Mix Examples-Quadratic Equations and Inequations Questions in English

(B) Given $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.
Note that $xy = 1$,so $y = \frac{1}{x}$.
First,calculate $x + y$:
$x + y = \frac{(\sqrt{5} + \sqrt{2})^2 + (\sqrt{5} - \sqrt{2})^2}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{(5 + 2 + 2\sqrt{10}) + (5 + 2 - 2\sqrt{10})}{5 - 2} = \frac{14}{3}$.
Next,calculate $x - y$:
$x - y = \frac{(\sqrt{5} + \sqrt{2})^2 - (\sqrt{5} - \sqrt{2})^2}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{(7 + 2\sqrt{10}) - (7 - 2\sqrt{10})}{3} = \frac{4\sqrt{10}}{3}$.
Now,evaluate $3x^2 - 3y^2 + 4xy = 3(x^2 - y^2) + 4(1) = 3(x - y)(x + y) + 4$.
Substituting the values:
$3 \times \left(\frac{4\sqrt{10}}{3}\right) \times \left(\frac{14}{3}\right) + 4 = \frac{56\sqrt{10}}{3} + 4 = \frac{56\sqrt{10} + 12}{3} = \frac{1}{3}[56\sqrt{10} + 12]$.

(B) Given the equation $(1 - ab)m^2 - (a^2 + b^2)m - (1 + ab) = 0$.
This implies $m(a^2 + b^2) + (m^2 + 1)ab = m^2 - 1$ ......$(i)$
Let $H_1, H_2, \dots, H_m$ be the $m$ harmonic means between $a$ and $b$.
The first harmonic mean $H_1 = \frac{(m + 1)ab}{a + mb}$ and the last harmonic mean $H_m = \frac{(m + 1)ab}{b + ma}$.
The difference $H_m - H_1 = (m + 1)ab \left[ \frac{1}{b + ma} - \frac{1}{a + mb} \right]$.
$H_m - H_1 = (m + 1)ab \left[ \frac{a + mb - b - ma}{(b + ma)(a + mb)} \right] = (m + 1)ab \frac{(m - 1)(b - a)}{(b + ma)(a + mb)}$.
$H_m - H_1 = \frac{(m^2 - 1)ab(b - a)}{m(a^2 + b^2) + (m^2 + 1)ab}$.
Using equation $(i)$,the denominator is $m^2 - 1$.
Thus,$H_m - H_1 = \frac{(m^2 - 1)ab(b - a)}{m^2 - 1} = ab(b - a)$.

(A) Let $r$ be the common ratio of the $G.P.$ $\alpha, \beta, \gamma, \delta$. Then $\beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.
From the sum and product of roots:
$\alpha + \beta = 1 \Rightarrow \alpha(1 + r) = 1$ $(i)$
$\alpha \beta = p \Rightarrow \alpha^2 r = p$ $(ii)$
$\gamma + \delta = 4 \Rightarrow \alpha r^2(1 + r) = 4$ $(iii)$
$\gamma \delta = q \Rightarrow \alpha^2 r^5 = q$ $(iv)$
Dividing $(iii)$ by $(i)$,we get $r^2 = 4$,so $r = \pm 2$.
If $r = 2$,then $\alpha(1+2) = 1 \Rightarrow \alpha = 1/3$ (not an integer).
If $r = -2$,then $\alpha(1-2) = 1 \Rightarrow \alpha = -1$.
Substituting $r = -2$ and $\alpha = -1$ into $(ii)$ and $(iv)$:
$p = (-1)^2(-2) = -2$
$q = (-1)^2(-2)^5 = -32$
Thus,$(p, q) = (-2, -32)$.

(A) Let $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$.
$y(x^2 + 2x + 3) = x^2 + 14x + 9$.
$(y - 1)x^2 + (2y - 14)x + (3y - 9) = 0$.
Since $x$ is real,the discriminant $D \ge 0$.
$D = (2y - 14)^2 - 4(y - 1)(3y - 9) \ge 0$.
$4(y - 7)^2 - 12(y - 1)(y - 3) \ge 0$.
$(y^2 - 14y + 49) - 3(y^2 - 4y + 3) \ge 0$.
$y^2 - 14y + 49 - 3y^2 + 12y - 9 \ge 0$.
$-2y^2 - 2y + 40 \ge 0$.
$y^2 + y - 20 \le 0$.
$(y + 5)(y - 4) \le 0$.
Thus,$-5 \le y \le 4$.
The maximum value is $4$ and the minimum value is $-5$.

(B) Let $y = \frac{x + 2}{2x^2 + 3x + 6}$.
Then $y(2x^2 + 3x + 6) = x + 2$,which simplifies to $2yx^2 + (3y - 1)x + (6y - 2) = 0$.
If $y = 0$,then $x = -2$,which is a real value.
If $y \neq 0$,for $x$ to be real,the discriminant $D \ge 0$.
$D = (3y - 1)^2 - 4(2y)(6y - 2) \ge 0$.
$9y^2 - 6y + 1 - 48y^2 + 16y \ge 0$.
$-39y^2 + 10y + 1 \ge 0$.
$39y^2 - 10y - 1 \le 0$.
Factoring the quadratic: $(13y + 1)(3y - 1) \le 0$.
This inequality holds for $y \in \left[ -\frac{1}{13}, \frac{1}{3} \right]$.
Since $y = 0$ is included in this interval,the range is $\left[ -\frac{1}{13}, \frac{1}{3} \right]$.

(A) Given $u = x^2 + 4y^2 + 9z^2 - 6yz - 3zx - 2xy$.
Multiply and divide by $2$:
$u = \frac{1}{2}(2x^2 + 8y^2 + 18z^2 - 12yz - 6zx - 4xy)$
Rearrange the terms to form perfect squares:
$u = \frac{1}{2}[(x^2 - 4xy + 4y^2) + (x^2 - 6zx + 9z^2) + (4y^2 - 12yz + 9z^2)]$
Simplify the expressions inside the brackets:
$u = \frac{1}{2}[(x - 2y)^2 + (x - 3z)^2 + (2y - 3z)^2]$
Since $x, y, z$ are real,the sum of squares is always non-negative. Thus,$u \geq 0$.

(D) Let $y = \frac{(x - a)(x - b)}{(x - c)}$.
Then $y(x - c) = x^2 - (a + b)x + ab$,which rearranges to $x^2 - (a + b + y)x + (ab + cy) = 0$.
For $x$ to be real,the discriminant $D$ must be $\ge 0$:
$D = (a + b + y)^2 - 4(ab + cy) \ge 0$.
Expanding this,we get $y^2 + 2y(a + b - 2c) + (a - b)^2 \ge 0$.
For the function to assume all real values $y$,this quadratic in $y$ must be non-negative for all $y$. However,since the coefficient of $y^2$ is positive,this quadratic will take all real values if its discriminant $D_y < 0$.
$D_y = [2(a + b - 2c)]^2 - 4(a - b)^2 < 0$.
$4(a + b - 2c)^2 - 4(a - b)^2 < 0$.
$(a + b - 2c)^2 - (a - b)^2 < 0$.
Using $A^2 - B^2 = (A-B)(A+B)$:
$(a + b - 2c - a + b)(a + b - 2c + a - b) < 0$.
$(2b - 2c)(2a - 2c) < 0$.
$4(b - c)(a - c) < 0$.
$(c - b)(c - a) < 0$.
This inequality holds if $c$ lies between $a$ and $b$,i.e.,$a < c < b$ or $b < c < a$.

(C) Let $y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4}$.
Multiplying both sides by $(x^2 + 3x + 4)$,we get $y(x^2 + 3x + 4) = x^2 - 3x + 4$.
Rearranging the terms,we get $(y - 1)x^2 + 3(y + 1)x + 4(y - 1) = 0$.
Since $x$ is real,the discriminant $D$ must be greater than or equal to $0$.
$D = [3(y + 1)]^2 - 4(y - 1)(4(y - 1)) \ge 0$.
$9(y + 1)^2 - 16(y - 1)^2 \ge 0$.
$(3(y + 1))^2 - (4(y - 1))^2 \ge 0$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(3y + 3 - 4y + 4)(3y + 3 + 4y - 4) \ge 0$.
$(-y + 7)(7y - 1) \ge 0$.
Multiplying by $-1$,we get $(y - 7)(7y - 1) \le 0$.
This inequality holds when $\frac{1}{7} \le y \le 7$.
Thus,the maximum value is $7$ and the minimum value is $\frac{1}{7}$.

(D) Let $y = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$.
Rearranging the equation,we get $y(x^2 + 2x - 7) = x^2 + 34x - 71$,which simplifies to $x^2(y - 1) + 2x(y - 17) + (71 - 7y) = 0$.
For $x$ to be real,the discriminant $D$ must be greater than or equal to $0$ $(D \ge 0)$.
$D = [2(y - 17)]^2 - 4(y - 1)(71 - 7y) \ge 0$.
Dividing by $4$,we get $(y - 17)^2 - (y - 1)(71 - 7y) \ge 0$.
$(y^2 - 34y + 289) - (71y - 7y^2 - 71 + 7y) \ge 0$.
$(y^2 - 34y + 289) - (-7y^2 + 78y - 71) \ge 0$.
$8y^2 - 112y + 360 \ge 0$.
Dividing by $8$,we get $y^2 - 14y + 45 \ge 0$.
Factoring the quadratic,we get $(y - 5)(y - 9) \ge 0$.
This inequality holds when $y \le 5$ or $y \ge 9$.
Therefore,the value of the expression does not lie between $5$ and $9$.

(C) From the graph,the parabola opens downwards,which implies that the coefficient of ${x^2}$ must be negative. Therefore,$a < 0$.
The roots of the quadratic equation $y = a{x^2} + bx + c$ are ${x_1}$ and ${x_2}$,where both ${x_1} > 0$ and ${x_2} > 0$.
The sum of the roots is given by ${x_1} + {x_2} = -\frac{b}{a}$.
Since both roots are positive,their sum is positive,so $-\frac{b}{a} > 0$,which implies $\frac{b}{a} < 0$.
This means that $a$ and $b$ must have opposite signs.
Since the parabola intersects the $X$-axis at two distinct points,the discriminant $D = {b^2} - 4ac > 0$,so ${b^2} > 4ac$.
Thus,both $(a)$ and $(b)$ are correct.

(C) Let $f(x) = x^3 - 3b^2x + 2c^3$. Since $f(x)$ is divisible by $x - a$ and $x - b$,we have $f(a) = 0$ and $f(b) = 0$.
$f(b) = b^3 - 3b^2(b) + 2c^3 = b^3 - 3b^3 + 2c^3 = -2b^3 + 2c^3 = 0$.
This implies $b^3 = c^3$,so $b = c$ (since $b, c$ are real).
Now,$f(a) = a^3 - 3b^2a + 2c^3 = 0$. Substituting $c = b$,we get $a^3 - 3ab^2 + 2b^3 = 0$.
Factoring the expression: $(a - b)(a^2 + ab - 2b^2) = 0$.
$(a - b)(a - b)(a + 2b) = 0$.
This gives $a = b$ or $a = -2b$.
Since $b = c$,the solutions are $a = b = c$ or $a = -2b = -2c$.

(A) Given $k = \frac{x^2 - x + 1}{x^2 + x + 1}$.
Rearranging the terms,we get $k(x^2 + x + 1) = x^2 - x + 1$.
$kx^2 + kx + k = x^2 - x + 1$.
$(k - 1)x^2 + (k + 1)x + (k - 1) = 0$.
Since $x$ is real,the discriminant $D \ge 0$.
$D = (k + 1)^2 - 4(k - 1)(k - 1) \ge 0$.
$(k + 1)^2 - 4(k - 1)^2 \ge 0$.
Using $a^2 - b^2 = (a - b)(a + b)$,we get $((k + 1) - 2(k - 1))((k + 1) + 2(k - 1)) \ge 0$.
$(-k + 3)(3k - 1) \ge 0$.
Multiplying by $-1$,we get $(k - 3)(3k - 1) \le 0$.
The roots are $k = \frac{1}{3}$ and $k = 3$.
Thus,$\frac{1}{3} \le k \le 3$.

(D) Given that $\tan \alpha$ and $\tan \beta$ are the roots of $x^2 + px + q = 0$.
By Vieta's formulas,$\tan \alpha + \tan \beta = -p$ and $\tan \alpha \tan \beta = q$.
Using the identity for $\tan(\alpha + \beta)$:
$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{-p}{1 - q} = \frac{p}{q - 1}$.
This confirms option $(b)$ is correct.
Now,consider the expression in $(a)$:
$E = \sin^2(\alpha + \beta) + p\sin(\alpha + \beta)\cos(\alpha + \beta) + q\cos^2(\alpha + \beta)$.
Divide by $\cos^2(\alpha + \beta)$ and multiply by $\cos^2(\alpha + \beta)$:
$E = \cos^2(\alpha + \beta) [\tan^2(\alpha + \beta) + p\tan(\alpha + \beta) + q]$.
Since $\cos^2(\alpha + \beta) = \frac{1}{1 + \tan^2(\alpha + \beta)}$,we substitute $\tan(\alpha + \beta) = \frac{p}{q - 1}$:
$E = \frac{1}{1 + (\frac{p}{q - 1})^2} [(\frac{p}{q - 1})^2 + p(\frac{p}{q - 1}) + q]$
$E = \frac{(q - 1)^2}{(q - 1)^2 + p^2} [\frac{p^2 + p^2(q - 1) + q(q - 1)^2}{(q - 1)^2}]$
$E = \frac{p^2 + p^2q - p^2 + q(q^2 - 2q + 1)}{p^2 + (q - 1)^2} = \frac{p^2q + q^3 - 2q^2 + q}{p^2 + (q - 1)^2} = \frac{q(p^2 + q^2 - 2q + 1)}{p^2 + (q - 1)^2} = \frac{q(p^2 + (q - 1)^2)}{p^2 + (q - 1)^2} = q$.
Thus,$(a)$ is also correct. Therefore,the correct option is $(d)$.

(D) Given the inequality $4x^2 - 16x + 15 < 0$.
Solving for $x$: $(2x - 3)(2x - 5) < 0$,which gives $\frac{3}{2} < x < \frac{5}{2}$.
The only integral solution in this interval is $x = 2$. Thus,$\tan \alpha = 2$.
The bisector of the first quadrant is the line $y = x$,which has a slope of $1$. Thus,$\cos \beta = 1$.
Since $\cos \beta = 1$,we have $\beta = 0$,so $\sin \beta = 0$.
Using the identity $\sin(\alpha + \beta)\sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta$:
$\sin^2 \alpha - \sin^2 \beta = \sin^2 \alpha - 0 = \sin^2 \alpha$.
Since $\tan \alpha = 2$,we have $\sin^2 \alpha = \frac{\tan^2 \alpha}{1 + \tan^2 \alpha} = \frac{2^2}{1 + 2^2} = \frac{4}{5}$.

(C) Let $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$.
$y(x^2 + 2x + 3) = x^2 + 14x + 9$
$x^2(y - 1) + 2x(y - 7) + (3y - 9) = 0$.
Since $x$ is real,the discriminant $D \geq 0$:
$D = [2(y - 7)]^2 - 4(y - 1)(3y - 9) \geq 0$
$4(y^2 - 14y + 49) - 4(3y^2 - 12y + 9) \geq 0$
$y^2 - 14y + 49 - 3y^2 + 12y - 9 \geq 0$
$-2y^2 - 2y + 40 \geq 0$
$y^2 + y - 20 \leq 0$
$(y + 5)(y - 4) \leq 0$.
Thus,$y$ lies between $-5$ and $4$.

(A) Given the inequality $27pqr \geq (p + q + r)^3$,we can rewrite it as $\sqrt[3]{pqr} \geq \frac{p + q + r}{3}$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,we know that $\frac{p + q + r}{3} \geq \sqrt[3]{pqr}$.
For both inequalities to hold simultaneously,we must have $p = q = r$.
Given the equation $3p + 4q + 5r = 12$,substitute $p = q = r = x$:
$3x + 4x + 5x = 12$ $\Rightarrow 12x = 12$ $\Rightarrow x = 1$.
Thus,$p = 1, q = 1, r = 1$.
Now,calculate $p^3 + q^4 + r^5 = 1^3 + 1^4 + 1^5 = 1 + 1 + 1 = 3$.

(B) By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for positive numbers $a, b, c/2, c/2$:
$\frac{a + b + c/2 + c/2}{4} \geq \sqrt[4]{a \cdot b \cdot \frac{c}{2} \cdot \frac{c}{2}}$
$\frac{a + b + c}{4} \geq \sqrt[4]{\frac{abc^2}{4}}$
Raising both sides to the power of $4$:
$\frac{(a + b + c)^4}{256} \geq \frac{abc^2}{4}$
$abc^2 \leq \frac{(a + b + c)^4}{64}$
Given the maximum value of $abc^2$ is $1/64$,we have $\frac{(a + b + c)^4}{64} = \frac{1}{64}$,which implies $a + b + c = 1$.
The equality holds when $a = b = c/2$.
Substituting $a = b = c/2$ into $a + b + c = 1$:
$c/2 + c/2 + c = 1 \implies 2c = 1 \implies c = 1/2$.
Then $a = b = 1/4$.

(A) Let $y = \frac{x^2 - 3x + 4}{x^2 + 3x + 4}$.
Multiplying both sides by $(x^2 + 3x + 4)$,we get $y(x^2 + 3x + 4) = x^2 - 3x + 4$.
Rearranging the terms,we get $(y - 1)x^2 + (3y + 3)x + (4y - 4) = 0$.
Since $x$ is a real number,the discriminant $D \geq 0$.
$D = (3y + 3)^2 - 4(y - 1)(4y - 4) \geq 0$.
$9(y + 1)^2 - 16(y - 1)^2 \geq 0$.
$(3(y + 1) - 4(y - 1))(3(y + 1) + 4(y - 1)) \geq 0$.
$(3y + 3 - 4y + 4)(3y + 3 + 4y - 4) \geq 0$.
$(-y + 7)(7y - 1) \geq 0$.
$(y - 7)(7y - 1) \leq 0$.
Thus,$\frac{1}{7} \leq y \leq 7$.
The maximum value is $7$ and the minimum value is $\frac{1}{7}$.

(D) Given that $\alpha + \beta$,$\alpha^2 + \beta^2$,and $\alpha^3 + \beta^3$ are in a geometric progression $(GP)$.
For three terms $x, y, z$ to be in $GP$,$y^2 = xz$.
Therefore,$(\alpha^2 + \beta^2)^2 = (\alpha + \beta)(\alpha^3 + \beta^3)$.
Expanding both sides: $\alpha^4 + \beta^4 + 2\alpha^2\beta^2 = \alpha^4 + \alpha\beta^3 + \beta\alpha^3 + \beta^4$.
Simplifying: $2\alpha^2\beta^2 = \alpha\beta(\alpha^2 + \beta^2)$.
Rearranging: $\alpha\beta(\alpha^2 + \beta^2 - 2\alpha\beta) = 0$.
This gives $\alpha\beta(\alpha - \beta)^2 = 0$.
Since $\alpha\beta = \frac{c}{a}$ and $(\alpha - \beta)^2 = \frac{(\alpha + \beta)^2 - 4\alpha\beta}{1} = \frac{b^2 - 4ac}{a^2} = \frac{\Delta}{a^2}$.
Substituting these values: $\frac{c}{a} \times \frac{\Delta}{a^2} = 0$.
Thus,$c\Delta = 0$.

(D) $1$. The graph is a downward-opening parabola,which implies that the coefficient of $x^2$ must be negative,so $a < 0$.
$2$. The graph intersects the $x$-axis at two distinct points $(x_1, 0)$ and $(x_2, 0)$,which means the quadratic equation $ax^2 + bx + c = 0$ has two distinct real roots. Therefore,the discriminant $D = b^2 - 4ac > 0$.
$3$. The graph intersects the $y$-axis below the origin,which means the $y$-intercept $c$ is negative,so $c < 0$.
$4$. Comparing these findings with the given options,none of the conditions $a > 0$,$b^2 - 4ac < 0$,or $c > 0$ are satisfied. Thus,the correct option is $D$.

(B) Let $f(x) = a^2x^2 + bx + 1$.
For $f(x) > 0$ for all $x \in R$,the parabola must open upwards and lie entirely above the $x$-axis.
This requires the coefficient of $x^2$ to be positive,i.e.,$a^2 > 0$ (which is true for $a \neq 0$),and the discriminant $D < 0$.
The discriminant $D = b^2 - 4(a^2)(1) = b^2 - 4a^2$.
Setting $D < 0$,we get $b^2 - 4a^2 < 0$,which implies $b^2 < 4a^2$.

(D) The given equations can be rewritten as:
$x^2 + 2bx + b^2 = 1 \implies (x + b)^2 = 1 \implies x = -b \pm 1$
$x^2 + 2ax + a^2 = 1 \implies (x + a)^2 = 1 \implies x = -a \pm 1$
Since each equation has exactly one root,this implies the discriminant is zero,but here the equations are already perfect squares. The condition that they share a root means one of the roots from the first set must equal one of the roots from the second set.
Given $a \neq b$,the roots are $\{-b+1, -b-1\}$ and $\{-a+1, -a-1\}$.
For them to share a root,we must have:
$-b + 1 = -a - 1 \implies a - b = -2$
$OR$
$-b - 1 = -a + 1 \implies a - b = 2$
Both cases are covered by the condition $|a - b| = 2$.
Since $|a - b| = 2$ implies $a - b = 2$ or $a - b = -2$,and $a - b + 2 = 0$ is equivalent to $a - b = -2$,all options $A, B,$ and $C$ are correct.

(B) Let $y = \frac{1 - x + x^2}{1 + x + x^2}$.
We can rewrite this as $y = \frac{(1 + x + x^2) - 2x}{1 + x + x^2} = 1 - \frac{2x}{1 + x + x^2}$.
For $x = 0$,$y = 1$.
For $x \neq 0$,$y = 1 - \frac{2}{\frac{1}{x} + 1 + x}$.
By $AM$-$GM$ inequality,for $x > 0$,$\frac{1}{x} + x \geq 2$,so $\frac{1}{x} + 1 + x \geq 3$.
Thus,$\frac{2}{\frac{1}{x} + 1 + x} \leq \frac{2}{3}$.
Therefore,$y = 1 - \frac{2}{\frac{1}{x} + 1 + x} \geq 1 - \frac{2}{3} = \frac{1}{3}$.
For $x < 0$,let $x = -t$ where $t > 0$. Then $y = \frac{1 + t + t^2}{1 - t + t^2} = \frac{1}{\frac{1 - t + t^2}{1 + t + t^2}}$.
Since the minimum value of the expression is $1/3$,the maximum value is $3$. Thus $y \geq 1/3$ for all $x$.
The minimum value is $1/3$.

(D) Given $p^2 + q^2 = 1$.
We know that $(p + q)^2 = p^2 + q^2 + 2pq = 1 + 2pq$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,$\frac{p^2 + q^2}{2} \geq \sqrt{p^2 q^2} = pq$.
Substituting the given value,$\frac{1}{2} \geq pq$.
Therefore,$(p + q)^2 = 1 + 2pq \leq 1 + 2(\frac{1}{2}) = 1 + 1 = 2$.
Taking the square root on both sides,$p + q \leq \sqrt{2}$.
Thus,the maximum value of $(p + q)$ is $\sqrt{2}$.

(C) Let $y = \frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$.
Rearranging the equation,we get $y(3x^2 + 9x + 7) = 3x^2 + 9x + 17$.
$3x^2(y - 1) + 9x(y - 1) + 7y - 17 = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = [9(y - 1)]^2 - 4(3(y - 1))(7y - 17) \geq 0$.
$81(y - 1)^2 - 12(y - 1)(7y - 17) \geq 0$.
Dividing by $3(y - 1)$ (assuming $y \neq 1$),we get $27(y - 1) - 4(7y - 17) \geq 0$.
$27y - 27 - 28y + 68 \geq 0$.
$-y + 41 \geq 0 \Rightarrow y \leq 41$.
Also,checking the denominator $3x^2 + 9x + 7$,its discriminant is $81 - 4(3)(7) = 81 - 84 = -3 < 0$,so the denominator is always positive.
Thus,the maximum value of $y$ is $41$.

(A) Let $t = x - [x] = \{x\}$. Since $0 \leq \{x\} < 1$,we have $0 \leq t < 1$.
The equation becomes $-3t^2 + 2t + a^2 = 0$,or $3t^2 - 2t - a^2 = 0$.
Solving for $t$ using the quadratic formula: $t = \frac{2 \pm \sqrt{4 - 4(3)(-a^2)}}{2(3)} = \frac{2 \pm \sqrt{4 + 12a^2}}{6} = \frac{1 \pm \sqrt{1 + 3a^2}}{3}$.
Since $t \geq 0$,we must take the positive root: $t = \frac{1 + \sqrt{1 + 3a^2}}{3}$.
For the equation to have no integral solution,$t$ must not be $0$ (because if $t=0$,then $x$ is an integer,which would be a solution). Thus,$t > 0$.
Also,we require $t < 1$ for a valid fractional part:
$\frac{1 + \sqrt{1 + 3a^2}}{3} < 1$ $\Rightarrow 1 + \sqrt{1 + 3a^2} < 3$ $\Rightarrow \sqrt{1 + 3a^2} < 2$.
Squaring both sides: $1 + 3a^2 < 4$ $\Rightarrow 3a^2 < 3$ $\Rightarrow a^2 < 1$.
This implies $-1 < a < 1$. Since $t > 0$,we check $a=0$: if $a=0$,$t = 1/3$,which is valid. However,if $a=0$,the equation is $-3t^2 + 2t = 0 \Rightarrow t(2-3t)=0$,so $t=0$ or $t=2/3$. $t=0$ gives an integral solution. Thus $a \neq 0$.
Therefore,$a \in (-1, 0) \cup (0, 1)$.

(D) Given $p^{2} + q^{2} = 1$ where $p, q > 0$.
We know the identity $(p+q)^{2} = p^{2} + q^{2} + 2pq$.
Substituting the given value: $(p+q)^{2} = 1 + 2pq$.
By the $AM \geq GM$ inequality,for positive real numbers $p^{2}$ and $q^{2}$,we have $\frac{p^{2} + q^{2}}{2} \geq \sqrt{p^{2}q^{2}} = pq$.
Since $p^{2} + q^{2} = 1$,we get $\frac{1}{2} \geq pq$,which implies $2pq \leq 1$.
Substituting this into the identity: $(p+q)^{2} = 1 + 2pq \leq 1 + 1 = 2$.
Taking the square root on both sides,we get $p+q \leq \sqrt{2}$.
Thus,the maximum value of $(p+q)$ is $\sqrt{2}$.

(A) Given $x = {(\sqrt{2} + 1)^{1/3}} - {(\sqrt{2} - 1)^{1/3}}$.
Using the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$,we have:
${x^3} = (\sqrt{2} + 1) - (\sqrt{2} - 1) - 3{(\sqrt{2} + 1)^{1/3}} \cdot {(\sqrt{2} - 1)^{1/3}} \cdot x$.
Since ${(\sqrt{2} + 1)^{1/3}} \cdot {(\sqrt{2} - 1)^{1/3}} = {((\sqrt{2} + 1)(\sqrt{2} - 1))^{1/3}} = {(2 - 1)^{1/3}} = 1^{1/3} = 1$,
we get ${x^3} = 2 - 3(1)x$.
Therefore,${x^3} + 3x = 2$.

(A) Let $f(x) = \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}}$.
Since the term under the square root must be non-negative,$x - 1 \ge 0$,so $x \ge 1$.
We can rewrite the expression as $\sqrt{(\sqrt{x-1})^2 + 1 + 2\sqrt{x-1}} + \sqrt{(\sqrt{x-1})^2 + 1 - 2\sqrt{x-1}}$.
This simplifies to $\sqrt{(\sqrt{x-1} + 1)^2} + \sqrt{(\sqrt{x-1} - 1)^2}$.
This is equal to $|\sqrt{x-1} + 1| + |\sqrt{x-1} - 1|$.
Since $\sqrt{x-1} + 1$ is always positive for $x \ge 1$,we have $(\sqrt{x-1} + 1) + |\sqrt{x-1} - 1|$.
Case $1$: If $1 \le x \le 2$,then $0 \le \sqrt{x-1} \le 1$,so $|\sqrt{x-1} - 1| = 1 - \sqrt{x-1}$.
Thus,$f(x) = \sqrt{x-1} + 1 + 1 - \sqrt{x-1} = 2$.
Case $2$: If $x > 2$,then $\sqrt{x-1} > 1$,so $|\sqrt{x-1} - 1| = \sqrt{x-1} - 1$.
Thus,$f(x) = \sqrt{x-1} + 1 + \sqrt{x-1} - 1 = 2\sqrt{x-1}$.

(A) Given $\alpha + \beta = p$ and $\alpha\beta = q$.
Let the roots of the required equation be $A$ and $B$.
$A = (\alpha^2 - \beta^2)(\alpha^3 - \beta^3) = (\alpha - \beta)(\alpha + \beta)(\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) = (\alpha - \beta)^2(\alpha + \beta)(\alpha^2 + \alpha\beta + \beta^2)$.
Since $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = p^2 - 4q$ and $\alpha^2 + \beta^2 + \alpha\beta = (\alpha + \beta)^2 - \alpha\beta = p^2 - q$,we have $A = (p^2 - 4q)p(p^2 - q) = p(p^4 - 5p^2q + 4q^2)$.
$B = \alpha^3\beta^2 + \alpha^2\beta^3 = \alpha^2\beta^2(\alpha + \beta) = q^2p$.
Sum of roots $S = A + B = p(p^4 - 5p^2q + 4q^2) + pq^2 = p(p^4 - 5p^2q + 5q^2)$.
Product of roots $P = A \times B = p(p^4 - 5p^2q + 4q^2) \times pq^2 = p^2q^2(p^4 - 5p^2q + 4q^2)$.
The quadratic equation is $x^2 - Sx + P = 0$.

(D) Given the quadratic equation $x^2 + 6x - 2 = 0$. Since $\alpha$ is a root,we have $\alpha^2 + 6\alpha - 2 = 0$,which implies $\alpha^2 = 2 - 6\alpha$.
For option $A$: $\alpha^2 + 5\alpha - 8 = (2 - 6\alpha) + 5\alpha - 8 = -\alpha - 6$. Since $\alpha + \beta = -6$,we have $\beta = -6 - \alpha$. Thus,option $A$ is correct.
For option $C$: $\frac{2\alpha^2 + 12\alpha - 6}{\alpha} = \frac{2(\alpha^2 + 6\alpha) - 6}{\alpha} = \frac{2(2) - 6}{\alpha} = \frac{-2}{\alpha}$. Since $\alpha\beta = -2$,we have $\beta = \frac{-2}{\alpha}$. Thus,option $C$ is correct.
For option $B$: From $\alpha\beta = -2$ and $\alpha + \beta = -6$,we have $\frac{\alpha + \beta}{\alpha\beta} = \frac{-6}{-2} = 3$,so $\frac{1}{\beta} + \frac{1}{\alpha} = 3$. This gives $\frac{\alpha + \beta}{\alpha\beta} = 3 \Rightarrow \frac{1}{\beta} = 3 - \frac{1}{\alpha} = \frac{3\alpha - 1}{\alpha}$,so $\beta = \frac{\alpha}{3\alpha - 1}$. Thus,option $B$ is correct.
Since all options are equivalent to $\beta$,the correct answer is $D$.

(C) $1$. Since the parabola opens downwards,the coefficient of $x^2$ must be negative,so $a < 0$.
$2$. The curve intersects the $y$-axis at $(0, c)$. From the figure,the intersection point is above the $x$-axis,so $c > 0$.
$3$. The vertex of the parabola is at $x = -b/(2a)$. From the figure,the vertex is to the left of the $y$-axis,so $-b/(2a) < 0$. Since $a < 0$,we have $-b/(2a) < 0 \Rightarrow b/a > 0$. Since $a < 0$,it follows that $b < 0$.
$4$. The curve intersects the $x$-axis at two distinct points,so the discriminant $D = b^2 - 4ac > 0$.
Comparing these with the options,$c > 0$ is correct.

(A) Let $y = x - 1$. Then $x = y + 1$. Since $x \ge 1$,$y \ge 0$.
Substituting this into the expression: $\sqrt{y + 1 + 2\sqrt{y}} + \sqrt{y + 1 - 2\sqrt{y}}$.
This simplifies to $\sqrt{(\sqrt{y} + 1)^2} + \sqrt{(\sqrt{y} - 1)^2}$.
$= |\sqrt{y} + 1| + |\sqrt{y} - 1|$.
Since $\sqrt{y} + 1$ is always positive,this is $\sqrt{y} + 1 + |\sqrt{y} - 1|$.
Case $1$: If $0 \le y \le 1$ (i.e.,$1 \le x \le 2$),then $|\sqrt{y} - 1| = 1 - \sqrt{y}$.
Expression $= \sqrt{y} + 1 + 1 - \sqrt{y} = 2$.
Case $2$: If $y > 1$ (i.e.,$x > 2$),then $|\sqrt{y} - 1| = \sqrt{y} - 1$.
Expression $= \sqrt{y} + 1 + \sqrt{y} - 1 = 2\sqrt{y} = 2\sqrt{x - 1}$.

(B) From the given quadratic equation,the sum of roots is $\sin \theta + \cos \theta = -(a - 1) = 1 - a$ .....$(1)$
The product of roots is $\sin \theta \cdot \cos \theta = \frac{1 - 2a}{4}$ .....$(2)$
Squaring equation $(1)$,we get $1 + 2 \sin \theta \cos \theta = (1 - a)^2$.
Substituting $(2)$ into this,$1 + 2(\frac{1 - 2a}{4}) = (1 - a)^2 \Rightarrow 1 + \frac{1 - 2a}{2} = 1 - 2a + a^2$.
Simplifying,$2 + 1 - 2a = 2 - 4a + 2a^2 \Rightarrow 2a^2 - 2a - 1 = 0$.
Solving for $a$,$a = \frac{2 \pm \sqrt{4 - 4(2)(-1)}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{1 \pm \sqrt{3}}{2}$.
Since $\sin \theta + \cos \theta = 1 - a > 0$ for $\theta \in (0, \frac{\pi}{2})$,we must have $1 - a > 0$,so $a < 1$. Thus,$a = \frac{1 - \sqrt{3}}{2}$.
Now,$\sin \theta \cos \theta = \frac{1 - 2(\frac{1 - \sqrt{3}}{2})}{4} = \frac{1 - 1 + \sqrt{3}}{4} = \frac{\sqrt{3}}{4}$.
So,$\sin 2\theta = 2 \sin \theta \cos \theta = \frac{\sqrt{3}}{2}$.
This gives $2\theta = 60^{\circ}$ or $120^{\circ}$,so $\theta = 30^{\circ}$ or $60^{\circ}$.
If $\theta = 30^{\circ}$,$\sin \theta = \frac{1}{2}$. If $\theta = 60^{\circ}$,$\sin \theta = \frac{\sqrt{3}}{2}$.
Maximum value of $a + \sin \theta = \frac{1 - \sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{1}{2}$.

(B) For the roots of $x^2 + bx + c = 0$ to be positive and distinct,we need:
$1$. Discriminant $D > 0$
$2$. Sum of roots $-b > 0$
$3$. Product of roots $c > 0$
Given $x^2 + (\sin \theta + \cos \theta)x + \frac{3}{8} = 0$:
Product of roots = $\frac{3}{8} > 0$ (Always true).
Sum of roots = $-(\sin \theta + \cos \theta) > 0 \implies \sin \theta + \cos \theta < 0$.
Discriminant $D = (\sin \theta + \cos \theta)^2 - 4(\frac{3}{8}) > 0$
$(\sin \theta + \cos \theta)^2 > \frac{3}{2}$
$1 + \sin 2\theta > \frac{3}{2} \implies \sin 2\theta > \frac{1}{2}$.
Solving $\sin 2\theta > \frac{1}{2}$ for $2\theta \in [0, 4\pi]$:
$\frac{\pi}{6} < 2\theta < \frac{5\pi}{6}$ or $\frac{13\pi}{6} < 2\theta < \frac{17\pi}{6}$.
$\frac{\pi}{12} < \theta < \frac{5\pi}{12}$ or $\frac{13\pi}{12} < \theta < \frac{17\pi}{12}$.
Check condition $\sin \theta + \cos \theta < 0$:
For $\theta \in (\frac{\pi}{12}, \frac{5\pi}{12})$,$\sin \theta + \cos \theta > 0$ (Both positive).
For $\theta \in (\frac{13\pi}{12}, \frac{17\pi}{12})$,$\sin \theta + \cos \theta < 0$ (Both negative).
Thus,the solution is $\theta \in (\frac{13\pi}{12}, \frac{17\pi}{12})$.

(D) For $x^2 - (a - 1)x + 3 = 0$ to have both roots positive:
$1$. Discriminant $D_1 \ge 0$ $\Rightarrow (a - 1)^2 - 12 \ge 0$ $\Rightarrow (a - 1)^2 \ge 12$ $\Rightarrow a - 1 \ge 2\sqrt{3}$ or $a - 1 \le -2\sqrt{3}$. Since $2\sqrt{3} \approx 3.46$,$a \ge 4.46$ or $a \le -2.46$.
$2$. Sum of roots $> 0$ $\Rightarrow a - 1 > 0$ $\Rightarrow a > 1$.
$3$. Product of roots $> 0 \Rightarrow 3 > 0$ (always true).
Combining these,$a \ge 5$ (for integral values).
For $x^2 + 3x + 6 - a = 0$ to have both roots negative:
$1$. Discriminant $D_2 \ge 0$ $\Rightarrow 3^2 - 4(6 - a) \ge 0$ $\Rightarrow 9 - 24 + 4a \ge 0$ $\Rightarrow 4a \ge 15$ $\Rightarrow a \ge 3.75$.
$2$. Sum of roots $< 0 \Rightarrow -3 < 0$ (always true).
$3$. Product of roots $> 0$ $\Rightarrow 6 - a > 0$ $\Rightarrow a < 6$.
Combining these,$3.75 \le a < 6$.
Intersection of both conditions: $a \in [5, 6)$,so $a = 5$.
Thus,there is only $1$ integral value of $a$.

(B) Given the inequality $256abcd \geq (a+b+c+d)^4$,we can rewrite it as $(abcd)^{\frac{1}{4}} \geq \frac{a+b+c+d}{4}$.
This is the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,which holds with equality if and only if $a = b = c = d$.
Given $3a + b + 2c + 5d = 11$,and substituting $a = b = c = d = k$,we get $3k + k + 2k + 5k = 11$.
$11k = 11 \Rightarrow k = 1$.
Thus,$a = b = c = d = 1$.
Substituting these values into the expression $a^3 + b + c^2 + 5d$,we get $1^3 + 1 + 1^2 + 5(1) = 1 + 1 + 1 + 5 = 8$.

(D) For the quadratic equation $(a + b)x^2 + 2bx + 1 = 0$ to have real and distinct roots,the discriminant $D$ must be greater than $0$.
$D = (2b)^2 - 4(a + b)(1) > 0$
$4b^2 - 4a - 4b > 0$
$b^2 - b > a$
Let $f(x, y) = y - (x^2 - x)$. For the point $(b, a)$,we evaluate $f(b, a) = a - (b^2 - b) = a - b^2 + b$.
Since $b^2 - b > a$,it follows that $a - b^2 + b < 0$.
This implies that the point $(b, a)$ lies inside the parabola $y = x^2 - x$ (or below the curve $y = x^2 - x$).
However,checking the options provided,the condition $a < b^2 - b$ corresponds to the region inside the parabola $y = x^2 - x$.

(D) Given the equation $x^8 - kx^2 + 3 = 0$,we can rewrite it as $k = \frac{x^8 + 3}{x^2} = x^6 + \frac{3}{x^2}$.
To find the minimum value of $k$,we use the $A.M. \geq G.M.$ inequality.
We can express $x^6 + \frac{3}{x^2}$ as $x^6 + \frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}$.
Applying $A.M. \geq G.M.$ for these four terms:
$\frac{x^6 + \frac{1}{x^2} + \frac{1}{x^2} + \frac{1}{x^2}}{4} \geq \sqrt[4]{x^6 \cdot \frac{1}{x^2} \cdot \frac{1}{x^2} \cdot \frac{1}{x^2}} = \sqrt[4]{1} = 1$.
Thus,$x^6 + \frac{3}{x^2} \geq 4$.
Therefore,the minimum value of $k$ is $4$.

(C) Let $\alpha$ and $\beta$ be the roots of the 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:
$\alpha \beta + \alpha + \beta + 1 = -2$
$(\alpha + 1)(\beta + 1) = -2$
Since $\alpha, \beta$ are integers,the possible pairs for $(\alpha + 1, \beta + 1)$ are $(1, -2), (-2, 1), (-1, 2), (2, -1)$.
Case $1$: $(\alpha + 1) = 1, (\beta + 1) = -2 \Rightarrow \alpha = 0, \beta = -3$. Then $\tan \theta = \alpha + \beta + 2 = 0 - 3 + 2 = -1$.
Case $2$: $(\alpha + 1) = -1, (\beta + 1) = 2 \Rightarrow \alpha = -2, \beta = 1$. Then $\tan \theta = \alpha + \beta + 2 = -2 + 1 + 2 = 1$.
For $\tan \theta = -1$ in $(0, 2\pi)$,$\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$.
For $\tan \theta = 1$ in $(0, 2\pi)$,$\theta = \frac{\pi}{4}, \frac{5\pi}{4}$.
The sum of all possible values of $\theta$ is $\frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi$.
Thus,$k = 4$.

(A) Since $\alpha$ and $\beta$ are roots of $x^2 + \alpha x + \beta = 0$,by Vieta's formulas:
$\alpha + \beta = -\alpha \implies \beta = -2\alpha$ .......$(1)$
$\alpha \beta = \beta$ .......$(2)$
From $(2)$,$\beta(\alpha - 1) = 0$. Since $\alpha \neq \beta$,$\beta$ cannot be $0$ (if $\beta=0$,then $\alpha=0$,but $\alpha \neq \beta$). Thus,$\alpha = 1$.
Substituting $\alpha = 1$ into $(1)$,we get $\beta = -2$.
Now,the inequality is $||y - (-2)| - 1| < 1$,which simplifies to $||y + 2| - 1| < 1$.
This implies $-1 < |y + 2| - 1 < 1$,so $0 < |y + 2| < 2$.
This means $-2 < y + 2 < 2$ and $y + 2 \neq 0$,so $-4 < y < 0$ and $y \neq -2$.
Thus,$y \in (-4, -2) \cup (-2, 0)$.
The integral values of $y$ are $-3$ and $-1$,which are exactly two values. Therefore,option $A$ is correct.

(A) Let the expression be $f(x, y) = \frac{x^2y^2 - 2x^2y + 2x^2 + 2xy - 2x + 1}{x^2y + x}$.
We can rewrite the numerator as $(xy + 1)^2 - 2x(xy + 1) + 2x^2$.
Thus,$f(x, y) = \frac{(xy + 1)^2 - 2x(xy + 1) + 2x^2}{x(xy + 1)}$.
Dividing each term by $x(xy + 1)$,we get $f(x, y) = \frac{xy + 1}{x} - 2 + \frac{2x}{xy + 1}$.
Let $u = \frac{xy + 1}{x}$. Since $x, y > 0$,$u = y + \frac{1}{x} > 0$.
Then $f(x, y) = u + \frac{2}{u} - 2$.
By the Arithmetic Mean-Geometric Mean Inequality $(AM \geq GM)$,$u + \frac{2}{u} \geq 2\sqrt{u \cdot \frac{2}{u}} = 2\sqrt{2}$.
Therefore,the minimum value $\lambda = 2\sqrt{2} - 2$.
Since $\sqrt{2} \approx 1.414$,$\lambda \approx 2(1.414) - 2 = 2.828 - 2 = 0.828$.
Thus,$0 < \lambda < 1$,which means $\lambda \in (0, 1)$.

(D) We need to solve the equation $|x - 2| + |x - 1| = x - 3$.
Case $1$: $x < 1$.
In this interval,$|x - 2| = -(x - 2) = -x + 2$ and $|x - 1| = -(x - 1) = -x + 1$.
The equation becomes $(-x + 2) + (-x + 1) = x - 3$,which simplifies to $-2x + 3 = x - 3$.
Solving for $x$,we get $3x = 6$,so $x = 2$.
However,$x = 2$ is not in the interval $x < 1$,so there is no solution in this case.
Case $2$: $1 \le x < 2$.
In this interval,$|x - 2| = -(x - 2) = -x + 2$ and $|x - 1| = x - 1$.
The equation becomes $(-x + 2) + (x - 1) = x - 3$,which simplifies to $1 = x - 3$.
Solving for $x$,we get $x = 4$.
However,$x = 4$ is not in the interval $1 \le x < 2$,so there is no solution in this case.
Case $3$: $x \ge 2$.
In this interval,$|x - 2| = x - 2$ and $|x - 1| = x - 1$.
The equation becomes $(x - 2) + (x - 1) = x - 3$,which simplifies to $2x - 3 = x - 3$.
Solving for $x$,we get $x = 0$.
However,$x = 0$ is not in the interval $x \ge 2$,so there is no solution in this case.
Since no case yields a valid solution,the equation has no solution.

(A) Let the roots of the first equation $x^2 + 2bx + b = 0$ be $x_1 = \cos^4 \theta + \alpha$ and $x_2 = \sin^4 \theta + \alpha$. The difference of the roots is $|x_1 - x_2| = \sqrt{D_1}/a_1 = \sqrt{(2b)^2 - 4b} = \sqrt{4b^2 - 4b}$.
Let the roots of the second equation $x^2 + 4x + 2 = 0$ be $y_1 = \cos^2 \theta + \beta$ and $y_2 = \sin^2 \theta + \beta$. The difference of the roots is $|y_1 - y_2| = \sqrt{D_2}/a_2 = \sqrt{4^2 - 4(2)} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.
Note that $|x_1 - x_2| = |\cos^4 \theta - \sin^4 \theta| = |(\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta)| = |\cos^2 \theta - \sin^2 \theta|$.
Also,$|y_1 - y_2| = |\cos^2 \theta - \sin^2 \theta|$.
Since the differences are equal,$\sqrt{4b^2 - 4b} = \sqrt{8}$.
Squaring both sides,$4b^2 - 4b = 8$,which simplifies to $b^2 - b - 2 = 0$.
Factoring the quadratic,$(b - 2)(b + 1) = 0$,so $b = 2$ or $b = -1$.
For the first equation $x^2 + 2bx + b = 0$ to have real roots,the discriminant $D = 4b^2 - 4b \ge 0$,which is satisfied for both $b=2$ and $b=-1$. However,checking the context of such problems,$b=2$ is the standard solution.

(NONE) Given the quadratic equation $3x^2 - 10x - 25 = 0$ with roots $\tan A$ and $\tan B$.
From the properties of roots,$\tan A + \tan B = \frac{10}{3}$ and $\tan A \tan B = -\frac{25}{3}$.
Using the formula $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan (A + B) = \frac{10/3}{1 - (-25/3)} = \frac{10/3}{28/3} = \frac{10}{28} = \frac{5}{14}$.
Let $S = 3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$.
Divide the entire expression by $\cos^2 (A + B)$:
$S = \cos^2 (A + B) [3 \tan^2 (A + B) - 10 \tan (A + B) - 25]$.
Since $\tan (A + B) = 5/14$,the term in the bracket is $3(5/14)^2 - 10(5/14) - 25 = 3(25/196) - 50/14 - 25 = 75/196 - 700/196 - 4900/196 = -5525/196$.
Alternatively,note that the expression is of the form $f(\tan(A+B)) \cdot \cos^2(A+B)$ where $f(x) = 3x^2 - 10x - 25$. Since $\tan(A+B) = 5/14$ is a root of $3x^2 - 10x - 25 = 0$,the value of the expression is $0$.

(B) Given the quadratic equation $9x^2 + 27x + 20 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-27 \pm \sqrt{27^2 - 4 \times 9 \times 20}}{2 \times 9} = \frac{-27 \pm \sqrt{729 - 720}}{18} = \frac{-27 \pm 3}{18}$.
Thus,the roots are $x_1 = \frac{-24}{18} = -\frac{4}{3}$ and $x_2 = \frac{-30}{18} = -\frac{5}{3}$.
Given $5 \cos A + 3 = 0$,we have $\cos A = -\frac{3}{5}$.
Then $\sec A = \frac{1}{\cos A} = -\frac{5}{3}$.
Since $\cos A$ is negative,$A$ is an obtuse angle in the second quadrant,so $\tan A$ is negative.
$\tan A = -\sqrt{\sec^2 A - 1} = -\sqrt{(-\frac{5}{3})^2 - 1} = -\sqrt{\frac{25}{9} - 1} = -\sqrt{\frac{16}{9}} = -\frac{4}{3}$.
Comparing the roots,the roots are $\sec A$ and $\tan A$.

(C) Given that $1$ is a root of $ax^2 + bx + c = 0$,we have $a(1)^2 + b(1) + c = 0$,which implies $a + b + c = 0$,or $c = -a - b$.
We want to find the intersection of $y = 4ax^2 + 3bx + 2c$ with the $x$-axis,so we set $y = 0$:
$4ax^2 + 3bx + 2c = 0$.
Substitute $c = -a - b$ into the equation:
$4ax^2 + 3bx + 2(-a - b) = 0$
$4ax^2 + 3bx - 2a - 2b = 0$
Rearranging the terms:
$4ax^2 - 2a + 3bx - 2b = 0$
$2a(2x^2 - 1) + b(3x - 2) = 0$.
Alternatively,using the condition $a+b+c=0$,we can write $c = -(a+b)$.
$y = 4ax^2 + 3bx + 2(-a-b) = 4ax^2 + 3bx - 2a - 2b$.
At $x=1$,$y = 4a + 3b - 2a - 2b = 2a + b$. This is not necessarily zero.
Let us re-evaluate the intersection $4ax^2 + 3bx + 2c = 0$ with $c = -a-b$:
$4ax^2 + 3bx - 2a - 2b = 0$.
For this to be a quadratic equation in $x$,the discriminant $D = (3b)^2 - 4(4a)(-2a-2b) = 9b^2 + 16a(2a+2b) = 9b^2 + 32a^2 + 32ab$.
This discriminant is not necessarily positive,zero,or negative without further constraints on $a$ and $b$. However,if we test specific values,e.g.,$a=1, b=-2, c=1$ (where $1-2+1=0$),then $y = 4x^2 - 6x + 2 = 2(2x^2 - 3x + 1) = 2(2x-1)(x-1)$. This intersects at $x=1$ and $x=1/2$ (two points).
If $a=1, b=-1, c=0$,then $y = 4x^2 - 3x = x(4x-3)$. This intersects at $x=0$ and $x=3/4$ (two points).
Thus,the curve intersects the $x$-axis at exactly two distinct points.

(D) Given the quadratic equation $(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$.
Sum of roots $\alpha + \beta = \frac{3}{m^2 + 1}$.
Product of roots $\alpha \beta = \frac{(m^2 + 1)^2}{m^2 + 1} = m^2 + 1$.
For the sum of roots to be maximum,the denominator $(m^2 + 1)$ must be minimum.
The minimum value of $m^2 + 1$ is $1$ (at $m = 0$).
Thus,$\alpha + \beta = \frac{3}{1} = 3$ and $\alpha \beta = 0^2 + 1 = 1$.
The absolute difference of the cubes of the roots is $|\alpha^3 - \beta^3| = |(\alpha - \beta)(\alpha^2 + \alpha \beta + \beta^2)|$.
We know $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = 3^2 - 4(1) = 9 - 4 = 5$,so $|\alpha - \beta| = \sqrt{5}$.
Also,$\alpha^2 + \alpha \beta + \beta^2 = (\alpha + \beta)^2 - \alpha \beta = 3^2 - 1 = 8$.
Therefore,$|\alpha^3 - \beta^3| = \sqrt{5} \times 8 = 8\sqrt{5}$.

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

(C) Given $\alpha, \beta$ are roots of $x^{2}+px+2=0$,so $\alpha+\beta = -p$ and $\alpha\beta = 2$.
The roots of $2x^{2}+2qx+1=0$ are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
Sum of roots: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-p}{2} = -q \Rightarrow p = 2q$.
Product of roots: $\frac{1}{\alpha\beta} = \frac{1}{2}$,which is consistent.
We need to evaluate $E = \left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$.
$E = \left(\frac{\alpha^{2}-1}{\alpha}\right)\left(\frac{\beta^{2}-1}{\beta}\right)\left(\frac{\alpha\beta+1}{\beta}\right)\left(\frac{\alpha\beta+1}{\alpha}\right) = \frac{(\alpha^{2}-1)(\beta^{2}-1)(\alpha\beta+1)^{2}}{(\alpha\beta)^{2}}$.
Since $\alpha^{2} = -p\alpha-2$ and $\beta^{2} = -p\beta-2$,then $\alpha^{2}-1 = -p\alpha-3$ and $\beta^{2}-1 = -p\beta-3$.
$E = \frac{(-p\alpha-3)(-p\beta-3)(2+1)^{2}}{2^{2}} = \frac{(p\alpha+3)(p\beta+3)(9)}{4} = \frac{9}{4}(p^{2}\alpha\beta + 3p(\alpha+\beta) + 9)$.
Substituting $\alpha\beta=2$ and $\alpha+\beta=-p$:
$E = \frac{9}{4}(2p^{2} - 3p^{2} + 9) = \frac{9}{4}(9-p^{2})$.