AP EAMCET 2016 Mathematics Question Paper with Answer and Solution

(B) Given equation: $12^{4+2x^2} = (24\sqrt{3})^{3x^2-2}$
Express $24\sqrt{3}$ in terms of base $12$:
$24\sqrt{3} = 12 \times 2 \times \sqrt{3} = 12 \times \sqrt{4} \times \sqrt{3} = 12 \times \sqrt{12} = 12^1 \times 12^{1/2} = 12^{3/2}$
Substituting this into the equation:
$12^{4+2x^2} = (12^{3/2})^{3x^2-2}$
$12^{4+2x^2} = 12^{\frac{3}{2}(3x^2-2)}$
Equating the exponents:
$4+2x^2 = \frac{3}{2}(3x^2-2)$
$8+4x^2 = 9x^2-6$
$9x^2-4x^2 = 8+6$
$5x^2 = 14$
$x^2 = \frac{14}{5}$
$x = \pm \sqrt{\frac{14}{5}}$

(A) Given equation: $|x|^2 - 5|x| - 24 = 0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2 - 5t - 24 = 0$.
Factoring the quadratic: $(t - 8)(t + 3) = 0$.
This gives $t = 8$ or $t = -3$.
Since $|x| \ge 0$,we reject $t = -3$.
Thus,$|x| = 8$,which implies $x = 8$ or $x = -8$.
The roots of the equation are $8$ and $-8$.
Product of the roots: $8 \times (-8) = -64$.
Sum of the roots: $8 + (-8) = 0$.
Therefore,the product and sum are $-64$ and $0$ respectively.

(B) Let the given equation be $x^3+\frac{1}{4} x^2-\frac{1}{16} x+\frac{1}{144}=0$ $(i)$.
If the roots of the new equation are $k$ times the roots of equation $(i)$,we replace $x$ with $\frac{x}{k}$ in equation $(i)$.
$\left(\frac{x}{k}\right)^3+\frac{1}{4}\left(\frac{x}{k}\right)^2-\frac{1}{16}\left(\frac{x}{k}\right)+\frac{1}{144}=0$
Multiplying the entire equation by $k^3$ to clear the denominators:
$x^3+\frac{k}{4} x^2-\frac{k^2}{16} x+\frac{k^3}{144}=0$
For the coefficients to be integers,$k$ must be such that $4$ divides $k$,$16$ divides $k^2$,and $144$ divides $k^3$.
Checking the options:
If $k=12$,then $\frac{k}{4} = \frac{12}{4} = 3$,$\frac{k^2}{16} = \frac{144}{16} = 9$,and $\frac{k^3}{144} = \frac{1728}{144} = 12$.
Since all coefficients $(1, 3, -9, 12)$ are integers,$k=12$ is a possible value.

(A) We have $(1-x+x^2-x^3)^4 = [(1-x) + x^2(1-x)]^4 = [(1-x)(1+x^2)]^4 = (1-x)^4(1+x^2)^4$.
Expanding both terms using the binomial theorem:
$(1-x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4$.
$(1+x^2)^4 = 1 + 4x^2 + 6x^4 + 4x^6 + x^8$.
We need the coefficient of $x^4$ in the product $(1 - 4x + 6x^2 - 4x^3 + x^4)(1 + 4x^2 + 6x^4 + 4x^6 + x^8)$.
The terms that result in $x^4$ are:
$(1 \times 6x^4) + (6x^2 \times 4x^2) + (x^4 \times 1) = 6x^4 + 24x^4 + 1x^4 = 31x^4$.
Therefore,the coefficient of $x^4$ is $31$.

(A) We have,
$\frac{(1+i)^{2016}}{(1-i)^{2014}} = \frac{(1+i)^{2014} \times (1+i)^2}{(1-i)^{2014}}$
$= \left(\frac{1+i}{1-i}\right)^{2014} \times (1+i)^2$
Since $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{2i}{2} = i$
So,the expression becomes $i^{2014} \times (1+2i+i^2)$
$= i^{2014} \times (2i)$
Since $i^4 = 1$,$i^{2014} = (i^4)^{503} \times i^2 = 1^{503} \times (-1) = -1$
Therefore,$-1 \times 2i = -2i$

(D) Given that $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9 z_1 z_2 + 4 z_1 z_3 + z_2 z_3| = 12$.
We know that $|z|^2 = z \bar{z}$,which implies $\bar{z} = \frac{|z|^2}{z}$.
Substituting $\bar{z}_1 = \frac{1}{z_1}$,$\bar{z}_2 = \frac{4}{z_2}$,and $\bar{z}_3 = \frac{9}{z_3}$ is not directly helpful here.
Instead,divide the given expression by $|z_1 z_2 z_3|$:
$|9 z_1 z_2 + 4 z_1 z_3 + z_2 z_3| = 12$
$|z_1 z_2 z_3| \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right| = 12$
Since $|z_1|=1, |z_2|=2, |z_3|=3$,we have $|z_1 z_2 z_3| = 1 \times 2 \times 3 = 6$.
$6 \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right| = 12 \implies \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right| = 2$.
Using $\bar{z} = \frac{|z|^2}{z}$,we have $\bar{z}_1 = \frac{1}{z_1}$,$\bar{z}_2 = \frac{4}{z_2}$,and $\bar{z}_3 = \frac{9}{z_3}$.
Thus,$|\bar{z}_1 + \bar{z}_2 + \bar{z}_3| = 2$.
Since $|\bar{z}| = |z|$,we get $|z_1 + z_2 + z_3| = 2$.

(C) The $n$th roots of unity are the roots of the equation $z^n - 1 = 0$.
We can factorize $z^n - 1$ as:
$z^n - 1 = (z - 1)(z - z_1)(z - z_2) \ldots (z - z_{n-1})$.
We also know that $z^n - 1 = (z - 1)(z^{n-1} + z^{n-2} + \ldots + z + 1)$.
Equating the two expressions for $z^n - 1$:
$(z - 1)(z^{n-1} + z^{n-2} + \ldots + z + 1) = (z - 1)(z - z_1)(z - z_2) \ldots (z - z_{n-1})$.
Dividing both sides by $(z - 1)$ (for $z \neq 1$):
$z^{n-1} + z^{n-2} + \ldots + z + 1 = (z - z_1)(z - z_2) \ldots (z - z_{n-1})$.
Taking the limit as $z \to 1$ or simply substituting $z = 1$ in the polynomial identity:
$1^{n-1} + 1^{n-2} + \ldots + 1 + 1 = (1 - z_1)(1 - z_2) \ldots (1 - z_{n-1})$.
Since there are $n$ terms on the left side (from $z^0$ to $z^{n-1}$),the sum is $n$.
Therefore,$(1 - z_1)(1 - z_2) \ldots (1 - z_{n-1}) = n$.

(C) Let the complex number be $z = x + iy$.
Given the equation $|z+3|^2 - |z-3|^2 = 15$.
Substituting $z = x + iy$,we get:
$|x + iy + 3|^2 - |x + iy - 3|^2 = 15$
$|(x+3) + iy|^2 - |(x-3) + iy|^2 = 15$
$(x+3)^2 + y^2 - ((x-3)^2 + y^2) = 15$
$(x^2 + 6x + 9 + y^2) - (x^2 - 6x + 9 + y^2) = 15$
$x^2 + 6x + 9 + y^2 - x^2 + 6x - 9 - y^2 = 15$
$12x = 15$
$x = \frac{15}{12} = \frac{5}{4}$
Since $x = \frac{5}{4}$ represents a vertical line in the complex plane,the locus is a straight line.

(B) The expression is the product of four consecutive integers: $(n+16)(n+17)(n+18)(n+19)$.
Let $k = n+16$. Then the expression becomes $k(k+1)(k+2)(k+3)$.
This is equivalent to $4! \times \binom{k+3}{4}$.
Since $\binom{k+3}{4}$ is always an integer for any positive integer $k$,the product of any $r$ consecutive integers is always divisible by $r!$.
Here,$r = 4$,so the expression is divisible by $4! = 4 \times 3 \times 2 \times 1 = 24$.
Thus,the greatest positive integer that divides the product for all $n$ is $24$.

(C) The given digits are $2, 3, 4, 5, 6$. There are $n = 5$ digits in total.
We need to form $4$-digit numbers without repetition.
The number of such $4$-digit numbers is $P(5, 4) = \frac{5!}{(5-4)!} = 120$.
In each place (thousands,hundreds,tens,units),each digit appears an equal number of times.
The number of times each digit appears in a specific place is $\frac{120}{5} = 24$ times.
The sum of the digits is $S = 2 + 3 + 4 + 5 + 6 = 20$.
The sum of the numbers is given by:
Sum $= 24 \times S \times (10^3 + 10^2 + 10^1 + 10^0)$
Sum $= 24 \times 20 \times (1111)$
Sum $= 480 \times 1111 = 533280$.

(C) For each element in the set $A$,there are $3$ possibilities:
$1$. The element is in subset $P$.
$2$. The element is in subset $Q$.
$3$. The element is in neither $P$ nor $Q$.
Since the subsets $P$ and $Q$ must be mutually disjoint,an element cannot be in both $P$ and $Q$.
Given that set $A$ has $n = 5$ elements,each of the $5$ elements has $3$ choices.
Therefore,the total number of ways to select the subsets $P$ and $Q$ is $3^n = 3^5 = 243$.

(C) We are given that $\cosh(x) = \frac{5}{4}$.
Using the identity for hyperbolic functions,$\cosh(3x) = 4\cosh^3(x) - 3\cosh(x)$.
Substituting the given value:
$\cosh(3x) = 4\left(\frac{5}{4}\right)^3 - 3\left(\frac{5}{4}\right)$
$= 4\left(\frac{125}{64}\right) - \frac{15}{4}$
$= \frac{125}{16} - \frac{15}{4}$
$= \frac{125 - 60}{16} = \frac{65}{16}$.

(A) Given that $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ where $\alpha \in \left(0, \frac{\pi}{16}\right)$.
Expanding the expression: $1 + \tan \alpha + \tan 4 \alpha + \tan \alpha \tan 4 \alpha = 2$.
Rearranging the terms: $\tan \alpha + \tan 4 \alpha = 1 - \tan \alpha \tan 4 \alpha$.
Dividing both sides by $(1 - \tan \alpha \tan 4 \alpha)$,we get: $\frac{\tan \alpha + \tan 4 \alpha}{1 - \tan \alpha \tan 4 \alpha} = 1$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have: $\tan(\alpha + 4 \alpha) = 1$.
This simplifies to: $\tan(5 \alpha) = 1$.
Since $\tan(\frac{\pi}{4}) = 1$,we have $5 \alpha = \frac{\pi}{4} + n\pi$.
For $\alpha \in \left(0, \frac{\pi}{16}\right)$,we take $n=0$,so $5 \alpha = \frac{\pi}{4}$,which gives $\alpha = \frac{\pi}{20}$.

(A) Given,$\cos \theta = \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}$.
Applying componendo and dividendo,we get:
$\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{1 + \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}}{1 - \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}}$
$\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{1 - \cos \alpha \cos \beta + \cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta - \cos \alpha + \cos \beta}$
$\frac{1 + \cos \theta}{1 - \cos \theta} = \frac{(1 + \cos \alpha)(1 - \cos \beta)}{(1 - \cos \alpha)(1 + \cos \beta)}$
Using half-angle formulas,$\frac{1 + \cos x}{1 - \cos x} = \cot^2 \frac{x}{2}$ and $\frac{1 - \cos x}{1 + \cos x} = \tan^2 \frac{x}{2}$:
$\cot^2 \frac{\theta}{2} = \cot^2 \frac{\alpha}{2} \tan^2 \frac{\beta}{2}$
Taking the reciprocal on both sides:
$\tan^2 \frac{\theta}{2} = \tan^2 \frac{\alpha}{2} \cot^2 \frac{\beta}{2}$
Taking the square root:
$\tan \frac{\theta}{2} = \tan \frac{\alpha}{2} \cot \frac{\beta}{2}$

(D) Let the expression be $E$.
$E = \frac{1+\sin 2 \alpha}{\cos 2 \alpha \tan (\alpha - \frac{3\pi}{4})} - \frac{1}{4} \sin 2 \alpha [\cot \frac{\alpha}{2} - \tan \frac{\alpha}{2}]$
Using $\tan(\alpha - \frac{3\pi}{4}) = \frac{\tan \alpha - (-1)}{1 + \tan \alpha (-1)} = \frac{\tan \alpha + 1}{1 - \tan \alpha} = \frac{\sin \alpha + \cos \alpha}{\cos \alpha - \sin \alpha}$.
Also,$\cot \frac{\alpha}{2} - \tan \frac{\alpha}{2} = \frac{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{\cos \alpha}{\frac{1}{2} \sin \alpha} = 2 \cot \alpha$.
Substituting these:
$E = \frac{(\cos \alpha + \sin \alpha)^2}{(\cos^2 \alpha - \sin^2 \alpha) \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}} - \frac{1}{4} (2 \sin \alpha \cos \alpha) (2 \cot \alpha)$
$E = \frac{(\cos \alpha + \sin \alpha)^2}{(\cos \alpha - \sin \alpha)(\cos \alpha + \sin \alpha) \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}} - \sin^2 \alpha$
$E = \frac{(\cos \alpha + \sin \alpha)^2}{(\cos \alpha + \sin \alpha)^2} - \sin^2 \alpha$
$E = 1 - \sin^2 \alpha = \cos^2 \alpha$.
Wait,re-evaluating the simplification:
$E = 1 - \sin^2 \alpha = \cos^2 \alpha$.
Given the options,let's re-check the identity $\cot(\frac{3\pi}{2} + \frac{\alpha}{2}) = -\tan \frac{\alpha}{2}$.
Correcting the final step: $1 - \sin^2 \alpha = \cos^2 \alpha$.
If the intended answer is $\sin^2 \alpha$,there might be a sign error in the problem statement or options. Based on standard trigonometric identities,the result is $\cos^2 \alpha$.

(B) Given that $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in $GP$.
Therefore,$\cos^2 \theta = \frac{1}{6} \sin \theta \cdot \tan \theta$.
$\cos^2 \theta = \frac{\sin^2 \theta}{6 \cos \theta} \implies 6 \cos^3 \theta = \sin^2 \theta$.
Since $\sin^2 \theta = 1 - \cos^2 \theta$,we have $6 \cos^3 \theta = 1 - \cos^2 \theta$.
$6 \cos^3 \theta + \cos^2 \theta - 1 = 0$.
Let $x = \cos \theta$. Then $6x^3 + x^2 - 1 = 0$.
By inspection,$x = \frac{1}{2}$ is a root since $6(\frac{1}{8}) + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0$.
Dividing $6x^3 + x^2 - 1$ by $(2x - 1)$,we get $3x^2 + 2x + 1 = 0$.
The discriminant of $3x^2 + 2x + 1$ is $D = 2^2 - 4(3)(1) = 4 - 12 = -8 < 0$,so there are no real roots for $x$ from this quadratic.
Thus,$\cos \theta = \frac{1}{2} = \cos \frac{\pi}{3}$.
The general solution is $\theta = 2n\pi \pm \frac{\pi}{3}$.

(B) In $\triangle ABC$,we use the Napier's analogy: $\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot \frac{A}{2}$.
Given $x = \tan \left(\frac{B-C}{2}\right) \tan \frac{A}{2}$,substituting the analogy gives $x = \frac{b-c}{b+c} \cot \frac{A}{2} \tan \frac{A}{2} = \frac{b-c}{b+c}$.
Similarly,$y = \frac{c-a}{c+a}$ and $z = \frac{a-b}{a+b}$.
Now,consider $\frac{1+x}{1-x} = \frac{1 + \frac{b-c}{b+c}}{1 - \frac{b-c}{b+c}} = \frac{b+c+b-c}{b+c-b+c} = \frac{2b}{2c} = \frac{b}{c}$.
Similarly,$\frac{1+y}{1-y} = \frac{c}{a}$ and $\frac{1+z}{1-z} = \frac{a}{b}$.
Multiplying these,we get $\left(\frac{1+x}{1-x}\right) \left(\frac{1+y}{1-y}\right) \left(\frac{1+z}{1-z}\right) = \frac{b}{c} \cdot \frac{c}{a} \cdot \frac{a}{b} = 1$.
This implies $(1+x)(1+y)(1+z) = (1-x)(1-y)(1-z)$.
Expanding both sides: $1 + (x+y+z) + (xy+yz+zx) + xyz = 1 - (x+y+z) + (xy+yz+zx) - xyz$.
Simplifying,we get $2(x+y+z) = -2xyz$,which means $x+y+z = -xyz$.

(D) Given that $x_1, x_2, x_3$ and $y_1, y_2, y_3$ are in geometric progression $(GP)$ with the same common ratio $r$.
Let $x_1 = a, x_2 = ar, x_3 = ar^2$ and $y_1 = b, y_2 = br, y_3 = br^2$.
The points are $A(a, b), B(ar, br), C(ar^2, br^2)$.
To check if they are collinear,we calculate the area of the triangle formed by these points:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$= \frac{1}{2} |a(br - br^2) + ar(br^2 - b) + ar^2(b - br)|$
$= \frac{1}{2} |abr(1 - r) + abr(r^2 - 1) + abr^2(1 - r)|$
$= \frac{1}{2} |abr - abr^2 + abr^3 - abr + abr^2 - abr^3| = 0$.
Since the area of the triangle is $0$,the points are collinear.

(C) Let the original coordinates be $(x, y)$ and the new coordinates be $(X, Y)$. The rotation transformation is given by:
$x = X \cos \frac{\pi}{6} - Y \sin \frac{\pi}{6} = \frac{\sqrt{3}X - Y}{2}$
$y = X \sin \frac{\pi}{6} + Y \cos \frac{\pi}{6} = \frac{X + \sqrt{3}Y}{2}$
Substituting these into the equation $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$:
$\sqrt{3} \left( \frac{\sqrt{3}X - Y}{2} \right)^2 - 4 \left( \frac{\sqrt{3}X - Y}{2} \right) \left( \frac{X + \sqrt{3}Y}{2} \right) + \sqrt{3} \left( \frac{X + \sqrt{3}Y}{2} \right)^2 = 0$
$\frac{\sqrt{3}}{4} (3X^2 - 2\sqrt{3}XY + Y^2) - \frac{4}{4} (\sqrt{3}X^2 + 3XY - XY - \sqrt{3}Y^2) + \frac{\sqrt{3}}{4} (X^2 + 2\sqrt{3}XY + 3Y^2) = 0$
Multiplying by $4$:
$3\sqrt{3}X^2 - 6XY + \sqrt{3}Y^2 - 4\sqrt{3}X^2 - 8XY + 4\sqrt{3}Y^2 + \sqrt{3}X^2 + 6XY + 3\sqrt{3}Y^2 = 0$
Combining like terms:
$(3\sqrt{3} - 4\sqrt{3} + \sqrt{3})X^2 + (-6 - 8 + 6)XY + (\sqrt{3} + 4\sqrt{3} + 3\sqrt{3})Y^2 = 0$
$0X^2 - 8XY + 8\sqrt{3}Y^2 = 0$
Dividing by $-8$:
$XY - \sqrt{3}Y^2 = 0$
Thus,the transformed equation is $\sqrt{3}Y^2 - XY = 0$.

(A) The vertices of the triangle are $A(a, b), B(a, c)$ and $C(d, c)$.
Since the $x$-coordinate of $A$ and $B$ is $a$,the side $AB$ is vertical. Since the $y$-coordinate of $B$ and $C$ is $c$,the side $BC$ is horizontal. Thus,$\triangle ABC$ is a right-angled triangle at vertex $B(a, c)$.
The orthocentre of a right-angled triangle is the vertex at which the right angle is formed. Therefore,the orthocentre is $B(a, c)$.
The centroid $G$ of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$G = \left(\frac{a+a+d}{3}, \frac{b+c+c}{3}\right) = \left(\frac{2 a+d}{3}, \frac{b+2 c}{3}\right)$.
The mid-point of the line segment joining the centroid $G\left(\frac{2 a+d}{3}, \frac{b+2 c}{3}\right)$ and the orthocentre $H(a, c)$ is:
$M = \left(\frac{\frac{2 a+d}{3}+a}{2}, \frac{\frac{b+2 c}{3}+c}{2}\right) = \left(\frac{2 a+d+3 a}{6}, \frac{b+2 c+3 c}{6}\right) = \left(\frac{5 a+d}{6}, \frac{b+5 c}{6}\right)$.

(B) Let the points be $A(3,-2,2)$ and $B(6,-17,-4)$. Let $P(2,3,4)$ divide $AB$ in the ratio $k:1$.
Using the section formula: $(2,3,4) = \left(\frac{6k+3}{k+1}, \frac{-17k-2}{k+1}, \frac{-4k+2}{k+1}\right)$.
Equating the $x$-coordinates: $2 = \frac{6k+3}{k+1}$ $\Rightarrow 2k+2 = 6k+3$ $\Rightarrow -4k = 1$ $\Rightarrow k = -\frac{1}{4}$.
The harmonic conjugate $Q$ divides $AB$ in the ratio $-k:1$,which is $\frac{1}{4}:1$ or $1:4$ externally.
The coordinates of $Q$ are given by $\left(\frac{1(6)+4(3)}{1+4}, \frac{1(-17)+4(-2)}{1+4}, \frac{1(-4)+4(2)}{1+4}\right)$.
Calculating these: $Q = \left(\frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5}\right) = \left(\frac{18}{5}, -5, \frac{4}{5}\right)$.

(C) The formula for the image $(x', y')$ of a point $(x_1, y_1)$ with respect to the line $ax+by+c=0$ is given by $\frac{x'-x_1}{a} = \frac{y'-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2}$.
For the point $(1,1)$ and line $x+y+5=0$, we have $a=1, b=1, c=5$.
$\frac{x'-1}{1} = \frac{y'-1}{1} = -2 \frac{1(1)+1(1)+5}{1^2+1^2} = -2 \frac{7}{2} = -7$.
Thus, $x'-1 = -7 \Rightarrow x' = -6$ and $y'-1 = -7 \Rightarrow y' = -6$.
The image point is $(-6, -6)$.
The distance $D$ from the origin $(0,0)$ to the point $(-6, -6)$ is calculated as:
$D = \sqrt{(-6-0)^2 + (-6-0)^2} = \sqrt{36+36} = \sqrt{72} = 6 \sqrt{2}$.

(D) Given that the lines $x+3y-9=0$,$4x+by-2=0$,and $2x-y-4=0$ are concurrent.
The condition for concurrency is that the determinant of the coefficients must be zero:
$\begin{vmatrix} 1 & 3 & -9 \\ 4 & b & -2 \\ 2 & -1 & -4 \end{vmatrix} = 0$
Expanding the determinant:
$1(-4b - 2) - 3(-16 + 4) - 9(-4 - 2b) = 0$
$-4b - 2 + 36 + 36 + 18b = 0$
$14b + 70 = 0$ $\Rightarrow 14b = -70$ $\Rightarrow b = -5$
Now,find the point of concurrency by solving $x+3y-9=0$ and $2x-y-4=0$:
From $2x-y-4=0$,we get $y = 2x-4$.
Substituting into $x+3y-9=0$: $x + 3(2x-4) - 9 = 0$ $\Rightarrow x + 6x - 12 - 9 = 0$ $\Rightarrow 7x = 21$ $\Rightarrow x = 3$.
Then $y = 2(3) - 4 = 2$. The point of concurrency is $(3, 2)$.
The required line passes through $(b, 0) = (-5, 0)$ and $(3, 2)$.
The equation is $y - 0 = \frac{2-0}{3 - (-5)}(x - (-5))$
$y = \frac{2}{8}(x+5)$ $\Rightarrow y = \frac{1}{4}(x+5)$ $\Rightarrow 4y = x+5$ $\Rightarrow x - 4y + 5 = 0$.

(A) Let the vertices of the triangle be $A = (a \cos k, a \sin k)$,$B = (b \sin k, -b \cos k)$,and $C = (1, 0)$.
Let $G(x, y)$ be the centroid of the triangle.
The coordinates of the centroid are given by $x = \frac{a \cos k + b \sin k + 1}{3}$ and $y = \frac{a \sin k - b \cos k + 0}{3}$.
From these,we have $3x - 1 = a \cos k + b \sin k$ ... $(i)$ and $3y = a \sin k - b \cos k$ ... $(ii)$.
Squaring and adding equations $(i)$ and $(ii)$:
$(3x - 1)^2 + (3y)^2 = (a \cos k + b \sin k)^2 + (a \sin k - b \cos k)^2$.
$(3x - 1)^2 + 9y^2 = a^2(\cos^2 k + \sin^2 k) + b^2(\sin^2 k + \cos^2 k) + 2ab \cos k \sin k - 2ab \sin k \cos k$.
Since $\sin^2 k + \cos^2 k = 1$,we get $(3x - 1)^2 + 9y^2 = a^2 + b^2$.
This is equivalent to $(1 - 3x)^2 + 9y^2 = a^2 + b^2$.

(A) Given lines are $x+y=1$ and $2y^2-xy-6x^2=0$.
Factorizing the second equation: $2y^2-4xy+3xy-6x^2=0$ $\Rightarrow 2y(y-2x)+3x(y-2x)=0$ $\Rightarrow (2y+3x)(y-2x)=0$.
Thus,the sides of the triangle are $L_1: x+y=1$,$L_2: y-2x=0$,and $L_3: 2y+3x=0$.
Finding vertices:
Intersection of $L_2$ and $L_3$ is $A(0,0)$.
Intersection of $L_1$ and $L_2$: $x+2x=1 \Rightarrow x=1/3, y=2/3$,so $B(1/3, 2/3)$.
Intersection of $L_1$ and $L_3$: $x+(-3x/2)=1$ $\Rightarrow -x/2=1$ $\Rightarrow x=-2, y=3$,so $C(-2, 3)$.
Altitude from $A(0,0)$ to $BC$ $(x+y=1)$: The slope of $BC$ is $-1$,so the altitude slope is $1$. Equation: $y-0=1(x-0) \Rightarrow x-y=0$.
Altitude from $C(-2,3)$ to $AB$ $(y-2x=0)$: The slope of $AB$ is $2$,so the altitude slope is $-1/2$. Equation: $y-3 = -1/2(x+2)$ $\Rightarrow 2y-6 = -x-2$ $\Rightarrow x+2y=4$.
Solving $x-y=0$ and $x+2y=4$: $y+2y=4$ $\Rightarrow 3y=4$ $\Rightarrow y=4/3$. Thus $x=4/3$.
The orthocentre is $\left(\frac{4}{3}, \frac{4}{3}\right)$.

(NONE) The given equation is $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$.
To find the point of intersection,we factor the quadratic expression. The homogeneous part is $2x^2 - 3xy - 2y^2 = (2x + y)(x - 2y)$.
Let the lines be $(2x + y + c_1) = 0$ and $(x - 2y + c_2) = 0$.
Expanding $(2x + y + c_1)(x - 2y + c_2) = 2x^2 - 3xy - 2y^2 + (2c_2 + c_1)x + (c_2 - 2c_1)y + c_1c_2 = 0$.
Comparing coefficients with $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$,we get $2c_2 + c_1 = 10$ and $c_2 - 2c_1 = 5$.
Solving these equations: $c_1 = 0$ and $c_2 = 5$.
So the lines are $2x + y = 0$ and $x - 2y + 5 = 0$.
The intersection point is found by solving $2x + y = 0$ and $x - 2y + 5 = 0$. Substituting $y = -2x$ into the second equation gives $x - 2(-2x) + 5 = 0$,so $5x = -5$,which means $x = -1$ and $y = 2$.
The line $L$ passes through $(0, 0)$ and $(-1, 2)$.
The slope of $L$ is $m_1 = \frac{2 - 0}{-1 - 0} = -2$.
Since $L$ is perpendicular to $kx + y + 3 = 0$,the slope of the second line is $m_2 = -k$.
For perpendicular lines,$m_1 \times m_2 = -1$.
$-2 \times (-k) = -1 \Rightarrow 2k = -1 \Rightarrow k = -\frac{1}{2}$.

(D) Given the circle $x^2+y^2=9$ and the line $x+y=3$.
To find the equation of the pair of lines joining the origin to the intersection points,we homogenize the equation of the circle using the line equation.
Since $x+y=3$,we have $\frac{x+y}{3} = 1$.
Substituting this into the circle equation:
$x^2+y^2 = 9(1)^2$
$x^2+y^2 = 9\left(\frac{x+y}{3}\right)^2$
$x^2+y^2 = 9\left(\frac{(x+y)^2}{9}\right)$
$x^2+y^2 = (x+y)^2$
$x^2+y^2 = x^2+y^2+2xy$
$2xy = 0$
$xy = 0$

(B) The given equation is $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$.
Factoring the homogeneous part $2x^2 - 3xy - 2y^2$,we get $(2x + y)(x - 2y) = 0$.
Thus,the lines are $2x + y + c_1 = 0$ and $x - 2y + c_2 = 0$.
Comparing with the original equation,we find the lines are $(2x + y)(x - 2y + 5) = 0$,which gives $2x + y = 0$ and $x - 2y + 5 = 0$.
The intersection point of these two lines is found by solving $2x + y = 0$ and $x - 2y + 5 = 0$.
From $y = -2x$,substituting into the second equation: $x - 2(-2x) + 5 = 0$ $\Rightarrow 5x = -5$ $\Rightarrow x = -1$.
Then $y = -2(-1) = 2$. The intersection point is $(-1, 2)$.
The line $L$ joins the origin $(0, 0)$ to $(-1, 2)$.
The slope of $L$ is $m_1 = \frac{2 - 0}{-1 - 0} = -2$.
Since $L$ is perpendicular to $kx + y + 3 = 0$,the product of their slopes is $-1$.
The slope of $kx + y + 3 = 0$ is $m_2 = -k$.
Therefore,$m_1 \times m_2 = -1$ $\Rightarrow (-2) \times (-k) = -1$ $\Rightarrow 2k = -1$ $\Rightarrow k = -\frac{1}{2}$.

(B) The lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are conjugate with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ if $r^2(a_1a_2 + b_1b_2) = (a_1g + b_1f - c_1)(a_2g + b_2f - c_2)$.
Given circle: $x^2 + y^2 - 2x - 2y + 1 = 0$,so $g = -1, f = -1, c = 1$.
The radius squared is $r^2 = g^2 + f^2 - c = (-1)^2 + (-1)^2 - 1 = 1$.
For the lines $kx + 2y - 4 = 0$ and $5x - 2y - 4 = 0$,we have $a_1 = k, b_1 = 2, c_1 = -4$ and $a_2 = 5, b_2 = -2, c_2 = -4$.
Substituting these values into the condition:
$1(k \times 5 + 2 \times (-2)) = (k(-1) + 2(-1) - (-4))(5(-1) + (-2)(-1) - (-4))$
$5k - 4 = (-k - 2 + 4)(-5 + 2 + 4)$
$5k - 4 = (-k + 2)(1)$
$5k - 4 = -k + 2$
$6k = 6$
$k = 1$.

(C) The equation of the line at $(1,1)$ is $x+y-2=0$. The slope of this line is $-1$.
Since the normal is perpendicular to the tangent,the slope of the normal is $1$.
Thus,$\tan \theta = 1$,which gives $\theta = \frac{\pi}{4}$.
Let the centre of the circle be $(h, k)$. The coordinates of the centre are given by $h = 1 \pm r \cos \theta$ and $k = 1 \pm r \sin \theta$.
Given $r = \sqrt{2}$ and $\theta = \frac{\pi}{4}$,we have:
$h = 1 \pm \sqrt{2} \cos \frac{\pi}{4} = 1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \pm 1$.
$k = 1 \pm \sqrt{2} \sin \frac{\pi}{4} = 1 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \pm 1$.
So,the possible centres are $(2, 2)$ or $(0, 0)$.
The equations of the circles are $x^2 + y^2 = 2$ or $(x-2)^2 + (y-2)^2 = 2$.
For the circle $x^2 + y^2 - 2 = 0$,the length of the tangent from $(1, 2)$ is $\sqrt{1^2 + 2^2 - 2} = \sqrt{1+4-2} = \sqrt{3}$.
For the circle $(x-2)^2 + (y-2)^2 - 2 = 0$,the length of the tangent from $(1, 2)$ is $\sqrt{(1-2)^2 + (2-2)^2 - 2} = \sqrt{(-1)^2 + 0^2 - 2} = \sqrt{1-2} = \sqrt{-1}$,which is not possible.
Thus,the length of the tangent is $\sqrt{3}$.

(A) The given equation of the circle is $x^2+y^2-2x-2y-3=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=-1$.
The centre of the circle is $O(-g, -f) = (1, 1)$.
Since the normal at any point on a circle always passes through the centre,the line segment $PQ$ is a diameter of the circle.
Therefore,the centre $O(1, 1)$ is the midpoint of the diameter $PQ$.
Let the coordinates of $Q$ be $(x, y)$.
Using the midpoint formula,we have:
$\frac{-1+x}{2} = 1 \implies -1+x = 2 \implies x = 3$
$\frac{2+y}{2} = 1 \implies 2+y = 2 \implies y = 0$
Thus,the coordinates of $Q$ are $(3, 0)$.

(A) Given circle: $x^2+y^2+4x-6y+4=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=-3, c=4$.
Centre $O = (-g, -f) = (-2, 3)$.
Radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+9-4} = \sqrt{9} = 3$.
Distance from origin $P(0,0)$ to centre $O(-2,3)$ is $d = \sqrt{(-2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.
Let $\alpha$ be the angle between the tangent and the line joining the origin to the centre.
In the right-angled triangle formed by the origin,the point of tangency,and the centre,$\sin \alpha = \frac{r}{d} = \frac{3}{\sqrt{13}}$.
Then $\cos \alpha = \sqrt{1-\sin^2 \alpha} = \sqrt{1-\frac{9}{13}} = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$.
Thus,$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/\sqrt{13}}{2/\sqrt{13}} = \frac{3}{2}$.
The angle between the two tangents is $2\alpha = 2 \tan^{-1} \left(\frac{3}{2}\right)$.

(A) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $x^2+y^2-4x-2y+4=0$: $2g(-2) + 2f(-1) = c+4 \Rightarrow -4g-2f = c+4$ $(i)$.
For $x^2+y^2-2x-4y+1=0$: $2g(-1) + 2f(-2) = c+1 \Rightarrow -2g-4f = c+1$ $(ii)$.
For $x^2+y^2+4x+2y+1=0$: $2g(2) + 2f(1) = c+1 \Rightarrow 4g+2f = c+1$ $(iii)$.
Adding $(i)$ and $(iii)$,we get $0 = 2c+5$,so $c = -\frac{5}{2}$.
Substituting $c$ into $(iii)$: $4g+2f = -\frac{5}{2}+1 = -\frac{3}{2} \Rightarrow 8g+4f = -3$ $(iv)$.
Substituting $c$ into $(ii)$: $-2g-4f = -\frac{5}{2}+1 = -\frac{3}{2} \Rightarrow 4g+8f = 3$ $(v)$.
Adding $(iv)$ and $(v)$: $12g+12f = 0 \Rightarrow g = -f$.
Substituting $g = -f$ into $(iv)$: $-8f+4f = -3$ $\Rightarrow -4f = -3$ $\Rightarrow f = \frac{3}{4}, g = -\frac{3}{4}$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{3}{4})^2 + (\frac{3}{4})^2 - (-\frac{5}{2})} = \sqrt{\frac{9}{16} + \frac{9}{16} + \frac{40}{16}} = \sqrt{\frac{58}{16}} = \sqrt{\frac{29}{8}}$.

(B) Given equation of the parabola:
$x^2-2x+3y-2=0$
Rearranging the terms to complete the square:
$x^2-2x = -3y+2$
Adding $1$ to both sides:
$x^2-2x+1 = -3y+2+1$
$(x-1)^2 = -3y+3$
$(x-1)^2 = -3(y-1)$
Comparing this with the standard form $(x-h)^2 = 4a(y-k)$,we get $4a = -3$,so $a = -\frac{3}{4}$.
The distance between the vertex $(h, k)$ and the focus is given by $|a|$.
Distance $= |-\frac{3}{4}| = \frac{3}{4}$.

(C) Given the parabola $y^2 = 4ax$,where $4a = 5$,so $a = \frac{5}{4}$.
Let the end points of the focal chord be $(x_1, y_1) = (at_1^2, 2at_1)$ and $(x_2, y_2) = (at_2^2, 2at_2)$.
For a focal chord,the condition is $t_1t_2 = -1$.
Then $x_1x_2 = (at_1^2)(at_2^2) = a^2(t_1t_2)^2 = a^2(-1)^2 = a^2$.
Also,$y_1y_2 = (2at_1)(2at_2) = 4a^2(t_1t_2) = 4a^2(-1) = -4a^2$.
We need to find $4x_1x_2 + y_1y_2$.
Substituting the values: $4(a^2) + (-4a^2) = 4a^2 - 4a^2 = 0$.
Since $a = \frac{5}{4}$,the result is $0$.

(A) In the expansion of $(1+x)^{2n}$,the middle term is $T_{n+1} = {}^{2n}C_n x^n$. Since the middle term is the greatest term,it must be greater than or equal to its adjacent terms $T_n$ and $T_{n+2}$.
$T_{n+1} \ge T_n \implies {}^{2n}C_n x^n \ge {}^{2n}C_{n-1} x^{n-1} \implies x \ge \frac{{}^{2n}C_{n-1}}{{}^{2n}C_n} = \frac{n}{n+1}$.
$T_{n+1} \ge T_{n+2} \implies {}^{2n}C_n x^n \ge {}^{2n}C_{n+1} x^{n+1} \implies x \le \frac{{}^{2n}C_n}{{}^{2n}C_{n+1}} = \frac{n+1}{n}$.
Thus,$x \in \left[\frac{n}{n+1}, \frac{n+1}{n}\right]$. Given the options,the interval is $\left(\frac{n}{n+1}, \frac{n+1}{n}\right)$.

(C) Given expression: $f(x) = \frac{3x}{(x-2)(x-1)}$.
Using partial fractions: $\frac{3x}{(x-2)(x-1)} = \frac{A}{x-2} + \frac{B}{x-1}$.
Solving for $A$ and $B$: $3x = A(x-1) + B(x-2)$.
For $x=1$,$3 = B(-1) \Rightarrow B = -3$.
For $x=2$,$6 = A(1) \Rightarrow A = 6$.
So,$f(x) = \frac{6}{x-2} - \frac{3}{x-1} = -\frac{6}{2(1-x/2)} + \frac{3}{1-x} = -3(1-x/2)^{-1} + 3(1-x)^{-1}$.
The expansion of $(1-u)^{-1}$ is valid for $|u| < 1$.
For $-3(1-x/2)^{-1}$,we need $|x/2| < 1 \Rightarrow |x| < 2$.
For $3(1-x)^{-1}$,we need $|x| < 1$.
The expansion is valid in the intersection of these intervals: $|x| < 2$ and $|x| < 1$,which is $|x| < 1$ or $-1 < x < 1$.

(A) Given the parametric equations of the ellipse:
$x = 3 \cos \theta$ and $y = 4 \sin \theta$.
Dividing by the coefficients,we get:
$\frac{x}{3} = \cos \theta$ and $\frac{y}{4} = \sin \theta$.
Squaring and adding these equations:
$\left(\frac{x}{3}\right)^2 + \left(\frac{y}{4}\right)^2 = \cos^2 \theta + \sin^2 \theta = 1$.
Thus,the equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{16} = 1$.
Here,$a^2 = 9$ and $b^2 = 16$. Since $b^2 > a^2$,the major axis is along the $y$-axis.
The distance between the foci is given by $2be$,where $e = \sqrt{1 - \frac{a^2}{b^2}}$.
Alternatively,the distance is $2 \sqrt{b^2 - a^2} = 2 \sqrt{16 - 9} = 2 \sqrt{7}$.

(B) Given equation: $9x^2 + 25y^2 - 36x + 50y - 164 = 0$
Rearranging terms: $9(x^2 - 4x) + 25(y^2 + 2y) = 164$
Completing the square: $9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 36 + 25$
$9(x-2)^2 + 25(y+1)^2 = 225$
Dividing by $225$: $\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1$
Here,$a^2 = 25$ and $b^2 = 9$,so $a = 5$ and $b = 3$.
Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
The equations of the latus rectum are $x - h = \pm ae$.
$x - 2 = \pm 5 \times \frac{4}{5} = \pm 4$.
$x = 2 + 4 = 6$ and $x = 2 - 4 = -2$.
Thus,the equations are $x - 6 = 0$ and $x + 2 = 0$.

(B) The equation of the line is $y=mx+2$ ... $(i)$
The equation of the hyperbola is $4x^2-9y^2=36$ ... $(ii)$
Dividing $(ii)$ by $36$,we get $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
For a line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the condition is $c^2 = a^2m^2 - b^2$.
Here,$a^2=9$,$b^2=4$,and $c=2$.
Substituting these values into the condition:
$2^2 = 9m^2 - 4$
$4 = 9m^2 - 4$
$9m^2 = 8$
$m^2 = \frac{8}{9}$
$m = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3}$

(C) Let $P(t, -1/t)$ be a point on the curve $xy = -1$. The equation of the tangent at $P$ is given by $x(-1/t) + y(t) = -2$,which simplifies to $y = x/t^2 + 2/t$.
For this line to be a tangent to the parabola $y^2 = 8x$ (where $a = 2$),the condition $c = a/m$ must be satisfied.
Here,$m = 1/t^2$ and $c = 2/t$.
Substituting these into the condition: $2/t = 2 / (1/t^2) \implies 2/t = 2t^2 \implies t^3 = 1 \implies t = 1$.
Substituting $t = 1$ into the tangent equation $y = x/t^2 + 2/t$,we get $y = x + 2$.

(A) Let $l = \lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$.
We can factor the numerator as follows:
$6^x - 3^x - 2^x + 1 = 3^x(2^x - 1) - 1(2^x - 1) = (2^x - 1)(3^x - 1)$.
Substituting this back into the limit:
$l = \lim _{x \rightarrow 0} \frac{(2^x - 1)(3^x - 1)}{x^2} = \lim _{x \rightarrow 0} \left( \frac{2^x - 1}{x} \right) \times \left( \frac{3^x - 1}{x} \right)$.
Using the standard limit formula $\lim _{x \rightarrow 0} \frac{a^x - 1}{x} = \log _e a$,we get:
$l = (\log _e 2) \times (\log _e 3)$.

(B) The first $50$ even natural numbers are $2, 4, 6, \ldots, 100$.
Mean,$\bar{x} = \frac{\sum_{i=1}^{50} x_i}{50} = \frac{2(1+2+3+\ldots+50)}{50} = \frac{2 \times \frac{50 \times 51}{2}}{50} = 51$.
Variance,$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$\sum x_i^2 = 2^2 + 4^2 + \ldots + 100^2 = 4(1^2 + 2^2 + \ldots + 50^2)$.
Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$\sum x_i^2 = 4 \times \frac{50(51)(101)}{6} = \frac{2 \times 50 \times 51 \times 101}{3} = 2 \times 50 \times 17 \times 101 = 171700$.
$\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

(C) Given,number of observations,$n = 10$.
Mean,$\bar{x} = 50$.
Sum of squares of deviations,$\sum_{i=1}^{n} (x_i - \bar{x})^2 = 250$.
Variance,$\sigma^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n} = \frac{250}{10} = 25$.
Standard deviation,$\sigma = \sqrt{25} = 5$.
Coefficient of variation ($C$.$V$.) is given by the formula: $\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100$.
$\text{C.V.} = \frac{5}{50} \times 100 = 0.1 \times 100 = 10$.

(D) In $\triangle ABC$,sides $a, b, c$ are in $GP$,so $b^2 = ac$.
Given $C - A = 60^{\circ}$.
Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Substituting $a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$,we get $\sin^2 B = \sin A \sin C$.
Using $2 \sin A \sin C = \cos(A - C) - \cos(A + C)$,we have $2 \sin^2 B = \cos(60^{\circ}) - \cos(180^{\circ} - B) = \frac{1}{2} + \cos B$.
$2(1 - \cos^2 B) = \frac{1}{2} + \cos B$.
$4 - 4 \cos^2 B = 1 + 2 \cos B$.
$4 \cos^2 B + 2 \cos B - 3 = 0$.
Solving for $\cos B$ using the quadratic formula: $\cos B = \frac{-2 \pm \sqrt{4 - 4(4)(-3)}}{2(4)} = \frac{-2 \pm \sqrt{52}}{8} = \frac{-2 \pm 2\sqrt{13}}{8} = \frac{-1 \pm \sqrt{13}}{4}$.
Since $B$ is an angle in a triangle and $\cos B > 0$ is required for the progression to hold with the given conditions,$\cos B = \frac{\sqrt{13}-1}{4}$.

(B) Given curves are $y^2+x^2=a^2 \sqrt{2}$ $(i)$ and $x^2-y^2=a^2$ $(ii)$.
Adding $(i)$ and $(ii)$,we get $2x^2 = a^2(\sqrt{2}+1) \Rightarrow x^2 = \frac{a^2(\sqrt{2}+1)}{2}$.
Subtracting $(ii)$ from $(i)$,we get $2y^2 = a^2(\sqrt{2}-1) \Rightarrow y^2 = \frac{a^2(\sqrt{2}-1)}{2}$.
For curve $(i)$,differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow m_1 = \frac{dy}{dx} = -\frac{x}{y}$.
For curve $(ii)$,differentiating with respect to $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow m_2 = \frac{dy}{dx} = \frac{x}{y}$.
The angle of intersection $\theta$ is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Substituting $m_1 = -\frac{x}{y}$ and $m_2 = \frac{x}{y}$:
$\tan \theta = |\frac{-\frac{x}{y} - \frac{x}{y}}{1 + (-\frac{x}{y})(\frac{x}{y})}| = |\frac{-2x/y}{1 - x^2/y^2}| = |\frac{-2xy}{y^2 - x^2}|$.
Since $x^2 = \frac{a^2(\sqrt{2}+1)}{2}$ and $y^2 = \frac{a^2(\sqrt{2}-1)}{2}$,then $y^2 - x^2 = -a^2$.
Also $x^2 y^2 = \frac{a^4(2-1)}{4} = \frac{a^4}{4} \Rightarrow xy = \frac{a^2}{2}$.
$\tan \theta = |\frac{-2(a^2/2)}{-a^2}| = |\frac{-a^2}{-a^2}| = 1$.
Therefore,$\theta = \frac{\pi}{4}$.

(B) Given,$A+B=\frac{\pi}{3}$.
Let $y = \tan A \tan B$.
Since $B = \frac{\pi}{3} - A$,we have $y = \tan A \tan(\frac{\pi}{3} - A)$.
Using the $AM$-$GM$ inequality for positive values $\tan A$ and $\tan B$:
$\frac{\tan A + \tan B}{2} \geq \sqrt{\tan A \tan B}$.
We know that $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B} = \frac{\sin(\pi/3)}{\cos A \cos B} = \frac{\sqrt{3}/2}{\cos A \cos B}$.
Alternatively,using the identity $\tan A \tan B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B) + \cos(A+B)}$.
For a fixed sum $A+B = \frac{\pi}{3}$,the product $\tan A \tan B$ is maximized when $A=B$.
Thus,$A = B = \frac{\pi}{6}$.
Maximum value $= \tan(\frac{\pi}{6}) \tan(\frac{\pi}{6}) = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.

(C) Given that $3$ out of $6$ vertices of a regular hexagon are chosen.
Total number of ways to choose $3$ vertices out of $6$ is given by $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
In a regular hexagon,only two equilateral triangles can be formed by joining its vertices,which are $\triangle ACE$ and $\triangle BDF$.
Thus,the number of favorable outcomes is $2$.
Therefore,the probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{20} = \frac{1}{10}$.

(C) The equations of the circles are $S_1: x^2+y^2-2x-4y+c=0$ and $S_2: x^2+y^2-4x-2y+4=0$.
For $S_1$,the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2+2^2-c} = \sqrt{5-c}$.
For $S_2$,the center $C_2 = (2, 1)$ and radius $r_2 = \sqrt{2^2+1^2-4} = 1$.
The distance between the centers $d = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
The angle $\theta$ between two circles is given by $\cos \theta = \left| \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} \right|$.
Given $\theta = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2}$.
Substituting the values: $\frac{1}{2} = \left| \frac{(5-c) + 1 - 2}{2 \cdot \sqrt{5-c} \cdot 1} \right| = \left| \frac{4-c}{2\sqrt{5-c}} \right|$.
Squaring both sides: $\frac{1}{4} = \frac{(4-c)^2}{4(5-c)} \Rightarrow 5-c = (4-c)^2$.
$5-c = 16 - 8c + c^2 \Rightarrow c^2 - 7c + 11 = 0$.
Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $c = \frac{7 \pm \sqrt{49-44}}{2} = \frac{7 \pm \sqrt{5}}{2}$.

(A) Let $f(x) = x^5+3x^3+4x+30$.
Taking the derivative with respect to $x$,we get $f'(x) = 5x^4+9x^2+4$.
Since $x^4 \ge 0$ and $x^2 \ge 0$ for all $x \in \mathbb{R}$,it follows that $5x^4+9x^2+4 \ge 4 > 0$ for all $x \in \mathbb{R}$.
Since $f'(x) > 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is strictly increasing on its domain.
$A$ strictly increasing continuous function can cross the $x$-axis at most once.
Since $\lim_{x \to -\infty} f(x) = -\infty$ and $\lim_{x \to \infty} f(x) = \infty$,by the Intermediate Value Theorem,there exists exactly one real root.

(B) Given system of equations:
$x-y+2z=4$
$3x+y+4z=6$
$x+y+z=1$
Let $\Delta = \begin{vmatrix} 1 & -1 & 2 \\ 3 & 1 & 4 \\ 1 & 1 & 1 \end{vmatrix}$
$= 1(1-4) - (-1)(3-4) + 2(3-1)$
$= 1(-3) + 1(-1) + 2(2) = -3 - 1 + 4 = 0$
Since $\Delta = 0$,we check $\Delta_1, \Delta_2, \Delta_3$:
$\Delta_1 = \begin{vmatrix} 4 & -1 & 2 \\ 6 & 1 & 4 \\ 1 & 1 & 1 \end{vmatrix} = 4(1-4) + 1(6-4) + 2(6-1) = 4(-3) + 1(2) + 2(5) = -12 + 2 + 10 = 0$
$\Delta_2 = \begin{vmatrix} 1 & 4 & 2 \\ 3 & 6 & 4 \\ 1 & 1 & 1 \end{vmatrix} = 1(6-4) - 4(3-4) + 2(3-6) = 1(2) - 4(-1) + 2(-3) = 2 + 4 - 6 = 0$
$\Delta_3 = \begin{vmatrix} 1 & -1 & 4 \\ 3 & 1 & 6 \\ 1 & 1 & 1 \end{vmatrix} = 1(1-6) + 1(3-6) + 4(3-1) = 1(-5) + 1(-3) + 4(2) = -5 - 3 + 8 = 0$
Since $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$,the system has infinitely many solutions.

(C) Given,$\angle A = 90^{\circ}$.
In a $\triangle ABC$,we know the property $r_2 + r_3 = 4R \cos^2 \frac{A}{2}$.
Substituting $\angle A = 90^{\circ}$:
$r_2 + r_3 = 4R \cos^2 \left(\frac{90^{\circ}}{2}\right) = 4R \cos^2 45^{\circ}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\cos^2 45^{\circ} = \frac{1}{2}$.
Thus,$r_2 + r_3 = 4R \times \frac{1}{2} = 2R$.
Now,calculating the expression:
$\cos^{-1}\left(\frac{R}{r_2+r_3}\right) = \cos^{-1}\left(\frac{R}{2R}\right) = \cos^{-1}\left(\frac{1}{2}\right) = 60^{\circ}$.

(A) Let $A = \left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$.
Applying $C_1 \rightarrow C_1 + C_2 + C_3$,we get:
$A = \left|\begin{array}{lll}a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b\end{array}\right| = (a+b+c) \left|\begin{array}{lll}1 & b & c \\ 1 & c & a \\ 1 & a & b\end{array}\right|$.
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$A = (a+b+c) \left|\begin{array}{ccc}1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c\end{array}\right|$.
Expanding along $C_1$:
$A = (a+b+c) [(c-b)(b-c) - (a-c)(a-b)]$
$A = (a+b+c) [-(b-c)^2 - (a^2 - ab - ac + bc)]$
$A = -(a+b+c) [a^2 + b^2 + c^2 - ab - bc - ca]$
$A = -\frac{1}{2}(a+b+c) [(a-b)^2 + (b-c)^2 + (c-a)^2]$.
Since $a, b, c$ are distinct positive real numbers,$(a+b+c) > 0$ and $[(a-b)^2 + (b-c)^2 + (c-a)^2] > 0$. Therefore,$A < 0$.

(A) Given,$x=\sin \left(2 \tan ^{-1} 2\right)$ and $y=\sin \left(\frac{1}{2} \tan ^{-1} \frac{4}{3}\right)$.
For $x$,let $\tan ^{-1} 2 = \alpha$,so $\tan \alpha = 2$.
Then $x = \sin(2\alpha) = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac{2(2)}{1 + 2^2} = \frac{4}{5} = 0.8$.
For $y$,let $\tan ^{-1} \frac{4}{3} = \beta$,so $\tan \beta = \frac{4}{3}$.
Using $\cos \beta = \frac{1}{\sqrt{1 + \tan^2 \beta}} = \frac{1}{\sqrt{1 + (16/9)}} = \frac{1}{\sqrt{25/9}} = \frac{3}{5}$.
Then $y = \sin(\beta/2) = \sqrt{\frac{1 - \cos \beta}{2}} = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \approx 0.447$.
Since $0.8 > 0.447$,we have $x > y$.

(D) We have $f(x) = \sqrt{\log_{0.5} x!}$.
For $f(x)$ to be defined,the expression inside the square root must be non-negative:
$\log_{0.5} x! \geq 0$.
Since the base of the logarithm is $0.5$ (which is between $0$ and $1$),the inequality sign reverses when we remove the logarithm:
$x! \leq (0.5)^0$.
$x! \leq 1$.
The factorial $x!$ is defined for non-negative integers $x$. We check the values:
For $x = 0$,$0! = 1 \leq 1$ (True).
For $x = 1$,$1! = 1 \leq 1$ (True).
For $x = 2$,$2! = 2 \not\leq 1$ (False).
Thus,the domain is $\{0, 1\}$.

(C) We have $f(x) = |x-1| + |x-2| + |x-3|$.
For the interval $2 < x < 3$:
$|x-1| = x-1$ (since $x > 1$)
$|x-2| = x-2$ (since $x > 2$)
$|x-3| = -(x-3) = 3-x$ (since $x < 3$)
Therefore,$f(x) = (x-1) + (x-2) + (3-x) = x - 2 + 3 - 2 = x$.
In the interval $(2, 3)$,the function $f(x) = x$ is strictly increasing.
Since $f(x) = x$,for every $x \in (2, 3)$,the range is $(2, 3)$.
Thus,the function is one-one and onto,which means it is a bijection.

(B) Given function,$f(x) = \begin{cases} 1-x, & x < 1 \\ (1-x)(2-x), & 1 \leq x \leq 2 \\ 3-x, & x > 2 \end{cases}$
For $f(x)$ to be continuous at $x=1$,we check the left-hand limit,right-hand limit,and the value of the function.
$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^-} (1-x) = 1-1 = 0$.
$\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (1-x)(2-x) = (1-1)(2-1) = 0 \times 1 = 0$.
$f(1) = (1-1)(2-1) = 0$.
Since $\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x) = f(1)$,the function is continuous at $x=1$.
Now,for $f(x)$ to be continuous at $x=2$,we check the limits.
$\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} (1-x)(2-x) = (1-2)(2-2) = (-1) \times 0 = 0$.
$\lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^+} (3-x) = 3-2 = 1$.
Since $\lim_{x \rightarrow 2^-} f(x) \neq \lim_{x \rightarrow 2^+} f(x)$,the function is discontinuous at $x=2$.

(A) Given the function,$f(x) = \begin{cases} x^2 + bx + c, & x < 1 \\ x, & x \geq 1 \end{cases}$.
For $f(x)$ to be differentiable at $x = 1$,it must be continuous at $x = 1$ and the left-hand derivative must equal the right-hand derivative.
First,for continuity at $x = 1$:
$\lim_{x \to 1^-} f(x) = f(1)$
$\lim_{x \to 1^-} (x^2 + bx + c) = 1$
$1 + b + c = 1 \Rightarrow b + c = 0$ (Equation $1$).
Next,for differentiability at $x = 1$,the derivatives must match:
$f'(x) = \begin{cases} 2x + b, & x < 1 \\ 1, & x > 1 \end{cases}$
$\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x)$
$2(1) + b = 1 \Rightarrow 2 + b = 1 \Rightarrow b = -1$.
Substituting $b = -1$ into Equation $1$:
$-1 + c = 0 \Rightarrow c = 1$.
Therefore,$b - c = -1 - 1 = -2$.

(C) Given $y = \log_2(\log_2 x)$.
Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:
$y = \frac{\log_e(\log_2 x)}{\log_e 2} = \frac{\log_e(\frac{\log_e x}{\log_e 2})}{\log_e 2}$.
$y = \frac{\log_e(\log_e x) - \log_e(\log_e 2)}{\log_e 2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 2)]$.
Since $\log_e(\log_e 2)$ is a constant,its derivative is $0$.
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x)$.
$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$.
$\frac{dy}{dx} = \frac{1}{x \log_e x \log_e 2}$.

(D) Given that $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$.
This implies that $f(x)$ can be written as $f(x)=(x-a)^2 g(x)$,where $g(a) \neq 0$.
First,we find the first derivative $f^{\prime}(x)$:
$f^{\prime}(x) = 2(x-a)g(x) + (x-a)^2 g^{\prime}(x)$.
Evaluating at $x=a$,we get $f^{\prime}(a) = 2(a-a)g(a) + (a-a)^2 g^{\prime}(a) = 0$.
Next,we find the second derivative $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = 2g(x) + 2(x-a)g^{\prime}(x) + 2(x-a)g^{\prime}(x) + (x-a)^2 g^{\prime \prime}(x) = 2g(x) + 4(x-a)g^{\prime}(x) + (x-a)^2 g^{\prime \prime}(x)$.
Evaluating at $x=a$,we get $f^{\prime \prime}(a) = 2g(a) + 4(a-a)g^{\prime}(a) + (a-a)^2 g^{\prime \prime}(a) = 2g(a)$.
Since $g(a) \neq 0$,it follows that $f^{\prime \prime}(a) \neq 0$.
Thus,$f(a)=0$,$f^{\prime}(a)=0$,and $f^{\prime \prime}(a) \neq 0$.

(A) Given $f(x) = x(x+3)(x-2)$.
Expanding the expression: $f(x) = x(x^2 + x - 6) = x^3 + x^2 - 6x$.
Differentiating with respect to $x$: $f^{\prime}(x) = 3x^2 + 2x - 6$.
We are given $f^{\prime}(c) = 10$,so $3c^2 + 2c - 6 = 10$.
Rearranging the equation: $3c^2 + 2c - 16 = 0$.
Factoring the quadratic equation: $3c^2 + 8c - 6c - 16 = 0$.
$c(3c + 8) - 2(3c + 8) = 0$.
$(3c + 8)(c - 2) = 0$.
Thus,$c = -\frac{8}{3}$ or $c = 2$.
Since we require $c \in (-1, 4)$,we reject $c = -\frac{8}{3}$ as it lies outside the interval.
Therefore,$c = 2$ is the correct value.

(A) Using the formula for integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = x^3$ and $v = e^{5x}$.
$\int x^3 e^{5x} dx = x^3 \frac{e^{5x}}{5} - \int 3x^2 \frac{e^{5x}}{5} dx = \frac{x^3 e^{5x}}{5} - \frac{3}{5} \int x^2 e^{5x} dx$.
Now,$\int x^2 e^{5x} dx = x^2 \frac{e^{5x}}{5} - \int 2x \frac{e^{5x}}{5} dx = \frac{x^2 e^{5x}}{5} - \frac{2}{5} \int x e^{5x} dx$.
And $\int x e^{5x} dx = x \frac{e^{5x}}{5} - \int 1 \frac{e^{5x}}{5} dx = \frac{x e^{5x}}{5} - \frac{e^{5x}}{25}$.
Substituting these back:
$\int x^3 e^{5x} dx = \frac{x^3 e^{5x}}{5} - \frac{3}{5} [\frac{x^2 e^{5x}}{5} - \frac{2}{5} (\frac{x e^{5x}}{5} - \frac{e^{5x}}{25})] + C$.
$= \frac{x^3 e^{5x}}{5} - \frac{3 x^2 e^{5x}}{25} + \frac{6 x e^{5x}}{125} - \frac{6 e^{5x}}{625} + C$.
$= \frac{e^{5x}}{625} [125 x^3 - 75 x^2 + 30 x - 6] + C$.
Since $5^4 = 625$,we have $f(x) = 125 x^3 - 75 x^2 + 30 x - 6$.

(A) Let $I = \int \frac{x}{(x^2+2x+2)^2} dx$.
Rewrite the numerator as $x = \frac{1}{2}(2x+2) - 1$.
Then $I = \frac{1}{2} \int \frac{2x+2}{(x^2+2x+2)^2} dx - \int \frac{1}{(x^2+2x+2)^2} dx$.
For the first integral,let $u = x^2+2x+2$,so $du = (2x+2)dx$.
$\frac{1}{2} \int u^{-2} du = -\frac{1}{2u} = -\frac{1}{2(x^2+2x+2)}$.
For the second integral,let $J = \int \frac{1}{((x+1)^2+1)^2} dx$.
Put $x+1 = \tan \theta$,so $dx = \sec^2 \theta d\theta$.
$J = \int \frac{\sec^2 \theta}{\sec^4 \theta} d\theta = \int \cos^2 \theta d\theta = \int \frac{1+\cos 2\theta}{2} d\theta = \frac{1}{2}(\theta + \frac{\sin 2\theta}{2}) + C$.
Since $\tan \theta = x+1$,$\theta = \tan^{-1}(x+1)$ and $\sin 2\theta = \frac{2\tan \theta}{1+\tan^2 \theta} = \frac{2(x+1)}{x^2+2x+2}$.
So $J = \frac{1}{2} \tan^{-1}(x+1) + \frac{x+1}{2(x^2+2x+2)} + C$.
Combining these,$I = -\frac{1}{2(x^2+2x+2)} - \frac{x+1}{2(x^2+2x+2)} - \frac{1}{2} \tan^{-1}(x+1) + C = -\frac{x+2}{2(x^2+2x+2)} - \frac{1}{2} \tan^{-1}(x+1) + C$.

(C) Let $I = \int \log \left(a^2+x^2\right) d x$.
Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = \log(a^2+x^2)$ and $v = 1$.
Then $u' = \frac{2x}{a^2+x^2}$ and $\int v dx = x$.
$I = x \log(a^2+x^2) - \int \frac{2x^2}{a^2+x^2} dx + C$.
$I = x \log(a^2+x^2) - 2 \int \frac{x^2+a^2-a^2}{a^2+x^2} dx + C$.
$I = x \log(a^2+x^2) - 2 \int (1 - \frac{a^2}{a^2+x^2}) dx + C$.
$I = x \log(a^2+x^2) - 2x + 2a^2 \int \frac{1}{a^2+x^2} dx + C$.
Since $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a})$,we get:
$I = x \log(a^2+x^2) - 2x + 2a^2 (\frac{1}{a} \tan^{-1}(\frac{x}{a})) + C$.
$I = x \log(a^2+x^2) - 2x + 2a \tan^{-1}(\frac{x}{a}) + C$.
Comparing with $h(x)+C$,we have $h(x) = x \log(a^2+x^2) - 2x + 2a \tan^{-1}(\frac{x}{a})$.

(A) Let $I = \int (\log x)^5 dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then $I = \int t^5 e^t dt$.
Using integration by parts repeatedly,$\int t^n e^t dt = e^t [t^n - n t^{n-1} + n(n-1) t^{n-2} - \dots + (-1)^n n!]$.
For $n = 5$,we have $I = e^t [t^5 - 5t^4 + 20t^3 - 60t^2 + 120t - 120] + C$.
Substituting $t = \log x$ back,we get $I = x [(\log x)^5 - 5(\log x)^4 + 20(\log x)^3 - 60(\log x)^2 + 120(\log x) - 120] + C$.
Comparing this with the given expression,we identify the coefficients: $A = 1, B = -5, C = 20, D = -60, E = 120, F = -120$.
Thus,$A + B + C + D + E + F = 1 - 5 + 20 - 60 + 120 - 120 = -44$.

(D) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2x} dx$.
We can split this into two integrals: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2-\cos 2x} dx + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{2-\cos 2x} dx$.
Let $f(x) = \frac{x}{2-\cos 2x}$. Since $f(-x) = \frac{-x}{2-\cos(-2x)} = -f(x)$,$f(x)$ is an odd function. Thus,$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) dx = 0$.
Now,$I = \frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx$. Since the integrand is an even function,$I = 2 \times \frac{\pi}{4} \int_{0}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{2-\cos 2x} dx$.
Using $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we get $I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1+\tan^2 x}{2(1+\tan^2 x) - (1-\tan^2 x)} dx = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 x}{1+3\tan^2 x} dx$.
Let $t = \tan x$,then $dt = \sec^2 x dx$. Limits change from $[0, \frac{\pi}{4}]$ to $[0, 1]$.
$I = \frac{\pi}{2} \int_{0}^{1} \frac{dt}{1+3t^2} = \frac{\pi}{2} \left[ \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}t) \right]_{0}^{1} = \frac{\pi}{2\sqrt{3}} \tan^{-1}(\sqrt{3}) = \frac{\pi}{2\sqrt{3}} \times \frac{\pi}{3} = \frac{\pi^2}{6\sqrt{3}}$.

(B) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{\sqrt{n^2-r^2}}$.
We can rewrite this as $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n-1} \frac{1}{n \sqrt{1-(\frac{r}{n})^2}}$.
By the definition of definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n-1} f(\frac{r}{n}) = \int_0^1 f(x) dx$.
Here,$f(x) = \frac{1}{\sqrt{1-x^2}}$.
So,$S = \int_0^1 \frac{1}{\sqrt{1-x^2}} dx$.
Evaluating the integral,we get $S = [\sin ^{-1} x]_0^1$.
$S = \sin ^{-1}(1) - \sin ^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

(D) Given curves are $y=\frac{x^2}{4 a}$ and $y=\frac{8 a^3}{x^2+4 a^2}$.
For the point of intersection,equate the two expressions:
$\frac{x^2}{4 a} = \frac{8 a^3}{x^2+4 a^2}$
$x^2(x^2+4 a^2) = 32 a^4$
$x^4+4 a^2 x^2 - 32 a^4 = 0$
$(x^2+8 a^2)(x^2-4 a^2) = 0$
Since $x^2 = -8 a^2$ is not possible,we have $x^2 = 4 a^2$,so $x = \pm 2 a$.
The area $A$ is given by $2 \int_0^{2 a} (\frac{8 a^3}{x^2+4 a^2} - \frac{x^2}{4 a}) dx$.
$A = 2 [8 a^3 \int_0^{2 a} \frac{1}{x^2+(2 a)^2} dx - \frac{1}{4 a} \int_0^{2 a} x^2 dx]$
$A = 2 [8 a^3 \cdot \frac{1}{2 a} \tan^{-1}(\frac{x}{2 a}) |_0^{2 a} - \frac{1}{4 a} \cdot \frac{x^3}{3} |_0^{2 a}]$
$A = 2 [4 a^2 \tan^{-1}(1) - \frac{1}{4 a} \cdot \frac{8 a^3}{3}]$
$A = 2 [4 a^2 \cdot \frac{\pi}{4} - \frac{2 a^2}{3}]$
$A = 2 a^2 (\pi - \frac{2}{3}) = a^2(2 \pi - \frac{4}{3})$.

(D) Given differential equation: $(2x - 4y + 3) \frac{dy}{dx} + (x - 2y + 1) = 0$.
Rearranging,we get $\frac{dy}{dx} = -\frac{x - 2y + 1}{2x - 4y + 3} = -\frac{(x - 2y) + 1}{2(x - 2y) + 3}$.
Let $v = x - 2y$. Then $\frac{dv}{dx} = 1 - 2 \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2} (1 - \frac{dv}{dx})$.
Substituting these into the equation: $\frac{1}{2} (1 - \frac{dv}{dx}) = -\frac{v + 1}{2v + 3}$.
$1 - \frac{dv}{dx} = -\frac{2v + 2}{2v + 3} \Rightarrow \frac{dv}{dx} = 1 + \frac{2v + 2}{2v + 3} = \frac{2v + 3 + 2v + 2}{2v + 3} = \frac{4v + 5}{2v + 3}$.
Separating variables: $\frac{2v + 3}{4v + 5} dv = dx$.
Multiply by $2$: $\frac{4v + 6}{4v + 5} dv = 2 dx$.
$\int (1 + \frac{1}{4v + 5}) dv = \int 2 dx$.
$v + \frac{1}{4} \log |4v + 5| = 2x + C$.
Substitute $v = x - 2y$: $(x - 2y) + \frac{1}{4} \log |4(x - 2y) + 5| = 2x + C$.
$\frac{1}{4} \log |4(x - 2y) + 5| = x + 2y + C$.
$\log |4(x - 2y) + 5| = 4x + 8y + C'$.

(C) Given differential equation: $(1+y^2) + (x - e^{\tan^{-1} y}) \frac{dx}{dy} = 0$.
Rearranging the terms: $(x - e^{\tan^{-1} y}) \frac{dx}{dy} = -(1+y^2)$,which implies $\frac{dx}{dy} = -\frac{1+y^2}{x - e^{\tan^{-1} y}}$.
This is not standard. Let us rewrite as: $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{e^{\tan^{-1} y}}{1+y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{e^{\tan^{-1} y}}{1+y^2}$.
Integrating Factor $(IF) = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.
Substituting the values: $x e^{\tan^{-1} y} = \int \frac{e^{\tan^{-1} y}}{1+y^2} \cdot e^{\tan^{-1} y} dy + C$.
$x e^{\tan^{-1} y} = \int \frac{e^{2 \tan^{-1} y}}{1+y^2} dy + C$.
Let $u = \tan^{-1} y$,then $du = \frac{1}{1+y^2} dy$.
$x e^{\tan^{-1} y} = \int e^{2u} du + C = \frac{1}{2} e^{2u} + C = \frac{1}{2} e^{2 \tan^{-1} y} + C$.
Multiplying by $2$: $2 x e^{\tan^{-1} y} = e^{2 \tan^{-1} y} + C$.

(D) Given: $|a|=3, |b|=4$ and the angle $\theta = 120^{\circ}$.
We know that $|4a+3b|^2 = (4a+3b) \cdot (4a+3b)$.
$= 16|a|^2 + 9|b|^2 + 24(a \cdot b)$.
$= 16|a|^2 + 9|b|^2 + 24|a||b| \cos \theta$.
Substitute the values: $16(3)^2 + 9(4)^2 + 24(3)(4) \cos(120^{\circ})$.
$= 16(9) + 9(16) + 288 \times (-1/2)$.
$= 144 + 144 - 144 = 144$.
Therefore,$|4a+3b| = \sqrt{144} = 12$.

(A) Given vectors are $a=2 \hat{i}+3 \hat{j}-5 \hat{k}$ and $b=m \hat{i}+n \hat{j}+12 \hat{k}$.
Since $a \times b = 0$,the vectors $a$ and $b$ are collinear (parallel).
For two parallel vectors $a = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $b = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$,the components are proportional:
$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$
Substituting the given values:
$\frac{2}{m} = \frac{3}{n} = \frac{-5}{12}$
From $\frac{2}{m} = \frac{-5}{12}$,we get $m = \frac{2 \times 12}{-5} = -\frac{24}{5}$.
From $\frac{3}{n} = \frac{-5}{12}$,we get $n = \frac{3 \times 12}{-5} = -\frac{36}{5}$.
Therefore,$(m, n) = \left(-\frac{24}{5}, -\frac{36}{5}\right)$.

(C) Given the equation: $a(\alpha \times \beta)+b(\beta \times \gamma)+c(\gamma \times \alpha)=0$.
Taking the dot product of the entire equation with vector $\gamma$:
$a(\alpha \times \beta) \cdot \gamma + b(\beta \times \gamma) \cdot \gamma + c(\gamma \times \alpha) \cdot \gamma = 0$.
Since the scalar triple product of vectors with repeated components is zero,we have $(\beta \times \gamma) \cdot \gamma = 0$ and $(\gamma \times \alpha) \cdot \gamma = 0$.
This simplifies the equation to: $a(\alpha \times \beta) \cdot \gamma = 0$,which is $a[\alpha \beta \gamma] = 0$.
Similarly,taking the dot product with $\alpha$ gives $b[\beta \gamma \alpha] = 0$,and with $\beta$ gives $c[\gamma \alpha \beta] = 0$.
Since at least one of $a, b, c$ is non-zero,the scalar triple product $[\alpha \beta \gamma]$ must be $0$.
Therefore,the vectors $\alpha, \beta, \gamma$ are coplanar.

(C) The given equation of the plane is $56x + 4y + 9z = 2016$.
Dividing by $2016$,we get the intercept form:
$\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$.
The plane meets the coordinate axes at $A(36, 0, 0)$,$B(0, 504, 0)$,and $C(0, 0, 224)$.
The centroid $G$ of $\triangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
$G = \left(\frac{36+0+0}{3}, \frac{0+504+0}{3}, \frac{0+0+224}{3}\right) = \left(12, 168, \frac{224}{3}\right)$.

(B) Given equations are:
$r = 3p + 4q$ $(i)$
$2r = p - 3q$ $(ii)$
From equation $(ii)$,we have $p = 2r + 3q$.
Substituting this value of $p$ in equation $(i)$:
$r = 3(2r + 3q) + 4q$
$r = 6r + 9q + 4q$
$r - 6r = 13q$
$-5r = 13q$
$r = -\frac{13}{5}q$
Since the scalar multiplier is negative,$r$ and $q$ have opposite directions.
Taking the magnitude on both sides:
$|r| = |-\frac{13}{5}q| = \frac{13}{5}|q| = 2.6|q|$
Since $2.6 > 2$,we have $|r| > 2|q|$.
Thus,$|r| > 2|q|$ and $r, q$ have opposite directions.

(B) Let the direction cosines of the line be $(l, m, n)$,where $l^2 + m^2 + n^2 = 1$.
The four diagonals of a cube are along the vectors $(\pm 1, \pm 1, \pm 1)$.
The unit vectors along the four diagonals are $\vec{d_1} = \frac{1}{\sqrt{3}}(1, 1, 1)$,$\vec{d_2} = \frac{1}{\sqrt{3}}(-1, 1, 1)$,$\vec{d_3} = \frac{1}{\sqrt{3}}(1, -1, 1)$,and $\vec{d_4} = \frac{1}{\sqrt{3}}(1, 1, -1)$.
The cosine of the angle $\theta$ between the line and a diagonal vector $\vec{d}$ is given by $|l \cdot d_x + m \cdot d_y + n \cdot d_z|$.
Thus,$\cos \alpha = \frac{1}{\sqrt{3}}|l+m+n|$,$\cos \beta = \frac{1}{\sqrt{3}}|-l+m+n|$,$\cos \gamma = \frac{1}{\sqrt{3}}|l-m+n|$,and $\cos \delta = \frac{1}{\sqrt{3}}|l+m-n|$.
Squaring these,we get $\cos^2 \alpha = \frac{1}{3}(l+m+n)^2$,$\cos^2 \beta = \frac{1}{3}(-l+m+n)^2$,$\cos^2 \gamma = \frac{1}{3}(l-m+n)^2$,and $\cos^2 \delta = \frac{1}{3}(l+m-n)^2$.
Summing these: $\sum \cos^2 \alpha = \frac{1}{3} [ (l+m+n)^2 + (-l+m+n)^2 + (l-m+n)^2 + (l+m-n)^2 ]$.
Expanding the squares: $\sum \cos^2 \alpha = \frac{1}{3} [ 4(l^2+m^2+n^2) ] = \frac{4}{3}(1) = \frac{4}{3}$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we have $\sum \sin^2 \alpha = 4 - \sum \cos^2 \alpha = 4 - \frac{4}{3} = \frac{8}{3}$.

(D) The given vector equation is $\vec{r} = (1+\lambda-\mu) \hat{i} + (2-\lambda) \hat{j} + (3-2 \lambda+2 \mu) \hat{k}$.
Rearranging the terms,we get:
$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} - 2\hat{k}) + \mu(-\hat{i} + 2\hat{k})$.
This is in the form $\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}$,where $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} - \hat{j} - 2\hat{k}$,and $\vec{c} = -\hat{i} + 2\hat{k}$.
The normal vector $\vec{n}$ to the plane is given by $\vec{n} = \vec{b} \times \vec{c}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -2 \\ -1 & 0 & 2 \end{vmatrix} = \hat{i}(-2 - 0) - \hat{j}(2 - 2) + \hat{k}(0 - 1) = -2\hat{i} - \hat{k}$.
The equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which implies $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
$\vec{a} \cdot \vec{n} = (\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (-2\hat{i} - \hat{k}) = (1)(-2) + (2)(0) + (3)(-1) = -2 - 3 = -5$.
So,$\vec{r} \cdot (-2\hat{i} - \hat{k}) = -5$,or $\vec{r} \cdot (2\hat{i} + \hat{k}) = 5$.
Substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$,we get $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + \hat{k}) = 5$,which simplifies to $2x + z = 5$.

(A) Let $A$ be the event that $A$ speaks the truth and $B$ be the event that $B$ speaks the truth.
Given probabilities are:
$P(A) = 75 \% = \frac{75}{100} = \frac{3}{4}$
$P(B) = 80 \% = \frac{80}{100} = \frac{4}{5}$
Their statements do not match if one speaks the truth and the other lies.
This can happen in two ways: ($A$ speaks the truth $AND$ $B$ lies) $OR$ ($A$ lies $AND$ $B$ speaks the truth).
Required probability $= P(A) \times P(\bar{B}) + P(\bar{A}) \times P(B)$
$= P(A) \times (1 - P(B)) + (1 - P(A)) \times P(B)$
$= \frac{3}{4} \times (1 - \frac{4}{5}) + (1 - \frac{3}{4}) \times \frac{4}{5}$
$= \frac{3}{4} \times \frac{1}{5} + \frac{1}{4} \times \frac{4}{5}$
$= \frac{3}{20} + \frac{4}{20} = \frac{7}{20}$

(A) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k !}$.
Here,$\lambda$ is the mean number of accidents per day.
Given that $10$ accidents occur in $50$ days,the average number of accidents per day is $\lambda = \frac{10}{50} = 0.2$.
We need to find the probability that three or more accidents occur in a day,which is $P(X \geq 3)$.
$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) + \dots$
This can be written as $\sum_{k=3}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}$ with $\lambda = 0.2$.
Thus,the correct expression is $\sum_{k=3}^{\infty} \frac{e^{-0.2} (0.2)^k}{k !}$.

(D) For a binomial distribution,the mean is given by $\mu = np = 4$ ... $(i)$
The variance is given by $\sigma^2 = npq = 2$ ... (ii)
Dividing equation (ii) by equation $(i)$,we get:
$\frac{npq}{np} = \frac{2}{4} \implies q = \frac{1}{2}$
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$
Substituting $p = \frac{1}{2}$ in equation $(i)$:
$n \times \frac{1}{2} = 4 \implies n = 8$
The probability of $X$ successes in a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$
For $k = 2$:
$P(X=2) = { }^8 C_2 \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^6 = 28 \times \left(\frac{1}{2}\right)^8 = \frac{28}{256} = \frac{7}{64}$