TS EAMCET 2023 Mathematics Question Paper with Answer and Solution

(B) The general term in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is given by $T_{r+1} = {}^{256}C_r (\sqrt{3})^{256-r} (\sqrt[8]{5})^r$.
This simplifies to $T_{r+1} = {}^{256}C_r (3)^{\frac{256-r}{2}} (5)^{\frac{r}{8}}$.
For the term to be an integer,both exponents $\frac{256-r}{2}$ and $\frac{r}{8}$ must be non-negative integers.
Since $0 \leq r \leq 256$,for $\frac{r}{8}$ to be an integer,$r$ must be a multiple of $8$,i.e.,$r \in \{0, 8, 16, \dots, 256\}$.
For these values of $r$,$\frac{256-r}{2} = 128 - \frac{r}{2}$ is also an integer because $r$ is a multiple of $8$ (and thus a multiple of $2$).
The number of such values of $r$ is given by the arithmetic progression $0, 8, 16, \dots, 256$.
Using the formula $a_n = a + (n-1)d$,we have $256 = 0 + (n-1)8$,which gives $n-1 = 32$,so $n = 33$.
Thus,there are $33$ integral terms.

(D) The given series is a geometric progression with the first term $a = 1$ and common ratio $r = (\cos \theta + i \sin \theta)$.
The $n^{th}$ term of a geometric progression is given by $T_n = a \cdot r^{n-1}$.
For the $10^{th}$ term $(n = 10)$,we have $T_{10} = 1 \cdot (\cos \theta + i \sin \theta)^{10-1} = (\cos \theta + i \sin \theta)^9$.
Using De Moivre's theorem,$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) = e^{i n \theta}$.
Substituting $\theta = \frac{\pi}{6}$ and $n = 9$:
$T_{10} = e^{i 9 (\frac{\pi}{6})} = e^{i \frac{3\pi}{2}}$.
Using Euler's formula $e^{i \phi} = \cos \phi + i \sin \phi$:
$T_{10} = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}) = 0 + i(-1) = -i$.

(B) Given $\left|\frac{z-i}{z i}\right|=2$.
Since $z=x iy$,we have $\left|\frac{x i(y-1)}{x i(y 1)}\right|=2$.
Squaring both sides,we get $\frac{x^2 (y-1)^2}{x^2 (y 1)^2}=4$.
$x^2 y^2-2y 1=4(x^2 y^2 2y 1)$.
$x^2 y^2-2y 1=4x^2 4y^2 8y 4$.
Rearranging the terms,we get $3x^2 3y^2 10y 3=0$.

(A) The total number of questions is $12$ ($3$ sections $\times$ $4$ questions each). The candidate must select $5$ questions such that at least one question is chosen from each section.
Possible distributions of $5$ questions across $3$ sections $(A, B, C)$ are:
Case $1$: $(2, 2, 1)$ in any order. The number of ways is $3 \times (^4C_2 \times ^4C_2 \times ^4C_1) = 3 \times (6 \times 6 \times 4) = 3 \times 144 = 432$.
Case $2$: $(3, 1, 1)$ in any order. The number of ways is $3 \times (^4C_3 \times ^4C_1 \times ^4C_1) = 3 \times (4 \times 4 \times 4) = 3 \times 64 = 192$.
Total number of ways = $432 + 192 = 624$.

(D) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
When the axes are rotated by an angle $\theta$,the new coordinates $(x', y')$ are related to the old coordinates $(x, y)$ by $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$.
Substituting these into the equation of the line:
$\frac{x' \cos \theta - y' \sin \theta}{a} + \frac{x' \sin \theta + y' \cos \theta}{b} = 1$
$x' \left( \frac{\cos \theta}{a} + \frac{\sin \theta}{b} \right) + y' \left( \frac{\cos \theta}{b} - \frac{\sin \theta}{a} \right) = 1$.
Comparing this with the intercept form of the line in the new axes,$\frac{x'}{p} + \frac{y'}{q} = 1$,we get:
$\frac{1}{p} = \frac{\cos \theta}{a} + \frac{\sin \theta}{b}$ and $\frac{1}{q} = \frac{\cos \theta}{b} - \frac{\sin \theta}{a}$.
Squaring and adding these equations:
$\frac{1}{p^2} + \frac{1}{q^2} = \left( \frac{\cos \theta}{a} + \frac{\sin \theta}{b} \right)^2 + \left( \frac{\cos \theta}{b} - \frac{\sin \theta}{a} \right)^2$
$= \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} + \frac{2 \sin \theta \cos \theta}{ab} + \frac{\cos^2 \theta}{b^2} + \frac{\sin^2 \theta}{a^2} - \frac{2 \sin \theta \cos \theta}{ab}$
$= \frac{\cos^2 \theta + \sin^2 \theta}{a^2} + \frac{\cos^2 \theta + \sin^2 \theta}{b^2}$
$= \frac{1}{a^2} + \frac{1}{b^2}$.
Thus,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

(B) The given series is $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$.
Adding $1$ to both sides,we get $1 + y = 1 + \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$.
This is of the form $(1 - x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$.
Comparing terms,we have $nx = \frac{3}{4}$ and $\frac{n(n+1)}{2}x^2 = \frac{3 \cdot 5}{4 \cdot 8} = \frac{15}{32}$.
From $nx = \frac{3}{4}$,we get $x = \frac{3}{4n}$.
Substituting into the second term: $\frac{n(n+1)}{2} \cdot \frac{9}{16n^2} = \frac{15}{32} \implies \frac{n+1}{n} \cdot \frac{9}{32} = \frac{15}{32} \implies \frac{n+1}{n} = \frac{15}{9} = \frac{5}{3}$.
Solving for $n$: $3n + 3 = 5n \implies 2n = 3 \implies n = \frac{3}{2}$.
Then $x = \frac{3}{4 \cdot (3/2)} = \frac{3}{6} = \frac{1}{2}$.
Thus,$1 + y = (1 - 1/2)^{-3/2} = (1/2)^{-3/2} = 2^{3/2} = 2\sqrt{2}$.
Squaring both sides: $(1 + y)^2 = (2\sqrt{2})^2 = 8$.
$1 + 2y + y^2 = 8 \implies y^2 + 2y - 7 = 0$.

(D) The equation of the ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$.
Let the point of tangency be $(x_0, y_0)$. The equation of the tangent at $(x_0, y_0)$ is $\frac{xx_0}{2} + yy_0 = 1$.
The intercepts on the axes are $A = (\frac{2}{x_0}, 0)$ and $B = (0, \frac{1}{y_0})$.
Let $(h, k)$ be the midpoint of the intercept $AB$. Then $h = \frac{1}{x_0}$ and $k = \frac{1}{2y_0}$,which implies $x_0 = \frac{1}{h}$ and $y_0 = \frac{1}{2k}$.
Since $(x_0, y_0)$ lies on the ellipse,we have $(\frac{1}{h})^2 + 2(\frac{1}{2k})^2 = 2$.
This simplifies to $\frac{1}{h^2} + \frac{2}{4k^2} = 2$,or $\frac{1}{h^2} + \frac{1}{2k^2} = 2$.
Replacing $(h, k)$ with $(x, y)$,we get $\frac{1}{x^2} + \frac{1}{2y^2} = 2$.

(B) Given the quadratic equation $4x^2 - 2x + k - 4 = 0$.
Let the roots be $\alpha$ and $\frac{1}{\alpha}$.
According to the property of quadratic equations $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Here,$a = 4$,$b = -2$,and $c = k - 4$.
Therefore,$\alpha \cdot \frac{1}{\alpha} = \frac{k - 4}{4}$.
$1 = \frac{k - 4}{4}$.
$4 = k - 4$.
$k = 8$.

(B) Given the equation $x^2+2x+2=0$.
Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4-8}}{2} = -1 \pm i$.
Let $\alpha = -1+i$ and $\beta = -1-i$.
In polar form,$\alpha = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})$ and $\beta = \sqrt{2}(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4})$.
Then $\alpha^{15} = (\sqrt{2})^{15} (\cos \frac{45\pi}{4} + i \sin \frac{45\pi}{4}) = 2^{7.5} (\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) = 2^{7.5} (-\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}})$.
Similarly,$\beta^{15} = (\sqrt{2})^{15} (\cos \frac{75\pi}{4} + i \sin \frac{75\pi}{4}) = 2^{7.5} (\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) = 2^{7.5} (-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})$.
Adding these,$\alpha^{15} + \beta^{15} = 2^{7.5} (-\frac{2}{\sqrt{2}}) = 2^{7.5} (-\sqrt{2}) = -2^8 = -256$.

(A) The given equation is $x^4+x^2+1=0$.
This can be factored as $(x^2+x+1)(x^2-x+1)=0$.
The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.
The roots of $x^2-x+1=0$ are $-\omega$ and $-\omega^2$.
Given $\alpha+\beta=-1$ and $\alpha^2=\beta$,we identify $\alpha=\omega$ and $\beta=\omega^2$ (since $\omega+\omega^2=-1$ and $\omega^2=\omega^2$).
Given $\gamma+\delta=1$ and $\gamma^2=-\delta$,we identify $\gamma=-\omega$ and $\delta=-\omega^2$ (since $-\omega-\omega^2=1$ and $(-\omega)^2 = \omega^2 = -(-\omega^2)$).
Now,calculate $\alpha^{2023}+\beta^{2023}+\gamma^{2022}+\delta^{2022}$:
$= \omega^{2023} + (\omega^2)^{2023} + (-\omega)^{2022} + (-\omega^2)^{2022}$
$= \omega^{2023} + \omega^{4046} + \omega^{2022} + \omega^{4044}$
Using $\omega^3=1$:
$= \omega^1 + \omega^1 + 1 + 1 = 2\omega + 2$.
Wait,re-evaluating the sum: $\omega^{2023} = \omega$,$\omega^{4046} = \omega^2$,$\omega^{2022} = 1$,$\omega^{4044} = 1$.
Sum $= \omega + \omega^2 + 1 + 1 = (-1) + 2 = 1$.

(C) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.
$\sin ^2 18^{\circ} = \left(\frac{\sqrt{5}-1}{4}\right)^2 = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.
$\cos ^2 36^{\circ} = \left(\frac{\sqrt{5}+1}{4}\right)^2 = \frac{5+1+2\sqrt{5}}{16} = \frac{6+2\sqrt{5}}{16} = \frac{3+\sqrt{5}}{8}$.
The sum of the roots is $\frac{3-\sqrt{5}}{8} + \frac{3+\sqrt{5}}{8} = \frac{6}{8} = \frac{3}{4}$.
The product of the roots is $\left(\frac{3-\sqrt{5}}{8}\right) \left(\frac{3+\sqrt{5}}{8}\right) = \frac{9-5}{64} = \frac{4}{64} = \frac{1}{16}$.
The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
$x^2 - \frac{3}{4}x + \frac{1}{16} = 0$.
Multiplying by $16$,we get $16x^2 - 12x + 1 = 0$.

(D) The equation $f(x)^{g(x)}=1$ is satisfied in the following cases:
Case $1$: $f(x)=1$
$x^2-7x+11=1 \implies x^2-7x+10=0 \implies (x-2)(x-5)=0 \implies x=2, 5$.
Case $2$: $g(x)=0$ and $f(x) \neq 0$
$x^2-6x-7=0 \implies (x-7)(x+1)=0 \implies x=7, -1$.
Case $3$: $f(x)=-1$ and $g(x)$ is an even integer
$x^2-7x+11=-1 \implies x^2-7x+12=0 \implies (x-3)(x-4)=0 \implies x=3, 4$.
Check $g(x)$ for $x=3, 4$:
For $x=3$,$g(3)=3^2-6(3)-7=9-18-7=-16$ (even,so $x=3$ is a solution).
For $x=4$,$g(4)=4^2-6(4)-7=16-24-7=-15$ (odd,so $x=4$ is not a solution).
The set of real values of $x$ is $\{2, 5, 7, -1, 3\}$.
The sum of these values is $2+5+7-1+3=16$.

(A) Given $x^2+ax+2=0$,we have $\alpha+\beta = -a$ and $\alpha\beta = 2$.
For $x^2-bx+c=0$,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$,so $\frac{1}{\alpha}+\frac{1}{\beta} = b$ and $\frac{1}{\alpha\beta} = c = \frac{1}{2}$.
Now,consider the expression:
$E = \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)$
$E = \left(\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}\right) \left(\alpha\beta - 1 - 1 + \frac{1}{\alpha\beta}\right)$
$E = \left(\alpha\beta + \frac{1}{\alpha\beta} + 2\right) \left(\alpha\beta + \frac{1}{\alpha\beta} - 2\right)$
$E = \left(\alpha\beta + \frac{1}{\alpha\beta}\right)^2 - 4$
Substituting $\alpha\beta = 2$:
$E = \left(2 + \frac{1}{2}\right)^2 - 4 = \left(\frac{5}{2}\right)^2 - 4 = \frac{25}{4} - 4 = \frac{9}{4}$.
Wait,let us re-evaluate the expression expansion:
$E = \left(\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}\right) \left(\alpha\beta - \frac{\alpha}{\beta} - \frac{\beta}{\alpha} + \frac{1}{\alpha\beta}\right)$
$E = \left(2 + 2 + \frac{1}{2}\right) \left(2 + \frac{1}{2} - \frac{\alpha^2+\beta^2}{\alpha\beta}\right) = \frac{9}{2} \left(\frac{5}{2} - \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\right)$
$E = \frac{9}{2} \left(\frac{5}{2} - \frac{a^2-4}{2}\right) = \frac{9}{2} \left(\frac{5-a^2+4}{2}\right) = \frac{9}{4}(9-a^2)$.

(B) Since $(x-2)$ is a factor of $x^2+ax+b$,we have $x=2$ as a root,so $(2)^2+2a+b=0$,which gives $2a+b=-4$ ... $(i)$.
Similarly,since $(x-2)$ is a factor of $x^2+cx+d$,we have $(2)^2+2c+d=0$,which gives $2c+d=-4$ ... $(ii)$.
Subtracting equation $(ii)$ from equation $(i)$:
$(2a+b) - (2c+d) = -4 - (-4)$
$2(a-c) + (b-d) = 0$
$b-d = -2(a-c)$
$b-d = 2(c-a)$
Therefore,$\frac{b-d}{c-a} = 2$.

(C) Let the common root be $\alpha$. The equations are $x^2-6x+a=0$ and $x^2-cx+6=0$.
Let the other roots be $\beta_1$ and $\beta_2$ respectively. Given $\beta_1 : \beta_2 = 4:3$,so $\beta_1 = 4k$ and $\beta_2 = 3k$ for some constant $k$.
From the properties of roots:
For the first equation: $\alpha + 4k = 6$ and $\alpha \cdot 4k = a$.
For the second equation: $\alpha + 3k = c$ and $\alpha \cdot 3k = 6$.
From $\alpha \cdot 3k = 6$,we have $k = \frac{2}{\alpha}$.
Substitute $k$ into the first equation: $\alpha + 4(\frac{2}{\alpha}) = 6 \Rightarrow \alpha + \frac{8}{\alpha} = 6$.
Multiplying by $\alpha$: $\alpha^2 - 6\alpha + 8 = 0$.
Factoring: $(\alpha - 4)(\alpha - 2) = 0$,so $\alpha = 4$ or $\alpha = 2$.
If $\alpha = 4$,then $4k = 6 - 4 = 2 \Rightarrow k = 0.5$. Then $\beta_1 = 4(0.5) = 2$ and $\beta_2 = 3(0.5) = 1.5$. Since $\beta_2$ must be an integer,this is rejected.
If $\alpha = 2$,then $4k = 6 - 2 = 4 \Rightarrow k = 1$. Then $\beta_1 = 4(1) = 4$ and $\beta_2 = 3(1) = 3$. Both are integers.
Thus,the common root is $2$.

(D) Let $\alpha$ be the non-zero common root of the equations $x^2+3x-2k=0$ and $x^2-2x-7k=0$.
Subtracting the two equations: $(x^2+3x-2k) - (x^2-2x-7k) = 0$ $\Rightarrow 5x+5k=0$ $\Rightarrow x=-k$.
Since $x=-k$ is a root,substitute it into the first equation: $(-k)^2+3(-k)-2k=0$ $\Rightarrow k^2-5k=0$ $\Rightarrow k(k-5)=0$.
Since the root is non-zero,$k \neq 0$,so $k=5$.
Substituting $k=5$ into the third equation: $5x^2+(5+2)x-(5+1)=0 \Rightarrow 5x^2+7x-6=0$.
Factoring the quadratic: $5x^2+10x-3x-6=0$ $\Rightarrow 5x(x+2)-3(x+2)=0$ $\Rightarrow (5x-3)(x+2)=0$.
The roots are $x=\frac{3}{5}$ and $x=-2$.
The positive root is $x=\frac{3}{5}$.

(B) Let the common root be $\alpha$.
Given equations are $ax^2-7x+c=0$ and $ax^2+5x-c=0$.
Since $\alpha$ is a common root,we have:
$a\alpha^2-7\alpha+c=0$ and $a\alpha^2+5\alpha-c=0$.
Subtracting the two equations:
$(ax^2-7x+c) - (ax^2+5x-c) = 0$ $\Rightarrow -12x + 2c = 0$ $\Rightarrow x = \frac{c}{6}$.
Thus,the common root $\alpha = \frac{c}{6}$.
Substituting $\alpha = \frac{c}{6}$ into $ax^2-7x+c=0$:
$a(\frac{c}{6})^2 - 7(\frac{c}{6}) + c = 0 \Rightarrow \frac{ac^2}{36} - \frac{7c}{6} + c = 0$.
Since $c \neq 0$,divide by $c$:
$\frac{ac}{36} - \frac{7}{6} + 1 = 0$ $\Rightarrow \frac{ac}{36} = \frac{1}{6}$ $\Rightarrow ac = 6$.
Given that $3$ is a root of $ax^2-7x+c=0$ other than the common root:
$a(3)^2 - 7(3) + c = 0$ $\Rightarrow 9a - 21 + c = 0$ $\Rightarrow c = 21 - 9a$.
Substitute $c = 21 - 9a$ into $ac = 6$:
$a(21 - 9a) = 6$ $\Rightarrow 21a - 9a^2 = 6$ $\Rightarrow 3a^2 - 7a + 2 = 0$.
Solving for $a$: $(3a-1)(a-2) = 0$,so $a = 2$ or $a = \frac{1}{3}$.
If $a = 2$,then $c = 21 - 9(2) = 3$. The equation $ax^2-7x+c=0$ becomes $2x^2-7x+3=0$.
Factoring: $(2x-1)(x-3) = 0$,so roots are $3$ and $\frac{1}{2}$.
The common root is $\frac{1}{2}$.

(C) For the first inequality: $x^2-7x+10 \geq 0$
$(x-2)(x-5) \geq 0$
Thus,$x \in (-\infty, 2] \cup [5, \infty)$.
For the second inequality: $2x+3-x^2 > 0$
$x^2-2x-3 < 0$
$(x-3)(x+1) < 0$
Thus,$x \in (-1, 3)$.
Taking the intersection of both intervals:
$(-\infty, 2] \cup [5, \infty) \cap (-1, 3) = (-1, 2]$.
Therefore,the set of all values of $x$ is $(-1, 2]$.

(D) Given the quadratic expression $f(x) = x^2+2px-2p+8 > 0$ for all real values of $x$.
For a quadratic $ax^2+bx+c > 0$ to be true for all $x \in R$,the conditions are $a > 0$ and the discriminant $D < 0$.
Here,$a = 1 > 0$,which is satisfied.
Now,calculate the discriminant $D = b^2 - 4ac < 0$:
$D = (2p)^2 - 4(1)(-2p+8) < 0$
$4p^2 + 8p - 32 < 0$
Divide by $4$:
$p^2 + 2p - 8 < 0$
Factorize the quadratic:
$(p+4)(p-2) < 0$
Using the sign scheme method,the expression is negative between the roots $p = -4$ and $p = 2$.
Therefore,the set of all possible values of $p$ is $p \in (-4, 2)$.

(B) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$.
We can rewrite the expression as:
$f(x) = \frac{(x^2+x+1) - 2x}{x^2+x+1} = 1 - \frac{2x}{x^2+x+1}$.
To minimize $f(x)$,we need to maximize the term $\frac{2x}{x^2+x+1}$.
For $x \neq 0$,we have $\frac{2x}{x^2+x+1} = \frac{2}{x + \frac{1}{x} + 1}$.
Since $x + \frac{1}{x} \geq 2$ for $x > 0$ and $x + \frac{1}{x} \leq -2$ for $x < 0$,the maximum value of $x + \frac{1}{x} + 1$ is not applicable here directly,but we analyze the range of $\frac{x}{x^2+x+1}$.
Let $y = \frac{x^2-x+1}{x^2+x+1} \implies y(x^2+x+1) = x^2-x+1 \implies (y-1)x^2 + (y+1)x + (y-1) = 0$.
For $x$ to be real,the discriminant $D \geq 0$:
$(y+1)^2 - 4(y-1)^2 \geq 0 \implies (y+1-2y+2)(y+1+2y-2) \geq 0 \implies (3-y)(3y-1) \geq 0$.
This implies $\frac{1}{3} \leq y \leq 3$.
Thus,the minimum value is $\frac{1}{3}$.

(B) Given the equation: $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^t + 4 = 0$.
Let $x = e^t$. Then the equation becomes $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0$.
Let the roots of the equation in $t$ be $t_1, t_2, t_3, t_4$.
Then the roots of the equation in $x$ are $x_1 = e^{t_1}, x_2 = e^{t_2}, x_3 = e^{t_3}, x_4 = e^{t_4}$.
By Vieta's formulas,the product of the roots of the polynomial $x^4 - 10x^3 + 29x^2 - 20x + 4 = 0$ is given by the constant term:
$x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 4$.
Substituting back $e^t$:
$e^{t_1} \cdot e^{t_2} \cdot e^{t_3} \cdot e^{t_4} = 4$.
$e^{(t_1 + t_2 + t_3 + t_4)} = 4$.
Taking the natural logarithm on both sides:
$t_1 + t_2 + t_3 + t_4 = \log_e 4$.
Since $4 = 2^2$,we have:
$t_1 + t_2 + t_3 + t_4 = \log_e 2^2 = 2\log_e 2$.

(D) Given the cubic equation: $x^3-14x^2+56x-64=0$.
By Vieta's formulas,for roots $\alpha, \beta, \gamma$:
$\alpha+\beta+\gamma=14$
$\alpha\beta+\beta\gamma+\gamma\alpha=56$
$\alpha\beta\gamma=64$
By testing small integer values,we find that $x=2$ is a root:
$2^3-14(2^2)+56(2)-64 = 8-56+112-64 = 0$.
Dividing the polynomial by $(x-2)$,we get:
$(x-2)(x^2-12x+32)=0$
$(x-2)(x-4)(x-8)=0$
The roots are $\alpha=2, \beta=4, \gamma=8$.
Since $\frac{4}{2} = \frac{8}{4} = 2$,the roots are in Geometric Progression $(GP)$.

(D) Given $x^3+3x^2-9x+\lambda = (x-\alpha)^2(x-\beta) = (x^2-2\alpha x+\alpha^2)(x-\beta) = x^3 - (2\alpha+\beta)x^2 + (2\alpha\beta+\alpha^2)x - \alpha^2\beta$.
Comparing coefficients:
$2\alpha+\beta = -3$ $(i)$
$2\alpha\beta+\alpha^2 = -9$ (ii)
$-\alpha^2\beta = \lambda$ (iii)
From $(i)$,$\beta = -3-2\alpha$.
Substitute into (ii): $2\alpha(-3-2\alpha) + \alpha^2 = -9$ $\Rightarrow -6\alpha - 4\alpha^2 + \alpha^2 = -9$ $\Rightarrow 3\alpha^2 + 6\alpha - 9 = 0$ $\Rightarrow \alpha^2 + 2\alpha - 3 = 0$.
Solving for $\alpha$: $(\alpha+3)(\alpha-1) = 0$,so $\alpha = 1$ or $\alpha = -3$.
If $\alpha = 1$,then $\beta = -3-2(1) = -5$. From (iii),$\lambda = -\alpha^2\beta = -(1)^2(-5) = 5$.
If $\alpha = -3$,then $\beta = -3-2(-3) = 3$. From (iii),$\lambda = -\alpha^2\beta = -(-3)^2(3) = -27$.
Thus,the values of $\lambda$ are $-27$ and $5$.

(B) Let the roots of the given equation $f(x) = x^4-2ax^3+4bx^2+8ax+16=0$ be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$.
The roots of the new equation are $p\alpha_1, p\alpha_2, p\alpha_3, p\alpha_4$.
Thus,the new equation is $f(\frac{x}{p}) = 0$.
Substituting $\frac{x}{p}$ into the equation:
$(\frac{x}{p})^4 - 2a(\frac{x}{p})^3 + 4b(\frac{x}{p})^2 + 8a(\frac{x}{p}) + 16 = 0$
Multiplying by $p^4$:
$x^4 - 2apx^3 + 4bp^2x^2 + 8ap^3x + 16p^4 = 0$
Since this is a reciprocal equation,the coefficient of $x^4$ must equal the constant term:
$1 = 16p^4$
$p^4 = \frac{1}{16}$
$p^2 = \frac{1}{4} \implies |p| = \frac{1}{2}$

(D) Given the equation $x^4-4x^3-7x^2+22x+24=0$.
By testing roots,we find the factors:
$(x+1)(x+2)(x-3)(x-4)=0$.
The roots are $x_1 = -1, x_2 = -2, x_3 = 3, x_4 = 4$.
We check the condition: the sum of two roots equals the sum of the other two.
$3 + (-1) = 2$ and $4 + (-2) = 2$.
The condition is satisfied.
The sum of the cubes of the roots is:
$(-1)^3 + (-2)^3 + (3)^3 + (4)^3 = -1 - 8 + 27 + 64 = 82$.

(C) Given the cubic equation: $x^3+4x^2-9x-36=0$
Factorizing the equation:
$x^2(x+4)-9(x+4)=0$
$(x^2-9)(x+4)=0$
$(x-3)(x+3)(x+4)=0$
The roots are $x = -4, -3, 3$.
Given the condition $\alpha < \beta < \gamma$,we have $\alpha = -4$,$\beta = -3$,and $\gamma = 3$.
Now,calculate $\alpha+2\beta+3\gamma$:
$\alpha+2\beta+3\gamma = -4 + 2(-3) + 3(3)$
$= -4 - 6 + 9$
$= -10 + 9 = -1$.

(B) Given the cubic equation $x^3-3x^2+3x+1=0$.
By Vieta's formulas,we have:
$\alpha+\beta+\gamma = 3$
$\alpha\beta+\beta\gamma+\gamma\alpha = 3$
$\alpha\beta\gamma = -1$
We need to find $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$.
Using the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$,let $a=\alpha\beta, b=\beta\gamma, c=\gamma\alpha$.
Then $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(\alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta)$.
Factor out $\alpha\beta\gamma$ from the second term:
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\beta+\alpha+\gamma)$.
Substitute the known values:
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (3)^2 - 2(-1)(3) = 9 + 6 = 15$.

(D) Let $f(x) = x^5-6x^4+11x^3-2x^2-12x+8$.
By testing small integer roots,we find that $x=2$ is a root.
Using synthetic division or polynomial division,we factor the expression:
$f(x) = (x-2)^3(x^2-1) = (x-2)^3(x-1)(x+1)$.
The roots are $x=2$ (with multiplicity $3$),$x=1$,and $x=-1$.
Since $\alpha$ is a multiple root,$\alpha=2$.
Substituting $\alpha=2$ into the expression $3\alpha^2-2\alpha+1$:
$3(2)^2-2(2)+1 = 3(4)-4+1 = 12-4+1 = 9$.

(A) Given the equation $18x^3-15x^2-4x+4=0$.
Let the roots be $\alpha, \alpha, \gamma$ where $\alpha > \gamma$.
From Vieta's formulas:
$2\alpha + \gamma = \frac{15}{18} = \frac{5}{6}$ $(i)$
$\alpha^2 + 2\alpha\gamma = \frac{-4}{18} = -\frac{2}{9}$ (ii)
$\alpha^2\gamma = -\frac{4}{18} = -\frac{2}{9}$ (iii)
From (ii) and (iii),$\alpha^2 + 2\alpha\gamma = \alpha^2\gamma$.
Since $\alpha \neq 0$,we have $\alpha + 2\gamma = \alpha\gamma$.
From $(i)$,$\gamma = \frac{5}{6} - 2\alpha$.
Substituting into $\alpha + 2\gamma = \alpha\gamma$:
$\alpha + 2(\frac{5}{6} - 2\alpha) = \alpha(\frac{5}{6} - 2\alpha)$
$\alpha + \frac{5}{3} - 4\alpha = \frac{5\alpha}{6} - 2\alpha^2$
$2\alpha^2 - 3\alpha + \frac{5}{3} = \frac{5\alpha}{6}$
$12\alpha^2 - 18\alpha + 10 = 5\alpha$
$12\alpha^2 - 23\alpha + 10 = 0$
$(3\alpha - 2)(4\alpha - 5) = 0$.
So,$\alpha = \frac{2}{3}$ or $\alpha = \frac{5}{4}$.
If $\alpha = \frac{2}{3}$,then $\gamma = \frac{5}{6} - 2(\frac{2}{3}) = \frac{5}{6} - \frac{4}{3} = -\frac{1}{2}$.
Since $\alpha > \gamma$ $(\frac{2}{3} > -\frac{1}{2})$,this is a valid solution.
Then $\alpha + \beta^2 + \gamma^3 = \frac{2}{3} + (\frac{2}{3})^2 + (-\frac{1}{2})^3 = \frac{2}{3} + \frac{4}{9} - \frac{1}{8} = \frac{48 + 32 - 9}{72} = \frac{71}{72}$.

(D) Given the cubic equation $2x^3+x^2-13x+6=0$.
By Vieta's formulas,for roots $\alpha, \beta, \gamma$:
$\alpha+\beta+\gamma = -\frac{1}{2}$
$\alpha\beta+\beta\gamma+\gamma\alpha = -\frac{13}{2}$
$\alpha\beta\gamma = -\frac{6}{2} = -3$
We use the identity: $\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - (\alpha\beta+\beta\gamma+\gamma\alpha))$
Note that $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (-\frac{1}{2})^2 - 2(-\frac{13}{2}) = \frac{1}{4} + 13 = \frac{53}{4}$.
Substituting these values:
$\alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)(\frac{53}{4} - (-\frac{13}{2})) + 3(-3)$
$= (-\frac{1}{2})(\frac{53}{4} + \frac{26}{4}) - 9$
$= (-\frac{1}{2})(\frac{79}{4}) - 9$
$= -\frac{79}{8} - \frac{72}{8} = -\frac{151}{8}$.

(C) Given the cubic equation $x^3+2x^2-x-2=0$.
Factorizing the equation:
$x^2(x+2)-1(x+2)=0$
$(x^2-1)(x+2)=0$
$(x-1)(x+1)(x+2)=0$
Thus,the roots are $\alpha=1, \beta=-1, \gamma=-2$.
We need to calculate $\alpha^6+\beta^6+\gamma^6$:
$\alpha^6+\beta^6+\gamma^6 = (1)^6 + (-1)^6 + (-2)^6$
$= 1 + 1 + 64$
$= 66$

(A) The given equation is $x^4+x^3-4x^2+x+1=0$.
Dividing by $x^2$,we get $(x^2 + \frac{1}{x^2}) + (x + \frac{1}{x}) - 4 = 0$.
Let $y = x + \frac{1}{x}$,then $y^2 - 2 + y - 4 = 0 \Rightarrow y^2 + y - 6 = 0$.
$(y+3)(y-2) = 0$,so $y = 2$ or $y = -3$.
If $x + \frac{1}{x} = 2$,then $x=1, 1$.
If $x + \frac{1}{x} = -3$,then $x^2 + 3x + 1 = 0$,so $x = \frac{-3 \pm \sqrt{5}}{2}$.
Let the roots be $r_1, r_2, r_3, r_4$. The sum of roots taken two at a time is $\sum r_i r_j = -4$.
When roots are diminished by $h$,the new roots are $r_i - h$.
The new coefficient of $x^2$ is $\sum (r_i - h)(r_j - h) = 0$.
This expands to $\sum r_i r_j - 3h \sum r_i + 6h^2 = 0$.
From the original equation,$\sum r_i = -1$ and $\sum r_i r_j = -4$.
Substituting these,$-4 - 3h(-1) + 6h^2 = 0 \Rightarrow 6h^2 + 3h - 4 = 0$.
The roots of this quadratic in $h$ are $\alpha$ and $\beta$.
Thus,$\alpha + \beta = -\frac{3}{6} = -\frac{1}{2}$ and $\alpha \beta = -\frac{4}{6} = -\frac{2}{3}$.
Then $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = (-\frac{1}{2})^2 - 4(-\frac{2}{3}) = \frac{1}{4} + \frac{8}{3} = \frac{3+32}{12} = \frac{35}{12}$.
Therefore,$12(\alpha - \beta)^2 = 12 \times \frac{35}{12} = 35$.

(A) Given the equation $x^3 + x^2 + x + 1 = 0$,the roots are $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha + \beta + \gamma = -1$
$\alpha \beta + \beta \gamma + \gamma \alpha = 1$
$\alpha \beta \gamma = -1$
$(i)$ $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha \beta + \beta \gamma + \gamma \alpha}{\alpha \beta \gamma} = \frac{1}{-1} = -1$. Thus,$(i)$ $\rightarrow$ $A$.
(ii) $\alpha^3 + \beta^3 + \gamma^3$: Since $\alpha, \beta, \gamma$ are roots,$\alpha^3 = -\alpha^2 - \alpha - 1$,$\beta^3 = -\beta^2 - \beta - 1$,$\gamma^3 = -\gamma^2 - \gamma - 1$.
Summing these: $\alpha^3 + \beta^3 + \gamma^3 = -(\alpha^2 + \beta^2 + \gamma^2) - (\alpha + \beta + \gamma) - 3$.
We know $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = (-1)^2 - 2(1) = 1 - 2 = -1$.
So,$\alpha^3 + \beta^3 + \gamma^3 = -(-1) - (-1) - 3 = 1 + 1 - 3 = -1$. Thus,(ii) $\rightarrow$ $A$.
(iii) $\alpha^4 + \beta^4 + \gamma^4$: Multiply the original equation by $x$: $x^4 + x^3 + x^2 + x = 0$.
Summing for roots: $(\alpha^4 + \beta^4 + \gamma^4) + (\alpha^3 + \beta^3 + \gamma^3) + (\alpha^2 + \beta^2 + \gamma^2) + (\alpha + \beta + \gamma) = 0$.
$(\alpha^4 + \beta^4 + \gamma^4) + (-1) + (-1) + (-1) = 0 \Rightarrow \alpha^4 + \beta^4 + \gamma^4 = 3$. Thus,(iii) $\rightarrow$ $D$.
(iv) $(\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 2(\alpha^2 + \beta^2 + \gamma^2) - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = 2(-1) - 2(1) = -2 - 2 = -4$. Thus,(iv) $\rightarrow$ $B$.
The correct match is $(i)$ $\rightarrow$ $A$,(ii) $\rightarrow$ $A$,(iii) $\rightarrow$ $D$,(iv) $\rightarrow$ $B$.

(C) Given the equation $k x^3 - 18 x^2 - 36 x + 8 = 0$.
Let the roots be $\alpha, \beta, \gamma$ which are in harmonic progression $(HP)$.
This implies that $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression $(AP)$.
Replacing $x$ with $\frac{1}{x}$ in the original equation,we get $k(\frac{1}{x})^3 - 18(\frac{1}{x})^2 - 36(\frac{1}{x}) + 8 = 0$,which simplifies to $8 x^3 - 36 x^2 - 18 x + k = 0$.
Let the roots of this new equation be $a-d, a, a+d$.
From the sum of roots,$(a-d) + a + (a+d) = -(\frac{-36}{8}) = \frac{9}{2}$ $\Rightarrow 3a = \frac{9}{2}$ $\Rightarrow a = \frac{3}{2}$.
Since $a = \frac{3}{2}$ is a root of $8 x^3 - 36 x^2 - 18 x + k = 0$,we substitute it:
$8(\frac{3}{2})^3 - 36(\frac{3}{2})^2 - 18(\frac{3}{2}) + k = 0$.
$8(\frac{27}{8}) - 36(\frac{9}{4}) - 27 + k = 0$.
$27 - 81 - 27 + k = 0$.
$k - 81 = 0 \Rightarrow k = 81$.

(B) Given $P(x) = 2x^4 + ax^3 + bx^2 + cx + d$ is a polynomial of degree $4$ with leading coefficient $2$.
We observe that for $x = 1, 2, 3, 4$,the values are $P(x) = x^2 + 3$.
Let $Q(x) = P(x) - (x^2 + 3)$.
Since $P(x)$ is a polynomial of degree $4$ with leading coefficient $2$,$Q(x)$ is also a polynomial of degree $4$ with leading coefficient $2$.
Since $P(1)=4, P(2)=7, P(3)=12, P(4)=19$,we have $Q(1)=0, Q(2)=0, Q(3)=0, Q(4)=0$.
Thus,$Q(x) = 2(x-1)(x-2)(x-3)(x-4)$.
Therefore,$P(x) = 2(x-1)(x-2)(x-3)(x-4) + x^2 + 3$.
To find $P(5)$,substitute $x = 5$:
$P(5) = 2(5-1)(5-2)(5-3)(5-4) + (5^2 + 3)$
$P(5) = 2(4)(3)(2)(1) + (25 + 3)$
$P(5) = 2(24) + 28 = 48 + 28 = 76$.

(D) Given inequations are $x^2-1 \leq 0$ and $x^2-x-2 \geq 0$.
For $x^2-1 \leq 0$:
$(x-1)(x+1) \leq 0$
This implies $x \in [-1, 1]$.
For $x^2-x-2 \geq 0$:
$(x-2)(x+1) \geq 0$
This implies $x \in (-\infty, -1] \cup [2, \infty)$.
The intersection of the two sets $x \in [-1, 1]$ and $x \in (-\infty, -1] \cup [2, \infty)$ is the single point $\{-1\}$.
Thus,the set of all values of $x$ is $\{-1\}$.

(C) The given series is $S = 1+i^2+i^4+i^6+\ldots+i^{2024}$.
We know that $i^2 = -1$,$i^4 = 1$,$i^6 = -1$,and so on.
This is a geometric progression with first term $a=1$,common ratio $r=i^2=-1$,and the number of terms $n = \frac{2024-0}{2} + 1 = 1013$.
The sum of a geometric series is $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values: $S = \frac{1((-1)^{1013} - 1)}{-1 - 1} = \frac{-1 - 1}{-2} = \frac{-2}{-2} = 1$.

(B) Let $z_1 = 1+\sqrt{3}i = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$.
Similarly,$z_2 = 1-\sqrt{3}i = 2(\cos \frac{\pi}{3} - i\sin \frac{\pi}{3}) = 2e^{-i\pi/3}$.
Then,$(1+\sqrt{3}i)^{2023} + (1-\sqrt{3}i)^{2023} = (2e^{i\pi/3})^{2023} + (2e^{-i\pi/3})^{2023}$.
$= 2^{2023} (e^{i2023\pi/3} + e^{-i2023\pi/3})$.
Since $2023 = 3 \times 674 + 1$,we have $\frac{2023\pi}{3} = 674\pi + \frac{\pi}{3}$.
Thus,$e^{i2023\pi/3} = e^{i(674\pi + \pi/3)} = e^{i\pi/3} = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.
Similarly,$e^{-i2023\pi/3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Sum $= 2^{2023} (\frac{1}{2} + i\frac{\sqrt{3}}{2} + \frac{1}{2} - i\frac{\sqrt{3}}{2}) = 2^{2023}(1) = 2^{2023}$.

(D) The given expression is an infinite geometric series: $\sum_{n=0}^{\infty}\left(\frac{i}{3}\right)^n = 1 + \frac{i}{3} + \left(\frac{i}{3}\right)^2 + \dots \infty$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$,where $a=1$ and $r=\frac{i}{3}$.
$S = \frac{1}{1-\frac{i}{3}} = \frac{1}{\frac{3-i}{3}} = \frac{3}{3-i}$.
To simplify,multiply the numerator and denominator by the conjugate $(3+i)$:
$S = \frac{3}{3-i} \times \frac{3+i}{3+i} = \frac{3(3+i)}{3^2 - i^2} = \frac{9+3i}{9 - (-1)} = \frac{9+3i}{10}$.

(C) The roots of the equation $x^2+x+1=0$ are $\omega$ and $\omega^2$,where $\omega$ is a complex cube root of unity.
Given $\alpha = \omega$ and $\beta = \omega^2$,we have $\alpha+\beta = -1$ and $\alpha\beta = 1$.
For any $n$,$\alpha^n+\beta^n = \omega^n+\omega^{2n}$.
If $n$ is a multiple of $3$,then $\omega^n = 1$ and $\omega^{2n} = 1$,so $\alpha^n+\beta^n = 1+1 = 2$.
If $n$ is not a multiple of $3$,then $\alpha^n+\beta^n = \omega^n+\omega^{2n} = -1$.
We need to evaluate $S = \sum_{n=1}^{12} (\alpha^n+\beta^n)^2$.
For $n=1, 2, 4, 5, 7, 8, 10, 11$ ($8$ terms),the value is $(-1)^2 = 1$.
For $n=3, 6, 9, 12$ ($4$ terms),the value is $(2)^2 = 4$.
Thus,$S = 8 \times (1) + 4 \times (4) = 8 + 16 = 24$.

(B) Given $z = (\alpha + i\beta)(2 + 7i) = (2\alpha - 7\beta) + i(7\alpha + 2\beta)$.
Since $z$ is purely imaginary,the real part must be zero:
$2\alpha - 7\beta = 0 \Rightarrow 2\alpha = 7\beta$.
Since $\alpha, \beta$ are integers,let $\alpha = 7k$ and $\beta = 2k$ for some non-zero integer $k$.
Then $|z|^2 = (\text{Re}(z))^2 + (\text{Im}(z))^2 = 0^2 + (7\alpha + 2\beta)^2$.
Substituting $\alpha = 7k$ and $\beta = 2k$:
$|z|^2 = (7(7k) + 2(2k))^2 = (49k + 4k)^2 = (53k)^2 = 2809k^2$.
For the minimum non-zero value,we take $k = 1$ (or $k = -1$):
$|z|^2 = 2809(1)^2 = 2809$.

(C) Let $Z = \frac{1+i \cos \theta}{1-2 i \cos \theta}$.
To make $Z$ purely real,we multiply the numerator and denominator by the conjugate of the denominator,$(1+2i \cos \theta)$:
$Z = \frac{(1+i \cos \theta)(1+2i \cos \theta)}{(1-2i \cos \theta)(1+2i \cos \theta)}$
$Z = \frac{1 + 2i \cos \theta + i \cos \theta + 2i^2 \cos^2 \theta}{1 + 4 \cos^2 \theta}$
Since $i^2 = -1$,we have:
$Z = \frac{(1 - 2 \cos^2 \theta) + i(3 \cos \theta)}{1 + 4 \cos^2 \theta}$
For $Z$ to be purely real,the imaginary part must be zero:
$\operatorname{Im}(Z) = \frac{3 \cos \theta}{1 + 4 \cos^2 \theta} = 0$
This implies $\cos \theta = 0$.
If $\cos \theta = 0$,then $\sin^2 \theta = 1 - \cos^2 \theta = 1$.
Substituting these values into the expression:
$\cos^3 \theta + \sin^2 \theta + \cos \theta + 1 = (0)^3 + 1 + 0 + 1 = 2$.

(B) Given $x+iy=\sqrt{\frac{3+i}{1+3i}}$.
Taking the modulus on both sides,we have $|x+iy| = \left|\sqrt{\frac{3+i}{1+3i}}\right|$.
$|x+iy| = \sqrt{\left|\frac{3+i}{1+3i}\right|}$.
Since $|z_1/z_2| = |z_1|/|z_2|$,we get $|x+iy| = \sqrt{\frac{|3+i|}{|1+3i|}} = \sqrt{\frac{\sqrt{3^2+1^2}}{\sqrt{1^2+3^2}}} = \sqrt{\frac{\sqrt{10}}{\sqrt{10}}} = \sqrt{1} = 1$.
We know that $|x+iy| = \sqrt{x^2+y^2}$,so $\sqrt{x^2+y^2} = 1$.
Squaring both sides,$x^2+y^2 = 1$.
Therefore,$\left(x^2+y^2\right)^2 = 1^2 = 1$.

(C) Let $Z = \sqrt{-5-12 i} + \sqrt{7+24 i}$.
We use the formula $\sqrt{a+ib} = \pm \left( \sqrt{\frac{|Z|+a}{2}} + i \frac{b}{|b|} \sqrt{\frac{|Z|-a}{2}} \right)$.
For $\sqrt{-5-12 i}$,$|Z| = \sqrt{(-5)^2 + (-12)^2} = 13$. Thus,$\sqrt{-5-12 i} = \pm \left( \sqrt{\frac{13-5}{2}} - i \sqrt{\frac{13+5}{2}} \right) = \pm(2-3i)$.
For $\sqrt{7+24 i}$,$|Z| = \sqrt{7^2 + 24^2} = 25$. Thus,$\sqrt{7+24 i} = \pm \left( \sqrt{\frac{25+7}{2}} + i \sqrt{\frac{25-7}{2}} \right) = \pm(4+3i)$.
Given $Z = \pm(2-3i) \pm(4+3i)$.
Possible values for $Z$ are $(2-3i) + (4+3i) = 6$,$(2-3i) - (4+3i) = -2-6i$,$-(2-3i) + (4+3i) = 2+6i$,and $-(2-3i) - (4+3i) = -6$.
Since $k$ is a negative real number,$k = -6$.

(C) Let $z = \frac{3+2 i \sin \theta}{1-2 i \sin \theta}$.
To make $z$ purely imaginary,its real part must be zero.
Multiply the numerator and denominator by the conjugate of the denominator $(1+2 i \sin \theta)$:
$z = \frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1-2 i \sin \theta)(1+2 i \sin \theta)}$
$z = \frac{3 + 6 i \sin \theta + 2 i \sin \theta + 4 i^2 \sin^2 \theta}{1 + 4 \sin^2 \theta}$
Since $i^2 = -1$,we have:
$z = \frac{(3 - 4 \sin^2 \theta) + i(8 \sin \theta)}{1 + 4 \sin^2 \theta}$
For $z$ to be purely imaginary,$\text{Re}(z) = 0$:
$\frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0$
$3 - 4 \sin^2 \theta = 0$
$\sin^2 \theta = \frac{3}{4}$
$\sin \theta = \pm \frac{\sqrt{3}}{2}$
This implies $\theta = n \pi \pm \frac{\pi}{3}$.

(D) The condition $|Z_1+Z_2|=|Z_1|+|Z_2|$ represents the triangle inequality becoming an equality.
This occurs if and only if the vectors representing $Z_1$ and $Z_2$ in the complex plane are in the same direction.
Therefore,the arguments (amplitudes) of $Z_1$ and $Z_2$ must be equal,i.e.,$\text{arg}(Z_1) = \text{arg}(Z_2)$.
Thus,the difference in their amplitudes is $\text{arg}(Z_1) - \text{arg}(Z_2) = 0$.

(D) Let $Z = \sin \frac{6 \pi}{5} + i(1 + \cos \frac{6 \pi}{5})$.
Using trigonometric identities $\sin 2\theta = 2\sin\theta\cos\theta$ and $1 + \cos 2\theta = 2\cos^2\theta$ with $\theta = \frac{3\pi}{5}$:
$Z = 2\sin \frac{3\pi}{5}\cos \frac{3\pi}{5} + i(2\cos^2 \frac{3\pi}{5})$
$Z = 2\cos \frac{3\pi}{5} (\sin \frac{3\pi}{5} + i\cos \frac{3\pi}{5})$
Since $\frac{3\pi}{5} = \frac{\pi}{2} + \frac{\pi}{10}$,we have $\sin \frac{3\pi}{5} = \cos \frac{\pi}{10}$ and $\cos \frac{3\pi}{5} = -\sin \frac{\pi}{10}$.
$Z = 2\cos \frac{3\pi}{5} (\cos \frac{\pi}{10} - i\sin \frac{\pi}{10})$
$Z = 2\cos \frac{3\pi}{5} (\cos(-\frac{\pi}{10}) + i\sin(-\frac{\pi}{10}))$
Since $\cos \frac{3\pi}{5} < 0$,the argument is $\pi - \frac{\pi}{10} = \frac{9\pi}{10}$.

(A) Let $z = x+iy$. The given expression is $\frac{(x-2)+i(y-1)}{(x+1)+i(y-2)}$.
To make this purely real,the imaginary part of the product of the numerator and the conjugate of the denominator must be zero.
Multiplying by the conjugate of the denominator: $\frac{[(x-2)+i(y-1)][(x+1)-i(y-2)]}{(x+1)^2+(y-2)^2}$.
The imaginary part is $(x+1)(y-1) - (x-2)(y-2) = 0$.
Expanding this: $(xy - x + y - 1) - (xy - 2x - 2y + 4) = 0$.
$xy - x + y - 1 - xy + 2x + 2y - 4 = 0$.
$x + 3y - 5 = 0$.
Since the denominator cannot be zero,$z \neq -(1-2i)$,which means $x \neq -1$ and $y \neq 2$.
Thus,the locus is the line $x+3y-5=0$ excluding the point $(-1,2)$.

(C) Given the equation $z^2-i=0$,we have $z^2=i$.
Expressing $i$ in polar form,$i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}$.
The roots are $z = \pm e^{i\frac{\pi}{4}}$.
Thus,the roots are $z_1 = e^{i\frac{\pi}{4}}$ and $z_2 = e^{i(\frac{\pi}{4} + \pi)} = e^{i\frac{5\pi}{4}}$.
Let $\alpha = e^{i\frac{\pi}{4}}$ and $\beta = e^{i\frac{5\pi}{4}}$.
Then $\operatorname{Arg} \alpha = \frac{\pi}{4}$ and $\operatorname{Arg} \beta = \frac{5\pi}{4}$.
Therefore,$|\operatorname{Arg} \beta - \operatorname{Arg} \alpha| = |\frac{5\pi}{4} - \frac{\pi}{4}| = |\pi| = \pi$.

(A) Let $Z = \frac{(1+i)^{2025}}{(1-i)^{2022}}$.
We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.
Substituting these values:
$Z = \frac{(\sqrt{2} e^{i\pi/4})^{2025}}{(\sqrt{2} e^{-i\pi/4})^{2022}}$
$Z = \frac{(\sqrt{2})^{2025} e^{i(2025\pi/4)}}{(\sqrt{2})^{2022} e^{-i(2022\pi/4)}}$
$Z = (\sqrt{2})^3 e^{i(2025\pi/4 + 2022\pi/4)}$
$Z = 2\sqrt{2} e^{i(4047\pi/4)}$
Since $4047\pi/4 = 1011\pi + 3\pi/4$,the principal argument is $\operatorname{Arg}(Z) = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$.

(C) The equation of the family of parabolas with focus at the origin and $X$-axis as its axis is given by $y^2 = 4a(x+a) = 4ax + 4a^2$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $a = \frac{1}{2} y \frac{dy}{dx}$.
Substituting the value of $a$ into the original equation:
$y^2 = 4 \left( \frac{1}{2} y \frac{dy}{dx} \right) x + 4 \left( \frac{1}{2} y \frac{dy}{dx} \right)^2$
$y^2 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$.
The highest order derivative is $\frac{dy}{dx}$,so the order $m = 1$.
The power of the highest order derivative is $2$,so the degree $n = 2$.
Therefore,$mn - m + n = (1 \times 2) - 1 + 2 = 2 - 1 + 2 = 3$.

(A) Let the plane be $ax + by + cz + d = 0$.
Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, -1, 3)$,the normal vector $\vec{n}$ to the plane is the vector from the origin to the foot of the perpendicular,which is $\vec{n} = (2 - 0)\hat{i} + (-1 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} - \hat{j} + 3\hat{k}$.
Thus,the equation of the plane is $2x - y + 3z = D$.
Since the point $(2, -1, 3)$ lies on the plane,we substitute these coordinates into the equation:
$2(2) - (-1) + 3(3) = D$
$4 + 1 + 9 = D$
$D = 14$.
Therefore,the equation of the plane is $2x - y + 3z - 14 = 0$.

(B) Let $y = \frac{x+3}{(x-1)(x+2)}$.
$(x-1)(x+2)y = x+3$
$y(x^2+x-2) = x+3$
$yx^2 + (y-1)x - (2y+3) = 0$.
Since $x \in R$,the discriminant $D \geq 0$:
$(y-1)^2 + 4y(2y+3) \geq 0$
$y^2 - 2y + 1 + 8y^2 + 12y \geq 0$
$9y^2 + 10y + 1 \geq 0$
$(9y+1)(y+1) \geq 0$.
The solution is $y \in (-\infty, -1] \cup [-\frac{1}{9}, \infty)$.
Thus,the range is $R - (-1, -\frac{1}{9})$.
Comparing with $R - (\alpha, \beta)$,we get $\alpha = -1$ and $\beta = -\frac{1}{9}$.
The line equation is $-x - \frac{1}{9}y + 1 = 0$,or $x + \frac{1}{9}y = 1$.
This can be written as $\frac{x}{1} + \frac{y}{9} = 1$.
The intercepts are $1$ and $9$.
The sum of the intercepts is $1 + 9 = 10$.

(C) Let $L = \lim _{n \rightarrow \infty}\left[\prod_{r=1}^n \left(1+\frac{r^2}{n^2}\right)\right]^{1 / n}$.
Taking the natural logarithm on both sides:
$\log L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}\right)$.
This is a Riemann sum,which can be expressed as a definite integral:
$\log L = \int_0^1 \log(1+x^2) dx$.
Using integration by parts,let $u = \log(1+x^2)$ and $dv = dx$:
$\int \log(1+x^2) dx = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx$.
$= x \log(1+x^2) - 2 \int \frac{x^2+1-1}{1+x^2} dx = x \log(1+x^2) - 2 \int \left(1 - \frac{1}{1+x^2}\right) dx$.
$= x \log(1+x^2) - 2x + 2 \tan^{-1} x$.
Evaluating from $0$ to $1$:
$\log L = [1 \cdot \log(2) - 2(1) + 2 \tan^{-1}(1)] - [0 - 0 + 0] = \log 2 - 2 + 2(\frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2}$.
$\log L = \log 2 + \frac{\pi-4}{2} = \log 2 + \log e^{\frac{\pi-4}{2}} = \log \left(2 e^{\frac{\pi-4}{2}}\right)$.
Therefore,$L = 2 e^{\frac{\pi-4}{2}}$.

(D) The given curve is $y = 2 - x - 3x^2$. The region is bounded by the $X$-axis $(y = 0)$,the $Y$-axis $(x = 0)$,and the line $x = -2$.
First,find the roots of the curve by setting $y = 0$:
$2 - x - 3x^2 = 0 \implies 3x^2 + x - 2 = 0 \implies (3x - 2)(x + 1) = 0$.
The roots are $x = -1$ and $x = 2/3$.
For $x \in [-2, -1]$,$y = 2 - x - 3x^2$ is negative (e.g.,at $x = -1.5$,$y = 2 + 1.5 - 3(2.25) = 3.5 - 6.75 = -3.25 < 0$).
For $x \in [-1, 0]$,$y = 2 - x - 3x^2$ is positive (e.g.,at $x = -0.5$,$y = 2 + 0.5 - 3(0.25) = 2.5 - 0.75 = 1.75 > 0$).
The total area is given by:
Area $= \int_{-2}^{-1} -(2 - x - 3x^2) dx + \int_{-1}^{0} (2 - x - 3x^2) dx$
$= \int_{-2}^{-1} (3x^2 + x - 2) dx + \int_{-1}^{0} (2 - x - 3x^2) dx$
$= [x^3 + \frac{x^2}{2} - 2x]_{-2}^{-1} + [2x - \frac{x^2}{2} - x^3]_{-1}^{0}$
$= [(-1 + 0.5 + 2) - (-8 + 2 + 4)] + [(0) - (-2 - 0.5 + 1)]$
$= [1.5 - (-2)] + [0 - (-1.5)]$
$= 3.5 + 1.5 = 5$.
Thus,the correct option is $D$.

(D) The required area $A$ is given by the integral $A = \int_0^{2\pi} |\sin 2x| \, dx$.
Since the function $f(x) = |\sin 2x|$ is periodic with period $\frac{\pi}{2}$,the area over $[0, 2\pi]$ consists of $4$ identical humps,each over an interval of length $\frac{\pi}{2}$.
Thus,$A = 4 \int_0^{\pi/2} \sin 2x \, dx$.
Evaluating the integral: $A = 4 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2}$.
$A = 4 \left( -\frac{1}{2} (\cos \pi - \cos 0) \right) = 4 \left( -\frac{1}{2} (-1 - 1) \right) = 4 \left( -\frac{1}{2} (-2) \right) = 4(1) = 4$.
Therefore,the area is $4$ square units.
Hence,option $D$ is correct.

(A) To find the area bounded by the curve $y=2x-x^2$ and the line $y=-x$,we first find their points of intersection by setting $2x-x^2 = -x$.
$2x-x^2+x = 0$
$3x-x^2 = 0$
$x(3-x) = 0$
So,the points of intersection are $x=0$ and $x=3$.
In the interval $[0, 3]$,the curve $y=2x-x^2$ lies above the line $y=-x$.
The area is given by the integral:
$\text{Area} = \int_0^3 ((2x-x^2) - (-x)) dx$
$= \int_0^3 (3x-x^2) dx$
$= \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_0^3$
$= \left( \frac{3(3)^2}{2} - \frac{(3)^3}{3} \right) - (0 - 0)$
$= \frac{27}{2} - \frac{27}{3}$
$= \frac{27}{2} - 9$
$= \frac{27-18}{2} = \frac{9}{2} \text{ sq. units.}$

(C) Given $A = \begin{bmatrix} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & -2 & 4 \\ 2 & 2 & -1 \\ -2 & 0 & 3 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} (1-2-1) & (2+0-2) & (-1+4+0) \\ (-1+0+2) & (-2+0+4) & (1+0+0) \\ (1-2+0) & (2+0+0) & (-1+4+0) \end{bmatrix} = \begin{bmatrix} -2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3 \end{bmatrix}$.
Next,calculate $A+B$:
$A+B = \begin{bmatrix} 1-3 & 2-2 & -1+4 \\ -1+2 & 0+2 & 2-1 \\ 1-2 & 2+0 & 0+3 \end{bmatrix} = \begin{bmatrix} -2 & 0 & 3 \\ 1 & 2 & 1 \\ -1 & 2 & 3 \end{bmatrix}$.
Comparing both results,we see that $A^2 = A+B$.

(A) Given the equation: $I+P+P^2+\ldots+P^{n}=0$ ... $(i)$
We can rewrite this as: $I+P+P^2+\ldots+P^{n-1} = -P^n$ ... $(ii)$
Now,multiply equation $(i)$ by $P^{-1}$ on the left side:
$P^{-1}(I+P+P^2+\ldots+P^{n}) = P^{-1}(0)$
$P^{-1}I + P^{-1}P + P^{-1}P^2 + \ldots + P^{-1}P^n = 0$
$P^{-1} + I + P + \ldots + P^{n-1} = 0$
From equation $(ii)$,we know that $I+P+\ldots+P^{n-1} = -P^n$.
Substituting this into the equation:
$P^{-1} + (-P^n) = 0$
Therefore,$P^{-1} = P^n$.

(D) Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
Given $AB = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} a & 2a+3b \\ c & 2c+3d \end{bmatrix}$.
Since $AB$ is a scalar matrix,$c = 0$ and $2a+3b = 0$,and $a = 2c+3d = 3d$.
Thus,$a = 3d$ and $b = -\frac{2}{3}a = -2d$.
So,$A = \begin{bmatrix} 3d & -2d \\ 0 & d \end{bmatrix}$.
Given $\det(3A) = 27$,we have $3^2 \det(A) = 27$,so $\det(A) = 3$.
$\det(A) = (3d)(d) - 0 = 3d^2 = 3$,which implies $d^2 = 1$. Assuming $d=1$,we get $A = \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix}$.
Then $A^2 = \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & -8 \\ 0 & 1 \end{bmatrix}$.
$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{3} \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1/3 & 2/3 \\ 0 & 1 \end{bmatrix}$.
Therefore,$3A^{-1} + A^2 = 3 \begin{bmatrix} 1/3 & 2/3 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 9 & -8 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} + \begin{bmatrix} 9 & -8 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 10 & -6 \\ 0 & 4 \end{bmatrix}$.

(A) Given $A = \begin{bmatrix} b & a & 0 \\ c & 0 & b \\ a & a & b \end{bmatrix}$ and $B = \begin{bmatrix} 0 & a & b \\ b & 0 & c \\ b & a & a \end{bmatrix}$.
Performing matrix multiplication $AB$:
$AB = \begin{bmatrix} b(0)+a(b)+0(b) & b(a)+a(0)+0(a) & b(b)+a(c)+0(a) \\ c(0)+0(b)+b(b) & c(a)+0(0)+b(a) & c(b)+0(c)+b(a) \\ a(0)+a(b)+b(b) & a(a)+a(0)+b(a) & a(b)+a(c)+b(a) \end{bmatrix} = \begin{bmatrix} ab & ab & b^2+ac \\ b^2 & ac+ab & bc+ab \\ ab+b^2 & a^2+ab & 2ab+ac \end{bmatrix}$.
Equating this to the given matrix $\begin{bmatrix} 2 & 2 & 7 \\ 1 & 8 & 5 \\ 3 & 6 & 10 \end{bmatrix}$:
From the element at $(2,1)$,$b^2 = 1$.
From the element at $(1,1)$,$ab = 2$.
From the element at $(1,3)$,$b^2+ac = 7 \implies 1+ac = 7 \implies ac = 6$.
Since $ab=2$ and $b^2=1$,we have $a = 2/b$. If $b=1$,$a=2$. If $b=-1$,$a=-2$.
Case $1$: $b=1, a=2$. Then $ac=6 \implies 2c=6 \implies c=3$.
Case $2$: $b=-1, a=-2$. Then $ac=6 \implies -2c=6 \implies c=-3$.
In both cases,$a^2+b^2+c^2 = (\pm 2)^2 + (\pm 1)^2 + (\pm 3)^2 = 4+1+9 = 14$.

(B) For matrix $A$ to be symmetric,$A = A^T$,which implies $a=b$,$c=d$,and $3=3$. Thus,$A = \begin{bmatrix} 1 & a & 3 \\ a & 2 & c \\ 3 & c & 4 \end{bmatrix}$.
For matrix $B$ to be skew-symmetric,$B = -B^T$,which implies diagonal elements are $0$. From $B_{13} = -B_{31}$,we get $b = -6$. From $B_{23} = -B_{32}$,we get $-7 = -c$,so $c = 7$.
Substituting these values,we get $a = b = -6$ and $d = c = 7$.
So,$A = \begin{bmatrix} 1 & -6 & 3 \\ -6 & 2 & 7 \\ 3 & 7 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 5 & -6 \\ -5 & 0 & -7 \\ 6 & 7 & 0 \end{bmatrix}$.
Now,calculating the product $AB$:
$AB = \begin{bmatrix} 1 & -6 & 3 \\ -6 & 2 & 7 \\ 3 & 7 & 4 \end{bmatrix} \begin{bmatrix} 0 & 5 & -6 \\ -5 & 0 & -7 \\ 6 & 7 & 0 \end{bmatrix} = \begin{bmatrix} 0+30+18 & 5+0+21 & -6+42+0 \\ 0-10+42 & -30+0+49 & 36-14+0 \\ 0-35+24 & 15+0+28 & -18-49+0 \end{bmatrix} = \begin{bmatrix} 48 & 26 & 36 \\ 32 & 19 & 22 \\ -11 & 43 & -67 \end{bmatrix}$.

(C) Given $A = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A^2 = 0$,all higher powers $A^n = 0$ for $n \geq 2$.
Given $f(x) = x + x^2 + x^3 + \ldots + x^{2023}$,we have $f(A) = A + A^2 + A^3 + \ldots + A^{2023}$.
Substituting $A^n = 0$ for $n \geq 2$,we get $f(A) = A + 0 + 0 + \ldots + 0 = A$.
Therefore,$f(A) + I = A + I$.
$f(A) + I = \begin{bmatrix} 0 & 3 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} 1(1)+2(3) & 1(2)+2(5) \\ 3(1)+5(3) & 3(2)+5(5) \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 18 & 31 \end{bmatrix}$.
Substitute $A^2$ and $A$ into the equation $\alpha A^2 - \beta A = 2I$:
$\alpha \begin{bmatrix} 7 & 12 \\ 18 & 31 \end{bmatrix} - \beta \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This gives the matrix equation:
$\begin{bmatrix} 7\alpha - \beta & 12\alpha - 2\beta \\ 18\alpha - 3\beta & 31\alpha - 5\beta \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$.
Comparing the elements,we get:
$7\alpha - \beta = 2$ $(i)$
$12\alpha - 2\beta = 0 \Rightarrow 6\alpha - \beta = 0$ $(ii)$
Subtracting $(ii)$ from $(i)$ gives $\alpha = 2$.
Substituting $\alpha = 2$ into $(ii)$,we get $6(2) - \beta = 0 \Rightarrow \beta = 12$.
Finally,$\alpha^2 + \beta = (2)^2 + 12 = 4 + 12 = 16$.

(B) We know that for any square matrix $P$,the trace of its transpose is equal to the trace of the matrix itself,i.e.,$\operatorname{Tr}(P) = \operatorname{Tr}(P^T)$.
Also,$\operatorname{Tr}(P - P^T) = \operatorname{Tr}(P) - \operatorname{Tr}(P^T) = 0$ and $\operatorname{Tr}(P + P^T) = \operatorname{Tr}(P) + \operatorname{Tr}(P^T) = 2\operatorname{Tr}(P)$.
Substituting these into the given equation:
$0 + 2\operatorname{Tr}(P) + \frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P)} + \operatorname{Tr}(P) \times \operatorname{Tr}(P) = 0$
Since $\operatorname{Tr}(P) \neq 0$,we have $\frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P)} = 1$.
Thus,the equation becomes: $2\operatorname{Tr}(P) + 1 + (\operatorname{Tr}(P))^2 = 0$.
This is a quadratic equation in terms of $\operatorname{Tr}(P)$:
$(\operatorname{Tr}(P))^2 + 2\operatorname{Tr}(P) + 1 = 0$
$(\operatorname{Tr}(P) + 1)^2 = 0$
$\operatorname{Tr}(P) = -1$.

(A) Given the equation: $(AB+BA)^{T}+(AB-BA)^{T}=2BA$
Using the property $(X+Y)^T = X^T + Y^T$,we get:
$(AB)^T + (BA)^T + (AB)^T - (BA)^T = 2BA$
$2(AB)^T = 2BA$
$(AB)^T = BA$
$B^T A^T = BA$
If $A$ and $B$ are symmetric,then $A^T = A$ and $B^T = B$. Substituting these,we get $BA = BA$,which is true.
If $A$ and $B$ are skew-symmetric,then $A^T = -A$ and $B^T = -B$. Substituting these,we get $(-B)(-A) = BA$,which is $BA = BA$,which is also true.
Thus,the condition holds if $A$ and $B$ are both symmetric or both skew-symmetric. Since the options provided are limited,the most appropriate conclusion based on the derivation is that the property holds for both cases.

(D) Given matrices: $P_{2 \times 3}$,$X_{4 \times 3}$,$Y_{4 \times 3}$.
$X^T$ has order $3 \times 4$.
$P^T$ has order $3 \times 2$.
$X^T Y$ has order $(3 \times 4) \times (4 \times 3) = 3 \times 3$.
$(X^T Y)^{-1}$ has order $3 \times 3$.
$P(X^T Y)^{-1}$ has order $(2 \times 3) \times (3 \times 3) = 2 \times 3$.
$P(X^T Y)^{-1} P^T$ has order $(2 \times 3) \times (3 \times 2) = 2 \times 2$.
Finally,the transpose of a $2 \times 2$ matrix remains $2 \times 2$.

(A) Given that $A$ is a symmetric matrix,we have $A^T = A$.
If the inverse $A^{-1}$ exists,we use the property $(A^{-1})^T = (A^T)^{-1}$.
Substituting $A^T = A$ into the equation,we get $(A^{-1})^T = (A)^{-1} = A^{-1}$.
Since the transpose of $A^{-1}$ is equal to $A^{-1}$,it follows that $A^{-1}$ is a symmetric matrix,provided it exists.

(B) The total number of $3 \times 3$ matrices with entries from $\{-1, 0, 1\}$ is $3^9 = 19683$. Since we are considering only non-zero matrices,the total number of possible matrices is $3^9 - 1 = 19682$.
$A$ skew-symmetric matrix $A$ satisfies $A^T = -A$. For a $3 \times 3$ matrix,this implies the diagonal elements must be $0$. The matrix takes the form:
$\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$
The entries $a, b, c$ can each be chosen from $\{-1, 0, 1\}$. Thus,there are $3^3 = 27$ such matrices.
Excluding the zero matrix (where $a=b=c=0$),the number of non-zero skew-symmetric matrices is $27 - 1 = 26$.
The probability is $\frac{26}{19682} = \frac{1}{757}$.

(A) Given $A = \begin{bmatrix} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{bmatrix}$ and $\det A = 190$.
Expanding the determinant along the first row:
$\det A = k(k^2 - 10) - 5(-2k - 25) + 2(4 + 5k) = 190$
$k^3 - 10k + 10k + 125 + 8 + 10k = 190$
$k^3 + 10k - 57 = 0$
By testing values,for $k=3$: $27 + 30 - 57 = 0$. Thus,$k=3$.
Substituting $k=3$,$A = \begin{bmatrix} 3 & 5 & 2 \\ 2 & -3 & 5 \\ 5 & 2 & -3 \end{bmatrix}$.
The cofactor matrix $C$ is calculated as:
$C_{11} = ((-3)(-3) - (5)(2)) = 9 - 10 = -1$
$C_{12} = -((2)(-3) - (5)(5)) = -(-6 - 25) = 31$
$C_{13} = ((2)(2) - (-3)(5)) = 4 + 15 = 19$
$C_{21} = -((5)(-3) - (2)(2)) = -(-15 - 4) = 19$
$C_{22} = ((3)(-3) - (2)(5)) = -9 - 10 = -19$
$C_{23} = -((3)(2) - (5)(5)) = -(6 - 25) = 19$
$C_{31} = ((5)(5) - (-3)(2)) = 25 + 6 = 31$
$C_{32} = -((3)(5) - (2)(2)) = -(15 - 4) = -11$
$C_{33} = ((3)(-3) - (5)(2)) = -9 - 10 = -19$
The adjoint matrix is the transpose of the cofactor matrix:
$\operatorname{Adj} A = C^T = \begin{bmatrix} -1 & 19 & 31 \\ 31 & -19 & -11 \\ 19 & 19 & -19 \end{bmatrix}$.

(C) We know that for a square matrix $M$ of order $n$,$|\operatorname{adj} M| = |M|^{n-1}$.
Given $A$ is a square matrix of order $n=3$.
First,consider the matrix $M = A^2$. The order of $M$ is $3$.
Then,$|\operatorname{adj} M| = |M|^{3-1} = |M|^2 = (|A|^2)^2 = |A|^4$.
Now,we need to find $|\operatorname{adj}(\operatorname{adj} A^2)|$.
Let $K = \operatorname{adj} A^2$. Then $|K| = |A|^4$.
Using the property $|\operatorname{adj} K| = |K|^{n-1}$,where $n=3$:
$|\operatorname{adj} K| = |K|^{3-1} = |K|^2$.
Substituting $|K| = |A|^4$:
$|\operatorname{adj}(\operatorname{adj} A^2)| = (|A|^4)^2 = |A|^8$.

(C) Let $A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}$. The adjoint of a matrix is the transpose of its cofactor matrix,$\text{adj}(A) = [C_{ij}]^T$.
The cofactors are calculated as follows:
$C_{11} = +\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = 1 - (-4) = 5$
$C_{12} = -\begin{vmatrix} -1 & -2 \\ 0 & 1 \end{vmatrix} = -(-1 - 0) = 1$
$C_{13} = +\begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} = -2 - 0 = -2$
$C_{21} = -\begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix} = -(0 - 4) = 4$
$C_{22} = +\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$
$C_{23} = -\begin{vmatrix} 1 & 0 \\ 0 & 2 \end{vmatrix} = -(2 - 0) = -2$
$C_{31} = +\begin{vmatrix} 0 & 2 \\ 1 & -2 \end{vmatrix} = 0 - 2 = -2$
$C_{32} = -\begin{vmatrix} 1 & 2 \\ -1 & -2 \end{vmatrix} = -(-2 - (-2)) = 0$
$C_{33} = +\begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = 1 - 0 = 1$
Thus,$\text{adj}(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 5 & 4 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{bmatrix}$.
Comparing this with the given matrix,we get $m = 4$ and $n = 1$.
Therefore,$m+n = 4+1 = 5$.

(B) Given $(A-2I)(A-3I) = O$.
Expanding the expression,we get $A^2 - 3A - 2A + 6I^2 = O$.
This simplifies to $A^2 - 5A + 6I = O$.
Rearranging to isolate $I$,we have $6I = 5A - A^2$.
Factoring out $A$,we get $6I = A(5I - A)$.
Multiplying by $A^{-1}$ on both sides,we get $6A^{-1} = 5I - A$,which implies $A^{-1} = \frac{5I - A}{6}$.
Now,substitute this into the expression $\frac{1}{5}A + \frac{6}{5}A^{-1}$:
$\frac{1}{5}A + \frac{6}{5} \left( \frac{5I - A}{6} \right) = \frac{1}{5}A + \frac{5I - A}{5} = \frac{1}{5}A + I - \frac{1}{5}A = I$.

(C) First,we find the determinant of $A$:
$|A| = -1(5 - 8) - (-3)(0 - 6) + (-2)(0 - 3) = -1(-3) + 3(-6) - 2(-3) = 3 - 18 + 6 = -9$.
Next,we find the matrix of cofactors $C_{ij}$:
$C_{11} = +(5 - 8) = -3$,$C_{12} = -(0 - 6) = 6$,$C_{13} = +(0 - 3) = -3$.
$C_{21} = -(-15 - (-8)) = -(-7) = 7$,$C_{22} = +(-5 - (-6)) = 1$,$C_{23} = -(-4 - (-9)) = -5$.
$C_{31} = +(-6 - (-2)) = -4$,$C_{32} = -(-2 - 0) = 2$,$C_{33} = +(-1 - 0) = -1$.
The adjoint matrix is the transpose of the cofactor matrix:
$\text{adj } A = \begin{bmatrix} -3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{|A|} \text{adj } A = -\frac{1}{9} \begin{bmatrix} -3 & 7 & -4 \\ 6 & 1 & 2 \\ -3 & -5 & -1 \end{bmatrix} = \begin{bmatrix} 1/3 & -7/9 & 4/9 \\ -6/9 & -1/9 & -2/9 \\ 3/9 & 5/9 & 1/9 \end{bmatrix}$.
Comparing this to the given form,we have $a_1 = 1/3$,$c_2 = 5/9$,and $b_3 = -2/9$.
Therefore,$a_1 + c_2 + b_3 = \frac{3}{9} + \frac{5}{9} - \frac{2}{9} = \frac{6}{9} = \frac{2}{3}$.

(D) Given $\operatorname{det}(A^2) = 25$.
Using the property $\operatorname{det}(A^n) = (\operatorname{det}(A))^n$,we have $(\operatorname{det}(A))^2 = 25$.
Now,calculate the determinant of matrix $A$:
$|A| = \begin{vmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{vmatrix}$.
Since this is an upper triangular matrix,the determinant is the product of the diagonal elements:
$|A| = 5 \times \alpha \times 5 = 25\alpha$.
Substituting this into the equation:
$(25\alpha)^2 = 25$.
$625\alpha^2 = 25$.
$\alpha^2 = \frac{25}{625} = \frac{1}{25}$.
Taking the square root on both sides:
$|\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5}$.

(D) To evaluate the determinant $\left|\begin{array}{ccc}\sqrt{3} & 2 \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5\end{array}\right|$,we expand along the first row:
$\Delta = \sqrt{3}(5 \times 5 - \sqrt{15} \times \sqrt{10}) - 2\sqrt{5}(\sqrt{15} \times 5 - 3 \times \sqrt{10}) + \sqrt{5}(\sqrt{15} \times \sqrt{15} - 3 \times 5)$
$\Delta = \sqrt{3}(25 - \sqrt{150}) - 2\sqrt{5}(5\sqrt{15} - 3\sqrt{10}) + \sqrt{5}(15 - 15)$
Since $\sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6}$,we have:
$\Delta = \sqrt{3}(25 - 5\sqrt{6}) - 2\sqrt{5}(5\sqrt{15} - 3\sqrt{10}) + 0$
$\Delta = 25\sqrt{3} - 5\sqrt{18} - 10\sqrt{75} + 6\sqrt{50}$
$\Delta = 25\sqrt{3} - 5(3\sqrt{2}) - 10(5\sqrt{3}) + 6(5\sqrt{2})$
$\Delta = 25\sqrt{3} - 15\sqrt{2} - 50\sqrt{3} + 30\sqrt{2}$
$\Delta = 15\sqrt{2} - 25\sqrt{3}$

(D) Given that $\omega$ is a cube root of unity,we know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Simplifying the powers of $\omega$ in the determinant:
$\omega^9 = (\omega^3)^3 = 1^3 = 1$
$\omega^{27} = (\omega^3)^9 = 1^9 = 1$
$\omega^{31} = \omega^{30} \cdot \omega = (\omega^3)^{10} \cdot \omega = 1^{10} \cdot \omega = \omega$
$\omega^{17} = \omega^{15} \cdot \omega^2 = (\omega^3)^5 \cdot \omega^2 = 1^5 \cdot \omega^2 = \omega^2$
$\omega^{30} = (\omega^3)^{10} = 1^{10} = 1$
$\omega^{41} = \omega^{39} \cdot \omega^2 = (\omega^3)^{13} \cdot \omega^2 = 1^{13} \cdot \omega^2 = \omega^2$
$\omega^{19} = \omega^{18} \cdot \omega = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$
Substituting these values into the determinant:
$\Delta = \left|\begin{array}{ccc}\omega+\omega^2 & \omega^2+1 & 1+\omega \\ 1+\omega & \omega+\omega^2 & \omega^2+1 \\ 1+\omega^2 & \omega^2+\omega & \omega+1\end{array}\right|$
Since $1+\omega+\omega^2 = 0$,we have $\omega+\omega^2 = -1$,$\omega^2+1 = -\omega$,and $1+\omega = -\omega^2$.
$\Delta = \left|\begin{array}{ccc}-1 & -\omega & -\omega^2 \\ -\omega^2 & -1 & -\omega \\ -\omega & -\omega^2 & -1\end{array}\right|$
Taking $-1$ common from each row:
$\Delta = (-1)^3 \left|\begin{array}{ccc}1 & \omega & \omega^2 \\ \omega^2 & 1 & \omega \\ \omega & \omega^2 & 1\end{array}\right|$
Using the property of cyclic determinants,this determinant is equal to $1(1-\omega^3) - \omega(\omega^2-\omega^2) + \omega^2(\omega^4-\omega) = 1(1-1) - 0 + \omega^2(\omega-\omega) = 0$.
Thus,the value is $0$.

(A) Given $f(x) = \left| \begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{array} \right|$.
Applying the column operation $C_3 \rightarrow C_3 - C_2$:
$f(x) = \left| \begin{array}{ccc} 1 & x & (x+1) - x \\ 2x & x(x-1) & (x+1)x - x(x-1) \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) - x(x-1)(x-2) \end{array} \right|$.
Simplifying the third column:
$(x+1) - x = 1$.
$(x+1)x - x(x-1) = x^2 + x - x^2 + x = 2x$.
$(x+1)x(x-1) - x(x-1)(x-2) = x(x-1) [ (x+1) - (x-2) ] = x(x-1) [ 3 ] = 3x(x-1)$.
Thus,$f(x) = \left| \begin{array}{ccc} 1 & x & 1 \\ 2x & x(x-1) & 2x \\ 3x(x-1) & x(x-1)(x-2) & 3x(x-1) \end{array} \right|$.
Since column $C_1$ and column $C_3$ are identical,the value of the determinant is $0$.
Therefore,$f(x) = 0$ for all $x$,which implies $f(100) = 0$.

(D) We evaluate the two determinants separately.
First,consider the determinant $D_1 = \left|\begin{array}{lll}2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3\end{array}\right|$.
Using the property $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right| = 3abc - a^3 - b^3 - c^3$,we get:
$D_1 = 3(2)(3)(5) - 2^3 - 3^3 - 5^3$
$D_1 = 90 - 8 - 27 - 125 = -70$.
Next,consider the determinant $D_2 = \left|\begin{array}{ccc}1 & 1 & 1 \\ 7 & 11 & 13 \\ 49 & 121 & 169\end{array}\right|$.
This is a Vandermonde determinant of the form $\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{array}\right| = (a-b)(b-c)(c-a)$,where $a=7, b=11, c=13$.
$D_2 = (7-11)(11-13)(13-7)$
$D_2 = (-4)(-2)(6) = 48$.
The required sum is $D_1 + D_2 = -70 + 48 = -22$.

(A) The given system of equations is:
$x + ky + 3z = -2$
$4x + 3y + kz = 14$
$2x + y + 2z = 3$
For the system to be solved by the matrix inversion method,the coefficient matrix $A$ must be invertible,which means the determinant $|A|$ must be non-zero $(|A| \neq 0)$.
$|A| = \begin{vmatrix} 1 & k & 3 \\ 4 & 3 & k \\ 2 & 1 & 2 \end{vmatrix}$
Expanding along the first row:
$|A| = 1(3 \times 2 - k \times 1) - k(4 \times 2 - k \times 2) + 3(4 \times 1 - 3 \times 2)$
$|A| = 1(6 - k) - k(8 - 2k) + 3(4 - 6)$
$|A| = 6 - k - 8k + 2k^2 - 6$
$|A| = 2k^2 - 9k$
Since $|A| \neq 0$:
$2k^2 - 9k \neq 0$
$k(2k - 9) \neq 0$
Therefore,$k \neq 0$ and $k \neq \frac{9}{2}$.

(B) Given the system of equations:
$3x - 2y + z = 5k$
$2x + 3y - 2z = -5k$
$x + 4y + 3z = k$
First,calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 3 & -2 & 1 \\ 2 & 3 & -2 \\ 1 & 4 & 3 \end{vmatrix} = 3(9 + 8) + 2(6 + 2) + 1(8 - 3) = 3(17) + 2(8) + 1(5) = 51 + 16 + 5 = 72$.
Next,calculate $D_3$ using Cramer's rule:
$D_3 = \begin{vmatrix} 3 & -2 & 5k \\ 2 & 3 & -5k \\ 1 & 4 & k \end{vmatrix} = k \begin{vmatrix} 3 & -2 & 5 \\ 2 & 3 & -5 \\ 1 & 4 & 1 \end{vmatrix} = k [3(3 + 20) + 2(2 + 5) + 5(8 - 3)] = k [3(23) + 2(7) + 5(5)] = k [69 + 14 + 25] = 108k$.
Since $z = \frac{D_3}{D} = 3$,we have:
$\frac{108k}{72} = 3$
$\frac{3k}{2} = 3$
$3k = 6$
$k = 2$.

(D) Given the system of linear equations:
$2x-3y+5z=12$ $(1)$
$5x+2y+3z=11$ $(2)$
$x+2y-3z=-3$ $(3)$
First,calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 2 & -3 & 5 \\ 5 & 2 & 3 \\ 1 & 2 & -3 \end{vmatrix} = 2(-6-6) + 3(-15-3) + 5(10-2) = 2(-12) + 3(-18) + 5(8) = -24 - 54 + 40 = -38$.
Now,calculate $D_1, D_2, D_3$ using Cramer's Rule:
$D_1 = \begin{vmatrix} 12 & -3 & 5 \\ 11 & 2 & 3 \\ -3 & 2 & -3 \end{vmatrix} = 12(-6-6) + 3(-33+9) + 5(22+6) = 12(-12) + 3(-24) + 5(28) = -144 - 72 + 140 = -76$.
$D_2 = \begin{vmatrix} 2 & 12 & 5 \\ 5 & 11 & 3 \\ 1 & -3 & -3 \end{vmatrix} = 2(-33+9) - 12(-15-3) + 5(-15-11) = 2(-24) - 12(-18) + 5(-26) = -48 + 216 - 130 = 38$.
$D_3 = \begin{vmatrix} 2 & -3 & 12 \\ 5 & 2 & 11 \\ 1 & 2 & -3 \end{vmatrix} = 2(-6-22) + 3(-15-11) + 12(10-2) = 2(-28) + 3(-26) + 12(8) = -56 - 78 + 96 = -38$.
Solving for $\alpha, \beta, \gamma$:
$\alpha = \frac{D_1}{D} = \frac{-76}{-38} = 2$.
$\beta = \frac{D_2}{D} = \frac{38}{-38} = -1$.
$\gamma = \frac{D_3}{D} = \frac{-38}{-38} = 1$.
Finally,calculate $2\alpha + 5\beta + 3\gamma$:
$2(2) + 5(-1) + 3(1) = 4 - 5 + 3 = 2$.

(A) The given system of linear equations is:
$x+y+z=5$
$x+2y+2z=6$
$x+3y+\lambda z=\mu$
The system is solvable by the Matrix Inversion Method if and only if the determinant of the coefficient matrix $A$ is non-zero $(|A| \neq 0)$.
The coefficient matrix $A$ is given by:
$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{bmatrix}$
Calculating the determinant $|A|$:
$|A| = 1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2)$
$|A| = 2\lambda - 6 - \lambda + 2 + 1$
$|A| = \lambda - 3$
For the system to be solvable by the Matrix Inversion Method,we require $|A| \neq 0$,which implies $\lambda - 3 \neq 0$,or $\lambda \neq 3$.
Since the matrix is invertible for any $\lambda \neq 3$,the system has a unique solution for any value of $\mu \in R$.
Therefore,the condition is $\lambda \neq 3, \mu \in R$.

(C) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.
For the given function,we must have $-1 \leq \log_2\left(\frac{x^2}{2}\right) \leq 1$.
Applying the base $2$ exponentiation to all parts,we get $2^{-1} \leq \frac{x^2}{2} \leq 2^1$.
This simplifies to $\frac{1}{2} \leq \frac{x^2}{2} \leq 2$.
Multiplying by $2$,we get $1 \leq x^2 \leq 4$.
Taking the square root,we find $x \in [-2, -1] \cup [1, 2]$.

(D) For the function $f(x)$ to be defined,the following conditions must be satisfied:
$1$. The expression inside the square root in the numerator must be non-negative: $\log_{10}\left(\frac{x}{x-2}\right) \geq 0$.
This implies $\frac{x}{x-2} \geq 10^0$,so $\frac{x}{x-2} \geq 1$.
$\frac{x}{x-2} - 1 \geq 0$ $\Rightarrow \frac{x - (x-2)}{x-2} \geq 0$ $\Rightarrow \frac{2}{x-2} > 0$.
This holds when $x-2 > 0$,i.e.,$x > 2$.
$2$. The expression inside the square root in the denominator must be strictly positive: $[x]^2 - 5[x] + 6 > 0$.
$([x]-2)([x]-3) > 0$.
This implies $[x] < 2$ or $[x] > 3$.
If $[x] < 2$,then $x < 2$. If $[x] > 3$,then $x \geq 4$.
$3$. Combining the conditions $x > 2$ and ($x < 2$ or $x \geq 4$),we get $x \in [4, \infty)$.

(C) The function is defined as $f(x) = \sqrt{\frac{|x|-x}{x-[x]}}$.
For the numerator $\sqrt{|x|-x}$ to be defined,we need $|x|-x \geq 0$,which implies $|x| \geq x$. This is true for all $x \in R$.
For the denominator $\sqrt{x-[x]}$ to be defined and non-zero,we need $x-[x] > 0$.
We know that $x-[x] = \{x\}$,where $\{x\}$ is the fractional part of $x$.
The condition $\{x\} > 0$ holds for all $x \notin Z$ (all real numbers except integers).
If $x \in Z$,then $\{x\} = 0$,which makes the denominator zero and the function undefined.
Therefore,the domain of the function is $R-Z$.

(A) For the combination ${ }^{n} C_{r}$ to be defined,we must have $n \geq r \geq 0$ and $n, r \in \mathbb{Z}_{\geq 0}$.
Here,$n = 16-x$ and $r = 2x-1$.
$1$) $r \geq 0 \implies 2x-1 \geq 0 \implies x \geq \frac{1}{2}$.
$2$) $n \geq r \implies 16-x \geq 2x-1 \implies 17 \geq 3x \implies x \leq \frac{17}{3} \approx 5.66$.
$3$) $n \geq 0 \implies 16-x \geq 0 \implies x \leq 16$.
Combining these,we get $\frac{1}{2} \leq x \leq 5.66$.
Since $n$ and $r$ must be non-negative integers,$x$ must be an integer such that $2x-1$ is a non-negative integer and $16-x$ is an integer $\geq 2x-1$.
The possible integer values for $x$ in the interval $[0.5, 5.66]$ are $x \in \{1, 2, 3, 4, 5\}$.

(B) We are given $f(x) = 8$.
Case $1$: For $x < 2$,$f(x) = x^2 - 4x + 3$.
Setting $x^2 - 4x + 3 = 8$,we get $x^2 - 4x - 5 = 0$.
Factoring gives $(x - 5)(x + 1) = 0$,so $x = 5$ or $x = -1$.
Since the condition is $x < 2$,only $x = -1$ is a valid solution.
Case $2$: For $x \geq 2$,$f(x) = x - 3$.
Setting $x - 3 = 8$,we get $x = 11$.
Since $11 \geq 2$,this is a valid solution.
Thus,the solutions are $x = -1$ and $x = 11$.
The total number of real numbers $x$ for which $f(x) = 8$ is $2$.

(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.
We can rewrite the function by completing the square:
$f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.
Since the domain is $x \in [2, \infty)$,the minimum value of $(x - 2)^2$ is $0$ (at $x = 2$).
As $x \rightarrow \infty$,$(x - 2)^2 \rightarrow \infty$.
Therefore,the range of $f(x)$ is $[0 + 1, \infty) = [1, \infty)$.

(B) Let $g(x) = x^4 - 2x^2 + 3 = (x^2 - 1)^2 + 2$.
Since $(x^2 - 1)^2 \geq 0$,the minimum value of $g(x)$ is $2$ (at $x^2 = 1$) and the maximum value is $\infty$.
Thus,the range of $g(x)$ is $[2, \infty)$.
Now,$f(x) = \log_{0.5}(g(x)) = \log_{1/2}(g(x)) = -\log_2(g(x))$.
Since $g(x) \in [2, \infty)$,we have $\log_2(g(x)) \in [\log_2 2, \log_2 \infty) = [1, \infty)$.
Multiplying by $-1$,we get $-\log_2(g(x)) \in (-\infty, -1]$.
Therefore,the range of $f(x)$ is $(-\infty, -1]$.

(B) Let $g(x) = -x^2-6x-5$. This is a downward-opening parabola.
The maximum value of $g(x)$ is given by $-\frac{D}{4a}$,where $D = b^2-4ac = (-6)^2 - 4(-1)(-5) = 36 - 20 = 16$.
The maximum value is $-\frac{16}{4(-1)} = 4$.
Thus,the range of $g(x)$ is $(-\infty, 4]$.
Since the function is $f(x) = -\sqrt{g(x)}$,the expression inside the square root must be non-negative,so $g(x) \in [0, 4]$.
Taking the square root,$\sqrt{g(x)} \in [0, 2]$.
Multiplying by $-1$,we get $f(x) \in [-2, 0]$.

(B) Given the function $f(x) = \frac{1}{x - |x|}$.
For $x \geq 0$,we have $|x| = x$,so the denominator $x - |x| = 0$. Thus,the function is undefined for $x \geq 0$.
For $x < 0$,we have $|x| = -x$,so the denominator becomes $x - (-x) = 2x$.
Therefore,$f(x) = \frac{1}{2x}$ for $x < 0$.
As $x$ ranges from $(-\infty, 0)$,the value of $2x$ ranges from $(-\infty, 0)$.
Consequently,the value of $\frac{1}{2x}$ ranges from $(-\infty, 0)$.
Thus,the range of the function is $(-\infty, 0)$.

(B) To find the range,we analyze the function in three intervals:
$1$. For $x < -1$,$f(x) = 2x - 3$. As $x \to -1^-$,$f(x) \to 2(-1) - 3 = -5$. Since $x < -1$,$f(x) < -5$. So,the range for this part is $(-\infty, -5)$.
$2$. For $-1 \leq x \leq 1$,$f(x) = 1 - x^2$. The minimum value is at $x = -1$ or $x = 1$,which is $1 - (1)^2 = 0$. The maximum value is at $x = 0$,which is $1 - 0 = 1$. So,the range for this part is $[0, 1]$.
$3$. For $x > 1$,$f(x) = 3x^2 + 2$. As $x \to 1^+$,$f(x) \to 3(1)^2 + 2 = 5$. Since $x > 1$,$f(x) > 5$. So,the range for this part is $(5, \infty)$.
Combining these,the total range is $(-\infty, -5) \cup [0, 1] \cup (5, \infty)$.

(D) function is a bijection if it is both one-to-one (injective) and onto (surjective).
$(a)$ $f(x) = \sqrt{\{x\}}$. Since $\{x\}$ is periodic with period $1$,$f(0.1) = f(1.1)$,so it is many-to-one.
$(b)$ $f(x) = -(x^2-4x+4)+1 = 1-(x-2)^2$. This is a parabola opening downwards,which is many-to-one.
$(c)$ $f(x) = \frac{1}{\sqrt{x-5}}$. The range is $(0, \infty)$,which is not equal to the codomain $R \setminus \{0\}$,so it is into.
$(d)$ $f(x) = \sqrt{16-x^2}$. For $x \in [0,4]$,$f(x)$ is strictly decreasing from $4$ to $0$. Thus,it is one-to-one and onto. Hence,it is a bijection.

(C) Given $f(x) = -|x|$.
First,we evaluate $(f \circ f \circ f)(x)$:
$f(f(f(x))) = f(f(-|x|)) = f(-|-|x||) = f(-|x|) = -|-|x|| = -|x| = f(x)$.
Wait,let us re-evaluate carefully:
$f(x) = -|x|$.
$f(f(x)) = f(-|x|) = -|-|x|| = -|x| = f(x)$.
Since $f(f(x)) = f(x)$,it follows that $(f \circ f \circ f)(x) = f(f(f(x))) = f(f(x)) = f(x)$.
Similarly,$(f \circ f \circ f)(-x) = f(f(f(-x))) = f(f(-x)) = f(-x)$.
Therefore,$(f \circ f \circ f)(x) + (f \circ f \circ f)(-x) = f(x) + f(-x)$.
Since $f(x) = -|x|$,we have $f(-x) = -|-x| = -|x| = f(x)$.
Thus,$f(x) + f(-x) = f(x) + f(x) = 2f(x)$.

(B) Given $f(x) = 3x-2$ and $g(x) = x^2+2$.
We need to find $(g \circ f)(x) + (f \circ g)(x)$.
First,$(g \circ f)(x) = g(f(x)) = (f(x))^2 + 2 = (3x-2)^2 + 2 = 9x^2 - 12x + 4 + 2 = 9x^2 - 12x + 6$.
Next,$(f \circ g)(x) = f(g(x)) = 3(g(x)) - 2 = 3(x^2+2) - 2 = 3x^2 + 6 - 2 = 3x^2 + 4$.
Adding these,$(g \circ f + f \circ g)(x) = (9x^2 - 12x + 6) + (3x^2 + 4) = 12x^2 - 12x + 10$.
Now,check the options by substituting $f(x)$ and $g(x)$:
$12g(x) - 4f(x) - 22 = 12(x^2+2) - 4(3x-2) - 22 = 12x^2 + 24 - 12x + 8 - 22 = 12x^2 - 12x + 10$.
Thus,the correct option is $B$.

(A) Given the function $f(x) = 2x + \sin x$.
To check for one-one,we find the derivative: $f'(x) = 2 + \cos x$.
Since $-1 \leq \cos x \leq 1$,it follows that $f'(x) = 2 + \cos x \geq 2 - 1 = 1 > 0$.
Since $f'(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing,which implies $f(x)$ is one-one.
To check for onto,we observe the range of $f(x)$. As $x \to \infty$,$f(x) \to \infty$,and as $x \to -\infty$,$f(x) \to -\infty$.
Since $f(x)$ is a continuous function,by the Intermediate Value Theorem,its range is $(-\infty, \infty) = R$.
Therefore,$f(x)$ is onto.
Thus,$f$ is one-one and onto.

(D) The function is defined as $f(x) = \begin{cases} 2x-3, & x < -2 \\ x^2-1, & -2 \leq x \leq 2 \\ 3x+2, & x > 2 \end{cases}$.
$1$. Checking for Injectivity (One-to-One):
$A$ function is injective if $f(x_1) = f(x_2) \implies x_1 = x_2$.
Consider the interval $[-2, 2]$,where $f(x) = x^2 - 1$.
We observe that $f(-1) = (-1)^2 - 1 = 0$ and $f(1) = (1)^2 - 1 = 0$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not injective.
$2$. Checking for Surjectivity (Onto):
$A$ function is surjective if its range equals its codomain $(R)$.
- For $x < -2$,$f(x) < 2(-2) - 3 = -7$. So,$f(x) \in (-\infty, -7)$.
- For $-2 \leq x \leq 2$,$f(x) = x^2 - 1$. The minimum value is $-1$ (at $x=0$) and the maximum is $f(-2) = f(2) = 3$. So,$f(x) \in [-1, 3]$.
- For $x > 2$,$f(x) > 3(2) + 2 = 8$. So,$f(x) \in (8, \infty)$.
The range of $f$ is $(-\infty, -7) \cup [-1, 3] \cup (8, \infty)$.
Since the range is not equal to the codomain $R$,the function is not surjective.
Conclusion: The function is neither injective nor surjective.

(B) The set $X$ consists of all $2 \times 2$ matrices with real entries. The function $f: X \rightarrow \mathbb{R}$ is defined as $f(A) = \det(A) = ad - bc$.
To check if $f$ is one-one,consider two different matrices $A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 2 & 0 \\ 0 & 0.5 \end{bmatrix}$.
Here,$f(A_1) = (1)(1) - (0)(0) = 1$ and $f(A_2) = (2)(0.5) - (0)(0) = 1$.
Since $f(A_1) = f(A_2)$ but $A_1 \neq A_2$,the function $f$ is not one-one.
To check if $f$ is onto,for any real number $k \in \mathbb{R}$,we can find a matrix $A = \begin{bmatrix} k & 0 \\ 0 & 1 \end{bmatrix} \in X$ such that $f(A) = (k)(1) - (0)(0) = k$.
Since every element in the codomain $\mathbb{R}$ has a pre-image in $X$,the function $f$ is onto.
Therefore,$f$ is onto but not one-one.

(C) The function is defined as $f(x) = |x-1| + |x-2|$.
We analyze the derivative $f^{\prime}(x)$ in different intervals:
$1$. For $x < 1$,$f(x) = -(x-1) - (x-2) = -2x + 3$,so $f^{\prime}(x) = -2$.
Since $-2023 < 1$,$f^{\prime}(-2023) = -2$.
$2$. For $1 < x < 2$,$f(x) = (x-1) - (x-2) = 1$,so $f^{\prime}(x) = 0$.
Since $1 < \frac{2024}{2023} < 2$,$f^{\prime}\left(\frac{2024}{2023}\right) = 0$.
$3$. For $x > 2$,$f(x) = (x-1) + (x-2) = 2x - 3$,so $f^{\prime}(x) = 2$.
Since $2023 > 2$,$f^{\prime}(2023) = 2$.
Summing these values: $f^{\prime}(-2023) + f^{\prime}\left(\frac{2024}{2023}\right) + f^{\prime}(2023) = -2 + 0 + 2 = 0$.