AP EAMCET 2022 Mathematics Question Paper with Answer and Solution

(D) The perpendicular bisectors of the sides of a triangle intersect at the circumcenter $O$.
Since the perpendicular bisectors of $AB$ and $AC$ are $x - y + 5 = 0$ and $x + 2y = 0$,their intersection point $O$ is found by solving these equations:
$x - y = -5$ and $x + 2y = 0$.
Subtracting the equations,we get $-3y = -5$,so $y = \frac{5}{3}$.
Then $x = -2y = -\frac{10}{3}$.
Thus,the circumcenter $O$ is $(-\frac{10}{3}, \frac{5}{3})$.
Since $O$ is the circumcenter,$OA = OB = OC$.
Let $B = (x_1, y_1)$. The reflection of $A(1, -2)$ about the line $x - y + 5 = 0$ gives $B$.
The formula for the reflection of $(x_0, y_0)$ about $ax + by + c = 0$ is $\frac{x - x_0}{a} = \frac{y - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.
For $A(1, -2)$ and $x - y + 5 = 0$:
$\frac{x_1 - 1}{1} = \frac{y_1 + 2}{-1} = -2 \frac{1 - (-2) + 5}{1^2 + (-1)^2} = -2 \frac{8}{2} = -8$.
So,$x_1 - 1 = -8 \implies x_1 = -7$ and $y_1 + 2 = 8 \implies y_1 = 6$. Thus,$B = (-7, 6)$.
Similarly,for $C(x_2, y_2)$,the reflection of $A(1, -2)$ about $x + 2y = 0$:
$\frac{x_2 - 1}{1} = \frac{y_2 + 2}{2} = -2 \frac{1 + 2(-2)}{1^2 + 2^2} = -2 \frac{-3}{5} = \frac{6}{5}$.
So,$x_2 - 1 = \frac{6}{5} \implies x_2 = \frac{11}{5}$ and $y_2 + 2 = \frac{12}{5} \implies y_2 = \frac{2}{5}$. Thus,$C = (\frac{11}{5}, \frac{2}{5})$.
The equation of line $BC$ passing through $(-7, 6)$ and $(\frac{11}{5}, \frac{2}{5})$ is:
$y - 6 = \frac{\frac{2}{5} - 6}{\frac{11}{5} - (-7)} (x - (-7)) = \frac{-\frac{28}{5}}{\frac{46}{5}} (x + 7) = -\frac{14}{23} (x + 7)$.
$23(y - 6) = -14(x + 7) \implies 23y - 138 = -14x - 98 \implies 14x + 23y - 40 = 0$.

(A) Given that $a, b, c$ are distinct positive real numbers such that $a^2+b^2+c^2=1$.
We know the algebraic identity: $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$.
Since $a, b, c > 0$,we have $(a+b+c)^2 > a^2+b^2+c^2$,which implies $a^2+b^2+c^2+2(ab+bc+ca) > a^2+b^2+c^2$.
This simplifies to $2(ab+bc+ca) > 0$,which is always true for positive numbers.
Also,we know that $(a-b)^2 + (b-c)^2 + (c-a)^2 > 0$ because $a, b, c$ are distinct.
Expanding this,we get $2(a^2+b^2+c^2) - 2(ab+bc+ca) > 0$.
Substituting $a^2+b^2+c^2=1$,we get $2(1) - 2(ab+bc+ca) > 0$,which means $2 > 2(ab+bc+ca)$,or $ab+bc+ca < 1$.
Thus,the value of $ab+bc+ca$ is less than $1$.

(C) Given $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3} = \frac{A x+B}{x^2+1} + \frac{C x+D}{\left(x^2+1\right)^2} + \frac{E x+F}{\left(x^2+1\right)^3}$.
Multiplying both sides by $(x^2+1)^3$,we get:
$x^4+24 x^2+28 = (A x+B)(x^2+1)^2 + (C x+D)(x^2+1) + (E x+F)$.
Expanding the right side:
$x^4+24 x^2+28 = (A x+B)(x^4+2x^2+1) + (C x^3+D x^2+C x+D) + E x+F$.
$x^4+24 x^2+28 = A x^5 + B x^4 + 2A x^3 + 2B x^2 + A x + B + C x^3 + D x^2 + C x + D + E x + F$.
Grouping the powers of $x$:
$x^4+24 x^2+28 = A x^5 + B x^4 + (2A+C) x^3 + (2B+D) x^2 + (A+C+E) x + (B+D+F)$.
Comparing coefficients on both sides:
$A = 0$
$B = 1$
$2A+C = 0 \Rightarrow C = 0$
$2B+D = 24$ $\Rightarrow 2(1)+D = 24$ $\Rightarrow D = 22$
$A+C+E = 0$ $\Rightarrow 0+0+E = 0$ $\Rightarrow E = 0$
$B+D+F = 28$ $\Rightarrow 1+22+F = 28$ $\Rightarrow F = 5$.
Therefore,$A+B+C+D+E+F = 0+1+0+22+0+5 = 28$.

(C) Given the partial fraction decomposition:
$\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$
Combining the terms on the right side:
$\frac{13x+43}{2x^2+17x+30} = \frac{A(x+6) + B(2x+5)}{(2x+5)(x+6)}$
Since the denominators are equal,we equate the numerators:
$13x + 43 = A(x+6) + B(2x+5)$
$13x + 43 = (A+2B)x + (6A+5B)$
Comparing the coefficients of $x$ and the constant terms:
$A + 2B = 13$ $(i)$
$6A + 5B = 43$ $(ii)$
Multiply equation $(i)$ by $6$:
$6A + 12B = 78$ $(iii)$
Subtract equation $(ii)$ from $(iii)$:
$(6A + 12B) - (6A + 5B) = 78 - 43$
$7B = 35 \Rightarrow B = 5$
Substitute $B = 5$ into equation $(i)$:
$A + 2(5) = 13$ $\Rightarrow A + 10 = 13$ $\Rightarrow A = 3$
Finally,calculate $A^2 + B^2$:
$A^2 + B^2 = 3^2 + 5^2 = 9 + 25 = 34$

(B) Given the partial fraction decomposition: $\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}$.
Since $x^3-1 = (x-1)(x^2+x+1)$,we have $2x^2+1 = A(x^2+x+1) + (Bx+C)(x-1)$.
Expanding the right side: $2x^2+1 = Ax^2 + Ax + A + Bx^2 - Bx + Cx - C$.
Grouping terms by powers of $x$: $2x^2 + 0x + 1 = (A+B)x^2 + (A-B+C)x + (A-C)$.
Comparing coefficients:
$1) A+B = 2$
$2) A-B+C = 0$
$3) A-C = 1$
From $(3)$,$C = A-1$. Substituting into $(2)$: $A-B+(A-1) = 0 \Rightarrow 2A-B = 1$.
Adding this to $(1)$: $(2A-B) + (A+B) = 1+2$ $\Rightarrow 3A = 3$ $\Rightarrow A = 1$.
Then $B = 2-A = 1$ and $C = A-1 = 0$.
Finally,$7A + 2B + C = 7(1) + 2(1) + 0 = 9$.

(B) The given expression is $\frac{x^3}{(2x-1)(x+2)(x-3)}$. Since the degree of the numerator is equal to the degree of the denominator,we perform polynomial division first.\\ The denominator is $(2x-1)(x^2-x-6) = 2x^3-2x^2-12x-x^2+x+6 = 2x^3-3x^2-11x+6$.\\ Dividing $x^3$ by $2x^3-3x^2-11x+6$,we get $A = 1/2$ as the quotient.\\ The expression becomes $\frac{1}{2} + \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x+2)(x-3)} = A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$.\\ Using the cover-up rule or substitution:\\ For $B$: Put $x = 1/2$,$B = \frac{\frac{3}{2}(1/4) + \frac{11}{2}(1/2) - 3}{(1/2+2)(1/2-3)} = \frac{3/8 + 11/4 - 3}{(5/2)(-5/2)} = \frac{(3+22-24)/8}{-25/4} = \frac{1/8}{-25/4} = -1/50$.\\ For $C$: Put $x = -2$,$C = \frac{\frac{3}{2}(4) + \frac{11}{2}(-2) - 3}{(-4-1)(-2-3)} = \frac{6-11-3}{(-5)(-5)} = \frac{-8}{25}$.\\ For $D$: Put $x = 3$,$D = \frac{\frac{3}{2}(9) + \frac{11}{2}(3) - 3}{(6-1)(3+2)} = \frac{27/2 + 33/2 - 3}{(5)(5)} = \frac{30-3}{25} = 27/25$.\\ Finally,$A+B+C = 1/2 - 1/50 - 8/25 = \frac{25-1-16}{50} = \frac{8}{50} = 4/25$.

(D) Given the partial fraction decomposition: $\frac{1}{(1-2x)^2(1-3x)} = \frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2}$.
Multiplying both sides by $(1-2x)^2(1-3x)$,we get: $1 = A(1-2x)^2 + B(1-2x)(1-3x) + C(1-3x)$.
Setting $x = \frac{1}{3}$: $1 = A(1 - \frac{2}{3})^2 = A(\frac{1}{3})^2 = \frac{A}{9} \implies A = 9$.
Setting $x = \frac{1}{2}$: $1 = C(1 - \frac{3}{2}) = C(-\frac{1}{2}) \implies C = -2$.
Setting $x = 0$: $1 = A(1)^2 + B(1)(1) + C(1) = A + B + C$.
Substituting $A = 9$ and $C = -2$: $1 = 9 + B - 2 \implies 1 = 7 + B \implies B = -6$.
Thus,the set is $\{9, -6, -2\}$.
The minimum value is $\min \{9, -6, -2\} = -6$.

(C) Given the expression: $\frac{x^3}{(2x-1)(x+2)(x-3)} = A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$.
First,perform polynomial division since the degree of the numerator is equal to the degree of the denominator. The denominator is $(2x-1)(x^2-x-6) = 2x^3 - 3x^2 - 11x + 6$. Dividing $x^3$ by $2x^3 - 3x^2 - 11x + 6$ gives $A = \frac{1}{2}$.
Now,equate the remainder: $\frac{x^3 - \frac{1}{2}(2x^3 - 3x^2 - 11x + 6)}{(2x-1)(x+2)(x-3)} = \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x+2)(x-3)} = \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$.
To find $C$,use the cover-up method by multiplying both sides by $(x+2)$ and setting $x = -2$:
$C = \left[ \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x-3)} \right]_{x=-2}$.
$C = \frac{\frac{3}{2}(4) + \frac{11}{2}(-2) - 3}{(2(-2)-1)(-2-3)} = \frac{6 - 11 - 3}{(-5)(-5)} = \frac{-8}{25}$.

(B) Given the partial fraction decomposition: $\frac{4 x^3+16 x+7}{\left(x^2+4\right)^2}=\frac{A x+B}{x^2+4}+\frac{C x+D}{\left(x^2+4\right)^2}$
Multiplying both sides by $(x^2+4)^2$,we get: $4 x^3+16 x+7 = (A x+B)(x^2+4) + (C x+D)$
Expanding the right side: $4 x^3+16 x+7 = A x^3 + B x^2 + 4Ax + 4B + Cx + D$
$4 x^3+16 x+7 = A x^3 + B x^2 + (4A+C)x + (4B+D)$
Comparing the coefficients of like powers of $x$:
$x^3$: $A = 4$
$x^2$: $B = 0$
$x^1$: $4A + C = 16$ $\Rightarrow 4(4) + C = 16$ $\Rightarrow 16 + C = 16$ $\Rightarrow C = 0$
Constant term: $4B + D = 7$ $\Rightarrow 4(0) + D = 7$ $\Rightarrow D = 7$
The values are $A=4, B=0, C=0, D=7$.
The non-zero values are $A$ and $D$.
Therefore,the number of non-zero values is $2$.

(D) Let $\alpha, \beta$ be the roots of $x^2+ax+b=0$ and $\gamma, \delta$ be the roots of $x^2+bx+a=0$.
From the relations between roots and coefficients:
$\alpha+\beta = -a, \alpha\beta = b$
$\gamma+\delta = -b, \gamma\delta = a$
Given that the difference between the roots is the same:
$|\alpha-\beta| = |\gamma-\delta|$
Squaring both sides:
$(\alpha-\beta)^2 = (\gamma-\delta)^2$
$(\alpha+\beta)^2 - 4\alpha\beta = (\gamma+\delta)^2 - 4\gamma\delta$
$(-a)^2 - 4b = (-b)^2 - 4a$
$a^2 - 4b = b^2 - 4a$
$a^2 - b^2 + 4a - 4b = 0$
$(a-b)(a+b) + 4(a-b) = 0$
$(a-b)(a+b+4) = 0$
Since $a \neq b$,we have $a-b \neq 0$.
Therefore,$a+b+4 = 0$.

(A) Given $(x+a)(x+1991)+1 = (x+b)(x+c)$.
Expanding the left side: $x^2 + (a+1991)x + 1991a + 1 = x^2 + (b+c)x + bc$.
Comparing coefficients,we have $b+c = a+1991$ and $bc = 1991a+1$.
Since $b$ and $c$ are roots of the quadratic equation $x^2 - (b+c)x + bc = 0$,the discriminant $D$ must be a perfect square,say $m^2$.
$D = (b+c)^2 - 4bc = (a+1991)^2 - 4(1991a+1) = m^2$.
$(a+1991)^2 - 4(1991a) - 4 = m^2$.
$a^2 + 2(1991)a + 1991^2 - 4(1991)a - 4 = m^2$.
$a^2 - 2(1991)a + 1991^2 - 4 = m^2$.
$(a-1991)^2 - 4 = m^2$.
$(a-1991)^2 - m^2 = 4$.
$(a-1991-m)(a-1991+m) = 4$.
Let $X = a-1991-m$ and $Y = a-1991+m$. Then $XY = 4$.
Since $Y-X = 2m$,$X$ and $Y$ must have the same parity. Since their product is $4$ (even),both must be even.
The possible pairs $(X, Y)$ are $(2, 2)$ and $(-2, -2)$.
Case $1$: $a-1991-m = 2$ and $a-1991+m = 2$. Adding gives $2(a-1991) = 4$ $\Rightarrow a-1991 = 2$ $\Rightarrow a = 1993$.
Case $2$: $a-1991-m = -2$ and $a-1991+m = -2$. Adding gives $2(a-1991) = -4$ $\Rightarrow a-1991 = -2$ $\Rightarrow a = 1989$.
Thus,$a \in \{1989, 1993\}$.

(D) The given quadratic equation is $x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0$.
For the roots to be distinct,the discriminant $D$ must be greater than $0$.
$D = b^2 - 4ac > 0$
$[-2(1 + 3m)]^2 - 4(1)(7(3 + 2m)) > 0$
$4(1 + 9m^2 + 6m) - 28(3 + 2m) > 0$
$4 + 36m^2 + 24m - 84 - 56m > 0$
$36m^2 - 32m - 80 > 0$
Dividing by $4$:
$9m^2 - 8m - 20 > 0$
Factoring the quadratic expression:
$(9m + 10)(m - 2) > 0$
The roots of the equation $9m^2 - 8m - 20 = 0$ are $m = 2$ and $m = -\frac{10}{9}$.
Thus,the inequality holds for $m \in (-\infty, -\frac{10}{9}) \cup (2, \infty)$.
Since $S$ is a subset of the set of real numbers $\mathbb{R}$ and the interval $(-\infty, -\frac{10}{9}) \cup (2, \infty)$ contains infinitely many real numbers,the number of elements in $S$ is infinite.

(D) Given the quadratic equation $x^2-10x-8=0$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\alpha^2 - 10\alpha - 8 = 0 \implies \alpha^2 - 8 = 10\alpha$
$\beta^2 - 10\beta - 8 = 0 \implies \beta^2 - 8 = 10\beta$
We need to evaluate $\frac{a_{10}-8a_8}{5a_9}$.
Substitute $a_n = \alpha^n - \beta^n$:
$\frac{(\alpha^{10}-\beta^{10}) - 8(\alpha^8-\beta^8)}{5(\alpha^9-\beta^9)}$
$= \frac{(\alpha^{10}-8\alpha^8) - (\beta^{10}-8\beta^8)}{5(\alpha^9-\beta^9)}$
$= \frac{\alpha^8(\alpha^2-8) - \beta^8(\beta^2-8)}{5(\alpha^9-\beta^9)}$
Using the relations $\alpha^2-8=10\alpha$ and $\beta^2-8=10\beta$:
$= \frac{\alpha^8(10\alpha) - \beta^8(10\beta)}{5(\alpha^9-\beta^9)}$
$= \frac{10\alpha^9 - 10\beta^9}{5(\alpha^9-\beta^9)}$
$= \frac{10(\alpha^9-\beta^9)}{5(\alpha^9-\beta^9)} = 2$.

(B) For the quadratic equation $x^2+(2m+1)x+m=0$ to have equal roots,the discriminant $D$ must be equal to $0$.
The discriminant is given by $D = b^2 - 4ac$.
Here,$a=1$,$b=(2m+1)$,and $c=m$.
Substituting these values into $D=0$:
$(2m+1)^2 - 4(1)(m) = 0$
$4m^2 + 4m + 1 - 4m = 0$
$4m^2 + 1 = 0$
$4m^2 = -1$
$m^2 = -\frac{1}{4}$
Since the square of any real number $m$ must be non-negative $(m^2 \ge 0)$,there are no real values of $m$ that satisfy this equation.
Therefore,the number of real values of $m$ is $0$.

(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\alpha^n$.
From the relation between roots and coefficients,we have:
$\alpha + \alpha^n = -\frac{b}{a}$ and $\alpha \cdot \alpha^n = \alpha^{n+1} = \frac{c}{a}$.
We need to evaluate the expression $E = (ac^n)^{1/(n+1)} + (a^nc)^{1/(n+1)}$.
Substituting $c = a\alpha^{n+1}$:
$E = (a(a\alpha^{n+1})^n)^{1/(n+1)} + (a^n(a\alpha^{n+1}))^{1/(n+1)}$
$E = (a^{n+1} \alpha^{n(n+1)})^{1/(n+1)} + (a^{n+1} \alpha^{n+1})^{1/(n+1)}$
$E = a\alpha^n + a\alpha$
$E = a(\alpha^n + \alpha)$
Substituting $\alpha^n + \alpha = -\frac{b}{a}$:
$E = a(-\frac{b}{a}) = -b$.

(A) Given $\frac{x_1^5+x_2^5}{x_1^3+x_2^3} = \frac{11}{3}$.
Using the identity $a^5+b^5 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)$,we have:
$\frac{(x_1^2+x_2^2)(x_1^3+x_2^3) - x_1^2x_2^2(x_1+x_2)}{x_1^3+x_2^3} = \frac{11}{3}$
$(x_1^2+x_2^2) - \frac{x_1^2x_2^2(x_1+x_2)}{(x_1+x_2)(x_1^2-x_1x_2+x_2^2)} = \frac{11}{3}$
Since $x_1^2+x_2^2 = 5$,we get $5 - \frac{x_1^2x_2^2}{5-x_1x_2} = \frac{11}{3}$.
Let $x_1x_2 = t$. Then $5 - \frac{t^2}{5-t} = \frac{11}{3}$.
$3(25 - 5t - t^2) = 55 - 11t \Rightarrow 3t^2 + 4t - 20 = 0$.
Solving for $t$,we get $t = 2$ or $t = -10/3$.
If $t = 2$,$(x_1+x_2)^2 = x_1^2+x_2^2 + 2x_1x_2 = 5 + 2(2) = 9$,so $x_1+x_2 = \pm 3$.
If $t = -10/3$,$(x_1+x_2)^2 = 5 + 2(-10/3) = -5/3 < 0$,which is impossible for real roots.
Thus,the quadratic equation is $x^2 - (x_1+x_2)x + x_1x_2 = 0$,which gives $x^2 \pm 3x + 2 = 0$.

(A) Given that $\alpha$ and $\beta$ are the roots of $a x^2+b x+c=0$.
Let the roots of the new equation be $t = \sqrt{5} \alpha$ and $t = \sqrt{5} \beta$.
Then $\alpha = \frac{t}{\sqrt{5}}$.
Substituting this into the original equation:
$a(\frac{t}{\sqrt{5}})^2 + b(\frac{t}{\sqrt{5}}) + c = 0$
$\frac{a t^2}{5} + \frac{b t}{\sqrt{5}} + c = 0$
Multiplying the entire equation by $5$:
$a t^2 + 5 \frac{b t}{\sqrt{5}} + 5 c = 0$
$a t^2 + \sqrt{5} b t + 5 c = 0$
Replacing $t$ with $x$,the required equation is $a x^2 + \sqrt{5} b x + 5 c = 0$.

(D) We know that $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$.
Expanding this,we get $2(a^2+b^2+c^2) - 2(ab+bc+ca) \geq 0$.
Since $a^2+b^2+c^2 = 1$,we have $2(1) - 2(ab+bc+ca) \geq 0$,which implies $ab+bc+ca \leq 1$.
Also,we know that $(a+b+c)^2 \geq 0$.
Expanding this,we get $a^2+b^2+c^2 + 2(ab+bc+ca) \geq 0$.
Substituting $a^2+b^2+c^2 = 1$,we get $1 + 2(ab+bc+ca) \geq 0$,which implies $ab+bc+ca \geq -\frac{1}{2}$.
Therefore,the range of $ab+bc+ca$ is $[-\frac{1}{2}, 1]$.
The set of extreme values is $\{\frac{-1}{2}, 1\}$.

(D) Let $\alpha$ and $\beta$ be the roots of the quadratic equation.
Given $\alpha+\beta=11$ and $\alpha^2+\beta^2=61$.
We know that $(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta$.
Substituting the values: $(11)^2 = 61 + 2\alpha\beta$.
$121 = 61 + 2\alpha\beta$ $\Rightarrow 2\alpha\beta = 60$ $\Rightarrow \alpha\beta = 30$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values: $x^2 - 11x + 30 = 0$.

(D) Given,$\alpha$ and $\beta$ are the roots of $2x^2 + 6x + k = 0$.
From the relation between roots and coefficients,$\alpha + \beta = -\frac{6}{2} = -3$ and $\alpha\beta = \frac{k}{2}$.
Now,consider the expression $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$.
Substituting the values,we get $\frac{(-3)^2 - 2(k/2)}{k/2} = \frac{9 - k}{k/2} = \frac{18 - 2k}{k} = \frac{18}{k} - 2$.
Since $k < 0$,let $f(k) = \frac{18}{k} - 2$.
As $k$ becomes more negative (i.e.,$k \to -\infty$),$\frac{18}{k} \to 0$,so $f(k) \to -2$.
For $k < 0$,the expression $\frac{18}{k} - 2$ is always less than $-2$.
Specifically,for $k < -18$,$\frac{18}{k}$ lies in the interval $(-1, 0)$,so $\left[\frac{18}{k} - 2\right] = -1 - 2 = -3$.
However,for $-18 \le k < 0$,the value of $\frac{18}{k}$ is $\le -1$.
The maximum value of the greatest integer function $\left[\frac{18}{k} - 2\right]$ for $k < 0$ is $-2$.

(A) Given that $\tan 30^{\circ}$ and $\tan 15^{\circ}$ are the roots of $x^2+ax+b=0$.
We know that $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$.
$\tan 15^{\circ} = \tan(45^{\circ}-30^{\circ}) = \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}} = \frac{1-1/\sqrt{3}}{1+1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.
Sum of roots: $-a = \tan 30^{\circ} + \tan 15^{\circ} = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}+1+3-\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{4}{\sqrt{3}(\sqrt{3}+1)}$.
So,$a = -\frac{4}{3+\sqrt{3}}$.
Product of roots: $b = \tan 30^{\circ} \cdot \tan 15^{\circ} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{\sqrt{3}-1}{3+\sqrt{3}}$.
Now,$1+a-b = 1 - \frac{4}{\sqrt{3}(\sqrt{3}+1)} - \frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)} = \frac{3+\sqrt{3}-4-\sqrt{3}+1}{\sqrt{3}(\sqrt{3}+1)} = \frac{0}{\sqrt{3}(\sqrt{3}+1)} = 0$.

(A) We use the method of partial fractions for the expression $\frac{x}{(x-1)(x^2+1)^2}$.
Let $\frac{x}{(x-1)(x^2+1)^2} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}$.
Multiplying by $(x-1)(x^2+1)^2$,we get $x = A(x^2+1)^2 + (Bx+C)(x-1)(x^2+1) + (Dx+E)(x-1)$.
Setting $x=1$,we get $1 = A(2)^2$,so $A = \frac{1}{4}$.
Comparing coefficients or substituting values for $x$,we find $B = -\frac{1}{4}$,$C = -\frac{1}{4}$,$D = -\frac{1}{2}$,and $E = \frac{1}{2}$.
Substituting these values,we have $\frac{x}{(x-1)(x^2+1)^2} = \frac{1}{4(x-1)} - \frac{x+1}{4(x^2+1)} + \frac{-x/2 + 1/2}{(x^2+1)^2}$.
This simplifies to $\frac{1}{4}\left[\frac{1}{x-1} - \frac{x+1}{x^2+1}\right] + \frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$.
Comparing this with the given equation,we get $y = \frac{1}{2}\left[\frac{1-x}{(x^2+1)^2}\right]$.

(B) Given equation: $|x-2|^2+|x-2|-2=0$
Let $y = |x-2|$. Since $|x-2| \geq 0$,we must have $y \geq 0$.
The equation becomes $y^2 + y - 2 = 0$.
Factoring the quadratic: $(y+2)(y-1) = 0$.
This gives $y = -2$ or $y = 1$.
Since $y \geq 0$,we discard $y = -2$.
Thus,$|x-2| = 1$.
This implies $x-2 = 1$ or $x-2 = -1$.
Solving for $x$: $x = 3$ or $x = 1$.
The sum of the real roots is $3 + 1 = 4$.

(B) Let $f(x) = (x-a)(x-c) + 2(x-b)(x-d)$.
Since $f(x)$ is a quadratic polynomial with a positive leading coefficient $(3x^2)$,the roots are real and distinct if the discriminant $D > 0$.
Alternatively,consider the values of $f(x)$ at $x=a, b, c, d$:
$f(a) = (a-a)(a-c) + 2(a-b)(a-d) = 2(a-b)(a-d)$. Since $a < b$ and $a < d$,$(a-b) < 0$ and $(a-d) < 0$,so $f(a) > 0$.
$f(b) = (b-a)(b-c) + 2(b-b)(b-d) = (b-a)(b-c)$. Since $b > a$ and $b < c$,$(b-a) > 0$ and $(b-c) < 0$,so $f(b) < 0$.
$f(c) = (c-a)(c-c) + 2(c-b)(c-d) = 2(c-b)(c-d)$. Since $c > b$ and $c < d$,$(c-b) > 0$ and $(c-d) < 0$,so $f(c) < 0$.
$f(d) = (d-a)(d-c) + 2(d-b)(d-d) = (d-a)(d-c)$. Since $d > a$ and $d > c$,$(d-a) > 0$ and $(d-c) > 0$,so $f(d) > 0$.
Since $f(a) > 0$ and $f(b) < 0$,there exists a root in $(a, b)$.
Since $f(c) < 0$ and $f(d) > 0$,there exists a root in $(c, d)$.
Thus,the equation has two real and distinct roots.

(C) Given the equation $3^{x+1}+3^{-x+1}=10$.
This can be rewritten as $3(3^x) + \frac{3}{3^x} = 10$.
Let $y = 3^x$. Then the equation becomes $3y + \frac{3}{y} = 10$.
Multiplying by $y$,we get $3y^2 - 10y + 3 = 0$.
Factoring the quadratic equation: $3y^2 - 9y - y + 3 = 0 \Rightarrow 3y(y-3) - 1(y-3) = 0$.
So,$(3y-1)(y-3) = 0$,which gives $y = 3$ or $y = \frac{1}{3}$.
Substituting back $y = 3^x$:
Case $1$: $3^x = 3^1 \Rightarrow x = 1$.
Case $2$: $3^x = 3^{-1} \Rightarrow x = -1$.
The positive real root is $x = 1$.
Therefore,there is only $1$ positive real root.

(B) Given the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$.
Let $t = \sqrt{\frac{x}{1-x}}$. For $t$ to be defined and real,we must have $\frac{x}{1-x} > 0$,which implies $x \in (0, 1)$.
The equation becomes $t + \frac{1}{t} = \frac{13}{6}$.
Multiplying by $6t$,we get $6t^2 - 13t + 6 = 0$.
Factoring the quadratic: $(2t - 3)(3t - 2) = 0$,so $t = \frac{3}{2}$ or $t = \frac{2}{3}$.
Case $1$: $\sqrt{\frac{x}{1-x}} = \frac{3}{2}$ $\Rightarrow \frac{x}{1-x} = \frac{9}{4}$ $\Rightarrow 4x = 9 - 9x$ $\Rightarrow 13x = 9$ $\Rightarrow x = \frac{9}{13}$.
Case $2$: $\sqrt{\frac{x}{1-x}} = \frac{2}{3}$ $\Rightarrow \frac{x}{1-x} = \frac{4}{9}$ $\Rightarrow 9x = 4 - 4x$ $\Rightarrow 13x = 4$ $\Rightarrow x = \frac{4}{13}$.
Both values $x = \frac{9}{13}$ and $x = \frac{4}{13}$ lie in the interval $(0, 1)$.
Thus,there are $2$ real roots.

(C) Given the equation: $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$
Rearranging terms: $4^{x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$
Since $2^{2x - 1} = \frac{4^{x}}{2}$,we have: $4^{x} + \frac{4^{x}}{2} = 3^{x} \cdot \sqrt{3} + \frac{3^{x}}{\sqrt{3}}$
$4^{x} \left(1 + \frac{1}{2}\right) = 3^{x} \left(\frac{3 + 1}{\sqrt{3}}\right)$
$4^{x} \left(\frac{3}{2}\right) = 3^{x} \left(\frac{4}{\sqrt{3}}\right)$
$\left(\frac{4}{3}\right)^{x} = \frac{4}{\sqrt{3}} \cdot \frac{2}{3} = \frac{8}{3\sqrt{3}}$
$\left(\frac{4}{3}\right)^{x} = \left(\frac{2}{\sqrt{3}}\right)^{3} = \left(\left(\frac{2}{\sqrt{3}}\right)^{2}\right)^{\frac{3}{2}} = \left(\frac{4}{3}\right)^{\frac{3}{2}}$
Comparing exponents,we get $x = \frac{3}{2}$.

(C) Given equation: $4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$
Rearranging terms: $2^{2x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$
Factor out common terms: $2^{2x}(1 + \frac{1}{2}) = 3^x(\sqrt{3} + \frac{1}{\sqrt{3}})$
Simplify: $2^{2x}(\frac{3}{2}) = 3^x(\frac{3 + 1}{\sqrt{3}}) = 3^x(\frac{4}{\sqrt{3}})$
Multiply both sides by $\sqrt{3}$: $2^{2x} \cdot 3 \cdot \sqrt{3} = 3^x \cdot 8$
Express in powers: $2^{2x} \cdot 3^{\frac{3}{2}} = 3^x \cdot 2^3$
Equating powers of $2$: $2x = 3 \Rightarrow x = \frac{3}{2}$
Equating powers of $3$: $x = \frac{3}{2}$
Thus,$x = \frac{3}{2}$.

(C) Given the equation $x^4-2x^3+x-380=0$.
By testing values,we find that $x=5$ is a root:
$5^4-2(5^3)+5-380 = 625-250+5-380 = 380-380 = 0$.
Next,we test $x=-4$:
$(-4)^4-2(-4)^3+(-4)-380 = 256+128-4-380 = 384-4-380 = 0$.
Since $x=5$ and $x=-4$ are roots,we can divide the polynomial by $(x-5)(x+4) = x^2-x-20$.
Performing polynomial division: $(x^4-2x^3+x-380) \div (x^2-x-20) = x^2-x+19$.
The remaining roots are found by solving $x^2-x+19=0$.
The discriminant $D = (-1)^2 - 4(1)(19) = 1 - 76 = -75$.
Since $D < 0$,the roots of $x^2-x+19=0$ are complex.
Thus,the only real roots are $5$ and $-4$.
The sum of the real roots is $5 + (-4) = 1$.

(A) Let the roots of the cubic equation $x^3-7x^2+36=0$ be $a, 2a,$ and $b$.
From the relation between roots and coefficients:
Sum of roots: $a + 2a + b = 7 \Rightarrow 3a + b = 7$ ... $(i)$
Sum of product of roots taken two at a time: $a(2a) + 2a(b) + b(a) = 0$ (since coefficient of $x$ is $0$)
$2a^2 + 3ab = 0 \Rightarrow a(2a + 3b) = 0$
Since $a$ cannot be $0$ (as $36 \neq 0$),we have $2a + 3b = 0 \Rightarrow b = -\frac{2a}{3}$.
Substitute $b$ in $(i)$: $3a - \frac{2a}{3} = 7$ $\Rightarrow \frac{7a}{3} = 7$ $\Rightarrow a = 3$.
Then $b = 7 - 3(3) = -2$.
The roots are $a=3, 2a=6, b=-2$.
The roots are $3, 6, -2$.
Thus,there is only $1$ negative root.

(B) Let $y = x^{1/3}$. Then the equation becomes $y^2 + y - 2 = 0$.
Factoring the quadratic equation: $(y + 2)(y - 1) = 0$.
This gives $y = 1$ or $y = -2$.
Since $y = x^{1/3}$,we have $x^{1/3} = 1 \Rightarrow x = 1^3 = 1$.
And $x^{1/3} = -2 \Rightarrow x = (-2)^3 = -8$.
The roots of the equation are $1$ and $-8$.
The sum of the squares of the roots is $(1)^2 + (-8)^2 = 1 + 64 = 65$.

(B) To find the remainder,we perform polynomial long division of $2 x^5-3 x^4+5 x^3-3 x^2+7 x-9$ by $x^2-x-3$:
$1$. Divide $2 x^5$ by $x^2$ to get $2 x^3$. Multiply $(x^2-x-3)$ by $2 x^3$ to get $2 x^5-2 x^4-6 x^3$. Subtracting this from the dividend gives $-x^4+11 x^3-3 x^2+7 x-9$.
$2$. Divide $-x^4$ by $x^2$ to get $-x^2$. Multiply $(x^2-x-3)$ by $-x^2$ to get $-x^4+x^3+3 x^2$. Subtracting this gives $10 x^3-6 x^2+7 x-9$.
$3$. Divide $10 x^3$ by $x^2$ to get $10 x$. Multiply $(x^2-x-3)$ by $10 x$ to get $10 x^3-10 x^2-30 x$. Subtracting this gives $4 x^2+37 x-9$.
$4$. Divide $4 x^2$ by $x^2$ to get $4$. Multiply $(x^2-x-3)$ by $4$ to get $4 x^2-4 x-12$. Subtracting this gives $41 x+3$.
Thus,the remainder is $41 x+3$.

(C) Given that $2, 3, 6$ are the roots of the polynomial $f(x) = x^3 + ax^2 + bx + c$.
By the factor theorem,we can write:
$f(x) = (x - 2)(x - 3)(x - 6)$
Expanding the right side:
$f(x) = (x^2 - 5x + 6)(x - 6)$
$f(x) = x^3 - 6x^2 - 5x^2 + 30x + 6x - 36$
$f(x) = x^3 - 11x^2 + 36x - 36$
Comparing this with $f(x) = x^3 + ax^2 + bx + c$,we get:
$a = -11$,$b = 36$,$c = -36$
Now,calculating $a - c$:
$a - c = -11 - (-36) = -11 + 36 = 25$

(C) Given equations are:
$y^2+z^2=3yz \Rightarrow \frac{y}{z}+\frac{z}{y}=3$ $(i)$
$z^2+x^2=8zx \Rightarrow \frac{z}{x}+\frac{x}{z}=8$ $(ii)$
$x^2+y^2=4xy \Rightarrow \frac{x}{y}+\frac{y}{x}=4$ $(iii)$
Multiplying $(i)$ and $(iii)$:
$\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right) = \frac{xy}{zy} + \frac{y^2}{xz} + \frac{z}{y} \cdot \frac{x}{y} + \frac{z}{y} \cdot \frac{y}{x} = \frac{x}{z} + \frac{y^2}{xz} + \frac{xz}{y^2} + \frac{z}{x} = 12$
$\left(\frac{x}{z}+\frac{z}{x}\right) + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$
Substituting the value from $(ii)$:
$8 + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$
$\frac{y^2}{xz}+\frac{xz}{y^2} = 12 - 8 = 4$

(B) Since $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,we perform polynomial division:
$ax^3 + bx + c = (x^2 + px + 1)(ax - ap) + (b - a + ap^2)x + (ap + c)$
For $x^2 + px + 1$ to be a factor,the remainder must be zero:
$1) \ ap + c = 0 \Rightarrow p = -\frac{c}{a} \dots (i)$
$2) \ b - a + ap^2 = 0$
Substituting $(i)$ into the second equation:
$b - a + a(-\frac{c}{a})^2 = 0$
$b - a + a(\frac{c^2}{a^2}) = 0$
$b - a + \frac{c^2}{a} = 0$
Multiplying by $a$:
$ab - a^2 + c^2 = 0$
$a^2 - c^2 = ab$

(C) Given equations are:
$y^2+z^2=a y z \Rightarrow \frac{y}{z}+\frac{z}{y}=a$ $(i)$
$z^2+x^2=b z x \Rightarrow \frac{z}{x}+\frac{x}{z}=b$ $(ii)$
$x^2+y^2=c x y \Rightarrow \frac{x}{y}+\frac{y}{x}=c$ $(iii)$
Multiplying $(i)$ and $(iii)$:
$(\frac{y}{z}+\frac{z}{y})(\frac{x}{y}+\frac{y}{x}) = ac$
$\frac{x}{z} + \frac{y^2}{zx} + \frac{zx}{y^2} + \frac{z}{x} = ac$
From $(ii)$,we know $\frac{z}{x}+\frac{x}{z}=b$. Substituting this into the expression:
$b + \frac{y^2}{zx} + \frac{zx}{y^2} = ac$
$\frac{y^2}{zx} + \frac{zx}{y^2} = ac-b$

(C) Given: $\frac{x^2+1}{x^4+4}=\frac{A x+B}{x^2-2 x+2}+\frac{C x+D}{x^2+2 x+2}$
Expanding the right side:
$\frac{x^2+1}{x^4+4} = \frac{(A x+B)(x^2+2 x+2)+(C x+D)(x^2-2 x+2)}{(x^2-2 x+2)(x^2+2 x+2)}$
Note that $(x^2-2 x+2)(x^2+2 x+2) = ((x^2+2)-2x)((x^2+2)+2x) = (x^2+2)^2 - (2x)^2 = x^4+4x^2+4-4x^2 = x^4+4$.
Equating the numerators:
$x^2+1 = (A+C)x^3 + (2A+B-2C+D)x^2 + (2A+2B+2C-2D)x + (2B+2D)$.
Comparing coefficients:
$1) A+C=0 \Rightarrow A=-C$
$2) 2A+B-2C+D=1$
$3) 2A+2B+2C-2D=0 \Rightarrow A+B+C-D=0$
$4) 2B+2D=1$
From $(1)$,$A+C=0$,so $(3)$ becomes $B-D=0 \Rightarrow B=D$.
Substituting $B=D$ into $(4)$,$2B+2B=1 \Rightarrow 4B=1 \Rightarrow B=D=\frac{1}{4}$.
From $(2)$,$2(A-C)+B+D=1$. Since $A=-C$,$2(2A)+2B=1 \Rightarrow 4A+2B=1$.
Substituting $B=\frac{1}{4}$,$4A + \frac{1}{2} = 1 \Rightarrow 4A = \frac{1}{2} \Rightarrow A = \frac{1}{8}$.
Thus $C = -\frac{1}{8}$.
Finally,$3A+2B+3C = 3(A+C) + 2B = 3(0) + 2(\frac{1}{4}) = \frac{1}{2} = 2D$.

(C) Let $f(x) = x^{12} - x^9 + x^4 - x + 1$.
We analyze the expression by grouping terms:
$f(x) = x^9(x^3 - 1) + x(x^3 - 1) + 1$.
Alternatively,consider the expression as $f(x) = x^{12} - x^9 + x^4 - x + 1$.
If $x \ge 1$,then $x^9(x^3 - 1) \ge 0$ and $x(x^3 - 1) \ge 0$,so $f(x) \ge 1 > 0$.
If $x \le 0$,then $f(x) = x^{12} + x^4 + (-x^9 - x) + 1$. Since $x \le 0$,$-x^9 \ge 0$ and $-x \ge 0$,so $f(x) > 0$.
If $0 < x < 1$,we can write $f(x) = x^{12} + x^4(1 - x^5) + (1 - x)$. Since $x^5 < 1$ and $x < 1$,all terms are positive,so $f(x) > 0$.
Since $f(x) > 0$ for all real $x$,the largest interval is $(-\infty, \infty)$.

(A) Given the equation $|1-i|^x=2^x$.
First,calculate the modulus of the complex number $1-i$:
$|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2} = 2^{\frac{1}{2}}$.
Substitute this into the equation:
$(2^{\frac{1}{2}})^x = 2^x$.
$2^{\frac{x}{2}} = 2^x$.
Since the bases are equal,equate the exponents:
$\frac{x}{2} = x$.
$x = 2x \Rightarrow x = 0$.
Thus,there is only $1$ integer solution,which is $x=0$.

(D) Given the sum $\sum_{k=0}^{40} i^k = x + iy$.
Since the sum of four consecutive powers of $i$ is zero,i.e.,$i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$,we can group the terms.
The sum has $41$ terms from $k=0$ to $k=40$.
$\sum_{k=0}^{40} i^k = i^0 + (i^1 + i^2 + i^3 + i^4) + \dots + (i^{37} + i^{38} + i^{39} + i^{40}) = 1 + 0 + \dots + 0 = 1$.
Thus,$x + iy = 1 + 0i$,which gives $x = 1$ and $y = 0$.
Substituting these values into the expression:
$x^{100} + x^{99}y + x^{242}y^2 + x^{97}y^3 = (1)^{100} + (1)^{99}(0) + (1)^{242}(0)^2 + (1)^{97}(0)^3 = 1 + 0 + 0 + 0 = 1$.

(C) Given expression: $i^{18}-3i^7+i^2(1+i^4)(i)^{22}$
We know that $i^2 = -1$,$i^3 = -i$,$i^4 = 1$.
$i^{18} = (i^4)^4 \times i^2 = (1)^4 \times (-1) = -1$
$i^7 = (i^4) \times i^3 = 1 \times (-i) = -i$
$i^{22} = (i^4)^5 \times i^2 = (1)^5 \times (-1) = -1$
Substituting these values:
$-1 - 3(-i) + (-1)(1+1)(-1)$
$= -1 + 3i + (-1)(2)(-1)$
$= -1 + 3i + 2$
$= 1 + 3i$

(A) Given $(x-iy)^{1/3} = a-ib$.
Cubing both sides,we get:
$x-iy = (a-ib)^3$
$x-iy = a^3 - (ib)^3 - 3a^2(ib) + 3a(ib)^2$
$x-iy = a^3 + ib^3 - 3a^2bi - 3ab^2$
$x-iy = (a^3 - 3ab^2) - i(3a^2b - b^3)$
Comparing real and imaginary parts:
$x = a^3 - 3ab^2 \implies \frac{x}{a} = a^2 - 3b^2$
$y = 3a^2b - b^3 \implies \frac{y}{b} = 3a^2 - b^2$
Adding these:
$\frac{x}{a} + \frac{y}{b} = (a^2 - 3b^2) + (3a^2 - b^2)$
$= 4a^2 - 4b^2 = 4(a^2 - b^2)$

(C) Given $x = -5 + 2 \sqrt{-4} = -5 + 4i$.
$x + 5 = 4i$
Squaring both sides:
$(x + 5)^2 = (4i)^2$
$x^2 + 10x + 25 = -16$
$x^2 + 10x + 41 = 0$
Now,divide the polynomial $P(x) = x^4 + 9x^3 + 35x^2 - x + 4$ by $x^2 + 10x + 41$:
$x^4 + 9x^3 + 35x^2 - x + 4 = (x^2 + 10x + 41)(x^2 - x + 4) - 160$
Since $x^2 + 10x + 41 = 0$,we have:
$P(x) = 0 \cdot (x^2 - x + 4) - 160 = -160$.

(B) Given $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$.
First,simplify the base fractions:
$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$.
Similarly,$\frac{1-i}{1+i} = \frac{1}{i} = -i$.
Substituting these into the expression:
$x+iy = (i)^3 - (-i)^3$.
$x+iy = -i - (-i^3) = -i - (i) = -2i$.
Comparing the real and imaginary parts:
$x = 0$ and $y = -2$.
Since $0 > -2$,we have $x > y$.

(A) Given: $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$
Multiply by the common denominator $(3+i)(3-i) = 3^2 + 1^2 = 10$:
$(x-1)(3-i) + (y-1)(3+i) = i(10)$
$3x - ix - 3 + i + 3y + iy - 3 - i = 10i$
$(3x + 3y - 6) + i(y - x) = 10i$
Comparing real and imaginary parts:
$3x + 3y - 6 = 0 \Rightarrow x + y = 2$
$y - x = 10$
Adding the two equations: $2y = 12 \Rightarrow y = 6$
Substituting $y = 6$ into $x + y = 2$: $x + 6 = 2 \Rightarrow x = -4$
Thus,$x = -4$ and $y = 6$.

(B) Let the complex number be $z = \sin \theta + i \cos \theta$.
The multiplicative inverse of $z$ is given by $\frac{1}{z}$.
$\frac{1}{z} = \frac{1}{\sin \theta + i \cos \theta}$.
To simplify,multiply the numerator and denominator by the conjugate $(\sin \theta - i \cos \theta)$:
$\frac{1}{z} = \frac{\sin \theta - i \cos \theta}{(\sin \theta + i \cos \theta)(\sin \theta - i \cos \theta)}$.
Using the identity $(a+ib)(a-ib) = a^2 + b^2$:
$\frac{1}{z} = \frac{\sin \theta - i \cos \theta}{\sin^2 \theta + \cos^2 \theta}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$\frac{1}{z} = \sin \theta - i \cos \theta$.
Thus,the multiplicative inverse is $(\sin \theta, -\cos \theta)$.

(A) Given: $\frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y}{3-i} = i$
Multiply by $(3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10$:
$((1+i)x-2i)(3-i) + (2-3i)y(3+i) = 10i$
$(3x - ix + 3ix - i^2x - 6i + 2i^2) + y(6 + 2i - 9i - 3i^2) = 10i$
$(3x - ix + 3ix + x - 6i - 2) + y(6 - 7i + 3) = 10i$
$(4x + 2ix - 2 - 6i) + y(9 - 7i) = 10i$
$(4x - 2 + 9y) + i(2x - 6 - 7y) = 10i$
Equating real and imaginary parts:
Real part: $4x + 9y - 2 = 0 \Rightarrow 4x + 9y = 2$
Imaginary part: $2x - 7y - 6 = 10 \Rightarrow 2x - 7y = 16$
Solving the system:
Multiply the second equation by $2$: $4x - 14y = 32$
Subtract from the first: $(4x + 9y) - (4x - 14y) = 2 - 32$
$23y = -30 \Rightarrow y = -30/23$
Substitute $y$ into $2x - 7y = 16$:
$2x - 7(-30/23) = 16 \Rightarrow 2x + 210/23 = 368/23$
$2x = 158/23 \Rightarrow x = 79/23$
Therefore,$x+y = 79/23 - 30/23 = 49/23$.

(D) Given,$(x-iy)^{1/3} = 2-i\sqrt{3}$
$\Rightarrow x-iy = (2-i\sqrt{3})^3$
Using the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:
$x-iy = 2^3 - 3(2^2)(i\sqrt{3}) + 3(2)(i\sqrt{3})^2 - (i\sqrt{3})^3$
$x-iy = 8 - 12i\sqrt{3} + 6(-3) - (-3i\sqrt{3})$
$x-iy = 8 - 12i\sqrt{3} - 18 + 3i\sqrt{3}$
$x-iy = -10 - 9i\sqrt{3}$
Comparing real and imaginary parts,we get $x = -10$ and $y = 9\sqrt{3}$.
Since the point $z = (x, y)$ lies on the line $\frac{x}{2} + \frac{y}{\sqrt{3}} = k$:
$\frac{-10}{2} + \frac{9\sqrt{3}}{\sqrt{3}} = k$
$-5 + 9 = k$
$k = 4$

(C) Given equation: $i z^3+z^2-z+i=0$
Factorizing the equation:
$i z^2(z-i) - 1(z-i) = 0$
$(z-i)(i z^2-1) = 0$
This gives two cases:
Case $1$: $z-i=0 \Rightarrow z=i$. Then $|z| = |i| = 1$.
Case $2$: $i z^2-1=0 \Rightarrow i z^2=1$.
Taking the modulus on both sides:
$|i z^2| = |1|$
$|i| \cdot |z|^2 = 1$
$1 \cdot |z|^2 = 1$ $\Rightarrow |z|^2 = 1$ $\Rightarrow |z| = 1$.
In both cases,$|z| = 1$.

(B) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $a = c$ and $b = -d$.
Given $z_1 = \sin x + i \cos 2x$ and $z_2 = \cos x - i \sin 2x$.
For these to be conjugates,we must have $\sin x = \cos x$ and $\cos 2x = -(-\sin 2x) = \sin 2x$.
From $\sin x = \cos x$,we get $\tan x = 1$,which implies $x = n\pi + \frac{\pi}{4}$.
From $\cos 2x = \sin 2x$,we get $\tan 2x = 1$,which implies $2x = m\pi + \frac{\pi}{4}$,or $x = \frac{m\pi}{2} + \frac{\pi}{8}$.
Since there is no value of $x$ that satisfies both conditions simultaneously,there is no solution.

(B) Given the curve $xy = 1$,we have $y = \frac{1}{x}$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.
The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^2}$.
The slope of the normal is the negative reciprocal of the tangent slope,which is $x^2$.
The slope of the line $ax + by + c = 0$ is $-\frac{a}{b}$.
Since the line is a normal to the curve,we equate the slopes: $x^2 = -\frac{a}{b}$.
Since $x^2$ is always non-negative for real $x$,we must have $-\frac{a}{b} > 0$,which implies $\frac{a}{b} < 0$.
This condition holds if $a$ and $b$ have opposite signs,i.e.,$(a > 0, b < 0)$ or $(a < 0, b > 0)$.

(A) Given curves are $y=a^x$ and $y=b^x$.
At the point of intersection,$a^x = b^x$,which implies $x=0$.
Thus,the curves intersect at the point $(0, 1)$.
Let $m_1$ be the slope of the tangent to $y=a^x$ at $x=0$.
$m_1 = \left. \frac{dy}{dx} \right|_{x=0} = \left. a^x \ln a \right|_{x=0} = \ln a$.
Let $m_2$ be the slope of the tangent to $y=b^x$ at $x=0$.
$m_2 = \left. \frac{dy}{dx} \right|_{x=0} = \left. b^x \ln b \right|_{x=0} = \ln b$.
The angle $\alpha$ between the curves is given by $\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,we get $\tan \alpha = \frac{\ln a - \ln b}{1 + \ln a \ln b}$.

(D) To decompose the fraction $\frac{x^4}{(x^2+1)(x^2+3)}$,we first perform long division since the degree of the numerator is equal to the degree of the denominator.
$\frac{x^4}{(x^2+1)(x^2+3)} = \frac{x^4}{x^4+4x^2+3} = \frac{(x^4+4x^2+3) - (4x^2+3)}{x^4+4x^2+3} = 1 - \frac{4x^2+3}{(x^2+1)(x^2+3)}$.
Now,we use partial fractions for $\frac{4x^2+3}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}$.
Thus,the expression becomes $1 + \frac{-Ax-B}{x^2+1} + \frac{-Cx-D}{x^2+3}$,which is of the form $1 + \frac{Px+Q}{x^2+1} + \frac{Rx+S}{x^2+3}$.

(D) Let $f(x) = x - \sin x$.
Then $f'(x) = 1 - \cos x$.
Since $\cos x \leq 1$ for all $x \in \mathbb{R}$,we have $f'(x) \geq 0$ for all $x$.
This means $f(x)$ is a non-decreasing function.
Since $f(0) = 0 - \sin(0) = 0$,it follows that $f(x) \geq 0$ for all $x \geq 0$.
Thus,$x - \sin x \geq 0$,which implies $\sin x \leq x$ for all $x \geq 0$.
For $x < 0$,let $x = -t$ where $t > 0$. Then $\sin(-t) = -\sin t$.
The inequality becomes $-\sin t \leq -t$,which simplifies to $\sin t \geq t$.
Since $\sin t < t$ for all $t > 0$,the inequality $\sin t \geq t$ is only satisfied at $t = 0$.
Therefore,the set of all $x$ for which $\sin x \leq x$ is $[0, \infty)$.

(B) Consider the function $f(x) = \frac{\sin x}{x}$ for $x \in (0, \pi/2)$.
Taking the derivative,$f'(x) = \frac{x \cos x - \sin x}{x^2}$.
Let $u(x) = x \cos x - \sin x$. Then $u'(x) = \cos x - x \sin x - \cos x = -x \sin x$.
Since $x \in (0, \pi/2)$,$u'(x) < 0$,which means $u(x)$ is a strictly decreasing function.
Since $u(0) = 0$ and $u(x)$ is decreasing,$u(x) < 0$ for all $x \in (0, \pi/2)$.
Thus,$f'(x) < 0$,meaning $f(x)$ is a strictly decreasing function on $(0, \pi/2)$.
As $x \to 0^+$,$f(x) \to 1$,and at $x = \pi/2$,$f(\pi/2) = \frac{\sin(\pi/2)}{\pi/2} = \frac{2}{\pi}$.
Since $f(x)$ is decreasing,for $0 < x < \pi/2$,we have $f(\pi/2) < f(x) < \lim_{x \to 0^+} f(x)$.
Therefore,$\frac{2}{\pi} < \frac{\sin x}{x} < 1$.
Hence,option $B$ is correct.

(B) For an equilateral triangle,the orthocenter,circumcenter,centroid,and incenter all coincide at the same point.
Given that the origin is the orthocenter,it is also the centroid of the triangle.
The centroid of a triangle with vertices $\vec{a}, \vec{b}, \vec{c}$ is given by $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$.
Since the centroid is the origin,we have:
$\frac{\vec{a}+\vec{b}+\vec{c}}{3} = \vec{0}$
$\vec{a}+\vec{b}+\vec{c} = \vec{0}$
Therefore,$\vec{a}+\vec{b} = -\vec{c}$.

(A) From the geometry of the quadrilateral,we have the vector relations:
$SQ = SP + PQ = (a-b) + a = 2a - b$
$SM = SQ + QM = (2a - b) + \frac{b}{2} = 2a - \frac{b}{2}$
$SM = \frac{1}{2}(4a - b)$
Comparing this with $SM = m(4a - b)$,we get $m = \frac{1}{2}$.
Given $SX = \frac{4}{5}SM$,we substitute the expression for $SM$:
$SX = \frac{4}{5} \cdot \frac{1}{2}(4a - b) = \frac{2}{5}(4a - b)$
Comparing this with $SX = n(4a - b)$,we get $n = \frac{2}{5}$.
Therefore,$m + n = \frac{1}{2} + \frac{2}{5} = \frac{5+4}{10} = \frac{9}{10}$.

(C) In $\triangle PAC$,by triangle law of vector addition,$\vec{PA} + \vec{AC} = \vec{PC}$ ...$(i)$
In $\triangle PBC$,by triangle law of vector addition,$\vec{PB} + \vec{BC} = \vec{PC}$ ...$(ii)$
Adding equations $(i)$ and $(ii)$:
$\vec{PA} + \vec{PB} + \vec{AC} + \vec{BC} = 2\vec{PC}$
Since $C$ is the mid-point of $AB$,we have $\vec{AC} = -\vec{BC}$,or $\vec{AC} + \vec{BC} = 0$.
Therefore,$\vec{PA} + \vec{PB} = 2\vec{PC}$.

(D) The direction cosines of a ray making angles $\alpha, \beta, \gamma$ with the coordinate axes satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given $\beta = 2\alpha$ and $\gamma = 3\alpha$,we have $\cos^2 \alpha + \cos^2 2\alpha + \cos^2 3\alpha = 1$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,the equation becomes:
$\frac{1 + \cos 2\alpha}{2} + \frac{1 + \cos 4\alpha}{2} + \frac{1 + \cos 6\alpha}{2} = 1$
$3 + \cos 2\alpha + \cos 4\alpha + \cos 6\alpha = 2$
$\cos 2\alpha + \cos 4\alpha + \cos 6\alpha = -1$
$(\cos 6\alpha + \cos 2\alpha) + \cos 4\alpha = -1$
$2 \cos 4\alpha \cos 2\alpha + \cos 4\alpha = -1$
$\cos 4\alpha (2 \cos 2\alpha + 1) = -1$.
Testing the options:
For $\alpha = \frac{\pi}{6}$: $\cos^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} + \cos^2 \frac{\pi}{2} = \frac{3}{4} + \frac{1}{4} + 0 = 1$.
For $\alpha = \frac{\pi}{4}$: $\cos^2 \frac{\pi}{4} + \cos^2 \frac{\pi}{2} + \cos^2 \frac{3\pi}{4} = \frac{1}{2} + 0 + \frac{1}{2} = 1$.
Thus,the possible values are $\frac{\pi}{6}$ and $\frac{\pi}{4}$.

(C) Let the point $P$ be $(at, \frac{a}{t})$.
The equation of the curve is $xy = a^2$.
Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
At point $P(at, \frac{a}{t})$,the slope of the tangent is $m = -\frac{a/t}{at} = -\frac{1}{t^2}$.
The equation of the tangent at $P$ is $y - \frac{a}{t} = -\frac{1}{t^2}(x - at)$.
$t^2y - at = -x + at$,which simplifies to $x + t^2y = 2at$.
To find the $x$-intercept,set $y = 0$: $x = 2at$. So,$A = (2at, 0)$.
To find the $y$-intercept,set $x = 0$: $t^2y = 2at$,so $y = \frac{2a}{t}$. So,$B = (0, \frac{2a}{t})$.
The area of the right-angled triangle $\triangle ABO$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2at) \times (\frac{2a}{t}) = 2a^2$ sq. units.

(C) Given the relation $f(x f(y)) = f(x y) + x$.
Since $f(x) - x = \lambda$,we have $f(x) = x + \lambda$.
Substituting this into the given functional equation:
$f(x(y + \lambda)) = (xy + \lambda) + x$
$x(y + \lambda) + \lambda = xy + \lambda + x$
$xy + x\lambda + \lambda = xy + \lambda + x$
Comparing the terms,we get $x\lambda = x$,which implies $\lambda = 1$.
Thus,$f(x) = x + 1$.
Now,we evaluate the limit:
$\lim _{x \rightarrow 0} \frac{(x + 1)^{\frac{1}{3}} - 1}{(x + 1)^{\frac{1}{2}} - 1}$
Using the standard limit formula $\lim _{u \rightarrow 1} \frac{u^n - 1}{u - 1} = n$:
$\lim _{x}$ ${\rightarrow 0} \frac{(1 + x)^{\frac{1}{3}} - 1}{(1 + x) - 1} \cdot \frac{(1 + x) - 1}{(1 + x)^{\frac{1}{2}} - 1} = \frac{1/3}{1/2} = \frac{2}{3}$.

(C) First,simplify the expression inside the limit: $\frac{1}{y-2} \left( \frac{1}{x} - \frac{1}{x+y-2} \right) = \frac{1}{y-2} \left( \frac{(x+y-2) - x}{x(x+y-2)} \right) = \frac{1}{y-2} \left( \frac{y-2}{x(x+y-2)} \right) = \frac{1}{x(x+y-2)}$.
Now,evaluate the limit as $y \to 2$: $\lim_{y \to 2} \frac{1}{x(x+y-2)} = \frac{1}{x(x+2-2)} = \frac{1}{x^2}$.
Finally,differentiate with respect to $x$: $\frac{d}{dx} \left( \frac{1}{x^2} \right) = \frac{d}{dx} (x^{-2}) = -2x^{-3} = \frac{-2}{x^3}$.

(D) The given expression is $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^4}{r^5+n^5}$.
Dividing the numerator and denominator of the general term by $n^5$,we get:
$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{(\frac{r}{n})^4}{(\frac{r}{n})^5+1}$.
This is a Riemann sum which can be expressed as the definite integral:
$\int_0^1 \frac{x^4}{1+x^5} dx$.
Let $u = 1+x^5$,then $du = 5x^4 dx$,or $x^4 dx = \frac{du}{5}$.
When $x=0, u=1$ and when $x=1, u=2$.
The integral becomes $\frac{1}{5} \int_1^2 \frac{1}{u} du = \frac{1}{5} [\ln |u|]_1^2 = \frac{1}{5} \ln 2$.
Using the property $a \ln b = \ln b^a$,we get $\ln 2^{1/5} = \ln \sqrt[5]{2}$.

(D) Let $L = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3+r^3}$.
We can rewrite the sum as:
$L = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3(1+(r/n)^3)} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{(r/n)^2}{1+(r/n)^3}$.
Using the definition of the definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f(\frac{r}{n}) = \int_0^1 f(x) dx$,we get:
$L = \int_0^1 \frac{x^2}{1+x^3} dx$.
Let $1+x^3 = t$,then $3x^2 dx = dt$,or $x^2 dx = \frac{dt}{3}$.
When $x=0, t=1$ and when $x=1, t=2$.
$L = \int_1^2 \frac{1}{t} \cdot \frac{dt}{3} = \frac{1}{3} [\log |t|]_1^2 = \frac{1}{3} (\log 2 - \log 1) = \frac{1}{3} \log 2$.
Using the property $a \log b = \log b^a$,we have $\frac{1}{3} \log 2 = \log 2^{1/3} = \log \sqrt[3]{2}$.

(B) Given,$p = 0.3, q = 1 - p = 0.7$.
Mean $(\mu) = np = 0.3n$.
Standard deviation $(\sigma) = \sqrt{npq} = \sqrt{n(0.3)(0.7)} = \sqrt{0.21n}$.
Given that $\mu = 3\sigma$.
Substituting the values,we get $0.3n = 3\sqrt{0.21n}$.
Dividing both sides by $3$,we get $0.1n = \sqrt{0.21n}$.
Squaring both sides,we get $(0.1n)^2 = 0.21n$.
$0.01n^2 = 0.21n$.
$n^2 = 21n$.
Since $n \neq 0$,we have $n = 21$.

(D) Given the equation: $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$.
Expanding the squares,we get: $(4 A^2+2 A B+2 B A+B^2)+(A^2-3 A B-3 B A+9 B^2)=5 A^2-2 A B+10 B^2$.
Combining like terms: $(4 A^2+A^2)+(B^2+9 B^2)+(2 A B-3 A B)+(2 B A-3 B A)=5 A^2-2 A B+10 B^2$.
This simplifies to: $5 A^2+10 B^2-A B-B A=5 A^2-2 A B+10 B^2$.
Subtracting $5 A^2+10 B^2$ from both sides,we get: $-A B-B A=-2 A B$.
Adding $A B$ to both sides,we get: $-B A=-A B$,which implies $A B=B A$.
Since $A$ and $B$ commute,$A B A B=A(B A) B=A(A B) B=(A A)(B B)=A^2 B^2$.

(B) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$.
We are given that $A^2 = I$.
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Calculating the product:
Row $1$: $(1)(1) + (0)(a) + (0)(b) = 1$,$(1)(0) + (0)(-1) + (0)(c) = 0$,$(1)(0) + (0)(0) + (0)(1) = 0$.
Row $2$: $(a)(1) + (-1)(a) + (0)(b) = 0$,$(a)(0) + (-1)(-1) + (0)(c) = 1$,$(a)(0) + (-1)(0) + (0)(1) = 0$.
Row $3$: $(b)(1) + (c)(a) + (1)(b) = 2b + ac$,$(b)(0) + (c)(-1) + (1)(c) = 0$,$(b)(0) + (c)(0) + (1)(1) = 1$.
Thus,$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2b + ac & 0 & 1 \end{bmatrix}$.
Equating this to the identity matrix $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$,we get $2b + ac = 0$.
Therefore,$b = -\frac{ac}{2}$.

(B) Given matrices are $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$.
First,we find the general form for $A^n$:
$A^2 = \left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$A^3 = \left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right] \left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 0 \\ 6 & 1\end{array}\right]$
By induction,$A^n = \left[\begin{array}{cc}1 & 0 \\ 2n & 1\end{array}\right]$. Thus,$A^6 = \left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right]$.
Next,we find the general form for $B^n$:
$B^2 = \left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 6 \\ 0 & 1\end{array}\right]$
$B^3 = \left[\begin{array}{ll}1 & 6 \\ 0 & 1\end{array}\right] \left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 9 \\ 0 & 1\end{array}\right]$
By induction,$B^n = \left[\begin{array}{cc}1 & 3n \\ 0 & 1\end{array}\right]$. Thus,$B^6 = \left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right]$.
Now,$A^6 + B^6 = \left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right] + \left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}2 & 18 \\ 12 & 2\end{array}\right]$.
Finally,$\operatorname{det}(A^6 + B^6) = (2 \times 2) - (18 \times 12) = 4 - 216 = -212$.

(C) Given $G(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Since $x+y=0$,we have $y = -x$.
Thus,$G(y) = G(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Now,$G(x)G(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Performing matrix multiplication:
Row $1$,Col $1$: $(\cos x)(\cos x) + (-\sin x)(-\sin x) + (0)(0) = \cos^2 x + \sin^2 x = 1$.
Row $1$,Col $2$: $(\cos x)(\sin x) + (-\sin x)(\cos x) + (0)(0) = 0$.
Row $2$,Col $1$: $(\sin x)(\cos x) + (\cos x)(-\sin x) + (0)(0) = 0$.
Row $2$,Col $2$: $(\sin x)(\sin x) + (\cos x)(\cos x) + (0)(0) = \sin^2 x + \cos^2 x = 1$.
Thus,$G(x)G(y) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$,which is the identity matrix.

(D) Given $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \cdot A$:
$A^2 = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^2 + 0 & 0 + 0 \\ x + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \\ x + 1 & 1 \end{bmatrix}$.
Next,calculate $A^3 = A^2 \cdot A$:
$A^3 = \begin{bmatrix} x^2 & 0 \\ x + 1 & 1 \end{bmatrix} \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} x^3 + 0 & 0 + 0 \\ x(x + 1) + 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} x^3 & 0 \\ x^2 + x + 1 & 1 \end{bmatrix}$.
Given $A^3 = B$,we have:
$\begin{bmatrix} x^3 & 0 \\ x^2 + x + 1 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$.
Comparing the corresponding elements:
$x^3 = 8 \implies x = 2$.
Also,$x^2 + x + 1 = 7 \implies x^2 + x - 6 = 0$.
Factoring the quadratic equation: $(x + 3)(x - 2) = 0$,which gives $x = 2$ or $x = -3$.
Since $x$ must satisfy both conditions,we take the common value,$x = 2$.

(D) Given $A = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix}$.
Calculating $AB$:
$AB = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} 3x & 4y \\ 5x & 6y \end{bmatrix}$.
Calculating $BA$:
$BA = \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 3x & 4x \\ 5y & 6y \end{bmatrix}$.
For $AB = BA$,we must have:
$3x = 3x$ (always true)
$4y = 4x \Rightarrow x = y$
$5x = 5y \Rightarrow x = y$
$6y = 6y$ (always true)
Thus,$AB = BA$ if and only if $x = y$.
Since $x, y \in \mathbb{N}$,we can choose any natural number $n$ such that $x = y = n$. Since there are infinitely many natural numbers,there exist infinitely many such matrices $B$.

(A) Given that $A$ is a singular matrix,its determinant must be zero,i.e.,$|A| = 0$.
$|A| = \begin{vmatrix} x & 1 & 2 \\ 2 & 4 & x \\ -3 & 3 & 2 \end{vmatrix} = 0$
Expanding along the first row:
$x(4 \times 2 - 3 \times x) - 1(2 \times 2 - (-3) \times x) + 2(2 \times 3 - (-3) \times 4) = 0$
$x(8 - 3x) - 1(4 + 3x) + 2(6 + 12) = 0$
$8x - 3x^2 - 4 - 3x + 36 = 0$
$-3x^2 + 5x + 32 = 0$
Multiplying by $-1$:
$3x^2 - 5x - 32 = 0$
For the quadratic equation $ax^2 + bx + c = 0$,the sum of roots $x_1 + x_2 = -b/a$ and the product of roots $x_1 x_2 = c/a$.
Here,$x_1 + x_2 = -(-5)/3 = 5/3$ and $x_1 x_2 = -32/3$.
Therefore,$x_1 + x_2 + x_1 x_2 = 5/3 - 32/3 = -27/3 = -9$.

(A) Given $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = x(0 - 1) - 2(0 - 2) + 1(2 - 2x)$
$|A| = -x + 4 + 2 - 2x = 6 - 3x$.
We are given $\det(A^3) = 125$.
Using the property $|A^n| = |A|^n$,we have $|A|^3 = 125$.
Taking the cube root on both sides,$|A| = 5$.
Substitute the value of $|A|$: $6 - 3x = 5$.
$3x = 6 - 5 = 1$.
$x = 1/3$.

(B) Given $A = nI$,where $I$ is the identity matrix of order $3 \times 3$.
Then $A^2 = (nI)^2 = n^2 I^2 = n^2 I = \begin{bmatrix} n^2 & 0 & 0 \\ 0 & n^2 & 0 \\ 0 & 0 & n^2 \end{bmatrix}$.
Next,$B^2 = \begin{bmatrix} 0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0 \end{bmatrix} = \begin{bmatrix} n^2 & 0 & 0 \\ 0 & n^2 & 0 \\ 0 & 0 & n^2 \end{bmatrix} = n^2 I$.
Also,$AB = (nI)B = n(IB) = nB = \begin{bmatrix} 0 & 0 & n^2 \\ 0 & n^2 & 0 \\ n^2 & 0 & 0 \end{bmatrix}$.
Now,$A^2 + B^2 + AB = n^2 I + n^2 I + nB = 2n^2 I + nB$.
Factoring out $n$,we get $n(2nI + B)$.

(C) Given $A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix}$.
We need to find $\operatorname{Tr}(A^2-A)$.
First,calculate $A-I$:
$A-I = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 3 \\ 1 & 6 & 9 \\ 2 & 3 & 6 \end{bmatrix}$.
Now,calculate $A(A-I)$:
$A^2-A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 3 \\ 1 & 6 & 9 \\ 2 & 3 & 6 \end{bmatrix}$.
The diagonal elements of the resulting matrix are:
$d_{11} = (1)(0) + (1)(1) + (3)(2) = 0 + 1 + 6 = 7$.
$d_{22} = (1)(1) + (7)(6) + (9)(3) = 1 + 42 + 27 = 70$.
$d_{33} = (2)(3) + (3)(9) + (7)(6) = 6 + 27 + 42 = 75$.
The trace is the sum of the diagonal elements:
$\operatorname{Tr}(A^2-A) = 7 + 70 + 75 = 152$.

(A) Given,$A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$.
The transpose of $A$ is $A^T = \begin{bmatrix} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{bmatrix}$.
Now,calculate the product $A A^T$:
$A A^T = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{bmatrix}$
$A A^T = \begin{bmatrix} (9+9+16) & (6+9+16) & (0+3+4) \\ (6+9+16) & (4+9+16) & (0+3+4) \\ (0+3+4) & (0+3+4) & (0+1+1) \end{bmatrix}$
$A A^T = \begin{bmatrix} 34 & 31 & 7 \\ 31 & 29 & 7 \\ 7 & 7 & 2 \end{bmatrix}$.
Since $(A A^T)^T = A A^T$,the matrix $A A^T$ is a symmetric matrix.

(C) Given that $A$ and $B$ are symmetric matrices,we have $A^{T} = A$ and $B^{T} = B$.
We are given $X = AB + BA$ and $Y = AB - BA$.
We need to find $(XY)^{T}$.
Using the property of transpose $(XY)^{T} = Y^{T} X^{T}$.
First,let us find $X^{T}$ and $Y^{T}$:
$X^{T} = (AB + BA)^{T} = (AB)^{T} + (BA)^{T} = B^{T}A^{T} + A^{T}B^{T} = BA + AB = X$.
$Y^{T} = (AB - BA)^{T} = (AB)^{T} - (BA)^{T} = B^{T}A^{T} - A^{T}B^{T} = BA - AB = -(AB - BA) = -Y$.
Now,$(XY)^{T} = Y^{T} X^{T}$.
Substituting the values of $Y^{T}$ and $X^{T}$,we get:
$(XY)^{T} = (-Y)(X) = -YX$.

(D) Given: $A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$
First,find the transpose of $A$: $A^T = \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix}$
Now,calculate $(A^T)^2$:
$(A^T)^2 = \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} (2)(2) + (-4)(-3) & (2)(-4) + (-4)(1) \\ (-3)(2) + (1)(-3) & (-3)(-4) + (1)(1) \end{bmatrix} = \begin{bmatrix} 16 & -12 \\ -9 & 13 \end{bmatrix}$
Next,calculate $(12A)^T$:
$12A = \begin{bmatrix} 24 & -36 \\ -48 & 12 \end{bmatrix} \Rightarrow (12A)^T = \begin{bmatrix} 24 & -48 \\ -36 & 12 \end{bmatrix}$
Finally,add the two matrices:
$(A^T)^2 + (12A)^T = \begin{bmatrix} 16 & -12 \\ -9 & 13 \end{bmatrix} + \begin{bmatrix} 24 & -48 \\ -36 & 12 \end{bmatrix} = \begin{bmatrix} 40 & -60 \\ -45 & 25 \end{bmatrix}$

(C) To find the adjoint of matrix $A$,we first find the cofactor of each element of $A$.
Let $C_{ij}$ be the cofactor of the element $a_{ij}$.
$C_{11} = +((-6)(4) - (5)(0)) = -24$
$C_{12} = -((2)(4) - (5)(5)) = -(8 - 25) = 17$
$C_{13} = +((2)(0) - (-6)(5)) = 30$
$C_{21} = -((-2)(4) - (2)(0)) = -(-8) = 8$
$C_{22} = +((1)(4) - (2)(5)) = 4 - 10 = -6$
$C_{23} = -((1)(0) - (-2)(5)) = -(0 + 10) = -10$
$C_{31} = +((-2)(5) - (2)(-6)) = -10 + 12 = 2$
$C_{32} = -((1)(5) - (2)(2)) = -(5 - 4) = -1$
$C_{33} = +((1)(-6) - (-2)(2)) = -6 + 4 = -2$
The cofactor matrix $C$ is $\begin{bmatrix} -24 & 17 & 30 \\ 8 & -6 & -10 \\ 2 & -1 & -2 \end{bmatrix}$.
The adjoint of $A$ is the transpose of the cofactor matrix,$\operatorname{Adj} A = C^T = \begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix}$.

(B) First,we calculate the determinant of matrix $A$:
$|A| = 1((-1)(1) - (3)(2)) - (-3)((-2)(-1) - (3)(3)) + 2((-2)(2) - (1)(3))$
$|A| = 1(-1 - 6) + 3(2 - 9) + 2(-4 - 3)$
$|A| = 1(-7) + 3(-7) + 2(-7) = -7 - 21 - 14 = -42$.
We know the property $A \cdot \operatorname{Adj} A = |A| I$,where $I$ is the identity matrix.
Therefore,$A^2 \operatorname{Adj} A = A(A \operatorname{Adj} A) = A(|A| I) = |A| A$.
Substituting the value of $|A|$,we get $A^2 \operatorname{Adj} A = -42 A$.

(A) Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{bmatrix}$. The adjoint of a matrix is the transpose of its cofactor matrix.
First,we calculate the cofactors $C_{ij}$ of matrix $A$:
$C_{11} = (-1)^{1+1}(3 \times 2 - 1 \times 1) = 5$
$C_{12} = (-1)^{1+2}(2 \times 2 - 3 \times 1) = -1$
$C_{13} = (-1)^{1+3}(2 \times 1 - 3 \times 3) = -7$
$C_{21} = (-1)^{2+1}(2 \times 2 - 3 \times 1) = -1$
$C_{22} = (-1)^{2+2}(1 \times 2 - 3 \times 3) = -7$
$C_{23} = (-1)^{2+3}(1 \times 1 - 3 \times 2) = 5$
$C_{31} = (-1)^{3+1}(2 \times 1 - 3 \times 3) = -7$
$C_{32} = (-1)^{3+2}(1 \times 1 - 2 \times 3) = 5$
$C_{33} = (-1)^{3+3}(1 \times 3 - 2 \times 2) = -1$
The cofactor matrix is $\begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}$.
The adjoint is the transpose of this matrix: $\text{adj}(A) = \begin{bmatrix} 5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1 \end{bmatrix}$.
Comparing this with $\begin{bmatrix} 5 & a & -7 \\ b & -7 & c \\ -7 & d & -1 \end{bmatrix}$,we get $a = -1, b = -1, c = 5, d = 5$.
Therefore,$a+b+c+d = -1 - 1 + 5 + 5 = 8$.

(B) Given $A = \frac{1}{7} \begin{bmatrix} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{bmatrix}$.
First,we check if $A$ is an orthogonal matrix by calculating $AA^T$.
$A^T = \frac{1}{7} \begin{bmatrix} 3 & -6 & -2 \\ -2 & -3 & 6 \\ 6 & 2 & 3 \end{bmatrix}$.
$AA^T = \frac{1}{49} \begin{bmatrix} 3 & -2 & 6 \\ -6 & -3 & 2 \\ -2 & 6 & 3 \end{bmatrix} \begin{bmatrix} 3 & -6 & -2 \\ -2 & -3 & 6 \\ 6 & 2 & 3 \end{bmatrix}$.
Calculating the product:
Row $1 \times$ Col $1 = (3)(3) + (-2)(-2) + (6)(6) = 9 + 4 + 36 = 49$.
Row $1 \times$ Col $2 = (3)(-6) + (-2)(-3) + (6)(2) = -18 + 6 + 12 = 0$.
Row $1 \times$ Col $3 = (3)(-2) + (-2)(6) + (6)(3) = -6 - 12 + 18 = 0$.
Similarly,all off-diagonal elements result in $0$ and diagonal elements result in $49$.
Thus,$AA^T = \frac{1}{49} (49I) = I$.
Since $AA^T = I$,it follows that $A^{-1} = A^T$.

(C) Given,$A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
First,we find the determinant of $A$:
$|A| = (\sin \alpha)(\sin \alpha) - (-\cos \alpha)(\cos \alpha) = \sin^2 \alpha + \cos^2 \alpha = 1$.
Next,we find the adjoint of $A$:
$\text{Adj } A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Since $A^{-1} = \frac{1}{|A|} \text{Adj } A$,we have:
$A^{-1} = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Now,calculate $A + A^{-1}$:
$A + A^{-1} = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} + \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix} = \begin{bmatrix} 2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha \end{bmatrix}$.
Given $A + A^{-1} = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we equate the corresponding elements:
$2 \sin \alpha = 1 \implies \sin \alpha = \frac{1}{2}$.
Thus,$\alpha = \frac{\pi}{6}$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}$.
First,calculate the product $M = A B^{-1}$:
$M = \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(0) & (1)(1) + (2)(2) \\ (-2)(1) + (1)(0) & (-2)(1) + (1)(2) \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ -2 & 0 \end{bmatrix}$.
Now,find the inverse $M^{-1} = (A B^{-1})^{-1}$.
The determinant $|M| = (1)(0) - (5)(-2) = 0 + 10 = 10$.
The adjoint $\text{adj}(M) = \begin{bmatrix} 0 & -5 \\ 2 & 1 \end{bmatrix}$.
Thus,$M^{-1} = \frac{1}{|M|} \text{adj}(M) = \frac{1}{10} \begin{bmatrix} 0 & -5 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -\frac{1}{2} \\ \frac{1}{5} & \frac{1}{10} \end{bmatrix}$.
Comparing this with $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,we get $a = 0, b = -\frac{1}{2}, c = \frac{1}{5}, d = \frac{1}{10}$.
Finally,calculate $2b + 5c + 10d$:
$2(-\frac{1}{2}) + 5(\frac{1}{5}) + 10(\frac{1}{10}) = -1 + 1 + 1 = 1$.

(A) Given the characteristic equation $A^3-6A^2+11A-6I=0$.
To find $A^{-1}$,multiply the entire equation by $A^{-1}$:
$A^{-1}(A^3-6A^2+11A-6I) = A^{-1}(0)$
$A^2-6A+11I-6A^{-1} = 0$
Rearranging the terms to solve for $A^{-1}$:
$6A^{-1} = A^2-6A+11I$
$A^{-1} = \frac{1}{6}(A^2-6A+11I)$

(A) Given,$A = \begin{bmatrix} 1 & 1 & a+1 \\ 1 & a+1 & 1 \\ a+1 & 1 & 1 \end{bmatrix}$.
Since $A$ is a non-invertible matrix,its determinant must be zero,i.e.,$|A| = 0$.
Calculating the determinant along the first row:
$|A| = 1((a+1)(1) - 1(1)) - 1(1(1) - 1(a+1)) + (a+1)(1(1) - (a+1)(a+1)) = 0$
$|A| = 1(a+1-1) - 1(1-a-1) + (a+1)(1-(a+1)^2) = 0$
$|A| = a + a + (a+1)(1 - (a^2 + 2a + 1)) = 0$
$|A| = 2a + (a+1)(-a^2 - 2a) = 0$
$|A| = 2a - a^3 - 2a^2 - a^2 - 2a = 0$
$-a^3 - 3a^2 = 0$
$-a^2(a+3) = 0$
Thus,the values of $a$ are $a = 0$ and $a = -3$.
The sum of all values of $a$ is $0 + (-3) = -3$.

(B) Given matrix $A = \begin{bmatrix} 1 & 2 & 1 & -1 \\ -1 & 2 & 3 & 5 \\ 0 & 1 & k & k \end{bmatrix}$.
We perform row operations to reduce the matrix to row echelon form.
Applying $R_2 \rightarrow R_2 + R_1$:
$A \sim \begin{bmatrix} 1 & 2 & 1 & -1 \\ 0 & 4 & 4 & 4 \\ 0 & 1 & k & k \end{bmatrix}$.
Applying $R_2 \rightarrow \frac{1}{4}R_2$:
$A \sim \begin{bmatrix} 1 & 2 & 1 & -1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & k & k \end{bmatrix}$.
Applying $R_3 \rightarrow R_3 - R_2$:
$A \sim \begin{bmatrix} 1 & 2 & 1 & -1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & k-1 & k-1 \end{bmatrix}$.
For the rank of the matrix to be $2$,the third row must be a zero row. Thus,$k-1 = 0$,which implies $k = 1$.
Now,we check which quadratic equation has $k=1$ as a root:
For option $B$: $x^2+x-2 = (1)^2 + (1) - 2 = 1 + 1 - 2 = 0$.
Thus,$k=1$ is a root of $x^2+x-2=0$.

(B) First,we find the rank of matrix $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 0 & -1 \end{bmatrix}$.
Calculating the determinant of $A$:
$|A| = 1(1(-1) - 2(0)) - 0(2(-1) - 2(1)) + 1(2(0) - 1(1)) = 1(-1) - 0 + 1(-1) = -1 - 1 = -2$.
Since $|A| \neq 0$,the rank of $A$ $(r_1)$ is $3$.
Next,we find the rank of matrix $B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & -8 \end{bmatrix}$.
This is a $2 \times 4$ matrix. The maximum possible rank is $2$.
We check for a non-zero $2 \times 2$ minor. Consider the minor formed by the last two columns:
$\begin{vmatrix} 3 & 4 \\ 6 & -8 \end{vmatrix} = (3)(-8) - (4)(6) = -24 - 24 = -48 \neq 0$.
Since there exists a non-zero $2 \times 2$ minor,the rank of $B$ $(r_2)$ is $2$.
Finally,$r_1 - r_2 = 3 - 2 = 1$.

(B) The given matrix is $A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 1 & k-1 \\ 0 & 0 & k-1 & 1 \end{bmatrix}$.
The rank of a matrix is the number of non-zero rows in its row-echelon form.
For the rank of $A$ to be $2$,the third row must become a zero row.
This requires the elements of the third row to be zero,i.e.,$k-1 = 0$ and $1 = 0$.
Since $1 = 0$ is a contradiction,it is impossible for the third row to be a zero row for any value of $k$.
Therefore,the rank of $A$ will always be $3$ for any $k \in R$ where $k \neq 1$.
If $k = 1$,the matrix becomes $A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$,which also has rank $3$.
Thus,there is no value of $k$ for which the rank of $A$ is $2$.

(C) To find the rank of the matrix,we convert it into row echelon form using elementary row operations:
$\begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 4 & 1 & 4 \end{bmatrix}$
Now,apply row operations to create zeros in the first column:
$\xrightarrow[R_3 \rightarrow R_3 - 4R_1]{R_2 \rightarrow R_2 - 2R_1} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{bmatrix}$
Next,apply row operation to create zeros in the second column:
$\xrightarrow{R_3 \rightarrow R_3 - R_2} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
The matrix is now in row echelon form. The number of non-zero rows is $2$.
Therefore,the rank of the matrix is $2$.

(B) Given matrix $A = \begin{bmatrix} 0 & 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 8 & 12 \\ 0 & 0 & 0 & 4 & 8 \end{bmatrix}$.
Perform row operation $R_2 \rightarrow R_2 - 2R_1$:
$A \sim \begin{bmatrix} 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 4 & 8 \end{bmatrix}$.
Perform row operation $R_3 \rightarrow R_3 - 2R_2$:
$A \sim \begin{bmatrix} 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$.
The number of non-zero rows in the row-echelon form is $2$.
Therefore,the rank of $A$ is $2$.

(D) Given matrix $A = \begin{bmatrix} 1 & -1 & 0 & -2 \\ -4 & 4 & 0 & 8 \\ -2 & 1 & 2 & 4 \end{bmatrix}$.
We apply elementary row operations to transform the matrix into row echelon form.
Step $1$: Apply $R_2 \rightarrow R_2 + 4R_1$ and $R_3 \rightarrow R_3 + 2R_1$.
$A \sim \begin{bmatrix} 1 & -1 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \end{bmatrix}$.
Step $2$: Swap $R_2$ and $R_3$ $(R_2 \leftrightarrow R_3)$.
$A \sim \begin{bmatrix} 1 & -1 & 0 & -2 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.
The matrix is now in row echelon form. The number of non-zero rows is $2$.
Therefore,the rank of matrix $A$ is $2$.

(B) We are given that $a_i^2+b_i^2+c_i^2=1$ and $a_i a_j+b_i b_j+c_i c_j=0$ for $i \neq j$.
This implies that the rows (or columns) of matrix $A$ are orthonormal vectors.
Specifically,if we consider $A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$,then $AA^T$ is the product of $A$ and its transpose.
The entry in the $i$-th row and $j$-th column of $AA^T$ is the dot product of the $i$-th row of $A$ and the $j$-th row of $A$.
Given the conditions,$AA^T = I$,where $I$ is the $3 \times 3$ identity matrix.
Therefore,$\det(AA^T) = \det(I) = 1$.

(C) Given $A = \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix}$.
The determinant of $A$ is $\det(A) = (\alpha^2)(-\alpha) - (5)(5) = -\alpha^3 - 25$.
We are given $\det(A^{10}) = 1024$.
Using the property $\det(A^n) = (\det A)^n$,we have $(\det A)^{10} = 1024$.
Since $1024 = 2^{10}$,we have $(\det A)^{10} = 2^{10}$,which implies $\det A = 2$ or $\det A = -2$.
Case $1$: $-\alpha^3 - 25 = 2 \Rightarrow -\alpha^3 = 27 \Rightarrow \alpha^3 = -27 \Rightarrow \alpha = -3$.
Case $2$: $-\alpha^3 - 25 = -2 \Rightarrow -\alpha^3 = 23 \Rightarrow \alpha^3 = -23 \Rightarrow \alpha = -\sqrt[3]{23}$.
Comparing with the given options,$\alpha = -3$ is the correct value.

(A) Given $A = \begin{bmatrix} 5 & \sin^2 \theta & \cos^2 \theta \\ -\sin^2 \theta & -5 & 1 \\ \cos^2 \theta & 1 & 5 \end{bmatrix}$.
Expanding the determinant along the first row:
$\det(A) = 5((-5)(5) - (1)(1)) - \sin^2 \theta((-\sin^2 \theta)(5) - (1)(\cos^2 \theta)) + \cos^2 \theta((-\sin^2 \theta)(1) - (-5)(\cos^2 \theta))$
$= 5(-25 - 1) - \sin^2 \theta(-5\sin^2 \theta - \cos^2 \theta) + \cos^2 \theta(-\sin^2 \theta + 5\cos^2 \theta)$
$= 5(-26) + 5\sin^4 \theta + \sin^2 \theta \cos^2 \theta - \sin^2 \theta \cos^2 \theta + 5\cos^4 \theta$
$= -130 + 5(\sin^4 \theta + \cos^4 \theta)$
Using the identity $\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2}\sin^2 2\theta$.
$\det(A) = -130 + 5(1 - \frac{1}{2}\sin^2 2\theta) = -130 + 5 - \frac{5}{2}\sin^2 2\theta = -125 - \frac{5}{2}\sin^2 2\theta$.
Since $\sin^2 2\theta \in [0, 1]$,the expression $-125 - \frac{5}{2}\sin^2 2\theta$ is maximized when $\sin^2 2\theta = 0$.
Therefore,the maximum value is $-125 - 0 = -125$.

(A) Let the first term of the arithmetic progression be $A$ and the common difference be $D$.
The terms are given by $a = A + 4D$,$b = A + 7D$,and $c = A + 12D$.
Consider the determinant $\Delta = \begin{vmatrix} a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1 \end{vmatrix}$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = \begin{vmatrix} a & 5 & 1 \\ b-a & 8-5 & 1-1 \\ c-a & 13-5 & 1-1 \end{vmatrix} = \begin{vmatrix} a & 5 & 1 \\ 3D & 3 & 0 \\ 8D & 8 & 0 \end{vmatrix}$.
Expanding along the third column:
$\Delta = 1 \cdot [(3D)(8) - (8D)(3)] = 1 \cdot [24D - 24D] = 0$.

(C) Given the matrix $A = \begin{bmatrix} -2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1 \end{bmatrix}$.
The determinant of $A$ is given by $\operatorname{det}(A) = -2(-1 - 3) - x(-x - 2) + 1(3x - 2) = 0$.
Simplifying the expression: $-2(-4) - x(-x - 2) + 3x - 2 = 0$.
$8 + x^2 + 2x + 3x - 2 = 0$.
$x^2 + 5x + 6 = 0$.
Factoring the quadratic equation: $(x + 2)(x + 3) = 0$.
The roots are $l = -2$ and $m = -3$ (or vice versa).
We need to find $l^3 - m^3$.
If $l = -2$ and $m = -3$,then $l^3 - m^3 = (-2)^3 - (-3)^3 = -8 - (-27) = -8 + 27 = 19$.
If $l = -3$ and $m = -2$,then $l^3 - m^3 = (-3)^3 - (-2)^3 = -27 - (-8) = -27 + 8 = -19$.
Since $19$ is an option,the correct value is $19$.

(A) Given matrices $A = \begin{bmatrix} 1 & b & c \\ b & 2 & 3 \\ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \\ -b & 0 & 2 \\ -c & -2 & 0 \end{bmatrix}$.
First,we calculate the sum $A+B$:
$A+B = \begin{bmatrix} 1+0 & b+b & c+c \\ b-b & 2+0 & 3+2 \\ c-c & 3-2 & 4+0 \end{bmatrix} = \begin{bmatrix} 1 & 2b & 2c \\ 0 & 2 & 5 \\ 0 & 1 & 4 \end{bmatrix}$.
Now,we find the determinant of the resulting matrix:
$\det(A+B) = \begin{vmatrix} 1 & 2b & 2c \\ 0 & 2 & 5 \\ 0 & 1 & 4 \end{vmatrix}$.
Expanding along the first column:
$\det(A+B) = 1 \times \begin{vmatrix} 2 & 5 \\ 1 & 4 \end{vmatrix} - 0 + 0 = 1 \times (2 \times 4 - 5 \times 1) = 1 \times (8 - 5) = 3$.

(D) Given,$a = \sin \frac{\pi}{6} = \frac{1}{2}$,$b = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $c = \cot \frac{\pi}{2} = 0$.
Substituting these values into the matrix $A$:
$A = \begin{bmatrix} (\frac{1}{2})^2 + 0^2 & (\frac{1}{2})^2 & (\frac{1}{2})^2 \\ (\frac{1}{\sqrt{2}})^2 & 0^2 + (\frac{1}{2})^2 & (\frac{1}{\sqrt{2}})^2 \\ 0^2 & 0^2 & (\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{2} \\ 0 & 0 & \frac{3}{4} \end{bmatrix}$.
Now,calculate the determinant $|A|$:
$|A| = \frac{1}{4} \left( \frac{1}{4} \cdot \frac{3}{4} - 0 \right) - \frac{1}{4} \left( \frac{1}{2} \cdot \frac{3}{4} - 0 \right) + \frac{1}{4} (0 - 0) = \frac{1}{4} \cdot \frac{3}{16} - \frac{1}{4} \cdot \frac{3}{8} = \frac{3}{64} - \frac{6}{64} = -\frac{3}{64}$.
Wait,let us re-evaluate the matrix $A$ construction:
$A = \begin{bmatrix} b^2+c^2 & a^2 & a^2 \\ b^2 & c^2+a^2 & b^2 \\ c^2 & c^2 & a^2+b^2 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/4 & 1/4 \\ 1/2 & 1/4 & 1/2 \\ 0 & 0 & 3/4 \end{bmatrix}$.
Expanding along the third row: $|A| = 0 - 0 + \frac{3}{4} \left( \frac{1}{4} \cdot \frac{1}{4} - \frac{1}{4} \cdot \frac{1}{2} \right) = \frac{3}{4} \left( \frac{1}{16} - \frac{2}{16} \right) = \frac{3}{4} \left( -\frac{1}{16} \right) = -\frac{3}{64} \neq 0$.
Re-checking the question values: $a^2 = 1/4$,$b^2 = 1/2$,$c^2 = 0$.
$A = \begin{bmatrix} 1/2 & 1/4 & 1/4 \\ 1/2 & 1/4 & 1/2 \\ 0 & 0 & 3/4 \end{bmatrix}$.
Since $|A| \neq 0$,it is a non-singular matrix.

(A) The given system of equations is $x+y+z=6$,$5x-y+2z=3$,and $2x+y-z=-5$.
In matrix form $AX=D$,where $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 5 & -1 & 2 \\ 2 & 1 & -1\end{array}\right]$,$X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$,and $D=\left[\begin{array}{c}6 \\ 3 \\ -5\end{array}\right]$.
First,we find the cofactor matrix $C$ of $A$:
$C_{11} = (-1)^{1+1} \begin{vmatrix} -1 & 2 \\ 1 & -1 \end{vmatrix} = 1-2 = -1$
$C_{12} = (-1)^{1+2} \begin{vmatrix} 5 & 2 \\ 2 & -1 \end{vmatrix} = -(-5-4) = 9$
$C_{13} = (-1)^{1+3} \begin{vmatrix} 5 & -1 \\ 2 & 1 \end{vmatrix} = 5-(-2) = 7$
$C_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -(-1-1) = 2$
$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -1-2 = -3$
$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = -(1-2) = 1$
$C_{31} = (-1)^{3+1} \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} = 2-(-1) = 3$
$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 1 \\ 5 & 2 \end{vmatrix} = -(2-5) = 3$
$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & 1 \\ 5 & -1 \end{vmatrix} = -1-5 = -6$
Thus,$\operatorname{adj}(A) = C^T = \left[\begin{array}{ccc}-1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6\end{array}\right]$.
Now,$(\operatorname{adj} A)D = \left[\begin{array}{ccc}-1 & 2 & 3 \\ 9 & -3 & 3 \\ 7 & 1 & -6\end{array}\right] \left[\begin{array}{c}6 \\ 3 \\ -5\end{array}\right] = \left[\begin{array}{c}-6+6-15 \\ 54-9-15 \\ 42+3+30\end{array}\right] = \left[\begin{array}{c}-15 \\ 30 \\ 75\end{array}\right]$.