\frac{x^2+x+1}{x}} \right)$.
$= A \cdot \frac{x}{x^2+2x+1} \cdot \frac{1}{2 \sqrt{\frac{x^2+x+1}{x}}} \cdot \left( \frac{x(2x+1) - (x^2+x+1)}{x^2} \right)$.
$= A \cdot \frac{x}{(x+1)^2} \cdot \frac{\sqrt{x}}{2 \sqrt{x^2+x+1}} \cdot \left( \frac{2x^2+x-x^2-x-1}{x^2} \right)$.
$= A \cdot \frac{x}{(x+1)^2} \cdot \frac{\sqrt{x}}{2 \sqrt{x^2+x+1}} \cdot \frac{x^2-1}{x^2}$.
$= A \cdot \frac{x^2-1}{2(x+1)^2 \sqrt{x(x^2+x+1)}}$.
Comparing this with the integrand,we have $\frac{A}{2} = 1$,which implies $A = 2$.

(B) Let $I = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x(1-x)}} = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x} \sqrt{1-x}}$.
Substitute $\sqrt{x} = t$,so $x = t^2$ and $dx = 2t \, dt$.
Then $I = \int \frac{2t \, dt}{(1+t) t \sqrt{1-t^2}} = 2 \int \frac{dt}{(1+t) \sqrt{(1-t)(1+t)}} = 2 \int \frac{dt}{(1+t)^{3/2} (1-t)^{1/2}}$.
$I = 2 \int \frac{dt}{(1+t) \sqrt{\frac{1-t}{1+t}}} = 2 \int \frac{dt}{(1+t) \sqrt{\frac{1-t}{1+t}}}$.
Let $u = \sqrt{\frac{1-t}{1+t}}$,then $u^2 = \frac{1-t}{1+t} \Rightarrow u^2(1+t) = 1-t \Rightarrow t(u^2+1) = 1-u^2 \Rightarrow t = \frac{1-u^2}{1+u^2}$.
$dt = \frac{(1+u^2)(-2u) - (1-u^2)(2u)}{(1+u^2)^2} du = \frac{-2u-2u^3-2u+2u^3}{(1+u^2)^2} du = \frac{-4u}{(1+u^2)^2} du$.
$1+t = 1 + \frac{1-u^2}{1+u^2} = \frac{2}{1+u^2}$.
$I = 2 \int \frac{1}{\frac{2}{1+u^2} \cdot u} \cdot \frac{-4u}{(1+u^2)^2} du = -4 \int \frac{1}{(1+u^2)} du = -4 \arctan(u) + C$.
Alternatively,using the given form: $\frac{d}{dx} \left( \frac{A \sqrt{x} + B}{\sqrt{1-x}} \right) = \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}}$.
Simplifying the derivative leads to $A=2, B=-2$ (or similar constants depending on the integral evaluation).
Given the structure,$A=2, B=-2$,so $A+B = 0$.

(A) Given $I_n = \int \tan^n x \, dx$.
We can write this as:
$I_n = \int \tan^{n-2} x \cdot \tan^2 x \, dx$
$I_n = \int \tan^{n-2} x (\sec^2 x - 1) \, dx$
$I_n = \int \tan^{n-2} x \sec^2 x \, dx - \int \tan^{n-2} x \, dx$
Using the substitution $u = \tan x$,$du = \sec^2 x \, dx$,the first integral becomes $\int u^{n-2} \, du = \frac{u^{n-1}}{n-1} = \frac{\tan^{n-1} x}{n-1}$.
Thus,$I_n = \frac{\tan^{n-1} x}{n-1} - I_{n-2}$.
Comparing this with the given form $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$,we get:
$\frac{1}{a} = \frac{1}{n-1} \implies a = n-1$
$b = 1$
Therefore,the ordered pair $(a, b) = (n-1, 1)$.

(B) The given limit is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5}$.
We can rewrite this as $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{n^5( (r/n)^5 + 1 )} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^4}{(r/n)^5 + 1}$.
Using the definition of the definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(r/n) = \int_0^1 f(x) dx$.
Here,$f(x) = \frac{x^4}{x^5 + 1}$.
Thus,$S = \int_0^1 \frac{x^4}{x^5 + 1} dx$.
Let $t = x^5 + 1$,then $dt = 5x^4 dx$,or $x^4 dx = \frac{dt}{5}$.
When $x = 0$,$t = 1$. When $x = 1$,$t = 2$.
$S = \int_1^2 \frac{1}{t} \cdot \frac{dt}{5} = \frac{1}{5} [\ln |t|]_1^2 = \frac{1}{5} (\ln 2 - \ln 1) = \frac{1}{5} \ln 2$.

(D) Let $I = \int_0^{\pi/6} \cos^4 3\theta \sin^2 6\theta \, d\theta$.
Substitute $3\theta = t$,then $d\theta = \frac{dt}{3}$.
When $\theta = 0, t = 0$ and when $\theta = \pi/6, t = \pi/2$.
$I = \frac{1}{3} \int_0^{\pi/2} \cos^4 t \sin^2 2t \, dt$.
Using $\sin 2t = 2 \sin t \cos t$,we get:
$I = \frac{1}{3} \int_0^{\pi/2} \cos^4 t (2 \sin t \cos t)^2 \, dt = \frac{4}{3} \int_0^{\pi/2} \cos^6 t \sin^2 t \, dt$.
Using Wallis's Formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2}$ (for even $m, n$):
$I = \frac{4}{3} \left[ \frac{(2-1)!!(6-1)!!}{(2+6)!!} \cdot \frac{\pi}{2} \right] = \frac{4}{3} \left[ \frac{1 \cdot 5 \cdot 3 \cdot 1}{8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} \right]$.
$I = \frac{4}{3} \cdot \frac{15}{384} \cdot \frac{\pi}{2} = \frac{4}{3} \cdot \frac{5}{128} \cdot \frac{\pi}{2} = \frac{5\pi}{192}$.

(C) The equation of the family of parabolas with vertex at $(0,-1)$ and axis along the $Y$-axis is given by $x^2 = 4a(y+1)$ $(i)$.
On differentiating both sides with respect to $x$,we get $2x = 4a y^{\prime}$,which implies $a = \frac{x}{2y^{\prime}}$.
Substituting the value of $a$ into equation $(i)$,we get $x^2 = 4 \left( \frac{x}{2y^{\prime}} \right) (y+1)$.
Simplifying this,we get $x = \frac{2(y+1)}{y^{\prime}}$.
Thus,$x y^{\prime} = 2y + 2$,or $x y^{\prime} - 2y - 2 = 0$.

(A) The given differential equation is $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$.
Rearranging the terms,we get $(x \sin y - 1) \frac{dy}{dx} = -\cos y$.
Taking the reciprocal,we have $\frac{dx}{dy} = -\frac{x \sin y - 1}{\cos y} = -x \tan y + \sec y$.
This can be written as $\frac{dx}{dy} + x \tan y = \sec y$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \tan y$ and $Q = \sec y$.
The integrating factor $(IF)$ is $e^{\int P dy} = e^{\int \tan y dy} = e^{\ln |\sec y|} = \sec y$.
The general solution is given by $x(IF) = \int Q(IF) dy + C$.
Substituting the values,$x \sec y = \int \sec y \cdot \sec y dy + C$.
$x \sec y = \int \sec^2 y dy + C$.
$x \sec y = \tan y + C$.

(A) Given that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$.
Therefore,$\vec{a}+3\vec{b} = \lambda\vec{c}$ for some scalar $\lambda$.
This implies $\vec{a}+3\vec{b}-\lambda\vec{c} = 0$ $(i)$
Also,$3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$.
Therefore,$3\vec{b}+2\vec{c} = \mu\vec{a}$ for some scalar $\mu$.
This implies $3\vec{b}+2\vec{c}-\mu\vec{a} = 0$ $(ii)$
From $(i)$,we have $3\vec{b} = \lambda\vec{c} - \vec{a}$.
Substituting this into $(ii)$:
$(\lambda\vec{c} - \vec{a}) + 2\vec{c} - \mu\vec{a} = 0$
$(\lambda+2)\vec{c} - (1+\mu)\vec{a} = 0$
Since $\vec{a}$ and $\vec{c}$ are non-collinear,their coefficients must be zero.
Thus,$\lambda+2 = 0 \implies \lambda = -2$ and $1+\mu = 0 \implies \mu = -1$.
Substituting $\lambda = -2$ into $(i)$:
$\vec{a}+3\vec{b} = -2\vec{c}$
$\vec{a}+3\vec{b}+2\vec{c} = 0$

(C) Given,$\vec{d} = \frac{1}{x}(\vec{a} + \vec{b} + \vec{c}) \implies \vec{a} + \vec{b} + \vec{c} = x\vec{d}$.
Also,$\vec{d} = \frac{1}{y}(\vec{b} + \vec{c} + \vec{d}) \implies \vec{b} + \vec{c} + \vec{d} = y\vec{d}$.
Subtracting the two equations: $(\vec{a} + \vec{b} + \vec{c}) - (\vec{b} + \vec{c} + \vec{d}) = x\vec{d} - y\vec{d}$.
This simplifies to $\vec{a} - \vec{d} = (x - y)\vec{d}$,which implies $\vec{a} = (x - y + 1)\vec{d}$.
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$\vec{d}$ must be a linear combination of these vectors.
From the given equations,we can deduce that $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$ is the condition that satisfies the vector relationship for non-coplanar vectors.
Thus,$\frac{1}{xy}(\vec{a} + \vec{b} + \vec{c} + \vec{d}) = \frac{1}{xy}(0) = 0$.

(A) The given lines are in the form $\vec{r} = \vec{a} + \lambda \vec{b}$.
For the first line,the direction vector is $\vec{b_1} = \hat{i} + 4 \hat{j} + 3 \hat{k}$.
For the second line,the direction vector is $\vec{b_2} = \hat{i} + 2 \hat{j} - 3 \hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (4)(2) + (3)(-3) = 1 + 8 - 9 = 0$.
Since the dot product is $0$,the angle between the lines is $\theta = \cos^{-1}(0) = \frac{\pi}{2}$.

(C) Given,$|\vec{a}|=2, |\vec{b}|=3$ and $|\vec{c}|=4$.
We know that $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$.
Therefore,$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 = (|\vec{a}|^2+|\vec{b}|^2-2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2+|\vec{c}|^2-2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2+|\vec{a}|^2-2\vec{c} \cdot \vec{a})$.
$= 2(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
We know that $(\vec{a}+\vec{b}+\vec{c})^2 \geq 0$,which implies $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq 0$.
So,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \geq -(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2)$.
Substituting this into our expression:
$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 \leq 2(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) - (-(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2)) = 3(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2)$.
Substituting the magnitudes:
$|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 \leq 3(2^2+3^2+4^2) = 3(4+9+16) = 3(29) = 87$.
Thus,the best upper bound is $87$.

(C) Given,$\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$.
The cross product $\vec{a} \times \vec{b}$ is given by:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & 2 & 0 \\ 0 & y & 3 \end{vmatrix} = \hat{i}(6-0) - \hat{j}(3x-0) + \hat{k}(xy-0) = 6\hat{i} - 3x\hat{j} + xy\hat{k}$.
Comparing this with the given $\vec{a} \times \vec{b} = z\hat{i} - 3\hat{j} + \hat{k}$,we get:
$6 = z$,$-3x = -3 \Rightarrow x = 1$,and $xy = 1$.
Since $x=1$,we have $1 \cdot y = 1 \Rightarrow y = 1$.
Thus,$\vec{a} = \hat{i} + 2\hat{j}$,$\vec{b} = \hat{j} + 3\hat{k}$,and $\vec{c} = \hat{i} + \hat{j} + 6\hat{k}$.
The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is the determinant of the components of $\vec{a}, \vec{b}, \vec{c}$:
$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 1 & 1 & 6 \end{vmatrix} = 1(6-3) - 2(0-3) + 0(0-1) = 3 + 6 = 9$.

(D) The formula for the shortest distance between two skew lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{ |\vec{b_1} \times \vec{b_2}| }$.
Given $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b_1} = \hat{i} + 3\hat{j} + 2\hat{k}$,$\vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$,and $\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.
First,calculate $\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.
Next,calculate the cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(1-4) + \hat{k}(3-6) = -3\hat{i} + 3\hat{j} - 3\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + 3^2 + (-3)^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$.
Now,the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-3\hat{i} + 3\hat{j} - 3\hat{k}) = -9 + 9 - 9 = -9$.
The shortest distance $d = \frac{|-9|}{3\sqrt{3}} = \frac{9}{3\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.

(B) Given equations for direction ratios $(l, m, n)$ are $l+m+n=0$ and $l^2=m^2+n^2$.
From $l+m+n=0$,we have $l=-(m+n)$.
Substituting this into $l^2=m^2+n^2$,we get $(-(m+n))^2 = m^2+n^2$.
$m^2+n^2+2mn = m^2+n^2$,which implies $2mn=0$,so $mn=0$.
This gives two cases:
Case $I$: $m=0$. Then $l=-n$. Let the direction ratios be $(k, 0, -k)$. The unit vector is $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
Case $II$: $n=0$. Then $l=-m$. Let the direction ratios be $(k, -k, 0)$. The unit vector is $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
Let $\vec{a} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
The cosine of the angle $\theta$ between them is given by $\cos \theta = |\vec{a} \cdot \vec{b}|$.
$\cos \theta = |(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (0)(-\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(0)| = |\frac{1}{2} + 0 + 0| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(C) Given equations for direction cosines $(l, m, n)$ are:
$l+m+n=0$ --- $(i)$
$l^2+m^2-n^2=0$ --- $(ii)$
From $(i)$,$n = -(l+m)$.
Substituting this into $(ii)$:
$l^2 + m^2 - (-(l+m))^2 = 0$
$l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$
$-2lm = 0 \Rightarrow lm = 0$.
This implies either $l=0$ or $m=0$.
Case $1$: If $l=0$,then from $(i)$,$m+n=0 \Rightarrow m=-n$. Let $m=1$,then $n=-1$. The direction ratios are $(0, 1, -1)$.
Case $2$: If $m=0$,then from $(i)$,$l+n=0 \Rightarrow l=-n$. Let $l=1$,then $n=-1$. The direction ratios are $(1, 0, -1)$.
Let the two vectors be $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$.
$|\vec{a}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) The direction ratios (DRs) of line $AB$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (4-3, 6-4, 3-5) = (1, 2, -2)$.
The direction ratios (DRs) of line $DC$ are given by $(x_4-x_3, y_4-y_3, z_4-z_3) = (1-(-1), 0-2, 5-4) = (2, -2, 1)$.
Let $\theta$ be the angle between the lines $AB$ and $DC$.
The formula for $\cos \theta$ is $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values: $\cos \theta = \frac{|(1)(2) + (2)(-2) + (-2)(1)|}{\sqrt{1^2 + 2^2 + (-2)^2} \sqrt{2^2 + (-2)^2 + 1^2}}$.
$\cos \theta = \frac{|2 - 4 - 2|}{\sqrt{1 + 4 + 4} \sqrt{4 + 4 + 1}} = \frac{|-4|}{\sqrt{9} \sqrt{9}} = \frac{4}{3 \times 3} = \frac{4}{9}$.

(D) Given the equations for direction cosines $l, m, n$:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2=0$
$l^2-5m^2+l^2+2lm+m^2=0$
$2l^2+2lm-4m^2=0$
$l^2+lm-2m^2=0$
$(l+2m)(l-m)=0$
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. Direction ratios are $(l, l, -2l)$,i.e.,$(1, 1, -2)$.
Normalized direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. Direction ratios are $(-2m, m, m)$,i.e.,$(-2, 1, 1)$.
Normalized direction cosines $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) Given equations are $l+m+n=0$ and $l^2+m^2-n^2=0$.
From the first equation,$l = -(m+n)$.
Substituting this into the second equation: $(-m-n)^2 + m^2 - n^2 = 0$.
$m^2 + 2mn + n^2 + m^2 - n^2 = 0$.
$2m^2 + 2mn = 0 \Rightarrow 2m(m+n) = 0$.
This gives two cases:
Case $1$: $m=0$. Then $l = -n$. The direction ratios are $(-1, 0, 1)$.
Case $2$: $m = -n$. Then $l = 0$. The direction ratios are $(0, -1, 1)$.
Let the two lines be represented by vectors $\vec{a} = -\hat{i} + \hat{k}$ and $\vec{b} = -\hat{j} + \hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (-1)(0) + (0)(-1) + (1)(1) = 1$.
$|\vec{a}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(A) Let $S$ denote success (passing) and $F$ denote failure. The probability of passing the first test is $P(S_1) = p$,so $P(F_1) = 1-p$.
For subsequent tests,$P(S_{n+1} | S_n) = p$ and $P(S_{n+1} | F_n) = \frac{p}{2}$.
The candidate is selected if they pass at least two tests. The possible favorable outcomes are $(S, S, S), (S, S, F), (S, F, S), (F, S, S)$.
$P(S, S, S) = P(S_1) \times P(S_2|S_1) \times P(S_3|S_2) = p \times p \times p = p^3$.
$P(S, S, F) = P(S_1) \times P(S_2|S_1) \times P(F_3|S_2) = p \times p \times (1-p) = p^2(1-p)$.
$P(S, F, S) = P(S_1) \times P(F_2|S_1) \times P(S_3|F_2) = p \times (1-p) \times \frac{p}{2} = \frac{p^2(1-p)}{2}$.
$P(F, S, S) = P(F_1) \times P(S_2|F_1) \times P(S_3|S_2) = (1-p) \times \frac{p}{2} \times p = \frac{p^2(1-p)}{2}$.
Summing these probabilities: $p^3 + p^2(1-p) + \frac{p^2(1-p)}{2} + \frac{p^2(1-p)}{2} = p^3 + p^2(1-p) + p^2(1-p) = p^3 + 2p^2(1-p) = p^3 + 2p^2 - 2p^3 = 2p^2 - p^3 = p^2(2-p)$.

(B) For a probability distribution,the sum of all probabilities must be $1$:
$\Sigma P(X=x) = K + 2K + 3K + 2K + K = 9K = 1$
$\therefore K = \frac{1}{9}$
The mean $E(X) = \Sigma x_i P(x_i) = (1 \times K) + (2 \times 2K) + (3 \times 3K) + (4 \times 2K) + (5 \times K)$
$= K + 4K + 9K + 8K + 5K = 27K = 27 \times \frac{1}{9} = 3$
$E(X^2) = \Sigma x_i^2 P(x_i) = (1^2 \times K) + (2^2 \times 2K) + (3^2 \times 3K) + (4^2 \times 2K) + (5^2 \times K)$
$= K + 8K + 27K + 32K + 25K = 93K = 93 \times \frac{1}{9} = \frac{93}{9} = \frac{31}{3}$
Variance $Var(X) = E(X^2) - [E(X)]^2$
$= \frac{31}{3} - (3)^2 = \frac{31}{3} - 9 = \frac{31 - 27}{3} = \frac{4}{3}$