AP EAMCET 2018 Mathematics Question Paper with Answer and Solution

(C) For the square root to be defined,$x^2+6x+5 \ge 0$,which implies $(x+5)(x+1) \ge 0$,so $x \in (-\infty, -5] \cup [-1, \infty)$.
For the inequality $\sqrt{x^2+6x+5} > (8-x)$ to hold,we must have $8-x < 0$ or ($8-x \ge 0$ and $x^2+6x+5 > (8-x)^2$).
Case $1$: $8-x < 0 \implies x > 8$. Since $x > 8$ satisfies $x \in [-1, \infty)$,this is a valid solution.
Case $2$: $8-x \ge 0 \implies x \le 8$. Squaring both sides: $x^2+6x+5 > 64-16x+x^2$.
$22x > 59 \implies x > \frac{59}{22}$.
Combining $x > \frac{59}{22}$ with $x \le 8$,we get $x \in (\frac{59}{22}, 8]$.
Combining Case $1$ and Case $2$: $x \in (\frac{59}{22}, 8] \cup (8, \infty) = (\frac{59}{22}, \infty)$.

(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2-4x+5=0$.
From the relation between roots and coefficients,we have $\alpha+\beta=4$ and $\alpha\beta=5$.
Let the new roots be $S_1 = \alpha^2+\beta$ and $S_2 = \alpha+\beta^2$.
The sum of the new roots is:
$S_1+S_2 = (\alpha^2+\beta)+(\alpha+\beta^2) = (\alpha^2+\beta^2)+(\alpha+\beta)$
$= ((\alpha+\beta)^2-2\alpha\beta)+(\alpha+\beta)$
$= (4^2-2(5))+4 = (16-10)+4 = 6+4 = 10$.
The product of the new roots is:
$S_1 \times S_2 = (\alpha^2+\beta)(\alpha+\beta^2) = \alpha^3+\alpha^2\beta^2+\alpha\beta+\beta^3$
$= (\alpha^3+\beta^3)+(\alpha\beta)^2+\alpha\beta$
$= ((\alpha+\beta)^3-3\alpha\beta(\alpha+\beta))+(\alpha\beta)^2+\alpha\beta$
$= (4^3-3(5)(4))+(5)^2+5$
$= (64-60)+25+5 = 4+30 = 34$.
The required quadratic equation is $x^2-(S_1+S_2)x+(S_1 \times S_2) = 0$.
Substituting the values,we get $x^2-10x+34=0$.

(B) Given equation is $(p^2+p-3)(p^2+p-2)-12=0$.
Let $y = p^2+p-2$.
Then the equation becomes $(y-1)y - 12 = 0$.
$y^2 - y - 12 = 0$.
$(y-4)(y+3) = 0$.
So,$y = 4$ or $y = -3$.
Case $1$: $p^2+p-2 = 4 \Rightarrow p^2+p-6 = 0$.
$(p+3)(p-2) = 0$,so $p = -3, 2$ (These are real roots).
Case $2$: $p^2+p-2 = -3 \Rightarrow p^2+p+1 = 0$.
For this quadratic,the discriminant $D = b^2 - 4ac = 1^2 - 4(1)(1) = -3 < 0$.
Thus,the roots are non-real.
The sum of the roots of $p^2+p+1=0$ is given by $-\frac{b}{a} = -\frac{1}{1} = -1$.

(A) Given equation: $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$.
$x=1$ is a root of the equation.
Dividing by $(x-1)$,we get $(x-1)(x^4-4 x^3+5 x^2-4 x+1)=0$.
For $x^4-4 x^3+5 x^2-4 x+1=0$,divide by $x^2$:
$x^2-4 x+5-\frac{4}{x}+\frac{1}{x^2}=0
$ $\Rightarrow (x^2+\frac{1}{x^2})-4(x+\frac{1}{x})+5=0
$ $\Rightarrow (x+\frac{1}{x})^2-2-4(x+\frac{1}{x})+5=0
$ $\Rightarrow (x+\frac{1}{x})^2-4(x+\frac{1}{x})+3=0$.
Let $y = x+\frac{1}{x}$,then $y^2-4y+3=0
$ $\Rightarrow (y-1)(y-3)=0
$ $\Rightarrow y=1, 3$.
Case $1$: $x+\frac{1}{x}=1 \Rightarrow x^2-x+1=0$ (roots are complex).
Case $2$: $x+\frac{1}{x}=3 \Rightarrow x^2-3x+1=0$ (roots are irrational).
Thus,$\alpha+\beta=3$ and $\alpha \beta=1$.
Substituting into $(\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0$:
$3x^2+2x-1=0
$ $\Rightarrow (3x-1)(x+1)=0
$ $\Rightarrow x=-1, \frac{1}{3}$.

(C) For the roots of the quadratic equation $x^2+2(a+b+c)x+3\lambda(ab+bc+ca)=0$ to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = [2(a+b+c)]^2 - 4(1)(3\lambda(ab+bc+ca)) \geq 0$
$4(a+b+c)^2 - 12\lambda(ab+bc+ca) \geq 0$
$(a+b+c)^2 \geq 3\lambda(ab+bc+ca)$
$a^2+b^2+c^2+2(ab+bc+ca) \geq 3\lambda(ab+bc+ca)$
Dividing by $(ab+bc+ca)$,we get:
$\frac{a^2+b^2+c^2}{ab+bc+ca} + 2 \geq 3\lambda$
Since $a, b, c$ are sides of a triangle,$a+b > c$,$b+c > a$,and $c+a > b$.
It is a known inequality that $a^2+b^2+c^2 < 2(ab+bc+ca)$.
Thus,$\frac{a^2+b^2+c^2}{ab+bc+ca} < 2$.
Substituting this into the inequality:
$3\lambda \leq \frac{a^2+b^2+c^2}{ab+bc+ca} + 2 < 2 + 2 = 4$
$3\lambda < 4 \Rightarrow \lambda < \frac{4}{3}$.

(A) From $xy+l(x+y)+m=0$,we have $x(y+l) = -(ly+m)$,so $x = -\frac{ly+m}{y+l}$. Substituting this into $x^2+\alpha x+\beta=0$ gives:
$\left(-\frac{ly+m}{y+l}\right)^2 + \alpha\left(-\frac{ly+m}{y+l}\right) + \beta = 0$
$(ly+m)^2 - \alpha(ly+m)(y+l) + \beta(y+l)^2 = 0$
$(l^2y^2 + m^2 + 2lmy) - \alpha(ly^2 + l^2y + my + ml) + \beta(y^2 + 2ly + l^2) = 0$
$(l^2 - \alpha l + \beta)y^2 + (2lm - \alpha l^2 - \alpha m + 2\beta l)y + (m^2 - \alpha ml + \beta l^2) = 0$
Since this equation has the same roots as $x^2 + \alpha x + \beta = 0$,the ratios of coefficients must be equal:
$\frac{l^2 - \alpha l + \beta}{1} = \frac{2lm - \alpha l^2 - \alpha m + 2\beta l}{\alpha} = \frac{m^2 - \alpha ml + \beta l^2}{\beta}$
Equating the first and third terms: $\beta(l^2 - \alpha l + \beta) = m^2 - \alpha ml + \beta l^2$
$\beta l^2 - \beta \alpha l + \beta^2 = m^2 - \alpha ml + \beta l^2$
$\beta^2 - \beta \alpha l - m^2 + \alpha ml = 0$
$\beta^2 - \beta m + \beta m - \beta \alpha l - m^2 + \alpha ml = 0$
$\beta(\beta - m) + (m - \alpha l)(\beta - m) = 0$
$(\beta - m)(\beta + m - \alpha l) = 0$
Thus,$\beta = m$ or $\beta = \alpha l - m$.

(C) Given the equation $x^3-2x^2+4x-1=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas,we have $\alpha+\beta+\gamma=2$,$\alpha\beta+\beta\gamma+\gamma\alpha=4$,and $\alpha\beta\gamma=1$.
Since $\alpha\beta\gamma=1$,we have $\beta\gamma=\frac{1}{\alpha}$,$\alpha\gamma=\frac{1}{\beta}$,and $\alpha\beta=\frac{1}{\gamma}$.
The roots of the new equation are $\beta\gamma+\frac{1}{\alpha} = \frac{1}{\alpha}+\frac{1}{\alpha} = \frac{2}{\alpha}$,$\alpha\gamma+\frac{1}{\beta} = \frac{2}{\beta}$,and $\alpha\beta+\frac{1}{\gamma} = \frac{2}{\gamma}$.
Let $y = \frac{2}{x}$,then $x = \frac{2}{y}$. Substituting this into the original equation: $(\frac{2}{y})^3 - 2(\frac{2}{y})^2 + 4(\frac{2}{y}) - 1 = 0$.
$\frac{8}{y^3} - \frac{8}{y^2} + \frac{8}{y} - 1 = 0$.
Multiplying by $-y^3$,we get $y^3 - 8y^2 + 8y - 8 = 0$.
Thus,the required equation is $x^3-8x^2+8x-8=0$.

(B) Given $f(x)$ is negative in $\left(-\infty, -\frac{5}{3}\right) \cup (3, \infty)$ and positive in $\left(-\frac{5}{3}, 3\right)$,we can write $f(x) = -k_1(x + \frac{5}{3})(x - 3)$ where $k_1 > 0$. This simplifies to $f(x) = k_1(x + \frac{5}{3})(3 - x)$.
Given $g(x)$ is negative in $\left(3, \frac{9}{2}\right)$ and positive elsewhere,we can write $g(x) = k_2(x - 3)(x - \frac{9}{2})$ where $k_2 > 0$.
Now,consider the product $P(x) = f(x)g(x) = k_1 k_2 (x + \frac{5}{3})(3 - x)(x - 3)(x - \frac{9}{2})$.
Since $(3 - x) = -(x - 3)$,we have $P(x) = -k_1 k_2 (x + \frac{5}{3})(x - 3)^2 (x - \frac{9}{2})$.
Let $K = k_1 k_2 > 0$. Then $P(x) = -K (x + \frac{5}{3})(x - 3)^2 (x - \frac{9}{2})$.
The roots are $x = -\frac{5}{3}, 3, \frac{9}{2}$. Note that $x = 3$ is a root of multiplicity $2$,so the sign does not change across $x = 3$.
For $x > \frac{9}{2}$,$P(x)$ is negative. For $3 < x < \frac{9}{2}$,$P(x)$ is positive. For $x < 3$ (and $x > -\frac{5}{3}$),$P(x)$ is positive.
Thus,in the interval $[0, 5]$,$P(x)$ is positive in $[0, 3) \cup (3, \frac{9}{2})$ and negative in $(\frac{9}{2}, 5]$. The correct option is $B$.

(A) Given that $ax^2 + bx + c = 0$ has no real roots,the discriminant $D = b^2 - 4ac < 0$.
Since $ax^2 + bx + c$ has no real roots,the expression $f(x) = ax^2 + bx + c$ must have the same sign for all $x \in \mathbb{R}$.
Given $4a + 2b + c > 0$,we have $f(2) > 0$,which implies $a > 0$ and $f(x) > 0$ for all $x \in \mathbb{R}$.
Since $f(x) > 0$ for all $x$,we have $f(0) = c > 0$ and $f(1) = a + b + c > 0$.
We are given $f(2) = 4a + 2b + c > 0$.
Now,consider $f(x) = ax^2 + bx + c$. Since $a > 0$,$f(x) > 0$ for all $x$.
Specifically,$f(1) = a + b + c > 0 \implies a + c > -b$.
Also,$f(0) = c > 0$.
Consider the expression $(c+a)(c+b)$.
Since $a > 0$ and $f(x) > 0$,we have $f(x) = a(x - \alpha)(x - \bar{\alpha}) > 0$.
Using the property $f(x) > 0$,we can evaluate $f(1) = a+b+c > 0$ and $f(0) = c > 0$.
From $ax^2+bx+c > 0$,we know $b^2 < 4ac$.
$(c+a)(c+b) = c^2 + bc + ac + ab$.
Since $b^2 < 4ac$,we can show that $(c+a)(c+b) > ab$.

(D) Let $y = \frac{x^2+4x+1}{x^2+x+1}$.
Then $y(x^2+x+1) = x^2+4x+1$,which implies $(y-1)x^2 + (y-4)x + (y-1) = 0$.
Since $x$ is real,the discriminant $D \ge 0$.
$D = (y-4)^2 - 4(y-1)^2 \ge 0$.
$(y-4)^2 - [2(y-1)]^2 \ge 0$.
$(y-4-2y+2)(y-4+2y-2) \ge 0$.
$(-y-2)(3y-6) \ge 0$.
$(y+2)(3y-6) \le 0$.
$(y+2)(y-2) \le 0$.
Thus,$-2 \le y \le 2$.
The minimum value is $-2$ and the maximum value is $2$.
The sum of the maximum and minimum values is $2 + (-2) = 0$.

(C) The given equation is $|x^2+2x-8| = -(x-2) = 2-x$.
Case $1$: $x^2+2x-8 \ge 0$.
$(x+4)(x-2) \ge 0 \implies x \in (-\infty, -4] \cup [2, \infty)$.
Then $x^2+2x-8 = 2-x \implies x^2+3x-10 = 0 \implies (x+5)(x-2) = 0$.
So $x = -5$ and $x = 2$. Both satisfy the condition.
Case $2$: $x^2+2x-8 < 0$.
$(x+4)(x-2) < 0 \implies x \in (-4, 2)$.
Then $-(x^2+2x-8) = 2-x \implies -x^2-2x+8 = 2-x \implies x^2+x-6 = 0 \implies (x+3)(x-2) = 0$.
So $x = -3$ and $x = 2$.
Since $x=2$ is not in $(-4, 2)$,we only take $x = -3$.
The distinct real solutions are $\{-5, 2, -3\}$.
Thus,there are $3$ distinct real solutions.

(A) The function $f(x) = x^2 + 2bx + 2c^2$ is a parabola opening upwards. Its minimum value occurs at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$.
The function $g(x) = -x^2 - 2cx + b^2$ is a parabola opening downwards. Its maximum value occurs at $x = -c$,which is $g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$.
Given that the minimum value of $f(x)$ is greater than the maximum value of $g(x)$:
$2c^2 - b^2 > c^2 + b^2$
$c^2 > 2b^2$
Taking the square root on both sides,we get $|c| > \sqrt{2}|b|$.

(B) We have two inequalities to solve:
$5x - 1 < (x + 1)^2$ and $(x + 1)^2 < 7x - 3$.
For the first inequality:
$5x - 1 < x^2 + 2x + 1$
$x^2 - 3x + 2 > 0$
$(x - 1)(x - 2) > 0$
This implies $x \in (-\infty, 1) \cup (2, \infty)$.
For the second inequality:
$(x + 1)^2 < 7x - 3$
$x^2 + 2x + 1 < 7x - 3$
$x^2 - 5x + 4 < 0$
$(x - 1)(x - 4) < 0$
This implies $x \in (1, 4)$.
Taking the intersection of both intervals:
$x \in ((-\infty, 1) \cup (2, \infty)) \cap (1, 4) = (2, 4)$.
The integers in the interval $(2, 4)$ is only $x = 3$.
Thus,there is only $1$ integral value.

(A) The function $f(x) = x^2 + 2bx + 2c^2$ is an upward-opening parabola. Its minimum value occurs at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$.
The function $g(x) = -x^2 - 2cx + b^2$ is a downward-opening parabola. Its maximum value occurs at $x = -c$,which is $g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$.
According to the problem,the minimum value of $f(x)$ is greater than the maximum value of $g(x)$:
$2c^2 - b^2 > c^2 + b^2$
$2c^2 - c^2 > b^2 + b^2$
$c^2 > 2b^2$.

(A) Given,$a, b$ and $c$ are the roots of the equation $x^3+qx+r=0$.
From Vieta's formulas:
$a+b+c=0$
$ab+bc+ca=q$
$abc=-r$
Since $a+b+c=0$,we have $(a+b+c)^2=0$.
$a^2+b^2+c^2+2(ab+bc+ca)=0$
$a^2+b^2+c^2+2(q)=0$
$a^2+b^2+c^2=-2q$
Now,consider the expression $(a-b)^2+(b-c)^2+(c-a)^2$:
$= a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca$
$= 2(a^2+b^2+c^2) - 2(ab+bc+ca)$
$= 2(-2q) - 2(q)$
$= -4q - 2q$
$= -6q$

(A) Let the roots of the cubic equation be $\alpha, -\alpha,$ and $\beta$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + (-\alpha) + \beta = 2p \Rightarrow \beta = 2p$.
Sum of roots taken two at a time: $\alpha(-\alpha) + (-\alpha)\beta + \alpha\beta = 3q$.
$-\alpha^2 - \alpha\beta + \alpha\beta = 3q \Rightarrow -\alpha^2 = 3q$.
Product of roots: $\alpha(-\alpha)\beta = 4r$.
$-\alpha^2 \beta = 4r$.
Substituting $\beta = 2p$ and $-\alpha^2 = 3q$ into the product equation:
$(3q)(2p) = 4r$.
$6pq = 4r$.
$r = \frac{6pq}{4} = \frac{3pq}{2}$.

(C) Given $f(x) = (x-a)(x-b) - \frac{a+b}{2} = x^2 - (a+b)x + ab - \frac{a+b}{2}$.
To find the minimum value,we find the vertex of the parabola. The derivative is $f'(x) = 2x - (a+b)$.
Setting $f'(x) = 0$,we get $x = \frac{a+b}{2}$.
The minimum value is $f\left(\frac{a+b}{2}\right) = \left(\frac{a+b}{2} - a\right)\left(\frac{a+b}{2} - b\right) - \frac{a+b}{2}$.
$f\left(\frac{a+b}{2}\right) = \left(\frac{b-a}{2}\right)\left(\frac{a-b}{2}\right) - \frac{a+b}{2} = -\frac{(a-b)^2}{4} - \frac{a+b}{2} = -\frac{a^2 - 2ab + b^2 + 2a + 2b}{4}$.
We can rewrite this as $f\left(\frac{a+b}{2}\right) = -\frac{(a+b)^2 - 4ab + 2(a+b)}{4} = -\frac{(a+b)^2}{4} + \frac{4ab - 2(a+b)}{4}$.
Since the roots of $f(x)=0$ are non-negative,let the roots be $x_1, x_2 \geq 0$. Then $x_1+x_2 = a+b \geq 0$ and $x_1x_2 = ab - \frac{a+b}{2} \geq 0$.
From $x_1x_2 \geq 0$,we have $ab \geq \frac{a+b}{2}$,which implies $4ab \geq 2(a+b)$,so $4ab - 2(a+b) \geq 0$.
Thus,the minimum value $f\left(\frac{a+b}{2}\right) = -\frac{(a+b)^2}{4} + \frac{4ab - 2(a+b)}{4} \geq -\frac{(a+b)^2}{4}$.

(B) Given roots are $z_1 = \sqrt{3} + \sqrt{27} = \sqrt{3} + 3\sqrt{3} = 4\sqrt{3}$ and $z_2 = \sqrt{2} + 5i$.
Since the coefficients are rational,irrational roots must occur in conjugate pairs.
For $z_1 = 4\sqrt{3}$,the conjugate is $-4\sqrt{3}$.
For $z_2 = \sqrt{2} + 5i$,the conjugate is $\sqrt{2} - 5i$.
However,for a polynomial with rational coefficients,if $\sqrt{a} + \sqrt{b}$ is a root,then $\pm\sqrt{a} \pm \sqrt{b}$ are also roots.
Here,$z_1 = 4\sqrt{3}$. The minimal polynomial for $4\sqrt{3}$ over $\mathbb{Q}$ is $x^2 - 48 = 0$,which has roots $4\sqrt{3}$ and $-4\sqrt{3}$.
For $z_2 = \sqrt{2} + 5i$,the minimal polynomial over $\mathbb{Q}$ is found by $(x - \sqrt{2})^2 = (5i)^2$,which is $x^2 - 2\sqrt{2}x + 2 = -25$,or $x^2 + 27 = 2\sqrt{2}x$.
Squaring both sides gives $(x^2 + 27)^2 = 8x^2$,which is $x^4 + 54x^2 + 729 = 8x^2$,or $x^4 + 46x^2 + 729 = 0$.
This polynomial has roots $\pm\sqrt{2} \pm 5i$.
Thus,the total degree is $2 + 4 = 6$.

(C) We have,$\frac{x^3}{(2x-1)(x-1)^2} = \frac{x^3}{2x^3-5x^2+4x-1}$.
Performing long division,we get:
$\frac{x^3}{2x^3-5x^2+4x-1} = \frac{1}{2} + \frac{1}{2} \left( \frac{5x^2-4x+1}{(2x-1)(x-1)^2} \right) \dots (i)$.
Now,let $\frac{5x^2-4x+1}{(2x-1)(x-1)^2} = \frac{B}{2x-1} + \frac{C}{x-1} + \frac{D}{(x-1)^2}$.
$5x^2-4x+1 = B(x-1)^2 + C(x-1)(2x-1) + D(2x-1)$.
Putting $x=1$,we get $5-4+1 = D(2-1) \Rightarrow D=2$.
Putting $x=1/2$,we get $5(1/4)-4(1/2)+1 = B(1/2-1)^2$ $\Rightarrow 5/4-2+1 = B(1/4)$ $\Rightarrow 1/4 = B/4$ $\Rightarrow B=1$.
Comparing coefficients of $x^2$: $5 = B + 2C$ $\Rightarrow 5 = 1 + 2C$ $\Rightarrow 2C = 4$ $\Rightarrow C=2$.
Substituting these into $(i)$:
$\frac{x^3}{(2x-1)(x-1)^2} = \frac{1}{2} + \frac{1}{2} \left( \frac{1}{2x-1} + \frac{2}{x-1} + \frac{2}{(x-1)^2} \right) = \frac{1}{2} + \frac{1/2}{2x-1} + \frac{1}{x-1} + \frac{1}{(x-1)^2}$.
Comparing with the given expression,$A=1/2, B=1/2, C=1, D=1$.
Then,$2A - 3B + 4C + 5D = 2(1/2) - 3(1/2) + 4(1) + 5(1) = 1 - 1.5 + 4 + 5 = 8.5 = \frac{17}{2}$.

(C) Given the cubic equation $x^3+0x^2+4x+1=0$.
Since $a, b,$ and $c$ are the roots,by Vieta's formulas:
$a+b+c = 0$
$ab+bc+ca = 4$
$abc = -1$
From $a+b+c=0$,we have $a+b = -c$,$b+c = -a$,and $c+a = -b$.
Substituting these into the expression:
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} = \frac{1}{-c} + \frac{1}{-a} + \frac{1}{-b}$
$= -(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
$= -(\frac{bc+ac+ab}{abc})$
$= -(\frac{4}{-1}) = 4$.

(A) Let the roots of the equation $x^3+p x^2+q x+r=0$ be $\alpha, \beta, \text{ and } \gamma$.
From the properties of roots,we have $\alpha+\beta+\gamma = -p$ $(i)$.
Given that the sum of two roots is zero,let $\beta+\gamma=0$.
Substituting this into $(i)$,we get $\alpha = -p$.
Since $\alpha$ is a root of the equation,it must satisfy $x^3+p x^2+q x+r=0$.
Substituting $x = -p$:
$(-p)^3 + p(-p)^2 + q(-p) + r = 0$
$-p^3 + p^3 - p q + r = 0$
$-p q + r = 0$
$r = p q$.

(D) Given: $z_1=1-2 i, z_2=1+i, z_3=3+4 i$.
We need to evaluate $\left(\frac{1}{z_1}+\frac{3}{z_2}\right) \frac{z_3}{z_2}$.
First,calculate the term inside the bracket:
$\frac{1}{1-2 i}+\frac{3}{1+i} = \frac{(1+i)+3(1-2 i)}{(1-2 i)(1+i)} = \frac{1+i+3-6 i}{1+i-2 i-2 i^2} = \frac{4-5 i}{1-i+2} = \frac{4-5 i}{3-i}$.
Now,multiply by $\frac{z_3}{z_2}$:
$\left(\frac{4-5 i}{3-i}\right) \left(\frac{3+4 i}{1+i}\right) = \frac{(4-5 i)(3+4 i)}{(3-i)(1+i)} = \frac{12+16 i-15 i-20 i^2}{3+3 i-i-i^2} = \frac{12+i+20}{3+2 i+1} = \frac{32+i}{4+2 i}$.
Rationalize the denominator:
$\frac{32+i}{4+2 i} \times \frac{4-2 i}{4-2 i} = \frac{128-64 i+4 i-2 i^2}{16-4 i^2} = \frac{128-60 i+2}{16+4} = \frac{130-60 i}{20} = \frac{13}{2}-3 i$.

(D) We have,$\frac{z-1}{z+i} = \frac{x+iy-1}{x+i(y+1)}$.
Multiplying the numerator and denominator by the conjugate of the denominator,$x-i(y+1)$,we get:
$\frac{(x-1)+iy}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)} = \frac{x(x-1) - i(x-1)(y+1) + ixy + y(y+1)}{x^2+(y+1)^2}$.
The real part is $\frac{x(x-1)+y(y+1)}{x^2+(y+1)^2}$.
Given that the real part is $1$,we have:
$\frac{x^2-x+y^2+y}{x^2+(y+1)^2} = 1$.
$x^2-x+y^2+y = x^2+y^2+2y+1$.
$-x+y = 2y+1$.
$x+y+1 = 0$.
Checking the options,for $(2016, -2017)$,we have $2016 + (-2017) + 1 = 2016 - 2017 + 1 = 0$.
Thus,the point $(2016, -2017)$ lies on the locus.

(B) Given $z = x + iy$,then $\bar{z} = x - iy$.
Given the equation $z^2 = (i \bar{z})^2$.
Expanding both sides: $z^2 = i^2 (\bar{z})^2$.
Since $i^2 = -1$,we have $z^2 = -(\bar{z})^2$,which implies $z^2 + (\bar{z})^2 = 0$.
Substitute $z = x + iy$ and $\bar{z} = x - iy$:
$(x + iy)^2 + (x - iy)^2 = 0$.
$(x^2 - y^2 + 2ixy) + (x^2 - y^2 - 2ixy) = 0$.
$2(x^2 - y^2) = 0$.
$x^2 - y^2 = 0$.
$x^2 = y^2$,which means $y = \pm x$.

(C) Given the equation $\alpha_1 + \alpha_2 z + \alpha_3 z^2 + \ldots + \alpha_n z^{n-1} + z^n = 0$.
Since $z = \cos \theta + i \sin \theta = e^{i \theta}$,we have $z^n = \cos n \theta + i \sin n \theta$.
Substituting $z$ into the equation: $\alpha_1 + \alpha_2 e^{i \theta} + \alpha_3 e^{i 2 \theta} + \ldots + \alpha_n e^{i (n-1) \theta} + e^{i n \theta} = 0$.
Multiply the entire equation by $e^{-i n \theta}$:
$\alpha_1 e^{-i n \theta} + \alpha_2 e^{-i (n-1) \theta} + \ldots + \alpha_n e^{-i \theta} + 1 = 0$.
Taking the real part of this equation:
$\alpha_1 \cos (-n \theta) + \alpha_2 \cos (-(n-1) \theta) + \ldots + \alpha_n \cos (-\theta) + 1 = 0$.
Since $\cos (-x) = \cos x$,this simplifies to:
$\alpha_1 \cos n \theta + \alpha_2 \cos (n-1) \theta + \ldots + \alpha_n \cos \theta + 1 = 0$.
Therefore,$\alpha_1 \cos n \theta + \alpha_2 \cos (n-1) \theta + \ldots + \alpha_n \cos \theta = -1$.

(A) Given,$13 e^{i \tan ^{-1} \frac{5}{12}} = a + i b$.
Using Euler's formula $e^{i \theta} = \cos \theta + i \sin \theta$,we have:
$13 [\cos(\tan ^{-1} \frac{5}{12}) + i \sin(\tan ^{-1} \frac{5}{12})] = a + i b$.
Let $\theta = \tan ^{-1} \frac{5}{12}$,then $\tan \theta = \frac{5}{12}$.
In a right-angled triangle with opposite side $5$ and adjacent side $12$,the hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
Thus,$\cos \theta = \frac{12}{13}$ and $\sin \theta = \frac{5}{13}$.
Substituting these values:
$13 [\frac{12}{13} + i \frac{5}{13}] = a + i b$.
$12 + 5i = a + i b$.
Comparing real and imaginary parts,we get $a = 12$ and $b = 5$.
Therefore,the ordered pair $(a, b) = (12, 5)$.

(D) Given $z = x + iy$. Then $\frac{z+1}{z+i} = \frac{(x+1) + iy}{x + i(y+1)}$.
To make this purely real,we multiply the numerator and denominator by the conjugate of the denominator: $\frac{(x+1) + iy}{x + i(y+1)} \times \frac{x - i(y+1)}{x - i(y+1)}$.
The denominator becomes $x^2 + (y+1)^2$,which is real.
The numerator is $(x+1)x + y(y+1) + i[xy - (x+1)(y+1)]$.
For the expression to be purely real,the imaginary part must be zero:
$xy - (x+1)(y+1) = 0$.
$xy - (xy + x + y + 1) = 0$.
$-x - y - 1 = 0 \Rightarrow x + y + 1 = 0$.
Since $z \neq -i$,the point $(0, -1)$ must be excluded.
Thus,the locus is $x+y+1=0, (x, y) \neq (0, -1)$.

(B) Given the complex equation:
$i z^3+4 z^2-z+4 i=0$
Grouping the terms:
$z^2(i z+4) - 1(z - 4i) = 0$
Wait,let us factorize correctly:
$i z^3 + 4 z^2 - z + 4 i = 0$
$z^2(i z + 4) + i(i z + 4) = 0$
$(i z + 4)(z^2 + i) = 0$
This gives $z = \frac{-4}{i} = 4i$ or $z^2 = -i$.
For $z^2 = -i$,we write $-i$ in polar form: $z^2 = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2})$.
Using De Moivre's theorem,$z = \cos(\frac{3\pi}{4} + k\pi) + i \sin(\frac{3\pi}{4} + k\pi)$ for $k=0, 1$.
For $k=0$,$z = \cos(\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$.
For $k=1$,$z = \cos(\frac{7\pi}{4}) + i \sin(\frac{7\pi}{4}) = \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$.
The magnitudes are $|4i| = 4$ and $|z| = \sqrt{(\pm \frac{1}{\sqrt{2}})^2 + (\mp \frac{1}{\sqrt{2}})^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = 1$.
Thus,the roots with minimum magnitude are $\pm(\frac{1-i}{\sqrt{2}})$.
Comparing with options,$\frac{1-i}{\sqrt{2}}$ is the correct choice.

(B) Given,$u+iv = \frac{3i}{(x+2)+iy}$.
Taking the reciprocal on both sides:
$\frac{1}{u+iv} = \frac{(x+2)+iy}{3i}$.
Multiply the numerator and denominator of the left side by the conjugate $(u-iv)$:
$\frac{u-iv}{u^2+v^2} = \frac{(x+2)+iy}{3i}$.
Multiply both sides by $3i$:
$(x+2)+iy = \frac{3i(u-iv)}{u^2+v^2} = \frac{3ui - 3vi^2}{u^2+v^2} = \frac{3v + 3ui}{u^2+v^2}$.
Equating the imaginary parts:
$y = \frac{3u}{u^2+v^2}$.

(C) Let $z = x + iy$. Then $\frac{z-i}{z-1} = \frac{x + i(y-1)}{(x-1) + iy}$.
To make this purely imaginary,we multiply by the conjugate of the denominator: $\frac{(x + i(y-1))((x-1) - iy)}{(x-1)^2 + y^2}$.
The real part is $\frac{x(x-1) + y(y-1)}{(x-1)^2 + y^2} = 0$.
This implies $x^2 - x + y^2 - y = 0$,which is $x^2 + y^2 - x - y = 0$.
Completing the square,we get $\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{\sqrt{2}}\right)^2$.
This represents a circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$.
Since the expression is undefined at $z = 1$ (i.e.,$(1, 0)$) and the numerator is zero at $z = i$ (i.e.,$(0, 1)$),these points must be excluded.

(A) Given that,$\alpha = \frac{a+ib}{1-c}$.
Taking the modulus squared on both sides,we get $|\alpha|^2 = \frac{a^2+b^2}{(1-c)^2} \quad \dots(i)$.
Given $a^2+b^2+c^2=c$,we have $a^2+b^2 = c-c^2 = c(1-c) \quad \dots(ii)$.
Substituting $(ii)$ into $(i)$,we get $|\alpha|^2 = \frac{c(1-c)}{(1-c)^2} = \frac{c}{1-c}$.
This implies $1+|\alpha|^2 = 1 + \frac{c}{1-c} = \frac{1-c+c}{1-c} = \frac{1}{1-c}$.
Thus,$1-c = \frac{1}{1+|\alpha|^2}$.
From $(ii)$,$a^2+b^2 = c(1-c) = (1-(1-c))(1-c) = (1-c) - (1-c)^2$.
Substituting $1-c = \frac{1}{1+|\alpha|^2}$,we get $a^2+b^2 = \frac{1}{1+|\alpha|^2} - \left(\frac{1}{1+|\alpha|^2}\right)^2 = \frac{1+|\alpha|^2-1}{(1+|\alpha|^2)^2} = \frac{|\alpha|^2}{(1+|\alpha|^2)^2}$.

(A) Given $z = \frac{1}{1 - \cos \theta + i \sin \theta}$.
Using trigonometric identities $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,we get:
$z = \frac{1}{2 \sin^2 \frac{\theta}{2} + i (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})}$
$z = \frac{1}{2 \sin \frac{\theta}{2} (\sin \frac{\theta}{2} + i \cos \frac{\theta}{2})}$
Since $\sin \frac{\theta}{2} + i \cos \frac{\theta}{2} = i (\cos \frac{\theta}{2} - i \sin \frac{\theta}{2}) = i e^{-i \theta/2}$,we have:
$z = \frac{1}{2 \sin \frac{\theta}{2} \cdot i e^{-i \theta/2}} = \frac{1}{2 \sin \frac{\theta}{2}} \cdot \frac{1}{i} e^{i \theta/2}$
Since $\frac{1}{i} = -i = e^{-i \pi/2}$,we get:
$z = \frac{1}{2 \sin \frac{\theta}{2}} e^{-i \pi/2} e^{i \theta/2} = \frac{1}{2} \operatorname{cosec} \frac{\theta}{2} e^{i (\theta/2 - \pi/2)}$
Thus,the modulus is $\frac{1}{2} \operatorname{cosec} \frac{\theta}{2}$ and the amplitude is $-(\frac{\pi}{2} - \frac{\theta}{2})$.

(B) Let $z = \sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$.
Using the trigonometric identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get:
$z = 2 \sin \frac{\pi}{10} \cos \frac{\pi}{10} + i(2 \sin^2 \frac{\pi}{10})$
$z = 2 \sin \frac{\pi}{10} (\cos \frac{\pi}{10} + i \sin \frac{\pi}{10})$
Since $2 \sin \frac{\pi}{10} > 0$,the complex number is in the polar form $r(\cos \theta + i \sin \theta)$ where $r = 2 \sin \frac{\pi}{10}$ and $\theta = \frac{\pi}{10}$.
Thus,the amplitude of the given complex number is $\frac{\pi}{10}$.

(C) Given $z = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.
$z^{101} = e^{i(101\pi/6)} = e^{i(16\pi + 5\pi/6)} = e^{i5\pi/6} = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{i}{2}$.
Also,$i^{103} = i^{100} \cdot i^3 = 1 \cdot (-i) = -i$.
Thus,$z^{101} + i^{103} = -\frac{\sqrt{3}}{2} + \frac{i}{2} - i = -\frac{\sqrt{3}}{2} - \frac{i}{2} = -\left(\frac{\sqrt{3}+i}{2}\right) = -z$.
Therefore,$\left(z^{101} + i^{103}\right)^{105} = (-z)^{105} = -z^{105}$.
Since $z = e^{i\pi/6}$,$z^{105} = e^{i(105\pi/6)} = e^{i(17\pi + 3\pi/6)} = e^{i(17\pi + \pi/2)} = e^{i17\pi} \cdot e^{i\pi/2} = (-1) \cdot i = -i$.
Wait,let us re-evaluate: $-z^{105} = -\left(e^{i\pi/6}\right)^{105} = -e^{i(17\pi + \pi/2)} = -(-1 \cdot i) = i$.
Actually,$z^3 = e^{i(3\pi/6)} = e^{i\pi/2} = i$. Thus,$-z^{105} = -(-z^3) = z^3$ is incorrect. Let us re-calculate: $(-z)^{105} = -z^{105} = -(e^{i\pi/6})^{105} = -e^{i(17\pi + \pi/2)} = -(-1 \cdot i) = i$. Since $z^3 = i$,the answer is $z^3$.

(C) Let $z = 1+\sin \theta+i \cos \theta$. We can write this as $z = 1+\cos(\frac{\pi}{2}-\theta) + i\sin(\frac{\pi}{2}-\theta)$.
Using the identities $1+\cos(2A) = 2\cos^2(A)$ and $\sin(2A) = 2\sin(A)\cos(A)$,where $A = \frac{\pi}{4}-\frac{\theta}{2}$,we get:
$z = 2\cos^2(\frac{\pi}{4}-\frac{\theta}{2}) + 2i\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})$
$z = 2\cos(\frac{\pi}{4}-\frac{\theta}{2}) [\cos(\frac{\pi}{4}-\frac{\theta}{2}) + i\sin(\frac{\pi}{4}-\frac{\theta}{2})]$
Then $z^n = 2^n \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) [\cos(\frac{n\pi}{4}-\frac{n\theta}{2}) + i\sin(\frac{n\pi}{4}-\frac{n\theta}{2})]$.
Similarly,the conjugate term is $\bar{z}^n = 2^n \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) [\cos(\frac{n\pi}{4}-\frac{n\theta}{2}) - i\sin(\frac{n\pi}{4}-\frac{n\theta}{2})]$.
Adding these two expressions:
$z^n + \bar{z}^n = 2^n \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) [2\cos(\frac{n\pi}{4}-\frac{n\theta}{2})]$
$= 2^{n+1} \cos^n(\frac{\pi}{4}-\frac{\theta}{2}) \cos(\frac{n\pi}{4}-\frac{n\theta}{2})$.

(A) Let $z = x + iy$, where $0 \leq x \leq 1$ and $0 \leq y \leq 1$.
Then $z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$.
Thus, $|e^{z^2}| = |e^{x^2 - y^2} \cdot e^{i(2xy)}| = e^{x^2 - y^2} \cdot |e^{i(2xy)}|$.
Since $|e^{i(2xy)}| = 1$, we have $|e^{z^2}| = e^{x^2 - y^2}$.
To maximize $e^{x^2 - y^2}$, we need to maximize the exponent $f(x, y) = x^2 - y^2$ subject to $0 \leq x \leq 1$ and $0 \leq y \leq 1$.
The maximum value of $x^2 - y^2$ occurs when $x$ is maximum $(x=1)$ and $y$ is minimum $(y=0)$.
Thus, the maximum value is $e^{1^2 - 0^2} = e^1 = e$.

(C) Given,$z = \frac{\sqrt{3} + i}{2} = \cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)$.
Using De Moivre's Theorem,$z^{101} = \cos \left(\frac{101\pi}{6}\right) + i \sin \left(\frac{101\pi}{6}\right)$.
Since $\frac{101\pi}{6} = 17\pi - \frac{\pi}{6}$,we have $z^{101} = \cos \left(17\pi - \frac{\pi}{6}\right) + i \sin \left(17\pi - \frac{\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right) = \frac{-\sqrt{3} + i}{2}$.
Also,$i^{103} = i^{100} \times i^3 = 1 \times (-i) = -i$.
Thus,$z^{101} + i^{103} = \frac{-\sqrt{3} + i}{2} - i = \frac{-\sqrt{3} - i}{2} = -\left(\frac{\sqrt{3} + i}{2}\right) = -z$.
Therefore,$\left(z^{101} + i^{103}\right)^{105} = (-z)^{105} = -z^{105}$.
$z^{105} = \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)^{105} = \cos \frac{105\pi}{6} + i \sin \frac{105\pi}{6} = \cos \frac{35\pi}{2} + i \sin \frac{35\pi}{2}$.
Since $\frac{35\pi}{2} = 18\pi - \frac{\pi}{2}$,we have $z^{105} = \cos \left(-\frac{\pi}{2}\right) + i \sin \left(-\frac{\pi}{2}\right) = 0 - i = -i$.
Finally,$-z^{105} = -(-i) = i$.

(B) Given the equation $x^{11}-x^6-x^5+1=0$.
Factorizing the expression:
$x^6(x^5-1) - 1(x^5-1) = 0$
$(x^6-1)(x^5-1) = 0$
This implies $x^6=1$ or $x^5=1$.
The roots are given by $x = \operatorname{cis}(\frac{2k\pi}{6})$ for $k \in \{0, 1, 2, 3, 4, 5\}$ or $x = \operatorname{cis}(\frac{2r\pi}{5})$ for $r \in \{0, 1, 2, 3, 4\}$.
For $k=1$,$x = \operatorname{cis}(\frac{2\pi}{6}) = \operatorname{cis}(\frac{\pi}{3})$.
Thus,$\operatorname{cis}(\frac{\pi}{3})$ is one of the complex roots.

(D) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$.
We need to evaluate the expression: $\frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}$.
First,simplify the first two terms:
$\frac{1}{1+2 \omega}+\frac{1}{2+\omega} = \frac{2+\omega+1+2 \omega}{(1+2 \omega)(2+\omega)} = \frac{3+3 \omega}{2+\omega+4 \omega+2 \omega^2} = \frac{3(1+\omega)}{2+5 \omega+2 \omega^2}$.
Since $1+\omega = -\omega^2$,the numerator is $-3 \omega^2$.
Also,$2+5 \omega+2 \omega^2 = 2(1+\omega^2)+5 \omega = 2(-\omega)+5 \omega = 3 \omega$.
So,the first two terms sum to $\frac{3(1+\omega)}{3 \omega} = \frac{1+\omega}{\omega} = \frac{-\omega^2}{\omega} = -\omega$.
Now,subtract the third term:
$-\frac{1}{1+\omega} = -\frac{1}{-\omega^2} = \frac{1}{\omega^2} = \omega$.
Thus,the total sum is $-\omega + \omega = 0$.

(B) Let the given expression be $E = \left[\frac{51+73 \omega+87 \omega^2}{73+87 \omega+51 \omega^2}+\frac{51+73 \omega+87 \omega^2}{87+51 \omega+73 \omega^2}\right]^{15}$.
Multiply the numerator of the first term by $\omega$ and divide by $\omega$:
$\frac{(51+73 \omega+87 \omega^2) \omega}{(73+87 \omega+51 \omega^2) \omega} = \frac{51 \omega+73 \omega^2+87 \omega^3}{73 \omega+87 \omega^2+51 \omega^3} = \frac{87+51 \omega+73 \omega^2}{73 \omega+87 \omega^2+51} = \frac{87+51 \omega+73 \omega^2}{\omega(73+87 \omega+51 \omega^2)}$.
Alternatively,observe the structure:
Let $A = 51+73 \omega+87 \omega^2$,$B = 73+87 \omega+51 \omega^2$,$C = 87+51 \omega+73 \omega^2$.
Note that $B = \omega^2 A$ and $C = \omega A$.
Thus,the expression becomes $\left[\frac{A}{\omega^2 A} + \frac{A}{\omega A}\right]^{15} = \left[\frac{1}{\omega^2} + \frac{1}{\omega}\right]^{15}$.
Since $\frac{1}{\omega^2} = \omega$ and $\frac{1}{\omega} = \omega^2$,we have $(\omega + \omega^2)^{15}$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Therefore,$(-1)^{15} = -1$.

(B) Let $S = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.
Note that $c+a \omega+b \omega^2 = \omega^2(c \omega + a \omega^2 + b) = \omega^2(b+c \omega+a \omega^2)$.
Thus,the expression becomes $\frac{a+b \omega+c \omega^2}{\omega^2(b+c \omega+a \omega^2)} + \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.
Factoring out $\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$,we get $\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} (\frac{1}{\omega^2} + 1) = \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} (\omega + 1)$.
Since $1+\omega+\omega^2 = 0$,we have $1+\omega = -\omega^2$.
However,evaluating the expression for specific values like $a=1, b=0, c=0$ gives $\frac{1}{\omega^2} + \frac{1}{\omega} = \omega + \omega^2 = -1$.
Thus,the value is $-1$.

(B) Given $a = \cos \left(\frac{8 \pi}{11}\right) + i \sin \left(\frac{8 \pi}{11}\right) = e^{i \frac{8 \pi}{11}}$.
Since $a^{11} = e^{i 8 \pi} = 1$,$a$ is an $11^{th}$ root of unity.
The sum of all $11^{th}$ roots of unity is $1 + a + a^2 + \dots + a^{10} = 0$.
Thus,$a + a^2 + a^3 + a^4 + a^5 + a^6 + a^7 + a^8 + a^9 + a^{10} = -1$.
Since $a^{11} = 1$,we have $a^{10} = \bar{a}$,$a^9 = \bar{a^2}$,$a^8 = \bar{a^3}$,$a^7 = \bar{a^4}$,and $a^6 = \bar{a^5}$.
Substituting these into the sum,we get $(a + \bar{a}) + (a^2 + \bar{a^2}) + (a^3 + \bar{a^3}) + (a^4 + \bar{a^4}) + (a^5 + \bar{a^5}) = -1$.
Using the property $z + \bar{z} = 2 \operatorname{Re}(z)$,we have $2 \operatorname{Re}(a) + 2 \operatorname{Re}(a^2) + 2 \operatorname{Re}(a^3) + 2 \operatorname{Re}(a^4) + 2 \operatorname{Re}(a^5) = -1$.
Therefore,$2 \operatorname{Re}(a + a^2 + a^3 + a^4 + a^5) = -1$,which implies $\operatorname{Re}(a + a^2 + a^3 + a^4 + a^5) = -\frac{1}{2}$.

(C) Let $z=x+iy$. Given $|z|^2+1=|z^2-1|$.
Substituting $z=x+iy$,we get $x^2+y^2+1 = |(x+iy)^2-1| = |x^2-y^2-1+2ixy|$.
Squaring both sides: $(x^2+y^2+1)^2 = (x^2-y^2-1)^2 + (2xy)^2$.
$(x^2+y^2+1)^2 - (x^2-y^2-1)^2 = 4x^2y^2$.
Using $a^2-b^2 = (a+b)(a-b)$:
$[(x^2+y^2+1) + (x^2-y^2-1)] \times [(x^2+y^2+1) - (x^2-y^2-1)] = 4x^2y^2$.
$(2x^2)(2y^2+2) = 4x^2y^2$.
$4x^2y^2 + 4x^2 = 4x^2y^2$.
$4x^2 = 0 \implies x=0$.
The locus $x=0$ represents the imaginary axis.

(A) Let $Z = x + iy$.
Then $\frac{Z-1}{Z+1} = \frac{(x-1) + iy}{(x+1) + iy}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(x+1) - iy$,we get:
$\frac{Z-1}{Z+1} = \frac{((x-1) + iy)((x+1) - iy)}{(x+1)^2 + y^2} = \frac{(x^2 - 1 + y^2) + i(y(x+1) - y(x-1))}{(x+1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + 2iy}{(x+1)^2 + y^2}$.
Given $\operatorname{Arg}\left(\frac{Z-1}{Z+1}\right) = \frac{\pi}{4}$,we have $\tan\left(\frac{\pi}{4}\right) = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{2y}{x^2 + y^2 - 1}$.
Since $\tan\left(\frac{\pi}{4}\right) = 1$,we have $1 = \frac{2y}{x^2 + y^2 - 1}$.
This simplifies to $x^2 + y^2 - 1 = 2y$,or $x^2 + y^2 - 2y - 1 = 0$.

(C) Given points are $A(3-i)$ and $B(2+i)$.
In the Argand plane,these correspond to coordinates $A(3, -1)$ and $B(2, 1)$.
The given equation is $|z-3+i|=|z-2-i|$.
This can be rewritten as $|z-(3-i)|=|z-(2+i)|$.
Let $P$ be the point representing the complex number $z$. Then the equation represents the set of points $P$ such that the distance from $P$ to $A$ is equal to the distance from $P$ to $B$,i.e.,$PA = PB$.
The locus of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of the line segment $AB$.

(B) $6$ men can be seated in a row in $6!$ ways. Now,in the $7$ gaps created by them,$6$ women can be arranged in $7_{P_6}$ ways.
$\therefore x = 6! \times 7_{P_6} = 6! \times 7!$
$6$ men can be seated in a circle in $(6-1)! = 5!$ ways. Now,in the $6$ gaps created by them,$6$ women can be arranged in $6!$ ways.
$\therefore y = 5! \times 6!$
Now,$x: y = (6! \times 7!) : (5! \times 6!) = 7! : 5!$
$\Rightarrow x: y = (7 \times 6 \times 5!) : 5! = 42: 1$

(B) To form a four-digit even number using the digits $0, 3, 5, 4$ without repetition,the last digit must be $0$ or $4$.
Case $1$: The last digit is $0$. The remaining $3$ positions can be filled by $3, 5, 4$ in $3! = 6$ ways. The numbers are $3540, 5340, 3450, 4350, 5430, 4530$. Their sum is $3540 + 5340 + 3450 + 4350 + 5430 + 4530 = 26640$.
Case $2$: The last digit is $4$. The first digit cannot be $0$. The possible numbers are $3054, 3504, 5034, 5304$. Their sum is $3054 + 3504 + 5034 + 5304 = 16896$.
Total sum $= 26640 + 16896 = 43536$.

(C) The word $SARANAM$ contains $7$ letters: $A, A, A, S, R, N, M$. The distinct letters are $\{A, S, R, N, M\}$.
We need to form $5$-letter words. The possible cases are:
$(i)$ All $5$ letters are different (using $S, A, R, N, M$):
Number of words $= 5! = 120$.
(ii) $2$ letters are alike (specifically $A$) and $3$ are different:
We choose $3$ letters from the remaining $4$ distinct letters $\{S, R, N, M\}$ in $^4C_3$ ways. The number of arrangements is $\frac{5!}{2!} = 60$.
Total words $= ^4C_3 \times 60 = 4 \times 60 = 240$.
(iii) $3$ letters are alike (specifically $A$) and $2$ are different:
We choose $2$ letters from the remaining $4$ distinct letters $\{S, R, N, M\}$ in $^4C_2$ ways. The number of arrangements is $\frac{5!}{3!} = 20$.
Total words $= ^4C_2 \times 20 = 6 \times 20 = 120$.
Total number of words $= 120 + 240 + 120 = 480$.

(A) The digits are $\{2, 3, 4, 5, 6\}$. Arranging them in decreasing order: $6, 5, 4, 3, 2$.
Total permutations possible = $5! = 120$.
We want the $89^{th}$ number when arranged in decreasing order.
Numbers starting with $6$: $4! = 24$ numbers (Ranks $1$ to $24$).
Numbers starting with $5$: $4! = 24$ numbers (Ranks $25$ to $48$).
Numbers starting with $4$: $4! = 24$ numbers (Ranks $49$ to $72$).
Numbers starting with $36$: $3! = 6$ numbers (Ranks $73$ to $78$).
Numbers starting with $35$: $3! = 6$ numbers (Ranks $79$ to $84$).
Numbers starting with $346$: $2! = 2$ numbers (Ranks $85$ to $86$).
Numbers starting with $345$: $2! = 2$ numbers (Ranks $87$ to $88$).
The $89^{th}$ number starts with $34265$ (Rank $89$) and the $90^{th}$ is $34256$ (Rank $90$).
Thus,the $89^{th}$ number is $34265$.

(D) Consider the range of numbers from $20001$ to $59999$. The total number of integers in this range is $59999 - 20001 + 1 = 39999$.
We can group these numbers into sets of $10$ consecutive integers of the form $(a_1 a_2 a_3 a_4 0, a_1 a_2 a_3 a_4 1, \dots, a_1 a_2 a_3 a_4 9)$.
In any such set of $10$ consecutive integers,the sum of the digits of $9$ numbers will have the same parity as the sum of the digits of the $10^{th}$ number,except for the last digit which cycles through $0$ to $9$.
Specifically,in any set of $10$ consecutive integers,exactly $5$ will have an even sum of digits and $5$ will have an odd sum of digits.
Since we have $39999$ numbers,we can form $3999$ full groups of $10$ (total $39990$ numbers) and there are $9$ remaining numbers ($59991$ to $59999$).
For the $39990$ numbers,exactly half have an even sum of digits: $39990 / 2 = 19995$.
For the remaining $9$ numbers ($59991$ to $59999$),the sum of the first four digits $(5+9+9+9 = 32)$ is even. Thus,the sum of all digits is even if the last digit is even $(0, 2, 4, 6, 8)$. In the range $59991$ to $59999$,the last digits are $1, 2, 3, 4, 5, 6, 7, 8, 9$. The even last digits are $2, 4, 6, 8$ (total $4$ numbers).
Total count $= 19995 + 4 = 19999$.

(B) For the quadratic equation $4^{\sec^2 \alpha} x^2 + 2x + (\beta^2 - \beta + \frac{1}{2}) = 0$ to have real roots,the discriminant $D$ must be $\geq 0$.
$D = (2)^2 - 4(4^{\sec^2 \alpha})(\beta^2 - \beta + \frac{1}{2}) \geq 0$
$4 - 4(4^{\sec^2 \alpha})(\beta^2 - \beta + \frac{1}{2}) \geq 0$
$4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \leq 1$
Since $\sec^2 \alpha \geq 1$,we have $4^{\sec^2 \alpha} \geq 4^1 = 4$.
Also,$\beta^2 - \beta + \frac{1}{2} = (\beta - \frac{1}{2})^2 + \frac{1}{4} \geq \frac{1}{4}$.
Multiplying these,$4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \geq 4 \times \frac{1}{4} = 1$.
For the inequality $4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \leq 1$ to hold,both expressions must be at their minimum values:
$4^{\sec^2 \alpha} = 4$ $\Rightarrow \sec^2 \alpha = 1$ $\Rightarrow \cos^2 \alpha = 1$.
$(\beta - \frac{1}{2})^2 = 0 \Rightarrow \beta = \frac{1}{2}$.
Therefore,$\cos^2 \alpha + \cos^{-1} \beta = 1 + \cos^{-1}(\frac{1}{2}) = 1 + \frac{\pi}{3}$.

(B) Given equations are:
$(1) \ x+y+z=3$
$(2) \ 2x+2y-z=3$
$(3) \ x+y-z=1$
Subtracting equation $(3)$ from equation $(1)$,we get:
$(x+y+z) - (x+y-z) = 3 - 1$
$2z = 2 \implies z = 1$
Substituting $z=1$ in equation $(1)$:
$x+y+1 = 3 \implies x+y = 2$
Since $(x_0, y_0, z_0)$ is a solution,we have $x_0+y_0 = 2$ and $z_0 = 1$.
Now,calculate $2x_0+2y_0+z_0$:
$2x_0+2y_0+z_0 = 2(x_0+y_0) + z_0$
$= 2(2) + 1 = 4 + 1 = 5$

(C) Given the function $f(x) = \begin{cases} 1+\frac{2x}{k}, & 0 \leq x < 1 \\ kx, & 1 \leq x < 2 \end{cases}$ where $k>0$.
We are given that $\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x)$.
Calculating the left-hand limit:
$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (1+\frac{2x}{k}) = 1+\frac{2}{k}$.
Calculating the right-hand limit:
$\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (kx) = k$.
Equating the two limits:
$1+\frac{2}{k} = k$
Multiplying by $k$ (since $k>0$):
$k+2 = k^2$
Rearranging the terms:
$k^2 - k - 2 = 0$
Factoring the quadratic equation:
$(k-2)(k+1) = 0$.
Since $k>0$,we have $k=2$.
Therefore,$k^2 = 2^2 = 4$.

(C) Given the function $f(x) = \begin{cases} 1 + 2x^k, & 0 \le x < 1 \\ kx, & 1 \le x < 2 \end{cases}$.
For the limit to exist at $x = 1$,we must have $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$.
Calculating the left-hand limit: $\lim_{x \to 1^-} (1 + 2x^k) = 1 + 2(1)^k = 1 + 2 = 3$.
Calculating the right-hand limit: $\lim_{x \to 1^+} (kx) = k(1) = k$.
Equating the two limits: $3 = k$.
Since $k = 3$,the value of $k^2$ is $3^2 = 9$.

(D) The given expression is $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r \sqrt{r}}{n^{5/2}}$.
This can be rewritten as $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^{3/2}}{n^{5/2}} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \left( \frac{r}{n} \right)^{3/2} \cdot \frac{1}{n}$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \sum_{r=1}^n f\left( \frac{r}{n} \right) \frac{1}{n} = \int_0^1 f(x) dx$.
Here,$f(x) = x^{3/2}$.
Thus,the limit is $\int_0^1 x^{3/2} dx = \left[ \frac{x^{5/2}}{5/2} \right]_0^1 = \frac{2}{5} (1 - 0) = \frac{2}{5}$.

(C) Let $A = \lim _{n \rightarrow \infty}\left[\prod_{r=1}^{n} \left(1+\frac{r^2}{n^2}\right)\right]^{\frac{1}{n}}$.
Taking the natural logarithm on both sides:
$\log A = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left(1+\frac{r^2}{n^2}\right)$.
Using the definition of the definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$,we get:
$\log A = \int_0^1 \log(1+x^2) dx$.
Integrating by parts,let $u = \log(1+x^2)$ and $dv = dx$. Then $du = \frac{2x}{1+x^2} dx$ and $v = x$.
$\int_0^1 \log(1+x^2) dx = [x \log(1+x^2)]_0^1 - \int_0^1 \frac{2x^2}{1+x^2} dx$.
$= [1 \cdot \log(2) - 0] - 2 \int_0^1 \left(\frac{1+x^2-1}{1+x^2}\right) dx$.
$= \log 2 - 2 \int_0^1 \left(1 - \frac{1}{1+x^2}\right) dx$.
$= \log 2 - 2 [x - \tan^{-1}(x)]_0^1$.
$= \log 2 - 2 [(1 - \tan^{-1}(1)) - (0 - 0)]$.
$= \log 2 - 2(1 - \frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2} = \log 2 + \frac{\pi-4}{2}$.
Since $\log A = \log 2 + \frac{\pi-4}{2}$,we have $A = e^{\log 2 + \frac{\pi-4}{2}} = 2 e^{\frac{\pi-4}{2}}$.

(B) The given determinant is $\Delta = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$.
Since $a, b, c$ are sides of a triangle,$a+b+c \neq 0$.
Thus,$a^2+b^2+c^2-ab-bc-ca = 0$,which implies $a=b=c$.
Since $a=b=c$,the triangle is equilateral,so $A=B=C=60^\circ$.
Then $\cos A \cos B + \cos B \cos C + \cos C \cos A = \cos 60^\circ \cos 60^\circ + \cos 60^\circ \cos 60^\circ + \cos 60^\circ \cos 60^\circ$.
$= \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.

(C) Given,$A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$.
First,we calculate the determinant $|A|$:
$|A| = 3((-3)(1) - (4)(-1)) - (-3)((2)(1) - (4)(0)) + 4((2)(-1) - (-3)(0))$
$|A| = 3(-3 + 4) + 3(2 - 0) + 4(-2 - 0) = 3(1) + 3(2) + 4(-2) = 3 + 6 - 8 = 1$.
Since $|A| = 1 \neq 0$,$A^{-1}$ exists.
Next,we find the adjoint of $A$ by calculating the cofactor matrix and transposing it.
The cofactor matrix $C$ is:
$C_{11} = (-1)^{1+1} \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = 1, C_{12} = (-1)^{1+2} \begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = -2, C_{13} = (-1)^{1+3} \begin{vmatrix} 2 & -3 \\ 0 & -1 \end{vmatrix} = -2$
$C_{21} = (-1)^{2+1} \begin{vmatrix} -3 & 4 \\ -1 & 1 \end{vmatrix} = -1, C_{22} = (-1)^{2+2} \begin{vmatrix} 3 & 4 \\ 0 & 1 \end{vmatrix} = 3, C_{23} = (-1)^{2+3} \begin{vmatrix} 3 & -3 \\ 0 & -1 \end{vmatrix} = 3$
$C_{31} = (-1)^{3+1} \begin{vmatrix} -3 & 4 \\ -3 & 4 \end{vmatrix} = 0, C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & 4 \\ 2 & 4 \end{vmatrix} = -4, C_{33} = (-1)^{3+3} \begin{vmatrix} 3 & -3 \\ 2 & -3 \end{vmatrix} = -3$
Thus,$\text{adj}(A) = C^T = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Since $|A| = 1$,$A^{-1} = \frac{\text{adj}(A)}{|A|} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Now,calculate $A^2 = A \cdot A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$.
Then,$A^3 = A^2 \cdot A = \begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{bmatrix}$.
Comparing $A^{-1}$ and $A^3$,we see that $A^{-1} = A^3$.

(A) The matrix is $M = \begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{bmatrix}$.
For the matrix to be non-singular,its determinant must be non-zero: $\det(M) \neq 0$.
Calculating the determinant: $\det(M) = 1(1 - \omega c) - a(\omega - \omega^2 c) + b(\omega^2 - \omega^2) = 1 - \omega c - a\omega + a\omega^2 c = 1 - \omega(a + c) + a\omega^2 c$.
Given $a, c \in \{\omega, \omega^2\}$,we test the combinations:
$1$. If $a = \omega, c = \omega$: $\det(M) = 1 - \omega(2\omega) + \omega(\omega^2)(\omega) = 1 - 2\omega^2 + \omega^4 = 1 - 2\omega^2 + \omega = 1 - 2\omega^2 + (-1 - \omega^2) = -3\omega^2 \neq 0$.
$2$. If $a = \omega^2, c = \omega^2$: $\det(M) = 1 - \omega(2\omega^2) + \omega^2(\omega^2)(\omega^2) = 1 - 2\omega^3 + \omega^6 = 1 - 2(1) + 1 = 0$.
$3$. If $a = \omega, c = \omega^2$: $\det(M) = 1 - \omega(\omega + \omega^2) + \omega(\omega^2)(\omega^2) = 1 - \omega(-1) + \omega^5 = 1 + \omega + \omega^2 = 0$.
$4$. If $a = \omega^2, c = \omega$: $\det(M) = 1 - \omega(\omega^2 + \omega) + \omega^2(\omega^2)(\omega) = 1 - \omega(-1) + \omega^5 = 1 + \omega + \omega^2 = 0$.
Since $b$ can be either $\omega$ or $\omega^2$ in the case where $\det(M) \neq 0$,and only the case $(a, c) = (\omega, \omega)$ yields a non-zero determinant,there are $2$ possible values for $b$ (namely $\omega$ and $\omega^2$).
Thus,the number of elements in $S$ is $2$.

(B) Given the equation $A^2 X = cAX$,we can rewrite it as $(A^2 - cA)X = 0$,which implies $A(A - cI)X = 0$. Since $X$ is a non-zero matrix,this implies that $c$ must be an eigenvalue of matrix $A$ or $A$ must be singular. More directly,the equation $AX = \lambda X$ defines the eigenvalues of $A$. If we let $AX = \lambda X$,then $A^2 X = A(AX) = A(\lambda X) = \lambda(AX) = \lambda^2 X$. Substituting this into the given equation $A^2 X = cAX$,we get $\lambda^2 X = c \lambda X$,which implies $\lambda^2 = c \lambda$ for any non-zero $X$. Thus,$\lambda(\lambda - c) = 0$. This means $c$ must be an eigenvalue of $A$.
To find the eigenvalues of $A$,we solve the characteristic equation $|A - \lambda I| = 0$:
$\begin{vmatrix} 5-\lambda & -3 & 0 \\ -3 & 5-\lambda & 0 \\ 0 & 0 & 2-\lambda \end{vmatrix} = 0$
$(2-\lambda) [(5-\lambda)^2 - (-3)^2] = 0$
$(2-\lambda) [\lambda^2 - 10\lambda + 25 - 9] = 0$
$(2-\lambda) (\lambda^2 - 10\lambda + 16) = 0$
$(2-\lambda) (\lambda - 8)(\lambda - 2) = 0$
The eigenvalues are $\lambda = 2, 2, 8$.
Since $c$ is an eigenvalue of $A$,the distinct values for $c$ are $2$ and $8$.
Therefore,there are $2$ distinct values of $c$.

(D) Given the matrix $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
First,calculate $A^2$: $A^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$.
This implies $A^2 + I = 0$,or $A^2 = -I$.
From this,$A^3 = A^2 \cdot A = -I \cdot A = -A$.
Also,$A^4 = (A^2)^2 = (-I)^2 = I$.
Now,let us check each option:
Option $A$: $A^3 - I = -A - I$ and $A(A - I) = A^2 - A = -I - A$. Thus,$A^3 - I = A(A - I)$ is correct.
Option $B$: $A^3 + I = -A + I$ and $A(A^3 - I) = A(-A - I) = -A^2 - A = I - A$. Thus,$A^3 + I = A(A^3 - I)$ is correct.
Option $C$: $A^4 - I = I - I = 0$ and $A^2 + I = -I + I = 0$. Thus,$A^4 - I = A^2 + I$ is correct.
Option $D$: $A^2 + I = -I + I = 0$ and $A(A^2 - I) = A(-I - I) = A(-2I) = -2A$. Since $0 \neq -2A$,this option is incorrect.

(D) Given,$A = \begin{bmatrix} k/2 & 0 & 0 \\ 0 & l/3 & 0 \\ 0 & 0 & m/4 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{bmatrix}$.
Since $A A^{-1} = I$,we have:
$\begin{bmatrix} k/2 & 0 & 0 \\ 0 & l/3 & 0 \\ 0 & 0 & m/4 \end{bmatrix} \begin{bmatrix} 1/2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1/4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Multiplying the diagonal elements,we get:
$\begin{bmatrix} k/4 & 0 & 0 \\ 0 & l/9 & 0 \\ 0 & 0 & m/16 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Comparing the corresponding elements:
$k/4 = 1 \implies k = 4$.
$l/9 = 1 \implies l = 9$.
$m/16 = 1 \implies m = 16$.
Therefore,$k+l+m = 4 + 9 + 16 = 29$.

(A) For a square matrix $A$ of order $n=3$,the property of the adjoint matrix is $|\operatorname{Adj} A| = |A|^{n-1}$.
Substituting $n=3$,we get $|\operatorname{Adj} A| = |A|^{3-1} = |A|^2$.
If $|A|=0$,then $|\operatorname{Adj} A| = 0^2 = 0$. Thus,statement $I$ is true.
For statement $II$,we know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix.
Taking the determinant on both sides,$|A \cdot A^{-1}| = |I|$.
Using the property $|AB| = |A||B|$,we get $|A| \cdot |A^{-1}| = 1$.
Since $|A| \neq 0$,we can divide by $|A|$,resulting in $|A^{-1}| = \frac{1}{|A|} = |A|^{-1}$. Thus,statement $II$ is true.
Therefore,both statements $I$ and $II$ are correct.

(A) We know that for any invertible matrix $A$,$\operatorname{Adj}(A) = |A| A^{-1}$.
Given that $P$ and $Q$ are invertible matrices,their product $QP$ is also an invertible matrix.
Therefore,$\operatorname{Adj}(QP) = |QP| (QP)^{-1}$.
Using the properties of determinants,$|QP| = |Q||P| = |P||Q|$.
Using the property of inverse matrices,$(QP)^{-1} = P^{-1} Q^{-1}$.
Thus,$\operatorname{Adj}(QP) = |P||Q| P^{-1} Q^{-1}$.
However,note that $\operatorname{Adj}(Q) = |Q| Q^{-1}$ and $\operatorname{Adj}(P) = |P| P^{-1}$.
Multiplying these,we get $\operatorname{Adj}(Q) \operatorname{Adj}(P) = (|Q| Q^{-1}) (|P| P^{-1}) = |Q||P| Q^{-1} P^{-1}$.
Since $\operatorname{Adj}(QP) = |QP| (QP)^{-1} = |Q||P| P^{-1} Q^{-1}$,and generally $P^{-1} Q^{-1} \neq Q^{-1} P^{-1}$,the standard identity is $\operatorname{Adj}(QP) = \operatorname{Adj}(P) \operatorname{Adj}(Q)$.

(A) Given the matrix equation $A^2+A+2I=0$.
Taking the determinant on both sides:
$A(A+I) = -2I$.
Taking the determinant of both sides:
$|A(A+I)| = |-2I|$.
Since $A$ is of order $3$,$|kI| = k^3|I| = k^3$.
$|A||A+I| = (-2)^3 |I| = -8(1) = -8$.
Since $|A||A+I| = -8$,it implies that $|A| \neq 0$ and $|A+I| \neq 0$.
Since $|A| \neq 0$,$A$ is a non-singular matrix.
Also,the determinant of a skew-symmetric matrix of odd order is always $0$.
Since $|A| \neq 0$,$A$ cannot be a skew-symmetric matrix.

(C) matrix has no inverse if its determinant is $0$.
Let $A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7 - t & -6 \end{bmatrix}$.
Setting $|A| = 0$:
$|A| = 1[5(-6) - t(7 - t)] - 3[2(-6) - 4t] + 2[2(7 - t) - 4(5)] = 0$
$|A| = 1[-30 - 7t + t^2] - 3[-12 - 4t] + 2[14 - 2t - 20] = 0$
$|A| = t^2 - 7t - 30 + 36 + 12t + 28 - 4t - 40 = 0$
$|A| = t^2 + t - 6 = 0$
Factoring the quadratic equation:
$(t + 3)(t - 2) = 0$
Thus,$t = -3$ or $t = 2$.

(B) Given matrix $A = \begin{bmatrix} 1 & -2 & 3 & -4 \\ 2 & 9 & 4 & 5 \\ 4 & 5 & 10 & -3 \\ 1 & 11 & -1 & 9 \end{bmatrix}$
Apply row operations to reduce the matrix to row-echelon form:
$R_2 \rightarrow R_2 - 2R_1$ gives $\begin{bmatrix} 1 & -2 & 3 & -4 \\ 0 & 13 & -2 & 13 \\ 4 & 5 & 10 & -3 \\ 1 & 11 & -1 & 9 \end{bmatrix}$
$R_3 \rightarrow R_3 - 4R_1$ gives $\begin{bmatrix} 1 & -2 & 3 & -4 \\ 0 & 13 & -2 & 13 \\ 0 & 13 & -2 & 13 \\ 1 & 11 & -1 & 9 \end{bmatrix}$
$R_4 \rightarrow R_4 - R_1$ gives $\begin{bmatrix} 1 & -2 & 3 & -4 \\ 0 & 13 & -2 & 13 \\ 0 & 13 & -2 & 13 \\ 0 & 13 & -4 & 13 \end{bmatrix}$
$R_3 \rightarrow R_3 - R_2$ gives $\begin{bmatrix} 1 & -2 & 3 & -4 \\ 0 & 13 & -2 & 13 \\ 0 & 0 & 0 & 0 \\ 0 & 13 & -4 & 13 \end{bmatrix}$
$R_4 \rightarrow R_4 - R_2$ gives $\begin{bmatrix} 1 & -2 & 3 & -4 \\ 0 & 13 & -2 & 13 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -2 & 0 \end{bmatrix}$
The number of non-zero rows in the row-echelon form is $3$.
Therefore,the rank of matrix $A$ is $3$.

(A) The given system of equations is:
$x+2y+z=1$
$x+3y+4z=k$
$x+5y+10z=k^2$
For the system to be consistent,the determinant of the coefficient matrix $D$ must be $0$ if the system has infinitely many solutions or a unique solution. Let us calculate $D$:
$D = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 3 & 4 \\ 1 & 5 & 10 \end{vmatrix} = 1(30-20) - 2(10-4) + 1(5-3) = 10 - 12 + 2 = 0$.
Since $D=0$,for the system to be consistent,we must have $D_1 = D_2 = D_3 = 0$. Let us calculate $D_1$:
$D_1 = \begin{vmatrix} 1 & 2 & 1 \\ k & 3 & 4 \\ k^2 & 5 & 10 \end{vmatrix} = 1(30-20) - 2(10k-4k^2) + 1(5k-3k^2) = 0$.
$10 - 20k + 8k^2 + 5k - 3k^2 = 0$.
$5k^2 - 15k + 10 = 0$.
Dividing by $5$,we get $k^2 - 3k + 2 = 0$.
$(k-1)(k-2) = 0$.
Thus,the real values of $k$ are $k=1$ and $k=2$. Let $A=1$ and $B=2$.
Therefore,$A+B = 1+2 = 3$.

(D) Let $\Delta = \left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|$.
Taking $5$ common from $R_1$,$7$ common from $R_2$,and $9$ common from $R_3$:
$\Delta = (5 \cdot 7 \cdot 9) \left|\begin{array}{lll}25 & 1 & 5 \\ 49 & 1 & 7 \\ 81 & 1 & 9\end{array}\right|$.
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = 315 \left|\begin{array}{lll}25 & 1 & 5 \\ 24 & 0 & 2 \\ 56 & 0 & 4\end{array}\right|$.
Expanding along the second column:
$\Delta = 315 \cdot (-1) \cdot \left|\begin{array}{ll}24 & 2 \\ 56 & 4\end{array}\right|$.
$\Delta = -315 \cdot (24 \cdot 4 - 56 \cdot 2) = -315 \cdot (96 - 112) = -315 \cdot (-16) = 5040$.
Since $7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040$,the result is $7!$.

(D) Given system of equations:
$x - 2y + 3z = 5$
$2x - 2y + z = 0$
$-x + 2y - 3z = 6$
Let $\Delta$ be the determinant of the coefficient matrix:
$\Delta = \begin{vmatrix} 1 & -2 & 3 \\ 2 & -2 & 1 \\ -1 & 2 & -3 \end{vmatrix}$
$= 1((-2)(-3) - (1)(2)) - (-2)((2)(-3) - (1)(-1)) + 3((2)(2) - (-2)(-1))$
$= 1(6 - 2) + 2(-6 + 1) + 3(4 - 2)$
$= 1(4) + 2(-5) + 3(2) = 4 - 10 + 6 = 0$
Since $\Delta = 0$,the system either has no solution or infinitely many solutions.
Now,calculate $\Delta_1$ by replacing the first column with the constants:
$\Delta_1 = \begin{vmatrix} 5 & -2 & 3 \\ 0 & -2 & 1 \\ 6 & 2 & -3 \end{vmatrix}$
$= 5((-2)(-3) - (1)(2)) - (-2)((0)(-3) - (1)(6)) + 3((0)(2) - (-2)(6))$
$= 5(6 - 2) + 2(0 - 6) + 3(0 + 12)$
$= 5(4) + 2(-6) + 3(12) = 20 - 12 + 36 = 44$
Since $\Delta = 0$ and $\Delta_1 \neq 0$,the system of equations has no solution.

(B) Given system of equations:
$x+y+z=3$ ... $(i)$
$2x+2y-z=3$ ... $(ii)$
$x+y-z=1$ ... $(iii)$
Adding equations $(i)$ and $(iii)$,we get:
$(x+y+z) + (x+y-z) = 3+1$
$2x+2y = 4$
$x+y = 2$
Substituting $x+y=2$ into equation $(i)$:
$2+z = 3$
$z = 1$
Now,we need to find the value of $2x_0+2y_0+z_0$ for a solution $(x_0, y_0, z_0)$:
$2x_0+2y_0+z_0 = 2(x_0+y_0) + z_0$
Since $x_0+y_0=2$ and $z_0=1$:
$= 2(2) + 1$
$= 4+1 = 5$

(B) The given system of equations is:
$x - ay - az = 0$
$-bx + y - bz = 0$
$-cx - cy + z = 0$
For a non-trivial solution,the determinant of the coefficient matrix must be zero:
$\Delta = \begin{vmatrix} 1 & -a & -a \\ -b & 1 & -b \\ -c & -c & 1 \end{vmatrix} = 0$
Expanding the determinant:
$1(1 - bc) + a(-b - bc) - a(bc + c) = 0$
$1 - bc - ab - abc - abc - ac = 0$
$1 - (ab + bc + ca) - 2abc = 0$
Divide by $(a+1)(b+1)(c+1)$:
Note that $(a+1)(b+1)(c+1) = abc + ab + bc + ca + a + b + c + 1$.
Alternatively,divide the determinant equation by $(a+1)(b+1)(c+1)$ or manipulate the system:
From $x = a(y+z)$,we get $x = a(x+y+z) - ax$,so $x(1+a) = a(x+y+z)$.
Thus,$\frac{x}{x+y+z} = \frac{a}{a+1}$.
Similarly,$\frac{y}{x+y+z} = \frac{b}{b+1}$ and $\frac{z}{x+y+z} = \frac{c}{c+1}$.
Adding these three equations:
$\frac{x+y+z}{x+y+z} = \frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1}$.
Since the solution is non-trivial,$x+y+z \neq 0$,so:
$1 = \frac{a}{a+1} + \frac{b}{b+1} + \frac{c}{c+1}$.

(D) For a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$.
$\left|\begin{array}{ccc} \sin 3 \theta & -1 & 1 \\ \cos 2 \theta & 4 & 3 \\ 2 & 7 & 7 \end{array}\right| = 0$
Expanding along the first row:
$\sin 3 \theta (28 - 21) - (-1) (7 \cos 2 \theta - 6) + 1 (7 \cos 2 \theta - 8) = 0$
$7 \sin 3 \theta + 7 \cos 2 \theta - 6 + 7 \cos 2 \theta - 8 = 0$
$7 \sin 3 \theta + 14 \cos 2 \theta - 14 = 0$
Dividing by $7$:
$\sin 3 \theta + 2 \cos 2 \theta - 2 = 0$
Using $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$ and $\cos 2 \theta = 1 - 2 \sin^2 \theta$:
$(3 \sin \theta - 4 \sin^3 \theta) + 2(1 - 2 \sin^2 \theta) - 2 = 0$
$3 \sin \theta - 4 \sin^3 \theta + 2 - 4 \sin^2 \theta - 2 = 0$
$-4 \sin^3 \theta - 4 \sin^2 \theta + 3 \sin \theta = 0$
$-\sin \theta (4 \sin^2 \theta + 4 \sin \theta - 3) = 0$
$-\sin \theta (2 \sin \theta - 1)(2 \sin \theta + 3) = 0$
Since $2 \sin \theta + 3 \neq 0$ for any real $\theta$,we have:
$\sin \theta = 0 \Rightarrow \theta = n \pi$
$\sin \theta = \frac{1}{2} \Rightarrow \theta = n \pi + (-1)^n \frac{\pi}{6}$
Thus,the set of values is $\theta = n \pi$ or $\theta = n \pi + (-1)^n \frac{\pi}{6}$.

(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.
Let $D$ be the determinant of the coefficient matrix:
$D = \begin{vmatrix} k & k+1 & k-1 \\ k-1 & k+2 & k \\ k+1 & k & k+2 \end{vmatrix} = 0$.
Applying column operations $C_1 \to C_1 + C_2 + C_3$:
$D = \begin{vmatrix} 3k & k+1 & k-1 \\ 3k+1 & k+2 & k \\ 3k+3 & k & k+2 \end{vmatrix} = 0$.
This simplifies to $D = (3k+1) \begin{vmatrix} 1 & k+1 & k-1 \\ 1 & k+2 & k \\ 1 & k & k+2 \end{vmatrix} = 0$.
Subtracting $R_1$ from $R_2$ and $R_3$:
$D = (3k+1) \begin{vmatrix} 1 & k+1 & k-1 \\ 0 & 1 & 1 \\ 0 & -1 & 3 \end{vmatrix} = 0$.
Expanding the determinant:
$(3k+1) [1(3 - (-1))] = 0$.
$(3k+1)(4) = 0$.
$3k+1 = 0 \implies k = -\frac{1}{3}$.
Wait,re-evaluating the determinant:
$D = k((k+2)(k+2) - k^2) - (k+1)((k-1)(k+2) - k(k+1)) + (k-1)(k(k-1) - (k+2)(k+1)) = 0$.
$D = k(k^2+4k+4-k^2) - (k+1)(k^2+k-2-k^2-k) + (k-1)(k^2-k-k^2-3k-2) = 0$.
$D = k(4k+4) - (k+1)(-2) + (k-1)(-4k-2) = 0$.
$4k^2+4k + 2k+2 - 4k^2-2k+4k+2 = 0$.
$8k + 4 = 0 \implies 8k = -4 \implies k = -\frac{1}{2}$.
Thus,the only possible value for $k$ is $-\frac{1}{2}$,and the sum of all possible values is $-\frac{1}{2}$.

(B) Let $f(y) = y^2-4y+6 = (y-2)^2+2$. Since $(y-2)^2 \geq 0$,we have $f(y) \geq 2$ for all $y \in R$.
For statement $I$: The identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ holds if and only if $x \in [-1, 1]$. Here,$f(y) \geq 2$,so $f(y)$ is never in $[-1, 1]$. Thus,statement $I$ is false.
For statement $II$: The identity $\sec^{-1}(x) + \operatorname{cosec}^{-1}(x) = \frac{\pi}{2}$ holds if and only if $|x| \geq 1$. Since $f(y) \geq 2$,the condition $|f(y)| \geq 1$ is satisfied for all $y \in R$. Thus,statement $II$ is true.

(D) We use the properties $\cot ^{-1}(-x) = \pi - \cot ^{-1} x$ and $\tan ^{-1}(-x) = -\tan ^{-1} x$ and $\cot ^{-1} x = \tan ^{-1} \frac{1}{x}$ for $x > 0$.
Given expression: $E = \tan ^{-1} 2 + \cot ^{-1}(-3) + \cot ^{-1} \frac{1}{3} + \tan ^{-1}\left(-\frac{1}{2}\right)$
Using the properties:
$E = \tan ^{-1} 2 + (\pi - \cot ^{-1} 3) + \tan ^{-1} 3 - \tan ^{-1} \frac{1}{2}$
Since $\cot ^{-1} 3 = \tan ^{-1} \frac{1}{3}$,we have:
$E = \pi + (\tan ^{-1} 2 - \tan ^{-1} \frac{1}{2}) + (\tan ^{-1} 3 - \tan ^{-1} \frac{1}{3})$
Using $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \frac{x-y}{1+xy}$:
$E = \pi + \tan ^{-1} \left( \frac{2 - 1/2}{1 + 2(1/2)} \right) + \tan ^{-1} \left( \frac{3 - 1/3}{1 + 3(1/3)} \right)$
$E = \pi + \tan ^{-1} \left( \frac{3/2}{2} \right) + \tan ^{-1} \left( \frac{8/3}{2} \right)$
$E = \pi + \tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{4}{3}$
Since $\tan ^{-1} \frac{4}{3} = \cot ^{-1} \frac{3}{4}$:
$E = \pi + (\tan ^{-1} \frac{3}{4} + \cot ^{-1} \frac{3}{4})$
Using $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$:
$E = \pi + \frac{\pi}{2} = \frac{3 \pi}{2}$

(A) Given: $\cos^{-1} \left( \frac{x}{2} \right) + \cos^{-1} \left( \frac{y}{3} \right) = \theta$
Using the identity $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$,we get:
$\cos^{-1} \left\{ \left( \frac{x}{2} \right) \left( \frac{y}{3} \right) - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{9}} \right\} = \theta$
$\Rightarrow \frac{xy}{6} - \sqrt{\frac{4-x^2}{4}} \sqrt{\frac{9-y^2}{9}} = \cos \theta$
$\Rightarrow \frac{xy}{6} - \frac{\sqrt{4-x^2} \sqrt{9-y^2}}{6} = \cos \theta$
$\Rightarrow xy - \sqrt{4-x^2} \sqrt{9-y^2} = 6 \cos \theta$
$\Rightarrow xy - 6 \cos \theta = \sqrt{4-x^2} \sqrt{9-y^2}$
Squaring both sides:
$(xy - 6 \cos \theta)^2 = (4-x^2)(9-y^2)$
$x^2 y^2 - 12xy \cos \theta + 36 \cos^2 \theta = 36 - 4y^2 - 9x^2 + x^2 y^2$
Canceling $x^2 y^2$ from both sides:
$-12xy \cos \theta + 36 \cos^2 \theta = 36 - 9x^2 - 4y^2$
Rearranging the terms:
$9x^2 + 4y^2 - 12xy \cos \theta = 36 - 36 \cos^2 \theta$
$9x^2 + 4y^2 - 12xy \cos \theta = 36(1 - \cos^2 \theta)$
$9x^2 + 4y^2 - 12xy \cos \theta = 36 \sin^2 \theta$

(A) The general term of the series is $T_k = \operatorname{Tan}^{-1} \frac{1}{k^2+k+1}$.
We can rewrite the argument as $\frac{1}{1+k(k+1)}$.
Using the identity $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \frac{x-y}{1+xy}$,we have:
$T_k = \operatorname{Tan}^{-1} (k+1) - \operatorname{Tan}^{-1} k$.
Summing from $k=1$ to $n$:
$S_n = \sum_{k=1}^{n} (\operatorname{Tan}^{-1} (k+1) - \operatorname{Tan}^{-1} k) = \operatorname{Tan}^{-1} (n+1) - \operatorname{Tan}^{-1} (1)$.
Using the formula $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \frac{x-y}{1+xy}$:
$S_n = \operatorname{Tan}^{-1} \frac{(n+1)-1}{1+(n+1)(1)} = \operatorname{Tan}^{-1} \frac{n}{n+2}$.
Thus,$\theta = \frac{n}{n+2}$.

(D) The function is $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$.
Factorizing the expression inside the logarithm:
$ax^3 + ax^2 + bx^2 + bx + cx + c = ax^2(x+1) + bx(x+1) + c(x+1) = (ax^2 + bx + c)(x+1)$.
Since $b^2 = 4ac$,the quadratic $ax^2 + bx + c$ can be written as $a(x + \frac{b}{2a})^2$.
Thus,$f(x) = \log \{a(x + \frac{b}{2a})^2(x+1)\}$.
For $f(x)$ to be defined,the argument of the logarithm must be strictly positive:
$a(x + \frac{b}{2a})^2(x+1) > 0$.
Given $a > 0$,the term $a(x + \frac{b}{2a})^2$ is always $\geq 0$. It is $0$ when $x = -\frac{b}{2a}$.
Therefore,we require $x+1 > 0$ and $x \neq -\frac{b}{2a}$.
This implies $x > -1$ and $x \neq -\frac{b}{2a}$.
The domain $D$ is $(-1, \infty) - \{-\frac{b}{2a}\}$.
This is equivalent to $R - (\{-\frac{b}{2a}\} \cup (-\infty, -1])$.
Thus,the correct option is $D$.

(A) For the function $f(x) = \sqrt{\frac{4-x^2}{[x]+2}}$ to be defined,we must have $\frac{4-x^2}{[x]+2} \geq 0$ and $[x]+2 \neq 0$.
This implies $\frac{x^2-4}{[x]+2} \leq 0$ and $[x] \neq -2$.
Case $1$: $x^2-4 \geq 0$ and $[x]+2 < 0$.
$x^2 \geq 4 \Rightarrow x \in (-\infty, -2] \cup [2, \infty)$.
$[x] < -2 \Rightarrow x < -2$.
Intersection: $x \in (-\infty, -2)$.
Case $2$: $x^2-4 \leq 0$ and $[x]+2 > 0$.
$x^2 \leq 4 \Rightarrow x \in [-2, 2]$.
$[x] > -2 \Rightarrow x \geq -1$.
Intersection: $x \in [-1, 2]$.
Combining both cases,the domain is $(-\infty, -2) \cup [-1, 2]$.

(C) For the function $f(x) = \left(1 + \frac{1}{x}\right)^x$ to be defined as a real-valued function,the base must be positive for all real $x$ in the domain.
We require $1 + \frac{1}{x} > 0$.
This simplifies to $\frac{x + 1}{x} > 0$.
Using the sign scheme (Wavy curve method) for the critical points $x = -1$ and $x = 0$:
For $x < -1$,$\frac{x+1}{x} > 0$ (e.g.,$x = -2 \implies \frac{-1}{-2} = 0.5 > 0$).
For $-1 < x < 0$,$\frac{x+1}{x} < 0$ (e.g.,$x = -0.5 \implies \frac{0.5}{-0.5} = -1 < 0$).
For $x > 0$,$\frac{x+1}{x} > 0$ (e.g.,$x = 1 \implies \frac{2}{1} = 2 > 0$).
Thus,the domain is $x \in (-\infty, -1) \cup (0, \infty)$.

(B) For the function $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$ to be defined,both terms must be defined.
For $\sqrt{2x - 1}$ to be defined,we must have $2x - 1 \geq 0$,which implies $x \geq \frac{1}{2}$.
For $\cos^{-1}\left(\frac{2x - 1}{3}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{2x - 1}{3} \leq 1$.
Multiplying by $3$,we get $-3 \leq 2x - 1 \leq 3$.
Adding $1$ to all parts,we get $-2 \leq 2x \leq 4$.
Dividing by $2$,we get $-1 \leq x \leq 2$.
The domain is the intersection of $x \geq \frac{1}{2}$ and $-1 \leq x \leq 2$.
Thus,the domain is $\left[\frac{1}{2}, 2\right]$.

(D) We know that any real number $x$ can be written as $x = [x] + \{x\}$,where $[x]$ is the greatest integer part and $\{x\}$ is the fractional part,with $0 \leq \{x\} < 1$.
Multiplying by $2$,we get $2x = 2[x] + 2\{x\}$.
Taking the greatest integer on both sides,$[2x] = [2[x] + 2\{x\}] = 2[x] + [2\{x\}]$.
Now,the function is $f(x) = [2x] - 2[x] = (2[x] + [2\{x\}]) - 2[x] = [2\{x\}]$.
Since $0 \leq \{x\} < 1$,we have $0 \leq 2\{x\} < 2$.
Therefore,$[2\{x\}]$ can take values based on the interval of $\{x\}$:
If $0 \leq \{x\} < \frac{1}{2}$,then $0 \leq 2\{x\} < 1$,so $[2\{x\}] = 0$.
If $\frac{1}{2} \leq \{x\} < 1$,then $1 \leq 2\{x\} < 2$,so $[2\{x\}] = 1$.
Thus,the range of $f$ is $\{0, 1\}$.

(A) Let $f(x) = y$. Since $f: R \rightarrow [-1, 1]$ is surjective,$y$ takes all values in $[-1, 1]$.
Given $\sin \left(g(x) - \frac{\pi}{3}\right) = \frac{y}{2} \sqrt{4 - y^2}$.
Let $y = 2 \sin \theta$. Since $y \in [-1, 1]$,$\sin \theta \in \left[-\frac{1}{2}, \frac{1}{2}\right]$,so $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$.
Then $\frac{y}{2} \sqrt{4 - y^2} = \sin \theta \sqrt{4 - 4 \sin^2 \theta} = \sin \theta \sqrt{4 \cos^2 \theta} = 2 \sin \theta \cos \theta = \sin(2 \theta)$.
Thus,$\sin \left(g(x) - \frac{\pi}{3}\right) = \sin(2 \theta)$.
Since $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$,$2 \theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$.
Therefore,$g(x) - \frac{\pi}{3} = 2 \theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$.
$g(x) \in \left[-\frac{\pi}{3} + \frac{\pi}{3}, \frac{\pi}{3} + \frac{\pi}{3}\right] = \left[0, \frac{2 \pi}{3}\right]$.
Since $g$ is surjective,the codomain $A = \left[0, \frac{2 \pi}{3}\right]$.

(D) Given the function $f(x) = \frac{1}{x^2+2x+2}$.
First,analyze the denominator: $x^2+2x+2 = (x+1)^2+1$.
Since $(x+1)^2 \ge 0$ for all $x \in R$,it follows that $(x+1)^2+1 \ge 1$.
Thus,the range of the denominator is $[1, \infty)$.
Taking the reciprocal,we have $0 < \frac{1}{(x+1)^2+1} \le 1$.
Therefore,for the function to be surjective,the codomain $A$ must be equal to the range,which is $(0, 1]$.

(A) Let $y = \frac{x^2+2x+1}{x^2+2x+7}$.
$y(x^2+2x+7) = x^2+2x+1$
$yx^2 + 2yx + 7y = x^2 + 2x + 1$
$(y-1)x^2 + 2(y-1)x + (7y-1) = 0$.
If $y=1$,then $0x^2 + 0x + 6 = 0$,which is impossible. Thus,$y \neq 1$.
For $x$ to be real,the discriminant $D \geq 0$:
$D = [2(y-1)]^2 - 4(y-1)(7y-1) \geq 0$
$4(y-1)^2 - 4(y-1)(7y-1) \geq 0$
$(y-1)[(y-1) - (7y-1)] \geq 0$
$(y-1)(-6y) \geq 0$
$6y(y-1) \leq 0$
This implies $y \in [0, 1]$.
Since $y \neq 1$,the range is $[0, 1)$.

(C) Step $1$: Analyze $A$. Let $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. Check $f(-x) = \frac{-x}{e^{-x}-1} + \frac{-x}{2} + 4 = \frac{-x e^x}{1-e^x} - \frac{x}{2} + 4 = \frac{x e^x}{e^x-1} - \frac{x}{2} + 4$. Since $\frac{e^x}{e^x-1} = 1 + \frac{1}{e^x-1}$,$f(-x) = x(1 + \frac{1}{e^x-1}) - \frac{x}{2} + 4 = x + \frac{x}{e^x-1} - \frac{x}{2} + 4 = \frac{x}{e^x-1} + \frac{x}{2} + 4 = f(x)$. Thus,$A$ is an even function $(II)$.
Step $2$: Analyze $B$. Let $f(x) = \tan^{-1}(\log|x+\sqrt{x^2+1}|)$. Since $x > 0$,the domain is restricted. However,considering the nature of the expression,it is neither odd nor even $(I)$.
Step $3$: Analyze $C$. For $3 < x < 5$,$|x-2| = x-2$,$|x-3| = x-3$,$|x-5| = -(x-5) = 5-x$. So $f(x) = (x-2) + (x-3) + (5-x) = x$. This is the identity function $(IV)$.
Step $4$: Analyze $D$. $f(x) = \sin 2x + \sin^2 x + \cos 3x$. This is a combination of trigonometric functions and is neither odd nor even $(I)$.
Wait,re-evaluating $A$: $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. This is a known even function. $C$ is identity. $B$ is neither. $D$ is neither. Checking options,$A-II$ is correct. $C-IV$ is correct. Thus,option $C$ is the correct match.

(A) Given,$g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2 x^2-5 x+2$.
Multiplying by $2$,we get $g(f(x))=4 x^2-10 x+4$.
Substituting $f(x)$ into $g(x)$,we have $(f(x))^2+(f(x))-2=4 x^2-10 x+4$.
Assuming $f(x)$ is a linear polynomial $f(x)=ax+b$,we get $(ax+b)^2+(ax+b)-2=4 x^2-10 x+4$.
Expanding the left side: $a^2 x^2+(2ab+a)x+(b^2+b-2)=4 x^2-10 x+4$.
Comparing coefficients:
$1) a^2=4 \Rightarrow a=2$ or $a=-2$.
$2) 2ab+a=-10$.
$3) b^2+b-2=4 \Rightarrow b^2+b-6=0 \Rightarrow (b+3)(b-2)=0 \Rightarrow b=-3$ or $b=2$.
If $a=2$,then $2(2)b+2=-10 \Rightarrow 4b=-12 \Rightarrow b=-3$. Thus $f(x)=2x-3$.
If $a=-2$,then $2(-2)b-2=-10 \Rightarrow -4b=-8 \Rightarrow b=2$. Thus $f(x)=-2x+2$.
Comparing with the given options,$f(x)=2x-3$ is the correct choice.

(C) Let $g(x) = f(f(x))$.
Given $f(x) = \begin{cases} 1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{cases}$.
For $0 \leq x \leq 2$,$f(x) = 1+x$. Since $0 \leq x \leq 2$,$1 \leq 1+x \leq 3$.
If $1 \leq 1+x \leq 2$ (i.e.,$0 \leq x \leq 1$),then $f(f(x)) = f(1+x) = 1+(1+x) = 2+x$.
If $2 < 1+x \leq 3$ (i.e.,$1 < x \leq 2$),then $f(f(x)) = f(1+x) = 3-(1+x) = 2-x$.
For $2 < x \leq 3$,$f(x) = 3-x$. Since $2 < x \leq 3$,$0 \leq 3-x < 1$.
Thus,$f(f(x)) = f(3-x) = 1+(3-x) = 4-x$.
So,$g(x) = \begin{cases} 2+x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \\ 4-x, & 2 < x \leq 3 \end{cases}$.
Checking continuity at $x=1$:
$LHL = \lim_{x \to 1^-} (2+x) = 3$,$RHL = \lim_{x \to 1^+} (2-x) = 1$. Since $LHL \neq RHL$,it is discontinuous at $x=1$.
Checking continuity at $x=2$:
$LHL = \lim_{x \to 2^-} (2-x) = 0$,$RHL = \lim_{x \to 2^+} (4-x) = 2$. Since $LHL \neq RHL$,it is discontinuous at $x=2$.
Therefore,$f(f(x))$ is discontinuous at $x=1$ and $x=2$.

(B) Given: $f: A \rightarrow B$ and $g: B \rightarrow C$.
Also,$g \circ f: A \rightarrow C$ is a bijection.
First,we prove $f$ is an injection:
Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$.
Then $g(f(x_1)) = g(f(x_2))$,which implies $(g \circ f)(x_1) = (g \circ f)(x_2)$.
Since $g \circ f$ is a bijection,it is injective,so $x_1 = x_2$.
Thus,$f$ is an injection.
Next,we prove $g$ is a surjection:
Let $z \in C$. Since $g \circ f: A \rightarrow C$ is a bijection,it is surjective.
Therefore,there exists $x \in A$ such that $(g \circ f)(x) = z$.
This means $g(f(x)) = z$.
Since $f(x) \in B$,let $y = f(x)$. Then $g(y) = z$ for some $y \in B$.
Thus,$g$ is a surjection.
Therefore,$f$ is an injection and $g$ is a surjection.

(C) Let $f(x) = a_0 x^n + a_1 x^{n-1} + \ldots + a_n$.
Given the functional equation $f(x) \cdot f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right)$.
This equation is satisfied by $f(x) = 1 \pm x^n$.
Given $f(4) = 257$,we have $1 \pm 4^n = 257$.
This implies $4^n = 256$,so $n = 4$.
Thus,$f(x) = 1 + x^4$.
Finally,$f(3) = 1 + 3^4 = 1 + 81 = 82$.

(C) Given $f(x) = x^3 + ax^2 + bx + c$,where $a = f^{\prime}(1)$,$b = f^{\prime \prime}(2)$,and $c = -f^{\prime \prime \prime}(3)$.
First,find the derivatives: $f^{\prime}(x) = 3x^2 + 2ax + b$,$f^{\prime \prime}(x) = 6x + 2a$,$f^{\prime \prime \prime}(x) = 6$.
Now,solve for constants:
$f^{\prime \prime \prime}(3) = 6$,so $c = -6$.
$f^{\prime \prime}(2) = 6(2) + 2a = 12 + 2a$. Since $b = f^{\prime \prime}(2)$,we have $b = 12 + 2a$.
$f^{\prime}(1) = 3(1)^2 + 2a(1) + b = 3 + 2a + b$. Since $a = f^{\prime}(1)$,we have $a = 3 + 2a + b$.
Substitute $b = 12 + 2a$ into the equation for $a$: $a = 3 + 2a + (12 + 2a) \Rightarrow a = 15 + 4a \Rightarrow -3a = 15 \Rightarrow a = -5$.
Then $b = 12 + 2(-5) = 2$.
So,$f(x) = x^3 - 5x^2 + 2x - 6$.
At $x=0$,$y = f(0) = -6$. The point is $(0, -6)$.
Slope $m = f^{\prime}(0) = 3(0)^2 - 10(0) + 2 = 2$.
Tangent at $(0, -6)$ is $y - (-6) = 2(x - 0) \Rightarrow y = 2x - 6$. $X$-intercept is $3$.
Normal at $(0, -6)$ is $y - (-6) = -\frac{1}{2}(x - 0) \Rightarrow y = -\frac{1}{2}x - 6$. $X$-intercept is $-12$.
The triangle is formed by the points $(0, -6)$,$(3, 0)$,and $(-12, 0)$.
Base length $= 3 - (-12) = 15$. Height $= |-6| = 6$.
Area $= \frac{1}{2} \times 15 \times 6 = 45$ sq. units.

(C) Given that $f(x + y) = f(x) + f(y)$ for all $x, y \in R$,this is Cauchy's functional equation,which implies $f(x) = cx$ for some constant $c$.
Since $f(1) = 7$,we have $c(1) = 7$,so $c = 7$.
Thus,$f(x) = 7x$.
Now,we are given $\sum_{r=1}^{n} f(r) = 14112$.
Substituting $f(r) = 7r$,we get $\sum_{r=1}^{n} 7r = 14112$.
$7 \sum_{r=1}^{n} r = 14112$.
$7 \cdot \frac{n(n + 1)}{2} = 14112$.
$\frac{n(n + 1)}{2} = \frac{14112}{7} = 2016$.
$n(n + 1) = 4032$.
Since $63 \times 64 = 4032$,we have $n = 63$.

(A) Given that the function $f(x)$ is continuous at $x = \frac{\pi}{4}$,we must have $f(\frac{\pi}{4}) = \lim_{x \to \frac{\pi}{4}} f(x)$.
Since $f(\frac{\pi}{4}) = k$,we evaluate the limit:
$k = \lim_{x \to \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$.
This is a $\frac{0}{0}$ indeterminate form. Applying $L$'$H$ôpital's Rule by differentiating the numerator and denominator with respect to $x$:
$k = \lim_{x \to \frac{\pi}{4}} \frac{\frac{d}{dx}(1-\sqrt{2} \sin x)}{\frac{d}{dx}(\pi-4 x)}$
$k = \lim_{x \to \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$
Substituting $x = \frac{\pi}{4}$:
$k = \frac{-\sqrt{2} \cos(\frac{\pi}{4})}{-4} = \frac{-\sqrt{2} \times \frac{1}{\sqrt{2}}}{-4} = \frac{-1}{-4} = \frac{1}{4}$.

(C) To find the points of discontinuity,we check the limits at the transition points $x = 0, 1, 2, 3$.
At $x = 0$: $f(0) = 3-2(0)^2 = 3$. $\lim_{x \to 0^-} f(x) = 3$ and $\lim_{x \to 0^+} f(x) = 2(0)+3 = 3$. Continuous.
At $x = 1$: $\lim_{x \to 1^-} f(x) = 2(1)+3 = 5$. $\lim_{x \to 1^+} f(x) = 2(1)^2-3(1) = -1$. Since $5 \neq -1$,$x = 1$ is a point of discontinuity.
At $x = 2$: $\lim_{x \to 2^-} f(x) = 2(2)^2-3(2) = 8-6 = 2$. $\lim_{x \to 2^+} f(x) = 2(2)-3 = 1$. Since $2 \neq 1$,$x = 2$ is a point of discontinuity.
At $x = 3$: $\lim_{x \to 3^-} f(x) = 2(3)-3 = 3$. $\lim_{x \to 3^+} f(x) = |3| = 3$. Continuous.
Thus,the points of discontinuity are $a = 2$ and $b = 1$ (given $a > b$).
Therefore,$3a - b = 3(2) - 1 = 6 - 1 = 5$.

(B) Given the function $f(x) = \begin{cases} [x], & \text{if } x < 2 \\ [x]-1, & \text{if } x \geq 2 \end{cases}$.
First,check continuity at $x = 2$:
$f(2) = [2] - 1 = 2 - 1 = 1$.
Left-hand limit $(LHL)$ at $x = 2$:
$\lim_{x \to 2^-} f(x) = \lim_{h \to 0} [2-h] = 1$.
Right-hand limit $(RHL)$ at $x = 2$:
$\lim_{x \to 2^+} f(x) = \lim_{h \to 0} ([2+h] - 1) = 2 - 1 = 1$.
Since $LHL = RHL = f(2)$,the function is continuous at $x = 2$.
For $x \in [1, 2)$,$f(x) = [x]$. The greatest integer function $[x]$ is continuous on any interval $[n, n+1)$ where $n \in \mathbb{Z}$. Thus,$f(x)$ is continuous on $[1, 2)$.
For $x \in [2, 3)$,$f(x) = [x] - 1$. Similarly,this is continuous on $[2, 3)$.
Since the function is also continuous at the junction $x = 2$,$f(x)$ is continuous on the interval $[1, 3)$.

(C) Since $f(x)$ is continuous on $R$,it must be continuous at $x=0$ and $x=2$.
At $x=0$,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
$\cos(0) = 3(0) + \alpha \implies 1 = \alpha$.
At $x=2$,$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$.
$3(2) + \alpha = \beta(2) + 3$.
Substituting $\alpha = 1$: $6 + 1 = 2\beta + 3 \implies 7 = 2\beta + 3 \implies 2\beta = 4 \implies \beta = 2$.
Therefore,$\alpha^2 + \beta^2 = (1)^2 + (2)^2 = 1 + 4 = 5$.

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Given $f(0) = a$.
Now,calculate the limit: $\lim_{x \to 0} (1+3x)^{\frac{4}{x}}$.
This is of the form $1^{\infty}$,which can be evaluated using the formula $\lim_{x \to 0} (1+f(x))^{g(x)} = e^{\lim_{x \to 0} f(x)g(x)}$.
Here,$f(x) = 3x$ and $g(x) = \frac{4}{x}$.
So,$\lim_{x \to 0} (1+3x)^{\frac{4}{x}} = e^{\lim_{x \to 0} (3x \cdot \frac{4}{x})} = e^{\lim_{x \to 0} 12} = e^{12}$.
Since the function is continuous,$a = e^{12}$.
Therefore,$\log a = \log(e^{12}) = 12 \log e = 12$ (assuming natural logarithm).

(A) Given: $f(x) = \begin{cases} \frac{\sin(a + 1)x + \sin x}{x}, & x < 0 \\ b, & x = 0 \\ \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}$
Since $f(x)$ is continuous on $R$,it must be continuous at $x = 0$.
Therefore,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = b$.
For the left-hand limit:
$\lim_{x \to 0^-} \frac{\sin(a + 1)x + \sin x}{x} = \lim_{x \to 0^-} \left( \frac{\sin(a + 1)x}{x} + \frac{\sin x}{x} \right) = (a + 1) + 1 = a + 2$.
So,$a + 2 = b$ . . . $(i)$.
For the right-hand limit:
$\lim_{x \to 0^+} \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1 + x} - 1)}{x \cdot \sqrt{x}} = \lim_{x \to 0^+} \frac{\sqrt{1 + x} - 1}{x}$.
Multiplying by the conjugate:
$\lim_{x \to 0^+} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)} = \lim_{x \to 0^+} \frac{1 + x - 1}{x(\sqrt{1 + x} + 1)} = \lim_{x \to 0^+} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{2}$.
Thus,$b = \frac{1}{2}$.
Substituting $b$ into $(i)$: $a + 2 = \frac{1}{2} \Rightarrow a = \frac{1}{2} - 2 = -\frac{3}{2}$.
Finally,$a + b = -\frac{3}{2} + \frac{1}{2} = -\frac{2}{2} = -1$.

(D) Given the functional equation $f(x+y)=f(x) \cdot f(y)$.
Setting $x=4$ and $y=0$,we get $f(4+0)=f(4) \cdot f(0)$,which implies $f(4)=f(4) \cdot f(0)$. Since $f(x)$ is differentiable,$f(x)$ is not identically zero,so $f(0)=1$.
By the definition of the derivative,$f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$.
Substituting $f(x+h)=f(x) \cdot f(h)$,we have $f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x)f(h)-f(x)}{h} = f(x) \lim_{h \rightarrow 0} \frac{f(h)-1}{h}$.
Since $f(0)=1$,this is $f^{\prime}(x) = f(x) \lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h} = f(x) \cdot f^{\prime}(0)$.
Given $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$,we substitute these into the equation $f^{\prime}(4) = f(4) \cdot f^{\prime}(0)$.
$24 = f(4) \cdot 3$.
Therefore,$f(4) = \frac{24}{3} = 8$.