QUADRATIC EQUATION Questions in English

(B) Given the quadratic equation $ax^2 + bx + c = 0$ with $a > 0, b < 0, c < 0$.
Sum of roots ($S$.$O$.$R$.) $= \alpha + \beta = -\frac{b}{a}$. Since $b < 0$ and $a > 0$,$-\frac{b}{a} > 0$,so $\alpha + \beta > 0$.
Product of roots ($P$.$O$.$R$.) $= \alpha \beta = \frac{c}{a}$. Since $c < 0$ and $a > 0$,$\frac{c}{a} < 0$,so $\alpha \beta < 0$.
Since the product of the roots is negative,one root must be positive and the other negative. Let $\alpha < 0$ and $\beta > 0$.
Given $\alpha + \beta > 0$,the positive root must have a larger absolute value than the negative root. Thus,$|\beta| > |\alpha|$.
Since $\beta > 0$,we have $\beta > |\alpha|$.
Therefore,$0 < |\alpha| < \beta$.

(D) Given the polynomial equation $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$ with roots $\alpha, \beta, \gamma, \delta$.
To find the sum of the reciprocals $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$,we can substitute $x = \frac{1}{y}$ into the original equation.
Substituting $x = \frac{1}{y}$ gives: $(\frac{1}{y})^4 - 100(\frac{1}{y})^3 + 2(\frac{1}{y})^2 + 4(\frac{1}{y}) + 10 = 0$.
Multiplying the entire equation by $y^4$,we get: $1 - 100y + 2y^2 + 4y^3 + 10y^4 = 0$,which rearranges to $10y^4 + 4y^3 + 2y^2 - 100y + 1 = 0$.
The roots of this new equation are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$.
The sum of the roots of a polynomial $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$ is given by $-\frac{a_{n-1}}{a_n}$.
Applying this to our new equation,the sum is $-\frac{4}{10} = -\frac{2}{5}$.

(D) Let $f(x) = (x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6)$.
We evaluate $f(x)$ at the given points:
$f(\alpha_1) = 0 + 3(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_4)(\alpha_1 - \alpha_6)$. Since $\alpha_1 < \alpha_2, \alpha_4, \alpha_6$,all three terms are negative. Thus,$f(\alpha_1) = 3(-)(-)(-) < 0$.
$f(\alpha_2) = (\alpha_2 - \alpha_1)(\alpha_2 - \alpha_3)(\alpha_2 - \alpha_5) + 0 = (+)(-)(-) = + > 0$.
Since $f(\alpha_1) < 0$ and $f(\alpha_2) > 0$,there is at least one real root in $(\alpha_1, \alpha_2)$.
$f(\alpha_3) = 0 + 3(\alpha_3 - \alpha_2)(\alpha_3 - \alpha_4)(\alpha_3 - \alpha_6) = 3(+)(-)(-) = + > 0$.
$f(\alpha_4) = (\alpha_4 - \alpha_1)(\alpha_4 - \alpha_3)(\alpha_4 - \alpha_5) + 0 = (+)(+)(-) = - < 0$.
Since $f(\alpha_3) > 0$ and $f(\alpha_4) < 0$,there is at least one real root in $(\alpha_3, \alpha_4)$.
$f(\alpha_5) = 0 + 3(\alpha_5 - \alpha_2)(\alpha_5 - \alpha_4)(\alpha_5 - \alpha_6) = 3(+)(+)(-) = - < 0$.
$f(\alpha_6) = (\alpha_6 - \alpha_1)(\alpha_6 - \alpha_3)(\alpha_6 - \alpha_5) + 0 = (+)(+)(+) = + > 0$.
Since $f(\alpha_5) < 0$ and $f(\alpha_6) > 0$,there is at least one real root in $(\alpha_5, \alpha_6)$.
As $x \to -\infty$,$f(x) \to -\infty$. Since $f(\alpha_1) < 0$,we check the interval $(-\infty, \alpha_1)$. The function is a cubic polynomial with a positive leading coefficient. It crosses the x-axis at least once in each of the intervals $(\alpha_1, \alpha_2)$,$(\alpha_3, \alpha_4)$,and $(\alpha_5, \alpha_6)$. Since it is a cubic,it has exactly $3$ real roots. Thus,there are no roots in $(-\infty, \alpha_1)$.

(B) Let $f(x) = ax^2 + 2bx + c$.
Given that $f(x) = 0$ has imaginary roots,the parabola $y = f(x)$ does not intersect the $x$-axis.
This means $f(x)$ is either always positive $(a > 0)$ or always negative $(a < 0)$.
We are given $4a + 4b + c < 0$.
Note that $f(2) = a(2)^2 + 2b(2) + c = 4a + 4b + c$.
Since $f(2) < 0$ and the function is always negative (because it has imaginary roots and at least one point $f(2)$ is negative),the entire parabola must lie below the $x$-axis.
Therefore,the $y$-intercept $f(0) = c$ must also be negative.
Thus,$c < 0$.

(A) For a quadratic equation $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant $D = b^2 - 4ac$.
Here, $a = 1$, $b = -\sqrt{13}$, and $c = 1$.
Calculating the discriminant: $D = (-\sqrt{13})^2 - 4(1)(1) = 13 - 4 = 9$.
Since $D > 0$ and $D$ is a perfect square, the roots are real, distinct, and rational if the coefficients are rational. However, since the coefficient $b = -\sqrt{13}$ is irrational, the roots are real and distinct.

(A) The given quadratic equation is $x^2 + x - n = 0$.
For the roots to be integers,the discriminant $D = b^2 - 4ac$ must be a perfect square of an odd integer.
$D = 1^2 - 4(1)(-n) = 1 + 4n$.
Let $1 + 4n = k^2$,where $k$ is an odd integer. Let $k = 2\lambda + 1$ for some integer $\lambda$.
Then $1 + 4n = (2\lambda + 1)^2 = 4\lambda^2 + 4\lambda + 1$.
$4n = 4\lambda^2 + 4\lambda \implies n = \lambda(\lambda + 1)$.
Given $n \in [5, 100]$,we have $5 \le \lambda(\lambda + 1) \le 100$.
For $\lambda = 2, n = 2(3) = 6$.
For $\lambda = 3, n = 3(4) = 12$.
For $\lambda = 4, n = 4(5) = 20$.
For $\lambda = 5, n = 5(6) = 30$.
For $\lambda = 6, n = 6(7) = 42$.
For $\lambda = 7, n = 7(8) = 56$.
For $\lambda = 8, n = 8(9) = 72$.
For $\lambda = 9, n = 9(10) = 90$.
For $\lambda = 10, n = 10(11) = 110$ (which is $> 100$).
Thus,there are $8$ possible values for $n$.

(A) Let $f(x) = x^2 + kx - 4$.
For the smaller root to lie in the interval $(-1, 2)$,the function must satisfy the condition that $f(-1) > 0$ and $f(2) < 0$.
$1$. $f(-1) = (-1)^2 + k(-1) - 4 = 1 - k - 4 = -k - 3$.
Since $f(-1) > 0$,we have $-k - 3 > 0$,which implies $k < -3$.
$2$. $f(2) = (2)^2 + k(2) - 4 = 4 + 2k - 4 = 2k$.
Since $f(2) < 0$,we have $2k < 0$,which implies $k < 0$.
Combining both conditions $k < -3$ and $k < 0$,we get $k < -3$.
Thus,the interval for $k$ is $(-\infty, -3)$.

(A) For the first equation $x^2 + 2bx + b = 0$,the roots are $r_1 = \cos^4 \theta + \alpha$ and $r_2 = \sin^4 \theta + \alpha$.
The difference of the roots is $|r_1 - r_2| = |(\cos^4 \theta + \alpha) - (\sin^4 \theta + \alpha)| = |\cos^4 \theta - \sin^4 \theta| = |(\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta)| = |\cos 2\theta|$.
From the quadratic formula,the difference of roots is $\sqrt{D}/a = \sqrt{(2b)^2 - 4(1)(b)} = \sqrt{4b^2 - 4b} = 2\sqrt{b^2 - b}$.
Thus,$|\cos 2\theta| = 2\sqrt{b^2 - b}$.
For the second equation $x^2 + 4x + 2 = 0$,the roots are $s_1 = \cos^2 \theta + \beta$ and $s_2 = \sin^2 \theta + \beta$.
The difference of the roots is $|s_1 - s_2| = |(\cos^2 \theta + \beta) - (\sin^2 \theta + \beta)| = |\cos^2 \theta - \sin^2 \theta| = |\cos 2\theta|$.
From the quadratic formula,the difference of roots is $\sqrt{D}/a = \sqrt{4^2 - 4(1)(2)} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.
Equating the two expressions for $|\cos 2\theta|$:
$2\sqrt{b^2 - b} = 2\sqrt{2}$
$\sqrt{b^2 - b} = \sqrt{2}$
$b^2 - b = 2$
$b^2 - b - 2 = 0$
$(b - 2)(b + 1) = 0$
So,$b = 2$ or $b = -1$. Since $b=2$ is an option,the correct answer is $2$.

(D) Let $f(x) = -x^2 + 5x - 10$ and $g(x) = (\frac{3}{2})^x$.
For the quadratic function $f(x) = -x^2 + 5x - 10$,the discriminant $D = b^2 - 4ac = 5^2 - 4(-1)(-10) = 25 - 40 = -15$.
The maximum value of $f(x)$ is given by $\frac{-D}{4a} = \frac{-(-15)}{4(-1)} = -\frac{15}{4} = -3.75$.
Since the coefficient of $x^2$ is negative,the parabola opens downwards,meaning $f(x) \leq -3.75$ for all real $x$.
For the exponential function $g(x) = (\frac{3}{2})^x$,we know that $g(x) > 0$ for all real $x$.
Since $f(x)$ is always negative $(f(x) \leq -3.75)$ and $g(x)$ is always positive $(g(x) > 0)$,there is no value of $x$ for which $f(x) = g(x)$.
Therefore,the equation has no real solutions.

(C) The roots of $x^2 - 2x - a^2 + 1 = 0$ are given by $(x - 1)^2 = a^2$,so $x = 1 \pm a$. Let the roots be $x_1 = 1 - a$ and $x_2 = 1 + a$.
Let $f(x) = x^2 - 2(a + 1)x + a(a - 1)$. For the roots of the first equation to lie between the roots of $f(x) = 0$,we must have $f(1 - a) < 0$ and $f(1 + a) < 0$.
First,$f(1 + a) = (1 + a)^2 - 2(a + 1)(1 + a) + a(a - 1) = (1 + a)^2 - 2(1 + a)^2 + a^2 - a = -(1 + a)^2 + a^2 - a = -(1 + 2a + a^2) + a^2 - a = -3a - 1$.
Setting $f(1 + a) < 0 \Rightarrow -3a - 1 < 0 \Rightarrow a > -1/3$.
Next,$f(1 - a) = (1 - a)^2 - 2(a + 1)(1 - a) + a(a - 1) = (a - 1)^2 + 2(a + 1)(a - 1) + a(a - 1) = (a - 1)[(a - 1) + 2(a + 1) + a] = (a - 1)(a - 1 + 2a + 2 + a) = (a - 1)(4a + 1)$.
Setting $f(1 - a) < 0 \Rightarrow (a - 1)(4a + 1) < 0 \Rightarrow a \in (-1/4, 1)$.
Taking the intersection of $a > -1/3$ and $a \in (-1/4, 1)$,we get $a \in (-1/4, 1)$.

(D) Given the quadratic equation $5x^2 - 3x - 1 = 0$,the sum of roots is $\alpha + \beta = \frac{3}{5}$ and the product of roots is $\alpha\beta = -\frac{1}{5}$.
The given series is $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\alpha^n + \beta^n}{n} x^n$.
This can be split into two logarithmic series: $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(\alpha x)^n}{n} + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(\beta x)^n}{n}$.
Using the expansion $\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$,we get $S = \ln(1 + \alpha x) + \ln(1 + \beta x)$.
$S = \ln((1 + \alpha x)(1 + \beta x)) = \ln(1 + x(\alpha + \beta) + \alpha\beta x^2)$.
Substituting the values: $S = \ln(1 + x(\frac{3}{5}) - \frac{1}{5}x^2) = \ln\left(\frac{5 + 3x - x^2}{5}\right)$.

(B) Let $\alpha$ be the common root of the equations $ax^2 + bx + c = 0$ and $bx^2 + cx + a = 0$.
Then,$a\alpha^2 + b\alpha + c = 0$ and $b\alpha^2 + c\alpha + a = 0$.
Using the method of cross-multiplication for these two equations:
$\frac{\alpha^2}{b(a) - c(c)} = \frac{\alpha}{c(b) - a(a)} = \frac{1}{a(c) - b(b)}$
$\frac{\alpha^2}{ab - c^2} = \frac{\alpha}{bc - a^2} = \frac{1}{ac - b^2}$
From the first and third parts: $\alpha^2 = \frac{ab - c^2}{ac - b^2}$
From the second and third parts: $\alpha = \frac{bc - a^2}{ac - b^2}$
Since $\alpha^2 = (\alpha)^2$,we have:
$\frac{ab - c^2}{ac - b^2} = \left( \frac{bc - a^2}{ac - b^2} \right)^2$
$(ab - c^2)(ac - b^2) = (bc - a^2)^2$
$a^2bc - ab^3 - ac^3 + b^2c^2 = b^2c^2 - 2a^2bc + a^4$
$a^4 + ab^3 + ac^3 = 3a^2bc$
Dividing both sides by $abc$ (since $a, b, c \neq 0$):
$\frac{a^3}{bc} + \frac{b^2}{c} + \frac{c^2}{b} = 3$ (This is not the target expression).
Actually,dividing $a^4 + ab^3 + ac^3 = 3a^2bc$ by $abc$ gives $\frac{a^3}{bc} + \frac{b^2}{c} + \frac{c^2}{b} = 3$. Wait,let's re-evaluate: $a(a^3 + b^3 + c^3) = 3a^2bc \Rightarrow \frac{a^3 + b^3 + c^3}{abc} = 3$.

(B) Let the roots of the equation $x^2 + (2 - \lambda) x + (10 - \lambda) = 0$ be $\alpha$ and $\beta$.
From the properties of roots,$\alpha + \beta = - (2 - \lambda) = \lambda - 2$ and $\alpha \beta = 10 - \lambda$.
The sum of the cubes of the roots is $S = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.
Substituting the values,$S = (\lambda - 2)^3 - 3(10 - \lambda)(\lambda - 2) = (\lambda - 2) [(\lambda - 2)^2 - 3(10 - \lambda)]$.
$S = (\lambda - 2) [\lambda^2 - 4\lambda + 4 - 30 + 3\lambda] = (\lambda - 2)(\lambda^2 - \lambda - 26) = \lambda^3 - 3\lambda^2 - 24\lambda + 52$.
To find the minimum,we differentiate $S$ with respect to $\lambda$: $\frac{dS}{d\lambda} = 3\lambda^2 - 6\lambda - 24 = 0$.
$3(\lambda^2 - 2\lambda - 8) = 0 \implies 3(\lambda - 4)(\lambda + 2) = 0$. Thus,$\lambda = 4$ or $\lambda = -2$.
Checking the second derivative: $\frac{d^2S}{d\lambda^2} = 6\lambda - 6$. For $\lambda = 4$,$6(4) - 6 = 18 > 0$ (minimum).
The difference of the roots is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta} = \sqrt{(\lambda - 2)^2 - 4(10 - \lambda)} = \sqrt{\lambda^2 - 4\lambda + 4 - 40 + 4\lambda} = \sqrt{\lambda^2 - 36}$.
For $\lambda = 4$,$|\alpha - \beta| = \sqrt{16 - 36} = \sqrt{-20} = i\sqrt{20} = 2i\sqrt{5}$. The magnitude is $|2i\sqrt{5}| = 2\sqrt{5}$.

(B) Given the equation: $\frac{1}{x+p} + \frac{1}{x+q} = \frac{1}{r}$
Simplifying the left side: $\frac{x+q+x+p}{(x+p)(x+q)} = \frac{1}{r}$
$\frac{2x+p+q}{x^2+(p+q)x+pq} = \frac{1}{r}$
Cross-multiplying: $r(2x+p+q) = x^2+(p+q)x+pq$
Rearranging into standard quadratic form $ax^2+bx+c=0$:
$x^2 + (p+q-2r)x + (pq-pr-qr) = 0$
Let the roots be $\alpha$ and $\beta$. Since the roots are equal in magnitude but opposite in sign,$\alpha = -\beta$,which implies $\alpha + \beta = 0$.
From the sum of roots formula,$\alpha + \beta = -(p+q-2r)$.
Setting this to zero: $p+q-2r = 0$,so $p+q = 2r$.
We need to find the sum of squares of the roots,$\alpha^2 + \beta^2$.
Since $\alpha = -\beta$,$\alpha^2 + \beta^2 = \alpha^2 + (-\alpha)^2 = 2\alpha^2$.
Also,$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
Since $\alpha+\beta = 0$,the sum of squares is $-2\alpha\beta$.
From the product of roots formula,$\alpha\beta = pq-pr-qr$.
Thus,$\alpha^2 + \beta^2 = -2(pq - pr - qr) = -2pq + 2pr + 2qr$.
Since $2r = p+q$,substitute $2pr+2qr = 2r(p+q) = (p+q)(p+q) = (p+q)^2$.
So,$\alpha^2 + \beta^2 = -2pq + (p+q)^2 = -2pq + p^2 + q^2 + 2pq = p^2 + q^2$.

(C) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$.
Given $p(0) = 1$,substituting $x = 0$ gives $c = 1$.
According to the Remainder Theorem,if $p(x)$ is divided by $(x - 1)$,the remainder is $p(1) = 4$.
Substituting $x = 1$ into $p(x) = ax^2 + bx + 1$,we get $a + b + 1 = 4$,which simplifies to $a + b = 3$.
Similarly,if $p(x)$ is divided by $(x + 1)$,the remainder is $p(-1) = 6$.
Substituting $x = -1$ into $p(x) = ax^2 + bx + 1$,we get $a - b + 1 = 6$,which simplifies to $a - b = 5$.
Adding the two equations $(a + b = 3)$ and $(a - b = 5)$,we get $2a = 8$,so $a = 4$.
Substituting $a = 4$ into $a + b = 3$,we get $4 + b = 3$,so $b = -1$.
Thus,the polynomial is $p(x) = 4x^2 - x + 1$.
To find $p(-2)$,substitute $x = -2$: $p(-2) = 4(-2)^2 - (-2) + 1 = 4(4) + 2 + 1 = 16 + 2 + 1 = 19$.

(C) Given the equation $2^{(x - 1)(x^2 + 5x - 50)} = 1$.
Since $2^0 = 1$,the exponent must be equal to $0$.
Therefore,$(x - 1)(x^2 + 5x - 50) = 0$.
Factoring the quadratic expression $x^2 + 5x - 50$,we get $(x + 10)(x - 5) = 0$.
So,the equation becomes $(x - 1)(x + 10)(x - 5) = 0$.
The real values of $x$ are $x = 1, x = -10, x = 5$.
The sum of these values is $1 + (-10) + 5 = -4$.

(C) Let $\alpha$ be the common root of the equations $x^2 + bx - 1 = 0$ and $x^2 + x + b = 0$.
Since $\alpha$ is a root,we have:
$\alpha^2 + b\alpha - 1 = 0$ --- $(1)$
$\alpha^2 + \alpha + b = 0$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(\alpha^2 + b\alpha - 1) - (\alpha^2 + \alpha + b) = 0$
$\alpha(b - 1) - (1 + b) = 0$
$\alpha(b - 1) = b + 1$
$\alpha = \frac{b + 1}{b - 1}$ (where $b \neq 1$)
Substitute $\alpha$ into equation $(2)$:
$\left(\frac{b + 1}{b - 1}\right)^2 + \left(\frac{b + 1}{b - 1}\right) + b = 0$
$(b + 1)^2 + (b + 1)(b - 1) + b(b - 1)^2 = 0$
$(b^2 + 2b + 1) + (b^2 - 1) + b(b^2 - 2b + 1) = 0$
$2b^2 + 2b + b^3 - 2b^2 + b = 0$
$b^3 + 3b = 0$
$b(b^2 + 3) = 0$
Since $b$ must be a real number for the root to be defined in the context of standard quadratic problems,we look for the magnitude. If $b^2 + 3 = 0$,then $b^2 = -3$,which gives $b = \pm i\sqrt{3}$.
Thus,$|b| = |\pm i\sqrt{3}| = \sqrt{3}$.

(A) Given equation: $\sqrt{2x + 1} - \sqrt{2x - 1} = 1$
Squaring both sides: $(\sqrt{2x + 1} - \sqrt{2x - 1})^2 = 1^2$
$(2x + 1) + (2x - 1) - 2\sqrt{(2x + 1)(2x - 1)} = 1$
$4x - 2\sqrt{4x^2 - 1} = 1$
Rearranging the terms: $4x - 1 = 2\sqrt{4x^2 - 1}$
Squaring both sides again: $(4x - 1)^2 = (2\sqrt{4x^2 - 1})^2$
$16x^2 - 8x + 1 = 4(4x^2 - 1)$
$16x^2 - 8x + 1 = 16x^2 - 4$
$-8x = -5$
$x = \frac{5}{8}$
Now,substitute $x = \frac{5}{8}$ into $\sqrt{4x^2 - 1}$:
$\sqrt{4(\frac{5}{8})^2 - 1} = \sqrt{4(\frac{25}{64}) - 1} = \sqrt{\frac{25}{16} - 1} = \sqrt{\frac{25 - 16}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

(B) Given the equation $2x^3 - 9x^2 + kx - 13 = 0$ with real coefficients,complex roots must occur in conjugate pairs.
Since $2 + 3i$ is a root,its conjugate $2 - 3i$ must also be a root.
Let the roots be $\alpha = 2 + 3i,$ $\beta = 2 - 3i,$ and $\gamma$ be the real root.
According to the relationship between roots and coefficients for a cubic equation $ax^3 + bx^2 + cx + d = 0,$ the product of the roots is given by $\alpha \beta \gamma = -\frac{d}{a}.$
Here,$a = 2$ and $d = -13,$ so the product of the roots is $\alpha \beta \gamma = -\frac{-13}{2} = \frac{13}{2}.$
Calculating the product of the complex roots: $\alpha \beta = (2 + 3i)(2 - 3i) = 2^2 - (3i)^2 = 4 + 9 = 13.$
Substituting these into the product formula: $13 \cdot \gamma = \frac{13}{2}.$
Solving for $\gamma,$ we get $\gamma = \frac{1}{2}.$
Thus,the real root exists and is equal to $\frac{1}{2}.$

(B) Given equation: $(a - 1)(x^4 + x^2 + 1) + (a + 1)(x^2 + x + 1)^2 = 0$.
Since $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$,we can factor out $(x^2 + x + 1)$:
$(x^2 + x + 1) [(a - 1)(x^2 - x + 1) + (a + 1)(x^2 + x + 1)] = 0$.
Simplifying the expression inside the bracket:
$(x^2 + x + 1) [ax^2 - ax + a - x^2 + x - 1 + ax^2 + ax + a + x^2 + x + 1] = 0$.
$(x^2 + x + 1) [2ax^2 + 2x + 2a] = 0$.
$2(x^2 + x + 1)(ax^2 + x + a) = 0$.
The quadratic $x^2 + x + 1 = 0$ has discriminant $D = 1 - 4 = -3 < 0$,so it has no real roots.
Thus,the real roots must come from $ax^2 + x + a = 0$.
For the roots to be real and distinct,we require $a \neq 0$ and discriminant $D > 0$.
$D = 1^2 - 4(a)(a) = 1 - 4a^2 > 0$.
$4a^2 < 1 \Rightarrow a^2 < 1/4 \Rightarrow |a| < 1/2$.
Since $a \neq 0$,the set of values is $a \in (-1/2, 0) \cup (0, 1/2)$.

(B) Given that the equations $ax^2 + bx + c = 0$ and $2x^2 + 3x + 4 = 0$ have a common root.
For two quadratic equations $a_1x^2 + b_1x + c_1 = 0$ and $a_2x^2 + b_2x + c_2 = 0$ to have a common root,the coefficients must be proportional if both roots are common,or we can use the condition for a single common root.
However,in this case,the discriminant of the second equation $2x^2 + 3x + 4 = 0$ is $D = b^2 - 4ac = 3^2 - 4(2)(4) = 9 - 32 = -23$.
Since $D < 0$,the roots of the second equation are complex conjugates. If one root is common,the other root must also be common because the coefficients $a, b, c$ are real.
Therefore,the two equations must be proportional:
$\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k$
This implies $a = 2k$,$b = 3k$,and $c = 4k$.
Thus,the ratio $a : b : c = 2k : 3k : 4k = 2 : 3 : 4$.

(A) Given that $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of $ax^2 + bx + 1 = 0$.
From the sum of roots: $\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a} \Rightarrow \frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha\beta}} = -\frac{b}{a}$.
From the product of roots: $\frac{1}{\sqrt{\alpha\beta}} = \frac{1}{a} \Rightarrow a = \sqrt{\alpha\beta}$.
Substituting $a$ in the sum of roots equation: $\frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha\beta}} = -\frac{b}{\sqrt{\alpha\beta}} \Rightarrow b = -(\sqrt{\alpha} + \sqrt{\beta})$.
The given equation is $x^2 + (b^3 - 3ab)x + a^3 = 0$.
Using the identity $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$,we have $b^3 = -(\sqrt{\alpha} + \sqrt{\beta})^3 = -(\alpha^{3/2} + \beta^{3/2} + 3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta}))$.
Also,$3ab = 3(\sqrt{\alpha\beta})(-(\sqrt{\alpha} + \sqrt{\beta})) = -3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta})$.
Thus,$b^3 - 3ab = -(\alpha^{3/2} + \beta^{3/2} + 3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta})) + 3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta}) = -(\alpha^{3/2} + \beta^{3/2})$.
And $a^3 = (\sqrt{\alpha\beta})^3 = \alpha^{3/2}\beta^{3/2}$.
The equation becomes $x^2 - (\alpha^{3/2} + \beta^{3/2})x + \alpha^{3/2}\beta^{3/2} = 0$.
The roots of this quadratic equation are $\alpha^{3/2}$ and $\beta^{3/2}$.

(D) Given the equation $x^2 - 4\sqrt{2}kx + 2e^{4\ln k} - 1 = 0$.
Since $e^{4\ln k} = k^4$, the equation becomes $x^2 - 4\sqrt{2}kx + 2k^4 - 1 = 0$.
From the properties of roots, $\alpha + \beta = 4\sqrt{2}k$ and $\alpha\beta = 2k^4 - 1$.
We are given $\alpha^2 + \beta^2 = 66$.
Using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$, we substitute the values:
$(4\sqrt{2}k)^2 = 66 + 2(2k^4 - 1)$
$32k^2 = 66 + 4k^4 - 2$
$4k^4 - 32k^2 + 64 = 0$
Dividing by $4$, we get $k^4 - 8k^2 + 16 = 0$, which is $(k^2 - 4)^2 = 0$.
Thus, $k^2 = 4$, so $k = 2$ or $k = -2$.
Now, $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta)$.
Substituting the known values: $\alpha^3 + \beta^3 = (4\sqrt{2}k)(66 - (2k^4 - 1)) = (4\sqrt{2}k)(67 - 2k^4)$.
If $k = 2$, $\alpha^3 + \beta^3 = (4\sqrt{2}(2))(67 - 2(16)) = 8\sqrt{2}(67 - 32) = 8\sqrt{2}(35) = 280\sqrt{2}$.
If $k = -2$, $\alpha^3 + \beta^3 = (4\sqrt{2}(-2))(67 - 2(16)) = -8\sqrt{2}(35) = -280\sqrt{2}$.
Given the options, the correct value is $-280\sqrt{2}$.

(C) Given equation: ${x^2} + |2x - 3| - 4 = 0$
Case $I$: If $x \ge \frac{3}{2}$,then $|2x - 3| = 2x - 3$.
The equation becomes: ${x^2} + 2x - 3 - 4 = 0 \Rightarrow {x^2} + 2x - 7 = 0$.
Roots are $x = \frac{-2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{-2 \pm \sqrt{32}}{2} = -1 \pm 2\sqrt{2}$.
Since $x \ge \frac{3}{2}$,we check the values: $-1 + 2\sqrt{2} \approx -1 + 2.828 = 1.828 > 1.5$ (Valid).
$-1 - 2\sqrt{2} \approx -3.828 < 1.5$ (Invalid).
So,$x_1 = 2\sqrt{2} - 1$.
Case $II$: If $x < \frac{3}{2}$,then $|2x - 3| = -(2x - 3) = -2x + 3$.
The equation becomes: ${x^2} - 2x + 3 - 4 = 0 \Rightarrow {x^2} - 2x - 1 = 0$.
Roots are $x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.
Since $x < \frac{3}{2}$,we check the values: $1 + \sqrt{2} \approx 2.414 > 1.5$ (Invalid).
$1 - \sqrt{2} \approx -0.414 < 1.5$ (Valid).
So,$x_2 = 1 - \sqrt{2}$.
The sum of the valid roots is $(2\sqrt{2} - 1) + (1 - \sqrt{2}) = \sqrt{2}$.

(A) Consider the equation $\sqrt{3 x^{2}+x+5}=x-3$.
For the square root to be defined and equal to a real number,the right-hand side must be non-negative,i.e.,$x-3 \geq 0$,which implies $x \geq 3$.
Squaring both sides of the equation:
$3 x^{2}+x+5=(x-3)^{2}$
$3 x^{2}+x+5=x^{2}-6 x+9$
$2 x^{2}+7 x-4=0$
Factoring the quadratic equation:
$2 x^{2}+8 x-x-4=0$
$2 x(x+4)-1(x+4)=0$
$(2 x-1)(x+4)=0$
This gives $x = \frac{1}{2}$ or $x = -4$.
Checking these values against the condition $x \geq 3$:
Neither $x = \frac{1}{2}$ nor $x = -4$ satisfies the condition $x \geq 3$.
Therefore,the given equation has no real solution.

(C) Let $f(x) = x^2 - (a + 1)x + a^2 + a - 8$.
For one root to be greater than $2$ and the other to be less than $2$,the value of the function at $x = 2$ must be negative,i.e.,$f(2) < 0$.
$f(2) = (2)^2 - (a + 1)(2) + a^2 + a - 8 < 0$.
$4 - 2a - 2 + a^2 + a - 8 < 0$.
$a^2 - a - 6 < 0$.
$(a - 3)(a + 2) < 0$.
This inequality holds when $-2 < a < 3$.

(C) The given quadratic equation is $x^2 + px + \frac{3p}{4} = 0.$
From the properties of roots,we have $\alpha + \beta = -p$ and $\alpha \beta = \frac{3p}{4}.$
We are given that $|\alpha - \beta| = \sqrt{10}.$
Squaring both sides,we get $(\alpha - \beta)^2 = 10.$
Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta,$ we substitute the values:
$(-p)^2 - 4\left(\frac{3p}{4}\right) = 10.$
$p^2 - 3p = 10.$
$p^2 - 3p - 10 = 0.$
Factoring the quadratic equation: $(p - 5)(p + 2) = 0.$
Thus,$p = 5$ or $p = -2.$
Therefore,$p \in \{-2, 5\}.$

(D) Given the inequality: $\frac{x - 5}{x^2 + 5x - 14} > 0$.
Factor the denominator: $x^2 + 5x - 14 = (x + 7)(x - 2)$.
So,the inequality is $\frac{x - 5}{(x + 7)(x - 2)} > 0$.
Using the wavy curve method (sign scheme),the critical points are $x = -7, 2, 5$.
The expression is positive in the intervals $(-7, 2) \cup (5, \infty)$.
The integers in the interval $(-7, 2)$ are $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.
The integers in the interval $(5, \infty)$ are $\{6, 7, 8, \dots\}$.
The least integral value $\alpha$ is $-6$.
Checking the options for $\alpha = -6$:
$(A)$ $(-6)^2 + 3(-6) - 4 = 36 - 18 - 4 = 14 \neq 0$.
$(B)$ $(-6)^2 - 5(-6) + 4 = 36 + 30 + 4 = 70 \neq 0$.
$(C)$ $(-6)^2 - 7(-6) + 6 = 36 + 42 + 6 = 84 \neq 0$.
$(D)$ $(-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0$.
Thus,$\alpha = -6$ satisfies option $(D)$.

(B) Given $\alpha^3 + \beta^3 = -p$ and $\alpha \beta = q$.
Let the roots of the required quadratic equation be $x_1 = \frac{\alpha^2}{\beta}$ and $x_2 = \frac{\beta^2}{\alpha}$.
The sum of the roots is $x_1 + x_2 = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta} = \frac{-p}{q}$.
The product of the roots is $x_1 \times x_2 = \frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha} = \alpha \beta = q$.
The required quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (\frac{-p}{q})x + q = 0$.
$x^2 + \frac{p}{q}x + q = 0$.
Multiplying by $q$,we get $qx^2 + px + q^2 = 0$.

(B) For the quadratic equation $(K - 2)x^2 + 8x + K + 4 = 0$ to have real and distinct roots,the discriminant $D$ must be greater than $0$.
$D = b^2 - 4ac = 8^2 - 4(K - 2)(K + 4) > 0$
$64 - 4(K^2 + 2K - 8) > 0$
$16 - (K^2 + 2K - 8) > 0$
$-K^2 - 2K + 24 > 0 \Rightarrow K^2 + 2K - 24 < 0$
$(K + 6)(K - 4) < 0 \Rightarrow -6 < K < 4$.
For the roots to be negative,the product of roots $\alpha\beta = \frac{c}{a} = \frac{K + 4}{K - 2} > 0$ and the sum of roots $\alpha + \beta = -\frac{b}{a} = -\frac{8}{K - 2} < 0$.
From $\frac{K + 4}{K - 2} > 0$,we get $K < -4$ or $K > 2$.
From $-\frac{8}{K - 2} < 0$,we get $\frac{8}{K - 2} > 0$,which implies $K > 2$.
Combining the conditions $-6 < K < 4$ and $K > 2$,we get $2 < K < 4$.
Among the given options,only $K = 3$ satisfies this condition.

(D) The given inequality is $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$.
This can be rearranged as:
$(a^2 p^2 - 2abp + b^2) + (b^2 p^2 - 2bpc + c^2) + (c^2 p^2 - 2cpd + d^2) \le 0$.
This simplifies to:
$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$.
Since the sum of squares of real numbers can only be non-negative,the only way for the sum to be less than or equal to $0$ is if each term is exactly $0$.
Therefore,$ap - b = 0$,$bp - c = 0$,and $cp - d = 0$.
This implies $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.
This is the definition of a Geometric Progression $(G.P.)$,where $a, b, c, d$ are in $G.P.$ with common ratio $p$.

(D) Given the quadratic equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$.
Let the roots be $x_1$ and $x_2$.
From the properties of roots,$x_1 + x_2 = \sin \alpha - 2$ and $x_1 x_2 = -(1 + \sin \alpha)$.
The sum of the squares of the roots is given by $S = x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2$.
Substituting the values: $S = (\sin \alpha - 2)^2 - 2(-(1 + \sin \alpha)) = \sin^2 \alpha - 4 \sin \alpha + 4 + 2 + 2 \sin \alpha = \sin^2 \alpha - 2 \sin \alpha + 6$.
To find the minimum value,we can rewrite this as $S = (\sin \alpha - 1)^2 + 5$.
Since the range of $\sin \alpha$ is $[-1, 1]$,the expression $( \sin \alpha - 1)^2$ is minimum when $\sin \alpha = 1$.
Thus,$\sin \alpha = 1$ implies $\alpha = \frac{\pi}{2}$.

(C) Given the quadratic equation $px^2 + qx + r = 0$ where $p, q, r \in \mathbb{R}$ and $r > p > 0.$
Since the roots $\alpha$ and $\beta$ are complex,they must be conjugates of each other,i.e.,$\beta = \bar{\alpha}.$
This implies $|\beta| = |\bar{\alpha}| = |\alpha|.$
From the properties of quadratic equations,the product of the roots is $\alpha \beta = \frac{r}{p}.$
Since $\beta = \bar{\alpha},$ we have $\alpha \bar{\alpha} = |\alpha|^2 = \frac{r}{p}.$
Given $r > p > 0,$ we have $\frac{r}{p} > 1.$
Therefore,$|\alpha|^2 > 1,$ which implies $|\alpha| > 1.$
Now,consider $|\alpha| + |\beta| = |\alpha| + |\alpha| = 2|\alpha|.$
Since $|\alpha| > 1,$ it follows that $2|\alpha| > 2.$
Thus,$|\alpha| + |\beta| > 2.$

(C) Given that $1$ is a root of the equation $ax^2 + bx + c = 0$.
Therefore,substituting $x = 1$ into the equation,we get $a(1)^2 + b(1) + c = 0$,which implies $a + b + c = 0$,or $a = -(b + c)$.
Now,consider the curve $y = 4ax^2 + 3bx + 2c$.
Since $a = -(b + c)$,we substitute this into the expression for $y$:
$y = 4(-(b + c))x^2 + 3bx + 2c$
$y = -4(b + c)x^2 + 3bx + 2c$
However,the problem states $1$ is a root of $ax^2 + bx + c = 0$,so $a + b + c = 0$. This means $c = -(a + b)$.
Substituting $c = -(a + b)$ into the curve equation:
$y = 4ax^2 + 3bx + 2(-(a + b)) = 4ax^2 + 3bx - 2a - 2b$.
Since $a + b + c = 0$,we have $b = -(a + c)$.
Actually,a simpler approach: Since $a+b+c=0$,then $c = -a-b$. The equation $ax^2+bx+c=0$ becomes $ax^2+bx-(a+b)=0$.
For the curve $y = 4ax^2 + 3bx + 2c$,at the $x-$axis,$y = 0$.
$4ax^2 + 3bx + 2c = 0$.
Since $a+b+c=0$,$c = -a-b$. Substituting this: $4ax^2 + 3bx - 2a - 2b = 0$.
If $x=1$,$4a + 3b - 2a - 2b = 2a + b$. This is not necessarily zero.
Let's re-evaluate: $a+b+c=0 \Rightarrow c = -a-b$.
$y = 4ax^2 + 3bx + 2(-a-b) = 4ax^2 + 3bx - 2a - 2b$.
At $x=1$,$y = 4a+3b-2a-2b = 2a+b$. This is not zero unless $2a+b=0$.
Wait,the condition $a+b+c=0$ implies $c = -a-b$. The curve is $y = 4ax^2 + 3bx + 2(-a-b) = 4ax^2 + 3bx - 2a - 2b$.
This is a quadratic equation in $x$. For it to intersect the $x-$axis at exactly one point,the discriminant $D$ must be $0$.
$D = (3b)^2 - 4(4a)(-2a-2b) = 9b^2 + 16a(2a+2b) = 9b^2 + 32a^2 + 32ab$.
This is not identically zero. Let's re-read: $1$ is a root,so $a+b+c=0$. The curve $y = 4ax^2 + 3bx + 2c$. At $x=1$,$y = 4a+3b+2c = 2a+b+(2a+2b+2c) = 2a+b+2(a+b+c) = 2a+b$. This is not necessarily $0$.
Actually,if $a+b+c=0$,then $c = -a-b$. The curve is $y = 4ax^2 + 3bx - 2a - 2b$. This is a parabola. It intersects the $x-$axis where $4ax^2 + 3bx - 2a - 2b = 0$. This is a quadratic equation. It will have roots $x = \frac{-3b \pm \sqrt{9b^2 - 16a(-2a-2b)}}{8a} = \frac{-3b \pm \sqrt{9b^2 + 32a^2 + 32ab}}{8a}$.
If $b = -2a$,then $c = -a - (-2a) = a$. The equation $ax^2 - 2ax + a = 0 \Rightarrow a(x-1)^2 = 0$,so $x=1$ is a root. The curve becomes $y = 4ax^2 - 6ax + 2a = 2a(2x^2 - 3x + 1) = 2a(2x-1)(x-1)$. This intersects at $x=1$ and $x=1/2$. Two points.
If $a=1, b=0, c=-1$,$y = 4x^2 - 2 = 0 \Rightarrow x^2 = 1/2 \Rightarrow x = \pm 1/\sqrt{2}$. Two points.
Thus,the curve intersects at exactly two distinct points.

(A) Given the quadratic equation $x^2 + 2x + 2 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-2 \pm \sqrt{4 - 8}}{2} = -1 \pm i$.
Let $\alpha = -1 + i$ and $\beta = -1 - i$.
In polar form,$\alpha = \sqrt{2} e^{i(3\pi/4)}$ and $\beta = \sqrt{2} e^{-i(3\pi/4)}$.
Then $\alpha^{15} + \beta^{15} = (\sqrt{2})^{15} [e^{i(45\pi/4)} + e^{-i(45\pi/4)}]$.
Using Euler's formula,this is $2^{7} \cdot \sqrt{2} \cdot 2 \cos(45\pi/4)$.
Since $45\pi/4 = 11\pi + \pi/4$,$\cos(45\pi/4) = \cos(11\pi + \pi/4) = -\cos(\pi/4) = -1/\sqrt{2}$.
Thus,$\alpha^{15} + \beta^{15} = 2^8 \cdot \sqrt{2} \cdot (-1/\sqrt{2}) = -2^8 = -256$.

(A) Let $f(x) = x^2 - mx + 4.$ For the roots to be real and distinct and lie in the interval $[1, 5],$ the following conditions must be satisfied:
$1$. Discriminant $D > 0: (-m)^2 - 4(1)(4) > 0 \Rightarrow m^2 - 16 > 0 \Rightarrow m \in (-\infty, -4) \cup (4, \infty).$
$2$. $f(1) > 0: 1^2 - m(1) + 4 > 0 \Rightarrow 5 - m > 0 \Rightarrow m < 5.$
$3$. $f(5) > 0: 5^2 - m(5) + 4 > 0 \Rightarrow 29 - 5m > 0 \Rightarrow m < 5.8.$
$4$. Vertex location: $1 < \frac{-b}{2a} < 5 \Rightarrow 1 < \frac{m}{2} < 5 \Rightarrow 2 < m < 10.$
Taking the intersection of all these conditions: $m \in (4, 5).$ Since $(4, 5)$ is given in option $A,$ it is the correct answer.

(C) For the roots of the quadratic equation $ax^2 + bx + c = 0$ to be rational,the discriminant $D = b^2 - 4ac$ must be a perfect square.
Here,$a = 6, b = -11, c = \alpha$.
$D = (-11)^2 - 4(6)(\alpha) = 121 - 24\alpha$.
Since $\alpha$ is a positive integer,$121 - 24\alpha \ge 0$,which implies $24\alpha \le 121$,so $\alpha \le 5.04$. Thus,$\alpha \in \{1, 2, 3, 4, 5\}$.
We check each value:
$1$. If $\alpha = 1, D = 121 - 24 = 97$ (not a perfect square).
$2$. If $\alpha = 2, D = 121 - 48 = 73$ (not a perfect square).
$3$. If $\alpha = 3, D = 121 - 72 = 49 = 7^2$ (perfect square).
$4$. If $\alpha = 4, D = 121 - 96 = 25 = 5^2$ (perfect square).
$5$. If $\alpha = 5, D = 121 - 120 = 1 = 1^2$ (perfect square).
The possible values for $\alpha$ are $3, 4, 5$. Therefore,there are $3$ such values.

(D) Let $f(x) = (c - 5)x^2 - 2cx + (c - 4)$.
For one root to lie in $(0, 2)$ and the other in $(2, 3)$,the value of $f(2)$ must have a sign opposite to the leading coefficient $(c - 5)$.
Case $I$: If $c - 5 > 0$ (i.e.,$c > 5$),then $f(2) < 0$.
$f(2) = (c - 5)(2)^2 - 2c(2) + (c - 4) = 4c - 20 - 4c + c - 4 = c - 24$.
So,$c - 24 < 0 \Rightarrow c < 24$.
Also,we need $f(0) > 0$ and $f(3) > 0$ for the roots to be in the specified intervals.
$f(0) = c - 4 > 0 \Rightarrow c > 4$.
$f(3) = (c - 5)(9) - 6c + (c - 4) = 9c - 45 - 6c + c - 4 = 4c - 49 > 0 \Rightarrow c > 12.25$.
Combining $c > 5$,$c < 24$,$c > 4$,and $c > 12.25$,we get $c \in (12.25, 24)$.
Integral values of $c$ are ${13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23}$.
Number of elements = $11$.
Case $II$: If $c - 5 < 0$ (i.e.,$c < 5$),then $f(2) > 0$.
$c - 24 > 0 \Rightarrow c > 24$,which contradicts $c < 5$. Thus,no solution here.
Therefore,$S = {13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23}$,and the number of elements is $11$.

(D) Let the roots of the quadratic equation $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$ be $\alpha$ and $\beta$.
The sum of the squares of the roots is given by $S = \alpha^2 + \beta^2$.
Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,we have:
$\alpha + \beta = -(3 - \lambda) = \lambda - 3$
$\alpha\beta = 2 - \lambda$
Substituting these into the expression for $S$:
$S = (\lambda - 3)^2 - 2(2 - \lambda)$
$S = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda$
$S = \lambda^2 - 4\lambda + 5$
To find the least value,we complete the square:
$S = (\lambda^2 - 4\lambda + 4) + 1 = (\lambda - 2)^2 + 1$
The expression $S$ attains its minimum value when $(\lambda - 2)^2 = 0$,which occurs at $\lambda = 2$.

(D) Let the roots of the quadratic equation $81x^2 + kx + 256 = 0$ be $\alpha$ and $\alpha^3$.
From the properties of quadratic equations,the product of the roots is given by $\alpha \cdot \alpha^3 = \frac{256}{81}$.
This simplifies to $\alpha^4 = \frac{256}{81}$,which gives $\alpha = \pm \frac{4}{3}$.
The sum of the roots is given by $\alpha + \alpha^3 = -\frac{k}{81}$.
Case $1$: If $\alpha = \frac{4}{3}$,then $\frac{4}{3} + (\frac{4}{3})^3 = -\frac{k}{81} \implies \frac{4}{3} + \frac{64}{27} = -\frac{k}{81} \implies \frac{36 + 64}{27} = -\frac{k}{81} \implies \frac{100}{27} = -\frac{k}{81} \implies k = -300$.
Case $2$: If $\alpha = -\frac{4}{3}$,then $-\frac{4}{3} - \frac{64}{27} = -\frac{k}{81} \implies -\frac{100}{27} = -\frac{k}{81} \implies k = 300$.
Comparing with the given options,the correct value is $-300$.

(C) The given quadratic equation is $x^2 \sin \theta - x(\sin \theta \cos \theta + 1) + \cos \theta = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:
$x = \frac{(\sin \theta \cos \theta + 1) \pm \sqrt{(\sin \theta \cos \theta + 1)^2 - 4 \sin \theta \cos \theta}}{2 \sin \theta}$
Since $(\sin \theta \cos \theta + 1)^2 - 4 \sin \theta \cos \theta = (\sin \theta \cos \theta - 1)^2$,we get:
$x = \frac{(\sin \theta \cos \theta + 1) \pm (\sin \theta \cos \theta - 1)}{2 \sin \theta}$
Taking the positive sign: $x = \frac{2 \sin \theta \cos \theta}{2 \sin \theta} = \cos \theta$.
Taking the negative sign: $x = \frac{2}{2 \sin \theta} = \csc \theta$.
Given $0 < \theta < 45^\circ$,we know $\cos \theta > \sin \theta$,so $\cos \theta < \csc \theta$. Thus,$\alpha = \cos \theta$ and $\beta = \csc \theta$.
The sum is $\sum_{n=0}^\infty \alpha^n + \sum_{n=0}^\infty (-\frac{1}{\beta})^n = \sum_{n=0}^\infty (\cos \theta)^n + \sum_{n=0}^\infty (-\sin \theta)^n$.
Using the infinite geometric series formula $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$:
Sum $= \frac{1}{1 - \cos \theta} + \frac{1}{1 - (-\sin \theta)} = \frac{1}{1 - \cos \theta} + \frac{1}{1 + \sin \theta}$.

(B) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
Given that $\lambda = \frac{\alpha}{\beta}$,the condition $\lambda + \frac{1}{\lambda} = 1$ implies $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1$.
This simplifies to $\frac{\alpha^2 + \beta^2}{\alpha\beta} = 1$,or $\alpha^2 + \beta^2 = \alpha\beta$.
Adding $\alpha\beta$ to both sides,we get $\alpha^2 + 2\alpha\beta + \beta^2 = 3\alpha\beta$,which means $(\alpha + \beta)^2 = 3\alpha\beta$.
From the quadratic equation $3m^2x^2 + m(m - 4)x + 2 = 0$,we have sum of roots $\alpha + \beta = -\frac{m(m - 4)}{3m^2} = -\frac{m - 4}{3m}$ and product of roots $\alpha\beta = \frac{2}{3m^2}$.
Substituting these into the equation $(\alpha + \beta)^2 = 3\alpha\beta$:
$\left(-\frac{m - 4}{3m}\right)^2 = 3 \left(\frac{2}{3m^2}\right)$
$\frac{(m - 4)^2}{9m^2} = \frac{2}{m^2}$
$(m - 4)^2 = 18$
$m^2 - 8m + 16 = 18$
$m^2 - 8m - 2 = 0$.
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $m = \frac{8 \pm \sqrt{64 - 4(1)(-2)}}{2} = \frac{8 \pm \sqrt{72}}{2} = \frac{8 \pm 6\sqrt{2}}{2} = 4 \pm 3\sqrt{2}$.
The least value of $m$ is $4 - 3\sqrt{2}$.

(C) For a quadratic expression $ax^2 + bx + c$ to be always positive for all $x \in R$,two conditions must be satisfied: $a > 0$ and $D < 0$.
$1$. Condition $a > 0$:
$(1 + 2m) > 0 \Rightarrow 2m > -1 \Rightarrow m > -\frac{1}{2}$.
$2$. Condition $D < 0$:
$D = [-2(1 + 3m)]^2 - 4(1 + 2m)(4(1 + m)) < 0$
$4(1 + 3m)^2 - 16(1 + 2m)(1 + m) < 0$
$(1 + 6m + 9m^2) - 4(1 + 3m + 2m^2) < 0$
$1 + 6m + 9m^2 - 4 - 12m - 8m^2 < 0$
$m^2 - 6m - 3 < 0$.
Solving $m^2 - 6m - 3 = 0$ using the quadratic formula: $m = \frac{6 \pm \sqrt{36 - 4(1)(-3)}}{2} = \frac{6 \pm \sqrt{48}}{2} = 3 \pm \sqrt{12} = 3 \pm 2\sqrt{3}$.
Since $\sqrt{12} \approx 3.46$,the roots are $3 - 3.46 = -0.46$ and $3 + 3.46 = 6.46$.
So,$3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$.
Combining with $m > -0.5$,the interval is $(-0.46, 6.46)$.
The integral values of $m$ are ${0, 1, 2, 3, 4, 5, 6}$.
There are $7$ such values.

(A) Given the quadratic equation $x^2 - 2x + 2 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i$.
Let $\alpha = 1 + i$ and $\beta = 1 - i$.
Now,$\frac{\alpha}{\beta} = \frac{1 + i}{1 - i} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{2i}{2} = i$.
Alternatively,$\frac{\alpha}{\beta} = \frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = i$ and if we take $\alpha = 1 - i$ and $\beta = 1 + i$,then $\frac{\alpha}{\beta} = -i$.
We need $(\frac{\alpha}{\beta})^n = 1$. If $\frac{\alpha}{\beta} = i$,then $i^n = 1$,which gives the least natural number $n = 4$.
If $\frac{\alpha}{\beta} = -i$,then $(-i)^n = 1$,which also gives the least natural number $n = 4$.
Thus,the least value of $n$ is $4$.

(C) Given equation: $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$.
Let $t = \sqrt{x}$,where $t > 0$. The equation becomes $|t - 2| + t(t - 4) + 2 = 0$.
$|t - 2| + t^2 - 4t + 2 = 0$.
We can rewrite $t^2 - 4t + 2$ as $(t^2 - 4t + 4) - 2 = (t - 2)^2 - 2$.
So,$|t - 2| + (t - 2)^2 - 2 = 0$.
Let $u = |t - 2|$. Since $u^2 = |t - 2|^2 = (t - 2)^2$,the equation becomes $u + u^2 - 2 = 0$.
$u^2 + u - 2 = 0$.
$(u + 2)(u - 1) = 0$.
This gives $u = -2$ or $u = 1$.
Since $u = |t - 2| \ge 0$,$u = -2$ is rejected.
Thus,$|t - 2| = 1$.
This implies $t - 2 = 1$ or $t - 2 = -1$.
$t = 3$ or $t = 1$.
Since $t = \sqrt{x}$,we have $\sqrt{x} = 3 \implies x = 9$ and $\sqrt{x} = 1 \implies x = 1$.
Both values satisfy $x > 0$.
The sum of the solutions is $9 + 1 = 10$.

(A) For the quadratic equation $ax^2 + bx + c = 0$ to have no real roots,the discriminant $D$ must be less than zero $(D < 0)$.
Here,$a = (1 + m^2)$,$b = -2(1 + 3m)$,and $c = (1 + 8m)$.
The discriminant $D = b^2 - 4ac = [-2(1 + 3m)]^2 - 4(1 + m^2)(1 + 8m)$.
$D = 4(1 + 9m^2 + 6m) - 4(1 + 8m + m^2 + 8m^3)$.
$D = 4(1 + 9m^2 + 6m - 1 - 8m - m^2 - 8m^3)$.
$D = 4(-8m^3 + 8m^2 - 2m) = -8m(4m^2 - 4m + 1) = -8m(2m - 1)^2$.
For no real roots,$-8m(2m - 1)^2 < 0$.
Since $(2m - 1)^2$ is always non-negative,for the expression to be negative,we must have $m > 0$ and $m \neq 1/2$.
Since there are infinitely many integers $m > 0$ (excluding $m = 1/2$ which is not an integer),there are infinitely many such integral values.

(C) Given that $p, q \in Q$ and $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$.
Since the coefficients are rational,irrational roots must occur in conjugate pairs.
Therefore,the other root is $2 + \sqrt{3}$.
Sum of roots $= (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.
From the equation $x^2 + px + q = 0$,the sum of roots is $-p$.
So,$-p = 4 \implies p = -4$.
Product of roots $= (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.
From the equation $x^2 + px + q = 0$,the product of roots is $q$.
So,$q = 1$.
Now,check the options by substituting $p = -4$ and $q = 1$:
$p^2 - 4q - 12 = (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0$.
Thus,$p^2 - 4q - 12 = 0$ is the correct relation.

(A) Given the quadratic equation $x^2 + x \sin \theta - 2 \sin \theta = 0$.
By Vieta's formulas,the sum of roots is $\alpha + \beta = -\sin \theta$ and the product of roots is $\alpha \beta = -2 \sin \theta$.
We need to evaluate the expression $E = \frac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$.
Simplifying the denominator: $\alpha^{-12} + \beta^{-12} = \frac{1}{\alpha^{12}} + \frac{1}{\beta^{12}} = \frac{\alpha^{12} + \beta^{12}}{(\alpha \beta)^{12}}$.
Substituting this into the expression: $E = \frac{\alpha^{12} + \beta^{12}}{\left(\frac{\alpha^{12} + \beta^{12}}{(\alpha \beta)^{12}}\right)(\alpha - \beta)^{24}} = \frac{(\alpha \beta)^{12}}{(\alpha - \beta)^{24}}$.
Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4 \alpha \beta$,we have $(\alpha - \beta)^{24} = ((\alpha + \beta)^2 - 4 \alpha \beta)^{12}$.
Thus,$E = \left[ \frac{\alpha \beta}{(\alpha + \beta)^2 - 4 \alpha \beta} \right]^{12}$.
Substituting the values of $\alpha + \beta$ and $\alpha \beta$: $E = \left[ \frac{-2 \sin \theta}{(-\sin \theta)^2 - 4(-2 \sin \theta)} \right]^{12} = \left[ \frac{-2 \sin \theta}{\sin^2 \theta + 8 \sin \theta} \right]^{12}$.
Since $\theta \in (0, \frac{\pi}{2})$,$\sin \theta \neq 0$,we can cancel $\sin \theta$: $E = \left( \frac{-2}{\sin \theta + 8} \right)^{12} = \frac{2^{12}}{(\sin \theta + 8)^{12}}$.

(B) Given inequality: $2^{\sqrt{\sin^2 x - 2\sin x + 5}} \cdot 4^{-\sin^2 y} \leq 1$
Rewrite the inequality as: $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 4^{\sin^2 y}$
Since $4 = 2^2$,we have: $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 2^{2\sin^2 y}$
Comparing the exponents: $\sqrt{(\sin x - 1)^2 + 4} \leq 2\sin^2 y$
We know that $(\sin x - 1)^2 \geq 0$,so $\sqrt{(\sin x - 1)^2 + 4} \geq \sqrt{4} = 2$.
Also,we know that $\sin^2 y \leq 1$,so $2\sin^2 y \leq 2$.
For the inequality $2 \leq \text{LHS} \leq \text{RHS} \leq 2$ to hold,both sides must be equal to $2$.
This implies $\sin x - 1 = 0 \Rightarrow \sin x = 1$ and $\sin^2 y = 1 \Rightarrow |\sin y| = 1$.
Thus,$\sin x = 1$ and $|\sin y| = 1$,which means $\sin x = |\sin y|$.

(D) Let $2^x = t$. Since $2^x > 0$,we must have $t > 0$.
The equation becomes $5 + |t - 1| = t(t - 2) = t^2 - 2t$.
Rearranging,we get $|t - 1| = t^2 - 2t - 5$.
Let $g(t) = |t - 1|$ and $f(t) = t^2 - 2t - 5$.
We look for intersections where $t > 0$.
Case $1$: $t \ge 1$,then $t - 1 = t^2 - 2t - 5 \Rightarrow t^2 - 3t - 4 = 0 \Rightarrow (t - 4)(t + 1) = 0$. Since $t > 0$,$t = 4$. This gives $2^x = 4 \Rightarrow x = 2$.
Case $2$: $0 < t < 1$,then $-(t - 1) = t^2 - 2t - 5 \Rightarrow -t + 1 = t^2 - 2t - 5 \Rightarrow t^2 - t - 6 = 0 \Rightarrow (t - 3)(t + 2) = 0$. Neither $t = 3$ nor $t = -2$ lies in the interval $(0, 1)$.
Thus,there is only $1$ real root.