TS EAMCET 2017 Mathematics Question Paper with Answer and Solution

(C) Given,$\alpha$ and $\beta$ are the roots of $x^2+2x+c=0$.
Sum of roots,$\alpha+\beta = -\frac{2}{1} = -2$.
Product of roots,$\alpha\beta = \frac{c}{1} = c$.
We know that $\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta)$.
This can be written as $\alpha^3+\beta^3 = (\alpha+\beta)[(\alpha+\beta)^2-3\alpha\beta]$.
Given $\alpha^3+\beta^3 = 4$,substituting the values:
$(-2)[(-2)^2 - 3c] = 4$.
$(-2)[4-3c] = 4$.
$4-3c = -2$.
$-3c = -6$.
$c = 2$.

(D) Let $f(x) = ax^2 + bx + c$.
Given $f(0) + f(1) = 0$,we have $c + (a + b + c) = 0$,which implies $a + b + 2c = 0$ $(i)$.
Given $f(-2) = 0$,we have $4a - 2b + c = 0$ (ii).
From $(i)$,$b = -a - 2c$. Substituting into (ii):
$4a - 2(-a - 2c) + c = 0$ $\Rightarrow 4a + 2a + 4c + c = 0$ $\Rightarrow 6a + 5c = 0$.
Let $a = 5k$,then $c = -6k$.
Substituting into $(i)$: $5k + b + 2(-6k) = 0$ $\Rightarrow 5k + b - 12k = 0$ $\Rightarrow b = 7k$.
Thus,$f(x) = k(5x^2 + 7x - 6) = k(5x^2 + 10x - 3x - 6) = k(5x(x + 2) - 3(x + 2)) = k(5x - 3)(x + 2)$.
The roots are $x = -2$ and $x = \frac{3}{5}$.
Therefore,$f\left(\frac{3}{5}\right) = 0$.

(B) Given equation: $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$
Rearranging the terms: $(x^2+\frac{1}{x^2}) - 8(x+\frac{1}{x}) + 9 = 0$
Let $t = x+\frac{1}{x}$. Then $t^2 = x^2+2+\frac{1}{x^2}$,so $x^2+\frac{1}{x^2} = t^2-2$.
Substituting into the equation: $(t^2-2) - 8t + 9 = 0$
$t^2 - 8t + 7 = 0$
$(t-7)(t-1) = 0$
Case $1$: $t = 7$ $\Rightarrow x+\frac{1}{x} = 7$ $\Rightarrow x^2-7x+1 = 0$. The discriminant $D = (-7)^2 - 4(1)(1) = 45 > 0$,so there are two real roots. The product of these roots is $c/a = 1$.
Case $2$: $t = 1$ $\Rightarrow x+\frac{1}{x} = 1$ $\Rightarrow x^2-x+1 = 0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 < 0$,so there are no real roots.
Thus,the product of all real roots is $1$.

(B) Given that $\alpha$ and $\beta$ are the roots of $a x^2+b x+c=0$.
$\alpha+\beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.
Let the roots of $p x^2+q x+r=0$ be $\gamma = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $\delta = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$.
Then $\alpha = \frac{1}{1+\gamma}$ and $\beta = \frac{1}{1+\delta}$.
Since $\alpha$ is a root of $a x^2+b x+c=0$,we have $a(\frac{1}{1+x})^2 + b(\frac{1}{1+x}) + c = 0$.
$a + b(1+x) + c(1+x)^2 = 0$.
$a + b + bx + c(1 + 2x + x^2) = 0$.
$c x^2 + (b+2c)x + (a+b+c) = 0$.
Comparing this with $p x^2+q x+r=0$,we get $r = a+b+c$.

(B) Given that the conjugate of $(x+iy)(1-2i)$ is $(1+i)$.
Let $z = (x+iy)(1-2i)$.
Then $\bar{z} = 1+i$.
Since $\bar{z} = (x-iy)(1+2i)$,we have $(x-iy)(1+2i) = 1+i$.
Therefore,$x-iy = \frac{1+i}{1+2i}$.
Taking the conjugate of both sides,we get $\overline{x-iy} = \overline{\left(\frac{1+i}{1+2i}\right)}$.
This simplifies to $x+iy = \frac{1-i}{1-2i}$.

(C) Consider the expression: $\left[\frac{\left(1+\cos \frac{\pi}{12}\right)+i \sin \frac{\pi}{12}}{\left(1+\cos \frac{\pi}{12}\right)-i \sin \frac{\pi}{12}}\right]^{72}$
Using the half-angle identities $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,we get:
$\left[\frac{2 \cos^2 \frac{\pi}{24} + i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}{2 \cos^2 \frac{\pi}{24} - i 2 \sin \frac{\pi}{24} \cos \frac{\pi}{24}}\right]^{72}$
Factoring out $2 \cos \frac{\pi}{24}$ from the numerator and denominator:
$\left[\frac{2 \cos \frac{\pi}{24} (\cos \frac{\pi}{24} + i \sin \frac{\pi}{24})}{2 \cos \frac{\pi}{24} (\cos \frac{\pi}{24} - i \sin \frac{\pi}{24})}\right]^{72} = \left(\frac{\cos \frac{\pi}{24} + i \sin \frac{\pi}{24}}{\cos \frac{\pi}{24} - i \sin \frac{\pi}{24}}\right)^{72}$
Using the property $\frac{e^{i\theta}}{e^{-i\theta}} = e^{i2\theta}$,we have:
$\left(e^{i \frac{\pi}{24}} / e^{-i \frac{\pi}{24}}\right)^{72} = (e^{i \frac{2\pi}{24}})^{72} = (e^{i \frac{\pi}{12}})^{72}$
$= e^{i \frac{72\pi}{12}} = e^{i 6\pi}$
By Euler's formula,$e^{i 6\pi} = \cos(6\pi) + i \sin(6\pi) = 1 + i(0) = 1$.

(A) Given equation: $(x-1)^3+64=0$
$\Rightarrow (x-1)^3 = -64$
$\Rightarrow (x-1)^3 = (-4)^3$
Let $y = x-1$,then $y^3 = (-4)^3$. The roots are $y = -4, -4\omega, -4\omega^2$,where $\omega$ is the complex cube root of unity.
Thus,$x-1 = -4, -4\omega, -4\omega^2$.
The roots are $x_1 = -3$,$x_2 = 1-4\omega$,and $x_3 = 1-4\omega^2$.
The complex roots are $x_2 = 1-4\omega$ and $x_3 = 1-4\omega^2$.
Sum of complex roots $= (1-4\omega) + (1-4\omega^2) = 2 - 4(\omega + \omega^2)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Sum $= 2 - 4(-1) = 2 + 4 = 6$.

(C) Let $z = x + iy$.
Substituting $z$ into the expression:
$\frac{2z+1}{iz+1} = \frac{2(x+iy)+1}{i(x+iy)+1} = \frac{(2x+1) + i(2y)}{(1-y) + ix}$.
To rationalize,multiply the numerator and denominator by the conjugate of the denominator,$(1-y) - ix$:
$\frac{[(2x+1) + i(2y)][(1-y) - ix]}{(1-y)^2 + x^2} = \frac{(2x+1)(1-y) + 2xy + i[2y(1-y) - x(2x+1)]}{(1-y)^2 + x^2}$.
The imaginary part is given as $-2$:
$\frac{2y - 2y^2 - 2x^2 - x}{(1-y)^2 + x^2} = -2$.
$2y - 2y^2 - 2x^2 - x = -2(1 - 2y + y^2 + x^2)$.
$2y - 2y^2 - 2x^2 - x = -2 + 4y - 2y^2 - 2x^2$.
Simplifying the equation:
$-x - 2y = -2$,or $x + 2y - 2 = 0$.
This is the equation of a straight line.

(C) The word $TRICK$ contains $5$ distinct letters: $T, R, I, C, K$.
We need to form a $5$-letter word such that $C$ is fixed in the middle position.
This leaves $4$ positions to be filled by the remaining $4$ letters $(T, R, I, K)$.
The number of ways to arrange $4$ distinct letters in $4$ positions is given by $4!$.
$4! = 4 \times 3 \times 2 \times 1 = 24$.
Therefore,there are $24$ possible ways.

(A) The total number of ways to choose the team is calculated by first selecting $5$ players from $10$ and then selecting $1$ captain from the remaining $5$ players.
This is given by the expression:
$({ }^{10} C_5) \times ({ }^5 C_1)$
$= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times 5$
$= 252 \times 5 = 1260$.

(D) The set is given by $S = \{2k \mid -9 \leq k \leq 10\}$.
Since $k$ ranges from $-9$ to $10$,the number of values for $k$ is $10 - (-9) + 1 = 20$.
Thus,the total number of elements in set $S$ is $20$.
The elements are $\{-18, -16, -14, -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$.
An integer is divisible by both $4$ and $6$ if it is divisible by $\text{lcm}(4, 6) = 12$.
We look for elements in $S$ that are multiples of $12$.
These are $\{-12, 0, 12\}$.
There are $3$ such favourable outcomes.
Therefore,the required probability is $\frac{3}{20}$.

(B) Consider the sum: $\sum_{k=1}^n k(k+2) = \sum_{k=1}^n (k^2 + 2k)$.
Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$:
$\sum_{k=1}^n k^2 + 2 \sum_{k=1}^n k = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.
Factor out $\frac{n(n+1)}{6}$:
$= \frac{n(n+1)}{6} [ (2n+1) + 6 ]$.
$= \frac{n(n+1)(2n+7)}{6}$.

(B) The general term in the expansion of $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$.
For the expansion of $(1+x)^{42}$,the coefficients are:
Coefficient of $(2r+1)^{\text{th}}$ term is ${^{42}C_{2r}}$.
Coefficient of $(r+1)^{\text{th}}$ term is ${^{42}C_r}$.
Given that these coefficients are equal:
${^{42}C_{2r}} = {^{42}C_r}$.
Using the property ${^nC_x} = {^nC_y}$,we have two cases:
Case $1$: $x = y$ $\Rightarrow 2r = r$ $\Rightarrow r = 0$.
Case $2$: $x + y = n$ $\Rightarrow 2r + r = 42$ $\Rightarrow 3r = 42$ $\Rightarrow r = 14$.
Since $r$ must be a positive integer in this context,$r = 14$ is the valid solution.

(B) The $p^{th}$ term in the expansion of $(1+x)^n$ is $T_p = { }^n C_{p-1} x^{p-1}$,so its coefficient is $p = { }^n C_{p-1}$.
The $(p+1)^{th}$ term is $T_{p+1} = { }^n C_p x^p$,so its coefficient is $q = { }^n C_p$.
We know the property of binomial coefficients: $\frac{{ }^n C_r}{{ }^n C_{r-1}} = \frac{n-r+1}{r}$.
Taking the ratio $\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
This implies $p \cdot q = p \cdot (n-p+1)$ is not the correct approach; rather,we use the given values directly.
Actually,the problem states the coefficients are $p$ and $q$. Let's re-evaluate: $p = { }^n C_{p-1}$ and $q = { }^n C_p$.
Using the identity ${ }^n C_{p-1} + { }^n C_p = { }^{n+1} C_p$,this does not directly yield $n+1$ unless specific conditions are met.
However,based on the standard problem type: $\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
Thus,$q = \frac{n-p+1}{p} \cdot p = n-p+1$.
Therefore,$p+q = p + (n-p+1) = n+1$.

(D) We are given that $\operatorname{cosec} \theta - \cot \theta = 2017$ $(i)$.
Using the identity $\operatorname{cosec}^2 \theta - \cot^2 \theta = 1$,we can write $(\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta) = 1$.
Therefore,$\operatorname{cosec} \theta + \cot \theta = \frac{1}{2017}$ (ii).
Adding $(i)$ and (ii):
$2 \operatorname{cosec} \theta = 2017 + \frac{1}{2017} > 0$,which implies $\operatorname{cosec} \theta > 0$.
Since $\operatorname{cosec} \theta > 0$,$\theta$ must lie in the $I$ or $II$ quadrant.
Subtracting $(i)$ from (ii):
$2 \cot \theta = \frac{1}{2017} - 2017 < 0$,which implies $\cot \theta < 0$.
Since $\cot \theta < 0$,$\theta$ must lie in the $II$ or $IV$ quadrant.
Since $\theta$ must satisfy both conditions ($\operatorname{cosec} \theta > 0$ and $\cot \theta < 0$),$\theta$ lies in the $II$ quadrant.

(D) Given,$\tan 20^{\circ}=\lambda$.
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,the expression becomes:
$\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}} = \tan(160^{\circ}-110^{\circ}) = \tan 50^{\circ}$.
However,we can also simplify the terms individually:
$\tan 160^{\circ} = \tan(180^{\circ}-20^{\circ}) = -\tan 20^{\circ} = -\lambda$.
$\tan 110^{\circ} = \tan(90^{\circ}+20^{\circ}) = -\cot 20^{\circ} = -\frac{1}{\lambda}$.
Substituting these into the expression:
$\frac{-\lambda - (-1/\lambda)}{1 + (-\lambda)(-1/\lambda)} = \frac{-\lambda + 1/\lambda}{1 + 1} = \frac{\frac{1-\lambda^2}{\lambda}}{2} = \frac{1-\lambda^2}{2\lambda}$.

(B) Given that $\tan \theta_1 = k \cot \theta_2$.
Since $\cot \theta_2 = \frac{1}{\tan \theta_2}$,we have $\tan \theta_1 = \frac{k}{\tan \theta_2}$,which implies $\tan \theta_1 \tan \theta_2 = k$.
Now,consider the expression $\frac{\cos (\theta_1 + \theta_2)}{\cos (\theta_1 - \theta_2)}$.
Using the expansion formulas,we get $\frac{\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}$.
Dividing the numerator and denominator by $\cos \theta_1 \cos \theta_2$,we obtain $\frac{1 - \tan \theta_1 \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2}$.
Substituting $\tan \theta_1 \tan \theta_2 = k$,the expression becomes $\frac{1-k}{1+k}$.

(D) We have,$\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$.
Using the logarithmic property $n \log a = \log a^n$,we get:
$\cosh ^{-1} x = \log _e(\sqrt{2}+1)^2$.
We know that $\cosh ^{-1} x = \log _e(x + \sqrt{x^2-1})$.
Therefore,$\log _e(x + \sqrt{x^2-1}) = \log _e(2 + 1 + 2\sqrt{2}) = \log _e(3 + 2\sqrt{2})$.
Comparing both sides,we have $x + \sqrt{x^2-1} = 3 + 2\sqrt{2}$.
Let $x = 3$,then $\sqrt{x^2-1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}$.
Thus,$3 + 2\sqrt{2} = 3 + 2\sqrt{2}$,which satisfies the equation.
Hence,$x = 3$.

(B) Given the equation $\cos 2 \theta = \sin \theta$.
Using the identity $\cos 2 \theta = 1 - 2 \sin^2 \theta$,we have:
$1 - 2 \sin^2 \theta = \sin \theta$
$2 \sin^2 \theta + \sin \theta - 1 = 0$
Factoring the quadratic equation:
$2 \sin^2 \theta + 2 \sin \theta - \sin \theta - 1 = 0$
$2 \sin \theta (\sin \theta + 1) - 1 (\sin \theta + 1) = 0$
$(\sin \theta + 1)(2 \sin \theta - 1) = 0$
This gives two cases:
$1) \sin \theta = -1 \Rightarrow \theta = \frac{3 \pi}{2}$
$2) \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}, \frac{5 \pi}{6}$
Since $\theta \in (0, 2 \pi)$,all three values $\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$ are valid solutions.
Thus,the total number of solutions is $3$.

(C) Given,$ABCD$ is a parallelogram with vertices $A(4,4,-1)$,$B(5,6,-1)$,$C(6,5,1)$ and $D(x, y, z)$.
We know that the diagonals of a parallelogram $ABCD$ bisect each other.
Therefore,the mid-point of $AC$ = mid-point of $BD$.
$\left(\frac{4+6}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right) = \left(\frac{x+5}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right)$
$\left(\frac{10}{2}, \frac{9}{2}, 0\right) = \left(\frac{x+5}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right)$
On comparing both sides,we get:
$\frac{x+5}{2} = \frac{10}{2}$ $\Rightarrow x+5 = 10$ $\Rightarrow x = 5$
$\frac{y+6}{2} = \frac{9}{2}$ $\Rightarrow y+6 = 9$ $\Rightarrow y = 3$
$\frac{z-1}{2} = 0$ $\Rightarrow z-1 = 0$ $\Rightarrow z = 1$
Thus,the vertex $D(x, y, z)$ is $(5, 3, 1)$.

(C) First,find the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$.
Multiplying the second equation by $3$,we get $9x + 6y + 15 = 0$.
Adding this to the first equation: $(5x - 6y - 1) + (9x + 6y + 15) = 0$ $\Rightarrow 14x + 14 = 0$ $\Rightarrow x = -1$.
Substituting $x = -1$ into $3x + 2y + 5 = 0$: $3(-1) + 2y + 5 = 0$ $\Rightarrow -3 + 2y + 5 = 0$ $\Rightarrow 2y = -2$ $\Rightarrow y = -1$.
The point of intersection is $(-1, -1)$.
The slope of the line $3x - 5y + 11 = 0$ is $m_1 = \frac{3}{5}$.
The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{5}{3}$.
The equation of the line passing through $(-1, -1)$ with slope $m_2 = -\frac{5}{3}$ is:
$(y - (-1)) = -\frac{5}{3}(x - (-1))
$ $\Rightarrow 3(y + 1) = -5(x + 1)
$ $\Rightarrow 3y + 3 = -5x - 5
$ $\Rightarrow 5x + 3y + 8 = 0$.

(D) Let the $X$-intercept be $a$ and the $Y$-intercept be $2a$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{2a} = 1$.
Multiplying by $2a$,we get $2x + y = 2a$,or $2x + y - 2a = 0$.
The perpendicular distance from the origin $(0, 0)$ to the line is given as $1$.
The formula for the distance from $(x_1, y_1)$ to $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values: $1 = \frac{|2(0) + 1(0) - 2a|}{\sqrt{2^2 + 1^2}}$.
$1 = \frac{|-2a|}{\sqrt{5}}$.
$|2a| = \sqrt{5}$,which means $2a = \pm \sqrt{5}$.
Substituting $2a$ back into the equation $2x + y = 2a$,we get $2x + y = \pm \sqrt{5}$.

(D) The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by the formula: $d = \left| \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right|$.
Given the point $(1, 1)$,the line $3x + 4y + c = 0$,and the distance $d = 7$:
$7 = \left| \frac{3(1) + 4(1) + c}{\sqrt{3^2 + 4^2}} \right|$
$7 = \left| \frac{7 + c}{\sqrt{9 + 16}} \right|$
$7 = \left| \frac{7 + c}{5} \right|$
$|7 + c| = 35$
This implies $7 + c = 35$ or $7 + c = -35$.
If $7 + c = 35$,then $c = 28$.
If $7 + c = -35$,then $c = -42$.
Thus,the possible values of $c$ are $28$ and $-42$.

(C) Let $P(x, y)$,$A=(5,3)$,$B=(3,-2)$.
Area of $\triangle PAB = \frac{1}{2} |x(3 - (-2)) - y(5 - 3) + 1(5(-2) - 3(3))| = 9$.
$\frac{1}{2} |5x - 2y - 19| = 9$.
$|5x - 2y - 19| = 18$.
$5x - 2y - 19 = 18$ or $5x - 2y - 19 = -18$.
$5x - 2y = 37$ or $5x - 2y = 1$.
These equations represent two lines with the same slope $m = \frac{5}{2}$,which are parallel lines.

(B) The given pair of lines $xy-x-y+1=0$ can be factored as $(x-1)(y-1)=0$.
This represents two lines: $x=1$ and $y=1$.
Since these lines are concurrent with $x+ay-3=0$,the point of intersection $(1,1)$ must satisfy the equation $x+ay-3=0$.
Substituting $(1,1)$ into $x+ay-3=0$,we get $1+a(1)-3=0$,which implies $a=2$.
Substituting $a=2$ into the second pair of lines,we get $2x^2-13xy-7y^2+x+23y-6=0$.
For a general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,the angle $\theta$ between the lines is given by $\tan \theta = \left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$.
Here $A=2, 2H=-13, B=-7$.
Thus,$\tan \theta = \left|\frac{2\sqrt{(-13/2)^2 - (2)(-7)}}{2-7}\right| = \left|\frac{2\sqrt{169/4 + 14}}{-5}\right| = \left|\frac{2\sqrt{225/4}}{-5}\right| = \left|\frac{2(15/2)}{-5}\right| = |-3| = 3$.
Since $\tan \theta = 3$,we have $\cos \theta = \frac{1}{\sqrt{1+3^2}} = \frac{1}{\sqrt{10}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$.

(A) Comparing the given equation with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=4, h=4, b=10, g=-4, f=-22$ and $c=14$.
To eliminate the first degree terms,the origin $(0,0)$ must be shifted to $(h_0, k_0)$ where $h_0 = \frac{bg-fh}{h^2-ab}$ and $k_0 = \frac{af-gh}{h^2-ab}$.
Calculating the denominator: $h^2-ab = 4^2 - (4)(10) = 16 - 40 = -24$.
Calculating $h_0$: $h_0 = \frac{(10)(-4) - (-22)(4)}{-24} = \frac{-40 + 88}{-24} = \frac{48}{-24} = -2$.
Calculating $k_0$: $k_0 = \frac{(4)(-22) - (-4)(4)}{-24} = \frac{-88 + 16}{-24} = \frac{-72}{-24} = 3$.
Thus,the origin must be shifted to $(-2, 3)$.

(C) Comparing the given equation with $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$,we get $a=2, b=2 \lambda, h=-5, g=\frac{5}{2}, f=-8, c=-3$.
Since the equation represents a pair of straight lines,the determinant condition is:
$\left|\begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=0$ $\Rightarrow \left|\begin{array}{ccc} 2 & -5 & 5/2 \\ -5 & 2 \lambda & -8 \\ 5/2 & -8 & -3 \end{array}\right|=0$.
Expanding the determinant: $2(-6 \lambda-64)+5(15+20)+\frac{5}{2}(40-5 \lambda)=0$.
$-12 \lambda-128+175+100-12.5 \lambda=0$ $\Rightarrow -24.5 \lambda = -147$ $\Rightarrow \lambda=6$.
Now,$b=2 \lambda = 12$. The point of intersection $(x, y)$ is given by $\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}\right)$.
Denominator $h^2-a b = (-5)^2 - (2)(12) = 25-24=1$.
$x = \frac{(12)(5/2) - (-8)(-5)}{1} = 30-40 = -10$.
$y = \frac{(2)(-8) - (5/2)(-5)}{1} = -16 + 12.5 = -3.5 = \frac{-7}{2}$.
Thus,the point of intersection is $\left(-10, \frac{-7}{2}\right)$.

(B) Given equation of the circle is $S \equiv x^2+y^2-13=0$.
On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.
The slope of the tangent $m$ at $(2,3)$ is $m = \left. \frac{dy}{dx} \right|_{(2,3)} = -\frac{2}{3}$.
Now,the point $\left(m, -\frac{1}{m}\right)$ becomes $\left(-\frac{2}{3}, \frac{3}{2}\right)$.
To check the position of this point with respect to the circle,we substitute it into the expression $S(x, y) = x^2+y^2-13$:
$S\left(-\frac{2}{3}, \frac{3}{2}\right) = \left(-\frac{2}{3}\right)^2 + \left(\frac{3}{2}\right)^2 - 13 = \frac{4}{9} + \frac{9}{4} - 13 = \frac{16 + 81 - 468}{36} = \frac{97 - 468}{36} = -\frac{371}{36}$.
Since $S\left(-\frac{2}{3}, \frac{3}{2}\right) < 0$,the point lies inside the circle.

(D) The given equation of the circle is $x^2+y^2-6x+4y=12$.
Completing the square,we get $(x-3)^2+(y+2)^2 = 12+9+4 = 25 = 5^2$.
The center is $(3, -2)$ and the radius $r=5$.
$A$ line parallel to $4x+3y+5=0$ is of the form $4x+3y+k=0$.
The distance from the center $(3, -2)$ to the tangent line $4x+3y+k=0$ must be equal to the radius $r=5$.
Using the distance formula,$\frac{|4(3)+3(-2)+k|}{\sqrt{4^2+3^2}} = 5$.
$\frac{|12-6+k|}{5} = 5
\Rightarrow |6+k| = 25$.
This gives $6+k = 25$ or $6+k = -25$.
So,$k = 19$ or $k = -31$.
The equations of the tangents are $4x+3y+19=0$ and $4x+3y-31=0$.

(A) The parametric equations of the given circle are $x = 12 \cos \theta$ and $y = 12 \sin \theta$.
For point $A$ with parameter $\theta = \frac{\pi}{3}$:
$x_A = 12 \cos \frac{\pi}{3} = 12 \times \frac{1}{2} = 6$
$y_A = 12 \sin \frac{\pi}{3} = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$
So,$A = (6, 6\sqrt{3})$.
For point $B$ with parameter $\theta = \frac{\pi}{6}$:
$x_B = 12 \cos \frac{\pi}{6} = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$
$y_B = 12 \sin \frac{\pi}{6} = 12 \times \frac{1}{2} = 6$
So,$B = (6\sqrt{3}, 6)$.
The length of the chord $AB$ is given by the distance formula:
$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$
$AB = \sqrt{(6\sqrt{3} - 6)^2 + (6 - 6\sqrt{3})^2}$
$AB = \sqrt{2 \times (6\sqrt{3} - 6)^2}$
$AB = \sqrt{2} \times |6\sqrt{3} - 6|$
$AB = 6\sqrt{2}(\sqrt{3} - 1)$
$AB = 6(\sqrt{6} - \sqrt{2})$.

(C) Let $S_1$ be the circle with centre $(2, 3)$ and radius $a$. The equation is $(x-2)^2 + (y-3)^2 = a^2$,which simplifies to $x^2 + y^2 - 4x - 6y + 13 - a^2 = 0$.
Let $S_2$ be the circle with centre $(5, 6)$ and radius $a$. The equation is $(x-5)^2 + (y-6)^2 = a^2$,which simplifies to $x^2 + y^2 - 10x - 12y + 61 - a^2 = 0$.
The radical axis is given by $S_1 - S_2 = 0$:
$(x^2 + y^2 - 4x - 6y + 13 - a^2) - (x^2 + y^2 - 10x - 12y + 61 - a^2) = 0$
$6x + 6y - 48 = 0 \Rightarrow x + y = 8$.
Since the circles cut orthogonally,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here $g_1 = -2, f_1 = -3, c_1 = 13 - a^2$ and $g_2 = -5, f_2 = -6, c_2 = 61 - a^2$.
$2(-2)(-5) + 2(-3)(-6) = (13 - a^2) + (61 - a^2)$
$20 + 36 = 74 - 2a^2$ $\Rightarrow 56 = 74 - 2a^2$ $\Rightarrow 2a^2 = 18$ $\Rightarrow a^2 = 9$ $\Rightarrow a = 3$.
Substituting $a = 3$ into the options:
$(A)$ $(9, 15) \Rightarrow 9 + 15 = 24 \neq 8$.
$(B)$ $(6, 3) \Rightarrow 6 + 3 = 9 \neq 8$.
$(C)$ $(3, 5) \Rightarrow 3 + 5 = 8$. This satisfies the equation.
$(D)$ $(3, 3) \Rightarrow 3 + 3 = 6 \neq 8$.
Thus,option $(C)$ is correct.

(D) The equations of the circles are:
$S_1: x^2+y^2-4x-6y+5=0$
$S_2: x^2+y^2-2x-4y-1=0$
$S_3: x^2+y^2-6x-2y=0$
To find the radical centre,we find the radical axes by subtracting the equations:
$S_1 - S_2 = 0$ $\Rightarrow (x^2+y^2-4x-6y+5) - (x^2+y^2-2x-4y-1) = 0$ $\Rightarrow -2x-2y+6=0$ $\Rightarrow x+y-3=0$ (Equation $i$)
$S_2 - S_3 = 0$ $\Rightarrow (x^2+y^2-2x-4y-1) - (x^2+y^2-6x-2y) = 0$ $\Rightarrow 4x-2y-1=0$ (Equation $ii$)
Solving equations $(i)$ and $(ii)$:
From $(i)$,$y = 3-x$.
Substitute into $(ii)$: $4x - 2(3-x) - 1 = 0$ $\Rightarrow 4x - 6 + 2x - 1 = 0$ $\Rightarrow 6x = 7$ $\Rightarrow x = \frac{7}{6}$.
Then $y = 3 - \frac{7}{6} = \frac{11}{6}$.
The radical centre is $(\frac{7}{6}, \frac{11}{6})$.
Checking the options,for option $D$: $18(\frac{7}{6}) - 12(\frac{11}{6}) + 1 = 3(7) - 2(11) + 1 = 21 - 22 + 1 = 0$.
Thus,the radical centre lies on the line $18x-12y+1=0$.

(B) Given equation: $y^2+6y-2x=-5$
Completing the square for $y$:
$y^2+6y+9 = 2x-5+9$
$(y+3)^2 = 2x+4$
$(y+3)^2 = 2(x+2)$
Comparing this with the standard form $(y-k)^2 = 4a(x-h)$,we get:
Vertex $(h, k) = (-2, -3)$. Thus,statement $I$ is true.
Here,$4a = 2$,so $a = \frac{1}{2}$.
The directrix of the parabola $(y-k)^2 = 4a(x-h)$ is given by $x = h-a$.
$x = -2 - \frac{1}{2} = -\frac{5}{2}$
$2x = -5 \Rightarrow 2x+5 = 0$.
Statement $II$ says the directrix is $y+3=0$,which is false.
Therefore,$I$ is true and $II$ is false.

(C) The equation of the line is $y = x + 4K$. For this to be a tangent to the parabola $y^2 = 8x$ (where $a=2$),the condition $c = a/m$ must be satisfied.
Here,$c = 4K$,$a = 2$,and $m = 1$.
So,$4K = 2/1 \implies 4K = 2 \implies K = 1/2$.
The point of contact $P$ is given by $(a/m^2, 2a/m) = (2/1^2, 2(2)/1) = (2, 4)$.
The equation of the normal to the parabola $y^2 = 4ax$ at $(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{2a}(x - x_1)$.
Substituting $x_1 = 2, y_1 = 4, a = 2$: $y - 4 = -\frac{4}{2(2)}(x - 2) \implies y - 4 = -1(x - 2) \implies x + y - 6 = 0$.
The point $(K, 2K)$ is $(1/2, 1)$.
The perpendicular distance from $(1/2, 1)$ to $x + y - 6 = 0$ is $d = \frac{|1/2 + 1 - 6|}{\sqrt{1^2 + 1^2}} = \frac{|3/2 - 6|}{\sqrt{2}} = \frac{|-9/2|}{\sqrt{2}} = \frac{9}{2\sqrt{2}}$.

(D) Given equation: $25x^2 + 100x + 4y^2 - 4y + 100 = 0$
Complete the squares: $25(x^2 + 4x + 4) + 4(y^2 - y + 1/4) = -100 + 100 + 1$
$25(x + 2)^2 + 4(y - 1/2)^2 = 1$
Divide by $1$: $\frac{(x + 2)^2}{(1/5)^2} + \frac{(y - 1/2)^2}{(1/2)^2} = 1$
Here $a^2 = 1/25$ and $b^2 = 1/4$. Since $b > a$,the major axis is vertical $(x = -2)$.
Eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1/25}{1/4}} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$.
The foci are $(h, k \pm be)$,where $(h, k) = (-2, 1/2)$.
Foci $= (-2, 1/2 \pm (1/2)(\sqrt{21}/5)) = (-2, 1/2 \pm \sqrt{21}/10) = (-2, \frac{5 \pm \sqrt{21}}{10})$.

(A) The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,and the end of the minor axis is $B(0, b)$.
Since $\triangle SBS^{\prime}$ is an isosceles right-angled triangle with the right angle at $B$,we have $SB = S^{\prime}B$ and $SB^2 + S^{\prime}B^2 = SS^{\prime 2}$.
Given $S(ae, 0)$,$S^{\prime}(-ae, 0)$,and $B(0, b)$,the distance $SB = \sqrt{(ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.
Similarly,$S^{\prime}B = \sqrt{(-ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.
The distance $SS^{\prime} = 2ae$.
Applying the Pythagorean theorem: $SB^2 + S^{\prime}B^2 = SS^{\prime 2}$.
$(a^2e^2 + b^2) + (a^2e^2 + b^2) = (2ae)^2$.
$2(a^2e^2 + b^2) = 4a^2e^2$.
$a^2e^2 + b^2 = 2a^2e^2$.
$b^2 = a^2e^2$.
Since $b^2 = a^2(1-e^2)$,we have $a^2(1-e^2) = a^2e^2$.
$1-e^2 = e^2$.
$2e^2 = 1$.
$e^2 = \frac{1}{2}$.
$e = \frac{1}{\sqrt{2}}$.

(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$.
For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$,we have $a^2 = 9$ and $b^2 = 4$.
The equation of the normal becomes $\frac{9x}{x_1} + \frac{4y}{y_1} = 9 + 4 = 13$.
Comparing this with the given line $x + y = -k$,we have the ratios:
$\frac{9/x_1}{1} = \frac{4/y_1}{1} = \frac{13}{-k}$.
This gives $x_1 = -\frac{9k}{13}$ and $y_1 = -\frac{4k}{13}$.
Since $(x_1, y_1)$ lies on the hyperbola,we substitute these values:
$\frac{(-9k/13)^2}{9} - \frac{(-4k/13)^2}{4} = 1$.
$\frac{81k^2}{169 \times 9} - \frac{16k^2}{169 \times 4} = 1$.
$\frac{9k^2}{169} - \frac{4k^2}{169} = 1$.
$\frac{5k^2}{169} = 1$.
$k^2 = \frac{169}{5}$.
$k = \pm \frac{13}{\sqrt{5}}$.

(A) We have,$\lim _{y \rightarrow 1}\left(\frac{1}{y^2-1}-\frac{2}{y^4-1}\right)$
$= \lim _{y \rightarrow 1}\left(\frac{y^2+1-2}{y^4-1}\right)$
$= \lim _{y \rightarrow 1}\left(\frac{y^2-1}{(y^2-1)(y^2+1)}\right)$
$= \lim _{y \rightarrow 1} \frac{1}{y^2+1}$
$= \frac{1}{1^2+1} = \frac{1}{2}$

(A) The variance of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2-1}{12}$.
For the first $n$ even numbers $(2, 4, 6, \dots, 2n)$,each term is $2$ times the corresponding natural number. Thus,the variance $A = 2^2 \times \frac{n^2-1}{12} = \frac{4(n^2-1)}{12} = \frac{n^2-1}{3}$.
For the first $n$ odd numbers $(1, 3, 5, \dots, 2n-1)$,these are obtained by subtracting $1$ from each of the first $n$ even numbers. Since variance is invariant under change of origin,the variance $B$ of the first $n$ odd numbers is the same as the variance of the first $n$ even numbers.
Therefore,$A = B$.

(A) Given,the mean $(\bar{x}) = 10$ and the number of observations $n = 8$.
The sum of the observations is $6 + 7 + 11 + 12 + 13 + \alpha + 12 + 16 = 77 + \alpha$.
Since $\bar{x} = \frac{\sum x_i}{n}$,we have $10 = \frac{77 + \alpha}{8}$.
$80 = 77 + \alpha \Rightarrow \alpha = 3$.
The data set is $6, 7, 11, 12, 13, 3, 12, 16$.
The mean deviation about the mean is $\frac{1}{n} \sum |x_i - \bar{x}|$.
$\text{MD}(\bar{x}) = \frac{|6-10| + |7-10| + |11-10| + |12-10| + |13-10| + |3-10| + |12-10| + |16-10|}{8}$.
$\text{MD}(\bar{x}) = \frac{|-4| + |-3| + |1| + |2| + |3| + |-7| + |2| + |6|}{8}$.
$\text{MD}(\bar{x}) = \frac{4 + 3 + 1 + 2 + 3 + 7 + 2 + 6}{8} = \frac{28}{8} = 3.5$.

(A) Let the sides of the triangle be $a = k$,$b = \sqrt{3}k$,and $c = 2k$.
Since $a^2 + b^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2 = c^2$,the triangle is a right-angled triangle with the hypotenuse $c = 2k$.
Let the angles opposite to sides $a, b, c$ be $A, B, C$ respectively.
Then $C = 90^{\circ}$.
Using the sine rule or trigonometric ratios in a right triangle:
$\sin A = \frac{a}{c} = \frac{k}{2k} = \frac{1}{2} \implies A = 30^{\circ}$.
$\sin B = \frac{b}{c} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \implies B = 60^{\circ}$.
Thus,the angles are $30^{\circ}, 60^{\circ}, 90^{\circ}$.
The ratio of the angles is $30^{\circ} : 60^{\circ} : 90^{\circ} = 1 : 2 : 3$.

(A) Given: $a=1, b=2, \angle C=60^{\circ}$.
$1$. Calculate the area $\Delta$ of the triangle:
$\Delta = \frac{1}{2} ab \sin C$
$\Delta = \frac{1}{2} \times 1 \times 2 \times \sin 60^{\circ}$
$\Delta = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$
$\Delta^2 = \frac{3}{4}$
$4 \Delta^2 = 3$
$2$. Calculate the side $c$ using the Law of Cosines:
$c^2 = a^2 + b^2 - 2ab \cos C$
$c^2 = 1^2 + 2^2 - 2(1)(2) \cos 60^{\circ}$
$c^2 = 1 + 4 - 4 \times \frac{1}{2}$
$c^2 = 5 - 2 = 3$
$3$. Calculate $4 \Delta^2 + c^2$:
$4 \Delta^2 + c^2 = 3 + 3 = 6$.

(D) Let $a = 13$,$b = 14$,and $c = 15$.
First,calculate the semi-perimeter $s$:
$s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21$.
Next,calculate the area of the triangle $\Delta$ using Heron's formula:
$\Delta = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$.
Now,calculate the circumradius $R = \frac{abc}{4\Delta}$:
$R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336} = \frac{65}{8}$.
Calculate the inradius $r = \frac{\Delta}{s}$:
$r = \frac{84}{21} = 4$.
Finally,compute $8R + r$:
$8R + r = 8 \times \left(\frac{65}{8}\right) + 4 = 65 + 4 = 69$.

(C) Given the partial fraction decomposition: $\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$.
Multiplying both sides by $(x^2+1)(x-2)$,we get: $x^2+5 = A(x^2+1) + (Bx+C)(x-2)$.
Expanding the right side: $x^2+5 = Ax^2 + A + Bx^2 - 2Bx + Cx - 2C$.
Grouping terms by powers of $x$: $x^2+5 = (A+B)x^2 + (C-2B)x + (A-2C)$.
Comparing coefficients on both sides:
$1$) $A+B = 1$
$2$) $C-2B = 0 \Rightarrow C = 2B$
$3$) $A-2C = 5$
Substitute $C=2B$ into equation $(3)$: $A - 2(2B) = 5 \Rightarrow A - 4B = 5$.
Now solve the system of equations:
$A+B = 1$
$A-4B = 5$
Subtracting the second from the first: $5B = -4 \Rightarrow B = -\frac{4}{5}$.
Then $A = 1 - B = 1 - (-4/5) = 9/5$.
And $C = 2B = 2(-4/5) = -8/5$.
Finally,$A+B+C = \frac{9}{5} - \frac{4}{5} - \frac{8}{5} = \frac{9-4-8}{5} = -\frac{3}{5}$.

(D) Given,$P(A)=0.6$,$P(B)=0.4$,and $P(A \cap B)=0$.
We need to find the probability that neither $A$ nor $B$ occurs,which is $P(\bar{A} \cap \bar{B})$.
By De Morgan's Law,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B})$.
We know that $P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the addition rule of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = 0.6 + 0.4 - 0 = 1.0$.
Therefore,$P(\overline{A \cup B}) = 1 - 1.0 = 0$.

(D) Total number of balls $= 5 + 3 + 4 = 12$.
Number of ways to draw $3$ balls from $12$ is ${}^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
Number of ways to draw $3$ balls of the same colour:
- $3$ red balls: ${}^{5}C_3 = 10$
- $3$ black balls: ${}^{3}C_3 = 1$
- $3$ white balls: ${}^{4}C_3 = 4$
Total ways for same colour $= 10 + 1 + 4 = 15$.
Probability of same colour $= \frac{15}{220} = \frac{3}{44}$.
Probability of not same colour $= 1 - \frac{3}{44} = \frac{41}{44}$.

(D) Given that $|\bar{a}| = 3$,$|\bar{b}| = 4$,and $|\bar{a}+\bar{b}| = 5$.
We know that $|\bar{a}+\bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + 2(\bar{a} \cdot \bar{b})$.
Substituting the values: $5^2 = 3^2 + 4^2 + 2(\bar{a} \cdot \bar{b})$.
$25 = 9 + 16 + 2(\bar{a} \cdot \bar{b}) \implies 25 = 25 + 2(\bar{a} \cdot \bar{b}) \implies \bar{a} \cdot \bar{b} = 0$.
This implies that $\bar{a}$ and $\bar{b}$ are perpendicular to each other.
Now,we need to find $|\bar{a}-\bar{b}|$.
$|\bar{a}-\bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 - 2(\bar{a} \cdot \bar{b})$.
$|\bar{a}-\bar{b}|^2 = 3^2 + 4^2 - 2(0) = 9 + 16 = 25$.
Therefore,$|\bar{a}-\bar{b}| = \sqrt{25} = 5$.

(C) Given curves are $x=y^2-2$ and $x=y$.
To find the points of intersection,we substitute $x=y$ into $x=y^2-2$:
$y = y^2 - 2$
$y^2 - y - 2 = 0$
$(y-2)(y+1) = 0$
So,$y=2$ and $y=-1$.
For $y=2$,$x=2$. For $y=-1$,$x=-1$.
The points of intersection are $(-1, -1)$ and $(2, 2)$.
The area $A$ is given by the integral of the right curve minus the left curve with respect to $y$ from $y=-1$ to $y=2$:
$A = \int_{-1}^{2} (y - (y^2 - 2)) \, dy$
$A = \int_{-1}^{2} (y - y^2 + 2) \, dy$
$A = \left[ \frac{y^2}{2} - \frac{y^3}{3} + 2y \right]_{-1}^{2}$
$A = \left( \frac{2^2}{2} - \frac{2^3}{3} + 2(2) \right) - \left( \frac{(-1)^2}{2} - \frac{(-1)^3}{3} + 2(-1) \right)$
$A = \left( 2 - \frac{8}{3} + 4 \right) - \left( \frac{1}{2} + \frac{1}{3} - 2 \right)$
$A = \left( 6 - \frac{8}{3} \right) - \left( \frac{3+2-12}{6} \right)$
$A = \frac{10}{3} - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$ square units.

(B) Any square matrix $A$ can be expressed as the sum of a symmetric matrix $B$ and a skew-symmetric matrix $C$,where $B = \frac{1}{2}(A + A^T)$ and $C = \frac{1}{2}(A - A^T)$.
Given $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4 \end{bmatrix}$.
Then $A^T = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & 0 & 4 \end{bmatrix}$.
Now,$C = \frac{1}{2}(A - A^T) = \frac{1}{2} \left( \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & 0 & 4 \end{bmatrix} \right)$.
$C = \frac{1}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0 \end{bmatrix}$.
Thus,the correct option is $B$.

(C) Let $A = \begin{bmatrix} x & x & x \\ x & x^2 & x \\ x & x & x+1 \end{bmatrix}$.
For the rank of $A$ to be $1$,all minors of order $2$ must be zero.
Consider the minor formed by the first two rows and first two columns: $\begin{vmatrix} x & x \\ x & x^2 \end{vmatrix} = x^3 - x^2 = x^2(x-1)$.
For this to be $0$,$x=0$ or $x=1$.
Case $1$: If $x=1$,$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}$. The minor $\begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = 2-1 = 1 \neq 0$. Thus,the rank is at least $2$. So $x=1$ is not the solution.
Case $2$: If $x=0$,$A = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. All minors of order $2$ are zero,and there is at least one non-zero element (the $1$ at $A_{33}$). Thus,the rank is $1$.
Therefore,the only possible value is $x=0$.

(B) Given,$\Delta = \left|\begin{array}{lll}1 & 5 & 6 \\ 0 & 1 & 7 \\ 0 & 0 & 1\end{array}\right|$.
Since this is an upper triangular matrix,the determinant is the product of the diagonal elements:
$\Delta = 1 \times 1 \times 1 = 1$.
Now,$\Delta^{\prime} = \left|\begin{array}{ccc}1 & 0 & 1 \\ 3 & 0 & 3 \\ 4 & 6 & 100\end{array}\right|$.
Expanding along the second column:
$\Delta^{\prime} = -0(300-12) + 0(100-4) - 6(3-3) = 0$.
Now,substitute $\Delta = 1$ and $\Delta^{\prime} = 0$ into the expression $(\Delta+\Delta^{\prime})^2-3(\Delta+\Delta^{\prime})+2$:
$= (1+0)^2 - 3(1+0) + 2$
$= 1^2 - 3 + 2$
$= 1 - 3 + 2 = 0$.
Thus,the correct option is $B$.

(B) Let $x = \frac{\sqrt{3}}{2}$ and $y = \sqrt{\frac{2}{3}}$.
Note that $x^2 + y^2 = \frac{3}{4} + \frac{2}{3} = \frac{9+8}{12} = \frac{17}{12} > 1$.
Since $x, y > 0$ and $x^2 + y^2 > 1$,we use the identity $\sin^{-1} x + \sin^{-1} y = \pi - \sin^{-1} (x \sqrt{1-y^2} + y \sqrt{1-x^2})$.
Substituting the values:
$\sin^{-1} \frac{\sqrt{3}}{2} + \sin^{-1} \sqrt{\frac{2}{3}} = \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2} \sqrt{1 - \frac{2}{3}} + \sqrt{\frac{2}{3}} \sqrt{1 - \frac{3}{4}} \right)$
$= \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2} \sqrt{\frac{1}{3}} + \sqrt{\frac{2}{3}} \sqrt{\frac{1}{4}} \right)$
$= \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2 \sqrt{3}} + \frac{\sqrt{2}}{\sqrt{3} \cdot 2} \right)$
$= \pi - \sin^{-1} \left( \frac{1}{2} + \frac{\sqrt{2}}{2 \sqrt{3}} \right)$
$= \pi - \sin^{-1} \left( \frac{\sqrt{3} + \sqrt{2}}{2 \sqrt{3}} \right)$.

(A) Given the function $f(x) = -3x - 3$.
To find the domain,we set $f(x)$ equal to each element in the range:
$(i)$ For $f(x) = 3$: $3 = -3x - 3$ $\Rightarrow 6 = -3x$ $\Rightarrow x = -2$.
(ii) For $f(x) = -6$: $-6 = -3x - 3$ $\Rightarrow -3 = -3x$ $\Rightarrow x = 1$.
(iii) For $f(x) = -9$: $-9 = -3x - 3$ $\Rightarrow -6 = -3x$ $\Rightarrow x = 2$.
(iv) For $f(x) = -18$: $-18 = -3x - 3$ $\Rightarrow -15 = -3x$ $\Rightarrow x = 5$.
Thus,the domain of $f$ is $\{-2, 1, 2, 5\}$.
Comparing this with the given options,$-1$ is not in the domain of $f$.

(A) We have a function $f: (-\infty, 0] \rightarrow [0, \infty)$ defined as $f(x) = x^2$.
Since each line parallel to the $x$-axis cuts the curve at most at one point,the function $f$ is one-one.
From the graph,it is clear that the range of $f$ is $[0, \infty)$,which is equal to the codomain.
Therefore,$f$ is onto.
Since $f$ is both one-one and onto,it is invertible.
The inverse function $f^{-1}$ maps the codomain of $f$ to the domain of $f$.
Thus,$f^{-1}: [0, \infty) \rightarrow (-\infty, 0]$.
Hence,the domain of $f^{-1}$ is $[0, \infty)$ and the range of $f^{-1}$ is $(-\infty, 0]$.

(D) For $f(x)$ to be continuous on $\mathbb{R}$,it must be continuous at $x=0, x=1,$ and $x=2$.
At $x=0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \implies \sin(0) = 0^2 + a^2 \implies a^2 = 0 \implies a = 0$.
At $x=1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \implies 1^2 + a^2 = b(1) + 2 \implies 1 + 0 = b + 2 \implies b = -1$.
At $x=2$: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \implies b(2) + 2 = 0 \implies 2(-1) + 2 = 0$,which is consistent.
Thus,$a = 0$ and $b = -1$.
Calculating $a+b+ab = 0 + (-1) + (0)(-1) = -1$.

(A) Given the equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Differentiating both sides with respect to $x$:
$\frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0$
$\frac{y}{b^2} \cdot \frac{dy}{dx} = -\frac{x}{a^2}$
$\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$
Now,differentiate again with respect to $x$ using the quotient rule:
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \frac{d}{dx} \left( \frac{x}{y} \right)$
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y(1) - x(\frac{dy}{dx})}{y^2} \right)$
Substitute $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$:
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y - x(-\frac{b^2 x}{a^2 y})}{y^2} \right)$
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y + \frac{b^2 x^2}{a^2 y}}{y^2} \right) = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 y^2 + b^2 x^2}{a^2 y^3} \right)$
Since $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $b^2 x^2 + a^2 y^2 = a^2 b^2$.
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 b^2}{a^2 y^3} \right) = -\frac{b^4}{a^2 y^3}$.

(D) Given the functional equation $f(x+y)=f(x) f(y)$.
By the definition of the derivative,$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$.
Using the given relation,$f(x+h) = f(x)f(h)$.
So,$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h)-1}{h}$.
Since $f(0+0) = f(0)f(0)$,we have $f(0) = f(0)^2$,so $f(0)=1$ (assuming $f(x) \neq 0$).
Thus,$f^{\prime}(0) = \lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{f(h)-1}{h} = 11$.
Therefore,$f^{\prime}(x) = f(x) \cdot 11$.
At $x=3$,$f^{\prime}(3) = f(3) \cdot 11 = 3 \cdot 11 = 33$.

(D) Given $g(x) = (f(2f(x)+2))^2$.
Applying the chain rule,we differentiate $g(x)$ with respect to $x$:
$g^{\prime}(x) = 2(f(2f(x)+2)) \cdot \frac{d}{dx}(f(2f(x)+2))$
$g^{\prime}(x) = 2(f(2f(x)+2)) \cdot f^{\prime}(2f(x)+2) \cdot \frac{d}{dx}(2f(x)+2)$
$g^{\prime}(x) = 2(f(2f(x)+2)) \cdot f^{\prime}(2f(x)+2) \cdot 2f^{\prime}(x)$
$g^{\prime}(x) = 4 \cdot f(2f(x)+2) \cdot f^{\prime}(2f(x)+2) \cdot f^{\prime}(x)$.
Now,substitute $x=0$:
$g^{\prime}(0) = 4 \cdot f(2f(0)+2) \cdot f^{\prime}(2f(0)+2) \cdot f^{\prime}(0)$.
Given $f(0)=-1$ and $f^{\prime}(0)=1$:
$g^{\prime}(0) = 4 \cdot f(2(-1)+2) \cdot f^{\prime}(2(-1)+2) \cdot (1)$
$g^{\prime}(0) = 4 \cdot f(0) \cdot f^{\prime}(0) \cdot 1$
$g^{\prime}(0) = 4 \cdot (-1) \cdot (1) \cdot 1 = -4$.

(B) Given equations of the curves are $x^2=8y$ $(i)$ and $xy=8$ $(ii)$.
To find the point of intersection,substitute $y = \frac{x^2}{8}$ from $(i)$ into $(ii)$:
$x\left(\frac{x^2}{8}\right) = 8 \Rightarrow x^3 = 64 \Rightarrow x = 4$.
Substituting $x=4$ into $(ii)$,we get $4y=8 \Rightarrow y=2$.
So,the point of intersection is $(4, 2)$.
For curve $(i)$,$x^2=8y \Rightarrow 2x = 8 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{4}$.
At $(4, 2)$,the slope $m_1 = \frac{4}{4} = 1$.
For curve $(ii)$,$xy=8 \Rightarrow x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
At $(4, 2)$,the slope $m_2 = -\frac{2}{4} = -\frac{1}{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{1 - (-1/2)}{1 + (1)(-1/2)} \right| = \left| \frac{3/2}{1/2} \right| = 3$.
Thus,$\theta = \tan^{-1}(3)$. Note: The angle between curves is typically defined as the acute angle,so $\tan^{-1}(3)$ is the standard representation.

(B) Given the curve equation: $(\frac{x}{a})^n+(\frac{y}{b})^n=2$.
On differentiating with respect to $x$,we get:
$\frac{n x^{n-1}}{a^n} + \frac{n y^{n-1}}{b^n} \frac{dy}{dx} = 0$.
$\Rightarrow \frac{dy}{dx} = -\frac{b^n x^{n-1}}{a^n y^{n-1}}$.
At the point $(a, b)$,the slope of the tangent is:
$(\frac{dy}{dx})_{(a, b)} = -\frac{b^n a^{n-1}}{a^n b^{n-1}} = -\frac{b}{a}$.
The equation of the tangent at $(a, b)$ is given by $y - y_1 = m(x - x_1)$:
$y - b = -\frac{b}{a}(x - a)$.
$ay - ab = -bx + ab$.
$bx + ay = 2ab$.
Dividing both sides by $ab$,we get:
$\frac{x}{a} + \frac{y}{b} = 2$.

(A) The curve $y=ax^3+bx+4$ passes through the point $(2, 14)$. Substituting these coordinates into the equation:
$14 = a(2)^3 + b(2) + 4$
$14 = 8a + 2b + 4$
$10 = 8a + 2b$
$5 = 4a + b$ --- $(i)$
The slope of the tangent to the curve is given by the derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = 3ax^2 + b$
At the point $(2, 14)$,the slope is $21$:
$21 = 3a(2)^2 + b$
$21 = 12a + b$ --- $(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(12a + b) - (4a + b) = 21 - 5$
$8a = 16$
$a = 2$
Substituting $a = 2$ into equation $(i)$:
$5 = 4(2) + b$
$5 = 8 + b$
$b = -3$
Thus,the values are $a = 2$ and $b = -3$.

(A) Given,$y=x^3-3 x^2+5$.
On differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = 3x^2 - 6x$.
For local maxima or local minima,we set $\frac{dy}{dx} = 0$.
$3x^2 - 6x = 0 \Rightarrow 3x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$.
Now,differentiating $\frac{dy}{dx}$ with respect to $x$,we get $\frac{d^2y}{dx^2} = 6x - 6$.
At $x = 0$,$\frac{d^2y}{dx^2} = 6(0) - 6 = -6 < 0$.
Since the second derivative is negative at $x = 0$,$x = 0$ is a point of local maxima.
At $x = 2$,$\frac{d^2y}{dx^2} = 6(2) - 6 = 6 > 0$.
Since the second derivative is positive at $x = 2$,$x = 2$ is a point of local minima.
Therefore,the local maximum is attained at $x = 0$.

(A) Given the equation: $\int f(x) \cos x \, dx = \frac{1}{2} [f(x)]^2 + C$.
Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus,we get:
$f(x) \cos x = \frac{d}{dx} \left( \frac{1}{2} [f(x)]^2 + C \right)$.
Applying the chain rule:
$f(x) \cos x = \frac{1}{2} \cdot 2 f(x) \cdot f'(x)$.
Simplifying the expression:
$f(x) \cos x = f(x) \cdot f'(x)$.
For $f(x) \neq 0$,we have $f'(x) = \cos x$.
Substituting $x = 0$:
$f'(0) = \cos(0) = 1$.

(A) Let $I = \int x^4 e^{2 x} d x$.
Using integration by parts,$\int u v d x = u \int v d x - \int (u' \int v d x) d x$.
Taking $u = x^4$ and $v = e^{2 x}$:
$I = x^4 \frac{e^{2 x}}{2} - \int 4 x^3 \frac{e^{2 x}}{2} d x = \frac{x^4 e^{2 x}}{2} - 2 \int x^3 e^{2 x} d x$.
Applying integration by parts again:
$\int x^3 e^{2 x} d x = x^3 \frac{e^{2 x}}{2} - \int 3 x^2 \frac{e^{2 x}}{2} d x = \frac{x^3 e^{2 x}}{2} - \frac{3}{2} \int x^2 e^{2 x} d x$.
Continuing the process:
$\int x^2 e^{2 x} d x = \frac{x^2 e^{2 x}}{2} - \int x e^{2 x} d x = \frac{x^2 e^{2 x}}{2} - (\frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4})$.
Substituting back into the expression for $I$:
$I = \frac{x^4 e^{2 x}}{2} - 2 [\frac{x^3 e^{2 x}}{2} - \frac{3}{2} (\frac{x^2 e^{2 x}}{2} - \frac{x e^{2 x}}{2} + \frac{e^{2 x}}{4})] + C$.
$I = \frac{x^4 e^{2 x}}{2} - x^3 e^{2 x} + \frac{3}{2} x^2 e^{2 x} - \frac{3}{2} x e^{2 x} + \frac{3}{4} e^{2 x} + C$.
$I = \frac{e^{2 x}}{4} (2 x^4 - 4 x^3 + 6 x^2 - 6 x + 3) + C$.

(B) We are given that $\int e^{2x} f^{\prime}(x) dx = g(x)$.
Let $I = \int (e^{2x} f(x) + e^{2x} f^{\prime}(x)) dx$.
This can be split into two integrals: $I = \int e^{2x} f(x) dx + \int e^{2x} f^{\prime}(x) dx$.
Using integration by parts for the first integral $\int e^{2x} f(x) dx$ (taking $f(x)$ as the first function and $e^{2x}$ as the second function):
$\int e^{2x} f(x) dx = f(x) \int e^{2x} dx - \int (f^{\prime}(x) \int e^{2x} dx) dx = \frac{1}{2} f(x) e^{2x} - \frac{1}{2} \int e^{2x} f^{\prime}(x) dx$.
Substituting this back into the expression for $I$:
$I = [\frac{1}{2} f(x) e^{2x} - \frac{1}{2} \int e^{2x} f^{\prime}(x) dx] + \int e^{2x} f^{\prime}(x) dx$.
$I = \frac{1}{2} e^{2x} f(x) + \frac{1}{2} \int e^{2x} f^{\prime}(x) dx$.
Since $\int e^{2x} f^{\prime}(x) dx = g(x)$,we have:
$I = \frac{1}{2} e^{2x} f(x) + \frac{1}{2} g(x) + C = \frac{1}{2} [e^{2x} f(x) + g(x)] + C$.

(B) $\int \frac{d x}{x\left(x^4+1\right)} = \int \frac{x^3 d x}{x^4\left(x^4+1\right)}$
Let $x^4 = t$,then $4x^3 dx = dt$,so $x^3 dx = \frac{dt}{4}$.
Substituting these into the integral:
$\int \frac{dt/4}{t(t+1)} = \frac{1}{4} \int \frac{dt}{t(t+1)}$
Using partial fractions:
$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$
Therefore:
$\frac{1}{4} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt = \frac{1}{4} (\log |t| - \log |t+1|) + C$
$= \frac{1}{4} \log \left| \frac{t}{t+1} \right| + C$
Substituting $t = x^4$ back:
$= \frac{1}{4} \log \left( \frac{x^4}{x^4+1} \right) + C$

(A) Let $I = \int_0^\pi \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x}$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^\pi \frac{(\pi - x) \, dx}{4 \cos^2(\pi - x) + 9 \sin^2(\pi - x)} = \int_0^\pi \frac{(\pi - x) \, dx}{4 \cos^2 x + 9 \sin^2 x}$.
Adding the two expressions for $I$:
$2I = \int_0^\pi \frac{\pi \, dx}{4 \cos^2 x + 9 \sin^2 x} = \pi \int_0^\pi \frac{\sec^2 x \, dx}{4 + 9 \tan^2 x}$.
Using the property $\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$ if $f(2a-x) = f(x)$:
$2I = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + 9 \tan^2 x} \Rightarrow I = \pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + 9 \tan^2 x}$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. Limits change from $[0, \pi/2]$ to $[0, \infty]$.
$I = \pi \int_0^\infty \frac{dt}{4 + 9t^2} = \frac{\pi}{9} \int_0^\infty \frac{dt}{(2/3)^2 + t^2}$.
Using $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})$:
$I = \frac{\pi}{9} \cdot \frac{3}{2} \left[ \tan^{-1}(\frac{3t}{2}) \right]_0^\infty = \frac{\pi}{6} \cdot \frac{\pi}{2} = \frac{\pi^2}{12}$.

(I-D, II-A, III-E, IV-C) The correct matches are as follows:
$I. \int_{-1}^1 x|x| dx = 0$ (since $f(x) = x|x|$ is an odd function,i.e.,$f(-x) = -x|-x| = -x|x| = -f(x)$). Thus,$I \rightarrow (d)$.
$II. \text{Let } I = \int_0^{\pi/2} \left(1 + \log \frac{4+3\sin x}{4+3\cos x}\right) dx$.
Using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^{\pi/2} \left(1 + \log \frac{4+3\cos x}{4+3\sin x}\right) dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi/2} \left(2 + \log \left(\frac{4+3\sin x}{4+3\cos x} \cdot \frac{4+3\cos x}{4+3\sin x}\right)\right) dx = \int_0^{\pi/2} (2 + \log 1) dx = \int_0^{\pi/2} 2 dx = \pi$.
Therefore,$I = \frac{\pi}{2}$. Thus,$II \rightarrow (a)$.
$III. \int_0^a f(x) dx = \int_0^a f(a-x) dx$ is a standard property of definite integrals. Thus,$III \rightarrow (e)$.
$IV. \int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx + \int_0^a f(x) dx$. Let $x = -t$ in the first integral,$dx = -dt$.
$int_{-a}^0 f(x) dx = \int_a^0 f(-t) (-dt) = \int_0^a f(-t) dt = \int_0^a f(-x) dx$.
So,$\int_{-a}^a f(x) dx = \int_0^a f(-x) dx + \int_0^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$. Thus,$IV \rightarrow (c)$.

(B) Given the equation of simple harmonic motion: $x = A \cos (nt + \alpha)$ ... $(i)$
Differentiating with respect to $t$:
$\frac{dx}{dt} = -A n \sin (nt + \alpha)$ ... (ii)
Differentiating again with respect to $t$:
$\frac{d^2x}{dt^2} = -An \frac{d}{dt} \sin (nt + \alpha)$
$\frac{d^2x}{dt^2} = -An^2 \cos (nt + \alpha)$
Substituting $x = A \cos (nt + \alpha)$ into the equation:
$\frac{d^2x}{dt^2} = -n^2 x$
Therefore,the differential equation is:
$\frac{d^2x}{dt^2} + n^2 x = 0$

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x+y}{x-y}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation:
$v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1+v}{1-v}$.
$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v - v + v^2}{1-v} = \frac{1+v^2}{1-v}$.
Separating the variables:
$\frac{1-v}{1+v^2} dv = \frac{dx}{x}$.
Integrating both sides:
$\int \frac{1}{1+v^2} dv - \int \frac{v}{1+v^2} dv = \int \frac{dx}{x}$.
$\tan^{-1}(v) - \frac{1}{2} \log(1+v^2) = \log|x| + C$.
Substituting $v = \frac{y}{x}$:
$\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + \frac{1}{2} \log\left(1 + \frac{y^2}{x^2}\right) + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + \frac{1}{2} \log\left(\frac{x^2+y^2}{x^2}\right) + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + \frac{1}{2} [\log(x^2+y^2) - \log(x^2)] + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + \frac{1}{2} \log(x^2+y^2) - \log|x| + C$.
$\tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2+y^2} + C$.

(C) Given the differential equation: $(y-3 x^2) d x+x d y=0$.
Rearranging the terms,we get: $y d x-3 x^2 d x+x d y=0$.
Grouping the terms $y d x$ and $x d y$,we have: $y d x+x d y=3 x^2 d x$.
Recognizing the product rule for differentiation,$d(x y) = y d x + x d y$,the equation becomes: $d(x y) = 3 x^2 d x$.
Integrating both sides: $\int d(x y) = \int 3 x^2 d x$.
This yields: $x y = x^3 + C$.
Dividing by $x$ (assuming $x \neq 0$),we get the solution: $y = x^2 + \frac{C}{x}$.

(A) Given that $|a|=1$ and $|b|=1$,and $\alpha$ is the angle between $a$ and $b$.
We know that $a \cdot b = |a||b| \cos \alpha = (1)(1) \cos \alpha = \cos \alpha$.
Since $a+b$ is a unit vector,we have $|a+b|=1$.
Squaring both sides,we get $|a+b|^2 = 1^2 = 1$.
Expanding the dot product,$(a+b) \cdot (a+b) = 1$.
$a \cdot a + a \cdot b + b \cdot a + b \cdot b = 1$.
Since $a \cdot a = |a|^2 = 1$ and $b \cdot b = |b|^2 = 1$,and $a \cdot b = b \cdot a = \cos \alpha$,we substitute these values:
$1 + \cos \alpha + \cos \alpha + 1 = 1$.
$2 + 2 \cos \alpha = 1$.
$2 \cos \alpha = 1 - 2$.
$2 \cos \alpha = -1$.
$\cos \alpha = -\frac{1}{2}$.

(C) Given that $a, b, c$ are unit vectors,so $|a|=|b|=|c|=1$.
Given $a+b+c=0$.
The angle between $a$ and $b$ is $\frac{\pi}{3}$.
Taking the cross product of $a+b+c=0$ with $a$:
$a \times (a+b+c) = a \times 0$
$a \times a + a \times b + a \times c = 0$
Since $a \times a = 0$,we have $a \times b = c \times a$.
Taking the magnitude,$|a \times b| = |c \times a|$.
Similarly,by taking the cross product with $b$,we get $|a \times b| = |b \times c|$.
Thus,$|a \times b| = |b \times c| = |c \times a|$.
Therefore,$|a \times b| + |b \times c| + |c \times a| = 3|a \times b|$.
Using the formula $|a \times b| = |a||b| \sin(\theta)$,where $\theta = \frac{\pi}{3}$:
$|a \times b| = 1 \times 1 \times \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.
So,$3|a \times b| = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$.

(B) Given $a = x \hat{i} + y \hat{j} + z \hat{k}$.
We need to evaluate the expression $E = (a \times \hat{i}) \cdot (\hat{i} + \hat{j}) + (a \times \hat{j}) \cdot (\hat{j} + \hat{k}) + (a \times \hat{k}) \cdot (\hat{k} + \hat{i})$.
Using the property of scalar triple product $[a, b, c] = (a \times b) \cdot c$,we can rewrite the expression as:
$E = [a, \hat{i}, \hat{i}] + [a, \hat{i}, \hat{j}] + [a, \hat{j}, \hat{j}] + [a, \hat{j}, \hat{k}] + [a, \hat{k}, \hat{k}] + [a, \hat{k}, \hat{i}]$.
Since the scalar triple product is zero if any two vectors are identical,we have $[a, \hat{i}, \hat{i}] = [a, \hat{j}, \hat{j}] = [a, \hat{k}, \hat{k}] = 0$.
Thus,$E = [a, \hat{i}, \hat{j}] + [a, \hat{j}, \hat{k}] + [a, \hat{k}, \hat{i}]$.
Using the cyclic property $[a, b, c] = a \cdot (b \times c)$,we get:
$E = a \cdot (\hat{i} \times \hat{j}) + a \cdot (\hat{j} \times \hat{k}) + a \cdot (\hat{k} \times \hat{i})$.
Since $\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,and $\hat{k} \times \hat{i} = \hat{j}$,we have:
$E = a \cdot \hat{k} + a \cdot \hat{i} + a \cdot \hat{j} = a \cdot (\hat{i} + \hat{j} + \hat{k})$.
Substituting $a = x \hat{i} + y \hat{j} + z \hat{k}$,we get:
$E = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = x + y + z$.

(A) We have,$a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=\hat{i}+3 \hat{j}+2 \hat{k}$.
Since $c$ is perpendicular to the plane of $a$ and $b$,$c$ is parallel to $a \times b$.
First,calculate $a \times b$:
$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(2 - (-9)) - \hat{j}(4 - (-3)) + \hat{k}(6 - 1) = 11 \hat{i} - 7 \hat{j} + 5 \hat{k}$.
The magnitude $|a \times b| = \sqrt{11^2 + (-7)^2 + 5^2} = \sqrt{121 + 49 + 25} = \sqrt{195}$.
Since $c$ is parallel to $a \times b$ and $|c|=2$,we have $c = \pm 2 \frac{a \times b}{|a \times b|} = \pm \frac{2}{\sqrt{195}} (11 \hat{i} - 7 \hat{j} + 5 \hat{k})$.
The volume of the parallelopiped is given by the scalar triple product $|[a, b, c]| = |(a \times b) \cdot c|$.
$|[a, b, c]| = |(a \times b) \cdot (\pm 2 \frac{a \times b}{|a \times b|})| = |\pm 2 \frac{|a \times b|^2}{|a \times b|}| = 2 |a \times b|$.
Substituting $|a \times b| = \sqrt{195}$,the volume is $2 \sqrt{195}$.

(A) Let $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. Since $a$ is a unit vector,$|a|^2 = a_1^2 + a_2^2 + a_3^2 = 1$.
Now,$a \times \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{i} = -a_2 \hat{k} + a_3 \hat{j}$.
So,$|a \times \hat{i}|^2 = a_2^2 + a_3^2$.
Similarly,$|a \times \hat{j}|^2 = a_1^2 + a_3^2$ and $|a \times \hat{k}|^2 = a_1^2 + a_2^2$.
Adding these,we get $|a \times \hat{i}|^2 + |a \times \hat{j}|^2 + |a \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2)$.
Since $a_1^2 + a_2^2 + a_3^2 = 1$,the sum is $2(1) = 2$.

(B) The equation of a plane passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ and perpendicular to a given plane $ax + by + cz = d$ can be found using the determinant form. The normal vector of the required plane is the cross product of the vector connecting the two points and the normal vector of the given plane.
Let the points be $A(1, -1, 6)$ and $B(0, 0, 7)$. The vector $\vec{AB} = (0-1, 0-(-1), 7-6) = (-1, 1, 1)$.
The normal vector of the plane $x - 2y + z = 6$ is $\vec{n_1} = (1, -2, 1)$.
The normal vector of the required plane is $\vec{n} = \vec{AB} \times \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1+2) - \hat{j}(-1-1) + \hat{k}(2-1) = 3\hat{i} + 2\hat{j} + 1\hat{k}$.
The equation of the plane passing through $(0, 0, 7)$ with normal vector $(3, 2, 1)$ is $3(x-0) + 2(y-0) + 1(z-7) = 0$,which simplifies to $3x + 2y + z = 7$.
Now,check the options by substituting the coordinates into the equation $3x + 2y + z = 7$:
For $(1, 1, 2)$: $3(1) + 2(1) + 2 = 3 + 2 + 2 = 7$. This satisfies the equation.

(D) Given vectors are $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$ and $\vec{c}=x\hat{i}+(x-2)\hat{j}-\hat{k}$.
Since the vectors are coplanar,their scalar triple product must be zero,i.e.,$[\vec{a} \ \vec{b} \ \vec{c}] = 0$.
This implies the determinant of the components is zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1((-1)(-1) - (2)(x-2)) - 1((1)(-1) - (2)(x)) + 1((1)(x-2) - (-1)(x)) = 0$
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$
$2x + 4 = 0$
$2x = -4$
$x = -2$

(C) For a probability distribution,the sum of all probabilities must be $1$:
$\sum P(X=x) = a + a + a + b + b + 0.3 = 1$
$3a + 2b + 0.3 = 1$
$3a + 2b = 0.7$ --- $(i)$
The mean of a random variable $X$ is given by $E(X) = \sum x_i P(x_i) = 4.2$:
$1(a) + 2(a) + 3(a) + 4(b) + 5(b) + 6(0.3) = 4.2$
$a + 2a + 3a + 4b + 5b + 1.8 = 4.2$
$6a + 9b = 4.2 - 1.8$
$6a + 9b = 2.4$
Dividing by $3$,we get:
$2a + 3b = 0.8$ --- $(ii)$
Now,solve the system of linear equations $(i)$ and $(ii)$:
Multiply $(i)$ by $3$ and $(ii)$ by $2$:
$9a + 6b = 2.1$ --- $(iii)$
$4a + 6b = 1.6$ --- $(iv)$
Subtract $(iv)$ from $(iii)$:
$5a = 0.5 \Rightarrow a = 0.1$
Substitute $a = 0.1$ into $(i)$:
$3(0.1) + 2b = 0.7$
$0.3 + 2b = 0.7$
$2b = 0.4 \Rightarrow b = 0.2$
Thus,$a = 0.1$ and $b = 0.2$.

(C) To find the variance of the random variable $X$,we use the formula: $\text{Var}(X) = E(X^2) - [E(X)]^2$.
First,we calculate the mean $E(X) = \sum P_i X_i$:
$E(X) = (0 \times 0.1) + (1 \times 0.4) + (2 \times 0.3) + (3 \times 0.2) + (4 \times 0) = 0 + 0.4 + 0.6 + 0.6 + 0 = 1.6$.
Next,we calculate $E(X^2) = \sum P_i X_i^2$:
$E(X^2) = (0^2 \times 0.1) + (1^2 \times 0.4) + (2^2 \times 0.3) + (3^2 \times 0.2) + (4^2 \times 0) = 0 + 0.4 + 1.2 + 1.8 + 0 = 3.4$.
Now,calculate the variance:
$\text{Var}(X) = 3.4 - (1.6)^2 = 3.4 - 2.56 = 0.84$.