AP EAMCET 2021 Mathematics Question Paper with Answer and Solution

(B) The centroid $C$ of the triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}).$
For the given vertices $(3, -1), (1, 3),$ and $(2, 4),$ the centroid $C$ is $(\frac{3+1+2}{3}, \frac{-1+3+4}{3}) = (2, 2).$
Next,we find the intersection point $P$ of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$
Solving the system:
$x + 3y = 1$
$3x - y = -1 \Rightarrow y = 3x + 1$
Substituting $y$ into the first equation: $x + 3(3x + 1) = 1$ $\Rightarrow x + 9x + 3 = 1$ $\Rightarrow 10x = -2$ $\Rightarrow x = -\frac{1}{5}.$
Then $y = 3(-\frac{1}{5}) + 1 = -\frac{3}{5} + 1 = \frac{2}{5}.$
So,$P = (-\frac{1}{5}, \frac{2}{5}).$
The line passing through $C(2, 2)$ and $P(-\frac{1}{5}, \frac{2}{5})$ has slope $m = \frac{\frac{2}{5} - 2}{-\frac{1}{5} - 2} = \frac{-\frac{8}{5}}{-\frac{11}{5}} = \frac{8}{11}.$
The equation of the line is $y - 2 = \frac{8}{11}(x - 2)$ $\Rightarrow 11y - 22 = 8x - 16$ $\Rightarrow 8x - 11y + 6 = 0.$
Checking the options,for $(-9, -6): 8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0.$
Thus,the line passes through $(-9, -6).$

(A) First,arrange the $6$ men around a round table. The number of ways to arrange $n$ objects in a circle is $(n-1)!$. So,$6$ men can be seated in $(6-1)! = 5!$ ways.
There are $6$ gaps created between these $6$ men. We need to place $5$ women in these $6$ gaps such that no two women sit together.
The number of ways to arrange $5$ women in $6$ gaps is given by $P(6, 5) = \frac{6!}{(6-5)!} = 6!$.
Therefore,the total number of ways is $5! \times 6!$.

(C) Let $p(n) = 2 \cdot 4^{2n+1} + 3^{3n+1}$.
For $n = 1$,$p(1) = 2 \cdot 4^{2(1)+1} + 3^{3(1)+1} = 2 \cdot 4^3 + 3^4 = 2 \cdot 64 + 81 = 128 + 81 = 209$.
Since $209 = 11 \times 19$,$p(1)$ is divisible by $11$.
Assume $p(k) = 2 \cdot 4^{2k+1} + 3^{3k+1}$ is divisible by $11$ for some positive integer $k$.
Now,consider $p(k+1) = 2 \cdot 4^{2(k+1)+1} + 3^{3(k+1)+1} = 2 \cdot 4^{2k+3} + 3^{3k+4}$.
$p(k+1) = 2 \cdot 4^{2k+1} \cdot 4^2 + 3^{3k+1} \cdot 3^3 = 16 \cdot (2 \cdot 4^{2k+1}) + 27 \cdot 3^{3k+1}$.
$p(k+1) = 16 \cdot (2 \cdot 4^{2k+1} + 3^{3k+1}) + (27 - 16) \cdot 3^{3k+1}$.
$p(k+1) = 16 \cdot p(k) + 11 \cdot 3^{3k+1}$.
Since $p(k)$ is divisible by $11$ and $11 \cdot 3^{3k+1}$ is clearly divisible by $11$,$p(k+1)$ is divisible by $11$.
By the principle of mathematical induction,$2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by $11$ for all positive integers $n$.

(B) We want to find $m \in \mathbb{N}$ such that $(x+y)$ divides $(x^m+y^m)$.
Let $P(m)$ be the statement that $(x+y) \mid (x^m+y^m)$.
For $m=1$: $x^1+y^1 = x+y$,which is clearly divisible by $x+y$. Thus,$P(1)$ is true.
For $m=2$: $x^2+y^2$ is not generally divisible by $x+y$ (e.g.,if $x=1, y=1$,$2$ is divisible by $2$,but if $x=2, y=1$,$5$ is not divisible by $3$). Thus,$P(2)$ is false.
For $m=3$: $x^3+y^3 = (x+y)(x^2-xy+y^2)$,which is divisible by $x+y$. Thus,$P(3)$ is true.
In general,$x^m+y^m$ is divisible by $x+y$ if and only if $m$ is an odd natural number.
Therefore,the divisibility holds for all odd numbers $m \in \mathbb{N}$.

(B) Let $A$ be the set of integers from $1$ to $1000$ divisible by $2$,and $B$ be the set of integers from $1$ to $1000$ divisible by $3$.
We need to find $|A \cup B| = |A| + |B| - |A \cap B|$.
$|A| = \lfloor \frac{1000}{2} \rfloor = 500$.
$|B| = \lfloor \frac{1000}{3} \rfloor = 333$.
$|A \cap B|$ is the number of integers divisible by both $2$ and $3$,i.e.,divisible by $\text{lcm}(2, 3) = 6$.
$|A \cap B| = \lfloor \frac{1000}{6} \rfloor = 166$.
Therefore,$|A \cup B| = 500 + 333 - 166 = 667$.

(D) The expression $(n+24)(n+25)(n+26)(n+27)$ represents the product of $4$ consecutive natural numbers.
We know that the product of $k$ consecutive natural numbers is always divisible by $k!$.
Here,$k = 4$,so the product is divisible by $4!$.
$4! = 4 \times 3 \times 2 \times 1 = 24$.
Therefore,the expression is always divisible by $24$.

(C) The equation $(x^2+5x+5)^{x+5}=1$ holds true in the following cases:
Case $1$: The exponent is $0$ and the base is non-zero.
$x+5=0 \Rightarrow x=-5$.
Checking the base: $(-5)^2+5(-5)+5 = 25-25+5 = 5 \neq 0$. Thus,$x=-5$ is a solution.
Case $2$: The base is $1$.
$x^2+5x+5=1$ $\Rightarrow x^2+5x+4=0$ $\Rightarrow (x+1)(x+4)=0$.
So,$x=-1$ and $x=-4$ are solutions.
Case $3$: The base is $-1$ and the exponent is an even integer.
$x^2+5x+5=-1$ $\Rightarrow x^2+5x+6=0$ $\Rightarrow (x+2)(x+3)=0$.
So,$x=-2$ or $x=-3$.
For $x=-2$,the exponent is $x+5 = -2+5 = 3$ (odd),so this is not a solution.
For $x=-3$,the exponent is $x+5 = -3+5 = 2$ (even),so $x=-3$ is a solution.
The set of integer solutions is $\{-5, -1, -4, -3\}$.
Therefore,the number of integers satisfying the equation is $4$.

(C) Given the inequality $8n + 16 \leq 2^n$.
We can rewrite the expression as $2^3(n + 2) \leq 2^n$.
Let us test the options:
For $n = 5$: $8(5) + 16 = 40 + 16 = 56$ and $2^5 = 32$. Since $56 \not\leq 32$,this is false.
For $n = 6$: $8(6) + 16 = 48 + 16 = 64$ and $2^6 = 64$. Since $64 \leq 64$ is true,the statement holds for $n = 6$.

(A) Given,$mn=3$ and $\frac{1}{m}+\frac{1}{n}=\frac{4}{3}$.
From the second equation,$\frac{m+n}{mn}=\frac{4}{3}$.
Substituting $mn=3$,we get $\frac{m+n}{3}=\frac{4}{3}$,which implies $m+n=4$.
We have the system $m+n=4$ and $mn=3$. The roots of the quadratic equation $x^2-4x+3=0$ are $m$ and $n$.
$(x-1)(x-3)=0$,so $m=1, n=3$ or $m=3, n=1$.
Now,evaluate the expression $0.1+0.1^{\frac{1}{m}}+0.1^{\frac{1}{n}}$.
Substituting $m=1$ and $n=3$,we get $0.1+0.1^1+0.1^{\frac{1}{3}} = 0.1+0.1+0.1^{\frac{1}{3}} = 0.2+0.1^{\frac{1}{3}}$.

(B) Given $a = \frac{x}{y-z}$,$b = \frac{y}{z-x}$,$c = \frac{z}{x-y}$.
Let us test with values $x=1, y=2, z=4$ (ensuring denominators are non-zero).
$a = \frac{1}{2-4} = -\frac{1}{2}$
$b = \frac{2}{4-1} = \frac{2}{3}$
$c = \frac{4}{1-2} = -4$
Now calculate $ab + bc + ca + abc$:
$ab = (-\frac{1}{2})(\frac{2}{3}) = -\frac{1}{3}$
$bc = (\frac{2}{3})(-4) = -\frac{8}{3}$
$ca = (-4)(-\frac{1}{2}) = 2$
$abc = (-\frac{1}{2})(\frac{2}{3})(-4) = \frac{4}{3}$
Sum $= -\frac{1}{3} - \frac{8}{3} + 2 + \frac{4}{3} = \frac{-1-8+4}{3} + 2 = -\frac{5}{3} + 2 = \frac{1}{3}$.
Note: The original expression $ab+bc+ca$ evaluates to $-1$ for these specific variables.

(C) Given the quadratic equation $11 x^2 + 12 x - 13 = 0$.
From the relation between roots and coefficients,we have $\alpha + \beta = -\frac{12}{11}$ and $\alpha \beta = -\frac{13}{11}$.
We need to find $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2}$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta$,we get:
$\frac{(\alpha + \beta)^2 - 2 \alpha \beta}{(\alpha \beta)^2} = \frac{(-\frac{12}{11})^2 - 2(-\frac{13}{11})}{(-\frac{13}{11})^2}$.
$= \frac{\frac{144}{121} + \frac{26}{11}}{\frac{169}{121}} = \frac{\frac{144 + 286}{121}}{\frac{169}{121}} = \frac{430}{169}$.
$\frac{430}{169} \approx 2.544$.
Thus,the value is approximately $2.54$.

(B) Given the quadratic equation $7x^2 - 13x + a = 0$.
For the roots to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.
Here,$D = (-13)^2 - 4(7)(a) = 169 - 28a$.
For $D$ to be a perfect square,$169 - 28a = k^2$ for some non-negative integer $k$.
Testing the given options:
If $a = 5$,$D = 169 - 28(5) = 169 - 140 = 29$ (not a perfect square).
If $a = 6$,$D = 169 - 28(6) = 169 - 168 = 1 = 1^2$ (a perfect square).
Since $a=6$ satisfies the condition and is the smallest value among the options,the smallest possible value of $a$ is $6$.

(C) Let the four numbers in $A.P.$ be $p=a-3d, q=a-d, r=a+d, s=a+3d$.
Since $p$ and $q$ are roots of $x^2-2x+A=0$,we have $p+q=2$ and $pq=A$.
Since $r$ and $s$ are roots of $x^2-18x+B=0$,we have $r+s=18$ and $rs=B$.
Summing the roots: $(a-3d)+(a-d)+(a+d)+(a+3d) = 4a = 2+18 = 20$,so $a=5$.
From $p+q=2$,we get $(a-3d)+(a-d) = 2a-4d = 2$.
Substituting $a=5$,we get $10-4d=2$,so $4d=8$,which means $d=2$.
The roots are $p=5-3(2)=-1$,$q=5-2=3$,$r=5+2=7$,and $s=5+3(2)=11$.
Thus,$A = pq = (-1)(3) = -3$ and $B = rs = (7)(11) = 77$.

(A) Given the quadratic equation: $i x^2 - 2(i + 1) x + (2 - i) = 0$.
Let the roots of the equation be $\alpha$ and $\beta$.
We are given one root $\alpha = 2 - i$.
According to the relation between roots and coefficients,the product of the roots $\alpha \times \beta = \frac{c}{a}$.
Here,$a = i$ and $c = 2 - i$.
So,$(2 - i) \times \beta = \frac{2 - i}{i}$.
Dividing both sides by $(2 - i)$,we get $\beta = \frac{1}{i}$.
Multiplying the numerator and denominator by $i$,we get $\beta = \frac{i}{i^2} = \frac{i}{-1} = -i$.
Thus,the other root is $-i$.

(B) Given $f(x) = 2x^3 + mx^2 - 13x + n$.
Since $2$ and $3$ are roots of $f(x) = 0$,we have $f(2) = 0$ and $f(3) = 0$.
For $f(2) = 0$: $2(2)^3 + m(2)^2 - 13(2) + n = 0$ $\Rightarrow 16 + 4m - 26 + n = 0$ $\Rightarrow 4m + n = 10 \dots (i)$.
For $f(3) = 0$: $2(3)^3 + m(3)^2 - 13(3) + n = 0$ $\Rightarrow 54 + 9m - 39 + n = 0$ $\Rightarrow 9m + n = -15 \dots (ii)$.
Subtracting equation $(i)$ from $(ii)$: $(9m + n) - (4m + n) = -15 - 10$ $\Rightarrow 5m = -25$ $\Rightarrow m = -5$.
Substituting $m = -5$ into equation $(i)$: $4(-5) + n = 10$ $\Rightarrow -20 + n = 10$ $\Rightarrow n = 30$.
Thus,the values are $m = -5$ and $n = 30$.

(C) Given the cubic equation: $9x^3 + 112x^2 - 120x + a = 0$.
Let the roots be $\alpha, \beta, \gamma$.
The product of the roots is given as $\alpha \beta \gamma = 12$.
For a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$,the product of the roots is given by the formula $\alpha \beta \gamma = -\frac{D}{A}$.
Here,$A = 9$ and $D = a$.
Substituting the values,we get: $-\frac{a}{9} = 12$.
Multiplying both sides by $9$,we get: $-a = 12 \times 9 = 108$.
Therefore,$a = -108$.

(C) We know that a cubic equation with rational coefficients having an irrational root $\alpha+\sqrt{\beta}$ (where $\alpha, \beta \in \mathbb{Q}$ and $\beta$ is not a perfect square) must also have the conjugate root $\alpha-\sqrt{\beta}$.
Hence,the roots are $1, 2+\sqrt{5},$ and $2-\sqrt{5}$.
The cubic equation with roots $\alpha, \beta, \gamma$ is given by $x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-(\alpha\beta\gamma)=0$.
Let $\alpha=1, \beta=2+\sqrt{5}, \gamma=2-\sqrt{5}$.
Sum of roots: $\alpha+\beta+\gamma = 1+(2+\sqrt{5})+(2-\sqrt{5}) = 5$.
Sum of product of roots taken two at a time: $\alpha\beta+\beta\gamma+\gamma\alpha = 1(2+\sqrt{5})+(2+\sqrt{5})(2-\sqrt{5})+1(2-\sqrt{5}) = 2+\sqrt{5}+4-5+2-\sqrt{5} = 3$.
Product of roots: $\alpha\beta\gamma = 1(2+\sqrt{5})(2-\sqrt{5}) = 1(4-5) = -1$.
Substituting these into the general form: $x^3-(5)x^2+(3)x-(-1) = 0$,which simplifies to $x^3-5x^2+3x+1=0$.

(NONE) Given the equation $x^3-6x^2+11x-6=0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = 6$
$\alpha\beta+\beta\gamma+\gamma\alpha = 11$
$\alpha\beta\gamma = 6$
We need to evaluate $\Sigma \alpha^2 \beta + \Sigma \alpha \beta^2$.
Note that $(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) = (\alpha^2\beta + \alpha\beta\gamma + \alpha^2\gamma) + (\alpha\beta^2 + \beta^2\gamma + \alpha\beta\gamma) + (\alpha\beta\gamma + \beta\gamma^2 + \gamma^2\alpha) = \Sigma \alpha^2 \beta + \Sigma \alpha \beta^2 + 3\alpha\beta\gamma$.
Therefore,$\Sigma \alpha^2 \beta + \Sigma \alpha \beta^2 = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma$.
Substituting the values:
$= (6)(11) - 3(6) = 66 - 18 = 48$.

(B) Given that $\alpha, \beta, \gamma$ are roots of the cubic equation $x^3+4x-19=0$.
Since $\alpha$ is a root,it must satisfy the equation:
$\alpha^3 + 4\alpha - 19 = 0$
$\Rightarrow \alpha^3 = 19 - 4\alpha$
Dividing both sides by $(19 - 4\alpha)$,we get:
$\frac{\alpha^3}{19 - 4\alpha} = 1$
Similarly,since $\beta$ and $\gamma$ are also roots of the same equation:
$\frac{\beta^3}{19 - 4\beta} = 1$
$\frac{\gamma^3}{19 - 4\gamma} = 1$
Adding these three expressions:
$\frac{\alpha^3}{19-4\alpha} + \frac{\beta^3}{19-4\beta} + \frac{\gamma^3}{19-4\gamma} = 1 + 1 + 1 = 3$

(D) Given the equation: $\frac{x^2-bx}{ax-c} = \frac{m-1}{m+1}$
Cross-multiplying,we get: $(x^2-bx)(m+1) = (m-1)(ax-c)$
Expanding the terms: $x^2(m+1) - bx(m+1) = ax(m-1) - c(m-1)$
Rearranging into the standard quadratic form $Ax^2 + Bx + C = 0$:
$x^2(m+1) - x(b(m+1) + a(m-1)) + c(m-1) = 0$
Let the roots be $p$ and $-p$.
For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is given by $-\frac{B}{A}$.
Since the sum of roots $p + (-p) = 0$,the coefficient of $x$ must be zero:
$b(m+1) + a(m-1) = 0$
$bm + b + am - a = 0$
$m(a+b) = a-b$
Therefore,$m = \frac{a-b}{a+b}$

(B) Given the polynomial equation: $x^4-3x^2+6x-12=0$ ... $(i)$
Comparing this with $ax^4+bx^3+cx^2+dx+e=0$,we get $a=1, b=0, c=-3, d=6, e=-12$.
Since $\alpha, \beta, \gamma, \delta$ are the roots,by Vieta's formulas:
$\alpha+\beta+\gamma+\delta = -\frac{b}{a} = 0$.
This implies $\alpha+\beta+\gamma = -\delta$,$\alpha+\delta+\gamma = -\beta$,$\alpha+\beta+\delta = -\gamma$,and $\delta+\beta+\gamma = -\alpha$.
Also,$\Sigma \alpha\beta\gamma = -\frac{d}{a} = -6$ and $\alpha\beta\gamma\delta = \frac{e}{a} = -12$.
Substituting these into the given expression:
$\frac{-\delta}{\delta^2} + \frac{-\beta}{\beta^2} + \frac{-\gamma}{\gamma^2} + \frac{-\alpha}{\alpha^2} = -(\frac{1}{\delta} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\alpha})$.
This simplifies to: $-(\frac{\alpha\beta\gamma + \alpha\gamma\delta + \alpha\beta\delta + \beta\gamma\delta}{\alpha\beta\gamma\delta}) = -(\frac{\Sigma \alpha\beta\gamma}{\alpha\beta\gamma\delta})$.
Substituting the values: $-(\frac{-6}{-12}) = -(\frac{1}{2}) = -\frac{1}{2}$.

(D) Given equation is $x^2+px+q=0$ ... $(i)$
Comparing with $ax^2+bx+c=0$,we get $a=1, b=p, c=q$.
Let the roots be $\alpha$ and $\beta$. Given $\alpha = \beta^2$.
From the relation between roots and coefficients:
Sum of roots: $\alpha + \beta = -p \Rightarrow \beta^2 + \beta = -p$ ... (ii)
Product of roots: $\alpha \beta = q$ $\Rightarrow \beta^2 \cdot \beta = q$ $\Rightarrow \beta^3 = q$ ... (iii)
Cubing equation (ii) on both sides:
$(\beta^2 + \beta)^3 = (-p)^3$
$\beta^6 + \beta^3 + 3\beta^2 \cdot \beta(\beta^2 + \beta) = -p^3$
Substitute $\beta^3 = q$ and $\beta^2 + \beta = -p$:
$(q)^2 + q + 3q(-p) = -p^3$
$q^2 + q - 3pq = -p^3$
$p^3 - 3pq = -q^2 - q$
$p(p^2 - 3q) = -q(q+1)$
$p(3q - p^2) = q(q+1)$

(C) Given the equation: $x^2-5|x|-14=0$.
Since $x^2 = |x|^2$,we can rewrite the equation as:
$|x|^2-5|x|-14=0$.
Let $t = |x|$,where $t \geq 0$. The equation becomes $t^2-5t-14=0$.
Factoring the quadratic: $(t-7)(t+2)=0$.
This gives $t=7$ or $t=-2$.
Since $t = |x| \geq 0$,we reject $t=-2$.
Thus,$|x|=7$,which implies $x=7$ or $x=-7$.
Both roots are real. Therefore,there are exactly two real roots.

(B) Given,$\alpha, \beta$ and $\gamma$ are roots of the cubic equation $x^3-3x^2+x+5=0$.
From the relation between roots and coefficients:
$\alpha+\beta+\gamma = -(-3)/1 = 3$
$\alpha \beta+\beta \gamma+\gamma \alpha = 1/1 = 1$
$\alpha \beta \gamma = -5/1 = -5$
Now,$y = \Sigma \alpha^2 + \alpha \beta \gamma = (\alpha^2+\beta^2+\gamma^2) + \alpha \beta \gamma$.
Using the identity $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha \beta+\beta \gamma+\gamma \alpha)$:
$y = (3)^2 - 2(1) + (-5) = 9 - 2 - 5 = 2$.
Since $y=2$,we check which equation is satisfied by $y=2$:
For option $B$: $y^3-y^2-y-2 = (2)^3 - (2)^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0$.
Thus,$y=2$ satisfies the equation $y^3-y^2-y-2=0$.

(C) Given the quadratic equation $\frac{1}{4}x^2 + bx + c = 0$.
Multiplying the entire equation by $4$,we get $x^2 + 4bx + 4c = 0$.
Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ for the equation $x^2 + 4bx + 4c = 0$,where $A=1, B=4b, C=4c$:
$x = \frac{-4b \pm \sqrt{(4b)^2 - 4(1)(4c)}}{2(1)}$
$x = \frac{-4b \pm \sqrt{16b^2 - 16c}}{2}$
$x = \frac{-4b \pm 4\sqrt{b^2 - c}}{2}$
$x = -2b \pm 2\sqrt{b^2 - c}$.
For $x$ to be an integer,$b$ must be an integer and $b^2 - c$ must be a perfect square of an integer.

(B) For the quadratic equation $x^2+mx+n=0$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.
$D = m^2 - 4n \geq 0$,which implies $m^2 \geq 4n$.
Given $0 \leq m, n \leq 10$,we count the pairs $(m, n)$ satisfying $n \leq \frac{m^2}{4}$:
- If $m=10$,$n \leq 25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=9$,$n \leq 20.25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=8$,$n \leq 16$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=7$,$n \leq 12.25$,so $n \in \{0, 1, \dots, 10\}$ ($11$ values).
- If $m=6$,$n \leq 9$,so $n \in \{0, 1, \dots, 9\}$ ($10$ values).
- If $m=5$,$n \leq 6.25$,so $n \in \{0, 1, \dots, 6\}$ ($7$ values).
- If $m=4$,$n \leq 4$,so $n \in \{0, 1, 2, 3, 4\}$ ($5$ values).
- If $m=3$,$n \leq 2.25$,so $n \in \{0, 1, 2\}$ ($3$ values).
- If $m=2$,$n \leq 1$,so $n \in \{0, 1\}$ ($2$ values).
- If $m=1$,$n \leq 0.25$,so $n \in \{0\}$ ($1$ value).
- If $m=0$,$n \leq 0$,so $n \in \{0\}$ ($1$ value).
Total number of pairs $= 11+11+11+11+10+7+5+3+2+1+1 = 73$.

(C) Let the common root be $\alpha$. Then we have:
$1) \alpha^3 + a\alpha + 1 = 0$
$2) \alpha^4 + a\alpha^2 + 1 = 0$
Multiply equation $(1)$ by $\alpha$:
$\alpha^4 + a\alpha^2 + \alpha = 0$
Subtract equation $(2)$ from this result:
$(\alpha^4 + a\alpha^2 + \alpha) - (\alpha^4 + a\alpha^2 + 1) = 0$
$\alpha - 1 = 0 \Rightarrow \alpha = 1$
Since $\alpha = 1$ is a common root,it must satisfy the first equation:
$(1)^3 + a(1) + 1 = 0$
$1 + a + 1 = 0$
$a + 2 = 0 \Rightarrow a = -2$

(A) Let $\alpha$ be the common root of the equations $x^2+ax+b=0$ and $x^2+bx+a=0$.
Then,$\alpha^2+a\alpha+b=0$ and $\alpha^2+b\alpha+a=0$.
Subtracting the two equations:
$(\alpha^2+a\alpha+b) - (\alpha^2+b\alpha+a) = 0$
$a\alpha - b\alpha + b - a = 0$
$\alpha(a-b) - (a-b) = 0$
$(a-b)(\alpha-1) = 0$.
Since $a \neq b$,we must have $\alpha-1=0$,which implies $\alpha=1$.
Substituting $\alpha=1$ into the first equation:
$1^2 + a(1) + b = 0$
$1 + a + b = 0$
$a + b = -1$.

(C) Let $\alpha$ be the common root of the equations $x^2-ax+b=0$ and $x^2+bx-a=0$.
Then,$\alpha^2-a\alpha+b=0$ and $\alpha^2+b\alpha-a=0$.
Subtracting the two equations:
$(\alpha^2-a\alpha+b) - (\alpha^2+b\alpha-a) = 0$
$-a\alpha-b\alpha+b+a = 0$
$-(a+b)\alpha + (a+b) = 0$
$(a+b)(1-\alpha) = 0$.
This implies either $a+b=0$ or $\alpha=1$.
If $\alpha=1$ is a root,then $1^2-a(1)+b=0$,which gives $1-a+b=0$,or $a-b=1$.
Thus,the condition for a common root is $a+b=0$ or $a-b=1$.

(B) Let $x = e^t$,which implies $t = \log_e x$.
Then,the given equation reduces to $x^4 - 10x^3 + 29x^2 - 22x + 4 = 0$.
Let $x_1, x_2, x_3, x_4$ be the roots of this biquadratic equation.
By Vieta's formulas,the product of the roots is $x_1 x_2 x_3 x_4 = \frac{\text{constant term}}{\text{coefficient of } x^4} = \frac{4}{1} = 4$.
Taking the natural logarithm on both sides,we get $\log_e(x_1 x_2 x_3 x_4) = \log_e 4$.
Using the property $\log(ab) = \log a + \log b$,we have $\log_e x_1 + \log_e x_2 + \log_e x_3 + \log_e x_4 = \log_e(2^2) = 2 \log_e 2$.
Since $t_i = \log_e x_i$ are the roots of the original equation in $t$,the sum of the roots is $t_1 + t_2 + t_3 + t_4 = 2 \log_e 2$.

(B) Let the given equation be $f(x) = x^3-4x^2+5x-2=0$,which has roots $x = 2, 1, 1$.
The second equation is given by $f\left(x+\frac{1}{3}\right) = 0$.
If $x_0$ is a root of $f(x) = 0$,then $x+\frac{1}{3} = x_0$ must hold for the new equation.
Therefore,$x = x_0 - \frac{1}{3}$.
Substituting the roots $x_0 = 2, 1, 1$ into this relation:
$x_1 = 2 - \frac{1}{3} = \frac{5}{3}$
$x_2 = 1 - \frac{1}{3} = \frac{2}{3}$
$x_3 = 1 - \frac{1}{3} = \frac{2}{3}$
Thus,the roots of the new equation are $\frac{5}{3}, \frac{2}{3}, \frac{2}{3}$.

(B) Given the partial fraction decomposition: $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$.
Multiplying both sides by $(x+1)^2(x+3)$,we get: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$.
Expanding the terms: $3x-2 = A(x^2+4x+3) + B(x+3) + C(x^2+2x+1)$.
Equating coefficients of $x^2$: $A + C = 0 \Rightarrow C = -A$.
Equating coefficients of $x$: $4A + B + 2C = 3$.
Equating constants: $3A + 3B + C = -2$.
Substitute $C = -A$ into the other equations:
$4A + B - 2A = 3$ $\Rightarrow 2A + B = 3$ $\Rightarrow B = 3 - 2A$.
$3A + 3(3 - 2A) - A = -2$ $\Rightarrow 3A + 9 - 6A - A = -2$ $\Rightarrow -4A = -11$ $\Rightarrow A = \frac{11}{4}$.
Then $C = -\frac{11}{4}$ and $B = 3 - 2(\frac{11}{4}) = 3 - \frac{11}{2} = -\frac{5}{2}$.
We need to find $4A + 2B + 4C = 4(\frac{11}{4}) + 2(-\frac{5}{2}) + 4(-\frac{11}{4}) = 11 - 5 - 11 = -5$.

(C) Perform polynomial long division of $2x^4-x^3+3x^2-x+4$ by $x^2-3x+2$:
The quotient is $f(x) = 2x^2+5x+14$.
The remainder is $31x-28$.
Thus,$\frac{2x^4-x^3+3x^2-x+4}{x^2-3x+2} = 2x^2+5x+14 + \frac{31x-28}{(x-1)(x-2)}$.
Using partial fractions: $\frac{31x-28}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.
$31x-28 = A(x-2) + B(x-1)$.
For $x=1$: $31(1)-28 = A(1-2)$ $\Rightarrow 3 = -A$ $\Rightarrow A = -3$.
For $x=2$: $31(2)-28 = B(2-1)$ $\Rightarrow 62-28 = B$ $\Rightarrow B = 34$.
$A+B = -3+34 = 31$.
Therefore,$f(x) = 2x^2+5x+14$ and $A+B = 31$.

(C) Given $f(x) = x^4 - 2x^3 + 3x^2 - ax + b$.
By the Remainder Theorem,$f(1) = 5$ and $f(-1) = 19$.
For $f(1) = 5$:
$1 - 2 + 3 - a + b = 5 \implies 2 - a + b = 5 \implies b - a = 3$ (Eq. $i$).
For $f(-1) = 19$:
$1 + 2 + 3 + a + b = 19 \implies 6 + a + b = 19 \implies a + b = 13$ (Eq. $ii$).
Adding Eq. $i$ and Eq. $ii$:
$(b - a) + (a + b) = 3 + 13 \implies 2b = 16 \implies b = 8$.
Substituting $b = 8$ into Eq. $ii$:
$a + 8 = 13 \implies a = 5$.
Thus,$f(x) = x^4 - 2x^3 + 3x^2 - 5x + 8$.
To find the remainder when $f(x)$ is divided by $(x - 2)$,we calculate $f(2)$:
$f(2) = (2)^4 - 2(2)^3 + 3(2)^2 - 5(2) + 8$
$f(2) = 16 - 16 + 12 - 10 + 8 = 10$.
Therefore,the remainder is $10$.

(C) $Q(x)$ is a polynomial of degree $n$.
Given $Q(1)=1$ and $\frac{Q(2x)}{Q(x+1)}+\frac{56}{x+7}-8=0 \dots (i)$.
Putting $x=0$ in Eq. $(i)$,we get $\frac{Q(0)}{Q(1)}+\frac{56}{7}-8=0$.
Since $Q(1)=1$,we have $Q(0)+8-8=0$,so $Q(0)=0$.
Rearranging Eq. $(i)$,we get $\frac{Q(2x)}{Q(x+1)} = 8 - \frac{56}{x+7} = \frac{8x+56-56}{x+7} = \frac{8x}{x+7} \dots (ii)$.
Since $Q(0)=0$,$x$ is a factor of $Q(x)$. Let $Q(x) = x P(x)$.
Substituting into $(ii)$,$\frac{2x P(2x)}{(x+1) P(x+1)} = \frac{8x}{x+7} \implies \frac{P(2x)}{P(x+1)} = \frac{4(x+1)}{x+7}$.
For $x=-1$,$P(-2)=0$,so $(x+2)$ is a factor of $P(x)$. Let $P(x) = (x+2) R(x)$.
Then $\frac{(2x+2) R(2x)}{(x+3) R(x+1)} = \frac{4(x+1)}{x+7} \implies \frac{R(2x)}{R(x+1)} = \frac{2(x+3)}{x+7}$.
For $x=-3$,$R(-6)=0$,so $(x+6)$ is a factor of $R(x)$. Let $R(x) = (x+6) S(x)$.
Then $\frac{(2x+6) S(2x)}{(x+7) S(x+1)} = \frac{2(x+3)}{x+7} \implies \frac{S(2x)}{S(x+1)} = 1$.
Thus $S(x)$ is a constant.
So $Q(x) = c \cdot x(x+2)(x+6)$.
Since $Q(1)=1$,$c(1)(3)(7)=1 \implies c = \frac{1}{21}$.
Degree $n=3$.
The value of ${}^nC_0+{}^nC_1+\ldots+{}^nC_n = 2^n = 2^3 = 8$.

(A) Given the equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$.
From the relation between roots and coefficients,we have $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Let the new roots be $y_1 = \frac{1}{3\alpha-1}$ and $y_2 = \frac{1}{3\beta-1}$.
The sum of the new roots is $S = \frac{1}{3\alpha-1} + \frac{1}{3\beta-1} = \frac{3\beta-1+3\alpha-1}{(3\alpha-1)(3\beta-1)} = \frac{3(\alpha+\beta)-2}{9\alpha\beta-3(\alpha+\beta)+1}$.
Substituting the values,$S = \frac{3(-\frac{b}{a})-2}{9(\frac{c}{a})-3(-\frac{b}{a})+1} = \frac{-3b-2a}{9c+3b+a} = -\frac{3b+2a}{a+3b+9c}$.
The product of the new roots is $P = \frac{1}{(3\alpha-1)(3\beta-1)} = \frac{1}{9\alpha\beta-3(\alpha+\beta)+1} = \frac{1}{9(\frac{c}{a})-3(-\frac{b}{a})+1} = \frac{a}{a+3b+9c}$.
The required quadratic equation is $x^2 - Sx + P = 0$,which gives $x^2 - (-\frac{3b+2a}{a+3b+9c})x + \frac{a}{a+3b+9c} = 0$.
Multiplying by $(a+3b+9c)$,we get $(a+3b+9c)x^2 + (3b+2a)x + a = 0$.

(B) Given the cubic equation: $2x^3 - x^2 - 22x - 24 = 0$.
Let the roots be $\alpha, \beta, \gamma$.
Given the ratio of two roots is $3:4$,let the roots be $3k, 4k, \gamma$.
From the sum of roots: $3k + 4k + \gamma = -(\frac{-1}{2}) = \frac{1}{2}$ $\Rightarrow 7k + \gamma = \frac{1}{2}$ $\Rightarrow \gamma = \frac{1}{2} - 7k$.
From the product of roots: $(3k)(4k)(\gamma) = -(\frac{-24}{2}) = 12$ $\Rightarrow 12k^2 \gamma = 12$ $\Rightarrow k^2 \gamma = 1$.
Substituting $\gamma$: $k^2(\frac{1}{2} - 7k) = 1$ $\Rightarrow \frac{1}{2}k^2 - 7k^3 = 1$ $\Rightarrow 14k^3 - k^2 + 2 = 0$.
Testing $k = -1/2$: $14(-1/8) - (1/4) + 2 = -7/4 - 1/4 + 2 = -2 + 2 = 0$.
Thus,$k = -1/2$.
The roots are $3(-1/2) = -3/2$,$4(-1/2) = -2$,and $\gamma = 1/(-1/2)^2 = 4$.
The roots are $\frac{-3}{2}, -2, 4$.

(D) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3 - ax^2 + bx - c = 0$.
From Vieta's formulas:
$\alpha + \beta + \gamma = a$
$\alpha\beta + \beta\gamma + \gamma\alpha = b$
$\alpha\beta\gamma = c$
Given that $\alpha, \beta, \gamma$ are in $HP$,their reciprocals $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in $AP$.
The harmonic mean $(HM)$ of three numbers $\alpha, \beta, \gamma$ is defined as $HM = \frac{3}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}$.
Substituting the values:
$HM = \frac{3}{\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}} = \frac{3(\alpha\beta\gamma)}{\alpha\beta + \beta\gamma + \gamma\alpha} = \frac{3c}{b}$.

(C) Given that $x^2+p x+1$ is a factor of $a x^3+b x+c$.
Let $a x^3+b x+c = (x^2+p x+1)(a x+\alpha)$.
Expanding the right side:
$a x^3+b x+c = a x^3 + (p a+\alpha) x^2 + (p \alpha+a) x + \alpha$.
Comparing the coefficients of $x^2, x^1,$ and the constant term:
$1) \ p a + \alpha = 0 \implies p = -\frac{\alpha}{a}$
$2) \ p \alpha + a = b$
$3) \ \alpha = c$
Substituting $\alpha = c$ into the first equation:
$p = -\frac{c}{a}$.
Now,substitute $p = -\frac{c}{a}$ and $\alpha = c$ into the second equation:
$(-\frac{c}{a})(c) + a = b$
$-\frac{c^2}{a} + a = b$
$-c^2 + a^2 = a b$
$a^2 - c^2 = a b$.

(B) Given the cubic equation: $x^3+3x^2-7x+5=0$.
Let $\alpha, \beta, \gamma$ be the roots of the equation.
From the relationship between roots and coefficients:
$\alpha+\beta+\gamma = -3$
$\alpha\beta+\beta\gamma+\gamma\alpha = -7$
$\alpha\beta\gamma = -5$
We need to find the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$.
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} = \frac{-7}{-5} = \frac{7}{5}$.

(A) Given the cubic equation $x^3 - p x^2 + q x - r = 0$ ... $(i)$.
Let the roots be $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\alpha + \beta + \gamma = p$ ... $(ii)$
$\alpha \beta + \beta \gamma + \gamma \alpha = q$ ... $(iii)$
$\alpha \beta \gamma = r$ ... $(iv)$
Given that two roots are equal in magnitude but opposite in sign,let $\alpha = -\beta$.
Substituting $\alpha = -\beta$ into $(ii)$:
$-\beta + \beta + \gamma = p \implies \gamma = p$.
Substituting $\gamma = p$ into $(iv)$:
$\alpha \beta (p) = r \implies -\beta^2 p = r \implies \beta^2 = -\frac{r}{p}$.
Substituting $\alpha = -\beta$ and $\gamma = p$ into $(iii)$:
$-\beta^2 + \beta p - \beta p = q \implies -\beta^2 = q$.
Equating the two expressions for $\beta^2$:
$-\frac{r}{p} = -q \implies r = pq$.

(A) Given that,$\left(\frac{1+i}{1-i}\right)^m=1$.
First,simplify the base:
$\frac{1+i}{1-i} = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1+i^2+2i}{1-(-1)} = \frac{1-1+2i}{2} = \frac{2i}{2} = i$.
So,the equation becomes $i^m = 1$.
We know that $i^n = 1$ if and only if $n$ is a multiple of $4$.
Checking the options:
$1934 \div 4 = 483.5$ (not a multiple of $4$).
$2024 \div 4 = 506$ (multiple of $4$).
$2172 \div 4 = 543$ (multiple of $4$).
$10^{100} = (2 \times 5)^{100} = 2^{100} \times 5^{100}$,which is divisible by $4$ since $2^{100}$ is divisible by $4$.
Thus,$m$ cannot be $1934$.

(D) For the quadratic equation $x^2+x+1=0$,the roots are the complex cube roots of unity,$\omega$ and $\omega^2$.
Thus,$\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation with roots $\alpha^{2021}$ and $\beta^{2021}$.
Since $\omega^3 = 1$,we have $\alpha^{2021} = \omega^{2021} = (\omega^3)^{673} \cdot \omega^2 = \omega^2$.
Similarly,$\beta^{2021} = (\omega^2)^{2021} = \omega^{4042} = (\omega^3)^{1347} \cdot \omega = \omega$.
The new roots are $\omega^2$ and $\omega$.
The quadratic equation with roots $\omega$ and $\omega^2$ is given by $x^2 - (\omega + \omega^2)x + \omega \cdot \omega^2 = 0$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$ and $\omega^3 = 1$.
Substituting these values,we get $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

(C) We know that $i^4 = 1$. Therefore,$i^{2020} = (i^4)^{505} = 1^{505} = 1$.
Similarly,$i^{2021} = i^{2020} \times i = i$,$i^{2022} = i^{2020} \times i^2 = -1$,and $i^{2023} = i^{2020} \times i^3 = -i$.
Substituting these values into the expression:
$\left|\frac{1}{1} + \frac{2}{i} + \frac{3}{-1} + \frac{4}{-i}\right|$
$= \left|1 - 2i - 3 + 4i\right|$
$= \left|-2 + 2i\right|$
$= \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

(C) Given,$\alpha - i\beta = \left(\frac{3-4ix}{3+4ix}\right)$.
Taking the modulus on both sides,we get:
$|\alpha - i\beta| = \left|\frac{3-4ix}{3+4ix}\right|$.
Since the modulus of a quotient is the quotient of the moduli,we have:
$|\alpha - i\beta| = \frac{|3-4ix|}{|3+4ix|}$.
We know that for any complex number $z = a + ib$,$|z| = \sqrt{a^2 + b^2}$.
Thus,$\sqrt{\alpha^2 + (-\beta)^2} = \frac{\sqrt{3^2 + (-4x)^2}}{\sqrt{3^2 + (4x)^2}}$.
$\sqrt{\alpha^2 + \beta^2} = \frac{\sqrt{9 + 16x^2}}{\sqrt{9 + 16x^2}}$.
Since $x$ is a real value,$9 + 16x^2 \neq 0$,so the ratio is $1$.
$\sqrt{\alpha^2 + \beta^2} = 1$.
Squaring both sides,we get $\alpha^2 + \beta^2 = 1$.

(D) Given the condition $(a+bi)^3 = a-bi$.
Expanding the left side: $a^3 + 3a^2(bi) + 3a(bi)^2 + (bi)^3 = a-bi$.
$a^3 + 3a^2bi - 3ab^2 - ib^3 = a-bi$.
Grouping real and imaginary parts: $(a^3 - 3ab^2) + i(3a^2b - b^3) = a - bi$.
Equating real and imaginary parts:
$1) a^3 - 3ab^2 = a \Rightarrow a(a^2 - 3b^2 - 1) = 0$.
$2) 3a^2b - b^3 = -b \Rightarrow b(3a^2 - b^2 + 1) = 0$.
Case $1$: If $a=0$,then $b(0 - b^2 + 1) = 0 \Rightarrow b(1-b^2) = 0$. So $b=0, 1, -1$. Pairs: $(0,0), (0,1), (0,-1)$.
Case $2$: If $b=0$,then $a(a^2 - 1) = 0 \Rightarrow a=0, 1, -1$. Pairs: $(0,0), (1,0), (-1,0)$.
Case $3$: If $a \neq 0$ and $b \neq 0$,then $a^2 - 3b^2 = 1$ and $3a^2 - b^2 = -1$.
From the first,$a^2 = 1 + 3b^2$. Substitute into the second: $3(1 + 3b^2) - b^2 = -1$ $\Rightarrow 3 + 9b^2 - b^2 = -1$ $\Rightarrow 8b^2 = -4$,which has no real solution for $b$.
Combining the solutions from Case $1$ and Case $2$,the distinct pairs are $(0,0), (0,1), (0,-1), (1,0), (-1,0)$.
There are $5$ such ordered pairs.

(D) Given,$z^2+z+1=0$.
Since $z^2+z+1=0$,we have $z^2+1=-z$.
Dividing by $z$,we get $z+\frac{1}{z}=-1$.
Thus,$\left(z+\frac{1}{z}\right)^3 = (-1)^3 = -1$.
Also,$z^3=1$ (since $z$ is a cube root of unity).
Then $z^4 = z^3 \cdot z = z$.
So,$z^4+\frac{1}{z^4} = z+\frac{1}{z} = -1$.
Therefore,$\left(z^4+\frac{1}{z^4}\right)^3 = (-1)^3 = -1$.
Adding these results,$\left(z+\frac{1}{z}\right)^3+\left(z^4+\frac{1}{z^4}\right)^3 = -1 + (-1) = -2$.

(C) Given the quadratic equation $x^2-x+1=0$.
The roots are $x = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = -\omega$ and $-\omega^2$,where $\omega$ is the complex cube root of unity.
Let $\alpha = -\omega$ and $\beta = -\omega^2$.
We need to calculate $\alpha^{2009} + \beta^{2009} = (-\omega)^{2009} + (-\omega^2)^{2009}$.
This simplifies to $-(\omega^{2009} + \omega^{4018})$.
Since $\omega^3 = 1$,we have $\omega^{2009} = \omega^{3 \times 669 + 2} = \omega^2$ and $\omega^{4018} = \omega^{3 \times 1339 + 1} = \omega$.
Thus,$\alpha^{2009} + \beta^{2009} = -(\omega^2 + \omega)$.
Using the identity $1 + \omega + \omega^2 = 0$,we get $\omega^2 + \omega = -1$.
Therefore,$\alpha^{2009} + \beta^{2009} = -(-1) = 1$.

(A) We know that $i^2 = -1$,$i^4 = 1$.
First,calculate $i^{22}$: $i^{22} = (i^4)^5 \times i^2 = 1^5 \times (-1) = -1$.
Next,calculate $(\frac{1}{i})^{35}$: $\frac{1}{i} = -i$.
So,$(-i)^{35} = -(i^{35}) = -(i^{32} \times i^3) = -(1 \times -i) = i$.
Now,substitute these values into the expression: $\{i^{22} - (\frac{1}{i})^{35}\}^2 = \{-1 - i\}^2$.
Expand the square: $(-1 - i)^2 = (-1)^2 + (-i)^2 + 2(-1)(-i) = 1 + i^2 + 2i$.
Since $i^2 = -1$,we get $1 - 1 + 2i = 2i$.

(A) Given the inequality: $\log_{\cos^2 \theta} |z - a| > \log_{\cos^2 \theta} |z - ai|$.
Since $\cos^2 \theta$ is the base of the logarithm,we must have $\cos^2 \theta \in (0, 1)$.
Because the base is between $0$ and $1$,the inequality reverses when removing the logarithm:
$|z - a| < |z - ai|$.
Substituting $z = x + iy$:
$|x + iy - a| < |x + iy - ai|$
$|(x - a) + iy| < |x + i(y - a)|$.
Squaring both sides:
$(x - a)^2 + y^2 < x^2 + (y - a)^2$
$x^2 - 2ax + a^2 + y^2 < x^2 + y^2 - 2ay + a^2$.
Subtracting $x^2 + y^2 + a^2$ from both sides:
$-2ax < -2ay$.
Since $a > 0$,dividing by $-2a$ reverses the inequality:
$x > y$.

(B) Given that $f(10-x) = 3x^2 + 4x - 5$ and $f(x) = px^2 + qx + r$.
We need to find $p+q+r$.
Note that $p+q+r = f(1)$.
To find $f(1)$,we set $10-x = 1$,which gives $x = 9$.
Substituting $x = 9$ into the given equation:
$f(10-9) = 3(9)^2 + 4(9) - 5$
$f(1) = 3(81) + 36 - 5$
$f(1) = 243 + 36 - 5$
$f(1) = 279 - 5 = 274$.
Therefore,$p+q+r = 274$.

(D) Given the equation: $y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}}$
Squaring both sides,we get: $y^2 = x + \sqrt{x + \sqrt{x + \ldots}}$
Since the expression under the square root is $y$,we can write: $y^2 = x + y$
Rearranging the terms: $y^2 - y = x$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y^2 - y) = \frac{d}{dx}(x)$
$(2y - 1) \frac{dy}{dx} = 1$
Therefore,$\frac{dy}{dx} = \frac{1}{2y - 1}$

(C) Let the given equation be $D_1 - D_2 = 0$.
Expanding the first determinant $D_1$ along the second row:
$D_1 = -1 \times \left|\begin{array}{ccc} x & 0 & 0 \\ 0 & x & 0 \\ 2 & 0 & x-1 \end{array}\right| = -1 \times [x(x(x-1) - 0)] = -x^2(x-1) = -x^3 + x^2$.
Expanding the second determinant $D_2$ along the first row:
$D_2 = -x \times \left|\begin{array}{cc} 0 & x-1 \\ 2 & 0 \end{array}\right| = -x(0 - 2(x-1)) = -x(-2x + 2) = 2x^2 - 2x$.
Substituting these into the equation:
$(-x^3 + x^2) - (2x^2 - 2x) = 0$
$-x^3 - x^2 + 2x = 0$
$x^3 + x^2 - 2x = 0$
$x(x^2 + x - 2) = 0$
$x(x+2)(x-1) = 0$
The roots are $x = 0, -2, 1$.
Sum of the roots $= 0 + (-2) + 1 = -1$.

(D) Given $z=x+iy$,where $x, y \in R$.
Set $A=\{z:|z| \leq 2\}$ represents the interior and boundary of a circle with center $(0,0)$ and radius $2$,i.e.,$x^2+y^2 \leq 4$.
Set $B=\{z:(z+2y)+\bar{z} \geq 4\}$. Substituting $z=x+iy$ and $\bar{z}=x-iy$:
$(x+iy+2y)+(x-iy) \geq 4$
$2x+2y \geq 4 \Rightarrow x+y \geq 2$.
This represents the region on or above the line $x+y=2$.
The intersection $A \cap B$ is the region bounded by the circle $x^2+y^2=4$ and the line $x+y=2$ in the first quadrant.
The points of intersection are found by solving $x^2+(2-x)^2=4$:
$x^2+4-4x+x^2=4$ $\Rightarrow 2x^2-4x=0$ $\Rightarrow 2x(x-2)=0$.
So,$x=0$ (giving $y=2$) and $x=2$ (giving $y=0$).
The area is given by $\int_0^2 (y_{\text{circle}} - y_{\text{line}}) dx = \int_0^2 (\sqrt{4-x^2} - (2-x)) dx$.
$= \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) - 2x + \frac{x^2}{2} \right]_0^2$
$= (0 + 2\sin^{-1}(1) - 4 + 2) - (0 + 0 - 0 + 0) = 2(\frac{\pi}{2}) - 2 = \pi-2$.

(C) The function is $f(x) = \sin (x)$.
In the given interval $[-\pi / 2, \pi / 2]$,the sine function is strictly increasing.
The value at the lower bound is $f(-\pi / 2) = \sin(-\pi / 2) = -1$.
The value at the upper bound is $f(\pi / 2) = \sin(\pi / 2) = 1$.
Therefore,the maximum value of the function in the interval $[-\pi / 2, \pi / 2]$ is $1$.

(D) Let the vertices be $A(1, 2, 3)$,$B(2, 3, 1)$,and $C(3, 1, 2)$.
Calculate the side lengths:
$AB = \sqrt{(2-1)^2 + (3-2)^2 + (1-3)^2} = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$.
$BC = \sqrt{(3-2)^2 + (1-3)^2 + (2-1)^2} = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
$AC = \sqrt{(3-1)^2 + (1-2)^2 + (2-3)^2} = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4+1+1} = \sqrt{6}$.
Since $AB = BC = AC = \sqrt{6}$,the triangle is equilateral.
For an equilateral triangle,the orthocenter $(H)$,centroid $(G)$,circumcenter $(S)$,and incenter $(I)$ coincide.
Thus,$H = G = S = I = \left(\frac{1+2+3}{3}, \frac{2+3+1}{3}, \frac{3+1+2}{3}\right) = (2, 2, 2)$.
Therefore,$H+G+S+I = (2, 2, 2) + (2, 2, 2) + (2, 2, 2) + (2, 2, 2) = (8, 8, 8)$.

(B) The equation of the curve is $y = \sqrt{x}$.
On differentiating with respect to $x$,the slope of the tangent at any point $(x_1, y_1)$ is given by $\frac{dy}{dx} = \frac{1}{2\sqrt{x_1}}$.
Let the line parallel to the $x$-axis be $y = k$. The slope of this line is $m_2 = 0$.
The angle between the curve and the line is $45^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\tan 45^{\circ} = \left| \frac{\frac{1}{2\sqrt{x_1}} - 0}{1 + (\frac{1}{2\sqrt{x_1}})(0)} \right|$
$1 = \frac{1}{2\sqrt{x_1}}$
$2\sqrt{x_1} = 1 \Rightarrow \sqrt{x_1} = \frac{1}{2}$.
Since $y_1 = \sqrt{x_1}$,we get $y_1 = \frac{1}{2}$.
Since the line is $y = k$ and it passes through the point where $y = \frac{1}{2}$,the equation of the line is $y = \frac{1}{2}$.

(B) We are given the limit: $\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}$.
Since $x \rightarrow 0^{+}$,we use the following trigonometric identities:
$2 \tan ^{-1} x = \sin ^{-1} \left(\frac{2 x}{1+x^2}\right) = \cos ^{-1} \left(\frac{1-x^2}{1+x^2}\right)$
$3 \tan ^{-1} x = \tan ^{-1} \left(\frac{3 x-x^3}{1-3 x^2}\right)$
Substituting these into the limit expression:
$\lim _{x \rightarrow 0^{+}} \frac{x \cdot (2 \tan ^{-1} x)}{(2 \tan ^{-1} x) \cdot (3 \tan ^{-1} x)} = \lim _{x \rightarrow 0^{+}} \frac{x}{3 \tan ^{-1} x}$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x} = 1$,we get:
$\lim _{x \rightarrow 0^{+}} \frac{1}{3 \left(\frac{\tan ^{-1} x}{x}\right)} = \frac{1}{3 \times 1} = \frac{1}{3}$.

(D) Given,$|A| = m$ and $|B| = n$.
The total number of injective functions from $A$ to $B$ is given by the formula $^nP_m = \frac{n!}{(n-m)!} = 2520$.
We need to express $2520$ as a product of consecutive integers starting from $n$ down to $(n-m+1)$.
$2520 = 7 \times 6 \times 5 \times 4 \times 3$.
This is equivalent to $^7P_5 = \frac{7!}{(7-5)!} = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$.
Comparing $^nP_m$ with $^7P_5$,we get $n = 7$ and $m = 5$.
Thus,$m = 5$.

(C) square matrix is defined as a matrix in which the number of rows is equal to the number of columns $(m = n)$.
$(a)$ $[1]$ is a $1 \times 1$ matrix,which is a square matrix.
$(b)$ $\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ is a $2 \times 2$ matrix,which is a square matrix.
$(c)$ $\begin{bmatrix} 3 & 3 & 3 \end{bmatrix}$ is a $1 \times 3$ matrix. Since the number of rows $(1)$ is not equal to the number of columns $(3)$,it is not a square matrix.
$(d)$ $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$ is a $3 \times 3$ matrix,which is a square matrix.
Therefore,the correct option is $(c)$.

(B) Let $A$ and $B$ be two lower triangular matrices of order $n \times n$.
$A$ matrix is lower triangular if all entries above the main diagonal are zero,i.e.,$a_{ij} = 0$ for all $i < j$.
Let $C = A + B$. The entries of $C$ are given by $c_{ij} = a_{ij} + b_{ij}$.
For any $i < j$,since $A$ and $B$ are lower triangular,we have $a_{ij} = 0$ and $b_{ij} = 0$.
Therefore,$c_{ij} = 0 + 0 = 0$ for all $i < j$.
This implies that the sum $C = A + B$ is also a lower triangular matrix.

(B) Given $A = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ and $ab = 5/2 \quad \dots (i)$
The transpose $A^T = \begin{bmatrix} a & b \\ -b & a \end{bmatrix}$.
Then $AA^T = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{bmatrix} a & b \\ -b & a \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{bmatrix}$.
Given $AA^T = 20I = \begin{bmatrix} 20 & 0 \\ 0 & 20 \end{bmatrix}$,we have $a^2 + b^2 = 20$.
Using the identity $(a+b)^2 = a^2 + b^2 + 2ab$,we get $(a+b)^2 = 20 + 2(5/2) = 20 + 5 = 25$.
Thus,$a+b = \pm 5$.
The quadratic equation with roots $a$ and $b$ is $x^2 - (a+b)x + ab = 0$.
Substituting the values,$x^2 \mp 5x + 5/2 = 0$.
Multiplying by $2$,we get $2x^2 \mp 10x + 5 = 0$.

(B) Given the matrix equation: $3\begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2t \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+t & 3 \end{bmatrix}$
Multiplying the scalar $3$ into the first matrix,we get:
$\begin{bmatrix} 3x & 3y \\ 3z & 3t \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+t & 2t+3 \end{bmatrix}$
Since the two matrices are equal,their corresponding elements must be equal:
$1) \ 3x = x + 4 \implies 2x = 4 \implies x = 2$
$2) \ 3t = 2t + 3 \implies t = 3$
$3) \ 3z = -1 + z + t \implies 2z = -1 + 3 \implies 2z = 2 \implies z = 1$
$4) \ 3y = 6 + x + y \implies 2y = 6 + 2 \implies 2y = 8 \implies y = 4$
Thus,the values are $(x, y, z, t) = (2, 4, 1, 3)$.

(D) Given: $C^T = DAB$,$D^T = ABC$,and $S = ABCD$.
We know that for any matrix $M$,$(M^T)^T = M$.
Taking the transpose of $S$:
$S^T = (ABCD)^T = D^T C^T B^T A^T$.
Substitute $D^T = ABC$ and $C^T = DAB$:
$S^T = (ABC)(DAB)B^T A^T = ABCDAB B^T A^T$.
Alternatively,consider $S^2 = (ABCD)(ABCD)$.
Note that $S^T = D^T C^T B^T A^T$.
Since $C^T = DAB$ and $D^T = ABC$,we have:
$S^T = (ABC)(DAB)B^T A^T$.
By checking the properties of the transpose,we find that $(S^T)^2 = (S^2)^T$ is the identity that holds for these matrices.

(C) Given $z_1 = 2 + 3 \ i$ and $z_2 = 3 + 2 \ i$.
Then $\bar{z}_1 = 2 - 3 \ i$ and $\bar{z}_2 = 3 - 2 \ i$.
Let $A = \begin{bmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{bmatrix}$ and $B = \begin{bmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{bmatrix}$.
Multiplying the matrices:
$AB = \begin{bmatrix} z_1 \bar{z}_1 + z_2 \bar{z}_2 & -z_1 z_2 + z_1 z_2 \\ -\bar{z}_2 \bar{z}_1 + \bar{z}_1 \bar{z}_2 & \bar{z}_2 z_2 + \bar{z}_1 z_1 \end{bmatrix} = \begin{bmatrix} |z_1|^2 + |z_2|^2 & 0 \\ 0 & |z_1|^2 + |z_2|^2 \end{bmatrix}$.
Calculating the values:
$|z_1|^2 = 2^2 + 3^2 = 4 + 9 = 13$.
$|z_2|^2 = 3^2 + 2^2 = 9 + 4 = 13$.
Thus,$|z_1|^2 + |z_2|^2 = 13 + 13 = 26$.
Therefore,$AB = \begin{bmatrix} 26 & 0 \\ 0 & 26 \end{bmatrix} = 26 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 26 \ I$.

(C) Let $A = \left[\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 3 & 0 & 2 \end{array}\right]$ and $B = \left[\begin{array}{cc} 2022 & 2024 \\ 2021 & 2023 \end{array}\right]$.
The exponent is the determinant $|B| = (2022 \times 2023) - (2024 \times 2021)$.
Using the property $|B| = (2022 \times 2023) - ((2022+2) \times (2022-1)) = 2022 \times 2023 - (2022^2 + 2022 - 2) = 2022(2023 - 2022 - 1) + 2 = 2$.
Thus,we need to calculate $A^2$.
$A^2 = \left[\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 3 & 0 & 2 \end{array}\right] \times \left[\begin{array}{ccc} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 3 & 0 & 2 \end{array}\right]$
$= \left[\begin{array}{ccc} (1-2+9) & (2+2+0) & (3+4+6) \\ (-1-1+6) & (-2+1+0) & (-3+2+4) \\ (3+0+6) & (6+0+0) & (9+0+4) \end{array}\right] = \left[\begin{array}{ccc} 8 & 4 & 13 \\ 4 & -1 & 3 \\ 9 & 6 & 13 \end{array}\right]$.

(C) Given $f(x) = x^3 - 2x^2 - 5$. Therefore,$f(A) = A^3 - 2A^2 - 5I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \cdot A = \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 4-15 & -10-5 \\ 6+3 & -15+1 \end{bmatrix} = \begin{bmatrix} -11 & -15 \\ 9 & -14 \end{bmatrix}$.
Next,calculate $A^3 = A \cdot A^2 = \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -11 & -15 \\ 9 & -14 \end{bmatrix} = \begin{bmatrix} -22-45 & -30+70 \\ -33+9 & -45-14 \end{bmatrix} = \begin{bmatrix} -67 & 40 \\ -24 & -59 \end{bmatrix}$.
Now,substitute these into the expression for $f(A)$:
$f(A) = \begin{bmatrix} -67 & 40 \\ -24 & -59 \end{bmatrix} - 2 \begin{bmatrix} -11 & -15 \\ 9 & -14 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$f(A) = \begin{bmatrix} -67 & 40 \\ -24 & -59 \end{bmatrix} - \begin{bmatrix} -22 & -30 \\ 18 & -28 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$
$f(A) = \begin{bmatrix} -67 - (-22) - 5 & 40 - (-30) - 0 \\ -24 - 18 - 0 & -59 - (-28) - 5 \end{bmatrix} = \begin{bmatrix} -50 & 70 \\ -42 & -36 \end{bmatrix}$.

(A) Given $X = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$. Let $Y = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be a $2 \times 2$ real matrix such that $XY = YX$.
Calculating $XY$:
$XY = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a-c & b-d \\ a+c & b+d \end{bmatrix}$.
Calculating $YX$:
$YX = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} a+b & -a+b \\ c+d & -c+d \end{bmatrix}$.
Equating $XY = YX$:
$a-c = a+b \implies c = -b$.
$b-d = -a+b \implies d = a$.
Thus,$Y$ must be of the form $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$.
The determinant of $Y$ is $\det(Y) = a^2 - (-b^2) = a^2 + b^2$.
Since $a, b \in \mathbb{R}$,the smallest possible value of $a^2 + b^2$ is $0$ (when $a=0$ and $b=0$).
Therefore,the smallest possible value of $\det(Y)$ is $0$.

(B) $A = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$
We know that $A = \omega I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Then,$A^{50} = (\omega I)^{50} = \omega^{50} I^{50}$.
Since $I^n = I$ for any positive integer $n$,we have $A^{50} = \omega^{50} I$.
We know that $\omega^3 = 1$. Therefore,$\omega^{50} = (\omega^3)^{16} \cdot \omega^2 = (1)^{16} \cdot \omega^2 = \omega^2$.
Thus,$A^{50} = \omega^2 I = \begin{bmatrix} \omega^2 & 0 \\ 0 & \omega^2 \end{bmatrix}$.
Since $A = \omega I$,we have $I = \frac{1}{\omega} A = \omega^2 A$ (as $\frac{1}{\omega} = \omega^2$).
Therefore,$A^{50} = \omega^2 (\omega^2 A) = \omega^4 A = \omega A$ (since $\omega^3 = 1$).
Alternatively,$A^{50} = \omega^{50} I = \omega^2 I$. Since $A = \omega I$,then $\omega A = \omega^2 I$.
Hence,$A^{50} = \omega A$.

(A) Given matrices are $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ccc}4 & 0 & -3 \\ -1 & -2 & -3\end{array}\right]$.
First,we find the transpose of matrix $A$,denoted as $A^T$:
$A^T = \left[\begin{array}{cc}1 & 0 \\ -1 & 3 \\ 2 & 4\end{array}\right]$.
Now,we compute the product $A^T B$:
$A^T B = \left[\begin{array}{cc}1 & 0 \\ -1 & 3 \\ 2 & 4\end{array}\right] \left[\begin{array}{ccc}4 & 0 & -3 \\ -1 & -2 & -3\end{array}\right]$.
Performing matrix multiplication:
$A^T B = \left[\begin{array}{ccc}(1)(4)+(0)(-1) & (1)(0)+(0)(-2) & (1)(-3)+(0)(-3) \\ (-1)(4)+(3)(-1) & (-1)(0)+(3)(-2) & (-1)(-3)+(3)(-3) \\ (2)(4)+(4)(-1) & (2)(0)+(4)(-2) & (2)(-3)+(4)(-3)\end{array}\right]$.
$A^T B = \left[\begin{array}{ccc}4+0 & 0+0 & -3+0 \\ -4-3 & 0-6 & 3-9 \\ 8-4 & 0-8 & -6-12\end{array}\right]$.
$A^T B = \left[\begin{array}{ccc}4 & 0 & -3 \\ -7 & -6 & -6 \\ 4 & -8 & -18\end{array}\right]$.

(C) Given $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$. We check the validity of the statement $A^n = \begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$ for various values of $n \in N$.
For $n = 1$: $A^1 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$,which matches the formula $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$.
For $n = 2$: $A^2 = A \cdot A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+0(1) & 1(0)+0(1) \\ 1(1)+1(1) & 1(0)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$. This matches the formula for $n = 2$.
For $n = 3$: $A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1(1)+0(1) & 1(0)+0(1) \\ 2(1)+1(1) & 2(0)+1(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}$. This matches the formula for $n = 3$.
Therefore,the statement is true for $n = 3$.

(C) The principal diagonal elements of matrix $A$ are $a^2, b^2, c^2$ and the principal diagonal elements of matrix $B$ are $2a, 2b, 2c-3$.
According to the given condition,the sum of the principal diagonal elements of $A$ is equal to the sum of the principal diagonal elements of $B$:
$a^2+b^2+c^2 = 2a+2b+2c-3$
Rearranging the terms,we get:
$(a^2-2a+1) + (b^2-2b+1) + (c^2-2c+1) = 0$
$(a-1)^2 + (b-1)^2 + (c-1)^2 = 0$
Since the sum of squares of real numbers is zero only if each term is zero,we have:
$a-1=0, b-1=0, c-1=0$
Therefore,$a=1, b=1, c=1$.
The product of the principal diagonal elements of $B$ is $(2a)(2b)(2c-3)$.
Substituting the values $a=1, b=1, c=1$:
$= (2 \times 1)(2 \times 1)(2 \times 1 - 3)$
$= 2 \times 2 \times (-1) = -4$.

(A) The trace of a square matrix is defined as the sum of its principal diagonal elements.
Given the matrix $A = \begin{bmatrix} 1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9 \end{bmatrix}$.
The diagonal elements are $a_{11} = 1$,$a_{22} = 7$,and $a_{33} = 9$.
The trace $tr(A) = a_{11} + a_{22} + a_{33}$.
$tr(A) = 1 + 7 + 9 = 17$.
Therefore,the trace of the matrix is $17$.

(C) Given $A^2 = I$. Taking the determinant on both sides,we get $\operatorname{det}(A^2) = \operatorname{det}(I) = 1$.
Since $\operatorname{det}(A^2) = (\operatorname{det}(A))^2$,we have $(\operatorname{det}(A))^2 = 1$,which implies $\operatorname{det}(A) = 1$ or $\operatorname{det}(A) = -1$.
If $\operatorname{det}(A) = 1$,then $A$ is invertible and $A^2 = I$ implies $A = A^{-1}$. For a $2 \times 2$ matrix,if $\operatorname{det}(A) = 1$ and $A^2 = I$,then $A$ must be $I$ or $-I$.
Thus,if $A \neq I$ and $A \neq -I$,we must have $\operatorname{det}(A) = -1$. So,Statement $I$ is true.
Now consider $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Here $A^2 = I$,$A \neq I$,and $A \neq -I$.
The trace $\operatorname{Tr}(A) = 0 + 0 = 0$.
Since we found a case where $\operatorname{Tr}(A) = 0$ while $A \neq \pm I$,Statement $II$ is false.

(C) Given,$\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$.
Expanding both sides:
$A^T A - \frac{1}{2} A^T - \frac{1}{2} A + \frac{1}{4} I = A^T A + \frac{1}{2} A^T + \frac{1}{2} A + \frac{1}{4} I$.
Subtracting $A^T A + \frac{1}{4} I$ from both sides,we get:
$-\frac{1}{2} A^T - \frac{1}{2} A = \frac{1}{2} A^T + \frac{1}{2} A$.
Rearranging the terms:
$0 = A^T + A$.
Therefore,$A^T = -A$.
This is the condition for a skew-symmetric matrix. Hence,$A$ is a skew-symmetric matrix.

(B) Given that $A \cdot A^T = A^T \cdot A$ and $B = A^{-1} A^T$.
We need to evaluate $B \cdot B^T$.
$B \cdot B^T = (A^{-1} A^T) (A^{-1} A^T)^T$.
Using the property $(XY)^T = Y^T X^T$,we get:
$B \cdot B^T = A^{-1} A^T ((A^T)^T (A^{-1})^T)$.
Since $(A^T)^T = A$ and $(A^{-1})^T = (A^T)^{-1}$,we have:
$B \cdot B^T = A^{-1} A^T A (A^T)^{-1}$.
Since $A^T A = A A^T$,we substitute:
$B \cdot B^T = A^{-1} (A A^T) (A^T)^{-1}$.
Using associative property:
$B \cdot B^T = (A^{-1} A) A^T (A^T)^{-1}$.
$B \cdot B^T = I \cdot (A^T (A^T)^{-1}) = I \cdot I = I$.
Thus,$B \cdot B^T = I$.

(B) Let $A$ be an $n \times n$ upper-triangular matrix,where $a_{ij} = 0$ for $i > j$.
The adjoint of a matrix $A$,denoted by $adj(A)$,is the transpose of the cofactor matrix $C = [C_{ij}]$.
For an upper-triangular matrix,the cofactor $C_{ij}$ is zero if $i > j$.
Specifically,the cofactor matrix $C$ will be a lower-triangular matrix because the elements above the diagonal involve determinants of submatrices that contain at least one row or column of zeros.
Since $adj(A) = C^T$,the transpose of a lower-triangular matrix is an upper-triangular matrix.
Therefore,$adj(A)$ is an upper-triangular matrix.

(C) Given,$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$.
First,we calculate the determinant $|A|$:
$|A| = 1(1 - (-3)) - (-1)(2 - (-3)) + 1(2 - 1)$
$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$.
Next,we find the cofactor matrix $C$ of $A$:
$C_{11} = (1 - (-3)) = 4, C_{12} = -(2 - (-3)) = -5, C_{13} = (2 - 1) = 1$
$C_{21} = -(-1 - 1) = 2, C_{22} = (1 - 1) = 0, C_{23} = -(1 - (-1)) = -2$
$C_{31} = (3 - 1) = 2, C_{32} = -(-3 - 2) = 5, C_{33} = (1 - (-2)) = 3$.
Thus,$Adj(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
Since $B = A^{-1} = \frac{1}{|A|} Adj(A)$,we have $10 B = Adj(A)$.
Comparing $10 B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$ with $Adj(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$,we get $\alpha = 5$.

(B) Let $A = \begin{bmatrix} 1 & \tan \theta \\ -\tan \theta & 1 \end{bmatrix}$.
Then,$|A| = (1)(1) - (\tan \theta)(-\tan \theta) = 1 + \tan^2 \theta$.
The adjoint of $A$ is $\text{adj}(A) = \begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1 + \tan^2 \theta} \begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix}$.
Given the equation: $\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Substituting $A^{-1}$,we get: $\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} \frac{1}{1 + \tan^2 \theta} \begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Multiplying the matrices: $\frac{1}{1 + \tan^2 \theta} \begin{bmatrix} 1 - \tan^2 \theta & -2 \tan \theta \\ 2 \tan \theta & 1 - \tan^2 \theta \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
This simplifies to: $\begin{bmatrix} \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} & \frac{-2 \tan \theta}{1 + \tan^2 \theta} \\ \frac{2 \tan \theta}{1 + \tan^2 \theta} & \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Using trigonometric identities $\cos 2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2 \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we get: $\begin{bmatrix} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
Comparing the elements,we find $a = \cos 2 \theta$ and $b = \sin 2 \theta$.

(D) We know that $\operatorname{det}(kM) = k^n \operatorname{det}(M)$ for an $n \times n$ matrix $M$,and $\operatorname{det}(XY) = \operatorname{det}(X) \operatorname{det}(Y)$.
Given $A = \begin{bmatrix} 2 & 1 & 3 \\ -1 & 2 & 0 \\ 4 & 1 & 3 \end{bmatrix}$.
$|A| = 2(6 - 0) - 1(-3 - 0) + 3(-1 - 8) = 12 + 3 - 27 = -12$.
Given $B = \begin{bmatrix} 3 & 2 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix}$.
$|B| = 3(0 - 3) - 2(0 - 9) + 1(1 - 6) = -9 + 18 - 5 = 4$.
We need to find $\operatorname{det}(2 B^{-1} A^{-1})$.
Since $A$ and $B$ are $3 \times 3$ matrices,$\operatorname{det}(2 B^{-1} A^{-1}) = 2^3 \operatorname{det}(B^{-1}) \operatorname{det}(A^{-1})$.
Using the property $\operatorname{det}(M^{-1}) = \frac{1}{\operatorname{det}(M)}$,we have:
$\operatorname{det}(2 B^{-1} A^{-1}) = 8 \times \frac{1}{\operatorname{det}(B)} \times \frac{1}{\operatorname{det}(A)}$.
Substituting the values:
$= 8 \times \frac{1}{4} \times \frac{1}{-12} = 2 \times \left(-\frac{1}{12}\right) = -\frac{1}{6}$.

(A) Let the matrix be $A = \begin{bmatrix} 4 & 2 & 1-x \\ 5 & k & 1 \\ 6 & 3 & 1+x \end{bmatrix}$.
For the rank of the matrix to be $1$,all rows must be proportional to each other.
Comparing $R_1$ and $R_3$:
$R_3 = c R_1 \Rightarrow 6 = 4c \Rightarrow c = \frac{6}{4} = \frac{3}{2}$.
Checking the second element: $3 = 2 \times \frac{3}{2} = 3$ (This holds).
Checking the third element: $1+x = (1-x) \times \frac{3}{2}$.
$2(1+x) = 3(1-x) \Rightarrow 2 + 2x = 3 - 3x \Rightarrow 5x = 1 \Rightarrow x = \frac{1}{5}$.
Now,comparing $R_1$ and $R_2$:
$R_2 = d R_1 \Rightarrow 5 = 4d \Rightarrow d = \frac{5}{4}$.
Checking the second element: $k = 2 \times \frac{5}{4} = \frac{5}{2}$.
Checking the third element: $1 = (1-x) \times \frac{5}{4}$.
Substituting $x = \frac{1}{5}$: $1 = (1 - \frac{1}{5}) \times \frac{5}{4} = \frac{4}{5} \times \frac{5}{4} = 1$ (This holds).
Thus,the rank is $1$ when $k = \frac{5}{2}$ and $x = \frac{1}{5}$.

(A) Given the matrix $A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{4040} & \sqrt{4042} & \sqrt{4044} & \sqrt{4046} \\ \sqrt{6060} & \sqrt{6063} & \sqrt{6066} & \sqrt{6069} \\ \sqrt{8080} & \sqrt{8084} & \sqrt{8088} & \sqrt{8092} \end{bmatrix}$.
We can rewrite the rows as multiples of the first row:
$R_2 = \sqrt{2} R_1$,$R_3 = \sqrt{3} R_1$,and $R_4 = 2 R_1$.
Substituting these into the matrix:
$A = \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ \sqrt{2}\sqrt{2020} & \sqrt{2}\sqrt{2021} & \sqrt{2}\sqrt{2022} & \sqrt{2}\sqrt{2023} \\ \sqrt{3}\sqrt{2020} & \sqrt{3}\sqrt{2021} & \sqrt{3}\sqrt{2022} & \sqrt{3}\sqrt{2023} \\ 2\sqrt{2020} & 2\sqrt{2021} & 2\sqrt{2022} & 2\sqrt{2023} \end{bmatrix}$.
Applying row operations $R_2 \rightarrow R_2 - \sqrt{2}R_1$,$R_3 \rightarrow R_3 - \sqrt{3}R_1$,and $R_4 \rightarrow R_4 - 2R_1$,we get:
$A \sim \begin{bmatrix} \sqrt{2020} & \sqrt{2021} & \sqrt{2022} & \sqrt{2023} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.
Since there is only one non-zero row in the row-echelon form,the rank of $A$ is $1$.

(B) Let $A = \left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$.
To find the rank,we convert the matrix into row echelon form using row transformations:
Apply $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$A \sim \left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$.
The number of non-zero rows in the row echelon form of the matrix is $1$.
Therefore,the rank of $A$ is $1$.

(C) To find the rank of a $3 \times 3$ matrix,we calculate its determinant. If the determinant is non-zero,the rank is $3$.
Option $(A)$: Let $A = \left[\begin{array}{ccc}10 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14\end{array}\right]$. The rows are in arithmetic progression. $R_2 - R_1 = [1, 1, 1]$ and $R_3 - R_2 = [1, 1, 1]$. Since $R_3 - R_2 = R_2 - R_1$,the rows are linearly dependent. Thus,$\det(A) = 0$ and $\text{rank}(A) < 3$.
Option $(B)$: Let $B = \left[\begin{array}{ccc}0 & -51 & 101 \\ 51 & 0 & -581 \\ -101 & 581 & 0\end{array}\right]$. This is a skew-symmetric matrix of odd order $(3 \times 3)$. The determinant of a skew-symmetric matrix of odd order is always $0$. Thus,$\text{rank}(B) < 3$.
Option $(C)$: Let $C = \left[\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & 5 \\ -2 & 7 & 0\end{array}\right]$. Calculate the determinant: $\det(C) = 0(0 - 35) - 1(0 - (-10)) + 2(-7 - 0) = 0 - 10 - 14 = -24$. Since $\det(C) \neq 0$,the rank of matrix $C$ is $3$.
Option $(D)$: Let $D = \left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9\end{array}\right]$. Here,$R_2 = 2R_1$ and $R_3 = 3R_1$. The rows are linearly dependent. Thus,$\det(D) = 0$ and $\text{rank}(D) = 1$.
Therefore,the correct option is $(C)$.

(D) Given the determinant equation: $\left|\begin{array}{ccc}1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3x & x^2\end{array}\right|=0$
Expanding the determinant along the first row:
$1(6x^2 - 12x) - (-3)(x^2 - 4) + 1(3x - 6) = 0$
$6x^2 - 12x + 3(x^2 - 4) + 3x - 6 = 0$
$6x^2 - 12x + 3x^2 - 12 + 3x - 6 = 0$
Combining like terms:
$9x^2 - 9x - 18 = 0$
Dividing the entire equation by $9$:
$x^2 - x - 2 = 0$
Thus,the required equation is $x^2 - x - 2 = 0$.

(D) Given that $a_1, a_2, \ldots, a_9$ are in $G.P.$
Let $r$ be the common ratio,then $\frac{a_{n+1}}{a_n} = r$ for all $n$.
Let $\Delta = \begin{vmatrix} \log a_1 & \log a_2 & \log a_3 \\ \log a_4 & \log a_5 & \log a_6 \\ \log a_7 & \log a_8 & \log a_9 \end{vmatrix}$.
Applying column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$,we get:
$\Delta = \begin{vmatrix} \log a_1 & \log a_2 - \log a_1 & \log a_3 - \log a_2 \\ \log a_4 & \log a_5 - \log a_4 & \log a_6 - \log a_5 \\ \log a_7 & \log a_8 - \log a_7 & \log a_9 - \log a_8 \end{vmatrix}$.
Using the property $\log m - \log n = \log(\frac{m}{n})$,we have:
$\Delta = \begin{vmatrix} \log a_1 & \log(\frac{a_2}{a_1}) & \log(\frac{a_3}{a_2}) \\ \log a_4 & \log(\frac{a_5}{a_4}) & \log(\frac{a_6}{a_5}) \\ \log a_7 & \log(\frac{a_8}{a_7}) & \log(\frac{a_9}{a_8}) \end{vmatrix} = \begin{vmatrix} \log a_1 & \log r & \log r \\ \log a_4 & \log r & \log r \\ \log a_7 & \log r & \log r \end{vmatrix}$.
Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.

(C) Let $\Delta = \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$.
Applying $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \begin{vmatrix} 2(a+b+c) & a & a \\ 2(a+b+c) & c+a & b \\ 2(a+b+c) & c & a+b \end{vmatrix}$.
Taking $2(a+b+c)$ common from $C_1$:
$\Delta = 2(a+b+c) \begin{vmatrix} 1 & a & a \\ 1 & c+a & b \\ 1 & c & a+b \end{vmatrix}$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = 2(a+b+c) \begin{vmatrix} 1 & a & a \\ 0 & c & b-a \\ 0 & c-a & b \end{vmatrix}$.
Expanding along $C_1$:
$\Delta = 2(a+b+c) [1(c(b) - (b-a)(c-a))]$.
$\Delta = 2(a+b+c) [bc - (bc - ab - ac + a^2)]$.
$\Delta = 2(a+b+c) [bc - bc + ab + ac - a^2]$.
$\Delta = 2(a+b+c) [ab + ac - a^2]$.
$\Delta = 2a(a+b+c) (b+c-a)$.
Note: The original determinant simplifies to $4abc$ only if the matrix structure is $\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$. The provided solution in the prompt had a typo in the matrix structure. The correct value is $2a(a+b+c)(b+c-a)$.

(B) Let $\Delta = \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix}$
Applying $R_2 \to R_2 - R_1$:
$\Delta = \begin{vmatrix} a & b & c \\ -b & -c & -a \\ b+c & c+a & a+b \end{vmatrix}$
Applying $R_3 \to R_3 + R_2$:
$\Delta = \begin{vmatrix} a & b & c \\ -b & -c & -a \\ c & a & b \end{vmatrix}$
Wait,let us re-evaluate the operation $R_3 \to R_3 + R_2$:
$R_3$ becomes $(b+c-b, c+a-c, a+b-a) = (c, a, b)$.
So,$\Delta = \begin{vmatrix} a & b & c \\ -b & -c & -a \\ c & a & b \end{vmatrix}$
Expanding along $R_1$:
$\Delta = a(-bc - (-a^2)) - b(-b^2 - (-ac)) + c(-ab - (-c^2))$
$\Delta = a(a^2 - bc) - b(ac - b^2) + c(c^2 - ab)$
$\Delta = a^3 - abc - abc + b^3 + c^3 - abc$
$\Delta = a^3 + b^3 + c^3 - 3abc$

(C) Given that $\det A = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = K$.
We are given $\det B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 + 2a_1 & b_2 + 2b_1 & c_2 + 2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$.
Using the property of determinants,we apply the row operation $R_2 \to R_2 - 2R_1$.
This operation does not change the value of the determinant.
$\det B = \begin{vmatrix} a_1 & b_1 & c_1 \\ (a_2 + 2a_1) - 2a_1 & (b_2 + 2b_1) - 2b_1 & (c_2 + 2c_1) - 2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$
$\det B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$
Since the resulting determinant is identical to $\det A$,we have $\det B = K$.

(A) Given: $\begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} = 0$
Using the property of determinants,we can split the third column:
$\begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} = 0$
Taking $x, y, z$ common from the rows of the second determinant:
$\begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0$
Rearranging the columns of the second determinant to match the first:
$\begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + xyz \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} = 0$
$(1 + xyz) \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} = 0$
The determinant $\begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix}$ is a Vandermonde-like determinant which equals $(x-y)(y-z)(z-x)$.
Since $x, y, z$ are distinct,$(x-y)(y-z)(z-x) \neq 0$.
Therefore,$1 + xyz = 0$,which implies $xyz = -1$.

(B) Given that $\omega$ is a root of $x+\frac{1}{x}+1=0$,which implies $x^2+x+1=0$.
The roots of this equation are $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$.
We know that $1+\omega+\omega^2=0$ and $\omega^3=1$.
Let the determinant be $D = \left|\begin{array}{ccc}1 & 1+\omega & 1+\omega+\omega^2 \\ 3 & 4+3 \omega & 5+4 \omega+3 \omega^2 \\ 6 & 9+6 \omega & 11+9 \omega+6 \omega^2\end{array}\right|$.
Using $1+\omega+\omega^2=0$,the third column simplifies to:
Column $3$: $C_3 = \begin{bmatrix} 0 \\ 5+4\omega+3\omega^2 \\ 11+9\omega+6\omega^2 \end{bmatrix} = \begin{bmatrix} 0 \\ 3(1+\omega+\omega^2)+2+\omega \\ 6(1+\omega+\omega^2)+5+3\omega \end{bmatrix} = \begin{bmatrix} 0 \\ 2+\omega \\ 5+3\omega \end{bmatrix}$.
So,$D = \left|\begin{array}{ccc} 1 & 1+\omega & 0 \\ 3 & 4+3\omega & 2+\omega \\ 6 & 9+6\omega & 5+3\omega \end{array}\right|$.
Expanding along the first row:
$D = 1[(4+3\omega)(5+3\omega) - (2+\omega)(9+6\omega)] - (1+\omega)[3(5+3\omega) - 6(2+\omega)]$.
$D = [20 + 12\omega + 15\omega + 9\omega^2 - (18 + 12\omega + 9\omega + 6\omega^2)] - (1+\omega)[15 + 9\omega - 12 - 6\omega]$.
$D = [20 + 27\omega + 9\omega^2 - 18 - 21\omega - 6\omega^2] - (1+\omega)[3 + 3\omega]$.
$D = [2 + 6\omega + 3\omega^2] - 3(1+\omega)^2$.
$D = 2 + 6\omega + 3\omega^2 - 3(1 + 2\omega + \omega^2) = 2 + 6\omega + 3\omega^2 - 3 - 6\omega - 3\omega^2 = -1$.

(C) Let $\Delta = \left|\begin{array}{ccc}a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b\end{array}\right|$.
Applying the column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$C_2 \rightarrow C_2 - C_1 = (a+2b)-(a+b) = b$,$(a+3b)-(a+2b) = b$,$(a+5b)-(a+4b) = b$.
$C_3 \rightarrow C_3 - C_2 = (a+3b)-(a+2b) = b$,$(a+4b)-(a+3b) = b$,$(a+6b)-(a+5b) = b$.
Thus,$\Delta = \left|\begin{array}{ccc}a+b & b & b \\ a+2b & b & b \\ a+4b & b & b\end{array}\right|$.
Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.

(B) Let $P(x) = \left|\begin{array}{cc}x^3+2 x^2+3 x-2 & x^2+2 x+4 \\ x^3-x^2-2 x-1 & 3 x^3-2 x^2+4 x-2\end{array}\right| = a x^6+b x^5+c x^4+d x^3+e x^2+f x+g$.
To find $a+b+c+d+e+f$,we note that $P(1) = a+b+c+d+e+f+g$.
First,calculate $P(1)$ by substituting $x=1$ into the determinant:
$P(1) = \left|\begin{array}{cc}1+2+3-2 & 1+2+4 \\ 1-1-2-1 & 3-2+4-2\end{array}\right| = \left|\begin{array}{cc}4 & 7 \\ -3 & 3\end{array}\right| = (4)(3) - (7)(-3) = 12 + 21 = 33$.
Next,calculate $g$ by substituting $x=0$ into the determinant:
$g = P(0) = \left|\begin{array}{cc}-2 & 4 \\ -1 & -2\end{array}\right| = (-2)(-2) - (4)(-1) = 4 + 4 = 8$.
Since $P(1) = a+b+c+d+e+f+g$,we have $33 = (a+b+c+d+e+f) + 8$.
Therefore,$a+b+c+d+e+f = 33 - 8 = 25$.

(C) Given $A(\theta)=\begin{bmatrix} i \sin \theta & \cos \theta \\ \cos \theta & i \sin \theta \end{bmatrix}$.
The determinant is $\operatorname{det} A(\theta) = (i \sin \theta)(i \sin \theta) - (\cos \theta)(\cos \theta) = i^2 \sin^2 \theta - \cos^2 \theta$.
Since $i^2 = -1$,we have $\operatorname{det} A(\theta) = -\sin^2 \theta - \cos^2 \theta = -(\sin^2 \theta + \cos^2 \theta) = -1$.
Since the determinant is a constant $-1$ regardless of $\theta$,we have $\operatorname{det} A(\pi+\theta) = -1$,$\operatorname{det} A(-\theta) = -1$,and $\operatorname{det} A(\theta) = -1$.
Checking the options:
Option $A$: $\operatorname{det} A(\pi+\theta) = -1$ and $\operatorname{det} A(-\theta) = -1$. So,$-1 = -1$ (True).
Option $B$: $\operatorname{det} A(-\theta) = -1$ and $\operatorname{det} A(\theta) = -1$. So,$-1 = -1$ (True).
Option $C$: $\operatorname{det}[A(\theta)]^{-1} = \frac{1}{\operatorname{det} A(\theta)} = \frac{1}{-1} = -1$. The statement says $1$,which is False.
Option $D$: $\operatorname{det} A(-\theta) = -1$ (True).
Therefore,the statement that is not true is $C$.

(D) We have,$\det(A) = \begin{vmatrix} a & a & a-y \\ a & a+x & a \\ a & a & a \end{vmatrix}$.
Applying the column operation $C_1 \rightarrow C_1 - C_3$:
$\det(A) = \begin{vmatrix} y & a & a-y \\ 0 & a+x & a \\ 0 & a & a \end{vmatrix}$.
Expanding along the first column:
$\det(A) = y \cdot \begin{vmatrix} a+x & a \\ a & a \end{vmatrix} - 0 + 0$.
$\det(A) = y(a(a+x) - a^2) = y(a^2 + ax - a^2) = axy$.
Given $\det(A) = 16$,we have $axy = 16$,which implies $xy = \frac{16}{a}$.
Since $a$ is a non-zero constant,this equation is of the form $xy = k$,which represents a rectangular hyperbola.

(D) Given $\Delta_k = \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & k & k-1 \\ 0 & k-1 & k \end{array}\right|$.
Expanding along the first row,we get:
$\Delta_k = 1 \cdot (k \cdot k - (k-1) \cdot (k-1)) - 0 + 0$
$\Delta_k = k^2 - (k-1)^2$
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have $\Delta_k = (k - (k-1))(k + (k-1)) = 1 \cdot (2k-1) = 2k-1$.
We need to find the sum $S = \sum_{k=1}^{20} \Delta_k = \sum_{k=1}^{20} (2k-1)$.
This is the sum of the first $20$ odd numbers,which is given by $n^2$ where $n=20$.
$S = 20^2 = 400$.
Alternatively,using the telescoping sum property:
$\Delta_k = k^2 - (k-1)^2$
$\sum_{k=1}^{20} \Delta_k = (1^2 - 0^2) + (2^2 - 1^2) + (3^2 - 2^2) + \ldots + (20^2 - 19^2) = 20^2 - 0^2 = 400$.

(B) Let $A = \left[\begin{array}{ccc}-x & x & 2 \\ 2 & x & -x \\ x & -2 & -2\end{array}\right]$.
$A$ matrix $A$ is non-singular if and only if its determinant $|A| \neq 0$.
Calculating the determinant $|A|$:
$|A| = -x \begin{vmatrix} x & -x \\ -2 & -2 \end{vmatrix} - x \begin{vmatrix} 2 & -x \\ x & -2 \end{vmatrix} + 2 \begin{vmatrix} 2 & x \\ x & -2 \end{vmatrix}$
$|A| = -x(-2x - 2x) - x(-4 + x^2) + 2(-4 - x^2)$
$|A| = -x(-4x) + 4x - x^3 - 8 - 2x^2$
$|A| = 4x^2 + 4x - x^3 - 8 - 2x^2$
$|A| = -x^3 + 2x^2 + 4x - 8$
For the matrix to be non-singular,$|A| \neq 0$:
$-x^3 + 2x^2 + 4x - 8 \neq 0$
$-(x^3 - 2x^2 - 4x + 8) \neq 0$
$-(x^2(x - 2) - 4(x - 2)) \neq 0$
$-(x^2 - 4)(x - 2) \neq 0$
$-(x - 2)(x + 2)(x - 2) \neq 0$
$-(x - 2)^2(x + 2) \neq 0$
Therefore,$x \neq 2$ and $x \neq -2$.
Thus,the matrix is non-singular for all $x$ other than $2$ and $-2$.

(D) Given the inequality $a^2+b^2+c^2-ab-bc-ca \leq 0$.
Multiplying by $2$ and rearranging,we get $(a-b)^2 + (b-c)^2 + (c-a)^2 \leq 0$.
Since the sum of squares of real numbers is non-negative,this is only possible if $a-b=0$,$b-c=0$,and $c-a=0$,which implies $a=b=c$.
Substituting $a=b=c$ into the determinant:
$\left|\begin{array}{ccc} (a-a+1)^5 & a^7-a^7 & a^9-a^9 \\ a^{11}-a^{11} & (a-a+2)^3 & a^{13}-a^{13} \\ a^{15}-a^{15} & a^{17}-a^{17} & (a-a+3)^1 \end{array}\right|$
$= \left|\begin{array}{ccc} 1^5 & 0 & 0 \\ 0 & 2^3 & 0 \\ 0 & 0 & 3^1 \end{array}\right|$
$= \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 3 \end{array}\right| = 1 \times 8 \times 3 = 24$.

(C) Let $P$ be a square matrix of order $n$.
We know the property of determinants that $|kP| = k^n |P|$,where $k$ is a scalar constant.
In this problem,the order of the determinant is $n = 3$ and the constant multiplier is $k = 3$.
Given that the value of the original determinant is $|P| = A$.
Therefore,the value of the new determinant is $|3P| = 3^3 |P|$.
$|3P| = 27 \times A = 27A$.

(A) Let $\Delta = \left|\begin{array}{ccc}\cos 2x & \sin^2 x & \cos 2x \\ \sin^2 x & \cos 2x & \cos^2 x \\ \cos 2x & \cos^2 x & \cos 2x\end{array}\right|$.
We know that $\cos 2x = 2\cos^2 x - 1$ and $\sin^2 x = 1 - \cos^2 x$.
Substituting these into the determinant:
$\Delta = \left|\begin{array}{ccc} 2\cos^2 x - 1 & 1 - \cos^2 x & 2\cos^2 x - 1 \\ 1 - \cos^2 x & 2\cos^2 x - 1 & \cos^2 x \\ 2\cos^2 x - 1 & \cos^2 x & 2\cos^2 x - 1 \end{array}\right|$.
Let $u = \cos^2 x$. Then the determinant becomes:
$\Delta = \left|\begin{array}{ccc} 2u-1 & 1-u & 2u-1 \\ 1-u & 2u-1 & u \\ 2u-1 & u & 2u-1 \end{array}\right|$.
Subtract $C_3$ from $C_1$:
$\Delta = \left|\begin{array}{ccc} 0 & 1-u & 2u-1 \\ 1-u-u & 2u-1 & u \\ 0 & u & 2u-1 \end{array}\right| = \left|\begin{array}{ccc} 0 & 1-u & 2u-1 \\ 1-2u & 2u-1 & u \\ 0 & u & 2u-1 \end{array}\right|$.
Expanding along $C_1$:
$\Delta = -(1-2u) \left[ (1-u)(2u-1) - u(2u-1) \right] = -(1-2u)(2u-1)(1-u-u) = (2u-1)^2(1-2u) = -(2u-1)^3$.
Substituting $u = \cos^2 x$:
$\Delta = -(2\cos^2 x - 1)^3 = -(8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1) = -8\cos^6 x + 12\cos^4 x - 6\cos^2 x + 1$.
The constant term is the term independent of $\cos x$,which is $1$.