TS EAMCET 2022 Mathematics Question Paper with Answer and Solution

(D) Given $a=5, b=6, c=9$. The circumradius $R = SB = \frac{27}{4 \sqrt{2}}$.
We know that $\sin 2C = 2 \sin C \cos C$.
Using the Sine Rule,$\sin C = \frac{c}{2R}$.
Using the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting these values:
$\sin 2C = 2 \times \left( \frac{c}{2R} \right) \times \left( \frac{a^2 + b^2 - c^2}{2ab} \right) = \frac{c}{R} \times \frac{a^2 + b^2 - c^2}{2ab}$.
$\sin 2C = \frac{9}{\frac{27}{4 \sqrt{2}}} \times \frac{5^2 + 6^2 - 9^2}{2 \times 5 \times 6}$.
$\sin 2C = \left( 9 \times \frac{4 \sqrt{2}}{27} \right) \times \frac{25 + 36 - 81}{60}$.
$\sin 2C = \left( \frac{4 \sqrt{2}}{3} \right) \times \left( \frac{-20}{60} \right) = \frac{4 \sqrt{2}}{3} \times \left( -\frac{1}{3} \right) = -\frac{4 \sqrt{2}}{9}$.

(D) First,calculate the semi-perimeter $S$:
$S = \frac{7+10+11}{2} = 14$
Calculate the area $\Delta$ using Heron's formula:
$\Delta = \sqrt{14(14-7)(14-10)(14-11)} = \sqrt{14 \times 7 \times 4 \times 3} = \sqrt{1176} = 14\sqrt{6}$
We have the formulas $R = \frac{abc}{4\Delta}$ and $r = \frac{\Delta}{S}$.
Therefore,$\frac{R}{r} = \frac{abc}{4\Delta} \times \frac{S}{\Delta} = \frac{abc \times S}{4\Delta^2}$.
Substitute the values:
$\frac{R}{r} = \frac{7 \times 10 \times 11 \times 14}{4 \times (14 \times 7 \times 4 \times 3)} = \frac{770 \times 14}{4 \times 1176} = \frac{10780}{4704} = \frac{55}{24}$.

(B) For set $A$,we require $x^2-8x+15 \geq 0$.
$(x-3)(x-5) \geq 0$,which gives $x \in (-\infty, 3] \cup [5, \infty)$.
For set $B$,we solve $\frac{x-3}{2x-5} - \frac{x-6}{2x-11} < 0$.
$\frac{(x-3)(2x-11) - (x-6)(2x-5)}{(2x-5)(2x-11)} < 0$.
$\frac{(2x^2-17x+33) - (2x^2-17x+30)}{(2x-5)(2x-11)} < 0$.
$\frac{3}{(2x-5)(2x-11)} < 0$.
This implies $(2x-5)(2x-11) < 0$,so $x \in \left(\frac{5}{2}, \frac{11}{2}\right)$.
Finally,$A \cap B = ((-\infty, 3] \cup [5, \infty)) \cap \left(\frac{5}{2}, \frac{11}{2}\right) = \left(\frac{5}{2}, 3\right] \cup \left[5, \frac{11}{2}\right)$.

(C) Given $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$.
Multiplying both sides by $(x+1)(2x^2+3)$,we get $3x+5 = A(2x^2+3) + (Bx+C)(x+1)$.
Setting $x = -1$: $3(-1)+5 = A(2(-1)^2+3) + 0 \implies 2 = 5A \implies A = \frac{2}{5}$.
Comparing coefficients of $x^2$: $0 = 2A + B \implies B = -2A = -2(\frac{2}{5}) = -\frac{4}{5}$.
Comparing constant terms: $5 = 3A + C \implies C = 5 - 3(\frac{2}{5}) = 5 - \frac{6}{5} = \frac{19}{5}$.
Thus,$f(x) = \frac{2}{5}x^3 - \frac{4}{5}x^2 + 7x + \frac{19}{5}$.
$f'(x) = \frac{6}{5}x^2 - \frac{8}{5}x + 7$.
$f'(-2) = \frac{6}{5}(4) - \frac{8}{5}(-2) + 7 = \frac{24}{5} + \frac{16}{5} + 7 = \frac{40}{5} + 7 = 8 + 7 = 15$.
Finally,$5C - f'(-2) = 5(\frac{19}{5}) - 15 = 19 - 15 = 4$.

(A) Given,$\frac{42-13x}{x^2+x-6} = \frac{A}{lx+m} + \frac{B}{px+q}$.
Factorizing the denominator: $x^2+x-6 = (x+3)(x-2)$.
So,$\frac{42-13x}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$.
Using partial fractions: $42-13x = A(x-2) + B(x+3)$.
For $x=2$: $42-26 = 5B \implies 16 = 5B \implies B = \frac{16}{5}$.
For $x=-3$: $42+39 = -5A \implies 81 = -5A \implies A = -\frac{81}{5}$.
Comparing with $\frac{A}{lx+m} + \frac{B}{px+q}$,we have $l=1, m=3, p=1, q=-2$ (satisfying $lm=3 > 0$ and $pq=-2 < 0$).
Then,$\frac{Alp}{Bmq} = \frac{(-\frac{81}{5}) \times 1 \times 1}{(\frac{16}{5}) \times 3 \times (-2)} = \frac{-81/5}{-96/5} = \frac{81}{96} = \frac{27}{32}$.

(C) The expression $2x^2 + 4x + 5 = 2(x^2 + 2x + 1) + 3 = 2(x+1)^2 + 3$. Since $(x+1)^2 \geq 0$,the minimum value is $3$ $(IV)$.
$(B)$ Let $y = \frac{x^2 + 4x + 1}{x^2 + x + 1}$. Then $y(x^2 + x + 1) = x^2 + 4x + 1 \implies (y-1)x^2 + (y-4)x + (y-1) = 0$. For $x$ to be real,$D \geq 0 \implies (y-4)^2 - 4(y-1)^2 \geq 0 \implies (y-4-2y+2)(y-4+2y-2) \geq 0 \implies (-y-2)(3y-6) \geq 0 \implies (y+2)(y-2) \leq 0 \implies -2 \leq y \leq 2$. The maximum value is $2$ $(III)$.
$(C)$ and $(D)$ Given $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$.
$x^2 + 1 \leq 3x^2 - 5x + 6 \implies 2x^2 - 5x + 5 \geq 0$ (Always true as $D = 25 - 40 < 0$).
$3x^2 - 5x + 6 \leq 2x^2 + 2 \implies x^2 - 5x + 4 \leq 0 \implies (x-1)(x-4) \leq 0 \implies 1 \leq x \leq 4$.
Thus $a = 1$ $(II)$ and $b = 4$ $(V)$.
Matching: $(A)$ $\rightarrow IV, (B)$ $\rightarrow III, (C)$ $\rightarrow V, (D)$ $\rightarrow II$. Correct option is $(C)$.

(D) For the function $f(x) = \log_2(2^x - 2) + \sqrt{1 - x}$ to be defined,both the logarithmic and square root terms must be real.
$1$. For the square root,we require $1 - x \geq 0$,which implies $x \leq 1$.
$2$. For the logarithm,we require the argument to be positive: $2^x - 2 > 0$,which implies $2^x > 2^1$,so $x > 1$.
Combining these two conditions,we need $x \leq 1$ $AND$ $x > 1$.
Since there is no real number $x$ that satisfies both $x \leq 1$ and $x > 1$ simultaneously,the set of such values is the empty set,denoted by $\phi$.

(C) Let $y = \frac{x^2-x+2}{x^2+x-2}$.
Rearranging the terms,we get $y(x^2+x-2) = x^2-x+2$.
$yx^2 + yx - 2y = x^2 - x + 2$.
$(y-1)x^2 + (y+1)x - (2y+2) = 0$.
For $x$ to be a real number,the discriminant $D \ge 0$.
$D = (y+1)^2 - 4(y-1)(-2y-2) \ge 0$.
$(y+1)^2 + 8(y-1)(y+1) \ge 0$.
$(y+1)[(y+1) + 8(y-1)] \ge 0$.
$(y+1)(y+1+8y-8) \ge 0$.
$(y+1)(9y-7) \ge 0$.
The critical points are $y = -1$ and $y = \frac{7}{9}$.
Testing the intervals,the inequality holds for $y \in (-\infty, -1] \cup \left[\frac{7}{9}, \infty\right)$.
Thus,the correct option is $C$.

(A) Let $h(x) = f(x) - g(x) = (A - L)x^2 + (B - M)x - N$.
Given:
$h(2) = 4(A - L) + 2(B - M) - N = 1$ ... $(i)$
$h(3) = 9(A - L) + 3(B - M) - N = 4$ ... $(ii)$
$h(4) = 16(A - L) + 4(B - M) - N = 9$ ... $(iii)$
Subtracting $(i)$ from $(ii)$:
$5(A - L) + (B - M) = 3$ ... $(iv)$
Subtracting $(ii)$ from $(iii)$:
$7(A - L) + (B - M) = 5$ ... $(v)$
Subtracting $(iv)$ from $(v)$:
$2(A - L) = 2 \Rightarrow A - L = 1$.
Substituting $A - L = 1$ in $(iv)$:
$5(1) + (B - M) = 3 \Rightarrow B - M = -2$.
Substituting $A - L = 1$ and $B - M = -2$ in $(i)$:
$4(1) + 2(-2) - N = 1$ $\Rightarrow 4 - 4 - N = 1$ $\Rightarrow N = -1$.
Thus,$h(x) = (1)x^2 + (-2)x - (-1) = x^2 - 2x + 1$.
Setting $h(x) = 0$:
$x^2 - 2x + 1 = 0$ $\Rightarrow (x - 1)^2 = 0$ $\Rightarrow x = 1$.

(B) We know the expansion for $\cosh(A \pm B) = \cosh A \cosh B \pm \sinh A \sinh B$.
Applying this to the given expression:
$\cosh(x + iy) = \cosh x \cosh(iy) + \sinh x \sinh(iy)$
$\cosh(x - iy) = \cosh x \cosh(iy) - \sinh x \sinh(iy)$
Subtracting the two equations:
$\cosh(x + iy) - \cosh(x - iy) = (\cosh x \cosh(iy) + \sinh x \sinh(iy)) - (\cosh x \cosh(iy) - \sinh x \sinh(iy))$
$= 2 \sinh x \sinh(iy)$
Since $\sinh(iy) = i \sin y$,we get:
$= 2i \sinh x \sin y$.

(D) Given the two families of curves $y^2=4ax$ and $x^2+\frac{y^2}{2}=c^2$.
For the first family $y^2=4ax$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$. Substituting $4a = \frac{y^2}{x}$,we get $2y \frac{dy}{dx} = \frac{y^2}{x}$,which implies $m_1 = \frac{dy}{dx} = \frac{y}{2x}$.
For the second family $x^2+\frac{y^2}{2}=c^2$,differentiating with respect to $x$ gives $2x + y \frac{dy}{dx} = 0$,which implies $m_2 = \frac{dy}{dx} = -\frac{2x}{y}$.
The product of the slopes is $m_1 \times m_2 = \left(\frac{y}{2x}\right) \times \left(-\frac{2x}{y}\right) = -1$.
Since the product of the slopes is $-1$,the curves intersect at a right angle. Thus,the angle between the curves is $\frac{\pi}{2}$.

(C) First,find the points of intersection of the curves $y^2=4x$ and $x^2+y^2=5$. Substituting $y^2=4x$ into the second equation gives $x^2+4x-5=0$.
Factoring the quadratic equation,we get $(x+5)(x-1)=0$,so $x=1$ or $x=-5$. Since $y^2=4x$,$x$ must be non-negative,so $x=1$.
For $x=1$,$y^2=4(1)=4$,which gives $y=2$ or $y=-2$. We consider the point $(1, 2)$.
For the curve $y^2=4x$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(1, 2)$,$m_1 = \frac{2}{2} = 1$.
For the curve $x^2+y^2=5$,differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$. At $(1, 2)$,$m_2 = -\frac{1}{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{1 - (-1/2)}{1 + (1)(-1/2)} \right| = \left| \frac{3/2}{1/2} \right| = 3$.
Thus,$|\tan \theta| = 3$.

(C) The given curve is $4x^2 + 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.
This is an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a^2 = 9$ $(a = 3)$ and $b^2 = 4$ $(b = 2)$.
The parametric coordinates of any point on the ellipse are $(a \cos \theta, b \sin \theta) = (3 \cos \theta, 2 \sin \theta)$.
The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $\theta$ is given by $ax \sec \theta - by \operatorname{cosec} \theta = a^2 - b^2$.
Substituting $a = 3$,$b = 2$,and $\theta = \frac{7\pi}{4}$:
$3x \sec(\frac{7\pi}{4}) - 2y \operatorname{cosec}(\frac{7\pi}{4}) = 9 - 4$
Since $\sec(\frac{7\pi}{4}) = \sec(2\pi - \frac{\pi}{4}) = \sec(\frac{\pi}{4}) = \sqrt{2}$ and $\operatorname{cosec}(\frac{7\pi}{4}) = \operatorname{cosec}(2\pi - \frac{\pi}{4}) = -\operatorname{cosec}(\frac{\pi}{4}) = -\sqrt{2}$:
$3x(\sqrt{2}) - 2y(-\sqrt{2}) = 5$
$3\sqrt{2}x + 2\sqrt{2}y = 5$
Thus,the equation of the normal is $3\sqrt{2}x + 2\sqrt{2}y - 5 = 0$.

(A) Given curves are $x^2+y^2=4$ and $y^2=3x$. Substituting $y^2=3x$ into the first equation:
$x^2+3x-4=0$
$(x+4)(x-1)=0$
Since $x$ must be non-negative for $y^2=3x$,we have $x=1$.
For $x=1$,$y^2=3$,so $y=\sqrt{3}$ (considering the point of intersection in the first quadrant).
Now,differentiate both curves with respect to $x$:
For $x^2+y^2=4$,$2x+2y\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$. At $(1, \sqrt{3})$,$m_1 = -\frac{1}{\sqrt{3}}$.
For $y^2=3x$,$2y\frac{dy}{dx}=3 \Rightarrow \frac{dy}{dx} = \frac{3}{2y}$. At $(1, \sqrt{3})$,$m_2 = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$.
$\tan \theta = \left|\frac{-\frac{1}{\sqrt{3}} - \frac{\sqrt{3}}{2}}{1 + (-\frac{1}{\sqrt{3}})(\frac{\sqrt{3}}{2})}\right| = \left|\frac{-\frac{2+3}{2\sqrt{3}}}{1-\frac{1}{2}}\right| = \left|\frac{-\frac{5}{2\sqrt{3}}}{\frac{1}{2}}\right| = \frac{5}{\sqrt{3}}$.

(C) Let $f(x) = -2x^2 + 3x + 1$.
To find the extreme value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = -4x + 3$.
Setting $f'(x) = 0$,we get $-4x + 3 = 0$,which implies $x = \frac{3}{4}$.
The extreme value $k$ is $f\left(\frac{3}{4}\right) = -2\left(\frac{3}{4}\right)^2 + 3\left(\frac{3}{4}\right) + 1 = -2\left(\frac{9}{16}\right) + \frac{9}{4} + 1 = -\frac{9}{8} + \frac{18}{8} + \frac{8}{8} = \frac{17}{8}$.
Now,we need to find the set of $x$ such that $kx^2 + 2x + 1 > 0$,where $k = \frac{17}{8}$.
Substituting $k$,we get $\frac{17}{8}x^2 + 2x + 1 > 0$.
Multiplying by $8$,we get $17x^2 + 16x + 8 > 0$.
The discriminant $D$ of this quadratic is $D = b^2 - 4ac = (16)^2 - 4(17)(8) = 256 - 544 = -288$.
Since $D < 0$ and the coefficient of $x^2$ $(17)$ is positive,the quadratic $17x^2 + 16x + 8$ is always positive for all real $x$.
Thus,the solution set is $(-\infty, \infty)$.

(D) Given the partial fraction decomposition: $\frac{x^2+3}{(x^2+1)(x^2+2)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}$.
Multiplying both sides by $(x^2+1)(x^2+2)$,we get: $x^2+3=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)$.
Expanding the right side: $x^2+3=Ax^3+2Ax+Bx^2+2B+Cx^3+Cx+Dx^2+D$.
Grouping the terms by powers of $x$: $x^2+3=(A+C)x^3+(B+D)x^2+(2A+C)x+(2B+D)$.
Comparing the coefficients on both sides:
$A+C=0$ (coefficient of $x^3$)
$B+D=1$ (coefficient of $x^2$)
$2A+C=0$ (coefficient of $x$)
$2B+D=3$ (constant term)
From $A+C=0$ and $2A+C=0$,subtracting the equations gives $A=0$,which implies $C=0$.
From $B+D=1$ and $2B+D=3$,subtracting the equations gives $B=2$. Substituting $B=2$ into $B+D=1$ gives $D=-1$.
Thus,$A=0, B=2, C=0, D=-1$.
Therefore,$A+B+C+D=0+2+0+(-1)=1$.

(C) Given the partial fraction decomposition: $\frac{x^2-3x+2}{(x-4)(x-3)^2} = \frac{A}{x-4} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$.
Multiply both sides by $(x-4)(x-3)^2$ to get: $x^2-3x+2 = A(x-3)^2 + B(x-3)(x-4) + C(x-4)$.
To find $A$,set $x=4$: $4^2 - 3(4) + 2 = A(4-3)^2 \Rightarrow 16 - 12 + 2 = A(1)^2 \Rightarrow A = 6$.
To find $C$,set $x=3$: $3^2 - 3(3) + 2 = C(3-4) \Rightarrow 9 - 9 + 2 = C(-1) \Rightarrow 2 = -C \Rightarrow C = -2$.
To find $B$,compare the coefficients of $x^2$ on both sides: $1 = A + B$. Since $A=6$,we have $1 = 6 + B \Rightarrow B = -5$.
Finally,calculate $A+B+C = 6 + (-5) + (-2) = 6 - 7 = -1$.

(D) Given the partial fraction decomposition: $\frac{x^2-2}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$.
Let $y = x^2$. The expression becomes $\frac{y-2}{(y+1)(y+3)} = \frac{B}{y+1} + \frac{D}{y+3}$ (since the numerator terms $Ax$ and $Cx$ must be $0$ for the equality to hold for all $x$).
To find $D$,we use the cover-up method by multiplying both sides by $(y+3)$ and setting $y = -3$:
$D = \left[ \frac{y-2}{y+1} \right]_{y=-3} = \frac{-3-2}{-3+1} = \frac{-5}{-2} = \frac{5}{2}$.

(C) Let $y = \frac{2x+1}{(x+1)^2(x-2)}$. Using partial fractions,we write: $\frac{2x+1}{(x+1)^2(x-2)} = \frac{a}{x-2} + \frac{b}{x+1} + \frac{c}{(x+1)^2}$.
Solving for constants: $2x+1 = a(x+1)^2 + b(x+1)(x-2) + c(x-2)$.
For $x=2$: $5 = a(3)^2 \Rightarrow a = \frac{5}{9}$.
For $x=-1$: $-1 = c(-3) \Rightarrow c = \frac{1}{3}$.
Comparing coefficients of $x^2$: $0 = a + b \Rightarrow b = -a = -\frac{5}{9}$.
So,$y = \frac{5/9}{x-2} - \frac{5/9}{x+1} + \frac{1/3}{(x+1)^2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -\frac{5/9}{(x-2)^2} + \frac{5/9}{(x+1)^2} - \frac{2/3}{(x+1)^3}$.
Comparing this with $\frac{A}{(x-2)^2} + \frac{B}{(x+1)^3} + \frac{C}{(x+1)^2}$,we find $A = -\frac{5}{9}$,$B = -\frac{2}{3}$,and $C = \frac{5}{9}$.
Therefore,$A+B+C = -\frac{5}{9} - \frac{2}{3} + \frac{5}{9} = -\frac{2}{3}$.

(D) Given the partial fraction decomposition: $\frac{3 x^2+ax+3}{(2 x+3)(x^2+2)}=\frac{3}{2 x+3}+\frac{Bx+C}{x^2+2}$
Equating the numerators after taking the common denominator:
$3 x^2+ax+3 = 3(x^2+2) + (Bx+C)(2x+3)$
$3 x^2+ax+3 = 3x^2 + 6 + 2Bx^2 + 3Bx + 2Cx + 3C$
$3 x^2+ax+3 = (3+2B)x^2 + (3B+2C)x + (6+3C)$
Comparing the coefficients of $x^2$,$x$,and the constant term:
$1) \ 3+2B = 3 \implies 2B = 0 \implies B = 0$
$2) \ 3B+2C = a \implies 3(0)+2C = a \implies 2C = a \implies C = \frac{a}{2}$
$3) \ 6+3C = 3 \implies 3C = -3 \implies C = -1$
Substituting $C = -1$ into $C = \frac{a}{2}$:
$-1 = \frac{a}{2} \implies a = -2$
Finally,calculating $a(B+C)$:
$a(B+C) = -2(0 + (-1)) = -2(-1) = 2$
Thus,the correct option is $D$.

(C) Given $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$.
Multiplying both sides by $(x-7)^3$,we get:
$2x^2-3x+5 = A(x-7)^2 + B(x-7) + C$
$2x^2-3x+5 = A(x^2-14x+49) + Bx - 7B + C$
$2x^2-3x+5 = Ax^2 + (B-14A)x + (49A-7B+C)$
Comparing the coefficients of $x^2$,$x$,and the constant term:
$A = 2$
$B-14A = -3$ $\Rightarrow B-14(2) = -3$ $\Rightarrow B-28 = -3$ $\Rightarrow B = 25$
$49A-7B+C = 5$ $\Rightarrow 49(2)-7(25)+C = 5$ $\Rightarrow 98-175+C = 5$ $\Rightarrow C-77 = 5$ $\Rightarrow C = 82$
Now,calculate $2A-3B+C$:
$2(2)-3(25)+82 = 4-75+82 = 11$.
Therefore,option $C$ is correct.

(D) Given the partial fraction decomposition: $\frac{x^2-x+1}{(x^2+1)(x^2+x+1)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+x+1}$
Multiplying both sides by $(x^2+1)(x^2+x+1)$,we get:
$x^2-x+1 = (Ax+B)(x^2+x+1) + (Cx+D)(x^2+1)$
$x^2-x+1 = Ax^3 + Ax^2 + Ax + Bx^2 + Bx + B + Cx^3 + Cx + Dx^2 + D$
$x^2-x+1 = (A+C)x^3 + (A+B+D)x^2 + (A+B+C)x + (B+D)$
Comparing coefficients on both sides:
$1) A+C = 0$
$2) A+B+D = 1$
$3) A+B+C = -1$
$4) B+D = 1$
From $(1)$,$C = -A$. Substituting into $(3)$: $A+B-A = -1 \Rightarrow B = -1$.
Using $B = -1$ in $(4)$: $-1+D = 1 \Rightarrow D = 2$.
Using $B = -1$ and $D = 2$ in $(2)$: $A-1+2 = 1 \Rightarrow A = 0$.
Then $C = -A = 0$.
Thus,$A=0, B=-1, C=0, D=2$.
Calculating $A+2B+C+2D = 0 + 2(-1) + 0 + 2(2) = -2 + 4 = 2$.

(D) Given the partial fraction decomposition: $\frac{x-2}{x^2(2x-3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x-3}$.
Multiplying both sides by $x^2(2x-3)$,we get: $x-2 = Ax(2x-3) + B(2x-3) + Cx^2$.
Setting $x = 0$: $-2 = B(-3) \Rightarrow B = \frac{2}{3}$.
Setting $x = \frac{3}{2}$: $\frac{3}{2} - 2 = C(\frac{3}{2})^2$ $\Rightarrow -\frac{1}{2} = C(\frac{9}{4})$ $\Rightarrow C = -\frac{2}{9}$.
Comparing the coefficients of $x^2$: $0 = 2A + C$ $\Rightarrow 2A = -C = \frac{2}{9}$ $\Rightarrow A = \frac{1}{9}$.
Now,calculate $2(A-C) = 2(\frac{1}{9} - (-\frac{2}{9})) = 2(\frac{1+2}{9}) = 2(\frac{3}{9}) = 2(\frac{1}{3}) = \frac{2}{3}$.
Since $B = \frac{2}{3}$,we have $2(A-C) = B$.

(B) $I \rightarrow |x|^2 - 4|x| + 3 < 0$
Let $t = |x|$,where $t \geq 0$. The inequality becomes $t^2 - 4t + 3 < 0$.
$(t - 1)(t - 3) < 0$,which implies $1 < t < 3$.
Since $t = |x|$,we have $1 < |x| < 3$.
This means $x \in (-3, -1) \cup (1, 3)$.
Thus,Statement $I$ is false.
$II \rightarrow x^2 - 8x + 15 > 0$
$(x - 3)(x - 5) > 0$.
The roots are $x = 3$ and $x = 5$. The inequality holds for $x < 3$ or $x > 5$.
Thus,Statement $II$ is true.

(A) Given points are $A(1, 3, 5)$,$B(2, 4, 6)$,and $C(4, 5, k)$.
First,calculate the squared distances between the points:
$AB^2 = (2-1)^2 + (4-3)^2 + (6-5)^2 = 1^2 + 1^2 + 1^2 = 3$
$BC^2 = (4-2)^2 + (5-4)^2 + (k-6)^2 = 4 + 1 + (k-6)^2 = k^2 - 12k + 41$
$AC^2 = (4-1)^2 + (5-3)^2 + (k-5)^2 = 9 + 4 + (k-5)^2 = k^2 - 10k + 38$
Case $1$: Right-angled at $A$ $(AB^2 + AC^2 = BC^2)$
$3 + k^2 - 10k + 38 = k^2 - 12k + 41$
$41 - 10k = 41 - 12k$
$2k = 0 \Rightarrow k = 0$
Case $2$: Right-angled at $B$ $(AB^2 + BC^2 = AC^2)$
$3 + k^2 - 12k + 41 = k^2 - 10k + 38$
$44 - 12k = 38 - 10k$
$6 = 2k \Rightarrow k = 3$
Case $3$: Right-angled at $C$ $(AC^2 + BC^2 = AB^2)$
$k^2 - 10k + 38 + k^2 - 12k + 41 = 3$
$2k^2 - 22k + 76 = 0$
$k^2 - 11k + 38 = 0$
The discriminant $D = (-11)^2 - 4(1)(38) = 121 - 152 = -31 < 0$. No real values for $k$.
Thus,the possible values for $k$ are $0$ and $3$. The number of possible values is $2$.

(D) Let the vertices of the tetrahedron be $A(1,2,3), B(2,-3,1), C(3,2,-1)$ and $D(a, b, c)$.
Since $G\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ is the centroid of the tetrahedron $ABCD$,we have:
$G = \frac{A+B+C+D}{4}$
$\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right) = \frac{(1+2+3+a, 2-3+2+b, 3+1-1+c)}{4}$
$\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right) = \frac{(6+a, 1+b, 3+c)}{4}$
Multiplying by $4$,we get:
$(10, 6, 9) = (6+a, 1+b, 3+c)$
Comparing the coordinates:
$6+a = 10 \Rightarrow a = 4$
$1+b = 6 \Rightarrow b = 5$
$3+c = 9 \Rightarrow c = 6$
Thus,$D = (4, 5, 6)$.
Now,we find the point $P$ that divides $GD$ in the ratio $1:2$ using the section formula:
$P = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$
Here $m=1, n=2$,$G = \left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ and $D = (4, 5, 6)$.
$x = \frac{1(4) + 2(5/2)}{1+2} = \frac{4+5}{3} = \frac{9}{3} = 3$
$y = \frac{1(5) + 2(3/2)}{1+2} = \frac{5+3}{3} = \frac{8}{3}$
$z = \frac{1(6) + 2(9/4)}{1+2} = \frac{6+9/2}{3} = \frac{21/2}{3} = \frac{7}{2}$
Therefore,the required point is $\left(3, \frac{8}{3}, \frac{7}{2}\right)$.

(B) Total number of balls $= 4 + 5 + 6 = 15$.
We need to draw $3$ balls of different colours,which means one red,one blue,and one yellow ball.
The number of ways to choose $1$ red,$1$ blue,and $1$ yellow ball is $\binom{4}{1} \times \binom{5}{1} \times \binom{6}{1} = 4 \times 5 \times 6 = 120$.
The total number of ways to draw $3$ balls from $15$ is $\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.
The required probability is $\frac{120}{455} = \frac{24}{91}$.

(C) When a pair of dice is thrown,the total number of outcomes is $n(S) = 6 \times 6 = 36$.
Prime numbers on a die are $\{2, 3, 5\}$. The outcomes where both dice show prime numbers are $\{(2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)\}$. There are $9$ such outcomes.
Probability of getting prime numbers on both dice in the first throw is $P(A) = \frac{9}{36} = \frac{1}{4}$.
Composite numbers on a die are $\{4, 6\}$ (Note: $1$ is neither prime nor composite). The outcomes where both dice show composite numbers are $\{(4,4), (4,6), (6,4), (6,6)\}$. There are $4$ such outcomes.
Probability of getting composite numbers on both dice in the second throw is $P(B) = \frac{4}{36} = \frac{1}{9}$.
Since the two throws are independent,the required probability is $P(A) \times P(B) = \frac{1}{4} \times \frac{1}{9} = \frac{1}{36}$.

(A) When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$.
On a single die,the prime numbers are $\{2, 3, 5\}$ and the composite numbers are $\{4, 6\}$. Note that $1$ is neither prime nor composite.
We need the probability of getting a prime number on one die and a composite number on the other.
The possible outcomes are:
$(2, 4), (2, 6), (3, 4), (3, 6), (5, 4), (5, 6)$ (prime on first,composite on second)
$(4, 2), (6, 2), (4, 3), (6, 3), (4, 5), (6, 5)$ (composite on first,prime on second)
Total favorable outcomes = $6 + 6 = 12$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{12}{36} = \frac{1}{3}$.

(C) Total balls = $3 + 6 = 9$. We draw $4$ balls. Total ways = $^9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
We want the probability of getting at least $2$ red balls.
This is $1 - [P(0 \text{ red}) + P(1 \text{ red})]$.
$P(0 \text{ red}) = \frac{^6C_0 \times ^3C_4}{126} = 0$ (since we cannot draw $4$ white balls).
$P(1 \text{ red}) = \frac{^6C_1 \times ^3C_3}{126} = \frac{6 \times 1}{126} = \frac{6}{126}$.
Probability of at least $2$ red balls = $1 - (0 + \frac{6}{126}) = 1 - \frac{6}{126} = \frac{120}{126} = \frac{20}{21}$.

(A) Total number of balls $= 9 + 4 = 13$.
Total ways to draw $3$ balls from $13$ is $^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286$.
The probability of getting at least one white ball is $1 - P(\text{no white ball})$.
If no white ball is drawn,all $3$ balls must be black.
Number of ways to draw $3$ black balls $= ^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
$P(\text{no white ball}) = \frac{84}{286} = \frac{42}{143}$.
$P(\text{at least one white ball}) = 1 - \frac{42}{143} = \frac{101}{143}$.
Therefore,option $A$ is correct.

(D) Total number of ways to draw $2$ cards from $52$ cards is given by $n(S) = ^{52}C_2 = \frac{52 \times 51}{2} = 1326$.
We need to find the probability of getting a king and a spade card.
There is only one card that is both a king and a spade (the King of Spades).
The other card can be any of the remaining $51$ cards.
However,the question asks for the probability of getting a king and a spade card simultaneously.
This means we need to select the King of Spades and one other card such that the set contains at least one king and at least one spade.
The favorable outcomes are: (King of Spades,any other King) or (King of Spades,any other Spade).
Number of other Kings = $3$. Number of other Spades = $12$.
Total favorable outcomes = $3 + 12 = 15$.
Probability = $\frac{15}{1326} = \frac{5}{442}$.

(D) When two dice are thrown,the total number of outcomes is $6 \times 6 = 36$.
We need to find the number of pairs $(x, y)$ such that $x^2 + y^2 \leq 25$,where $x, y \in \{1, 2, 3, 4, 5, 6\}$.
Let us list the favorable outcomes for each value of $x$:
If $x = 1$,then $1^2 + y^2 \leq 25 \implies y^2 \leq 24$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes).
If $x = 2$,then $2^2 + y^2 \leq 25 \implies y^2 \leq 21$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes).
If $x = 3$,then $3^2 + y^2 \leq 25 \implies y^2 \leq 16$. Possible $y$ values are $\{1, 2, 3, 4\}$ ($4$ outcomes).
If $x = 4$,then $4^2 + y^2 \leq 25 \implies y^2 \leq 9$. Possible $y$ values are $\{1, 2, 3\}$ ($3$ outcomes).
If $x = 5$,then $5^2 + y^2 \leq 25 \implies y^2 \leq 0$. No possible $y$ values since $y \geq 1$.
If $x = 6$,then $6^2 + y^2 \leq 25$. No possible $y$ values.
Total favorable outcomes $= 4 + 4 + 4 + 3 = 15$.
The probability is $\frac{15}{36} = \frac{5}{12}$.
Thus,option $(d)$ is correct.

(C) Let $P(A)$ be the probability that the first person hits the target and $P(B)$ be the probability that the second person hits the target.
Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{1}{5}$.
The target is hit if at least one of them hits the target.
It is easier to calculate the probability that the target is not hit at all.
The probability that the first person misses is $P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that the second person misses is $P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$.
Since they attempt independently,the probability that both miss is $P(A' \cap B') = P(A') \times P(B') = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}$.
The probability that the target is hit is $1 - P(A' \cap B') = 1 - \frac{3}{5} = \frac{2}{5}$.
Thus,the correct option is $C$.

(B) Total number of ways to draw $2$ cards from $52$ is $^{52}C_2 = \frac{52 \times 51}{2} = 1326$.
Prime numbers in a suit are $2, 3, 5, 7$ (total $4$ cards per suit,$16$ in total).
Multiples of $5$ in a suit are $5$ and $10$ (total $2$ cards per suit,$8$ in total).
Note that $5$ is both a prime number and a multiple of $5$.
Let $A$ be the set of prime cards ($16$ cards) and $B$ be the set of multiples of $5$ ($8$ cards).
The intersection $A \cap B$ contains the four $5$s.
We want one card from $A$ and one from $B$.
Case $1$: One card is $5$ and the other is a prime other than $5$. Number of ways = $^4C_1 \times ^{12}C_1 = 4 \times 12 = 48$.
Case $2$: One card is $5$ and the other is a multiple of $5$ other than $5$ (i.e.,$10$). Number of ways = $^4C_1 \times ^4C_1 = 4 \times 4 = 16$.
Case $3$: One card is a prime other than $5$ and the other is a multiple of $5$ other than $5$. Number of ways = $^{12}C_1 \times ^4C_1 = 12 \times 4 = 48$.
Total favorable outcomes = $48 + 16 + 48 = 112$.
Wait,re-evaluating: The selection is $1$ prime and $1$ multiple of $5$.
Favorable outcomes = $(16 \times 8) - (\text{overlap}) = 128 - 4 = 124$.
Probability = $\frac{124}{1326} = \frac{62}{663}$.
Thus,option $(b)$ is correct.