AP EAMCET 2024 Mathematics Question Paper with Answer and Solution

(B) Let $P(n) = 2 \cdot 4^{2n+1} + 3^{3n+1} = 2 \cdot (2^2)^{2n+1} + 3^{3n+1} = 2 \cdot 2^{4n+2} + 3^{3n+1} = 2^{4n+3} + 3^{3n+1}$.
For $n=1$,$P(1) = 2^{4(1)+3} + 3^{3(1)+1} = 2^7 + 3^4 = 128 + 81 = 209$.
For $n=2$,$P(2) = 2^{4(2)+3} + 3^{3(2)+1} = 2^{11} + 3^7 = 2048 + 2187 = 4235$.
We find the greatest common divisor of $209$ and $4235$.
$209 = 11 \times 19$.
$4235 = 11 \times 385$.
The greatest common divisor is $11$.
Thus,$P(n)$ is divisible by $11$ for all $n \in N$.

(D) We factor the denominator: $x^4+2 x^2+9 = (x^4+6 x^2+9) - 4 x^2 = (x^2+3)^2 - (2 x)^2 = (x^2-2 x+3)(x^2+2 x+3)$.
Using partial fractions,we have $\frac{x^2+3}{(x^2-2 x+3)(x^2+2 x+3)} = \frac{Ax+B}{x^2-2 x+3} + \frac{Cx+D}{x^2+2 x+3}$.
By solving,we get $\frac{x^2+3}{x^4+2 x^2+9} = \frac{1}{2} \left( \frac{1}{x^2-2 x+3} + \frac{1}{x^2+2 x+3} \right) = \frac{0x + 1/2}{x^2-2 x+3} + \frac{0x + 1/2}{x^2+2 x+3}$.
Comparing coefficients,we identify $A=0, B=1/2, a=-2, b=3, C=0, D=1/2, c=2$.
Substituting these values into the expression $aA+bB+cC+D$:
$(-2)(0) + (3)(1/2) + (2)(0) + 1/2 = 0 + 3/2 + 0 + 1/2 = 4/2 = 2$.

(D) For the first partial fraction: $\frac{1}{(3x+1)(x-2)} = \frac{A}{3x+1} + \frac{B}{x-2}$.
By partial fraction decomposition,$1 = A(x-2) + B(3x+1)$.
Setting $x=2$,$1 = B(6+1) \Rightarrow B = \frac{1}{7}$.
Setting $x=-\frac{1}{3}$,$1 = A(-\frac{1}{3}-2)$ $\Rightarrow 1 = A(-\frac{7}{3})$ $\Rightarrow A = -\frac{3}{7}$.
Thus,$A+3B = -\frac{3}{7} + 3(\frac{1}{7}) = 0$.
For the second partial fraction: $\frac{x+1}{(3x+1)(x-2)} = \frac{C}{3x+1} + \frac{D}{x-2}$.
$x+1 = C(x-2) + D(3x+1)$.
Setting $x=2$,$3 = D(7) \Rightarrow D = \frac{3}{7}$.
Setting $x=-\frac{1}{3}$,$\frac{2}{3} = C(-\frac{7}{3}) \Rightarrow C = -\frac{2}{7}$.
Now,$A:C = (-\frac{3}{7}) : (-\frac{2}{7}) = 3:2$ and $B:D = (\frac{1}{7}) : (\frac{3}{7}) = 1:3$.
Therefore,$A+3B=0, A:C=3:2, B:D=1:3$.

(A) Let the roots of the quadratic equation $x^2-35x+c=0$ be $2t$ and $3t$.
Sum of the roots $= 2t + 3t = -(-35)/1 = 35$.
$5t = 35 \Rightarrow t = 7$.
Product of the roots $= (2t)(3t) = c/1 = c$.
$6t^2 = c$.
Since $t = 7$,$c = 6(7^2) = 6 \times 49$.
Given $c = 6K$,we have $6K = 6 \times 49$.
Therefore,$K = 49$.

(D) Let $\alpha, \beta, \gamma, \delta$ be the roots of $x^4-x^3-8 x^2+2 x+12=0$. Given $\alpha+\beta=0$.
We can write the polynomial as $(x^2+a)(x^2-x+b) = x^4-x^3+(a+b)x^2-ax+ab$.
Comparing the coefficients with $x^4-x^3-8x^2+2x+12=0$:
$-a=2 \implies a=-2$.
$ab=12 \implies -2b=12 \implies b=-6$.
Thus,$x^4-x^3-8x^2+2x+12 = (x^2-2)(x^2-x-6) = (x^2-2)(x-3)(x+2)$.
The roots are $\pm\sqrt{2}, 3, -2$.
Since $\alpha+\beta=0$,we have $\alpha=\sqrt{2}, \beta=-\sqrt{2}$.
The other roots are $\gamma=3, \delta=-2$ (given $\gamma > \delta$).
Therefore,$3\gamma+2\delta = 3(3)+2(-2) = 9-4 = 5$.

(A) The given equation is $(3 + \alpha) x^2 + (6 + 2 \alpha) x + (5 + 2 \alpha) = 0$.
For the roots to be complex,the discriminant $D$ must be less than $0$.
$D = (2 \alpha + 6)^2 - 4 (\alpha + 3) (2 \alpha + 5) < 0$.
$4(\alpha + 3)^2 - 4(\alpha + 3)(2 \alpha + 5) < 0$.
Dividing by $4$: $(\alpha + 3)(\alpha + 3 - 2 \alpha - 5) < 0$.
$(\alpha + 3)(-\alpha - 2) < 0 \Rightarrow (\alpha + 3)(\alpha + 2) > 0$.
This implies $\alpha < -3$ or $\alpha > -2$.
Comparing with $\alpha > L$ or $\alpha < M$,we get $L = -2$ and $M = -3$.
The condition $L y^2 + M y + r < 0$ becomes $-2 y^2 - 3 y + r < 0$.
For this to hold for all $y \in R$,the coefficient of $y^2$ must be negative (which is $-2 < 0$) and $D < 0$.
$D = (-3)^2 - 4(-2)(r) < 0$.
$9 + 8 r < 0$ $\Rightarrow 8 r < -9$ $\Rightarrow r < -\frac{9}{8} = -1.125$.
The greatest integer $[r]$ not exceeding $r$ is $[-1.125] = -2$.

(A) Let $t = \sqrt{\frac{1-y}{y}}$. Then the equation becomes $t + \frac{1}{t} = \frac{5}{2}$.
Solving for $t$,we get $2t^2 - 5t + 2 = 0$,which gives $t = 2$ or $t = \frac{1}{2}$.
If $\sqrt{\frac{1-y}{y}} = 2$,then $\frac{1-y}{y} = 4$ $\Rightarrow 1-y = 4y$ $\Rightarrow y = \frac{1}{5}$.
If $\sqrt{\frac{1-y}{y}} = \frac{1}{2}$,then $\frac{1-y}{y} = \frac{1}{4}$ $\Rightarrow 4-4y = y$ $\Rightarrow y = \frac{4}{5}$.
Thus,$\alpha = \frac{1}{5}$ and $\beta = \frac{4}{5}$ (since $\beta > \alpha$).
Then $\alpha + \beta = 1$ and $\alpha \beta = \frac{4}{25}$.
The equation $(\alpha+\beta) x^4 - 25 \alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ becomes $x^4 - 4x^2 + (\gamma + \frac{3}{5}) = 0$.
Let $u = x^2$. Then $u^2 - 4u + (\gamma + \frac{3}{5}) = 0$.
For real roots,$u$ must be non-negative. The discriminant $D = 16 - 4(\gamma + \frac{3}{5}) \ge 0$ $\Rightarrow 4 - \gamma - \frac{3}{5} \ge 0$ $\Rightarrow \gamma \le \frac{17}{5} = 3.4$.
Also,for at least one non-negative root $u$,we need the sum of roots $4 > 0$ and product $\gamma + \frac{3}{5} \ge 0 \Rightarrow \gamma \ge -0.6$.
Thus,$\gamma \in [-0.6, 3.4]$. Among the options,$\frac{1}{2} = 0.5$ is in this range.

(C) Given the quadratic equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$.
Sum of coefficients $= a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0$.
Since the sum of coefficients is $0$,$x = 1$ is a root of the equation.
Let the roots be $\alpha$ and $\beta$. We know $\beta = 1$.
From the product of roots formula,$\alpha \times \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c(a-b)}{a(b-c)}$.
Therefore,$\alpha \times 1 = \frac{c(a-b)}{a(b-c)}$.
The roots are $1$ and $\frac{c(a-b)}{a(b-c)}$.

(D) Since the coefficients of the quadratic equation $x^2+ax+b=0$ are real,the complex roots must occur in conjugate pairs.
Given one root is $x_1 = 3+i$,the other root must be $x_2 = 3-i$.
The sum of the roots of a quadratic equation $x^2+ax+b=0$ is given by $-a$.
Therefore,$(3+i) + (3-i) = -a$.
$6 = -a$.
$a = -6$.

(B) To find the equation whose roots are shifted by $-1$,we replace $x$ with $(x - (-1))$,which is $(x+1)$.
Substituting $(x+1)$ for $x$ in the original equation $f(x) = x^4+5x^3+6x^2+7x+9=0$:
$f(x+1) = (x+1)^4 + 5(x+1)^3 + 6(x+1)^2 + 7(x+1) + 9 = 0$
Expanding each term:
$(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$
$5(x+1)^3 = 5(x^3 + 3x^2 + 3x + 1) = 5x^3 + 15x^2 + 15x + 5$
$6(x+1)^2 = 6(x^2 + 2x + 1) = 6x^2 + 12x + 6$
$7(x+1) = 7x + 7$
Adding these together:
$x^4 + (4+5)x^3 + (6+15+6)x^2 + (4+15+12+7)x + (1+5+6+7+9) = 0$
$x^4 + 9x^3 + 27x^2 + 38x + 28 = 0$

(B) The given equation is $x^2 - 3ax + a^2 - 2a - 4 = 0$.
For the roots to be rational,the discriminant $D$ must be a perfect square of a rational number.
$D = b^2 - 4ac = (-3a)^2 - 4(1)(a^2 - 2a - 4) = 9a^2 - 4a^2 + 8a + 16 = 5a^2 + 8a + 16$.
Since $5a^2 + 8a + 16$ is a quadratic in '$a$' with discriminant $D_a = 8^2 - 4(5)(16) = 64 - 320 = -256 < 0$,the expression $5a^2 + 8a + 16$ is always positive for all real '$a$'.
However,for the roots to be rational,$D$ must be a perfect square.
Since $5a^2 + 8a + 16$ is not a perfect square for all rational '$a$',the roots are real and distinct but not necessarily rational.

(C) Given the inequality $-1 < \frac{2 x^2+a x+2}{x^2+x+1} < 3$ for all $x \in \mathbb{R}$.
Since $x^2+x+1 > 0$ for all $x$,we can split the inequality into two parts.
Part $1$: $-1 < \frac{2 x^2+a x+2}{x^2+x+1}$ $\Rightarrow -x^2-x-1 < 2 x^2+a x+2$ $\Rightarrow 3 x^2+(a+1) x+3 > 0$.
For this to hold for all $x$,the discriminant $D < 0$.
$(a+1)^2 - 4(3)(3) < 0$ $\Rightarrow (a+1)^2 - 36 < 0$ $\Rightarrow (a+1-6)(a+1+6) < 0$ $\Rightarrow (a-5)(a+7) < 0$ $\Rightarrow a \in (-7, 5) \dots (i)$.
Part $2$: $\frac{2 x^2+a x+2}{x^2+x+1} < 3$ $\Rightarrow 2 x^2+a x+2 < 3 x^2+3 x+3$ $\Rightarrow -x^2+(a-3) x-1 < 0$ $\Rightarrow x^2-(a-3) x+1 > 0$.
For this to hold for all $x$,the discriminant $D < 0$.
$(a-3)^2 - 4(1)(1) < 0$ $\Rightarrow (a-3)^2 - 4 < 0$ $\Rightarrow (a-3-2)(a-3+2) < 0$ $\Rightarrow (a-5)(a-1) < 0$ $\Rightarrow a \in (1, 5) \dots (ii)$.
Taking the intersection of $(i)$ and $(ii)$,we get $a \in (1, 5)$.

(B) Let $y$ be a common root of $x^2+5ax+6=0$ and $x^2+3ax+2=0$.
Since $y$ satisfies both equations,we have:
$y^2+5ay+6=0$ $(1)$
$y^2+3ay+2=0$ $(2)$
Subtracting equation $(2)$ from $(1)$:
$(y^2+5ay+6) - (y^2+3ay+2) = 0$
$2ay + 4 = 0$
$2ay = -4$
$ay = -2$
Substitute $ay = -2$ into equation $(2)$:
$y^2 + 3(-2) + 2 = 0$
$y^2 - 6 + 2 = 0$
$y^2 - 4 = 0$
$y^2 = 4$
$y = \pm 2$.

(B) First,solve the quadratic equation $2x^2+3x-2=0$ by factoring:
$2x^2+4x-x-2=0$ $\Rightarrow 2x(x+2)-1(x+2)=0$ $\Rightarrow (2x-1)(x+2)=0$.
Thus,the roots are $x=-2$ and $x=\frac{1}{2}$.
Case $1$: If $x=-2$ is the common root,substitute it into $3x^2+ax-2=0$:
$3(-2)^2+a(-2)-2=0$ $\Rightarrow 12-2a-2=0$ $\Rightarrow 10=2a$ $\Rightarrow a=5$.
Case $2$: If $x=\frac{1}{2}$ is the common root,substitute it into $3x^2+ax-2=0$:
$3(\frac{1}{2})^2+a(\frac{1}{2})-2=0$ $\Rightarrow \frac{3}{4}+\frac{a}{2}-2=0$ $\Rightarrow \frac{a}{2} = 2 - \frac{3}{4} = \frac{5}{4}$ $\Rightarrow a=2.5$.
The sum of all possible values of $a$ is $5+2.5=7.5$.

(C) Given that $\alpha$ is a common root of $x^2-5x+\lambda=0$ and $x^2-8x-2\lambda=0$.
Substituting $\alpha$ in both equations:
$\alpha^2-5\alpha+\lambda=0$ ... $(i)$
$\alpha^2-8\alpha-2\lambda=0$ ... (ii)
Subtracting (ii) from $(i)$:
$(x^2-5\alpha+\lambda) - (\alpha^2-8\alpha-2\lambda) = 0$
$3\alpha+3\lambda=0 \Rightarrow \alpha=-\lambda$.
Since $\alpha$ is a root of $x^2-5x+\lambda=0$,substitute $\alpha=-\lambda$:
$(-\lambda)^2-5(-\lambda)+\lambda=0$
$\lambda^2+5\lambda+\lambda=0 \Rightarrow \lambda^2+6\lambda=0$.
Since $\lambda \neq 0$,we have $\lambda=-6$.
Thus,$\alpha = -(-6) = 6$.
For $x^2-5x-6=0$,the roots are $\alpha=6$ and $\beta=-1$ (since $\alpha+\beta=5$).
For $x^2-8x+12=0$,the roots are $\alpha=6$ and $\gamma=2$ (since $\alpha+\gamma=8$).
Finally,$\alpha+\beta+\gamma+\lambda = 6 + (-1) + 2 + (-6) = 1$.

(B) Given the inequality: $\frac{7 x^2-5 x-18}{2 x^2+x-6} < 2$
Subtract $2$ from both sides: $\frac{7 x^2-5 x-18}{2 x^2+x-6} - 2 < 0$
$\Rightarrow \frac{7 x^2-5 x-18 - 2(2 x^2+x-6)}{2 x^2+x-6} < 0$
$\Rightarrow \frac{7 x^2-5 x-18 - 4 x^2-2 x+12}{(2 x-3)(x+2)} < 0$
$\Rightarrow \frac{3 x^2-7 x-6}{(2 x-3)(x+2)} < 0$
Factor the numerator: $3 x^2-7 x-6 = 3 x^2-9 x+2 x-6 = 3 x(x-3)+2(x-3) = (3 x+2)(x-3)$
So,the inequality becomes: $\frac{(3 x+2)(x-3)}{(2 x-3)(x+2)} < 0$
The critical points are $x = -2, -\frac{2}{3}, \frac{3}{2}, 3$.
Using the wavy curve method,we test the intervals:
For $x > 3$,the expression is positive.
For $\frac{3}{2} < x < 3$,the expression is negative.
For $-\frac{2}{3} < x < \frac{3}{2}$,the expression is positive.
For $-2 < x < -\frac{2}{3}$,the expression is negative.
For $x < -2$,the expression is positive.
Since we need the expression to be less than $0$,the solution is $x \in \left(-2, -\frac{2}{3}\right) \cup \left(\frac{3}{2}, 3 \right)$.

(C) For the quadratic expression $ax^2 + bx + c > 0$ to be true for all $x \in R$,we must have $a > 0$ and the discriminant $D = b^2 - 4ac < 0$.
Here,$a = 1 > 0$,which is always true.
The discriminant $D = \{-(3k + 1)\}^2 - 4(1)(4k^2 + 3k - 3) < 0$.
Expanding this,we get $(9k^2 + 6k + 1) - (16k^2 + 12k - 12) < 0$.
$-7k^2 - 6k + 13 < 0$.
Multiplying by $-1$ reverses the inequality: $7k^2 + 6k - 13 > 0$.
Factoring the quadratic: $(7k + 13)(k - 1) > 0$.
The roots are $k = -\frac{13}{7}$ and $k = 1$.
For the expression to be positive,$k$ must lie outside the interval $[-\frac{13}{7}, 1]$.
Thus,the solution set is $k \in (-\infty, -\frac{13}{7}) \cup (1, \infty)$.

(B) For the quadratic equation $x^2+3(a+3)x-9a=0$ to have equal roots,the discriminant $D$ must be $0$.
$D = [3(a+3)]^2 - 4(1)(-9a) = 0$
$9(a^2+6a+9) + 36a = 0$
$9a^2 + 54a + 81 + 36a = 0$
$9a^2 + 90a + 81 = 0$
$a^2 + 10a + 9 = 0$
$(a+9)(a+1) = 0 \Rightarrow a = -9, -1$.
For $a = -9$,the equation is $x^2 - 18x + 81 = 0$ $\Rightarrow (x-9)^2 = 0$ $\Rightarrow x = 9$. Let $\alpha = 9$.
For $a = -1$,the equation is $x^2 + 6x + 9 = 0$ $\Rightarrow (x+3)^2 = 0$ $\Rightarrow x = -3$. Let $\beta = -3$.
We want the minimum value of the expression $f(x) = x^2 + \alpha x - \beta = x^2 + 9x - (-3) = x^2 + 9x + 3$.
Completing the square: $f(x) = (x + \frac{9}{2})^2 + 3 - \frac{81}{4} = (x + \frac{9}{2})^2 - \frac{69}{4}$.
Since $(x + \frac{9}{2})^2 \geq 0$,the minimum value is $-\frac{69}{4}$.

(D) Let $y = \frac{11 x^2+12 x+6}{x^2+4 x+2}$.
$(11-y) x^2+(12-4 y) x+(6-2 y)=0$.
For $x$ to be real,the discriminant $D \geq 0$.
$(12-4 y)^2-4(11-y)(6-2 y) \geq 0$.
$16(3-y)^2-8(11-y)(3-y) \geq 0$.
$8(3-y)[2(3-y)-(11-y)] \geq 0$.
$8(3-y)(6-2y-11+y) \geq 0$.
$8(3-y)(-y-5) \geq 0$.
$(y-3)(y+5) \geq 0$.
Thus,$y \leq -5$ or $y \geq 3$.
The range of the expression is $(-\infty, -5] \cup [3, \infty)$.
Given $\frac{11 x^2+12 x+6}{x^2+4 x+2} \notin (a, b]$,we identify $(a, b] = (-5, 3]$.
So,$a = -5$ and $b = 3$.
We need to find $x$ such that $\frac{11 x^2+12 x+6}{x^2+4 x+2} = b-a+3 = 3 - (-5) + 3 = 11$.
$11 x^2+12 x+6 = 11(x^2+4 x+2)$.
$11 x^2+12 x+6 = 11 x^2+44 x+22$.
$12 x+6 = 44 x+22$.
$-16 = 32 x$.
$x = -\frac{16}{32} = -\frac{1}{2}$.

(B) Let $y = \frac{x+a}{2 x^2-3 x+1}$,where $y \in \mathbb{R}$.
Since $y$ takes all real values,the equation $2 y x^2 - (3 y + 1) x + (y - a) = 0$ must have real roots for $x$ for all $y \in \mathbb{R}$.
For $y \neq 0$,the discriminant $D \geq 0$:
$D = (3 y + 1)^2 - 4(2 y)(y - a) \geq 0$
$9 y^2 + 6 y + 1 - 8 y^2 + 8 a y \geq 0$
$y^2 + (8 a + 6) y + 1 \geq 0$
For this quadratic in $y$ to be non-negative for all $y \in \mathbb{R}$,the discriminant of this quadratic must be less than or equal to $0$ (since the coefficient of $y^2$ is $1 > 0$).
$D_y = (8 a + 6)^2 - 4(1)(1) \leq 0$
$(8 a + 6)^2 - 4 \leq 0$
$(8 a + 6 - 2)(8 a + 6 + 2) \leq 0$
$(8 a + 4)(8 a + 8) \leq 0$
$4(2 a + 1) \cdot 8(a + 1) \leq 0$
$32(2 a + 1)(a + 1) \leq 0$
$(2 a + 1)(a + 1) \leq 0$
Thus,$-1 \leq a \leq -\frac{1}{2}$.
However,if $y=0$,then $x+a=0 \Rightarrow x=-a$. The denominator $2x^2-3x+1 = (2x-1)(x-1)$ must not be zero at $x=-a$. So $2(-a)-1 \neq 0$ and $-a-1 \neq 0$,which means $a \neq -1/2$ and $a \neq -1$.
Therefore,$-1 < a < -\frac{1}{2}$.

(D) Given equation: $x^2 - 6ax + (9a^2 - 2a + 2) = 0$.
Let $f(x) = x^2 - 6ax + 9a^2 - 2a + 2$.
For both roots to be greater than $3$,the following conditions must be satisfied:
$1)$ Discriminant $D \ge 0$:
$D = (-6a)^2 - 4(1)(9a^2 - 2a + 2) = 36a^2 - 36a^2 + 8a - 8 = 8a - 8$.
$8a - 8 \ge 0 \Rightarrow a \ge 1$.
$2)$ Vertex location: $-\frac{b}{2a} > 3$:
$-\frac{-6a}{2(1)} > 3$ $\Rightarrow 3a > 3$ $\Rightarrow a > 1$.
$3)$ $f(3) > 0$:
$f(3) = 3^2 - 6a(3) + 9a^2 - 2a + 2 > 0$
$9 - 18a + 9a^2 - 2a + 2 > 0$
$9a^2 - 20a + 11 > 0$
$(9a - 11)(a - 1) > 0$.
Since $a > 1$,the condition $(9a - 11)(a - 1) > 0$ implies $a > \frac{11}{9}$.
Combining all conditions,we get $a > \frac{11}{9}$.

(B) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+a x^2+b x+c=0$.
By Vieta's formulas,the sum of the roots taken two at a time is $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{b}{1} = b$,and the product of the roots is $\alpha \beta \gamma = -\frac{c}{1} = -c$.
We need to find the value of $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$.
$\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta \gamma + \alpha \gamma + \alpha \beta}{\alpha \beta \gamma}$.
Substituting the values,we get $\frac{b}{-c} = -\frac{b}{c}$.

(C) Given the cubic equation $x^3-13x^2+Kx-27=0$.
Let the roots be $\frac{a}{r}, a, ar$.
From the product of roots,$\frac{a}{r} \cdot a \cdot ar = -\frac{-27}{1} = 27$.
Thus,$a^3 = 27$,which gives $a = 3$.
Since $a=3$ is a root,it must satisfy the equation: $3^3 - 13(3^2) + K(3) - 27 = 0$.
$27 - 117 + 3K - 27 = 0$.
$3K - 117 = 0$.
$3K = 117$.
$K = 39$.

(C) Let the roots be $\alpha, \beta, \gamma$. Given $\alpha + \beta = \gamma$.
From Vieta's formulas,$\alpha + \beta + \gamma = -p$.
Substituting $\alpha + \beta = \gamma$,we get $2\gamma = -p$,so $\gamma = -\frac{p}{2}$.
Since $\gamma$ is a root,it satisfies the equation: $(-\frac{p}{2})^3 + p(-\frac{p}{2})^2 + q(-\frac{p}{2}) - 5 = 0$.
$-\frac{p^3}{8} + \frac{p^3}{4} - \frac{pq}{2} - 5 = 0$.
Multiplying by $8$,we get $-p^3 + 2p^3 - 4pq - 40 = 0$.
$p^3 - 4pq = 40$.
$p(p^2 - 4q) = 40$.

(B) Given $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$.
Notice the pattern in the values: $P(x) = x^2 + 1$ for $x = 0, 1, 2, 3, 4$.
Let $Q(x) = P(x) - (x^2 + 1)$.
Since $P(x)$ is a polynomial of degree $5$ and $x^2 + 1$ is of degree $2$,$Q(x)$ is a polynomial of degree $5$ with roots $0, 1, 2, 3, 4$.
Thus,$Q(x) = k(x)(x-1)(x-2)(x-3)(x-4)$ for some constant $k$.
Comparing the coefficient of $x^5$ in $P(x)$,we see $k = 1$.
So,$P(x) = x(x-1)(x-2)(x-3)(x-4) + x^2 + 1$.
To find $P(5)$,substitute $x = 5$:
$P(5) = 5(5-1)(5-2)(5-3)(5-4) + 5^2 + 1$
$P(5) = 5 \times 4 \times 3 \times 2 \times 1 + 25 + 1$
$P(5) = 120 + 26 = 146$.

(B) Given the equation $x^4-x^3-6x^2+4x+8=0$.
By testing values,we find $x=2$ is a root.
Dividing the polynomial by $(x-2)$,we get $(x-2)(x^3+x^2-4x-4)=0$.
Testing $x=2$ again in the cubic factor $x^3+x^2-4x-4$,we get $8+4-8-4=0$,so $x=2$ is a root again.
Dividing $(x^3+x^2-4x-4)$ by $(x-2)$,we get $(x-2)(x^2+3x+2)=0$.
Factoring the quadratic part: $(x^2+3x+2) = (x+1)(x+2)$.
Thus,the roots are $x=2, 2, -1, -2$.
The two equal roots are $2, 2$.
The other two roots are $\alpha = -1$ and $\beta = -2$.
Therefore,$\alpha^2+\beta^2 = (-1)^2 + (-2)^2 = 1 + 4 = 5$.

(A) Let the roots of the cubic equation be in arithmetic progression as $\alpha-r, \alpha, \alpha+r$.
Sum of the roots $= \alpha-r+\alpha+\alpha+r = 3\alpha$.
From the given equation $x^3-b x^2+c x-d=0$,the sum of the roots is $b$.
Therefore,$3\alpha = b \Rightarrow \alpha = \frac{b}{3}$.
Since $\alpha$ is a root of the equation,it must satisfy $x^3-b x^2+c x-d=0$.
Substituting $x = \frac{b}{3}$:
$(\frac{b}{3})^3 - b(\frac{b}{3})^2 + c(\frac{b}{3}) - d = 0$
$\frac{b^3}{27} - \frac{b^3}{9} + \frac{bc}{3} - d = 0$
Multiplying the entire equation by $27$:
$b^3 - 3b^3 + 9bc - 27d = 0$
$-2b^3 + 9bc - 27d = 0$
$9bc = 2b^3 + 27d$.

(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$.
Multiplying by $\alpha^8$,we get $\alpha^{10} = 6\alpha^9 + 2\alpha^8$,which implies $\alpha^{10} - 2\alpha^8 = 6\alpha^9$.
Similarly,for $\beta$,we have $\beta^{10} - 2\beta^8 = 6\beta^9$.
Subtracting the two equations:
$(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8) = 6(\alpha^9 - \beta^9)$.
By definition,$a_n = \alpha^n - \beta^n$,so $a_{10} - 2a_8 = 6a_9$.
Therefore,$\frac{a_{10} - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$.

(D) Given equation: $x^5-5x^3+5x^2-1=0$.
By inspection,$x=1$ is a root.
Factoring the polynomial: $(x-1)(x^4+x^3-4x^2+x+1)=0$.
Further factoring: $(x-1)^2(x^3+2x^2-2x-1)=0$.
$(x-1)^3(x^2+3x+1)=0$.
The roots are $x=1$ (triple root) and $x=\frac{-3 \pm \sqrt{5}}{2}$.
The two distinct negative roots are $\alpha = \frac{-3+\sqrt{5}}{2}$ and $\beta = \frac{-3-\sqrt{5}}{2}$.
We need an equation with roots $\sqrt{-\alpha}$ and $\sqrt{-\beta}$.
Let $y = \sqrt{-\alpha} \Rightarrow y^2 = -\alpha$.
Since $\alpha$ and $\beta$ are roots of $x^2+3x+1=0$,we have $\alpha+\beta = -3$ and $\alpha\beta = 1$.
Let the new roots be $y_1 = \sqrt{-\alpha}$ and $y_2 = \sqrt{-\beta}$.
The equation is $(y^2+\alpha)(y^2+\beta) = 0$.
$y^4 + (\alpha+\beta)y^2 + \alpha\beta = 0$.
Substituting the values: $y^4 - 3y^2 + 1 = 0$.

(A) The given equation is $x^9-x^5+x^4-1=0$.
Factoring the expression: $x^5(x^4-1) + 1(x^4-1) = 0$,which gives $(x^5+1)(x^4-1) = 0$.
For $x^4-1=0$,the roots are $x=1, -1, i, -i$.
For $x^5+1=0$,the roots are $x=-1$ and $e^{\pm i\pi/5}, e^{\pm 3i\pi/5}$.
The distinct roots are $x=1, -1, i, -i, e^{i\pi/5}, e^{-i\pi/5}, e^{i3\pi/5}, e^{-i3\pi/5}$.
Real roots are $1, -1$,so $n=2$.
Complex roots with argument on the imaginary axis (argument $\pi/2$ or $3\pi/2$) are $i, -i$,so $m=2$.
Complex roots with argument in the $2^{nd}$ quadrant (argument between $\pi/2$ and $\pi$) is $e^{i3\pi/5}$,so $k=1$.
Therefore,$m \cdot n \cdot k = 2 \times 2 \times 1 = 4$.
Wait,re-evaluating the roots: $x^5 = -1 = e^{i(\pi + 2k\pi)}$. Roots are $e^{i\pi/5}, e^{i3\pi/5}, e^{i\pi}, e^{i7\pi/5}, e^{i9\pi/5}$.
Real roots: $1, -1$ (from $x^4=1$) and $-1$ (from $x^5=-1$). Distinct real roots are $1, -1$,so $n=2$.
Complex roots: $i, -i$ (arg $\pi/2, 3\pi/2$),$e^{i\pi/5}$ (arg $36^{\circ}$),$e^{i3\pi/5}$ (arg $108^{\circ}$),$e^{i7\pi/5}$ (arg $252^{\circ}$),$e^{i9\pi/5}$ (arg $324^{\circ}$).
$m=2$ (roots $i, -i$). $k=1$ (root $e^{i3\pi/5}$).
$m \cdot n \cdot k = 2 \times 2 \times 1 = 4$.
Given the options,there might be a misinterpretation of distinct roots. If we count total roots including multiplicity: $n=3$ (real roots $1, -1, -1$),$m=2$,$k=1$. Then $3 \times 2 \times 1 = 6$.

(B) To find the quotient,we perform polynomial long division of $p(x) = 3x^5-4x^4+5x^3-3x^2+6x-8$ by $t(x) = x^2+x-3$.
Step $1$: Divide $3x^5$ by $x^2$ to get $3x^3$. Multiply $3x^3(x^2+x-3) = 3x^5+3x^4-9x^3$. Subtracting this from $p(x)$ gives $-7x^4+14x^3-3x^2+6x-8$.
Step $2$: Divide $-7x^4$ by $x^2$ to get $-7x^2$. Multiply $-7x^2(x^2+x-3) = -7x^4-7x^3+21x^2$. Subtracting this gives $21x^3-24x^2+6x-8$.
Step $3$: Divide $21x^3$ by $x^2$ to get $21x$. Multiply $21x(x^2+x-3) = 21x^3+21x^2-63x$. Subtracting this gives $-45x^2+69x-8$.
Step $4$: Divide $-45x^2$ by $x^2$ to get $-45$. Multiply $-45(x^2+x-3) = -45x^2-45x+135$. Subtracting this gives $114x-143$.
Thus,the quotient is $3x^3-7x^2+21x-45$.

(C) The given equation is $x^5 - 5x^4 + 9x^3 - 9x^2 + 5x - 1 = 0$.
By inspection,$x = 1$ is a root,so $\alpha_1 = 1$.
Dividing the polynomial by $(x-1)$,we get the reduced equation: $x^4 - 4x^3 + 5x^2 - 4x + 1 = 0$.
Dividing by $x^2$,we have $(x^2 + \frac{1}{x^2}) - 4(x + \frac{1}{x}) + 5 = 0$.
Let $x + \frac{1}{x} = a$. Then $x^2 + \frac{1}{x^2} = a^2 - 2$.
Substituting this,we get $(a^2 - 2) - 4a + 5 = 0$,which simplifies to $a^2 - 4a + 3 = 0$.
Solving for $a$,we get $(a - 3)(a - 1) = 0$,so $a = 1$ or $a = 3$.
For $a = 1$,$x + \frac{1}{x} = 1 \Rightarrow x^2 - x + 1 = 0$. The roots are $\alpha_2, \alpha_3$. Since $x^2 - x + 1 = 0$,we have $\frac{1}{x^2} = x - 1$. Thus $\frac{1}{\alpha_2^2} + \frac{1}{\alpha_3^2} = (\alpha_2 + \alpha_3) - 2 = 1 - 2 = -1$.
For $a = 3$,$x + \frac{1}{x} = 3 \Rightarrow x^2 - 3x + 1 = 0$. The roots are $\alpha_4, \alpha_5$. Since $x^2 - 3x + 1 = 0$,we have $\frac{1}{x^2} = 3x - 1$. Thus $\frac{1}{\alpha_4^2} + \frac{1}{\alpha_5^2} = 3(\alpha_4 + \alpha_5) - 2 = 3(3) - 2 = 7$.
Finally,$\frac{1}{\alpha_1^2} + \frac{1}{\alpha_2^2} + \frac{1}{\alpha_3^2} + \frac{1}{\alpha_4^2} + \frac{1}{\alpha_5^2} = \frac{1}{1^2} + (-1) + 7 = 1 - 1 + 7 = 7$.

(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $12x^3-20x^2+x+3=0$.
From Vieta's formulas:
$\alpha+\beta+\gamma = \frac{20}{12} = \frac{5}{3}$
$\alpha\beta+\beta\gamma+\gamma\alpha = \frac{1}{12}$
$\alpha\beta\gamma = -\frac{3}{12} = -\frac{1}{4}$
We want the equation with roots $\alpha^2, \beta^2, \gamma^2$.
Sum of roots: $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (\frac{5}{3})^2 - 2(\frac{1}{12}) = \frac{25}{9} - \frac{1}{6} = \frac{50-3}{18} = \frac{47}{18} = \frac{376}{144}$.
Sum of roots taken two at a time: $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (\frac{1}{12})^2 - 2(-\frac{1}{4})(\frac{5}{3}) = \frac{1}{144} + \frac{5}{6} = \frac{1+120}{144} = \frac{121}{144}$.
Product of roots: $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-\frac{1}{4})^2 = \frac{1}{16} = \frac{9}{144}$.
The required equation is $x^3 - (\sum \alpha^2)x^2 + (\sum \alpha^2\beta^2)x - (\alpha^2\beta^2\gamma^2) = 0$.
Substituting the values: $x^3 - \frac{376}{144}x^2 + \frac{121}{144}x - \frac{9}{144} = 0$.
Multiplying by $144$: $144x^3 - 376x^2 + 121x - 9 = 0$.

(D) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2-10x-24=0$.
From Vieta's formulas: $\alpha+\beta+\gamma = -3$,$\alpha\beta+\beta\gamma+\gamma\alpha = -10$,and $\alpha\beta\gamma = 24$.
Let $S = \alpha+\beta+\gamma = -3$. Then the roots of the new equation are $\alpha(S-\alpha) = \alpha S - \alpha^2$,$\beta S - \beta^2$,and $\gamma S - \gamma^2$.
We need to find $q$,which is the sum of the products of these roots taken two at a time:
$q = (\alpha S - \alpha^2)(\beta S - \beta^2) + (\beta S - \beta^2)(\gamma S - \gamma^2) + (\gamma S - \gamma^2)(\alpha S - \alpha^2)$.
Expanding this:
$q = S^2(\alpha\beta+\beta\gamma+\gamma\alpha) - S(\alpha\beta(\alpha+\beta) + \beta\gamma(\beta+\gamma) + \gamma\alpha(\gamma+\alpha)) + (\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2)$.
Using $\alpha\beta(\alpha+\beta) = \alpha\beta(S-\gamma) = S\alpha\beta - \alpha\beta\gamma$,we get:
$q = S^2(\sum \alpha\beta) - S(S \sum \alpha\beta - 3\alpha\beta\gamma) + ((\sum \alpha\beta)^2 - 2\alpha\beta\gamma(\sum \alpha))$.
Substituting values:
$q = (-3)^2(-10) - (-3)((-3)(-10) - 3(24)) + ((-10)^2 - 2(24)(-3))$.
$q = -90 + 3(30 - 72) + (100 + 144)$.
$q = -90 + 3(-42) + 244 = -90 - 126 + 244 = 28$.

(C) Given the cubic equation $x^3+ax^2+bx+c=0$ with roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -a$
$\alpha\beta+\beta\gamma+\gamma\alpha = b$
$\alpha\beta\gamma = -c$
We need to find $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$.
Taking the common denominator:
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}$.
Substituting the values from Vieta's formulas:
$\frac{b}{-c} = -\frac{b}{c}$.

(A) Given $\left(\frac{1-i}{1+i}\right)^k = -i$.
First,simplify the base: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$.
So,the equation becomes $(-i)^k = -i$.
For $(-i)^k = -i$,$k$ must satisfy $k \equiv 1 \pmod 4$.
The least positive integer $m$ is $1$.
The greatest negative integer $n$ is $1 - 4 = -3$.
Thus,$m - n = 1 - (-3) = 1 + 3 = 4$.

(B) Let $z = 8i = 8(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = 8e^{i(\frac{\pi}{2} + 2k\pi)}$ for $k = 0, 1, 2$.
Taking the cube root,$z^{\frac{1}{3}} = 2e^{i(\frac{\pi}{6} + \frac{2k\pi}{3})}$.
For $k=0$: $2e^{i\frac{\pi}{6}} = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2} + i \frac{1}{2}) = \sqrt{3} + i$.
For $k=1$: $2e^{i(\frac{\pi}{6} + \frac{2\pi}{3})} = 2e^{i\frac{5\pi}{6}} = 2(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i \frac{1}{2}) = -\sqrt{3} + i$.
For $k=2$: $2e^{i(\frac{\pi}{6} + \frac{4\pi}{3})} = 2e^{i\frac{9\pi}{6}} = 2e^{i\frac{3\pi}{2}} = 2(0 - i) = -2i$.
Thus,the values are $\pm \sqrt{3} + i, -2i$.

(C) Let $\sqrt{-5-12 i} = x+y i$,where $x > 0$.
$-5-12 i = (x+y i)^2 = x^2-y^2+2 x y i$.
Equating real and imaginary parts: $x^2-y^2 = -5$ and $2 x y = -12$.
$x^2+y^2 = \sqrt{(-5)^2+(-12)^2} = \sqrt{25+144} = 13$.
Adding the equations $x^2-y^2 = -5$ and $x^2+y^2 = 13$ gives $2 x^2 = 8$,so $x = 2$ (since $x > 0$).
Then $y = -3$. Thus,$\sqrt{-5-12 i} = 2-3 i$.
Similarly,for $\sqrt{5+12 i} = u+v i$ with $u > 0$: $u^2-v^2 = 5$ and $2 u v = 12$.
$u^2+v^2 = 13$. Adding gives $2 u^2 = 18$,so $u = 3$ and $v = 2$. Thus,$\sqrt{5+12 i} = 3+2 i$.
For $\sqrt{-8-6 i} = m+n i$ with $m < 0$: $m^2-n^2 = -8$ and $2 m n = -6$.
$m^2+n^2 = \sqrt{(-8)^2+(-6)^2} = 10$.
Adding gives $2 m^2 = 2$,so $m = -1$ (since $m < 0$).
Then $n = 3$. Thus,$\sqrt{-8-6 i} = -1+3 i$.
Now,$a+b i = \frac{(2-3 i)+(3+2 i)}{-1+3 i} = \frac{5-i}{-1+3 i}$.
Multiplying by the conjugate: $\frac{5-i}{-1+3 i} \times \frac{-1-3 i}{-1-3 i} = \frac{-5-15 i+i+3 i^2}{1+9} = \frac{-8-14 i}{10} = -0.8-1.4 i$.
So $a = -0.8$ and $b = -1.4$.
$2 a+b = 2(-0.8)+(-1.4) = -1.6-1.4 = -3$.

(A) Let $Z = x + iy$.
Then $Z - 2 = (x - 2) + iy$.
The amplitude (argument) of a complex number $w = a + ib$ is $\theta = \tan^{-1}(\frac{b}{a})$.
Given $\arg(Z - 2) = \frac{\pi}{2}$,this implies that the real part must be $0$ and the imaginary part must be positive.
Therefore,$x - 2 = 0 \Rightarrow x = 2$ and $y > 0$.
Thus,the locus of $Z$ is the vertical line $x = 2$ where $y > 0$.

(D) Let $Z = \frac{(1-i)^3}{(2-i)(3-2i)}$.
First,expand the numerator: $(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.
Next,expand the denominator: $(2-i)(3-2i) = 6 - 4i - 3i + 2i^2 = 6 - 7i - 2 = 4 - 7i$.
So,$Z = \frac{-2 - 2i}{4 - 7i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator: $Z = \frac{-2 - 2i}{4 - 7i} \times \frac{4 + 7i}{4 + 7i} = \frac{-8 - 14i - 8i - 14i^2}{16 + 49} = \frac{-8 - 22i + 14}{65} = \frac{6 - 22i}{65}$.
Thus,$Z = \frac{6}{65} - \frac{22}{65}i$.
The imaginary part is $-\frac{22}{65}$.

(A) Let $\sqrt{7+24 i} = x+iy$. Squaring both sides,we get $7+24 i = (x^2-y^2) + 2xyi$.
Equating the real and imaginary parts,we have $x^2-y^2 = 7$ and $2xy = 24$,which implies $xy = 12$ or $y = \frac{12}{x}$.
Substituting $y$ into the first equation: $x^2 - (\frac{12}{x})^2 = 7 \Rightarrow x^2 - \frac{144}{x^2} = 7$.
Multiplying by $x^2$: $x^4 - 7x^2 - 144 = 0$.
Let $u = x^2$,then $u^2 - 7u - 144 = 0$. Factoring gives $(u-16)(u+9) = 0$.
Since $x$ is real,$x^2 = 16$,so $x = \pm 4$.
If $x = 4$,$y = 3$. If $x = -4$,$y = -3$.
Thus,the square roots are $\pm(4+3i)$.

(D) Given,$|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$.
Since $1+\omega+\omega^2=0$,we have $-1-\omega = \omega^2$.
Substituting this into the second equation: $|\omega^2 Z + \omega^2| = a$.
Since $|\omega^2| = 1$,this simplifies to $|Z+1| = a$.
We can rewrite this as $|(Z-1)+2| = a$.
By the triangle inequality,$|Z-1+2| \leq |Z-1| + |2|$.
Given $|Z-1| \leq 2$,we have $a = |Z+1| \leq |Z-1| + 2 \leq 2 + 2 = 4$.
Also,the modulus $a$ must be non-negative,so $a \geq 0$.
Thus,the range of possible values for $a$ is $0 \leq a \leq 4$.

(D) Given the equation $z^3+\bar{z}=0$,where $z=x+iy$ is a complex number.
Taking the modulus on both sides: $|z^3| = |-\bar{z}|$.
Since $|z^3| = |z|^3$ and $|-\bar{z}| = |z|$,we have $|z|^3 = |z|$.
This implies $|z|(|z|^2-1) = 0$.
Case $1$: $|z|=0$,which gives the solution $z=0$.
Case $2$: $|z|^2=1$,which implies $z\bar{z}=1$,so $\bar{z}=\frac{1}{z}$.
Substituting $\bar{z}=\frac{1}{z}$ into the original equation: $z^3 + \frac{1}{z} = 0$,which simplifies to $z^4+1=0$.
The equation $z^4 = -1$ has $4$ distinct roots.
Including the solution $z=0$ from Case $1$,the total number of distinct solutions is $1 + 4 = 5$.

(A) Let $z = x + iy$. The condition that $\frac{z-2i}{z-2}$ is purely imaginary implies that its real part is $0$.
Alternatively,$\frac{z-2i}{z-2} + \overline{\left(\frac{z-2i}{z-2}\right)} = 0$.
$\frac{z-2i}{z-2} + \frac{\bar{z}+2i}{\bar{z}-2} = 0$.
$(z-2i)(\bar{z}-2) + (\bar{z}+2i)(z-2) = 0$.
$z\bar{z} - 2z - 2i\bar{z} + 4i + z\bar{z} - 2\bar{z} + 2iz - 4i = 0$.
$2|z|^2 - 2(z+\bar{z}) + 2i(z-\bar{z}) = 0$.
$|z|^2 - (z+\bar{z}) + i(z-\bar{z}) = 0$.
Substituting $z = x+iy$,we get $x^2 + y^2 - 2x - 2y = 0$.
$(x-1)^2 + (y-1)^2 = 2$.
This represents a circle with center $(1, 1)$ and radius $r = \sqrt{2}$.
The circle passes through the origin $(0,0)$.
The entire circle lies in the first quadrant.
Area $= \pi r^2 = \pi(\sqrt{2})^2 = 2\pi$.

(C) Let $z = \frac{3-2 i \sin \theta}{1+2 i \sin \theta}$.
To rationalize,multiply the numerator and denominator by the conjugate of the denominator $(1-2 i \sin \theta)$:
$z = \frac{(3-2 i \sin \theta)(1-2 i \sin \theta)}{(1+2 i \sin \theta)(1-2 i \sin \theta)}$
$z = \frac{3 - 6 i \sin \theta - 2 i \sin \theta + 4 i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
Since $i^2 = -1$,we have:
$z = \frac{3 - 4 \sin^2 \theta - 8 i \sin \theta}{1 + 4 \sin^2 \theta}$
For $z$ to be a purely imaginary number,the real part must be zero:
$\frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0$
$3 - 4 \sin^2 \theta = 0$
$\sin^2 \theta = \frac{3}{4} = \left(\frac{\sqrt{3}}{2}\right)^2$
$\sin^2 \theta = \sin^2 \left(\frac{\pi}{3}\right)$
Therefore,the general solution is $\theta = n \pi \pm \frac{\pi}{3}$.

(C) First,calculate the product of the complex numbers:
$(4-3i)(2+3i) = 8 + 12i - 6i - 9i^2 = 8 + 6i + 9 = 17 + 6i$.
Then,multiply by $(1+4i)$:
$(17+6i)(1+4i) = 17 + 68i + 6i + 24i^2 = 17 + 74i - 24 = -7 + 74i$.
The complex conjugate of a complex number $z = a + bi$ is $\bar{z} = a - bi$.
Therefore,the complex conjugate of $-7 + 74i$ is $-7 - 74i$.

(B) Let $z = \frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}$
Numerator: $(1+i \sqrt{3})(-\sqrt{3}-i) = -\sqrt{3} - i - 3i - i^2 \sqrt{3} = -\sqrt{3} - 4i + \sqrt{3} = -4i$
Denominator: $(1-i)(-i) = -i + i^2 = -i - 1 = -(1+i)$
So,$z = \frac{-4i}{-(1+i)} = \frac{4i}{1+i}$
Multiply numerator and denominator by the conjugate $(1-i)$:
$z = \frac{4i(1-i)}{(1+i)(1-i)} = \frac{4i - 4i^2}{1 - i^2} = \frac{4i + 4}{1 + 1} = \frac{4+4i}{2} = 2+2i$
Since $z = 2+2i$ lies in the $I^{st}$ quadrant,the argument is given by $\theta = \tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.

(D) Given $z=x+iy$ and $x^2+y^2=1$, we can write $z=e^{i\phi}$ where $\phi = \tan^{-1}(\frac{y}{x})$.
$z_1 = ze^{i\theta} = e^{i\phi}e^{i\theta} = e^{i(\phi+\theta)}$.
Then $z_1^{2n} = e^{i2n(\phi+\theta)}$.
Consider the expression $\frac{z_1^{2n}-1}{z_1^{2n}+1} = \frac{e^{i2n(\phi+\theta)}-1}{e^{i2n(\phi+\theta)}+1}$.
Multiply numerator and denominator by $e^{-in(\phi+\theta)}$:
$= \frac{e^{in(\phi+\theta)} - e^{-in(\phi+\theta)}}{e^{in(\phi+\theta)} + e^{-in(\phi+\theta)}} = \frac{2i \sin(n(\phi+\theta))}{2 \cos(n(\phi+\theta))}$.
$= i \tan(n(\phi+\theta)) = i \tan(n(\theta+\tan^{-1}(\frac{y}{x})))$.

(D) Given $Z = \cos \theta + i \sin \theta$.
By De Moivre's Theorem,$Z^{2n} = \cos(2n\theta) + i \sin(2n\theta)$.
Substituting this into the expression:
$\frac{1 + Z^{2n}}{1 - Z^{2n}} = \frac{1 + \cos(2n\theta) + i \sin(2n\theta)}{1 - (\cos(2n\theta) + i \sin(2n\theta))} = \frac{1 + \cos(2n\theta) + i \sin(2n\theta)}{1 - \cos(2n\theta) - i \sin(2n\theta)}$.
Using half-angle identities $1 + \cos(2A) = 2 \cos^2 A$ and $1 - \cos(2A) = 2 \sin^2 A$,and $\sin(2A) = 2 \sin A \cos A$:
$= \frac{2 \cos^2(n\theta) + 2i \sin(n\theta) \cos(n\theta)}{2 \sin^2(n\theta) - 2i \sin(n\theta) \cos(n\theta)} = \frac{2 \cos(n\theta) [\cos(n\theta) + i \sin(n\theta)]}{2i \sin(n\theta) [-i \sin(n\theta) + \cos(n\theta)]}$.
$= \frac{\cos(n\theta)}{i \sin(n\theta)} = -i \cot(n\theta)$.

(A) Given that $(r, \theta) = r(\cos \theta + i \sin \theta) = re^{i\theta}$.
Thus,$x = e^{i\alpha}$,$y = e^{i\beta}$,and $z = e^{i\gamma}$.
Given $x + y + z = 0$,we know that for complex numbers,if $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.
Dividing by $xyz$,we get $\frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} = 3$.
Substituting the exponential forms:
$\frac{e^{2i\alpha}}{e^{i\beta}e^{i\gamma}} + \frac{e^{2i\beta}}{e^{i\alpha}e^{i\gamma}} + \frac{e^{2i\gamma}}{e^{i\alpha}e^{i\beta}} = 3$.
$e^{i(2\alpha - \beta - \gamma)} + e^{i(2\beta - \alpha - \gamma)} + e^{i(2\gamma - \alpha - \beta)} = 3$.
Taking the real part of both sides:
$\cos(2\alpha - \beta - \gamma) + \cos(2\beta - \alpha - \gamma) + \cos(2\gamma - \alpha - \beta) = 3$.

(C) Given $x+y=20$. We want to maximize $f(x, y) = x^3 y$.
Using the $AM$-$GM$ inequality for the terms $\frac{x}{3}, \frac{x}{3}, \frac{x}{3}, y$:
$\frac{\frac{x}{3} + \frac{x}{3} + \frac{x}{3} + y}{4} \geq \sqrt[4]{\left(\frac{x}{3}\right)^3 y}$
$\frac{x+y}{4} \geq \sqrt[4]{\frac{x^3 y}{27}}$
$\frac{20}{4} \geq \sqrt[4]{\frac{x^3 y}{27}} \Rightarrow 5 \geq \sqrt[4]{\frac{x^3 y}{27}}$
$5^4 \geq \frac{x^3 y}{27} \Rightarrow x^3 y \leq 27 \times 625 = 16875$.
So,$k = 16875$.
The maximum occurs when $\frac{x}{3} = y$.
Since $x+y=20$,we have $3y+y=20$ $\Rightarrow 4y=20$ $\Rightarrow y=5=\beta$ and $x=15=\alpha$.
Now,$\frac{k}{\alpha^2 \beta^2} = \frac{27 \times 625}{15^2 \times 5^2} = \frac{27 \times 625}{225 \times 25} = \frac{16875}{5625} = 3$.
Since $\frac{\alpha}{\beta} = \frac{15}{5} = 3$,the correct option is $\frac{\alpha}{\beta}$.

(C) Since $A$ is a singular matrix,its determinant is zero: $\left| \begin{smallmatrix} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{smallmatrix} \right| = 0$.
Expanding along the first row: $2(5k - 2) - 3(5 - 8) + 4(1 - 4k) = 0$.
$10k - 4 + 9 + 4 - 16k = 0$.
$-6k + 9 = 0$ $\Rightarrow 6k = 9$ $\Rightarrow k = \frac{3}{2}$.
The roots of the required quadratic equation are $\alpha = k = \frac{3}{2}$ and $\beta = \frac{1}{k} = \frac{2}{3}$.
The quadratic equation is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
$x^2 - (\frac{3}{2} + \frac{2}{3})x + (\frac{3}{2} \times \frac{2}{3}) = 0$.
$x^2 - (\frac{9+4}{6})x + 1 = 0$.
$x^2 - \frac{13}{6}x + 1 = 0$.
Multiplying by $6$,we get $6x^2 - 13x + 6 = 0$.

(B) Given the equation $a^2 x^4 + b^2 y^4 = c^6$.
Applying the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for the two terms $a^2 x^4$ and $b^2 y^4$:
$\frac{a^2 x^4 + b^2 y^4}{2} \geq \sqrt{(a^2 x^4)(b^2 y^4)}$
Substituting the given value $a^2 x^4 + b^2 y^4 = c^6$:
$\frac{c^6}{2} \geq \sqrt{a^2 b^2 x^4 y^4}$
$\frac{c^6}{2} \geq ab x^2 y^2$
$x^2 y^2 \leq \frac{c^6}{2ab}$
Taking the square root on both sides:
$xy \leq \frac{c^3}{\sqrt{2ab}}$
Thus,the maximum value of $xy$ is $\frac{c^3}{\sqrt{2ab}}$.

(D) Let $f(x) = \cot x$.
Given $1^{\prime} = 0.0175$ radians,then $2^{\prime} = 0.035$ radians.
The derivative is $f^{\prime}(x) = -\operatorname{cosec}^2 x$.
Using the linear approximation formula $f(a+h) \approx f(a) + h f^{\prime}(a)$:
$f(45^{\circ} + 2^{\prime}) \approx \cot(45^{\circ}) + (0.035) \times (-\operatorname{cosec}^2(45^{\circ}))$.
Since $\cot(45^{\circ}) = 1$ and $\operatorname{cosec}(45^{\circ}) = \sqrt{2}$,then $\operatorname{cosec}^2(45^{\circ}) = 2$.
$f(45^{\circ} + 2^{\prime}) \approx 1 - 0.035 \times 2 = 1 - 0.07 = 0.93$.

(D) Given: $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$
Taking $\cos$ on both sides:
$\cos(\cos ^{-1} 2x + \cos ^{-1} 3x) = \cos(\frac{\pi}{3})$
Using the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$(2x)(3x) - \sqrt{1-(2x)^2} \sqrt{1-(3x)^2} = \frac{1}{2}$
$6x^2 - \frac{1}{2} = \sqrt{1-4x^2} \sqrt{1-9x^2}$
Squaring both sides:
$(6x^2 - \frac{1}{2})^2 = (1-4x^2)(1-9x^2)$
$36x^4 - 6x^2 + \frac{1}{4} = 1 - 13x^2 + 36x^4$
$7x^2 = \frac{3}{4}$
$x^2 = \frac{3}{28}$
Then $4x^2 = 4 \times \frac{3}{28} = \frac{3}{7} = \frac{a}{b}$
Thus,$a = 3$ and $b = 7$.
$a + b = 3 + 7 = 10$.

(C) Given the equation: $\sin \left(2 \tan ^{-1} \frac{3}{4}\right)=\cos \left(2 \tan ^{-1} x\right)$
Using the formula $\sin(2 \tan^{-1} \theta) = \frac{2\theta}{1+\theta^2}$ and $\cos(2 \tan^{-1} x) = \frac{1-x^2}{1+x^2}$:
$\frac{2 \times \frac{3}{4}}{1+\left(\frac{3}{4}\right)^2} = \frac{1-x^2}{1+x^2}$
$\frac{\frac{3}{2}}{1+\frac{9}{16}} = \frac{1-x^2}{1+x^2}$
$\frac{\frac{3}{2}}{\frac{25}{16}} = \frac{1-x^2}{1+x^2}$
$\frac{3}{2} \times \frac{16}{25} = \frac{24}{25} = \frac{1-x^2}{1+x^2}$
$24(1+x^2) = 25(1-x^2)$
$24 + 24x^2 = 25 - 25x^2$
$49x^2 = 1$
$x^2 = \frac{1}{49}$
$x = \frac{1}{7}$

(C) In $\triangle PQR$,$L, M, N$ are midpoints of $PQ, QR, RP$ respectively.
By the midpoint theorem,$\overrightarrow{LM} = \frac{1}{2} \overrightarrow{PR}$,$\overrightarrow{MN} = \frac{1}{2} \overrightarrow{PQ}$,and $\overrightarrow{NL} = \frac{1}{2} \overrightarrow{QR}$.
Also,$\overrightarrow{QM} = \frac{1}{2} \overrightarrow{QR} = \overrightarrow{NL}$,$\overrightarrow{LN} = \frac{1}{2} \overrightarrow{PR} = \overrightarrow{ML}$,$\overrightarrow{RN} = \frac{1}{2} \overrightarrow{RP} = -\frac{1}{2} \overrightarrow{PR} = -\overrightarrow{ML}$.
Substituting these into the expression:
$\overrightarrow{QM} + \overrightarrow{LN} + \overrightarrow{ML} + \overrightarrow{RN} - \overrightarrow{MN} - \overrightarrow{QL}$
$= \overrightarrow{NL} + \overrightarrow{ML} + \overrightarrow{ML} - \overrightarrow{ML} - \overrightarrow{MN} - \overrightarrow{QL}$
Since $\overrightarrow{QL} = -\overrightarrow{LQ} = -\overrightarrow{MN}$,the expression simplifies to $\vec{0}$.

(B) The slope of the line passing through $(0,3)$ and $(5,-2)$ is $m = \frac{-2-3}{5-0} = \frac{-5}{5} = -1$.
The equation of the line is $y - 3 = -1(x - 0)$,which simplifies to $x + y - 3 = 0$ or $y = -x + 3$.
For the line to be tangent to the curve $y = \frac{c}{x+1}$,the derivative of the curve must equal the slope of the line: $\frac{dy}{dx} = \frac{-c}{(x+1)^2} = -1$.
This implies $c = (x+1)^2$.
Substituting $y = \frac{c}{x+1}$ into the line equation: $\frac{c}{x+1} = -x + 3$.
Since $c = (x+1)^2$,we have $\frac{(x+1)^2}{x+1} = -x + 3$,which simplifies to $x + 1 = -x + 3$.
Solving for $x$: $2x = 2$,so $x = 1$.
Substituting $x = 1$ into $c = (x+1)^2$,we get $c = (1+1)^2 = 4$.

(C) Let the line pass through $A(0, 2, 4)$ and $B(3, 0, 1)$. The direction ratios of the line $AB$ are $(3-0, 0-2, 1-4) = (3, -2, -3)$.
The equation of the line $AB$ is $\frac{x-0}{3} = \frac{y-2}{-2} = \frac{z-4}{-3} = k$.
Any point $C$ on the line is $(3k, -2k+2, -3k+4)$.
Since $PC$ is perpendicular to $AB$,the direction ratios of $PC$ are $(3k+1, -2k+1, -3k+4)$.
The dot product of the direction ratios of $PC$ and $AB$ is zero:
$3(3k+1) + (-2)(-2k+1) + (-3)(-3k+4) = 0$
$9k + 3 + 4k - 2 + 9k - 12 = 0$
$22k - 11 = 0 \Rightarrow k = \frac{1}{2}$.
The coordinates of $C$ are $(\frac{3}{2}, 1, \frac{5}{2})$.
The perpendicular distance $PC$ is $\sqrt{(\frac{3}{2} - (-1))^2 + (1-1)^2 + (\frac{5}{2} - 0)^2} = \sqrt{(\frac{5}{2})^2 + 0^2 + (\frac{5}{2})^2} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}}$.

(C) The equation of the curve is $x^{2/3} + y^{2/3} = a^{2/3}$ $\dots(i)$.
Any point on the curve can be represented as $(a \cos^3 \theta, a \sin^3 \theta)$.
The slope of the tangent is $\frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
The equation of the tangent is $y - a \sin^3 \theta = -\tan \theta (x - a \cos^3 \theta)$,which simplifies to $x \sin \theta + y \cos \theta = a \sin \theta \cos \theta = \frac{a}{2} \sin 2\theta$.
The perpendicular distance $p_1$ from the origin $(0,0)$ to the tangent is $p_1 = \left| \frac{-\frac{a}{2} \sin 2\theta}{\sqrt{\sin^2 \theta + \cos^2 \theta}} \right| = \frac{a}{2} \sin 2\theta$,so $2p_1 = a \sin 2\theta$ $\dots(ii)$.
The slope of the normal is $\cot \theta$. The equation of the normal is $y - a \sin^3 \theta = \cot \theta (x - a \cos^3 \theta)$,which simplifies to $x \cos \theta - y \sin \theta = a \cos 2\theta$.
The perpendicular distance $p_2$ from the origin to the normal is $p_2 = \left| \frac{-a \cos 2\theta}{\sqrt{\cos^2 \theta + \sin^2 \theta}} \right| = a \cos 2\theta$ $\dots(iii)$.
Squaring and adding $(ii)$ and $(iii)$ gives $(2p_1)^2 + p_2^2 = a^2 \sin^2 2\theta + a^2 \cos^2 2\theta = a^2$.
Thus,$4p_1^2 + p_2^2 = a^2$. Comparing this with $k_1 p_1^2 + k_2 p_2^2 = a^2$,we get $k_1 = 4$ and $k_2 = 1$.
Therefore,$k_1 + k_2 = 4 + 1 = 5$.

(A) The equation of line $AB$ passing through $A(1, 0, 2)$ and $B(2, 1, 0)$ is $\frac{x-1}{2-1} = \frac{y-0}{1-0} = \frac{z-2}{0-2} = \lambda$ $\Rightarrow \frac{x-1}{1} = \frac{y}{1} = \frac{z-2}{-2} = \lambda$.
Any point on $AB$ is $P(\lambda+1, \lambda, -2\lambda+2)$.
The equation of line $CD$ passing through $C(2, -5, 3)$ and $D(0, 3, 2)$ is $\frac{x-2}{0-2} = \frac{y+5}{3+5} = \frac{z-3}{2-3} = \mu$ $\Rightarrow \frac{x-2}{-2} = \frac{y+5}{8} = \frac{z-3}{-1} = \mu$.
Any point on $CD$ is $Q(-2\mu+2, 8\mu-5, -\mu+3)$.
For intersection,$P = Q$:
$\lambda+1 = -2\mu+2 \Rightarrow \lambda + 2\mu = 1$
$\lambda = 8\mu-5 \Rightarrow \lambda - 8\mu = -5$
Subtracting the equations: $10\mu = 6 \Rightarrow \mu = \frac{3}{5}$.
Substituting $\mu$: $\lambda = 8(\frac{3}{5}) - 5 = \frac{24-25}{5} = -\frac{1}{5}$.
Wait,checking intersection: $\lambda+1 = -2(3/5)+2 = 4/5$,$\lambda = -1/5$.
Re-calculating: $\lambda+2\mu=1$ and $\lambda-8\mu=-5$. Subtracting: $10\mu=6 \Rightarrow \mu=0.6$. $\lambda = 1 - 1.2 = -0.2$.
Using $\lambda = -0.2$: $x = 0.8, y = -0.2, z = 2.4$.
Check $CD$: $x = -2(0.6)+2 = 0.8, y = 8(0.6)-5 = -0.2, z = -0.6+3 = 2.4$.
$P(0.8, -0.2, 2.4) = (4/5, -1/5, 12/5)$.
$a+b+c = (4-1+12)/5 = 15/5 = 3$.

(B) Let $x = \sinh^{-1}(\sqrt{8})$ and $y = \cosh^{-1}(5)$.
Then $\sinh x = \sqrt{8}$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = 3$.
Thus $e^x = \cosh x + \sinh x = 3 + \sqrt{8} = 3 + 2\sqrt{2}$.
For $y = \cosh^{-1}(5)$,$\cosh y = 5$ and $\sinh y = \sqrt{\cosh^2 y - 1} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$.
Thus $e^y = \cosh y + \sinh y = 5 + 2\sqrt{6}$.
We need to evaluate $\cosh(x + y) = \frac{e^{x+y} + e^{-(x+y)}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$.
Note that $e^{-x} = \frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}$ and $e^{-y} = \frac{1}{5 + 2\sqrt{6}} = 5 - 2\sqrt{6}$.
$e^x e^y = (3 + 2\sqrt{2})(5 + 2\sqrt{6}) = 15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}$.
$e^{-x} e^{-y} = (3 - 2\sqrt{2})(5 - 2\sqrt{6}) = 15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3}$.
$\cosh(x + y) = \frac{(15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}) + (15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3})}{2} = \frac{30 + 16\sqrt{3}}{2} = 15 + 8\sqrt{3}$.

(A) Let $x = \cos \theta$. As $x \rightarrow 1$,$\theta \rightarrow 0$.
Substituting this into the limit:
$\lim _{\theta \rightarrow 0} \frac{\sqrt{\cos \theta}-1}{\theta^2}$
Multiply the numerator and denominator by $(\sqrt{\cos \theta}+1)$:
$= \lim _{\theta \rightarrow 0} \frac{\cos \theta - 1}{\theta^2(\sqrt{\cos \theta}+1)}$
Using the identity $\cos \theta - 1 = -2 \sin^2(\frac{\theta}{2})$:
$= \lim _{\theta \rightarrow 0} \frac{-2 \sin^2(\frac{\theta}{2})}{\theta^2(\sqrt{\cos \theta}+1)}$
$= \lim _{\theta}$ ${\rightarrow 0} -2 \cdot \left(\frac{\sin(\frac{\theta}{2})}{\frac{\theta}{2} \cdot 2}\right)^2 \cdot \frac{1}{\sqrt{\cos \theta}+1}$
$= -2 \cdot \frac{1}{4} \cdot \frac{1}{1+1} = -2 \cdot \frac{1}{4} \cdot \frac{1}{2} = -\frac{1}{4}$.

(D) Let $y = \lim _{n \rightarrow \infty} \left[ \prod_{r=1}^n \left(1 + \frac{r^2}{n^2} \right) \right]^{\frac{1}{n}}$.
Taking the natural logarithm on both sides:
$\log y = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1 + \left(\frac{r}{n}\right)^2 \right)$.
This is a Riemann sum,which can be expressed as the definite integral:
$\log y = \int_0^1 \log(1+x^2) dx$.
Using integration by parts,let $u = \log(1+x^2)$ and $dv = dx$:
$\log y = [x \log(1+x^2)]_0^1 - \int_0^1 \frac{2x^2}{1+x^2} dx$.
$\log y = \log 2 - 2 \int_0^1 \left(1 - \frac{1}{1+x^2} \right) dx$.
$\log y = \log 2 - 2 [x - \tan^{-1} x]_0^1$.
$\log y = \log 2 - 2 (1 - \frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2}$.
Thus,$y = e^{\log 2 - 2 + \frac{\pi}{2}} = 2 e^{\frac{\pi}{2} - 2}$.
Comparing with $ae^b$,we get $a = 2$ and $b = \frac{\pi}{2} - 2$.
Therefore,$a + b = 2 + \frac{\pi}{2} - 2 = \frac{\pi}{2}$.

(B) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
Evaluating the limit: $\lim_{x \to 0} \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$.
This is a $1^{\infty}$ indeterminate form.
Using the formula $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x)-1]}$:
$\lim_{x \to 0} f(x) = e^{\lim_{x \to 0} \frac{1}{x} \left( \frac{1+x}{1-x} - 1 \right)}$
$= e^{\lim_{x \to 0} \frac{1}{x} \left( \frac{1+x - (1-x)}{1-x} \right)}$
$= e^{\lim_{x \to 0} \frac{1}{x} \left( \frac{2x}{1-x} \right)}$
$= e^{\lim_{x \to 0} \frac{2}{1-x}}$
$= e^{\frac{2}{1-0}} = e^2$.
Therefore,$f(0) = e^2$.

(D) Let $\Delta = \left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|$.
Option $(A)$: $\left|\begin{array}{ccc}a+1 & b+1 & c+1 \\ a^2+1 & b^2+1 & c^2+1 \\ 1 & 1 & 1\end{array}\right| = \left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right| + \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right| = \Delta + 0 = \Delta$.
Option $(B)$: Applying $C_1 \rightarrow C_1-C_2$ and $C_2 \rightarrow C_2-C_3$,we get $\left|\begin{array}{ccc}a-b & b-c & c \\ a^2-b^2 & b^2-c^2 & c^2 \\ 0 & 0 & 1\end{array}\right|$. Expanding along $R_3$,we get $(a-b)(b^2-c^2) - (b-c)(a^2-b^2) = (a-b)(b-c)(b+c) - (b-c)(a-b)(a+b) = (a-b)(b-c)(b+c-a-b) = (a-b)(b-c)(c-a)$,which is the value of $\Delta$.
Option $(C)$: $\left|\begin{array}{ccc}a(a+1) & b(b+1) & c(c+1) \\ a+1 & b+1 & c+1 \\ -1 & -1 & -1\end{array}\right|$. Adding $R_2$ to $R_1$,we get $\left|\begin{array}{ccc}a^2+2a+1 & b^2+2b+1 & c^2+2c+1 \\ a+1 & b+1 & c+1 \\ -1 & -1 & -1\end{array}\right| = \left|\begin{array}{ccc}(a+1)^2 & (b+1)^2 & (c+1)^2 \\ a+1 & b+1 & c+1 \\ -1 & -1 & -1\end{array}\right|$. This simplifies to $\Delta$.
Option $(D)$: The determinant $\left|\begin{array}{ccc}a+b & b+c & c+a \\ a^2+b^2 & b^2+c^2 & c^2+a^2 \\ 2 & 2 & 2\end{array}\right|$ is not equal to $\Delta$.

(A) For the matrix to be singular,its determinant must be equal to $0$.
Expanding the determinant along the second column:
$\begin{vmatrix} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2x-1 \end{vmatrix} = (x+3) \begin{vmatrix} x-2 & 1 \\ 2 & 2x-1 \end{vmatrix} = 0$
$(x+3) [(x-2)(2x-1) - 2] = 0$
$(x+3) [2x^2 - x - 4x + 2 - 2] = 0$
$(x+3) [2x^2 - 5x] = 0$
$x(x+3)(2x-5) = 0$
The roots are $x = -3, 0, \frac{5}{2}$.
Given $\alpha < \beta < \gamma$,we have $\alpha = -3$,$\beta = 0$,and $\gamma = \frac{5}{2}$.
Now,calculate $2\alpha + 3\beta + 4\gamma$:
$2(-3) + 3(0) + 4(\frac{5}{2}) = -6 + 0 + 10 = 4$.

(A) Given the determinant equation: $\left|\begin{array}{ccc}x^2+2x & x+2 & 1 \\ 2x+1 & x-1 & 1 \\ x+2 & -1 & 1\end{array}\right|=0$
Applying row operations $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$:
$\left|\begin{array}{ccc}x^2+2x & x+2 & 1 \\ 1-x^2 & -3 & 0 \\ -x^2-x+2 & -x-3 & 0\end{array}\right|=0$
Expanding along the third column:
$1 \cdot [(1-x^2)(-x-3) - (-3)(-x^2-x+2)] = 0$
$(1-x^2)(-x-3) + 3(-x^2-x+2) = 0$
$-x-3+x^3+3x^2-3x^2-3x+6 = 0$
$x^3-4x+3 = 0$
Factoring the cubic equation: $(x-1)(x^2+x-3) = 0$
The roots are $x=1$ and $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-3)}}{2} = \frac{-1 \pm \sqrt{13}}{2}$.
The positive roots are $x=1$ and $x = \frac{\sqrt{13}-1}{2}$.
Sum of positive roots $= 1 + \frac{\sqrt{13}-1}{2} = \frac{2+\sqrt{13}-1}{2} = \frac{1+\sqrt{13}}{2}$.

(D) Given $3 A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$. Since $A A^{T} = I$,we have $(3 A)(3 A)^{T} = 9 I$.
$(3 A)(3 A)^{T} = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$.
Multiplying the matrices,the element at $(1, 3)$ is $1(a) + 2(2) + 2(b) = a + 4 + 2b = 0$.
The element at $(2, 3)$ is $2(a) + 1(2) - 2(b) = 2a + 2 - 2b = 0$.
From $2a - 2b = -2$,we get $a - b = -1$,so $a = b - 1$.
Substituting into $a + 2b = -4$,we get $(b - 1) + 2b = -4 \Rightarrow 3b = -3 \Rightarrow b = -1$.
Then $a = -1 - 1 = -2$.
Thus,$\frac{a}{b} + \frac{b}{a} = \frac{-2}{-1} + \frac{-1}{-2} = 2 + \frac{1}{2} = \frac{5}{2}$.

(D) Given $A = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix}$. Let $B = \begin{bmatrix} x & y & z \\ a & b & c \\ u & v & w \end{bmatrix}$.
Since $AB = BA$,we perform matrix multiplication on both sides.
Calculating $AB = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix} \begin{bmatrix} x & y & z \\ a & b & c \\ u & v & w \end{bmatrix} = \begin{bmatrix} a+2u & b+2v & c+2w \\ 2x+3a & 2y+3b & 2z+3c \\ 4x+3u & 4y+3v & 4z+3w \end{bmatrix}$.
Calculating $BA = \begin{bmatrix} x & y & z \\ a & b & c \\ u & v & w \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 2y+4z & x+3y & 2x+3z \\ 2b+4c & a+3b & 2a+3c \\ 2v+4w & u+3v & 2u+3w \end{bmatrix}$.
By comparing the elements of $AB$ and $BA$,we check the given options.
Testing option $D$: $B = \begin{bmatrix} 9 & -3 & -6 \\ -6 & -8 & 4 \\ -12 & 4 & -2 \end{bmatrix}$.
Substituting these values into the matrix multiplication confirms that $AB = BA$ holds true for this matrix.

(C) Given that $A$ is a $3 \times 3$ matrix and $|A| = K$.
First,consider the matrix $(A - A^T)$.
Since $(A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T)$,the matrix $(A - A^T)$ is a skew-symmetric matrix.
For any skew-symmetric matrix of odd order $n$,the determinant is $0$. Since $n = 3$ is odd,$|A - A^T| = 0$.
Next,consider the matrix $(AA^T)$.
Using the property of determinants $|XY| = |X||Y|$ and $|A^T| = |A|$,we have $|AA^T| = |A||A^T| = |A||A| = K \cdot K = K^2$.
Therefore,the sum of the determinants is $|AA^T| + |A - A^T| = K^2 + 0 = K^2$.

(D) Given that $A$ is a $4 \times 4$ matrix,so $n = 4$.
The adjoint matrix $P = \operatorname{adj}(A)$.
We know that $|\operatorname{adj}(A)| = |A|^{n-1}$,so $|P| = |A|^{4-1} = |A|^3$.
Given $|P| = |\frac{A}{2}|$. Since $A$ is a $4 \times 4$ matrix,$|\frac{A}{2}| = \frac{1}{2^4} |A| = \frac{1}{16} |A|$.
Equating the two expressions: $|A|^3 = \frac{1}{16} |A|$.
This implies $|A|^3 - \frac{1}{16} |A| = 0$,so $|A|(|A|^2 - \frac{1}{16}) = 0$.
Since $|A| \neq 0$ (as $A^{-1}$ exists),$|A|^2 = \frac{1}{16}$,which gives $|A| = \pm \frac{1}{4}$.
Finally,$|A^{-1}| = \frac{1}{|A|} = \frac{1}{\pm 1/4} = \pm 4$.

(C) For any square matrix $B$ of order $n$,the property of the adjoint matrix is $|\operatorname{Adj}(B)| = |B|^{n-1}$.
In the given Reason $(R)$,it is stated that $|\operatorname{Adj}(B)| = |B|^n$,which is incorrect because the exponent should be $n-1$.
For Assertion $(A)$,given $B$ is a $3 \times 3$ matrix $(n=3)$ and $|B|=6$,we have $|\operatorname{Adj}(B)| = |B|^{3-1} = |B|^2 = 6^2 = 36$.
Thus,Assertion $(A)$ is true,but Reason $(R)$ is false.

(A) Given that $P$ and $Q$ are $3 \times 3$ matrices,we have $|PQ| = |P||Q| = 1$.
Since $|P| = 9$,we get $|Q| = \frac{1}{9}$.
We need to find $|\text{adj}(P \cdot \text{adj}(3Q))|$.
Using the property $|\text{adj}(A)| = |A|^{n-1}$ for an $n \times n$ matrix,here $n=3$,so $|\text{adj}(A)| = |A|^2$.
Thus,$|\text{adj}(P \cdot \text{adj}(3Q))| = |P \cdot \text{adj}(3Q)|^2 = |P|^2 \cdot |\text{adj}(3Q)|^2$.
Since $|\text{adj}(3Q)| = |3Q|^{3-1} = |3Q|^2 = (3^3 |Q|)^2 = (27 |Q|)^2$.
Substituting $|Q| = \frac{1}{9}$,we get $|\text{adj}(3Q)| = (27 \times \frac{1}{9})^2 = 3^2 = 9$.
Now,$|\text{adj}(P \cdot \text{adj}(3Q))| = |P|^2 \cdot (9)^2 = 9^2 \cdot 9^2 = 9^4$.

(A) We know that $\operatorname{Adj} A = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix}$,where $C_{ij}$ is the cofactor of the element $a_{ij}$.
Given $\operatorname{Adj} A = \begin{bmatrix} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{bmatrix}$.
Comparing the elements:
$C_{11} = 7 \Rightarrow \begin{vmatrix} 2 & b \\ 1 & 3 \end{vmatrix} = 7 \Rightarrow 6 - b = 7 \Rightarrow b = -1$.
$C_{13} = 1 \Rightarrow \begin{vmatrix} 1 & 2 \\ c & 1 \end{vmatrix} = 1 \Rightarrow 1 - 2c = 1 \Rightarrow c = 0$.
$C_{33} = 5 \Rightarrow \begin{vmatrix} a & 1 \\ 1 & 2 \end{vmatrix} = 5 \Rightarrow 2a - 1 = 5 \Rightarrow 2a = 6 \Rightarrow a = 3$.
Thus,$a^2 + b^2 + c^2 = (3)^2 + (-1)^2 + (0)^2 = 9 + 1 + 0 = 10$.

(B) Given the equation $ABCDE = I$.
Multiply both sides by $A^{-1}$ from the left: $A^{-1}(ABCDE) = A^{-1}I \Rightarrow BCDE = A^{-1}$.
Multiply both sides by $B^{-1}$ from the left: $B^{-1}(BCDE) = B^{-1}A^{-1} \Rightarrow CDE = B^{-1}A^{-1}$.
Multiply both sides by $E^{-1}$ from the right: $(CDE)E^{-1} = B^{-1}A^{-1}E^{-1} \Rightarrow CD = B^{-1}A^{-1}E^{-1}$.
Multiply both sides by $D^{-1}$ from the right: $(CD)D^{-1} = B^{-1}A^{-1}E^{-1}D^{-1} \Rightarrow C = B^{-1}A^{-1}E^{-1}D^{-1}$.
Taking the inverse of both sides: $C^{-1} = (B^{-1}A^{-1}E^{-1}D^{-1})^{-1}$.
Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we get $C^{-1} = (D^{-1})^{-1}(E^{-1})^{-1}(A^{-1})^{-1}(B^{-1})^{-1} = DEAB$.

(C) Given that $n \geq 2$ and $a_{ij} = i + j$.
Case-$1$: Let $n = 2$.
$A = \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} \Rightarrow |A| = (2)(4) - (3)(3) = 8 - 9 = -1 \neq 0$.
Since the determinant is non-zero,the rank of $A$ is $2$.
Case-$2$: Let $n = 3$.
$A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$,we get:
$A \sim \begin{bmatrix} 2 & 3 & 4 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$.
Since $R_2$ and $R_3$ are identical,the rank is $2$.
For any $n > 2$,the rows $R_i$ follow the pattern $R_i = (i+1, i+2, \dots, i+n)$.
Note that $R_3 - R_2 = R_2 - R_1 = (1, 1, \dots, 1)$.
Thus,$R_3 - 2R_2 + R_1 = 0$,which means the rows are linearly dependent for $n \geq 3$.
Therefore,the rank of $A$ is $2$ for all $n \geq 2$.

(C) homogeneous system of linear equations $AX = 0$ has only the trivial solution $(x = 0, y = 0, z = 0)$ if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| \neq 0$.
For a $3 \times 3$ matrix,if the determinant is non-zero,the matrix is non-singular and has full rank.
Since the matrix $A$ is of order $3 \times 3$ and $|A| \neq 0$,the rank of the matrix $A$ is $3$.

(C) To find the roots $\alpha, \beta, \gamma$,we expand the determinant: $\left|\begin{array}{ccc} 1-x & -2 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0$.
Expanding along the first row:
$(1-x)[(4-x)(1-x) - 4] - (-2)[-2(1-x) - (-2)] + 1[4 - (4-x)] = 0$.
$(1-x)[4 - 4x - x + x^2 - 4] + 2[-2 + 2x + 2] + 1[4 - 4 + x] = 0$.
$(1-x)[x^2 - 5x] + 2[2x] + x = 0$.
$x^2 - 5x - x^3 + 5x^2 + 4x + x = 0$.
$-x^3 + 6x^2 = 0$,which implies $x^3 - 6x^2 = 0$.
Comparing this with the cubic equation $ax^3 + bx^2 + cx + d = 0$,we have $a=1, b=-6, c=0, d=0$.
For a cubic equation,the sum of roots taken two at a time is given by $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$.
Substituting the values,$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{0}{1} = 0$.

(D) Given determinant: $\Delta = \left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$
Applying column operations $C_1 \rightarrow C_1 - C_2$ and $C_3 \rightarrow C_3 - C_2$:
$\Delta = \left|\begin{array}{ccc}0 & 1 & 0 \\ a^2-b^2 & b^2 & c^2-b^2 \\ a^3-b^3 & b^3 & c^3-b^3\end{array}\right|$
Taking common factors $(a-b)$ from $C_1$ and $(c-b)$ from $C_3$:
$\Delta = (a-b)(c-b) \left|\begin{array}{ccc}0 & 1 & 0 \\ a+b & b^2 & c+b \\ a^2+ab+b^2 & b^3 & c^2+bc+b^2\end{array}\right|$
Applying $C_3 \rightarrow C_3 - C_1$:
$\Delta = (a-b)(c-b) \left|\begin{array}{ccc}0 & 1 & 0 \\ a+b & b^2 & c-a \\ a^2+ab+b^2 & b^3 & c^2-a^2+bc-ab\end{array}\right|$
Since $c^2-a^2+bc-ab = (c-a)(c+a) + b(c-a) = (c-a)(a+b+c)$,we take $(c-a)$ common from $C_3$:
$\Delta = (a-b)(c-b)(c-a) \left|\begin{array}{ccc}0 & 1 & 0 \\ a+b & b^2 & 1 \\ a^2+ab+b^2 & b^3 & a+b+c\end{array}\right|$
Expanding along $R_1$:
$\Delta = (a-b)(c-b)(c-a) \cdot (-1) \cdot [(a+b)(a+b+c) - (a^2+ab+b^2)]$
$\Delta = (a-b)(b-c)(c-a) [a^2+ab+ac+ab+b^2+bc - a^2-ab-b^2]$
$\Delta = (a-b)(b-c)(c-a)(ab+bc+ca)$.

(B) Let $\Delta = \left|\begin{array}{ccc} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{array}\right|$.
Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$,we get:
$\Delta = \left|\begin{array}{ccc} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2a & b \\ 2(a+b+c) & a & c+a+2b \end{array}\right|$.
Taking $2(a+b+c)$ common from $C_1$:
$\Delta = 2(a+b+c) \left|\begin{array}{ccc} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{array}\right|$.
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = 2(a+b+c) \left|\begin{array}{ccc} 1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c \end{array}\right|$.
Expanding along $C_1$,we get:
$\Delta = 2(a+b+c) \cdot (1) \cdot [(a+b+c)(a+b+c) - 0] = 2(a+b+c)^3$.

(A) Let $B = A - A^{T}$.
The transpose of $B$ is $B^{T} = (A - A^{T})^{T} = A^{T} - (A^{T})^{T} = A^{T} - A = -(A - A^{T}) = -B$.
Since $B^{T} = -B$,the matrix $B = A - A^{T}$ is a skew-symmetric matrix.
For any skew-symmetric matrix $B$ of odd order $n$,the determinant is given by $\det(B) = (-1)^{n} \det(B^{T})$.
Since $B^{T} = -B$,we have $\det(B) = (-1)^{n} \det(-B) = (-1)^{n} (-1)^{n} \det(B) = (-1)^{2n} \det(B) = \det(B)$.
However,specifically for odd order $n=3$,$\det(B) = \det(B^{T}) = \det(-B) = (-1)^{3} \det(B) = -\det(B)$.
This implies $2 \det(B) = 0$,so $\det(B) = 0$.
Therefore,$\det(A - A^{T}) = 0$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$.
First,we find the characteristic equation of $A$ using the formula $A^2 - \text{tr}(A)A + |A|I = 0$.
The trace of $A$ is $\text{tr}(A) = 1 + (-5) = -4$.
The determinant of $A$ is $|A| = (1)(-5) - (2)(-2) = -5 + 4 = -1$.
Thus,the characteristic equation is $A^2 - (-4)A + (-1)I = 0$,which simplifies to $A^2 + 4A - I = 0$,or $A^2 + 4A = I$.
Multiplying by $2$,we get $2A^2 + 8A = 2I$.
Comparing this with the given equation $\alpha A^2 + \beta A = 2I$,we identify $\alpha = 2$ and $\beta = 8$.
Therefore,$\alpha + \beta = 2 + 8 = 10$.

(D) Given $A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 1+0+6 & 0+0+4 & 2+0+8 \\ 2+2+9 & 0+1+6 & 4+3+12 \\ 3+4+12 & 0+2+8 & 6+6+16 \end{bmatrix} = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix}$.
Now,calculate $5A = 5 \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix}$.
And $6I = 6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$.
Finally,compute $A^2 - 5A + 6I$:
$A^2 - 5A + 6I = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \begin{bmatrix} 7-5+6 & 4-0+0 & 10-10+0 \\ 13-10+0 & 7-5+6 & 19-15+0 \\ 19-15+0 & 10-10+0 & 28-20+6 \end{bmatrix} = \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix}$.

(C) Given,$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5\end{array}\right|$,$\Delta_1=\left|\begin{array}{ccc}5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5\end{array}\right|$ and $X=\left[\begin{array}{l}\alpha \\ 2 \\ \beta\end{array}\right]$.
By Cramer's rule,$x = \frac{\Delta_1}{\Delta}$.
First,calculate $\Delta = 1(-5-2) - 1(10+2) + 1(2-1) = -7 - 12 + 1 = -18$.
Next,calculate $\Delta_1 = 5(-5-2) - 1(20-22) + 1(4+11) = 5(-7) - 1(-2) + 15 = -35 + 2 + 15 = -18$.
Thus,$\alpha = x = \frac{\Delta_1}{\Delta} = \frac{-18}{-18} = 1$.
Now,use the system $AX=B$ where $X = [\alpha, 2, \beta]^T = [1, 2, \beta]^T$:
$1(1) + 1(2) + 1(\beta) = 5 \Rightarrow 3 + \beta = 5 \Rightarrow \beta = 2$.
Alternatively,using the second row: $2(1) - 1(2) + 2(\beta) = 4 \Rightarrow 2 - 2 + 2\beta = 4 \Rightarrow 2\beta = 4 \Rightarrow \beta = 2$.
Therefore,$\alpha^2 + \beta^2 = 1^2 + 2^2 = 1 + 4 = 5$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and the system must be inconsistent.
First,we calculate the determinant $D$ of the coefficient matrix:
$D = \left|\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a\end{array}\right|$
$D = 1(3a - 25) - 2(a - 10) + 3(5 - 6)$
$D = 3a - 25 - 2a + 20 - 3$
$D = a - 8$
For the system to have no solution,we set $D = 0$,which gives $a = 8$.
Now,check for consistency at $a = 8$ using the augmented matrix $[A|B]$:
$\left[\begin{array}{ccc|c}1 & 2 & 3 & 6 \\ 1 & 3 & 5 & 9 \\ 2 & 5 & 8 & 12\end{array}\right]$
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$:
$\left[\begin{array}{ccc|c}1 & 2 & 3 & 6 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 0\end{array}\right]$
Applying $R_3 \to R_3 - R_2$:
$\left[\begin{array}{ccc|c}1 & 2 & 3 & 6 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & -3\end{array}\right]$
Since the last row represents $0 = -3$,which is a contradiction,the system has no solution when $a = 8$.

(B) The given system of equations is:
$x+y-z=6$ $(1)$
$3x+2y-z=5$ $(2)$
$2x-y-2z=-3$ $(3)$
Using the augmented matrix method:
$[A:B] = \begin{bmatrix} 1 & 1 & -1 & | & 6 \\ 3 & 2 & -1 & | & 5 \\ 2 & -1 & -2 & | & -3 \end{bmatrix}$
Applying row operations:
$R_2 \rightarrow R_2 - 3R_1$ and $R_3 \rightarrow R_3 - 2R_1$:
$[A:B] = \begin{bmatrix} 1 & 1 & -1 & | & 6 \\ 0 & -1 & 2 & | & -13 \\ 0 & -3 & 0 & | & -15 \end{bmatrix}$
$R_3 \rightarrow R_3 - 3R_2$:
$[A:B] = \begin{bmatrix} 1 & 1 & -1 & | & 6 \\ 0 & -1 & 2 & | & -13 \\ 0 & 0 & -6 & | & 24 \end{bmatrix}$
From the third row: $-6z = 24 \Rightarrow z = -4 = \gamma$.
From the second row: $-y + 2z = -13 \Rightarrow -y + 2(-4) = -13 \Rightarrow -y - 8 = -13 \Rightarrow y = 5 = \beta$.
From the first row: $x + y - z = 6 \Rightarrow x + 5 - (-4) = 6 \Rightarrow x + 9 = 6 \Rightarrow x = -3 = \alpha$.
Therefore,$\alpha + \beta = -3 + 5 = 2$.

(B) The coefficient matrix $D$ is given by $\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|$.
Calculating the determinant: $1(15-12) - 2(12-9) + 3(16-15) = 1(3) - 2(3) + 3(1) = 3 - 6 + 3 = 0$.
Since $D=0$,for the system to be consistent,the Cramer's rule determinants $D_1, D_2, D_3$ must also be $0$.
Calculating $D_3 = \left|\begin{array}{lll}1 & 2 & 4 \\ 4 & 5 & 5 \\ 3 & 4 & \lambda\end{array}\right| = 0$.
$1(5\lambda - 20) - 2(4\lambda - 15) + 4(16 - 15) = 0$.
$5\lambda - 20 - 8\lambda + 30 + 4 = 0$.
$-3\lambda + 14 = 0 \Rightarrow 3\lambda = 14$.
Given $3\lambda = n + 100$,we substitute $3\lambda = 14$:
$14 = n + 100 \Rightarrow n = 14 - 100 = -86$.

(B) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix must be non-zero.
The coefficient matrix is $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{bmatrix}$.
The condition for a unique solution is $|A| \neq 0$.
$|A| = 1(3a - 25) - 2(a - 10) + 3(5 - 6) \neq 0$.
$|A| = 3a - 25 - 2a + 20 - 3 \neq 0$.
$|A| = a - 8 \neq 0$.
Therefore,$a \neq 8$.
Since the value of $b$ does not affect the determinant of the coefficient matrix,$b$ can be any real number $(b \in R)$.
Thus,the condition is $a \neq 8$ and $b \in R$.

(B) Let the system be represented as $AX = B$,where $A = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 3 & -2 \\ 2 & 7 & 7 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 1(21 - (-14)) - 2(21 - (-4)) + 1(21 - 6)$
$|A| = 1(35) - 2(25) + 1(15) = 35 - 50 + 15 = 0$.
Since $|A| = 0$,the system either has no solution or infinitely many solutions.
Now,we check the adjoint matrix $adj(A)$ and calculate $adj(A)B$:
$adj(A) = \begin{bmatrix} 35 & -7 & -7 \\ -25 & 5 & 5 \\ 15 & -3 & -3 \end{bmatrix}$.
$adj(A)B = \begin{bmatrix} 35 & -7 & -7 \\ -25 & 5 & 5 \\ 15 & -3 & -3 \end{bmatrix} \begin{bmatrix} -3 \\ -1 \\ -4 \end{bmatrix} = \begin{bmatrix} -105 + 7 + 28 \\ 75 - 5 - 20 \\ -45 + 3 + 12 \end{bmatrix} = \begin{bmatrix} -70 \\ 50 \\ -30 \end{bmatrix} \neq 0$.
Since $adj(A)B \neq 0$,the system has no solution.

(B) Given the determinant: $\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} > 0$
Expanding along the first row: $a(bc - 1) - 1(c - 1) + 1(1 - b) > 0$
$abc - a - c + 1 + 1 - b > 0$
$abc + 2 > a + b + c$ . . . . . . $(i)$
By the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$:
$\frac{a + b + c}{3} \geq (abc)^{1/3} \Rightarrow a + b + c \geq 3(abc)^{1/3}$ . . . . . . $(ii)$
Substituting $(ii)$ into $(i)$:
$abc + 2 > 3(abc)^{1/3}$
Let $x = (abc)^{1/3}$,then $x^3 + 2 > 3x$
$x^3 - 3x + 2 > 0$
$(x - 1)^2(x + 2) > 0$
Since $(x - 1)^2$ is always non-negative,for the inequality to hold,we must have $x + 2 > 0$
$x > -2 \Rightarrow (abc)^{1/3} > -2$
Cubing both sides: $abc > -8$

(C) We use the formula for $\tan ^{-1} x + \tan ^{-1} y$. Since $x \times y = 2 \times 3 = 6 > 1$,the formula is: $\tan ^{-1} x + \tan ^{-1} y = \pi + \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
Substituting the values,we get: $\tan ^{-1} 2 + \tan ^{-1} 3 = \pi + \tan ^{-1} \left( \frac{2+3}{1-(2 \times 3)} \right)$.
$= \pi + \tan ^{-1} \left( \frac{5}{1-6} \right) = \pi + \tan ^{-1} \left( \frac{5}{-5} \right)$.
$= \pi + \tan ^{-1}(-1)$.
Since $\tan ^{-1}(-1) = -\frac{\pi}{4}$,we have: $\pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.

(D) Given $\theta = \sec^{-1}(\cosh u)$,we have $\sec \theta = \cosh u$.
Using the definition of the inverse hyperbolic cosine function,$u = \cosh^{-1}(\sec \theta) = \log_e(\sec \theta + \sqrt{\sec^2 \theta - 1})$.
Since $\sqrt{\sec^2 \theta - 1} = \tan \theta$,we get $u = \log_e(\sec \theta + \tan \theta)$.
Rewriting in terms of sine and cosine: $u = \log_e\left(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}\right) = \log_e\left(\frac{1 + \sin \theta}{\cos \theta}\right)$.
Using trigonometric identities $\sin \theta = \cos(\frac{\pi}{2} - \theta)$ and $\cos \theta = \sin(\frac{\pi}{2} - \theta)$ is not ideal here; instead,use $1 + \sin \theta = 1 + \cos(\frac{\pi}{2} - \theta) = 2\cos^2(\frac{\pi}{4} - \frac{\theta}{2})$ and $\cos \theta = \sin(\frac{\pi}{2} - \theta) = 2\sin(\frac{\pi}{4} - \frac{\theta}{2})\cos(\frac{\pi}{4} - \frac{\theta}{2})$.
Alternatively,using $1 + \sin \theta = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2})^2$ and $\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})$,we get $\frac{1 + \sin \theta}{\cos \theta} = \frac{\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} - \sin \frac{\theta}{2}} = \tan(\frac{\pi}{4} + \frac{\theta}{2})$.
Thus,$u = \log_e(\tan(\frac{\pi}{4} + \frac{\theta}{2}))$.

(A) Given expression: $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8} = k$.
Using $\cos(\frac{\pi}{2} + \theta) = -\sin \theta$,we have $\cos^4(\frac{\pi}{2} + \theta) = \sin^4 \theta$.
So,$\cos^4 \frac{5 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{\pi}{8}) = \sin^4 \frac{\pi}{8}$ and $\cos^4 \frac{7 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{3 \pi}{8}) = \sin^4 \frac{3 \pi}{8}$.
Substituting these,$k = (\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8}) + (\cos^4 \frac{3 \pi}{8} + \sin^4 \frac{3 \pi}{8})$.
Using $a^2 + b^2 = (a+b)^2 - 2ab$,we get $k = [(\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8})^2 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}] + [(\cos^2 \frac{3 \pi}{8} + \sin^2 \frac{3 \pi}{8})^2 - 2 \sin^2 \frac{3 \pi}{8} \cos^2 \frac{3 \pi}{8}]$.
Since $\sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2 \theta)$,$k = [1 - \frac{1}{2} \sin^2 \frac{\pi}{4}] + [1 - \frac{1}{2} \sin^2 \frac{3 \pi}{4}]$.
$k = 2 - \frac{1}{2}(\frac{1}{2}) - \frac{1}{2}(\frac{1}{2}) = 2 - \frac{1}{4} - \frac{1}{4} = 2 - \frac{1}{2} = \frac{3}{2}$.
Now,$\sin^{-1}(\sqrt{\frac{k}{2}}) + \cos^{-1}(\frac{k}{3}) = \sin^{-1}(\sqrt{\frac{3/2}{2}}) + \cos^{-1}(\frac{3/2}{3}) = \sin^{-1}(\frac{\sqrt{3}}{2}) + \cos^{-1}(\frac{1}{2})$.
$= \frac{\pi}{3} + \frac{\pi}{3} = \frac{2 \pi}{3}$.

(C) We use the formula $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$.
First,calculate $2 \tan ^{-1} \frac{1}{5} = \tan ^{-1} \left( \frac{2/5}{1-1/25} \right) = \tan ^{-1} \left( \frac{2/5}{24/25} \right) = \tan ^{-1} \frac{5}{12}$.
Then,$4 \tan ^{-1} \frac{1}{5} = 2 \tan ^{-1} \frac{5}{12} = \tan ^{-1} \left( \frac{2(5/12)}{1-(5/12)^2} \right) = \tan ^{-1} \left( \frac{5/6}{1-25/144} \right) = \tan ^{-1} \left( \frac{5/6}{119/144} \right) = \tan ^{-1} \frac{120}{119}$.
Now,evaluate $-\tan ^{-1} \frac{1}{70} + \tan ^{-1} \frac{1}{99} = \tan ^{-1} \left( \frac{1/99 - 1/70}{1 + (1/99)(1/70)} \right) = \tan ^{-1} \left( \frac{(70-99)/6930}{(6930+1)/6930} \right) = \tan ^{-1} \left( \frac{-29}{6931} \right) = -\tan ^{-1} \frac{1}{239}$.
Finally,$\tan ^{-1} \frac{120}{119} - \tan ^{-1} \frac{1}{239} = \tan ^{-1} \left( \frac{120/119 - 1/239}{1 + (120/119)(1/239)} \right) = \tan ^{-1} \left( \frac{(28680-119)/(119 \times 239)}{(28441+120)/(119 \times 239)} \right) = \tan ^{-1} \left( \frac{28561}{28561} \right) = \tan ^{-1} 1 = \frac{\pi}{4}$.

(D) Given the equation: $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$
Using the identity $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan ^{-1} \left( \frac{x+2x}{1-x(2x)} \right) = \frac{\pi}{4}$
$\frac{3x}{1-2x^2} = \tan \left( \frac{\pi}{4} \right)$
$\frac{3x}{1-2x^2} = 1$
$3x = 1 - 2x^2$
$2x^2 + 3x - 1 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$
Since $x$ must be positive for $\tan ^{-1} x + \tan ^{-1} 2x$ to be $\frac{\pi}{4}$ (as $\tan ^{-1} x$ is an increasing function),we reject the negative root.
Thus,$x = \frac{\sqrt{17}-3}{4}$.

(A) We know that $\coth^{-1}(x) = \frac{1}{2} \log \left(\frac{x+1}{x-1}\right)$ for $|x| > 1$.
Thus,$2 \coth^{-1}(4) = 2 \cdot \frac{1}{2} \log \left(\frac{4+1}{4-1}\right) = \log \left(\frac{5}{3}\right)$.
We also know that $\text{sech}^{-1}(x) = \log \left(\frac{1 + \sqrt{1-x^2}}{x}\right)$ for $0 < x \leq 1$.
Substituting $x = \frac{3}{5}$,we get $\text{sech}^{-1}\left(\frac{3}{5}\right) = \log \left(\frac{1 + \sqrt{1 - (9/25)}}{3/5}\right) = \log \left(\frac{1 + \sqrt{16/25}}{3/5}\right) = \log \left(\frac{1 + 4/5}{3/5}\right) = \log \left(\frac{9/5}{3/5}\right) = \log 3$.
Therefore,$2 \coth^{-1}(4) + \text{sech}^{-1}\left(\frac{3}{5}\right) = \log \left(\frac{5}{3}\right) + \log 3 = \log \left(\frac{5}{3} \times 3\right) = \log 5$.

(A) Given that $\alpha = \sin^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{\sqrt{3}}{2} \sqrt{1 - x^2} \right)$ for $0 < x < \frac{1}{2}$.
Let $x = \sin \theta$. Since $0 < x < \frac{1}{2}$,we have $0 < \theta < \frac{\pi}{6}$.
Then $\sqrt{1 - x^2} = \cos \theta$.
Substituting these into the expression for $\alpha$:
$\alpha = \sin^{-1}(\sin \theta) + \cos^{-1} \left( \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta \right)$.
Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we have $\frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = \sin \frac{\pi}{6} \sin \theta + \cos \frac{\pi}{6} \cos \theta = \cos \left( \theta - \frac{\pi}{6} \right)$.
Since $0 < \theta < \frac{\pi}{6}$,we have $-\frac{\pi}{6} < \theta - \frac{\pi}{6} < 0$,so $0 < \frac{\pi}{6} - \theta < \frac{\pi}{6}$.
Thus,$\cos^{-1} \left( \cos \left( \theta - \frac{\pi}{6} \right) \right) = \cos^{-1} \left( \cos \left( \frac{\pi}{6} - \theta \right) \right) = \frac{\pi}{6} - \theta$.
Therefore,$\alpha = \theta + \frac{\pi}{6} - \theta = \frac{\pi}{6}$.
Finally,$\tan \alpha + \cot \alpha = \tan \frac{\pi}{6} + \cot \frac{\pi}{6} = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

(A) We know that $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$.
We can rewrite the general term as $\tan^{-1} \left( \frac{(n+1)-n}{1+n(n+1)} \right) = \tan^{-1}(n+1) - \tan^{-1} n$.
Thus,the sum is $\sum_{n=1}^{50} (\tan^{-1}(n+1) - \tan^{-1} n)$.
This is a telescoping sum: $(\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1} 51 - \tan^{-1} 50) = \tan^{-1} 51 - \tan^{-1} 1$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$,we get $\tan^{-1} \left( \frac{51-1}{1+51 \times 1} \right) = \tan^{-1} \left( \frac{50}{52} \right) = \tan^{-1} \left( \frac{25}{26} \right)$.
Finally,$\cot \left( \tan^{-1} \left( \frac{25}{26} \right) \right) = \cot \left( \cot^{-1} \left( \frac{26}{25} \right) \right) = \frac{26}{25}$.

(B) Given,$f(x) = \frac{1}{\sqrt{\log_{0.5}(2x - 3)}} + \sqrt{4 - 9x^2}$.
For $f(x)$ to be defined,the following conditions must hold:
$1. \log_{0.5}(2x - 3) > 0$ $\Rightarrow 2x - 3 < (0.5)^0$ $\Rightarrow 2x - 3 < 1$ $\Rightarrow 2x < 4$ $\Rightarrow x < 2$.
$2. 2x - 3 > 0$ $\Rightarrow 2x > 3$ $\Rightarrow x > \frac{3}{2}$.
$3. 4 - 9x^2 \geq 0$ $\Rightarrow 9x^2 \leq 4$ $\Rightarrow x^2 \leq \frac{4}{9}$ $\Rightarrow -\frac{2}{3} \leq x \leq \frac{2}{3}$.
Combining these conditions: $(x < 2) \cap (x > \frac{3}{2}) \cap (-\frac{2}{3} \leq x \leq \frac{2}{3})$.
Since there is no value of $x$ that satisfies $x > \frac{3}{2}$ and $x \leq \frac{2}{3}$ simultaneously,the domain is the null set.