QUADRATIC EQUATION Questions in English

(D) Given the quadratic equation $ax^2 + bx + c = 0$ with $a < 0, b > 0, c > 0$.
Multiplying by $-1$,we get $-ax^2 - bx - c = 0$,where the coefficients are now $A = -a > 0, B = -b < 0, C = -c < 0$.
The product of the roots is $\alpha \beta = \frac{C}{A} = \frac{-c}{-a} = \frac{c}{a}$. Since $c > 0$ and $a < 0$,the product $\alpha \beta < 0$,which implies the roots have opposite signs.
The sum of the roots is $\alpha + \beta = -\frac{B}{A} = -\frac{-b}{-a} = -\frac{b}{a}$. Since $b > 0$ and $a < 0$,the ratio $\frac{b}{a} < 0$,so $\alpha + \beta = -(\text{negative value}) > 0$.
Since the sum of the roots is positive and the product is negative,the root with the larger absolute value must be positive.
Given $\alpha < \beta$,it follows that $\alpha$ is the negative root and $\beta$ is the positive root.
Since the sum $\alpha + \beta > 0$,we have $|\beta| > |\alpha|$,which means $\beta > |\alpha|$ is false,but rather the positive root $\beta$ has a larger magnitude than the negative root $\alpha$,i.e.,$|\beta| > |\alpha|$.
Thus,$\alpha < 0 < |\alpha| < \beta$.

(B) Given the equation $x^3 + qx - r = 0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas:
$\alpha + \beta + \gamma = 0$
$\alpha \beta + \beta \gamma + \gamma \alpha = q$
$\alpha \beta \gamma = r$
We need to find the equation with roots $y_1 = \beta \gamma + \frac{1}{\alpha}$,$y_2 = \gamma \alpha + \frac{1}{\beta}$,and $y_3 = \alpha \beta + \frac{1}{\gamma}$.
Since $\alpha \beta \gamma = r$,we have $\beta \gamma = \frac{r}{\alpha}$,$\gamma \alpha = \frac{r}{\beta}$,and $\alpha \beta = \frac{r}{\gamma}$.
Thus,the roots are $y_1 = \frac{r}{\alpha} + \frac{1}{\alpha} = \frac{r+1}{\alpha}$,$y_2 = \frac{r+1}{\beta}$,and $y_3 = \frac{r+1}{\gamma}$.
Let $y = \frac{r+1}{x}$,then $x = \frac{r+1}{y}$.
Substitute $x$ into the original equation:
$(\frac{r+1}{y})^3 + q(\frac{r+1}{y}) - r = 0$
$\frac{(r+1)^3}{y^3} + \frac{q(r+1)}{y} - r = 0$
Multiply by $y^3$:
$(r+1)^3 + q(r+1)y^2 - ry^3 = 0$
$ry^3 - q(r+1)y^2 - (r+1)^3 = 0$.
Replacing $y$ with $x$,we get $rx^3 - q(r+1)x^2 - (r+1)^3 = 0$.

(C) Let $f(x) = x^2 - 3x + a$. Since $\alpha < 1 < \beta$,the value of the quadratic function at $x = 1$ must be negative because the parabola opens upwards.
$f(1) < 0$
$1^2 - 3(1) + a < 0$
$1 - 3 + a < 0$
$-2 + a < 0$
$a < 2$
Thus,$a \in (-\infty, 2)$.
Note: The condition $D > 0$ is automatically satisfied when $Af(1) < 0$ for a quadratic with a positive leading coefficient,as $f(1) < 0$ implies the function takes negative values,meaning it must cross the $x$-axis at two distinct points.

(A) Let the roots of the equation $x^3 + 3x^2 - 1 = 0$ be $\alpha, \beta,$ and $\gamma$.
From Vieta's formulas,the product of the roots is $\alpha \beta \gamma = -(\frac{-1}{1}) = 1$.
Therefore,$\alpha \beta = \frac{1}{\gamma}$.
Let $y = \alpha \beta = \frac{1}{\gamma}$,which implies $\gamma = \frac{1}{y}$.
Since $\gamma$ is a root of the original equation $x^3 + 3x^2 - 1 = 0$,we substitute $x = \gamma = \frac{1}{y}$ into the equation:
$(\frac{1}{y})^3 + 3(\frac{1}{y})^2 - 1 = 0$.
Multiplying the entire equation by $y^3$,we get:
$1 + 3y - y^3 = 0$.
Rearranging the terms,we get $y^3 - 3y - 1 = 0$.
Replacing $y$ with $x$,the required equation is $x^3 - 3x - 1 = 0$.

(A) The given quadratic equation is $x^2 - r^2(r + 1)x + r^5 = 0$.
Let the roots be $\alpha_r$ and $\beta_r$. From the properties of roots,we have $\alpha_r + \beta_r = r^2(r + 1) = r^3 + r^2$ and $\alpha_r \beta_r = r^5$.
Since $r^5 = r^2 \cdot r^3$,the roots are $\alpha_r = r^2$ and $\beta_r = r^3$ (given $\alpha_r < \beta_r$ for $r > 1$).
We need to calculate $S = \sum_{r=1}^n (3\alpha_r + 2\beta_r) = 3\sum_{r=1}^n r^2 + 2\sum_{r=1}^n r^3$.
Using the standard summation formulas $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^n r^3 = \frac{n^2(n+1)^2}{4}$,we get:
$S = 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n^2(n+1)^2}{4} = \frac{n(n+1)(2n+1)}{2} + \frac{n^2(n+1)^2}{2}$.
Factoring out $\frac{n(n+1)}{2}$,we get $S = \frac{n(n+1)}{2} [ (2n+1) + n(n+1) ] = \frac{n(n+1)}{2} [ 2n + 1 + n^2 + n ] = \frac{n(n+1)}{2} (n^2 + 3n + 1)$.

(A) Given the cubic equation $x^3 - 2x^2 + 3x - 2 = 0$.
By inspection,$x = 1$ is a root because $1 - 2 + 3 - 2 = 0$.
Dividing the polynomial by $(x - 1)$,we get $(x - 1)(x^2 - x + 2) = 0$.
Let $\alpha = 1$. Then $\beta$ and $\gamma$ are roots of $x^2 - x + 2 = 0$.
From the quadratic equation,$\beta + \gamma = 1$ and $\beta\gamma = 2$.
We need to evaluate $S = \frac{\alpha\beta}{\alpha + \beta} + \frac{\alpha\gamma}{\alpha + \gamma} + \frac{\beta\gamma}{\beta + \gamma}$.
Substituting $\alpha = 1$: $S = \frac{\beta}{1 + \beta} + \frac{\gamma}{1 + \gamma} + \frac{\beta\gamma}{\beta + \gamma}$.
Since $\beta + \gamma = 1$,the last term is $\frac{2}{1} = 2$.
For the first two terms: $\frac{\beta(1 + \gamma) + \gamma(1 + \beta)}{(1 + \beta)(1 + \gamma)} = \frac{\beta + \beta\gamma + \gamma + \beta\gamma}{1 + \beta + \gamma + \beta\gamma} = \frac{(\beta + \gamma) + 2\beta\gamma}{1 + (\beta + \gamma) + \beta\gamma}$.
Substituting the values: $\frac{1 + 2(2)}{1 + 1 + 2} = \frac{5}{4}$.
Thus,$S = \frac{5}{4} + 2 = \frac{5 + 8}{4} = \frac{13}{4}$.

(D) Given the polynomial $P(x) = x^2 + ax + b$ has factors $(x - a)(x - b)$.
Expanding the factors: $(x - a)(x - b) = x^2 - (a + b)x + ab$.
Comparing this with $P(x) = x^2 + ax + b$,we get:
$a = -(a + b) \Rightarrow 2a + b = 0$ (Equation $1$)
$b = ab$ (Equation $2$)
From Equation $2$,$b(a - 1) = 0$,so $b = 0$ or $a = 1$.
Case $1$: If $b = 0$,then from Equation $1$,$2a + 0 = 0 \Rightarrow a = 0$. Thus,$P(x) = x^2$. Then $P(2) = 2^2 = 4$.
Case $2$: If $a = 1$,then from Equation $1$,$2(1) + b = 0 \Rightarrow b = -2$. Thus,$P(x) = x^2 + x - 2$. Then $P(2) = 2^2 + 2 - 2 = 4$.
In both cases,$P(2) = 4$.

(B) Given equation: $\ln(\ln x) = \log_x e$
Using the property $\log_x e = \frac{1}{\ln x}$,the equation becomes $\ln(\ln x) = \frac{1}{\ln x}$.
Let $\ln x = t$. For $\ln x$ to be defined,$x > 0$. For $\ln(\ln x)$ to be defined,$\ln x > 0$,so $x > 1$. Also,for $\log_x e$ to be defined,$x > 0$ and $x \neq 1$.
Substituting $t = \ln x$,we get $\ln t = \frac{1}{t}$,or $t \ln t = 1$.
Let $f(t) = t \ln t$. We want to find the number of solutions for $f(t) = 1$ where $t > 0$.
$f'(t) = \ln t + 1$. Setting $f'(t) = 0$ gives $t = 1/e$.
At $t = 1/e$,$f(1/e) = (1/e) \ln(1/e) = -1/e \approx -0.368$.
As $t \to 0^+$,$f(t) \to 0$. As $t \to \infty$,$f(t) \to \infty$.
Since $f(t)$ is continuous and strictly increasing for $t > 1/e$,and $f(1) = 0 < 1$ while $f(e) = e > 1$,by the Intermediate Value Theorem,there exists exactly one solution for $t$ in the interval $(1, e)$.
Since $t = \ln x$ is a one-to-one function,there is exactly one solution for $x$.

(B) Given $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$.
This simplifies to $\frac{x_1 x_2 + x_2 x_3 + x_3 x_1}{x_1 x_2 x_3} = \frac{1}{x_1 + x_2 + x_3}$.
$(x_1 + x_2 + x_3)(x_1 x_2 + x_2 x_3 + x_3 x_1) = x_1 x_2 x_3$.
Let $x_1, x_2, x_3$ be roots of $t^3 + \alpha t^2 + \beta t + \gamma = 0$,where $\alpha = -(x_1+x_2+x_3)$,$\beta = (x_1 x_2 + x_2 x_3 + x_3 x_1)$,and $\gamma = -x_1 x_2 x_3$.
The equation becomes $(-\alpha)(\beta) = -(-\gamma) \Rightarrow \gamma = \alpha \beta$.
So,$t^3 + \alpha t^2 + \beta t + \alpha \beta = 0 \Rightarrow t^2(t + \alpha) + \beta(t + \alpha) = 0 \Rightarrow (t + \alpha)(t^2 + \beta) = 0$.
The roots are $t_1 = -\alpha$,$t_2 = \sqrt{-\beta}$,$t_3 = -\sqrt{-\beta}$.
Thus,$x_1 = -(x_1+x_2+x_3) \Rightarrow x_2 + x_3 = 0 \Rightarrow x_3 = -x_2$.
Now,check the condition $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$.
For $x_3 = -x_2$,the expression becomes $\frac{1}{x_1^n + x_2^n + (-x_2)^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{(-x_2)^n}$.
If $n$ is odd,$(-x_2)^n = -x_2^n$,so $\frac{1}{x_1^n + x_2^n - x_2^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} - \frac{1}{x_2^n} \Rightarrow \frac{1}{x_1^n} = \frac{1}{x_1^n}$,which is true.

(D) Given the equation $x^3 - x - 2 = 0$,since $\alpha, \beta, \gamma$ are roots,we have $\alpha^3 = \alpha + 2$,$\beta^3 = \beta + 2$,and $\gamma^3 = \gamma + 2$.
Multiplying by $\alpha^2$,we get $\alpha^5 = \alpha^3 + 2\alpha^2 = (\alpha + 2) + 2\alpha^2 = 2\alpha^2 + \alpha + 2$.
Similarly,$\beta^5 = 2\beta^2 + \beta + 2$ and $\gamma^5 = 2\gamma^2 + \gamma + 2$.
Summing these,$\sum \alpha^5 = 2\sum \alpha^2 + \sum \alpha + 6$.
From the equation $x^3 + 0x^2 - x - 2 = 0$,we have $\sum \alpha = 0$ and $\sum \alpha\beta = -1$.
Using the identity $\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta$,we get $\sum \alpha^2 = (0)^2 - 2(-1) = 2$.
Substituting these values: $\sum \alpha^5 = 2(2) + 0 + 6 = 4 + 6 = 10$.

(B) Let $f(t) = t^2 + 5mt - 3m + 1$,where $t = \{x\}$. Since $x \in R$,$t$ takes all values in the interval $[0, 1)$.
We are given $f(t) < 0$ for all $t \in [0, 1)$.
Since $f(t)$ is a quadratic in $t$ with a positive leading coefficient,the parabola opens upwards.
For $f(t) < 0$ to hold for all $t \in [0, 1)$,the values at the endpoints must satisfy $f(0) \le 0$ and $f(1) \le 0$.
Evaluating at $t = 0$: $f(0) = -3m + 1 < 0 \implies 3m > 1 \implies m > 1/3$.
Evaluating at $t = 1$: $f(1) = 1 + 5m - 3m + 1 = 2m + 2 < 0 \implies 2m < -2 \implies m < -1$.
Since there is no value of $m$ that satisfies both $m > 1/3$ and $m < -1$,there are no such integral values of $m$.

(B) Given the quadratic function $f(x) = ax^2 + 2bx + c$.
We are given that $f(x) = 0$ has imaginary roots,which implies the discriminant $D = (2b)^2 - 4ac < 0$,or $4b^2 - 4ac < 0$,which simplifies to $b^2 < ac$.
We are also given that $4a + 4b + c < 0$. Note that $f(2) = a(2)^2 + 2b(2) + c = 4a + 4b + c$. Thus,$f(2) < 0$.
Since $f(x)$ has imaginary roots,the parabola $y = f(x)$ does not intersect the $x$-axis. This means $f(x)$ is either always positive (if $a > 0$) or always negative (if $a < 0$).
If $a > 0$,then $f(x) > 0$ for all $x$,which contradicts $f(2) < 0$. Therefore,we must have $a < 0$.
Since $a < 0$ and $f(x)$ is always negative,the entire parabola lies below the $x$-axis. The $y$-intercept is $f(0) = c$. Since the entire graph is below the $x$-axis,the $y$-intercept must be negative. Therefore,$c < 0$.

(A) Let $f(x) = x^2 - (2a + 3)x + a^2 + 3a$. For both roots to lie in $(0, 4)$,the following conditions must be satisfied:
$1$. Discriminant $D \ge 0$: $D = (2a + 3)^2 - 4(a^2 + 3a) = 4a^2 + 12a + 9 - 4a^2 - 12a = 9$. Since $9 > 0$,the roots are always real.
$2$. $0 < \text{Vertex} < 4$: $0 < \frac{2a + 3}{2} < 4 \implies 0 < 2a + 3 < 8 \implies -3 < 2a < 5 \implies -1.5 < a < 2.5$.
$3$. $f(0) > 0$: $a^2 + 3a > 0 \implies a(a + 3) > 0 \implies a \in (-\infty, -3) \cup (0, \infty)$.
$4$. $f(4) > 0$: $16 - 4(2a + 3) + a^2 + 3a > 0 \implies 16 - 8a - 12 + a^2 + 3a > 0 \implies a^2 - 5a + 4 > 0 \implies (a - 1)(a - 4) > 0 \implies a \in (-\infty, 1) \cup (4, \infty)$.
Taking the intersection of all conditions: $a \in (0, 1)$.
The only integer value in this interval is none. However,checking the boundary conditions for roots to be in $(0, 4)$,we find no integer $a$ satisfies the condition. Thus,the number of integral values is $0$.

(C) For the equation $ax^2 + bx + c = 0$,the roots are $x_1 = \alpha - \beta$ and $x_2 = \gamma - \delta$.
The difference of the roots is $|x_1 - x_2| = |(\alpha - \beta) - (\gamma - \delta)| = |\alpha - \beta - \gamma + \delta| = \frac{\sqrt{D_1}}{|a|}$,where $D_1 = b^2 - 4ac$.
For the equation $Ax^2 + Bx + C = 0$,the roots are $y_1 = \alpha + \delta$ and $y_2 = \beta + \gamma$.
The difference of the roots is $|y_1 - y_2| = |(\alpha + \delta) - (\beta + \gamma)| = |\alpha - \beta + \delta - \gamma| = \frac{\sqrt{D_2}}{|A|}$,where $D_2 = B^2 - 4AC$.
Since $|\alpha - \beta - \gamma + \delta| = |\alpha - \beta + \delta - \gamma|$,the absolute differences of the roots are equal.
Therefore,$\frac{\sqrt{D_1}}{|a|} = \frac{\sqrt{D_2}}{|A|}$.
Rearranging the terms,we get $\left| \frac{a}{A} \right| = \sqrt{\frac{D_1}{D_2}}$.

(D) The function $y = |f(|x|)|$ is non-derivable at $5$ points if the quadratic equation $x^2 - x + k - 2 = 0$ has two distinct positive roots.
For the roots to be real and distinct,the discriminant $D > 0$:
$D = (-1)^2 - 4(1)(k - 2) > 0$
$1 - 4k + 8 > 0$
$9 - 4k > 0 \Rightarrow k < 9/4$.
For the roots to be positive,the product of roots $\alpha\beta > 0$ and the sum of roots $\alpha + \beta > 0$:
Product $\alpha\beta = k - 2 > 0 \Rightarrow k > 2$.
Sum $\alpha + \beta = 1 > 0$ (which is always true).
Combining these conditions,we get $2 < k < 9/4$.
Thus,the set of values is $(2, 9/4)$.

(C) Let $\alpha$ and $\beta$ be the integral roots of the quadratic equation $x^2 + (2 - \tan \theta)x - (1 + \tan \theta) = 0$.
From the properties of roots,we have:
$\alpha + \beta = \tan \theta - 2$ $(1)$
$\alpha \beta = -\tan \theta - 1$ $(2)$
Adding $(1)$ and $(2)$,we get:
$\alpha + \beta + \alpha \beta = -3$
Adding $1$ to both sides to factorize:
$\alpha \beta + \alpha + \beta + 1 = -3 + 1$
$(\alpha + 1)(\beta + 1) = -2$
Since $\alpha$ and $\beta$ are integers,$(\alpha + 1)$ and $(\beta + 1)$ must be integer factors of $-2$. The possible pairs for $(\alpha + 1, \beta + 1)$ are $(-1, 2), (2, -1), (1, -2), (-2, 1)$.
Case $1$: $\alpha + 1 = -1$ and $\beta + 1 = 2 \Rightarrow \alpha = -2, \beta = 1$. Then $\tan \theta = \alpha + \beta + 2 = -2 + 1 + 2 = 1$.
Case $2$: $\alpha + 1 = 1$ and $\beta + 1 = -2 \Rightarrow \alpha = 0, \beta = -3$. Then $\tan \theta = \alpha + \beta + 2 = 0 - 3 + 2 = -1$.
For $\tan \theta = 1$ in $(0, 2\pi)$,$\theta = \frac{\pi}{4}, \frac{5\pi}{4}$.
For $\tan \theta = -1$ in $(0, 2\pi)$,$\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$.
The sum of all possible values of $\theta$ is $\frac{\pi}{4} + \frac{5\pi}{4} + \frac{3\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$.
Thus,$k = 4$.

(B) Given the equation: $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$.
Rearranging the terms: $2\cos(ax + b) = x - \frac{x^2 + 5}{2}$.
Multiply by $-1/2$: $-\cos(ax + b) = \frac{x^2 + 5}{4} - \frac{x}{2} = \frac{x^2 - 2x + 5}{4}$.
This simplifies to: $-\cos(ax + b) = \frac{(x - 1)^2}{4} + 1$.
We know that the range of $-\cos(ax + b)$ is $[-1, 1]$.
Also,for the right side,since $(x - 1)^2 \ge 0$,we have $\frac{(x - 1)^2}{4} + 1 \ge 1$.
For the equation to have a solution,both sides must be equal to $1$.
Thus,$\frac{(x - 1)^2}{4} + 1 = 1 \Rightarrow x = 1$.
And $-\cos(a(1) + b) = 1 \Rightarrow \cos(a + b) = -1$.
Therefore,$a + b = (2k + 1)\pi$ for $k \in I$. For $k=0$,$a + b = \pi$.

(C) Let $a^2(a + p) = b^2(b + p) = c^2(c + p) = k$.
This implies that $a, b, c$ are the roots of the cubic equation $x^2(x + p) = k$,which can be rewritten as $x^3 + px^2 - k = 0$.
For a cubic equation of the form $Ax^3 + Bx^2 + Cx + D = 0$,the sum of the product of roots taken two at a time is given by $\frac{C}{A}$.
In the equation $x^3 + px^2 + 0x - k = 0$,we have $A = 1, B = p, C = 0, D = -k$.
Therefore,$bc + ca + ab = \frac{C}{A} = \frac{0}{1} = 0$.

(A) Let $f(x) = \frac{3}{x - a^3} + \frac{5}{x - a^5} + \frac{7}{x - a^7}$.
Given $a > 1$,we have $a^3 < a^5 < a^7$.
As $x \to (a^3)^+$,$f(x) \to +\infty$. As $x \to (a^5)^-$,$f(x) \to -\infty$. Thus,there is a root in $(a^3, a^5)$.
As $x \to (a^5)^+$,$f(x) \to +\infty$. As $x \to (a^7)^-$,$f(x) \to -\infty$. Thus,there is a root in $(a^5, a^7)$.
Since $a > 1$,all values in the intervals $(a^3, a^5)$ and $(a^5, a^7)$ are positive.
Therefore,the equation has two real and positive roots.

(B) From the graph,the parabola opens upwards,so $a > 0$.
The vertex is at $x = -\frac{b}{2a}$,which is on the negative side of the $y$-axis,so $-\frac{b}{2a} < 0$. Since $a > 0$,we have $b > 0$.
The $y$-intercept is at $c$,and the graph intersects the $y$-axis below the $x$-axis,so $c < 0$.
The graph intersects the $x$-axis at two distinct points,so the discriminant $D = b^2 - 4ac > 0$.
Now,let's evaluate the options:
$A$: $abc = (+)(+)(-) = - < 0$. This is correct.
$B$: $ac^2bD = (+)(+)(+)(+) = + > 0$. Thus,$ac^2bD < 0$ is incorrect.
$C$: $\frac{a^2c}{b^2D} = \frac{(+)(+)(-)}{(+)(+)} = - < 0$. This is correct.
$D$: $bD = (+)(+) = + > 0$. This is correct.
Therefore,the incorrect statement is $B$.

(B) Given $P(x) = x^3 - ax^2 + bx + c$ has integral roots $\alpha, \beta, \gamma$.
Thus,$P(x) = (x - \alpha)(x - \beta)(x - \gamma)$.
Given $P(6) = 3$,we have $(6 - \alpha)(6 - \beta)(6 - \gamma) = 3$.
Since $\alpha, \beta, \gamma$ are integers,$(6 - \alpha), (6 - \beta),$ and $(6 - \gamma)$ must be integer factors of $3$.
The factors of $3$ are $\{1, -1, 3, -3\}$.
Let $x_1 = 6 - \alpha, x_2 = 6 - \beta, x_3 = 6 - \gamma$. Then $x_1 x_2 x_3 = 3$.
Also,$a = \alpha + \beta + \gamma = (6 - x_1) + (6 - x_2) + (6 - x_3) = 18 - (x_1 + x_2 + x_3)$.
Possible sets for $(x_1, x_2, x_3)$ such that their product is $3$:
$1)$ $(1, 1, 3) \Rightarrow x_1 + x_2 + x_3 = 5 \Rightarrow a = 18 - 5 = 13$.
$2)$ $(1, -1, -3) \Rightarrow x_1 + x_2 + x_3 = -3 \Rightarrow a = 18 - (-3) = 21$.
$3)$ $(-1, -1, 3) \Rightarrow x_1 + x_2 + x_3 = 1 \Rightarrow a = 18 - 1 = 17$.
Comparing with the options,$a$ cannot be $15$.

(A) Let the roots of the quadratic equation $x^2 + (k + 1)x + \lambda = 0$ be $\alpha$ and $\alpha^2$.
From the properties of roots,the product of the roots is $\alpha \cdot \alpha^2 = \alpha^3 = \lambda$.
The sum of the roots is $\alpha + \alpha^2 = -(k + 1)$.
Cubing the sum equation: $(\alpha + \alpha^2)^3 = (-(k + 1))^3$.
$\alpha^3 + (\alpha^2)^3 + 3\alpha \cdot \alpha^2(\alpha + \alpha^2) = -(k + 1)^3$.
Substituting $\alpha^3 = \lambda$ and $\alpha + \alpha^2 = -(k + 1)$:
$\lambda + \lambda^2 + 3\lambda(-(k + 1)) = -(k + 1)^3$.
$\lambda^2 + \lambda - 3\lambda(k + 1) = -(k + 1)^3$.
For the roots to be squares of each other,we consider specific cases:
$1$. If $\alpha = 0$,then $\lambda = 0$. The equation becomes $x^2 + (k + 1)x = 0$. Roots are $0, -(k+1)$. For $0^2 = -(k+1)$,$k = -1$.
$2$. If $\alpha = 1$,then $\lambda = 1$. The equation becomes $x^2 + (k + 1)x + 1 = 0$. Roots are $1, 1$. Sum $1+1 = -(k+1) \implies k = -3$.
$3$. If $\alpha = \omega$ (cube root of unity),then $\alpha^2 = \omega^2$. The equation is $x^2 + x + 1 = 0$. Comparing with $x^2 + (k+1)x + \lambda = 0$,we get $k+1 = 1 \implies k = 0$ and $\lambda = 1$.
The possible values of $k$ are $-1, -3, 0$.
The sum of these values is $(-1) + (-3) + 0 = -4$.

(C) Given the equations $2ax^2 - 3bx + 4c = 0$ and $3x^2 - 4x + 5 = 0$.
Since the second equation $3x^2 - 4x + 5 = 0$ has a discriminant $D = (-4)^2 - 4(3)(5) = 16 - 60 = -44 < 0$,it has imaginary roots.
If two quadratic equations have a common root and the coefficients are real,and one equation has imaginary roots,then both equations must have the same roots.
Thus,the ratios of the coefficients must be equal:
$\frac{2a}{3} = \frac{-3b}{-4} = \frac{4c}{5} = k$
$\frac{2a}{3} = \frac{3b}{4} = \frac{4c}{5} = k$
From this,we get $a = \frac{3k}{2}$,$b = \frac{4k}{3}$,and $c = \frac{5k}{4}$.
Now,calculate $\frac{a + b}{c}$:
$\frac{a + b}{c} = \frac{\frac{3k}{2} + \frac{4k}{3}}{\frac{5k}{4}} = \frac{\frac{9k + 8k}{6}}{\frac{5k}{4}} = \frac{17k}{6} \times \frac{4}{5k} = \frac{17 \times 2}{3 \times 5} = \frac{34}{15}$.

(A) Given that $\alpha + \beta = 3$ and $|\alpha - \beta| = 4$.
Squaring the second equation,we get $(\alpha - \beta)^2 = 16$.
We know the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.
Substituting the known values: $16 = (3)^2 - 4\alpha\beta$.
$16 = 9 - 4\alpha\beta$.
$4\alpha\beta = 9 - 16 = -7$.
So,$\alpha\beta = -\frac{7}{4}$.
The quadratic equation with roots $\alpha$ and $\beta$ is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Substituting the values: $x^2 - 3x - \frac{7}{4} = 0$.
Multiplying the entire equation by $4$,we get $4x^2 - 12x - 7 = 0$.

(B) Given the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.
This can be rewritten as $\frac{1}{2}(a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2] = 0$.
Since $a, b, c$ are distinct,$(a - b)^2 + (b - c)^2 + (c - a)^2 \neq 0$,therefore $a + b + c = 0$.
For the quadratic equation $ax^2 + bx + c = 0$,if we substitute $x = 1$,we get $a(1)^2 + b(1) + c = a + b + c$.
Since $a + b + c = 0$,$x = 1$ is one root of the equation.
Let the roots be $\alpha$ and $\beta$. We know that the product of roots $\alpha \cdot \beta = c/a$.
Since $\alpha = 1$,the other root $\beta = c/a$.

(B) Let $t = \sqrt{\log_3(x) - 1}$. Since the square root is defined,$\log_3(x) - 1 \ge 0$,so $\log_3(x) \ge 1$,which implies $x \ge 3$. Also,$t \ge 0$.
Then $\log_3(x) = t^2 + 1$.
The inequality becomes $t + \frac{\frac{1}{2} \cdot 3 \log_3(x)}{-1} + 2 > 0$.
Substituting $\log_3(x) = t^2 + 1$:
$t - \frac{3}{2}(t^2 + 1) + 2 > 0$
$t - \frac{3}{2}t^2 - \frac{3}{2} + 2 > 0$
$t - \frac{3}{2}t^2 + \frac{1}{2} > 0$
Multiply by $-2$: $3t^2 - 2t - 1 < 0$.
Factoring the quadratic: $(3t + 1)(t - 1) < 0$.
This holds for $-\frac{1}{3} < t < 1$. Since $t \ge 0$,we have $0 \le t < 1$.
Substituting back $t = \sqrt{\log_3(x) - 1}$:
$0 \le \sqrt{\log_3(x) - 1} < 1$
$0 \le \log_3(x) - 1 < 1$
$1 \le \log_3(x) < 2$
$3^1 \le x < 3^2$
$3 \le x < 9$.
The integers satisfying this are $3, 4, 5, 6, 7, 8$. There are $6$ such integers.

(A) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix,denoted by $\Delta$,must be non-zero $(\Delta \neq 0)$.
The coefficient matrix is:
$A = \begin{bmatrix} 1 & 1 & 1 \\ 5 & -1 & \alpha \\ 2 & 3 & -1 \end{bmatrix}$
Calculating the determinant $\Delta$:
$\Delta = 1((-1)(-1) - (3)(\alpha)) - 1((5)(-1) - (2)(\alpha)) + 1((5)(3) - (2)(-1))$
$\Delta = 1(1 - 3\alpha) - 1(-5 - 2\alpha) + 1(15 + 2)$
$\Delta = 1 - 3\alpha + 5 + 2\alpha + 17$
$\Delta = 23 - \alpha$
For a unique solution,$\Delta \neq 0$,which implies $23 - \alpha \neq 0$,or $\alpha \neq 23$.
Since the condition depends only on $\alpha$ and is independent of $\beta$,the existence of a unique solution depends on $\alpha$ only.

(D) The given equation is $x^4 + x^3 = 2(x^2 + 3ax + 2a^2)$.
Rearranging the terms,we get $x^4 + x^3 - 2x^2 - 6ax - 4a^2 = 0$.
This can be factored as $(x^2 + 2x + 2a)(x^2 - x - 2a) = 0$.
For the equation to have four real solutions,both quadratic factors must have real roots.
For $x^2 + 2x + 2a = 0$,the discriminant $D_1 = 2^2 - 4(1)(2a) = 4 - 8a \geq 0$,which implies $a \leq \frac{1}{2}$.
For $x^2 - x - 2a = 0$,the discriminant $D_2 = (-1)^2 - 4(1)(-2a) = 1 + 8a \geq 0$,which implies $a \geq -\frac{1}{8}$.
Combining these conditions,the set of values for $a$ is $[ - \frac{1}{8}, \frac{1}{2} ]$.

(C) The equation is $x_1 + x_2 = 100$,where $x_1, x_2 \in \mathbb{N}$.
Total number of natural solutions is given by the stars and bars formula $\binom{n-1}{k-1} = \binom{100-1}{2-1} = \binom{99}{1} = 99$.
We need to exclude cases where $x_1$ or $x_2$ are multiples of $5$.
If $x_1$ is a multiple of $5$,then $x_1 \in \{5, 10, 15, \dots, 95\}$. There are $19$ such values.
If $x_1 = 5k$,then $5k + x_2 = 100 \implies x_2 = 100 - 5k = 5(20-k)$.
Since $x_2$ must be a natural number,$20-k \ge 1 \implies k \le 19$. Thus,there are $19$ solutions where $x_1$ is a multiple of $5$.
Similarly,there are $19$ solutions where $x_2$ is a multiple of $5$.
Note that if both $x_1$ and $x_2$ are multiples of $5$,then $x_1 = 5k$ and $x_2 = 5m$,so $5k + 5m = 100 \implies k + m = 20$.
The number of natural solutions for $k+m=20$ is $\binom{20-1}{2-1} = 19$.
By the Principle of Inclusion-Exclusion,the number of solutions where at least one is a multiple of $5$ is $19 + 19 - 19 = 19$.
Therefore,the number of solutions where neither is a multiple of $5$ is $99 - 19 = 80$.

(B) Let $2^x = t$. Since $x > 0$,we have $t > 1$. The equation becomes $t^2 - (k + 2)t + 2k = 0$.
Factoring the quadratic: $(t - k)(t - 2) = 0$.
So,the roots are $t_1 = k$ and $t_2 = 2$.
For the original equation to have exactly one positive root,we analyze the roots of $t$:
$1$. One root $t_2 = 2$ is already greater than $1$,which corresponds to $x = \log_2(2) = 1$,which is a positive root.
$2$. For the equation to have exactly one positive root,the other root $t_1 = k$ must not yield a positive $x$. Since $t = 2^x$,$t > 0$ is required. If $t_1 = k \le 1$,then $x = \log_2(k) \le 0$.
$3$. If $k = 2$,the roots are $t = 2, 2$. Then $x = 1$ (only one distinct positive root). Thus $k=2$ is a solution.
$4$. If $k \le 1$,the roots are $t_1 \le 1$ and $t_2 = 2$. $t_1 \le 1$ gives $x \le 0$,and $t_2 = 2$ gives $x = 1$. This satisfies the condition.
Combining these,the set of values is $(-\infty, 1] \cup \{2\}$.

(A) The given equation is $xyz = 2^5 \times 3^2 \times 5^2$.
We need to find the number of natural number solutions $(x, y, z)$.
Let $x = 2^{a_1} 3^{b_1} 5^{c_1}$,$y = 2^{a_2} 3^{b_2} 5^{c_2}$,and $z = 2^{a_3} 3^{b_3} 5^{c_3}$,where $a_i, b_i, c_i \ge 0$.
Since $x, y, z$ must be natural numbers,we require $x, y, z \ge 1$.
For the exponents of prime factors,we have the following equations:
$a_1 + a_2 + a_3 = 5$ (where $a_i \ge 0$)
$b_1 + b_2 + b_3 = 2$ (where $b_i \ge 0$)
$c_1 + c_2 + c_3 = 2$ (where $c_i \ge 0$)
Using the stars and bars formula,the number of non-negative integer solutions for $x_1 + x_2 + \dots + x_k = n$ is $\binom{n+k-1}{k-1}$.
For $a_i$: $\binom{5+3-1}{3-1} = \binom{7}{2} = 21$.
For $b_i$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.
For $c_i$: $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.
The total number of solutions is $21 \times 6 \times 6 = 756$.

(A) Let $y = \sec^{-1}x$. The inequality becomes $(y - 4)(y - 1)(y - 2) \ge 0$.
Since the domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,the range is $[0, \pi/2) \cup (\pi/2, \pi]$.
However,note that $y = \sec^{-1}x$ implies $y \in [0, \pi]$ and $y \neq \pi/2$.
Since $y \le \pi \approx 3.14$,the term $(y - 4)$ is always negative because $y < 4$.
Dividing the inequality by $(y - 4)$,the sign of the inequality reverses: $(y - 1)(y - 2) \le 0$.
This holds for $y \in [1, 2]$.
Since $y = \sec^{-1}x$,we have $1 \le \sec^{-1}x \le 2$.
Taking the secant of all parts,we get $\sec 1 \le x \le \sec 2$.
Thus,the solution set is $[\sec 1, \sec 2]$.

(A) Let $f(x) = |x + |2x - 1|| - |x - |2x - 1||$.
We know that $|a+b| - |a-b| = 2 \text{sgn}(b) \min(|a|, |b|)$ is not directly applicable,but we can analyze the function $f(x)$ by cases:
Case $1$: $x \ge 1$. Then $|2x-1| = 2x-1$. So $f(x) = |x + 2x - 1| - |x - (2x - 1)| = |3x - 1| - |-x + 1| = (3x - 1) - (x - 1) = 2x$.
Case $2$: $1/2 \le x < 1$. Then $|2x-1| = 2x-1$. So $f(x) = |x + 2x - 1| - |x - (2x - 1)| = |3x - 1| - |-x + 1| = (3x - 1) - (-x + 1) = 4x - 2$.
Case $3$: $0 \le x < 1/2$. Then $|2x-1| = 1 - 2x$. So $f(x) = |x + 1 - 2x| - |x - (1 - 2x)| = |1 - x| - |3x - 1| = (1 - x) - (1 - 3x) = 2x$.
Case $4$: $x < 0$. Then $|2x-1| = 1 - 2x$. So $f(x) = |x + 1 - 2x| - |x - (1 - 2x)| = |1 - x| - |3x - 1| = (1 - x) - (1 - 3x) = 2x$.
Thus,$f(x) = 2x$ for $x < 1/2$,$f(x) = 4x - 2$ for $1/2 \le x < 1$,and $f(x) = 2x$ for $x \ge 1$.
The graph of $f(x)$ increases from $-\infty$ to $1$ at $x=1/2$,then increases from $0$ to $2$ as $x$ goes from $1/2$ to $1$,then continues to increase for $x > 1$.
For $K$ to have exactly three solutions,$K$ must be such that the horizontal line $y=K$ intersects the graph at three points. Looking at the behavior,there is no such $K$ that yields exactly three solutions. Thus,the number of such positive integral values is $0$.

(D) Given equations are:
$xy = \frac{1}{9}$
$x(y + 1) = \frac{7}{9}$
$y(x + 1) = \frac{5}{18}$
We need to find the value of $(x + 1)(y + 1)$.
Expanding the expression,we get:
$(x + 1)(y + 1) = xy + x + y + 1$
From the given equations:
$x(y + 1) = xy + x = \frac{7}{9}$
$y(x + 1) = yx + y = \frac{5}{18}$
Adding these two equations:
$(xy + x) + (yx + y) = \frac{7}{9} + \frac{5}{18}$
$2xy + x + y = \frac{14}{18} + \frac{5}{18} = \frac{19}{18}$
Since $xy = \frac{1}{9}$,we have $2(\frac{1}{9}) + x + y = \frac{19}{18}$
$\frac{2}{9} + x + y = \frac{19}{18}$
$x + y = \frac{19}{18} - \frac{4}{18} = \frac{15}{18} = \frac{5}{6}$
Now,substitute the values into $(x + 1)(y + 1) = xy + (x + y) + 1$:
$(x + 1)(y + 1) = \frac{1}{9} + \frac{5}{6} + 1$
$= \frac{2}{18} + \frac{15}{18} + \frac{18}{18} = \frac{35}{18}$

(B) Let $\alpha$ be the common root of the two equations.
Then,$\alpha^2 + p\alpha + 2q = 0$ and $\alpha^2 + q\alpha + 2p = 0$.
Subtracting the second equation from the first,we get:
$(\alpha^2 + p\alpha + 2q) - (\alpha^2 + q\alpha + 2p) = 0$
$\alpha(p - q) - 2(p - q) = 0$
$(p - q)(\alpha - 2) = 0$
Since it is given that $p \ne q$,we must have $\alpha - 2 = 0$,which implies $\alpha = 2$.
Substituting $\alpha = 2$ into the first equation:
$(2)^2 + p(2) + 2q = 0$
$4 + 2p + 2q = 0$
$2(p + q) = -4$
$p + q = -2$

(A) Given the quadratic equation $x^2 - \sqrt{2}x + 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm \sqrt{-2}}{2} = \frac{\sqrt{2} \pm i\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \pm i\frac{1}{\sqrt{2}}$.
These roots can be written in polar form as $\alpha = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = e^{i\pi/4}$ and $\beta = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) = e^{-i\pi/4}$.
Now,we need to find $\alpha^{50} + \beta^{50}$.
$\alpha^{50} = (e^{i\pi/4})^{50} = e^{i50\pi/4} = e^{i25\pi/2} = e^{i(12\pi + \pi/2)} = e^{i\pi/2} = i$.
$\beta^{50} = (e^{-i\pi/4})^{50} = e^{-i50\pi/4} = e^{-i25\pi/2} = e^{-i(12\pi + \pi/2)} = e^{-i\pi/2} = -i$.
Therefore,$\alpha^{50} + \beta^{50} = i + (-i) = 0$.

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Dividing by $a$,we get $\frac{2b}{a} = 1 + \frac{c}{a}$ ... $(i)$.
From the properties of roots,$\alpha + \beta = -\frac{b}{a} = 15$,which implies $\frac{b}{a} = -15$.
Substituting $\frac{b}{a} = -15$ into equation $(i)$:
$2(-15) = 1 + \frac{c}{a} \Rightarrow -30 = 1 + \frac{c}{a} \Rightarrow \frac{c}{a} = -31$.
Since $\alpha\beta = \frac{c}{a}$,we have $\alpha\beta = -31$.

(B) Let the roots of $x^2 + ax + b = 0$ be $\alpha$ and $\beta$. Then,$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = a^2 - 4b$.
Let the roots of $x^2 + bx + a = 0$ be $\alpha'$ and $\beta'$. Then,$(\alpha' - \beta')^2 = (\alpha' + \beta')^2 - 4\alpha'\beta' = b^2 - 4a$.
Given that the difference between the roots is equal,we have $(\alpha - \beta)^2 = (\alpha' - \beta')^2$.
Therefore,$a^2 - 4b = b^2 - 4a$.
Rearranging the terms,we get $a^2 - b^2 = 4b - 4a$.
$(a - b)(a + b) = -4(a - b)$.
Since $a \ne b$,we can divide both sides by $(a - b)$,which gives $a + b = -4$.

(B) Let $f(x) = x^2 - 2ax + a^2 - 6a$.
For $f(x) \leqslant 0$ to hold for all $x \in [1, 2]$,the values of the function at the endpoints must satisfy $f(1) \leqslant 0$ and $f(2) \leqslant 0$.
Step $1$: Solve $f(1) \leqslant 0$.
$f(1) = 1 - 2a + a^2 - 6a = a^2 - 8a + 1 \leqslant 0$.
The roots of $a^2 - 8a + 1 = 0$ are $a = \frac{8 \pm \sqrt{64 - 4}}{2} = 4 \pm \sqrt{15}$.
Thus,$a \in [4 - \sqrt{15}, 4 + \sqrt{15}]$ ... $(i)$.
Step $2$: Solve $f(2) \leqslant 0$.
$f(2) = 4 - 4a + a^2 - 6a = a^2 - 10a + 4 \leqslant 0$.
The roots of $a^2 - 10a + 4 = 0$ are $a = \frac{10 \pm \sqrt{100 - 16}}{2} = 5 \pm \sqrt{21}$.
Thus,$a \in [5 - \sqrt{21}, 5 + \sqrt{21}]$ ... $(ii)$.
Step $3$: Find the intersection of $(i)$ and $(ii)$.
Since $5 - \sqrt{21} \approx 5 - 4.58 = 0.42$ and $4 - \sqrt{15} \approx 4 - 3.87 = 0.13$,the lower bound is $5 - \sqrt{21}$.
Since $4 + \sqrt{15} \approx 4 + 3.87 = 7.87$ and $5 + \sqrt{21} \approx 5 + 4.58 = 9.58$,the upper bound is $4 + \sqrt{15}$.
Therefore,$a \in [5 - \sqrt{21}, 4 + \sqrt{15}]$.

(A) Given that $\alpha$ and $\beta$ are roots of the equation $x^2 + \alpha x + \beta = 0$.
By Vieta's formulas:
Sum of roots: $\alpha + \beta = -\alpha \implies 2\alpha + \beta = 0$ .......$(1)$
Product of roots: $\alpha \beta = \beta \implies \beta(\alpha - 1) = 0$ .......$(2)$
From $(2)$,either $\beta = 0$ or $\alpha = 1$.
Case $I$: If $\beta = 0$,then from $(1)$,$2\alpha = 0 \implies \alpha = 0$. But the problem states $\alpha \neq \beta$,so this is rejected.
Case $II$: If $\alpha = 1$,then from $(1)$,$2(1) + \beta = 0 \implies \beta = -2$. Here $\alpha \neq \beta$ holds.
Now substitute $\alpha = 1$ and $\beta = -2$ into the inequality $| |y - (-2)| - 1 | < 1$:
$| |y + 2| - 1 | < 1$
$-1 < |y + 2| - 1 < 1$
$0 < |y + 2| < 2$
This implies $|y + 2| < 2$ and $|y + 2| \neq 0$.
$-2 < y + 2 < 2$ and $y + 2 \neq 0$.
$-4 < y < 0$ and $y \neq -2$.
Thus,$y \in (-4, -2) \cup (-2, 0)$.
Checking option $D$: For $x^2 + x - 2 > 0$,the roots are $x = -2, 1$. The parabola opens upward,so $x^2 + x - 2 > 0$ for $x \in (-\infty, -2) \cup (1, \infty)$. This does not hold for $x \in [-1, 0]$.

(A) Let $f(x) = (p - 5)x^2 - 2px + (p - 4)$.
For the roots to be positive,one less than $2$ and the other between $2$ and $3$,the parabola must open upwards,so $p - 5 > 0 \Rightarrow p > 5$.
Given the conditions,we must have $f(0) > 0$,$f(2) < 0$,and $f(3) > 0$.
$1$) $f(0) = p - 4 > 0 \Rightarrow p > 4$.
$2$) $f(2) = (p - 5)(4) - 2p(2) + (p - 4) = 4p - 20 - 4p + p - 4 = p - 24 < 0 \Rightarrow p < 24$.
$3$) $f(3) = (p - 5)(9) - 2p(3) + (p - 4) = 9p - 45 - 6p + p - 4 = 4p - 49 > 0 \Rightarrow p > \frac{49}{4}$.
Combining $p > 5$,$p > 4$,$p < 24$,and $p > \frac{49}{4}$,we get the interval $\left( \frac{49}{4}, 24 \right)$.

(B) Let $S = a + b + c + d + e$. The given system of equations can be written as:
$S + a = 6$
$S + b = 12$
$S + c = 24$
$S + d = 48$
$S + e = 96$
Summing these five equations,we get:
$5S + (a + b + c + d + e) = 6 + 12 + 24 + 48 + 96$
$5S + S = 186$
$6S = 186$
$S = 31$
Now,substitute $S = 31$ into the third equation $S + c = 24$:
$31 + c = 24$
$c = 24 - 31 = -7$
Therefore,$|c| = |-7| = 7$.

(C) Given the equation $x^3 + 8x + 1 = 0$,since $a, b, c$ are roots,we have $x^3 = -(8x + 1)$.
Thus,$8x + 1 = -x^3$.
Substituting this into the expression:
$\frac{bc}{(-b^3)(-c^3)} + \frac{ac}{(-a^3)(-c^3)} + \frac{ab}{(-a^3)(-b^3)}$
$= \frac{bc}{b^3c^3} + \frac{ac}{a^3c^3} + \frac{ab}{a^3b^3}$
$= \frac{1}{b^2c^2} + \frac{1}{a^2c^2} + \frac{1}{a^2b^2}$
$= \frac{a^2 + b^2 + c^2}{a^2b^2c^2}$
Using Vieta's formulas for $x^3 + 0x^2 + 8x + 1 = 0$:
Sum of roots $\Sigma a = a + b + c = 0$.
Sum of roots taken two at a time $\Sigma ab = ab + bc + ca = 8$.
Product of roots $abc = -1$.
Now,$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 0^2 - 2(8) = -16$.
Also,$(abc)^2 = (-1)^2 = 1$.
Therefore,the value is $\frac{-16}{1} = -16$.

(A) The given quadratic equation is $x^2 - 2ax + (a^2 - 4) = 0$.
We can rewrite this as $x^2 - 2ax + (a-2)(a+2) = 0$.
Factoring the quadratic equation,we get $(x - (a-2))(x - (a+2)) = 0$.
Thus,the roots are $x_1 = a-2$ and $x_2 = a+2$.
Since $a-2 < a+2$,the smaller root is $a-2$ and the larger root is $a+2$.
We are given the conditions: $a-2 < 1$ and $a+2 > 6$.
From $a-2 < 1$,we get $a < 3$.
From $a+2 > 6$,we get $a > 4$.
Since there is no value of $a$ that satisfies both $a < 3$ and $a > 4$ simultaneously,the number of integral values of $a$ is $0$.

(C) For a cubic polynomial $y = ax^3 + bx^2 + cx + d$ to be symmetric about a vertical line $x = k$,the cubic term must vanish,meaning $a = 0$.
If $a = 0$,the equation becomes $y = bx^2 + cx + d$,which is a parabola.
$A$ parabola $y = bx^2 + cx + d$ is symmetric about the vertical line $x = -\frac{c}{2b}$.
Given that the graph is symmetric about $x = k$,we have $k = -\frac{c}{2b}$.
Rearranging this gives $k + \frac{c}{2b} = 0$.
Since $a = 0$,we can write $a + \frac{c}{2b} + k = 0$.

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$.
We can write the roots in polar form: $\alpha = 1 + i\sqrt{3} = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$ and $\beta = 1 - i\sqrt{3} = 2 \left( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} \right)$.
Using De Moivre's Theorem,$\alpha^n = 2^n \left( \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3} \right)$ and $\beta^n = 2^n \left( \cos \frac{n\pi}{3} - i \sin \frac{n\pi}{3} \right)$.
Adding these,$\alpha^n + \beta^n = 2^n \left( \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3} + \cos \frac{n\pi}{3} - i \sin \frac{n\pi}{3} \right)$.
$\alpha^n + \beta^n = 2^n \left( 2 \cos \frac{n\pi}{3} \right) = 2^{n+1} \cos \frac{n\pi}{3}$.

(C) Let $u = x - 1$. Since $x - 1$ is under the square root,we must have $u \geq 0$.
The equation becomes $\sqrt{u + 4 - 4\sqrt{u}} + \sqrt{u + 9 - 6\sqrt{u}} = 1$.
This simplifies to $\sqrt{(\sqrt{u} - 2)^2} + \sqrt{(\sqrt{u} - 3)^2} = 1$,which is $|\sqrt{u} - 2| + |\sqrt{u} - 3| = 1$.
Let $y = \sqrt{u}$. Then $|y - 2| + |y - 3| = 1$.
We know that $|y - a| + |y - b| = |b - a|$ if and only if $y$ lies between $a$ and $b$ (inclusive).
Here,$|y - 2| + |y - 3| = |3 - 2| = 1$.
Therefore,$2 \leq y \leq 3$.
Substituting back $y = \sqrt{u}$,we get $2 \leq \sqrt{u} \leq 3$.
Squaring the inequality,$4 \leq u \leq 9$.
Since $u = x - 1$,we have $4 \leq x - 1 \leq 9$,which gives $5 \leq x \leq 10$.
Thus,the solution is $x \in [5, 10]$.

(B) For the quadratic equation $x^2 - \alpha x + \alpha + 1 = 0$ to have integral roots,its discriminant $D$ must be a perfect square.
$D = \alpha^2 - 4(1)(\alpha + 1) = \alpha^2 - 4\alpha - 4$.
Let $D = k^2$ for some non-negative integer $k$.
$\alpha^2 - 4\alpha - 4 = k^2$
$(\alpha^2 - 4\alpha + 4) - 8 = k^2$
$(\alpha - 2)^2 - k^2 = 8$
$(\alpha - 2 - k)(\alpha - 2 + k) = 8$.
Since $8$ is the product of two integers,we consider the factors of $8$: $(1, 8), (2, 4), (-1, -8), (-2, -4)$.
Case $1$: $\alpha - 2 - k = 2$ and $\alpha - 2 + k = 4$. Adding these gives $2(\alpha - 2) = 6 \Rightarrow \alpha - 2 = 3 \Rightarrow \alpha = 5$.
Case $2$: $\alpha - 2 - k = -4$ and $\alpha - 2 + k = -2$. Adding these gives $2(\alpha - 2) = -6 \Rightarrow \alpha - 2 = -3 \Rightarrow \alpha = -1$.
Case $3$: $\alpha - 2 - k = 4$ and $\alpha - 2 + k = 2$. Adding these gives $2(\alpha - 2) = 6 \Rightarrow \alpha = 5$.
Case $4$: $\alpha - 2 - k = -2$ and $\alpha - 2 + k = -4$. Adding these gives $2(\alpha - 2) = -6 \Rightarrow \alpha = -1$.
Thus,the distinct integral values of $\alpha$ are $5$ and $-1$.
The sum of these values is $5 + (-1) = 4$.

(B) Let $P(x) = x^3 - 2x^2 + 4x + 5074$. Since $r_1, r_2, r_3$ are the roots of $P(x) = 0$,we can write the polynomial as:
$P(x) = (x - r_1)(x - r_2)(x - r_3)$
We want to find the value of $(r_1 + 2)(r_2 + 2)(r_3 + 2)$.
Notice that $(r_1 + 2)(r_2 + 2)(r_3 + 2) = -(-r_1 - 2)(-r_2 - 2)(-r_3 - 2) = -((-2 - r_1)(-2 - r_2)(-2 - r_3))$.
Alternatively,consider $P(-2)$:
$P(-2) = (-2 - r_1)(-2 - r_2)(-2 - r_3) = (-1)^3 (2 + r_1)(2 + r_2)(2 + r_3) = -(r_1 + 2)(r_2 + 2)(r_3 + 2)$.
Now,calculate $P(-2)$ using the given equation:
$P(-2) = (-2)^3 - 2(-2)^2 + 4(-2) + 5074$
$P(-2) = -8 - 2(4) - 8 + 5074$
$P(-2) = -8 - 8 - 8 + 5074 = -24 + 5074 = 5050$.
Since $P(-2) = -(r_1 + 2)(r_2 + 2)(r_3 + 2)$,we have:
$5050 = -(r_1 + 2)(r_2 + 2)(r_3 + 2)$
Therefore,$(r_1 + 2)(r_2 + 2)(r_3 + 2) = -5050$.

(A) The given quadratic equation is $(x - a)(x - 10) + 1 = 0$.
Expanding this,we get $x^{2} - (10 + a)x + 10a + 1 = 0$.
For the roots to be integers,the discriminant $D$ must be a perfect square.
$D = b^{2} - 4ac = (10 + a)^{2} - 4(10a + 1)$.
$D = 100 + 20a + a^{2} - 40a - 4$.
$D = a^{2} - 20a + 96$.
Completing the square,$D = (a - 10)^{2} - 4$.
Let $D = k^{2}$ for some integer $k \ge 0$.
Then $k^{2} = (a - 10)^{2} - 4$,which implies $(a - 10)^{2} - k^{2} = 4$.
$(a - 10 - k)(a - 10 + k) = 4$.
Since the difference between these two factors is $2k$ (an even number),both factors must be even.
The possible pairs of factors of $4$ are $(2, 2)$ or $(-2, -2)$.
Case $1$: $a - 10 - k = 2$ and $a - 10 + k = 2$.
Adding these,$2(a - 10) = 4 \Rightarrow a - 10 = 2 \Rightarrow a = 12$.
Case $2$: $a - 10 - k = -2$ and $a - 10 + k = -2$.
Adding these,$2(a - 10) = -4 \Rightarrow a - 10 = -2 \Rightarrow a = 8$.
Thus,the integral values of $a$ are $8$ and $12$.