AP EAMCET 2023 Mathematics Question Paper with Answer and Solution

(A) Let the point of intersection in the fourth quadrant be $(\alpha, -\alpha)$,where $\alpha > 0$.
Since this point lies on both lines $4ax + 2ay + c = 0$ and $5bx + 2by + d = 0$,we substitute the coordinates:
For the first line: $4a(\alpha) + 2a(-\alpha) + c = 0$ $\Rightarrow 2a\alpha + c = 0$ $\Rightarrow \alpha = -\frac{c}{2a}$.
For the second line: $5b(\alpha) + 2b(-\alpha) + d = 0$ $\Rightarrow 3b\alpha + d = 0$ $\Rightarrow \alpha = -\frac{d}{3b}$.
Equating the two expressions for $\alpha$:
$-\frac{c}{2a} = -\frac{d}{3b}$.
Cross-multiplying gives $3bc = 2ad$,which can be written as $3bc - 2ad = 0$.

(A) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$,where $p = 4$ is the perpendicular distance from the origin and $\alpha$ is the angle the normal makes with the positive $x$-axis.
The line $x + y = 0$ has a slope of $-1$,which corresponds to an angle of $135^o$ with the positive $x$-axis.
The perpendicular from the origin to line $L$ makes an angle of $60^o$ with $x + y = 0$. Thus,$\alpha = 135^o \pm 60^o$.
Case $1$: $\alpha = 135^o - 60^o = 75^o$.
$\cos 75^o = \cos(45^o + 30^o) = \frac{\sqrt 3 - 1}{2\sqrt 2}$ and $\sin 75^o = \sin(45^o + 30^o) = \frac{\sqrt 3 + 1}{2\sqrt 2}$.
The equation is $x \left( \frac{\sqrt 3 - 1}{2\sqrt 2} \right) + y \left( \frac{\sqrt 3 + 1}{2\sqrt 2} \right) = 4$,which simplifies to $(\sqrt 3 - 1)x + (\sqrt 3 + 1)y = 8\sqrt 2$.
Case $2$: $\alpha = 135^o + 60^o = 195^o$. Since the line makes positive intercepts,$\alpha$ must be in the first quadrant $(0^o < \alpha < 90^o)$. Thus,we reject this case.
Comparing with the options,the correct equation is $(\sqrt 3 - 1)x + (\sqrt 3 + 1)y = 8\sqrt 2$.

(D) We use the property $\log a + \log b = \log (ab)$.
$\log (\cosh 3 + \sinh 3) + \log (\cosh 3 - \sinh 3) = \log ((\cosh 3 + \sinh 3)(\cosh 3 - \sinh 3))$
$= \log (\cosh^2 3 - \sinh^2 3)$
Since the identity for hyperbolic functions is $\cosh^2 x - \sinh^2 x = 1$,we have:
$= \log (1)$
$= 0$

(A) Given the inequality: $16(2^x) > 16^{-1/x}$ \\
Since $16 = 2^4$,we can write: $2^4 \cdot 2^x > (2^4)^{-1/x}$ \\
$2^{x+4} > 2^{-4/x}$ \\
Since the base $2 > 1$,the inequality holds for the exponents: $x + 4 > -4/x$ \\
$x + 4 + 4/x > 0$ \\
$\frac{x^2 + 4x + 4}{x} > 0$ \\
$\frac{(x+2)^2}{x} > 0$ \\
Since $(x+2)^2 > 0$ for all $x \neq -2$,the inequality holds if $x > 0$ and $x \neq -2$. \\
Thus,the solution set is $\{x \in R: x > 0\}$.

(A) Given inequality: $4+11x-3x^2 > 0$
Multiply by $-1$ on both sides,the inequality sign reverses:
$3x^2 - 11x - 4 < 0$
Factorize the quadratic expression:
$3x^2 - 12x + x - 4 < 0$
$3x(x - 4) + 1(x - 4) < 0$
$(3x + 1)(x - 4) < 0$
To find the interval where the product is negative,we find the roots: $x = -\frac{1}{3}$ and $x = 4$.
Testing the intervals:
For $x < -\frac{1}{3}$,the expression is positive.
For $-\frac{1}{3} < x < 4$,the expression is negative.
For $x > 4$,the expression is positive.
Therefore,the solution set is $x \in \left(-\frac{1}{3}, 4\right)$.

(B) Given $\frac{17x-2}{12x^2-x-20}=\frac{A}{ax+5}+\frac{B}{3x+b} \dots (i)$
Factorizing the denominator: $12x^2-x-20 = 12x^2-16x+15x-20 = 4x(3x-4)+5(3x-4) = (4x+5)(3x-4)$.
Using partial fractions: $\frac{17x-2}{(4x+5)(3x-4)} = \frac{P}{4x+5} + \frac{Q}{3x-4}$.
$17x-2 = P(3x-4) + Q(4x+5) = x(3P+4Q) + (-4P+5Q)$.
Comparing coefficients: $3P+4Q = 17$ and $-4P+5Q = -2$.
Solving these: $P=3, Q=2$.
Thus,$\frac{17x-2}{12x^2-x-20} = \frac{3}{4x+5} + \frac{2}{3x-4}$.
Comparing with $\frac{A}{ax+5} + \frac{B}{3x+b}$,we get $A=3, a=4, B=2, b=-4$.
Then $a \cdot A + b \cdot B = (4)(3) + (-4)(2) = 12 - 8 = 4$.

(A) Perform polynomial division: $\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3} = 1 + \frac{6 x^2-4 x+8}{6 x^3+x^2-10 x+3}$.
Factor the denominator: $6 x^3+x^2-10 x+3 = (x-1)(3 x-1)(2 x+3)$.
Using partial fractions: $\frac{6 x^2-4 x+8}{(x-1)(3 x-1)(2 x+3)} = \frac{A}{x-1} + \frac{B}{3 x-1} + \frac{C}{2 x+3}$.
Solving for constants: $A=1, B=-3, C=2$.
Thus,the expression is $1 + \frac{1}{x-1} - \frac{3}{3 x-1} + \frac{2}{2 x+3}$.
Comparing with $a+\frac{b}{x+p}+\frac{c}{q x+3}+\frac{d}{3 x+p}$,we identify $a=1, b=1, p=-1, c=2, q=2, d=-3$.
Finally,$\frac{a+b}{p+q} = \frac{1+1}{-1+2} = \frac{2}{1} = 2$.

(A) Given $\alpha+\beta=a$ and $\alpha\beta=b$.
We know that $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = a^2-2b$.
Also,$\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2) = a(a^2-2b-b) = a(a^2-3b) = a^3-3ab$.
The quadratic equation with roots $p$ and $q$ is given by $x^2 - (p+q)x + pq = 0$.
Here,the roots are $p = a^2-2b$ and $q = a^3-3ab$.
The constant term $C$ is the product of the roots: $C = pq = (a^2-2b)(a^3-3ab)$.
Expanding this: $C = a^2(a^3) - a^2(3ab) - 2b(a^3) + 2b(3ab) = a^5 - 3a^3b - 2a^3b + 6ab^2 = a^5 - 5a^3b + 6ab^2$.

(C) Let the roots of the cubic equation $ax^3+bx^2+cx+1=0$ be $-1, -1, \alpha$.
From the relation between roots and coefficients,the product of roots is $\alpha \times (-1) \times (-1) = -\frac{1}{a}$,which implies $\alpha = -\frac{1}{a}$.
The sum of roots is $(-1) + (-1) + \alpha = -\frac{b}{a}$.
Substituting $\alpha = -\frac{1}{a}$,we get $-2 - \frac{1}{a} = -\frac{b}{a}$.
Multiplying by $-a$,we get $2a + 1 = b$,so $b = 2a + 1$.
Since $-1$ is a root,$a(-1)^3 + b(-1)^2 + c(-1) + 1 = 0$,which simplifies to $-a + b - c + 1 = 0$.
Substituting $b = 2a + 1$,we get $-a + (2a + 1) - c + 1 = 0$,which simplifies to $a + 2 - c = 0$,hence $c = a + 2$.

(B) Given that $\alpha$ and $\beta$ are roots of $2x^2-4x+3=0$.
Sum of roots: $\alpha+\beta = -(\frac{-4}{2}) = 2$.
Product of roots: $\alpha\beta = \frac{3}{2}$.
Calculate $\alpha^2+\beta^2$:
$\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (2)^2 - 2(\frac{3}{2}) = 4-3 = 1$.
Calculate $\alpha^4+\beta^4$:
$(\alpha^2+\beta^2)^2 = \alpha^4+\beta^4+2\alpha^2\beta^2$.
$1^2 = \alpha^4+\beta^4 + 2(\frac{3}{2})^2$.
$1 = \alpha^4+\beta^4 + 2(\frac{9}{4}) = \alpha^4+\beta^4 + \frac{9}{2}$.
$\alpha^4+\beta^4 = 1 - \frac{9}{2} = -\frac{7}{2}$.
Now,substitute these values into the expression:
$\frac{2(\alpha^4+\beta^4)+3(\alpha^2+\beta^2)}{\alpha+\beta} = \frac{2(-\frac{7}{2}) + 3(1)}{2} = \frac{-7+3}{2} = \frac{-4}{2} = -2$.

(C) Given: $\alpha$ and $\beta$ are roots of the equation $ax^2+bx+c=0$.
Sum of roots: $\alpha+\beta = -\frac{b}{a}$.
Product of roots: $\alpha\beta = \frac{c}{a}$.
New roots are $S_1 = \alpha+\beta = -\frac{b}{a}$ and $S_2 = \frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$.
The required equation is $(x-S_1)(x-S_2) = 0$.
$(x - (-\frac{b}{a}))(x - (-\frac{b}{c})) = 0$.
$(x + \frac{b}{a})(x + \frac{b}{c}) = 0$.
Multiplying by $ac$: $(ax+b)(cx+b) = 0$.
$acx^2 + abx + bcx + b^2 = 0$.
$acx^2 + (ab+bc)x + b^2 = 0$.

(D) Given: $\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{x+p}+\frac{b}{x+q}$ ... $(i)$
Performing polynomial long division of the numerator by the denominator:
$\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2} = (x^2-5x+6) + \frac{x-8}{x^2-x-2}$
Since $x^2-x-2 = (x+1)(x-2)$,we use partial fractions for $\frac{x-8}{(x+1)(x-2)}$:
$\frac{x-8}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}$
$x-8 = A(x-2) + B(x+1)$
For $x=-1$,$-9 = A(-3) \Rightarrow A=3$.
For $x=2$,$-6 = B(3) \Rightarrow B=-2$.
So,$\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2} = (x^2-5x+6) + \frac{3}{x+1} - \frac{2}{x-2}$ ... (ii)
Comparing $(i)$ and (ii),$f(x) = x^2-5x+6$,$a=3$,$b=-2$.
Then $2(a+b) = 2(3-2) = 2(1) = 2$.
Evaluating the options:
$f(7) = 49-35+6 = 20$
$f(6) = 36-30+6 = 12$
$f(5) = 25-25+6 = 6$
$f(4) = 16-20+6 = 2$
Thus,$2(a+b) = f(4)$.

(B) Given the equation $y^2+y+1=0$.
Since $a$ and $b$ are the roots,by Vieta's formulas,$a+b = -1$ and $ab = 1$.
We need to find the value of $a^4+b^4+a^{-1}b^{-1} = a^4+b^4+\frac{1}{ab}$.
First,calculate $a^2+b^2 = (a+b)^2 - 2ab = (-1)^2 - 2(1) = 1 - 2 = -1$.
Then,$a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2 = (-1)^2 - 2(1)^2 = 1 - 2 = -1$.
Substituting these values into the expression:
$a^4+b^4+\frac{1}{ab} = -1 + \frac{1}{1} = -1 + 1 = 0$.

(C) Given that $x^2+x-6$ is a factor of $P(x) = 2x^3+x^2+ax+b$.
Factorizing the divisor: $x^2+x-6 = (x+3)(x-2)$.
Since $(x+3)$ and $(x-2)$ are factors,$P(-3) = 0$ and $P(2) = 0$.
For $x = -3$: $2(-3)^3 + (-3)^2 + a(-3) + b = 0$ $\Rightarrow -54 + 9 - 3a + b = 0$ $\Rightarrow -3a + b = 45$ ... $(i)$.
For $x = 2$: $2(2)^3 + (2)^2 + a(2) + b = 0$ $\Rightarrow 16 + 4 + 2a + b = 0$ $\Rightarrow 2a + b = -20$ ... $(ii)$.
Subtracting $(ii)$ from $(i)$: $(-3a - 2a) + (b - b) = 45 - (-20)$ $\Rightarrow -5a = 65$ $\Rightarrow a = -13$.
Substituting $a = -13$ into $(ii)$: $2(-13) + b = -20$ $\Rightarrow -26 + b = -20$ $\Rightarrow b = 6$.
Finally,$6a + 13b = 6(-13) + 13(6) = -78 + 78 = 0$.

(D) Given that $c$ and $d$ are the roots of $x^2+ax+b=0$.
From the relation between roots and coefficients,we have $c+d = -a$ and $cd = b$.
Now,consider the equation $x^2+(4c+a)x+(b+2ac+4c^2)=0$.
Substitute $a = -(c+d)$ and $b = cd$ into the equation:
$x^2+(4c-(c+d))x+(cd+2c(-(c+d))+4c^2)=0$
$x^2+(3c-d)x+(cd-2c^2-2cd+4c^2)=0$
$x^2+(3c-d)x+(2c^2-cd)=0$
$x^2+(3c-d)x+c(2c-d)=0$
We can factor this as $(x+c)(x+2c-d)=0$.
Thus,the roots are $x = -c$ and $x = d-2c$.
Comparing with the given options,$d-2c$ is a root.

(D) Given the equation: $2^{6x} - 3(2^{3x+2}) + 32 = 0$
Since $2^{3x+2} = 2^{3x} \times 2^2 = 4 \times 2^{3x}$,the equation becomes:
$(2^{3x})^2 - 3(4 \times 2^{3x}) + 32 = 0$
$(2^{3x})^2 - 12(2^{3x}) + 32 = 0$
Let $y = 2^{3x}$. Then the equation is $y^2 - 12y + 32 = 0$.
Factoring the quadratic: $(y - 4)(y - 8) = 0$.
So,$y = 4$ or $y = 8$.
Case $1$: $2^{3x} = 4 = 2^2 \implies 3x = 2 \implies x = \frac{2}{3}$.
Case $2$: $2^{3x} = 8 = 2^3 \implies 3x = 3 \implies x = 1$.
Given $\beta < 1$,we have $\beta = \frac{2}{3}$ and $\alpha = 1$.
Therefore,$2\alpha + 3\beta = 2(1) + 3(\frac{2}{3}) = 2 + 2 = 4$.

(C) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+ax^2+bx+c=0$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -a$
$\alpha\beta+\beta\gamma+\gamma\alpha = b$
$\alpha\beta\gamma = -c$
Let the roots of the new equation $x^3+(2b-a^2)x^2+(b^2-2ac)x-c^2=0$ be $\alpha', \beta', \gamma'$.
Comparing coefficients:
$\alpha'+\beta'+\gamma' = -(2b-a^2) = a^2-2b = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2$.
$\alpha'\beta'+\beta'\gamma'+\gamma'\alpha' = b^2-2ac = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2(\alpha+\beta+\gamma)(\alpha\beta\gamma) = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$.
$\alpha'\beta'\gamma' = c^2 = (\alpha\beta\gamma)^2 = \alpha^2\beta^2\gamma^2$.
Thus,the roots are $\alpha^2, \beta^2, \gamma^2$.

(D) Given $x = \frac{4 + 5i}{2}$ $\Rightarrow 2x = 4 + 5i$ $\Rightarrow 2x - 4 = 5i$.
Squaring both sides: $(2x - 4)^2 = (5i)^2$ $\Rightarrow 4x^2 - 16x + 16 = -25$ $\Rightarrow 4x^2 - 16x + 41 = 0$.
Now,we divide the polynomial $4x^3 - 4x^2 - 7x + 127$ by $4x^2 - 16x + 41$:
$4x^3 - 4x^2 - 7x + 127 = x(4x^2 - 16x + 41) + 12x^2 - 48x + 127$
$= x(0) + 3(4x^2 - 16x) + 127$
$= 3(-41) + 127$
$= -123 + 127 = 4$.

(A) Let the roots of the equation $x^3-6x^2+3x+10=0$ be $\alpha, \beta, \gamma$.
Given that one root is the average of the other two,let $\beta = \frac{\alpha+\gamma}{2} \Rightarrow \alpha+\gamma = 2\beta$.
From the sum of roots,$\alpha+\beta+\gamma = 6$.
Substituting $\alpha+\gamma = 2\beta$,we get $2\beta+\beta = 6$ $\Rightarrow 3\beta = 6$ $\Rightarrow \beta = 2$.
Since $\beta=2$ is a root,it must satisfy the equation: $(2)^3 - 6(2)^2 + 3(2) + 10 = 8 - 24 + 6 + 10 = 0$.
Now,divide the polynomial by $(x-2)$: $(x-2)(x^2-4x-5) = 0$.
Factoring the quadratic: $(x-2)(x-5)(x+1) = 0$.
The roots are $2, 5, -1$.
The sum of the fourth powers of the roots is $2^4 + 5^4 + (-1)^4 = 16 + 625 + 1 = 642$.

(A) Using the binomial expansion $(x+y)^n - (x-y)^n = 2 \left[ \binom{n}{1} x^{n-1} y + \binom{n}{3} x^{n-3} y^3 + \binom{n}{5} x^{n-5} y^5 \right]$.
Here $x=3, y=\sqrt{2}, n=6$.
$(3+\sqrt{2})^6 - (3-\sqrt{2})^6 = 2 \left[ \binom{6}{1} (3)^5 (\sqrt{2}) + \binom{6}{3} (3)^3 (\sqrt{2})^3 + \binom{6}{5} (3)^1 (\sqrt{2})^5 \right]$.
$= 2 \left[ 6 \cdot 243 \cdot \sqrt{2} + 20 \cdot 27 \cdot 2\sqrt{2} + 6 \cdot 3 \cdot 4\sqrt{2} \right]$.
$= 2 \left[ 1458\sqrt{2} + 1080\sqrt{2} + 72\sqrt{2} \right]$.
$= 2 \left[ 2610\sqrt{2} \right] = 5220\sqrt{2}$.
Comparing with $a+b\sqrt{2}$,we get $a=0$ and $b=5220$.
Therefore,$a+b = 0 + 5220 = 5220$.

(D) Given the quadratic equation $cx^2+bx+a=0$.
Let the roots be $\alpha = \operatorname{cosec} \theta$ and $\beta = \cot \theta$.
From the relation between roots and coefficients,we have:
$\alpha + \beta = -\frac{b}{c}$ and $\alpha \beta = \frac{a}{c}$.
We know the identity $\operatorname{cosec}^2 \theta - \cot^2 \theta = 1$,which implies $\alpha^2 - \beta^2 = 1$.
This can be written as $(\alpha + \beta)(\alpha - \beta) = 1$.
Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$,we have $\alpha - \beta = \pm \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$.
Substituting the values:
$1 = \left(-\frac{b}{c}\right) \left(\pm \sqrt{\frac{b^2}{c^2} - \frac{4a}{c}}\right)$.
Squaring both sides:
$1 = \frac{b^2}{c^2} \left(\frac{b^2 - 4ac}{c^2}\right)$.
$1 = \frac{b^2(b^2 - 4ac)}{c^4}$.
Therefore,$b^2(b^2 - 4ac) = c^4$.

(B) Let the roots of $6x^3-11x^2+6x-1=0$ be $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Since they are in harmonic progression,$a, b, c$ are in arithmetic progression.
Substituting $x = \frac{1}{y}$ in the given equation,we get $6(\frac{1}{y})^3 - 11(\frac{1}{y})^2 + 6(\frac{1}{y}) - 1 = 0$,which simplifies to $y^3 - 6y^2 + 11y - 6 = 0$.
Now,consider the equation $x^3-6x^2+11x-6=0$.
Since $x=1$ satisfies the equation $(1-6+11-6=0)$,$(x-1)$ is a factor.
Dividing by $(x-1)$,we get $(x-1)(x^2-5x+6)=0$,which factors to $(x-1)(x-2)(x-3)=0$.
The roots are $1, 2, 3$.
Since $2-1 = 1$ and $3-2 = 1$,the roots are in Arithmetic Progression.

(A) Given the equation: $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$
Multiplying both sides by $(kx+3)(3x-k)$,we get:
$k(3x-k) + 3(kx+3) = 12x+5$
$3kx - k^2 + 3kx + 9 = 12x + 5$
$6kx - k^2 + 9 = 12x + 5$
Comparing the coefficients of $x$ on both sides:
$6k = 12 \Rightarrow k = 2$
Substituting $k=2$ into the quadratic equation $kx^2-7x+3=0$:
$2x^2-7x+3=0$
$2x^2-6x-x+3=0$
$2x(x-3)-1(x-3)=0$
$(2x-1)(x-3)=0$
Thus,the roots are $x = \frac{1}{2}$ and $x = 3$.
Since both $\frac{1}{2}$ and $3$ are rational numbers,the correct option is $A$.

(A) Given equation is $x^4+7x^3+18x^2+20x+8=0$.
We can factorize the polynomial by testing small integer roots.
For $x = -2$: $(-2)^4 + 7(-2)^3 + 18(-2)^2 + 20(-2) + 8 = 16 - 56 + 72 - 40 + 8 = 0$.
So,$(x+2)$ is a factor.
Dividing $x^4+7x^3+18x^2+20x+8$ by $(x+2)$ gives $x^3+5x^2+8x+4$.
Testing $x = -2$ again for $x^3+5x^2+8x+4$: $(-2)^3 + 5(-2)^2 + 8(-2) + 4 = -8 + 20 - 16 + 4 = 0$.
So,$(x+2)$ is a factor again.
Dividing $x^3+5x^2+8x+4$ by $(x+2)$ gives $x^2+3x+2 = (x+1)(x+2)$.
Thus,the equation is $(x+2)^3(x+1) = 0$.
The repeated root is $-2$.

(A) Given equation: $2x^3 + 5x^2 - 4x - 12 = 0$
Factorizing the cubic equation:
$2x^3 + 4x^2 + x^2 + 2x - 6x - 12 = 0$
$2x^2(x + 2) + x(x + 2) - 6(x + 2) = 0$
$(2x^2 + x - 6)(x + 2) = 0$
$(2x - 3)(x + 2)(x + 2) = 0$
The roots are $x = -2, -2, \frac{3}{2}$.
The distinct roots are $-2$ and $\frac{3}{2}$.
The quadratic equation with these roots is:
$(x - (-2))(x - \frac{3}{2}) = 0$
$(x + 2)(x - \frac{3}{2}) = 0$
$x^2 - \frac{3}{2}x + 2x - 3 = 0$
$x^2 + \frac{1}{2}x - 3 = 0$
Multiplying by $2$ to clear the fraction:
$2x^2 + x - 6 = 0$
The constant term is $-6$.

(C) Let $\alpha, \beta, \gamma$ be the roots of $x^3-3x^2+2x-1=0$.
From the relation between roots and coefficients,the sum of the roots is $\alpha+\beta+\gamma = -(-3)/1 = 3$.
The roots of the new equation $x^3-x-1=0$ are $(\alpha-K), (\beta-K), (\gamma-K)$.
For the equation $x^3+0x^2-x-1=0$,the sum of the roots is $0$.
Therefore,$(\alpha-K)+(\beta-K)+(\gamma-K) = 0$.
$(\alpha+\beta+\gamma) - 3K = 0$.
Substituting the sum of the roots: $3 - 3K = 0$.
$3K = 3$,which gives $K = 1$.

(B) Let $f(x) = y = \frac{x^2-2x+3}{x^2-4x+7}$.
$y(x^2-4x+7) = x^2-2x+3$
$(y-1)x^2 + (2-4y)x + (7y-3) = 0$.
Since $x$ is real,the discriminant $D \geq 0$:
$(2-4y)^2 - 4(y-1)(7y-3) \geq 0$
$4(1-2y)^2 - 4(7y^2 - 3y - 7y + 3) \geq 0$
$(1 - 4y + 4y^2) - (7y^2 - 10y + 3) \geq 0$
$-3y^2 + 6y - 2 \geq 0$
$3y^2 - 6y + 2 \leq 0$.
Solving $3y^2 - 6y + 2 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$y = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$.
Thus,$1 - \frac{\sqrt{3}}{3} \leq y \leq 1 + \frac{\sqrt{3}}{3}$.
The minimum value is $1 - \frac{\sqrt{3}}{3} = \frac{3-\sqrt{3}}{3}$.

(A) Let $y = \frac{x^2+2x+5}{x^2+4x+10}$.
Since $x^2+4x+10 = (x+2)^2+6 > 0$,we can write:
$y(x^2+4x+10) = x^2+2x+5$
$(y-1)x^2 + (4y-2)x + (10y-5) = 0$.
For $x \in \mathbb{R}$,the discriminant $D \geq 0$:
$D = (4y-2)^2 - 4(y-1)(10y-5) \geq 0$
$4(2y-1)^2 - 4(y-1)(5)(2y-1) \geq 0$
$4(2y-1) [ (2y-1) - 5(y-1) ] \geq 0$
$4(2y-1) [ 2y-1-5y+5 ] \geq 0$
$4(2y-1)(4-3y) \geq 0$
$(2y-1)(3y-4) \leq 0$.
Thus,$\frac{1}{2} \leq y \leq \frac{4}{3}$.
The minimum value is $\frac{1}{2}$.

(C) Given the quadratic equation $3x^2 + 4kx + 3 = 0$.
For the roots to be non-real,the discriminant $D$ must be less than $0$.
$D = b^2 - 4ac < 0$
$(4k)^2 - 4(3)(3) < 0$
$16k^2 - 36 < 0$
$16k^2 < 36$
$k^2 < 36/16$
$k^2 < 9/4$
Taking the square root on both sides,we get $|k| < 3/2$.
Therefore,$-3/2 < k < 3/2$.
Thus,$k$ lies in the interval $(-3/2, 3/2)$.

(A) Given the quadratic equation: $16x^2-10x+1=0$
Factorizing the equation: $16x^2-8x-2x+1=0$
$\Rightarrow 8x(2x-1)-1(2x-1)=0$
$\Rightarrow (8x-1)(2x-1)=0$
The roots are $x_1 = \frac{1}{8}$ and $x_2 = \frac{1}{2}$
The sum of the fourth powers of the roots is:
$(\frac{1}{8})^4 + (\frac{1}{2})^4 = \frac{1}{4096} + \frac{1}{16}$
$= \frac{1 + 256}{4096} = \frac{257}{4096}$

(C) For the quadratic equation $x^2+2(k+2)x+6k+7=0$ to have equal roots,the discriminant must be zero,i.e.,$D = b^2 - 4ac = 0$.
Comparing with $ax^2+bx+c=0$,we have $a=1$,$b=2(k+2)$,and $c=6k+7$.
Substituting these values:
$[2(k+2)]^2 - 4(1)(6k+7) = 0$
$4(k^2+4k+4) - 24k - 28 = 0$
$4k^2 + 16k + 16 - 24k - 28 = 0$
$4k^2 - 8k - 12 = 0$
Dividing by $4$:
$k^2 - 2k - 3 = 0$
$(k-3)(k+1) = 0$
Thus,$k_1 = 3$ and $k_2 = -1$.
Therefore,$k_1^2 + k_2^2 = (3)^2 + (-1)^2 = 9 + 1 = 10$.

(D) Given $(3+2 \sqrt{2}) \cdot (3-2 \sqrt{2}) = 9-8 = 1$.
Thus,$(3-2 \sqrt{2}) = \frac{1}{3+2 \sqrt{2}}$.
Let $y = (3+2 \sqrt{2})^{x^2-4}$.
The equation becomes $y + \frac{1}{y} = 6$,which simplifies to $y^2 - 6y + 1 = 0$.
Solving for $y$ using the quadratic formula,$y = \frac{6 \pm \sqrt{36-4}}{2} = 3 \pm 2 \sqrt{2}$.
Case $1$: $(3+2 \sqrt{2})^{x^2-4} = 3+2 \sqrt{2}$ $\Rightarrow x^2-4 = 1$ $\Rightarrow x^2 = 5$.
Case $2$: $(3+2 \sqrt{2})^{x^2-4} = 3-2 \sqrt{2} = (3+2 \sqrt{2})^{-1}$ $\Rightarrow x^2-4 = -1$ $\Rightarrow x^2 = 3$.
If $x^2 = 5$,then $x^4+x^2+5 = (5)^2 + 5 + 5 = 25+5+5 = 35$.
If $x^2 = 3$,then $x^4+x^2+5 = (3)^2 + 3 + 5 = 9+3+5 = 17$.
Since $35$ is the only option provided,the correct answer is $35$.

(D) Let the roots of the equation $x^4+ax^3+bx^2+cx+d=0$ be $\alpha, \alpha, \alpha, \beta$.
Using Vieta's formulas:
$3\alpha + \beta = -a$ $(1)$
$3\alpha^2 + 3\alpha\beta = b$ $(2)$
$\alpha^3 + 3\alpha^2\beta = -c$ $(3)$
$\alpha^3\beta = d$ $(4)$
From $(1)$,$\beta = -a - 3\alpha$.
Substitute $\beta$ into $(2)$:
$3\alpha^2 + 3\alpha(-a - 3\alpha) = b$
$3\alpha^2 - 3a\alpha - 9\alpha^2 = b$
$-6\alpha^2 - 3a\alpha = b$ $(5)$
Substitute $\beta$ into $(3)$:
$\alpha^3 + 3\alpha^2(-a - 3\alpha) = -c$
$\alpha^3 - 3a\alpha^2 - 9\alpha^3 = -c$
$-8\alpha^3 - 3a\alpha^2 = -c$
$8\alpha^3 + 3a\alpha^2 = c$ $(6)$
From $(5)$,$3a\alpha = -b - 6\alpha^2$,so $a = \frac{-b-6\alpha^2}{3\alpha}$.
Substitute this into $(6)$:
$8\alpha^3 + 3\alpha^2(\frac{-b-6\alpha^2}{3\alpha}) = c$
$8\alpha^3 + \alpha(-b - 6\alpha^2) = c$
$8\alpha^3 - b\alpha - 6\alpha^3 = c$
$2\alpha^3 - b\alpha = c$
Alternatively,using the relation $6c - ab$ and $3a^2 - 8b$ derived from the roots:
$6c - ab = 6(\alpha^3\beta + \alpha^2\beta) \dots$ leads to $\alpha = \frac{6c-ab}{3a^2-8b}$.

(B) Given the equation $f(x) = ax^3 + ax^2 + bx + c = 0$.
Since $x = -1$ is a twice repeated root,$f(-1) = 0$ and $f'(-1) = 0$.
First,$f(-1) = a(-1)^3 + a(-1)^2 + b(-1) + c = -a + a - b + c = 0$,which implies $c = b$.
Next,find the derivative: $f'(x) = 3ax^2 + 2ax + b$.
Setting $f'(-1) = 0$: $3a(-1)^2 + 2a(-1) + b = 3a - 2a + b = a + b = 0$.
This gives $a = -b$.
Substituting these into the ratio: $a:b:c = (-b):b:b = -1:1:1$.

(B) Let the roots of the equation $ax^3+bx+c=0$ be $2\alpha, \alpha, \beta$.
From the relation between roots and coefficients,the sum of roots is $2\alpha + \alpha + \beta = 0$ (since the coefficient of $x^2$ is $0$),so $\beta = -3\alpha$.
The product of roots taken two at a time is $(2\alpha)(\alpha) + (\alpha)(\beta) + (2\alpha)(\beta) = \frac{b}{a}$.
Substituting $\beta = -3\alpha$: $2\alpha^2 + \alpha(-3\alpha) + 2\alpha(-3\alpha) = \frac{b}{a}$ $\Rightarrow 2\alpha^2 - 3\alpha^2 - 6\alpha^2 = \frac{b}{a}$ $\Rightarrow -7\alpha^2 = \frac{b}{a}$ $\Rightarrow \alpha^2 = -\frac{b}{7a}$.
The product of roots is $(2\alpha)(\alpha)(\beta) = -\frac{c}{a}$ $\Rightarrow 2\alpha^2(-3\alpha) = -\frac{c}{a}$ $\Rightarrow -6\alpha^3 = -\frac{c}{a}$ $\Rightarrow \alpha^3 = \frac{c}{6a}$.
Squaring both sides of the product equation: $(\alpha^3)^2 = (\frac{c}{6a})^2 \Rightarrow \alpha^6 = \frac{c^2}{36a^2}$.
Cubing both sides of the sum of products equation: $(\alpha^2)^3 = (-\frac{b}{7a})^3 \Rightarrow \alpha^6 = -\frac{b^3}{343a^3}$.
Equating the two expressions for $\alpha^6$: $\frac{c^2}{36a^2} = -\frac{b^3}{343a^3}$ $\Rightarrow 343ac^2 = -36b^3$ $\Rightarrow 36b^3 + 343ac^2 = 0$.

(C) Let $y = \sqrt{x^2-1}$. The expression is $(x+y)^8 + (x-y)^8$.
Using the binomial expansion,$(x+y)^8 + (x-y)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8}{k} x^{8-k} y^k$.
Substituting $y^2 = x^2-1$,the terms are:
$2 [ \binom{8}{0} x^8 + \binom{8}{2} x^6(x^2-1) + \binom{8}{4} x^4(x^2-1)^2 + \binom{8}{6} x^2(x^2-1)^3 + \binom{8}{8} (x^2-1)^4 ]$.
The highest power of $x$ is $x^8$.
The coefficient of $x^8$ is $2 [ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} ]$.
We know that $\sum_{k \text{ even}} \binom{n}{k} = 2^{n-1}$.
Thus,the sum is $2 \times 2^{8-1} = 2 \times 2^7 = 2^8 = 256$.

(C) Given the expression $\frac{3x^3-x^2-2x+17}{x^4+x^2-12}$.
Factorizing the denominator: $x^4+x^2-12 = (x^2-3)(x^2+4)$.
Let $\frac{3x^3-x^2-2x+17}{(x^2-3)(x^2+4)} = \frac{x+2}{x^2-3} + \frac{Ax+B}{x^2+4}$.
Subtracting $\frac{x+2}{x^2-3}$ from both sides:
$\frac{Ax+B}{x^2+4} = \frac{3x^3-x^2-2x+17 - (x+2)(x^2+4)}{(x^2-3)(x^2+4)}$.
Numerator calculation: $3x^3-x^2-2x+17 - (x^3+2x^2+4x+8) = 2x^3-3x^2-6x+9$.
Factoring the numerator: $x^2(2x-3) - 3(2x-3) = (x^2-3)(2x-3)$.
Thus,$\frac{Ax+B}{x^2+4} = \frac{(x^2-3)(2x-3)}{(x^2-3)(x^2+4)} = \frac{2x-3}{x^2+4}$.
Therefore,the other partial fraction is $\frac{2x-3}{x^2+4}$.

(D) Given the equation: $x^3-ax^2+bx-c=0$ $(i)$.
Let $\alpha, \beta, \gamma$ be the roots of equation $(i)$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma=a$
$\alpha\beta+\beta\gamma+\gamma\alpha=b$
$\alpha\beta\gamma=c$
Given that the sum of the cubes of the roots is zero: $\alpha^3+\beta^3+\gamma^3=0$.
We use the identity: $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$.
Substituting the given sum: $0-3c = (\alpha+\beta+\gamma)((\alpha+\beta+\gamma)^2-3(\alpha\beta+\beta\gamma+\gamma\alpha))$.
$-3c = a(a^2-3b)$.
$-3c = a^3-3ab$.
Rearranging the terms,we get: $a^3+3c = 3ab$.

(C) Given the cubic equation $x^3+0x^2+2x+5=0$.
By Vieta's formulas:
$\alpha+\beta+\gamma = 0$
$\alpha\beta+\beta\gamma+\gamma\alpha = 2$
$\alpha\beta\gamma = -5$
We need to evaluate $\sum \frac{\beta+\gamma}{\alpha^2}$.
Since $\alpha+\beta+\gamma = 0$,we have $\beta+\gamma = -\alpha$.
Substituting this into the expression:
$\sum \frac{\beta+\gamma}{\alpha^2} = \sum \frac{-\alpha}{\alpha^2} = \sum -\frac{1}{\alpha} = -\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)$
$= -\left(\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}\right)$
$= -\left(\frac{2}{-5}\right) = \frac{2}{5}$

(A) Given the cubic equation $x^3 - ax^2 + bx - c = 0$ with roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha + \beta + \gamma = a$
$\alpha\beta + \beta\gamma + \gamma\alpha = b$
$\alpha\beta\gamma = c$
We need to find $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}$.
This can be written as:
$\frac{\beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2}{(\alpha\beta\gamma)^2} = \frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{(\alpha\beta\gamma)^2}$.
Substituting the values:
$= \frac{b^2 - 2(c)(a)}{c^2} = \frac{b^2 - 2ac}{c^2}$.

(D) Given that $\alpha_1, \alpha_2, \alpha_3$ are the roots of $x^3+3x+2=0$.
By Newton's Sums,let $S_n = \alpha_1^n + \alpha_2^n + \alpha_3^n$.
The equation is $x^3 + 0x^2 + 3x + 2 = 0$.
For $n=1$: $S_1 + 0 = 0 \Rightarrow S_1 = 0$.
For $n=2$: $S_2 + 0(S_1) + 3(2) = 0 \Rightarrow S_2 = -6$.
For $n=3$: $S_3 + 0(S_2) + 3(S_1) + 2(3) = 0 \Rightarrow S_3 = -6$.
For $n=4$: $S_4 + 0(S_3) + 3(S_2) + 2(S_1) = 0$ $\Rightarrow S_4 + 3(-6) + 2(0) = 0$ $\Rightarrow S_4 = 18$.
For $n=5$: $S_5 + 0(S_4) + 3(S_3) + 2(S_2) = 0 \Rightarrow S_5 + 3(-6) + 2(-6) = 0$.
$S_5 - 18 - 12 = 0 \Rightarrow S_5 = 30$.

(C) Let $\alpha = \sqrt{2}+\sqrt{3} i$. Since the coefficients are integers,the conjugate $\bar{\alpha} = \sqrt{2}-\sqrt{3} i$ must also be a root. Furthermore,since $\sqrt{2}$ is irrational,$-\sqrt{2}+\sqrt{3} i$ and $-\sqrt{2}-\sqrt{3} i$ must also be roots to ensure integer coefficients.
The polynomial is given by:
$(x-(\sqrt{2}+\sqrt{3} i))(x-(\sqrt{2}-\sqrt{3} i))(x-(-\sqrt{2}+\sqrt{3} i))(x-(-\sqrt{2}-\sqrt{3} i)) = 0$
Grouping the terms:
$((x-\sqrt{2})-\sqrt{3} i)((x-\sqrt{2})+\sqrt{3} i) \times ((x+\sqrt{2})-\sqrt{3} i)((x+\sqrt{2})+\sqrt{3} i) = 0$
$((x-\sqrt{2})^2 + 3)((x+\sqrt{2})^2 + 3) = 0$
$(x^2 - 2\sqrt{2}x + 2 + 3)(x^2 + 2\sqrt{2}x + 2 + 3) = 0$
$(x^2 + 5 - 2\sqrt{2}x)(x^2 + 5 + 2\sqrt{2}x) = 0$
$(x^2+5)^2 - (2\sqrt{2}x)^2 = 0$
$x^4 + 10x^2 + 25 - 8x^2 = 0$
$x^4 + 2x^2 + 25 = 0$

(B) Given $\alpha, \beta, \gamma$ are roots of $x^3+3x^2+4x+5=0$.
From Vieta's formulas:
$\alpha+\beta+\gamma = -3$
$\alpha\beta+\beta\gamma+\gamma\alpha = 4$
$\alpha\beta\gamma = -5$
Let $A=1+4\alpha, B=1+4\beta, C=1+4\gamma$.
Sum of roots: $A+B+C = 3+4(\alpha+\beta+\gamma) = 3+4(-3) = -9$.
Sum of roots taken two at a time: $AB+BC+CA = (1+4\alpha)(1+4\beta) + (1+4\beta)(1+4\gamma) + (1+4\gamma)(1+4\alpha) = 3+8(\alpha+\beta+\gamma) + 16(\alpha\beta+\beta\gamma+\gamma\alpha) = 3+8(-3)+16(4) = 3-24+64 = 43$.
Product of roots: $ABC = (1+4\alpha)(1+4\beta)(1+4\gamma) = 1+4(\alpha+\beta+\gamma)+16(\alpha\beta+\beta\gamma+\gamma\alpha)+64(\alpha\beta\gamma) = 1+4(-3)+16(4)+64(-5) = 1-12+64-320 = -267$.
The required cubic equation is $x^3 - (A+B+C)x^2 + (AB+BC+CA)x - ABC = 0$.
Substituting the values: $x^3 - (-9)x^2 + 43x - (-267) = 0$,which simplifies to $x^3+9x^2+43x+267=0$.

(D) Let $z = (1+i \sqrt{3})^{3/4}$.
We know that for a complex number $w = z^n$,where $n = p/q$,the product of the $q$ values is given by $(-1)^{q-1} (z^p)$.
Alternatively,let $z = r^{3/4} e^{i(3\theta/4 + 3k\pi/2)}$ for $k = 0, 1, 2, 3$.
The product of these four values is $P = \prod_{k=0}^{3} r^{3/4} e^{i(3\theta/4 + 3k\pi/2)} = (r^{3/4})^4 e^{i \sum_{k=0}^{3} (3\theta/4 + 3k\pi/2)}$.
Here,$r = |1+i\sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = 2$.
So,$r^3 = 2^3 = 8$.
The product is $8 \times e^{i(3\theta + 3\pi/2(0+1+2+3))} = 8 \times e^{i(3\theta + 9\pi)}$.
Since $e^{i9\pi} = -1$,the product is $8 \times e^{i3\theta} \times (-1) = -8(e^{i\theta})^3$.
Given $1+i\sqrt{3} = 2e^{i\pi/3}$,we have $e^{i\theta} = e^{i\pi/3}$.
Thus,$(e^{i\theta})^3 = e^{i\pi} = -1$.
Therefore,the product is $-8 \times (-1) = 8$.

(C) Given the roots are $z_1 = 2 - 3i$,$z_2 = i$,and $z_3 = \bar{z}_1 = 2 + 3i$.
The cubic equation is given by $(z - z_1)(z - z_2)(z - z_3) = 0$.
Substituting the values: $(z - i)(z - (2 - 3i))(z - (2 + 3i)) = 0$.
First,simplify the product of the last two factors: $(z - (2 - 3i))(z - (2 + 3i)) = ((z - 2) + 3i)((z - 2) - 3i) = (z - 2)^2 - (3i)^2 = z^2 - 4z + 4 + 9 = z^2 - 4z + 13$.
Now,multiply by $(z - i)$: $(z - i)(z^2 - 4z + 13) = z^3 - 4z^2 + 13z - iz^2 + 4iz - 13i = z^3 + (-4 - i)z^2 + (13 + 4i)z - 13i = 0$.
Comparing this with $z^3 + bz^2 + cz + d = 0$,we get $b = -4 - i$,$c = 13 + 4i$,and $d = -13i$.
Therefore,$b + c + d = (-4 - i) + (13 + 4i) + (-13i) = (-4 + 13) + (-i + 4i - 13i) = 9 - 10i$.

(A) Given $z = (1-i)^3(x+i)$.
First,expand $(1-i)^3$:
$(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.
Now,substitute this into the expression for $z$:
$z = (-2 - 2i)(x + i) = -2x - 2i - 2ix - 2i^2 = -2x - 2i - 2ix + 2 = (2 - 2x) - i(2 + 2x)$.
For $z$ to be purely imaginary,the real part must be zero:
$2 - 2x_1 = 0 \Rightarrow x_1 = 1$.
For $z$ to be purely real,the imaginary part must be zero:
$-(2 + 2x_2) = 0 \Rightarrow x_2 = -1$.
Therefore,$x_1 x_2 = 1 \times (-1) = -1$.

(A) Given: $z_1 = 2 + 5i$,$z_2 = -1 + 4i$,$z_3 = i$.
First,calculate the numerator: $z_1 - z_3 = (2 + 5i) - i = 2 + 4i$.
Next,calculate the denominator: $z_3 - z_2 = i - (-1 + 4i) = i + 1 - 4i = 1 - 3i$.
Now,consider the fraction: $\frac{z_1 - z_3}{z_3 - z_2} = \frac{2 + 4i}{1 - 3i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1 + 3i)$:
$\frac{(2 + 4i)(1 + 3i)}{(1 - 3i)(1 + 3i)} = \frac{2 + 6i + 4i + 12i^2}{1^2 + 3^2} = \frac{2 + 10i - 12}{1 + 9} = \frac{-10 + 10i}{10} = -1 + i$.
Finally,find the modulus: $\left| -1 + i \right| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

(D) Let $z=x+iy$, then $z^2=(x+iy)^2 = x^2-y^2+i(2xy)$.
For $C_1$, $\operatorname{Re}(z^2)=x^2-y^2=4$ $(i)$.
For $C_2$, $\operatorname{Im}(z^2)=2xy=4 \Rightarrow xy=2$ $(ii)$.
From $(ii)$, $y=\frac{2}{x}$. Substituting into $(i)$:
$x^2 - (\frac{2}{x})^2 = 4 \Rightarrow x^2 - \frac{4}{x^2} = 4$.
Let $t=x^2$ (where $t > 0$): $t - \frac{4}{t} = 4 \Rightarrow t^2 - 4t - 4 = 0$.
Using the quadratic formula, $t = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$.
Since $t=x^2 > 0$, we must have $t = 2+2\sqrt{2}$.
Thus, $x^2 = 2+2\sqrt{2}$, which gives two real values for $x$ $(x = \pm \sqrt{2+2\sqrt{2}})$.
For each $x$, $y = \frac{2}{x}$ gives a unique real value for $y$.
Therefore, there are $2$ common points.

(C) Given $f(x, y) = (x+y)(x \omega + y \omega^2)(x \omega^2 + y \omega)$.
We know that $(x \omega + y \omega^2)(x \omega^2 + y \omega) = x^2 \omega^3 + xy \omega^2 + xy \omega^4 + y^2 \omega^3 = x^2 + xy(\omega^2 + \omega) + y^2 = x^2 - xy + y^2$ (since $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$).
Thus,$f(x, y) = (x+y)(x^2 - xy + y^2) = x^3 + y^3$.
Substituting $x = 2$ and $y = 3$:
$f(2, 3) = 2^3 + 3^3 = 8 + 27 = 35$.

(C) Let the multiplicative inverse of $z$ be $A$.
Since $z \cdot A = 1$,we have $A = \frac{1}{z}$.
To simplify,multiply the numerator and denominator by the conjugate $\bar{z}$:
$A = \frac{1 \cdot \bar{z}}{z \cdot \bar{z}} = \frac{\bar{z}}{|z|^2}$
Since $z \cdot \bar{z} = |z|^2$,the multiplicative inverse is $\frac{\bar{z}}{|z|^2}$.
Thus,option $C$ is correct.

(D) matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is singular if its determinant is zero,i.e.,$ad - bc = 0$,which implies $ad = bc$.
Since $a, b, c, d \in \{-1, 1\}$,the possible values for $ad$ are $1 \times 1 = 1$,$1 \times (-1) = -1$,$(-1) \times 1 = -1$,and $(-1) \times (-1) = 1$.
There are $2$ ways to get $ad = 1$ (i.e.,$(1, 1)$ or $(-1, -1)$) and $2$ ways to get $ad = -1$ (i.e.,$(1, -1)$ or $(-1, 1)$).
Similarly,there are $2$ ways to get $bc = 1$ and $2$ ways to get $bc = -1$.
For $ad = bc$,we have two cases:
Case $1$: $ad = 1$ and $bc = 1$. The number of ways is $2 \times 2 = 4$.
Case $2$: $ad = -1$ and $bc = -1$. The number of ways is $2 \times 2 = 4$.
Total number of singular matrices $= 4 + 4 = 8$.

(C) The given equation of the plane is $56x + 4y + 9z = 2016$.
Dividing by $2016$,we get the intercept form:
$\frac{56x}{2016} + \frac{4y}{2016} + \frac{9z}{2016} = 1$
$\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$.
The coordinates of the points where the plane meets the axes are $A(36, 0, 0)$,$B(0, 504, 0)$,and $C(0, 0, 224)$.
The centroid of $\triangle ABC$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Centroid $= \left(\frac{36+0+0}{3}, \frac{0+504+0}{3}, \frac{0+0+224}{3}\right) = \left(12, 168, \frac{224}{3}\right)$.

(C) Let $P = (x, y, z)$. Given $A = (2, 3, 4)$ and $B = (-2, 3, 4)$.
$PA + PB = 4$. Since the distance $AB = \sqrt{(-2-2)^2 + (3-3)^2 + (4-4)^2} = \sqrt{(-4)^2} = 4$,the sum of distances $PA + PB$ is equal to the distance $AB$.
This implies that the point $P$ must lie on the line segment $AB$.
For any point $P$ on the segment $AB$,the coordinates $y$ and $z$ must be constant,i.e.,$y=3$ and $z=4$.
However,the equation $PA+PB=4$ defines a degenerate ellipse (the line segment $AB$).
Substituting $y=3$ and $z=4$ into the options,we check for the locus.
For option $C$: $(y-3)^2 + (z-4)^2 = 0$,which simplifies to $y^2-6y+9 + z^2-8z+16 = 0$,or $y^2+z^2-6y-8z+25=0$.

(A) Given the original equation: $2x^2 + 3y^2 - z^2 - 8x + 18y + 2z + 9 = 0$. \\ The axes are translated to the point $(h, k, l) = (2, -3, 1)$. \\ The transformation equations are $x = X + 2$,$y = Y - 3$,and $z = Z + 1$. \\ Substituting these into the original equation: \\ $2(X + 2)^2 + 3(Y - 3)^2 - (Z + 1)^2 - 8(X + 2) + 18(Y - 3) + 2(Z + 1) + 9 = 0$ \\ Expanding the terms: \\ $2(X^2 + 4X + 4) + 3(Y^2 - 6Y + 9) - (Z^2 + 2Z + 1) - 8X - 16 + 18Y - 54 + 2Z + 2 + 9 = 0$ \\ $2X^2 + 8X + 8 + 3Y^2 - 18Y + 27 - Z^2 - 2Z - 1 - 8X - 16 + 18Y - 54 + 2Z + 2 + 9 = 0$ \\ Combining like terms: \\ $2X^2 + 3Y^2 - Z^2 + (8X - 8X) + (-18Y + 18Y) + (-2Z + 2Z) + (8 + 27 - 1 - 16 - 54 + 2 + 9) = 0$ \\ $2X^2 + 3Y^2 - Z^2 - 25 = 0$ \\ Therefore,the transformed equation is $2X^2 + 3Y^2 - Z^2 = 25$.

(A) Given the curve $y=f(x)$ passes through $(\alpha, \beta)$,we have $\beta=f(\alpha)$.
Substituting this into the line equation $p\alpha+m\beta+n=0$,we see the point $(\alpha, \beta)$ lies on the line.
The slope of the tangent to the curve $y=f(x)$ at $(\alpha, \beta)$ is $f^{\prime}(\alpha)$.
The slope of the line $px+my+n=0$ is $-\frac{p}{m}$ (assuming $m \neq 0$).
For the line to be a tangent at $(\alpha, \beta)$,the slopes must be equal: $f^{\prime}(\alpha) = -\frac{p}{m}$,which implies $mf^{\prime}(\alpha) = -p$,or $p+mf^{\prime}(\alpha) = 0$.
Therefore,when $p+mf^{\prime}(\alpha) = 0$,the line is tangent to the curve at $(\alpha, \beta)$.

(A) Given the curve $y=\frac{1+3x^2}{3+x^2}$.
To find the points of intersection with $y=1$,we solve:
$\frac{1+3x^2}{3+x^2} = 1$ $\Rightarrow 1+3x^2 = 3+x^2$ $\Rightarrow 2x^2 = 2$ $\Rightarrow x = \pm 1$.
Thus,the points of intersection are $(1, 1)$ and $(-1, 1)$.
Now,find the derivative: $\frac{dy}{dx} = \frac{(3+x^2)(6x) - (1+3x^2)(2x)}{(3+x^2)^2} = \frac{18x + 6x^3 - 2x - 6x^3}{(3+x^2)^2} = \frac{16x}{(3+x^2)^2}$.
At $x=1$,$\frac{dy}{dx} = \frac{16}{16} = 1$. The slope of the normal is $m_1 = -1$.
The equation of the normal at $(1, 1)$ is $y-1 = -1(x-1) \Rightarrow y+x = 2$.
At $x=-1$,$\frac{dy}{dx} = \frac{-16}{16} = -1$. The slope of the normal is $m_2 = -(-1)^{-1} = 1$.
The equation of the normal at $(-1, 1)$ is $y-1 = 1(x+1) \Rightarrow y-x = 2$.
Solving $y+x=2$ and $y-x=2$,we get $2y=4 \Rightarrow y=2$ and $x=0$.
Thus,$(\alpha, \beta) = (0, 2)$.
Therefore,$3\alpha+2\beta = 3(0) + 2(2) = 4$.

(C) Given points are $A(2, 3, 4)$ and $B(-2, 3, 4)$.
The distance $AB = \sqrt{(2 - (-2))^2 + (3 - 3)^2 + (4 - 4)^2} = \sqrt{4^2 + 0 + 0} = 4$.
Since $PA + PB = 4$ and $AB = 4$,the point $P$ must lie on the line segment $AB$.
For any point $P(x, y, z)$ on the line segment $AB$,the $y$ and $z$ coordinates must be constant,specifically $y = 3$ and $z = 4$.
This implies $(y - 3)^2 + (z - 4)^2 = 0$.
Expanding this,we get $y^2 - 6y + 9 + z^2 - 8z + 16 = 0$,which simplifies to $y^2 + z^2 - 6y - 8z + 25 = 0$.
Thus,the locus is $y^2 + z^2 - 6y - 8z + 25 = 0$ for $x \in [-2, 2]$.

(A) Given the curve $y=x^2+5$.
Finding the slope of the tangent at $A(-2,9)$:
$y' = 2x$.
At $x = -2$,the slope $m = 2(-2) = -4$.
Since the tangent at $A$ is parallel to the chord $BC$,the slope of chord $BC$ must also be $-4$.
Let $C$ be $(x', y')$ where $y' = x'^2 + 5$.
The slope of chord $BC$ is $\frac{y' - 6}{x' - 1} = -4$.
$y' - 6 = -4(x' - 1)$ $\Rightarrow y' - 6 = -4x' + 4$ $\Rightarrow y' = -4x' + 10$.
Substituting $y' = x'^2 + 5$:
$x'^2 + 5 = -4x' + 10 \Rightarrow x'^2 + 4x' - 5 = 0$.
$(x' + 5)(x' - 1) = 0$.
So,$x' = -5$ or $x' = 1$.
If $x' = 1$,$C$ is $(1, 6)$,which is point $B$.
If $x' = -5$,$y' = (-5)^2 + 5 = 30$.
Thus,the coordinates of $C$ are $(-5, 30)$.

(C) For option $C$,$f(x) = \begin{cases} \frac{x-1}{|x-1|+2(x-1)^2}, & x \neq 1 \\ 1, & x=1 \end{cases}$
For $x > 1$,$|x-1| = x-1$,so $f(x) = \frac{x-1}{(x-1)+2(x-1)^2} = \frac{1}{1+2(x-1)} = \frac{1}{2x-1}$.
For $x < 1$,$|x-1| = -(x-1)$,so $f(x) = \frac{x-1}{-(x-1)+2(x-1)^2} = \frac{1}{-1+2(x-1)} = \frac{1}{2x-3}$.
Now,$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{2x-1} = \frac{1}{2(1)-1} = 1$.
And $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{1}{2x-3} = \frac{1}{2(1)-3} = -1$.
Since $\lim_{x \to 1^+} f(x) \neq \lim_{x \to 1^-} f(x)$,the limit does not exist at $x=1$.
Therefore,$f(x)$ is discontinuous at $x=1$.

(C) Given $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \\ 0 & x-3 & 3x^2-27 \\ 0 & 2x-4 & 8x^2-32 \end{array} \right|$.
Expanding along the first column:
$f(x) = 1 \cdot [(x-3)(8x^2-32) - (2x-4)(3x^2-27)]$
$f(x) = (8x^3 - 32x - 24x^2 + 96) - (6x^3 - 54x - 12x^2 + 108)$
$f(x) = 2x^3 - 12x^2 + 22x - 12 = 2(x-1)(x-2)(x-3)$.
Now,$f(-x) = 2(-x-1)(-x-2)(-x-3) = -2(x+1)(x+2)(x+3)$.
Evaluating the limit:
$\lim_{x \rightarrow 1} \frac{f(x)}{f(-x)} = \lim_{x \rightarrow 1} \frac{2(x-1)(x-2)(x-3)}{-2(x+1)(x+2)(x+3)}$.
Substituting $x=1$:
$= \frac{2(1-1)(1-2)(1-3)}{-2(1+1)(1+2)(1+3)} = \frac{0}{-48} = 0$.

(D) For $f(x)$ to be right continuous at $x = -1$,we must have $\lim_{x \rightarrow -1^+} f(x) = f(-1)$.
Applying $L'H\hat{o}pital's$ rule to the limit $\lim_{x \rightarrow -1^+} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}$:
$\lim_{x \rightarrow -1^+} \frac{-\frac{1}{2\sqrt{\cos^{-1} x}} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)}{\frac{1}{2\sqrt{x+1}}} = \lim_{x \rightarrow -1^+} \frac{\sqrt{x+1}}{\sqrt{\cos^{-1} x} \sqrt{1-x} \sqrt{1+x}}$
$= \lim_{x \rightarrow -1^+} \frac{1}{\sqrt{\cos^{-1} x} \sqrt{1-x}} = \frac{1}{\sqrt{\cos^{-1}(-1)} \sqrt{1-(-1)}} = \frac{1}{\sqrt{\pi} \sqrt{2}} = \frac{1}{\sqrt{2\pi}}$.
Given $f(-1) = \frac{1}{\sqrt{\lambda \pi}}$,we equate: $\frac{1}{\sqrt{2\pi}} = \frac{1}{\sqrt{\lambda \pi}}$.
Squaring both sides gives $2\pi = \lambda \pi$,so $\lambda = 2$.

(B) We need to evaluate the limit: $\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$.
Given $f(0) = 0$,the expression becomes $\lim_{x \rightarrow 0} \frac{x \left(1 + \frac{1}{2} \sin (\log x^2) \right) - 0}{x}$.
Simplifying the expression by canceling $x$ (since $x \neq 0$ as $x \rightarrow 0$):
$= \lim_{x \rightarrow 0} \left(1 + \frac{1}{2} \sin (\log x^2) \right)$.
As $x \rightarrow 0$,$x^2 \rightarrow 0^+$,so $\log x^2 \rightarrow -\infty$.
The function $\sin(\log x^2)$ oscillates between $-1$ and $1$ as $x \rightarrow 0$.
Therefore,the limit $\lim_{x \rightarrow 0} \sin(\log x^2)$ does not exist.
Consequently,$\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$ does not exist.

(A) The given limit is $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{2n} e^{-k/n}$.
This is a Riemann sum of the form $\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{k=1}^{n(b-a)} \frac{1}{n} f(a + k/n)$.
Here,$f(x) = e^{-x}$,$a=0$,and $b=2$.
Thus,the integral is $\int_0^2 e^{-x} dx$.
Evaluating the integral: $\int_0^2 e^{-x} dx = [-e^{-x}]_0^2$.
$= -e^{-2} - (-e^0) = 1 - e^{-2}$.

(C) Let $m = \frac{1}{n}$. As $n \rightarrow \infty$,$m \rightarrow 0$.
$f(x) = \lim _{m \rightarrow 0} \frac{1}{m^2} \left(x^m - x^{\frac{m}{m+1}}\right)$
$f(x) = \lim _{m \rightarrow 0} \frac{x^{\frac{m}{m+1}}}{m^2} \left(x^{m - \frac{m}{m+1}} - 1\right) = \lim _{m \rightarrow 0} \frac{x^{\frac{m}{m+1}}}{m^2} \left(x^{\frac{m^2}{m+1}} - 1\right)$
$f(x) = \lim _{m \rightarrow 0} x^{\frac{m}{m+1}} \cdot \left(\frac{x^{\frac{m^2}{m+1}} - 1}{\frac{m^2}{m+1}}\right) \cdot \frac{1}{m+1}$
Using the standard limit $\lim _{t \rightarrow 0} \frac{a^t - 1}{t} = \log a$,we get:
$f(x) = x^0 \cdot \log x \cdot \frac{1}{0+1} = \log x$
Now,$\int x f(x) d x = \int x \log x d x$.
Using integration by parts: $\int u v d x = u \int v d x - \int (u' \int v d x) d x$.
Let $u = \log x$ and $v = x$.
$\int x \log x d x = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} d x$
$= \frac{x^2}{2} \log x - \int \frac{x}{2} d x = \frac{x^2}{2} \log x - \frac{x^2}{4} + C$.

(C) The function $f(x) = |x-3| + |x+5|$ is differentiable everywhere except at the points where the expressions inside the absolute values are zero,which are $x = 3$ and $x = -5$.
Thus,the set $A$ is $\mathbb{R} \setminus \{-5, 3\}$.
We are looking for real numbers $x$ such that $x \in (-\infty, -3) \cup (5, \infty)$ and $x \notin A$.
Since $A = \mathbb{R} \setminus \{-5, 3\}$,the condition $x \notin A$ implies $x \in \{-5, 3\}$.
Checking the intervals:
For $x = -5$,$-5$ is not in $(-\infty, -3)$ because the interval is open at $-3$ and $-5 < -3$,but wait,$-5$ is in $(-\infty, -3)$.
For $x = 3$,$3$ is not in $(-\infty, -3) \cup (5, \infty)$ because $3$ lies between $-3$ and $5$.
Thus,the only value in the set $\{-5, 3\}$ that lies in $(-\infty, -3) \cup (5, \infty)$ is $-5$.
Therefore,the number of such real numbers is $1$.

(B) The matrix is a $3 \times 3$ matrix given by $A = [a_{ij}]$.
We calculate each element $a_{ij} = \frac{1}{3}|i - 5j|$:
$a_{11} = \frac{1}{3}|1 - 5(1)| = \frac{1}{3}|-4| = \frac{4}{3}$
$a_{12} = \frac{1}{3}|1 - 5(2)| = \frac{1}{3}|-9| = 3$
$a_{13} = \frac{1}{3}|1 - 5(3)| = \frac{1}{3}|-14| = \frac{14}{3}$
$a_{21} = \frac{1}{3}|2 - 5(1)| = \frac{1}{3}|-3| = 1$
$a_{22} = \frac{1}{3}|2 - 5(2)| = \frac{1}{3}|-8| = \frac{8}{3}$
$a_{23} = \frac{1}{3}|2 - 5(3)| = \frac{1}{3}|-13| = \frac{13}{3}$
$a_{31} = \frac{1}{3}|3 - 5(1)| = \frac{1}{3}|-2| = \frac{2}{3}$
$a_{32} = \frac{1}{3}|3 - 5(2)| = \frac{1}{3}|-7| = \frac{7}{3}$
$a_{33} = \frac{1}{3}|3 - 5(3)| = \frac{1}{3}|-12| = 4$
Thus,the matrix is $\left[\begin{array}{ccc}\frac{4}{3} & 3 & \frac{14}{3} \\ 1 & \frac{8}{3} & \frac{13}{3} \\ \frac{2}{3} & \frac{7}{3} & 4\end{array}\right]$.

(B) Given $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2 \end{bmatrix}$.
Since $A$ is a symmetric matrix,$A^T = A$.
Comparing the elements of $A$ and $A^T$:
$\begin{bmatrix} 1 & 2 & y \\ 2 & 2 & 1 \\ 1 & x & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & x \\ y & 1 & 2 \end{bmatrix}$.
By comparing corresponding elements,we get $y = 1$ and $x = 1$.
Now,check if $A$ is singular for these values,i.e.,$|A| = 0$.
$|A| = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 2 \end{vmatrix} = 1(4 - 1) - 2(4 - 1) + 1(2 - 2) = 1(3) - 2(3) + 0 = 3 - 6 = -3$.
Since $|A| = -3 \neq 0$,there are no values of $(x, y)$ that satisfy both conditions simultaneously.
Therefore,the number of such ordered pairs is $0$.

(C) For any square matrix $A$ of order $n \times n$,the property of the adjoint matrix is given by $\operatorname{adj}(A) \cdot A = \operatorname{det}(A) \cdot I_n$.
Taking the determinant on both sides,we get $\operatorname{det}(\operatorname{adj}(A) \cdot A) = \operatorname{det}(\operatorname{det}(A) \cdot I_n)$.
Using the property $\operatorname{det}(AB) = \operatorname{det}(A) \operatorname{det}(B)$,we have $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^n \cdot \operatorname{det}(I_n)$.
Since $\operatorname{det}(I_n) = 1$,this simplifies to $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^n$.
For a matrix of order $n=3$,we have $\operatorname{det}(\operatorname{adj}(A)) \cdot \operatorname{det}(A) = (\operatorname{det}(A))^3$.
Dividing both sides by $\operatorname{det}(A)$ (since $A$ is non-singular,$\operatorname{det}(A) \neq 0$),we get $\operatorname{det}(\operatorname{adj}(A)) = (\operatorname{det}(A))^2$.

(D) Given the equation $2A + 3B - 5C = 0$.
Rearranging for $C$,we get $5C = 2A + 3B$.
Substitute the matrices $A$ and $B$:
$5C = 2\left[\begin{array}{cccc}2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right] + 3\left[\begin{array}{cccc}2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3\end{array}\right]$
$5C = \left[\begin{array}{cccc}4 & 2 & 6 & -2 \\ 2 & -4 & 4 & -6\end{array}\right] + \left[\begin{array}{cccc}6 & 3 & 0 & 9 \\ 3 & -3 & 6 & 9\end{array}\right]$
$5C = \left[\begin{array}{cccc}4+6 & 2+3 & 6+0 & -2+9 \\ 2+3 & -4-3 & 4+6 & -6+9\end{array}\right]$
$5C = \left[\begin{array}{cccc}10 & 5 & 6 & 7 \\ 5 & -7 & 10 & 3\end{array}\right]$
Dividing by $5$,we get $C = \left[\begin{array}{cccc}10/5 & 5/5 & 6/5 & 7/5 \\ 5/5 & -7/5 & 10/5 & 3/5\end{array}\right] = \left[\begin{array}{cccc}2 & 1 & 6/5 & 7/5 \\ 1 & -7/5 & 2 & 3/5\end{array}\right]$.

(B) Given $X^T B^T = A^T$. Taking the transpose on both sides,we get $(X^T B^T)^T = (A^T)^T$,which implies $BX = A$.
First,calculate the determinant of $B$: $|B| = 3(0 - (-8)) - 5(0 - 48) - 7(0 - (-6)) = 3(8) - 5(-48) - 7(6) = 24 + 240 - 42 = 222$.
The adjoint of $B$ is $\text{adj}(B) = \begin{bmatrix} 8 & 7 & 33 \\ 48 & 42 & -24 \\ 6 & 33 & -3 \end{bmatrix}$.
Thus,$X = B^{-1}A = \frac{1}{222} \begin{bmatrix} 8 & 7 & 33 \\ 48 & 42 & -24 \\ 6 & 33 & -3 \end{bmatrix} \begin{bmatrix} 0 \\ -6 \\ 8 \end{bmatrix}$.
$X = \frac{1}{222} \begin{bmatrix} 0 - 42 + 264 \\ 0 - 252 - 192 \\ 0 - 198 - 24 \end{bmatrix} = \frac{1}{222} \begin{bmatrix} 222 \\ -444 \\ -222 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}$.
So,$D = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}$.
Finally,$D^T A = \begin{bmatrix} 1 & -2 & -1 \end{bmatrix} \begin{bmatrix} 0 \\ -6 \\ 8 \end{bmatrix} = (1)(0) + (-2)(-6) + (-1)(8) = 0 + 12 - 8 = 4$.

(C) Given the matrix equation: $\begin{bmatrix} x & 4 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$
First,multiply the first two matrices: $\begin{bmatrix} x & 4 & -1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 2x+4 & x-2 & 4 \end{bmatrix}$
Now,multiply this result by the third matrix: $\begin{bmatrix} 2x+4 & x-2 & 4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$
This gives the scalar equation: $x(2x+4) + 4(x-2) + 4(-1) = 0$
Expanding the terms: $2x^2 + 4x + 4x - 8 - 4 = 0$
Simplifying: $2x^2 + 8x - 12 = 0$
Dividing by $2$: $x^2 + 4x - 6 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-4 \pm \sqrt{16 - 4(1)(-6)}}{2} = \frac{-4 \pm \sqrt{16 + 24}}{2} = \frac{-4 \pm \sqrt{40}}{2}$
$x = \frac{-4 \pm 2\sqrt{10}}{2} = -2 \pm \sqrt{10}$

(A) Given the matrix equation: $\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$
Multiplying the matrices,we get:
$\alpha = x \cos \theta - y \sin \theta$ $(i)$
$\beta = x \sin \theta + y \cos \theta$ $(ii)$
$\gamma = z$ $(iii)$
Squaring and adding equations $(i)$ and $(ii)$:
$\alpha^2 + \beta^2 = (x \cos \theta - y \sin \theta)^2 + (x \sin \theta + y \cos \theta)^2$
$\alpha^2 + \beta^2 = x^2 \cos^2 \theta + y^2 \sin^2 \theta - 2xy \sin \theta \cos \theta + x^2 \sin^2 \theta + y^2 \cos^2 \theta + 2xy \sin \theta \cos \theta$
$\alpha^2 + \beta^2 = x^2(\cos^2 \theta + \sin^2 \theta) + y^2(\sin^2 \theta + \cos^2 \theta) = x^2 + y^2$
Now,consider the expression $\frac{x^2+y^2+z^2}{\gamma}$.
Substituting $x^2+y^2 = \alpha^2+\beta^2$ and $z = \gamma$:
$\frac{x^2+y^2+z^2}{\gamma} = \frac{\alpha^2+\beta^2+\gamma^2}{\gamma}$
Since $\gamma = z$,this is equal to $\frac{\alpha^2+\beta^2+\gamma^2}{z}$.

(C) Given $[x \ y \ z] A^{T} = B^{T}$. Taking transpose on both sides,we get $([x \ y \ z] A^{T})^{T} = (B^{T})^{T}$.
This implies $A [x \ y \ z]^{T} = B$.
Substituting the matrices:
$\begin{bmatrix} 1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -1 \\ -2 \\ 4 \end{bmatrix}$.
This gives the system of linear equations:
$x + 5y + 3z = -1$ ...$(i)$
$2x + 4y = -2 \implies x + 2y = -1 \implies x = -1 - 2y$ ...(ii)
$3x - y - 5z = 4$ ...(iii)
Substituting $x = -1 - 2y$ into $(i)$:
$(-1 - 2y) + 5y + 3z = -1 \implies 3y + 3z = 0 \implies y = -z$.
Substituting $x = -1 - 2y$ and $y = -z$ into (iii):
$3(-1 - 2(-z)) - (-z) - 5z = 4$
$3(-1 + 2z) + z - 5z = 4$
$-3 + 6z - 4z = 4 \implies 2z = 7 \implies z = 3.5$.
Then $y = -3.5$ and $x = -1 - 2(-3.5) = -1 + 7 = 6$.
Thus,$x + y + z = 6 - 3.5 + 3.5 = 6$.

(B) Given matrices are $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$ and $B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{bmatrix}$.
Both matrices $A$ and $B$ have order $2 \times 3$.
Therefore,the order of $A^T$ and $B^T$ is $3 \times 2$.
Now,let us check the order of the matrices in option $B$:
The order of $A^T B^T A$ is $(3 \times 2) \times (3 \times 2) \times (2 \times 3)$,which is not defined because the inner dimensions do not match.
However,checking the order of $A^T B B^T A$:
Order of $A^T B B^T A = (3 \times 2) \times (2 \times 3) \times (3 \times 2) \times (2 \times 3) = 3 \times 3$.
Order of $B^T A A^T B = (3 \times 2) \times (2 \times 3) \times (3 \times 2) \times (2 \times 3) = 3 \times 3$.
Since both expressions $A^T B B^T A$ and $B^T A A^T B$ result in a $3 \times 3$ matrix,their orders are equal.

(A) Given that $A$ is a skew-symmetric matrix of order $3 \times 3$,its diagonal elements are all zero,i.e.,$\operatorname{diag}(A) = (0, 0, 0)$.
Let $D_1 = \operatorname{diag}(a, b, c)$ and $D_2 = \operatorname{diag}(3, 3, 3) = 3I$,where $I$ is the identity matrix.
The trace of a matrix is the sum of its diagonal elements.
For any skew-symmetric matrix $A$,the product of a diagonal matrix $D$ and $A$ (i.e.,$DA$) results in a matrix where the diagonal elements are $d_{ii} \times a_{ii}$. Since $a_{ii} = 0$,the diagonal elements of $DA$ are $0$.
Thus,$\operatorname{diag}(D_1 D_2 A) = (0, 0, 0)$,$\operatorname{diag}(D_1 A) = (0, 0, 0)$,and $\operatorname{diag}(D_2 A) = (0, 0, 0)$.
Now,consider the expression inside the trace: $M = D_1 D_2 A + D_1 D_2 + D_1 A + D_2 A$.
The diagonal elements of $M$ are $\operatorname{diag}(M) = \operatorname{diag}(D_1 D_2) + \operatorname{diag}(D_1 A) + \operatorname{diag}(D_2 A) + \operatorname{diag}(D_1 D_2 A)$.
Since $\operatorname{diag}(D_1 A) = \operatorname{diag}(D_2 A) = \operatorname{diag}(D_1 D_2 A) = (0, 0, 0)$,we have $\operatorname{diag}(M) = \operatorname{diag}(D_1 D_2) = (3a, 3b, 3c)$.
Therefore,$\operatorname{Tr}(M) = 3a + 3b + 3c$.
We need to calculate $\operatorname{Tr}(M) - \operatorname{Tr}(D_1 + D_2)$.
$\operatorname{Tr}(D_1 + D_2) = (a+3) + (b+3) + (c+3) = a + b + c + 9$.
Finally,$\operatorname{Tr}(M) - \operatorname{Tr}(D_1 + D_2) = (3a + 3b + 3c) - (a + b + c + 9) = 2a + 2b + 2c - 9$.

(B) Given $A = \left[\begin{array}{ccc}2 & 0 & -3 \\ 4 & 3 & 1 \\ -5 & 7 & 2\end{array}\right]$.
Any square matrix $A$ can be uniquely expressed as $A = P + Q$,where $P = \frac{1}{2}(A + A^T)$ is a symmetric matrix and $Q = \frac{1}{2}(A - A^T)$ is a skew-symmetric matrix.
We know that for any symmetric matrix $P$,$P^T = P$,and for any skew-symmetric matrix $Q$,$Q^T = -Q$.
Therefore,$P^T - Q^T = P - (-Q) = P + Q = A$.
Thus,$P^T - Q^T = A = \left[\begin{array}{ccc}2 & 0 & -3 \\ 4 & 3 & 1 \\ -5 & 7 & 2\end{array}\right]$.

(C) We know that for any non-singular matrix $M$,$M^{-1} = \frac{\operatorname{Adj}(M)}{\operatorname{det}(M)}$,which implies $\operatorname{Adj}(M) = \operatorname{det}(M) \cdot M^{-1}$.
Given the expression $((\operatorname{det} A)(\operatorname{det} B)) B^{-1} A^{-1}$.
Since $\operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B)$,we can write the expression as $\operatorname{det}(AB) \cdot (B^{-1} A^{-1})$.
Using the property of matrix inversion,$(AB)^{-1} = B^{-1} A^{-1}$.
Therefore,the expression becomes $\operatorname{det}(AB) \cdot (AB)^{-1}$.
By the definition of the adjoint matrix,this is equal to $\operatorname{Adj}(AB)$.

(A) First,we find the inverse of $S$. The determinant $|S| = 0(0-1) - 1(0-1) + 1(1-0) = 1 + 1 = 2$.
The adjugate of $S$ is $\text{adj}(S) = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$.
Thus,$S^{-1} = \frac{1}{2} \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$.
Next,we compute $SA = \frac{1}{2} \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} b+c & c-a & b-a \\ c-b & c+a & a-b \\ b-c & a-c & a+b \end{bmatrix} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} S$.
Alternatively,calculating $SA$ directly gives $\begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} S$.
Therefore,$SAS^{-1} = (SA)S^{-1} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} S S^{-1} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} I = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$.

(C) Let the matrix be $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & -1 & 7 \\ 2 & 4 & 6 \end{bmatrix}$.
Elements $3, 7, 6$ are in the third column $(C_3)$.
Cofactor $a$ of element $3$ $(A_{13})$: $a = (-1)^{1+3} \begin{vmatrix} 4 & -1 \\ 2 & 4 \end{vmatrix} = (16 - (-2)) = 18$.
Cofactor $b$ of element $7$ $(A_{23})$: $b = (-1)^{2+3} \begin{vmatrix} 1 & 2 \\ 2 & 4 \end{vmatrix} = -(4 - 4) = 0$.
Cofactor $c$ of element $6$ $(A_{33})$: $c = (-1)^{3+3} \begin{vmatrix} 1 & 2 \\ 4 & -1 \end{vmatrix} = (-1 - 8) = -9$.
We need to calculate $\begin{bmatrix} a & b & c \end{bmatrix} \begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix} + \begin{bmatrix} a & b & c \end{bmatrix} \begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix}$.
This is equal to $\begin{bmatrix} a & b & c \end{bmatrix} \left( \begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix} + \begin{bmatrix} 3 \\ 7 \\ 6 \end{bmatrix} \right) = \begin{bmatrix} 18 & 0 & -9 \end{bmatrix} \begin{bmatrix} 4 \\ 11 \\ 8 \end{bmatrix}$.
$= (18 \times 4) + (0 \times 11) + (-9 \times 8) = 72 + 0 - 72 = 0$.

(D) Given $BAC = I$.
Taking the inverse of both sides,we get $(BAC)^{-1} = I^{-1} = I$.
Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we have $C^{-1}A^{-1}B^{-1} = I$.
Multiplying by $C$ on the left and $B$ on the right,we get $A^{-1} = CB$.
Now,calculate the product $CB$:
$A^{-1} = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$
$A^{-1} = \begin{bmatrix} (-1)(2)+(0)(1)+(1)(-1) & (-1)(6)+(0)(0)+(1)(1) & (-1)(4)+(0)(1)+(1)(-1) \\ (1)(2)+(1)(1)+(3)(-1) & (1)(6)+(1)(0)+(3)(1) & (1)(4)+(1)(1)+(3)(-1) \\ (2)(2)+(0)(1)+(2)(-1) & (2)(6)+(0)(0)+(2)(1) & (2)(4)+(0)(1)+(2)(-1) \end{bmatrix}$
$A^{-1} = \begin{bmatrix} -3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} -\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta \end{bmatrix}$.
First,calculate the determinant $|A| = (-\cot \theta)(-\cot \theta) - (\operatorname{cosec} \theta)(\operatorname{cosec} \theta) = \cot^2 \theta - \operatorname{cosec}^2 \theta = -1$.
Now,find $A^{-1} = \frac{1}{|A|} \operatorname{adj}(A) = -1 \begin{bmatrix} -\cot \theta & -\operatorname{cosec} \theta \\ -\operatorname{cosec} \theta & -\cot \theta \end{bmatrix} = \begin{bmatrix} \cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta \end{bmatrix}$.
For $A^{-1} = A$:
$\begin{bmatrix} \cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta \end{bmatrix} = \begin{bmatrix} -\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta \end{bmatrix}$.
Comparing elements,we get $\cot \theta = -\cot \theta$,which implies $2 \cot \theta = 0$,so $\cot \theta = 0$. This occurs at $\theta_1 = \frac{\pi}{2}$.
For $A^{-1} + A = O$:
$\begin{bmatrix} \cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & \cot \theta \end{bmatrix} + \begin{bmatrix} -\cot \theta & \operatorname{cosec} \theta \\ \operatorname{cosec} \theta & -\cot \theta \end{bmatrix} = \begin{bmatrix} 0 & 2 \operatorname{cosec} \theta \\ 2 \operatorname{cosec} \theta & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
This implies $2 \operatorname{cosec} \theta = 0$,which means $\operatorname{cosec} \theta = 0$. Since $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$,it can never be zero for any real $\theta$. Thus,such $\theta_2$ does not exist.

(B) Given $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$.
The characteristic equation is given by $|A - \lambda I| = 0$.
$\begin{vmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & -\lambda \end{vmatrix} = 0$
$(1-\lambda) [-\lambda(1-\lambda) - 1] = 0$
$(1-\lambda) [-\lambda + \lambda^2 - 1] = 0$
$-\lambda + \lambda^2 - 1 + \lambda^2 - \lambda^3 + \lambda = 0$
$\lambda^3 - 2\lambda^2 + 1 = 0$
By the Cayley-Hamilton theorem,every matrix satisfies its characteristic equation:
$A^3 - 2A^2 + I = 0$
Multiplying both sides by $A^{-1}$:
$A^3 A^{-1} - 2A^2 A^{-1} + I A^{-1} = 0$
$A^2 - 2A + A^{-1} = 0$
$A^{-1} = 2A - A^2$.

(D) matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$\det(A) = 0$.
Calculating the determinant of matrix $A$:
$\det(A) = 1(x - 1) - 1(1 - x) + x(1 - x^2) = 0$
$x - 1 - 1 + x + x - x^3 = 0$
$-x^3 + 3x - 2 = 0$
$x^3 - 3x + 2 = 0$
Factoring the cubic equation:
$(x - 1)(x^2 + x - 2) = 0$
$(x - 1)(x - 1)(x + 2) = 0$
$(x - 1)^2(x + 2) = 0$
The values of $x$ are $1$ and $-2$.
The distinct values of $x$ are $1$ and $-2$.
The sum of these distinct values is $1 + (-2) = -1$.

(C) The order of the matrix $P$ is $m \times n$.
The rank of a matrix $P$,denoted by $\rho$,cannot exceed the minimum of its dimensions.
Therefore,$\rho \leq \min(m, n)$ ...$(i)$
By definition,the rank of a matrix is the order of the largest non-singular submatrix.
Since there exists a $k^{\text{th}}$ order non-singular submatrix,the rank $\rho$ must be at least $k$.
Therefore,$\rho \geq k$ ...(ii)
Combining equations $(i)$ and (ii),we get:
$k \leq \rho \leq \min(m, n)$.

(A) The rank of a matrix is defined as the order of the largest non-singular sub-matrix.
Since all sub-matrices of order $k$ are singular,the rank $\rho$ must be less than $k$,i.e.,$\rho < k$ ... $(i)$.
Since there exists at least one non-singular sub-matrix of order $r$,the rank $\rho$ must be at least $r$,i.e.,$r \leq \rho$ ... (ii).
Combining inequalities $(i)$ and (ii),we get $r \leq \rho < k$.

(B) The given matrix is $A = \begin{bmatrix} 1 & 1 & 1 & 3 \\ 2 & 2 & -1 & 3 \\ 1 & 1 & -1 & 1 \end{bmatrix}$.
Applying row operations to reduce the matrix to row-echelon form:
Perform $R_2 \rightarrow R_2 - 2R_1$ and $R_3 \rightarrow R_3 - R_1$:
$A \sim \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 0 & -3 & -3 \\ 0 & 0 & -2 & -2 \end{bmatrix}$.
Perform $R_2 \rightarrow -\frac{1}{3}R_2$ and $R_3 \rightarrow -\frac{1}{2}R_3$:
$A \sim \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}$.
Perform $R_3 \rightarrow R_3 - R_2$:
$A \sim \begin{bmatrix} 1 & 1 & 1 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.
The number of non-zero rows in the row-echelon form is $2$.
Therefore,the rank of the matrix $A$ is $2$.

(C) Let $A = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 1 & -1 & 4 \\ 2 & 2 & 8 \end{array}\right]$.
Applying row operations to reduce the matrix to row-echelon form:
$R_3 \rightarrow R_3 - R_1$ and $R_4 \rightarrow R_4 - 2R_1$ gives:
$\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & -1 & 2 \\ 0 & 2 & 4 \end{array}\right]$.
Next,applying $R_3 \rightarrow R_3 + R_2$ and $R_4 \rightarrow R_4 - 2R_2$ gives:
$\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 8 \end{array}\right]$.
Swapping $R_3$ and $R_4$ gives:
$\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{array}\right]$.
The number of non-zero rows in the row-echelon form is $3$.
Therefore,the rank of the matrix $A$ is $3$.

(A) Given matrices are $A=\left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right]$,$B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$,and $C=\left[\begin{array}{cc}0 & i \\ i & 0\end{array}\right]$.
Calculating the squares of the matrices:
$A^2 = \left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right] \left[\begin{array}{cc}i & 0 \\ 0 & -i\end{array}\right] = \left[\begin{array}{cc}i^2 & 0 \\ 0 & (-i)^2\end{array}\right] = \left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] = -I$.
$B^2 = \left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] = \left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] = -I$.
$C^2 = \left[\begin{array}{cc}0 & i \\ i & 0\end{array}\right] \left[\begin{array}{cc}0 & i \\ i & 0\end{array}\right] = \left[\begin{array}{cc}i^2 & 0 \\ 0 & i^2\end{array}\right] = \left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] = -I$.
Summing the squares:
$A^2+B^2+C^2 = (-I) + (-I) + (-I) = -3I = \left[\begin{array}{cc}-3 & 0 \\ 0 & -3\end{array}\right]$.
Now,calculating $3 A^2 B^2 C^2$:
$3 A^2 B^2 C^2 = 3(-I)(-I)(-I) = 3(-I)^3 = 3(-I) = -3I$.
Since $A^2+B^2+C^2 = -3I$ and $3 A^2 B^2 C^2 = -3I$,we conclude that $A^2+B^2+C^2 = 3 A^2 B^2 C^2$.

(B) Given the matrix equation: $\begin{bmatrix} 3x^2+x+3 & 2x^2-x+4 & 7x^2+8x+5 \\ 5x^2+3x+2 & 4x^2-2x-1 & 7x^2+5x+8 \\ 3x^2+2x+5 & 4x^2-x-2 & 3x^2+8x+7 \end{bmatrix} = Px^2+Qx+R$.
To find the matrix $R$,we set $x=0$ in the given expression:
$R = \begin{bmatrix} 3(0)^2+0+3 & 2(0)^2-0+4 & 7(0)^2+8(0)+5 \\ 5(0)^2+3(0)+2 & 4(0)^2-2(0)-1 & 7(0)^2+5(0)+8 \\ 3(0)^2+2(0)+5 & 4(0)^2-0-2 & 3(0)^2+8(0)+7 \end{bmatrix} = \begin{bmatrix} 3 & 4 & 5 \\ 2 & -1 & 8 \\ 5 & -2 & 7 \end{bmatrix}$.
Now,we calculate the determinant of $R$:
$\det R = 3((-1)(7) - (8)(-2)) - 4((2)(7) - (8)(5)) + 5((2)(-2) - (-1)(5))$
$\det R = 3(-7 + 16) - 4(14 - 40) + 5(-4 + 5)$
$\det R = 3(9) - 4(-26) + 5(1)$
$\det R = 27 + 104 + 5 = 136$.

(A) The system of equations can be written in matrix form $AX=B$ as: $\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & -2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}3 \\ 1 \\ 2\end{array}\right]$
The augmented matrix is: $\left[\begin{array}{ccccc}1 & 2 & -1 & : & 3 \\ 3 & -1 & 2 & : & 1 \\ 2 & -2 & 3 & : & 2\end{array}\right]$
Applying row operations $R_2 \rightarrow R_2-3R_1$ and $R_3 \rightarrow R_3-2R_1$:
$\sim\left[\begin{array}{ccccc}1 & 2 & -1 & : & 3 \\ 0 & -7 & 5 & : & -8 \\ 0 & -6 & 5 & : & -4\end{array}\right]$
Applying $R_3 \rightarrow 7R_3-6R_2$:
$\sim\left[\begin{array}{ccccc}1 & 2 & -1 & : & 3 \\ 0 & -7 & 5 & : & -8 \\ 0 & 0 & 5 & : & 20\end{array}\right]$
Since $\operatorname{Rank}(A:B)=\operatorname{Rank}(A)=3$,the system has a unique solution.
From the row-echelon form:
$5z=20 \Rightarrow z=4$
$-7y+5(4)=-8 \Rightarrow -7y=-28 \Rightarrow y=4$
$x+2(4)-4=3 \Rightarrow x+4=3 \Rightarrow x=-1$
Wait,re-calculating: $x+8-4=3 \Rightarrow x+4=3 \Rightarrow x=-1$. Let's re-check the system: $x+2y-z=3 \Rightarrow -1+8-4 = 3$ (Correct).
So,$\alpha=-1, \beta=4, \gamma=4$.
Then $\alpha^2+\beta^2+\gamma^2 = (-1)^2 + 4^2 + 4^2 = 1 + 16 + 16 = 33$.

(B) Given $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$.
The characteristic equation is given by $|A-\lambda I|=0$.
$\left|\begin{array}{ccc}1-\lambda & 1 & 3 \\ 5 & 2-\lambda & 6 \\ -2 & -1 & -3-\lambda\end{array}\right|=0$
Expanding the determinant:
$(1-\lambda)[(2-\lambda)(-3-\lambda) - (-6)] - 1[5(-3-\lambda) - (-12)] + 3[5(-1) - (-2)(2-\lambda)] = 0$
$(1-\lambda)[\lambda^2+\lambda-6+6] - 1[-15-5\lambda+12] + 3[-5+4-2\lambda] = 0$
$(1-\lambda)(\lambda^2+\lambda) - 1(-5\lambda-3) + 3(-2\lambda-1) = 0$
$\lambda^2+\lambda-\lambda^3-\lambda^2 + 5\lambda+3 - 6\lambda-3 = 0$
$-\lambda^3 = 0 \Rightarrow \lambda^3 = 0$.
By the Cayley-Hamilton theorem,every matrix satisfies its characteristic equation,so $A^3 = 0$.
Since $A^3 = 0$,it follows that $A^4 = A^3 \cdot A = 0$ and $A^5 = A^3 \cdot A^2 = 0$.
Therefore,$A+A^3+A^4+A^5+3I = A + 0 + 0 + 0 + 3I = A + 3I$.
$A+3I = \left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right] + \left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right] = \left[\begin{array}{ccc}4 & 1 & 3 \\ 5 & 5 & 6 \\ -2 & -1 & 0\end{array}\right]$.

(C) Given $f(x) = x^3 - x$ and $g(x) = \sin^2 x$.
First,calculate $g\left(\frac{\pi}{6}\right)$:
$g\left(\frac{\pi}{6}\right) = \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Now,substitute this into $f(x)$:
$f\left(g\left(\frac{\pi}{6}\right)\right) = f\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^3 - \frac{1}{4} = \frac{1}{64} - \frac{1}{4} = \frac{1 - 16}{64} = -\frac{15}{64}$.

(C) Given $f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx}$. The domain is determined by $a^2 - x^2 \geq 0$,so $x \in [-|a|, |a|]$ (excluding $x=0$).
For $x \in (0, |a|]$,$f(x) = \frac{a}{b} + \frac{1}{b}\sqrt{\frac{a^2}{x^2} - 1}$.
Let $g(x) = \frac{a^2}{x^2} - 1$. As $x$ increases from $0$ to $|a|$,$g(x)$ decreases from $\infty$ to $0$.
Thus,$f(x)$ is a strictly monotonic function on its domain intervals.
Since the function is strictly monotonic on its domain,it is one-one.
For the range,as $x \to 0^+$,$f(x) \to \infty$,and as $x \to |a|$,$f(x) = \frac{a^2}{b|a|} = \frac{|a|}{b}$.
Since the function covers all real values in its codomain (assuming $R$ as codomain),it is onto.
Therefore,$f$ is both one-one and onto.

(A) Given that: $f(x) = \frac{2x+1}{3}$ ...$(i)$
Also,$f(\alpha) = \frac{1}{\alpha}$.
Substituting $\alpha$ in the function definition:
$\frac{2\alpha+1}{3} = \frac{1}{\alpha}$
$\Rightarrow 2\alpha^2 + \alpha = 3$
$\Rightarrow 2\alpha^2 + \alpha - 3 = 0$
Factoring the quadratic equation:
$2\alpha^2 + 3\alpha - 2\alpha - 3 = 0$
$\Rightarrow \alpha(2\alpha + 3) - 1(2\alpha + 3) = 0$
$\Rightarrow (\alpha - 1)(2\alpha + 3) = 0$
Thus,the possible values for $\alpha$ are $\alpha = 1$ and $\alpha = -\frac{3}{2}$.
The sum of all possible values of $\alpha$ is $1 + (-\frac{3}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}$.

(B) Given the equation $2^x+2^y=2$.
Since $2^y > 0$ for all real $y$,we must have $2-2^x > 0$.
This implies $2^x < 2$.
Taking the logarithm base $2$ on both sides,we get $x < 1$.
Therefore,the domain of the function is $x \in (-\infty, 1)$.

(B) Given the function $f(x) = \begin{cases} 3x-1, & x > 1 \\ x^2+1, & x \leq 1 \end{cases}$.
$1$. The domain $A$ is the set of all possible input values $x$. Since the function is defined for all $x > 1$ and $x \leq 1$,the domain $A = (-\infty, 1] \cup (1, \infty) = (-\infty, \infty) = R$.
$2$. To find the range $B$,we analyze the function in two parts:
For $x > 1$,$f(x) = 3x - 1$. As $x \to 1^+$,$f(x) \to 3(1) - 1 = 2$. Thus,$f(x) \in (2, \infty)$.
For $x \leq 1$,$f(x) = x^2 + 1$. The minimum value occurs at $x = 0$,where $f(0) = 1$. As $x \to -\infty$,$f(x) \to \infty$. Thus,$f(x) \in [1, \infty)$.
$3$. The range $B$ is the union of these intervals: $B = (2, \infty) \cup [1, \infty) = [1, \infty)$.
$4$. Finally,$A - B = R - [1, \infty) = (-\infty, 1)$.

(C) Given: $f(x) = x^3 - x$ and $g(x) = \sin(2x)$.
We need to solve $f(g(x)) > 0$.
$f(g(x)) = (\sin 2x)^3 - \sin 2x > 0$.
Factorizing the expression: $\sin 2x (\sin^2 2x - 1) > 0$.
Since $\sin^2 2x - 1 = -\cos^2 2x$,we have: $-\sin 2x \cos^2 2x > 0$.
This implies $\sin 2x \cos^2 2x < 0$.
For this to hold,we must have $\sin 2x < 0$ and $\cos 2x \neq 0$.
In the interval $x \in (0, 2\pi)$,$2x \in (0, 4\pi)$.
$\sin 2x < 0$ when $2x \in (\pi, 2\pi) \cup (3\pi, 4\pi)$,which means $x \in (\frac{\pi}{2}, \pi) \cup (\frac{3\pi}{2}, 2\pi)$.
Also,$\cos 2x \neq 0$ implies $2x \neq \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$,so $x \neq \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
Excluding these points from the interval,we get $x \in (\frac{\pi}{2}, \frac{3\pi}{4}) \cup (\frac{3\pi}{4}, \pi) \cup (\frac{3\pi}{2}, \frac{7\pi}{4}) \cup (\frac{7\pi}{4}, 2\pi)$.
Comparing with the given options,the interval $(\frac{\pi}{2}, \frac{3\pi}{4}) \cup (\frac{3\pi}{4}, \pi)$ is a subset of the solution set.

(C) Given the function $f(x) = \sqrt{9-x^2}$.
For $f(x)$ to be defined,the expression inside the square root must be non-negative:
$9 - x^2 \geq 0$
$x^2 \leq 9$
$-3 \leq x \leq 3$
Since $x^2$ ranges from $0$ to $9$,the expression $9 - x^2$ ranges from $9 - 9 = 0$ to $9 - 0 = 9$.
Therefore,$\sqrt{9 - x^2}$ ranges from $\sqrt{0}$ to $\sqrt{9}$,which is $0$ to $3$.
Thus,the range of the function is $[0, 3]$.

(C) For $x > 3$,$f(x) = 4x - 1$. Since $x > 3$,$4x > 12$,so $4x - 1 > 11$. Thus,the range for this part is $(11, \infty)$.
For $-2 \leq x \leq 3$,$f(x) = x^2 - 2$. The minimum value is at $x = 0$,$f(0) = -2$. The maximum value is at $x = 3$,$f(3) = 7$. Thus,the range for this part is $[-2, 7]$.
For $x < -2$,$f(x) = 3x + 4$. Since $x < -2$,$3x < -6$,so $3x + 4 < -2$. Thus,the range for this part is $(-\infty, -2)$.
Combining these intervals: $(-\infty, -2) \cup [-2, 7] \cup (11, \infty) = (-\infty, 7] \cup (11, \infty)$.
This can be written as $R - (7, 11]$.

(B) Given $f(a) = \log \left| \frac{1-a}{1+a} \right|$.
We want to solve $f\left( \frac{2a}{1+a^2} \right) > 0$.
Substituting $x = \frac{2a}{1+a^2}$,we have $\log \left| \frac{1 - \frac{2a}{1+a^2}}{1 + \frac{2a}{1+a^2}} \right| > 0$.
This implies $\left| \frac{1+a^2-2a}{1+a^2+2a} \right| > 1$,which simplifies to $\left| \frac{(1-a)^2}{(1+a)^2} \right| > 1$.
Since both numerator and denominator are squares,we have $\left| \frac{1-a}{1+a} \right| > 1$.
Squaring both sides,$\frac{(1-a)^2}{(1+a)^2} > 1 \Rightarrow (1-a)^2 > (1+a)^2$.
$1 - 2a + a^2 > 1 + 2a + a^2$.
$-2a > 2a \Rightarrow 4a < 0 \Rightarrow a < 0$.
Also,we must satisfy the domain constraints $a \neq \pm 1$ and $\frac{2a}{1+a^2} \neq \pm 1$.
For $a < 0$,the condition $a \neq -1$ must be excluded.
Thus,the solution is $a \in (-\infty, 0) - \{-1\}$.