Mix Examples-Quadratic Equations and Inequations Questions in English

(C) Given the roots of $x^{2}-3x+p=0$ are $\alpha$ and $\beta$,and the roots of $x^{2}-6x+q=0$ are $\gamma$ and $\delta$.
Since $\alpha, \beta, \gamma, \delta$ are in geometric progression,let them be $a, ar, ar^{2}, ar^{3}$.
From the first equation,$\alpha+\beta = a+ar = 3$ and $\alpha\beta = a^{2}r = p$.
From the second equation,$\gamma+\delta = ar^{2}+ar^{3} = 6$ and $\gamma\delta = a^{2}r^{5} = q$.
Dividing the sum of roots of the second equation by the sum of roots of the first equation: $\frac{ar^{2}(1+r)}{a(1+r)} = \frac{6}{3} \implies r^{2} = 2$.
Now,calculate $p$ and $q$ in terms of $a$ and $r$: $p = a^{2}r$ and $q = a^{2}r^{5} = a^{2}r(r^{2})^{2} = p(2)^{2} = 4p$.
The ratio $\frac{2q+p}{2q-p} = \frac{2(4p)+p}{2(4p)-p} = \frac{8p+p}{8p-p} = \frac{9p}{7p} = \frac{9}{7}$.

(C) Given the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$,where $x \in \mathbb{R} \setminus \{1, 2\}$.
Simplifying the equation: $\frac{2(x-2)-(x-1)}{(x-1)(x-2)} = \frac{2}{k} \Rightarrow \frac{x-3}{x^2-3x+2} = \frac{2}{k}$.
This gives $k(x-3) = 2(x^2-3x+2)$.
If $x=3$,then $0 = 2(9-9+2) = 4$,which is impossible. Thus $x \neq 3$.
For $x \neq 3$,we have $k = \frac{2(x^2-3x+2)}{x-3} = 2 \left( \frac{x^2-3x+2}{x-3} \right) = 2 \left( x + \frac{2}{x-3} \right) = 2 \left( (x-3) + \frac{2}{x-3} + 3 \right)$.
Let $f(x) = (x-3) + \frac{2}{x-3} + 3$.
Using the $AM$-$GM$ inequality,for $x > 3$,$(x-3) + \frac{2}{x-3} \geq 2\sqrt{2}$,so $f(x) \geq 3 + 2\sqrt{2}$.
For $x < 3$,$(x-3) + \frac{2}{x-3} = -(|x-3| + \frac{2}{|x-3|}) \leq -2\sqrt{2}$,so $f(x) \leq 3 - 2\sqrt{2}$.
Thus,$k = 2f(x) \in (-\infty, 6-4\sqrt{2}] \cup [6+4\sqrt{2}, \infty)$.
The equation has no real roots if $k$ lies in the gap: $k \in (6-4\sqrt{2}, 6+4\sqrt{2})$.
Approximating $\sqrt{2} \approx 1.414$,$6-4(1.414) = 6-5.656 = 0.344$ and $6+4(1.414) = 6+5.656 = 11.656$.
So $k \in (0.344, 11.656)$.
The integral values of $k$ are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.
The sum of these values is $\frac{11 \times 12}{2} = 66$.

(C) For the quadratic expression $x^{2}+2(a+4)x-5a+64 > 0$ to be true for all $x \in \mathbb{R}$,its discriminant $D$ must be less than $0$.
$D = [2(a+4)]^{2} - 4(1)(-5a+64) < 0$
$4(a^{2}+8a+16) + 20a - 256 < 0$
$4a^{2} + 32a + 64 + 20a - 256 < 0$
$4a^{2} + 52a - 192 < 0$
Dividing by $4$,we get $a^{2} + 13a - 48 < 0$.
Factoring the quadratic,we get $(a+16)(a-3) < 0$.
This implies $a \in (-16, 3)$.
Since $a$ must be an integer in the range $[-5, 30]$,the possible values for $a$ are $\{-5, -4, -3, -2, -1, 0, 1, 2\}$.
The number of favorable values is $8$.
The total number of integers in the range $[-5, 30]$ is $30 - (-5) + 1 = 36$.
Therefore,the required probability is $\frac{8}{36} = \frac{2}{9}$.

(B) Given the equation $x^{2} + 3^{1/4}x + 3^{1/2} = 0$.
Since $\alpha$ is a root,$\alpha^{2} + 3^{1/2} = -3^{1/4}\alpha$.
Squaring both sides: $(\alpha^{2} + 3^{1/2})^{2} = (3^{1/4}\alpha)^{2} = 3^{1/2}\alpha^{2}$.
$\alpha^{4} + 2 \cdot 3^{1/2}\alpha^{2} + 3 = 3^{1/2}\alpha^{2}$.
$\alpha^{4} + 3^{1/2}\alpha^{2} + 3 = 0$.
Multiply by $(\alpha^{2} - 3^{1/2})$: $(\alpha^{2} - 3^{1/2})(\alpha^{4} + 3^{1/2}\alpha^{2} + 3) = 0$.
This is the form $(a-b)(a^{2}+ab+b^{2}) = a^{3}-b^{3}$,so $\alpha^{6} - (3^{1/2})^{3} = 0$.
$\alpha^{6} = 3^{3/2} = 3 \sqrt{3}$.
Then $\alpha^{12} = (3 \sqrt{3})^{2} = 9 \times 3 = 27 = 3^{3}$.
Since $\alpha^{12} = 27$,then $\alpha^{96} = (\alpha^{12})^{8} = (27)^{8} = (3^{3})^{8} = 3^{24}$.
Similarly,$\beta^{12} = 27$ and $\beta^{96} = 3^{24}$.
The expression is $\alpha^{96}(\alpha^{12}-1) + \beta^{96}(\beta^{12}-1) = 3^{24}(27-1) + 3^{24}(27-1)$.
$= 3^{24}(26) + 3^{24}(26) = 2 \times 26 \times 3^{24} = 52 \times 3^{24}$.

(B) For the equation $ax^{2}-2bx+15=0$,since it has a repeated root $\alpha$,the discriminant must be zero:
$D = (-2b)^{2} - 4(a)(15) = 0 \implies 4b^{2} = 60a \implies b^{2} = 15a$.
Also,the root $\alpha = -\frac{-2b}{2a} = \frac{b}{a}$.
Substituting $a = \frac{b^{2}}{15}$ into $\alpha = \frac{b}{a}$,we get $\alpha = \frac{b}{b^{2}/15} = \frac{15}{b}$.
Since $\alpha$ is a root of $x^{2}-2bx+21=0$,we have:
$(\frac{15}{b})^{2} - 2b(\frac{15}{b}) + 21 = 0$
$\frac{225}{b^{2}} - 30 + 21 = 0 \implies \frac{225}{b^{2}} = 9 \implies b^{2} = 25$.
For the equation $x^{2}-2bx+21=0$,the sum of roots $\alpha+\beta = 2b$ and the product $\alpha\beta = 21$.
We need to find $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta$.
$\alpha^{2}+\beta^{2} = (2b)^{2} - 2(21) = 4b^{2} - 42$.
Substituting $b^{2} = 25$:
$\alpha^{2}+\beta^{2} = 4(25) - 42 = 100 - 42 = 58$.

(C) Given $p+q=3$ and $p^{4}+q^{4}=369$.
We need to find $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2} = \left(\frac{p+q}{pq}\right)^{-2} = \frac{(pq)^{2}}{(p+q)^{2}} = \frac{(pq)^{2}}{3^{2}} = \frac{(pq)^{2}}{9}$.
We know that $p^{2}+q^{2} = (p+q)^{2}-2pq = 9-2pq$.
Also,$p^{4}+q^{4} = (p^{2}+q^{2})^{2}-2p^{2}q^{2} = (9-2pq)^{2}-2(pq)^{2} = 369$.
Expanding this,we get $81+4(pq)^{2}-36pq-2(pq)^{2} = 369$.
$2(pq)^{2}-36pq+81 = 369 \implies 2(pq)^{2}-36pq-288 = 0$.
Dividing by $2$,we get $(pq)^{2}-18pq-144 = 0$.
Factoring the quadratic equation,$(pq-24)(pq+6) = 0$.
So,$pq=24$ or $pq=-6$.
If $pq=24$,then $p^{2}+q^{2} = 9-2(24) = 9-48 = -39$,which is impossible for real numbers $p$ and $q$.
Thus,$pq=-6$.
Substituting $pq=-6$ into our expression: $\frac{(pq)^{2}}{9} = \frac{(-6)^{2}}{9} = \frac{36}{9} = 4$.

(D) Given equations are:
$x^{2}-4 \lambda x+5=0$ with roots $\alpha, \beta$ $\implies \alpha+\beta=4 \lambda$ and $\alpha \beta=5$.
$x^{2}-(3 \sqrt{2}+2 \sqrt{3}) x+(7+3 \lambda \sqrt{3})=0$ with roots $\alpha, \gamma$ $\implies \alpha+\gamma=3 \sqrt{2}+2 \sqrt{3}$ and $\alpha \gamma=7+3 \lambda \sqrt{3}$.
Subtracting the sum of roots: $(\alpha+\gamma)-(\alpha+\beta) = (3 \sqrt{2}+2 \sqrt{3}) - 4 \lambda \implies \gamma-\beta = 3 \sqrt{2}+2 \sqrt{3}-4 \lambda$.
Given $\beta+\gamma=3 \sqrt{2}$.
Adding the two equations: $2 \gamma = 6 \sqrt{2}+2 \sqrt{3}-4 \lambda \implies \gamma = 3 \sqrt{2}+\sqrt{3}-2 \lambda$.
Subtracting: $2 \beta = 4 \lambda - 2 \sqrt{3} \implies \beta = 2 \lambda - \sqrt{3}$.
Since $\alpha+\beta=4 \lambda$,we have $\alpha = 4 \lambda - (2 \lambda - \sqrt{3}) = 2 \lambda + \sqrt{3}$.
Using $\alpha \beta = 5$: $(2 \lambda + \sqrt{3})(2 \lambda - \sqrt{3}) = 5 \implies 4 \lambda^{2}-3 = 5 \implies 4 \lambda^{2}=8 \implies \lambda^{2}=2$.
Since $\alpha = 2 \lambda + \sqrt{3}$,$\beta = 2 \lambda - \sqrt{3}$,and $\gamma = 3 \sqrt{2}+2 \sqrt{3}-\alpha = 3 \sqrt{2}+2 \sqrt{3}-(2 \lambda+\sqrt{3}) = 3 \sqrt{2}+\sqrt{3}-2 \lambda$.
We need $(\alpha+2 \beta+\gamma)^{2} = (\alpha+\beta+\beta+\gamma)^{2} = (4 \lambda + 3 \sqrt{2})^{2}$.
Since $\lambda^{2}=2$,let $\lambda = \sqrt{2}$ (assuming positive root for consistency): $(4 \sqrt{2}+3 \sqrt{2})^{2} = (7 \sqrt{2})^{2} = 49 \times 2 = 98$.

(A) Let $f(x) = x^2 + bx + c$. Since the leading coefficient is $1$ and $f(0) = p$,we have $c = p$. Thus,$f(x) = x^2 + bx + p$.
Given $f(1) = 1 + b + p = \frac{1}{3}$,so $b = \frac{1}{3} - 1 - p = -\frac{2}{3} - p$.
Let $\alpha$ be the common root of $f(x) = 0$ and $f(f(f(f(x)))) = 0$. Since $f(\alpha) = 0$,we have $f(f(f(f(\alpha)))) = f(f(f(0))) = f(f(p)) = 0$.
This implies that $f(p)$ must be a root of $f(x) = 0$,so $f(f(p)) = 0$ means $f(p) = \alpha$ or $f(p) = \beta$,where $\alpha, \beta$ are roots of $f(x) = 0$.
Since $f(x) = (x-\alpha)(x-\beta)$,we have $f(0) = \alpha \beta = p$.
Also,$f(1) = (1-\alpha)(1-\beta) = \frac{1}{3}$.
If $f(p) = \alpha$,then $(p-\alpha)(p-\beta) = \alpha$. Substituting $p = \alpha \beta$,we get $(\alpha \beta - \alpha)(\alpha \beta - \beta) = \alpha$.
$\alpha(\beta-1) \beta(\alpha-1) = \alpha$. Since $\alpha \neq 0$,$(\beta-1)(\alpha-1)\beta = 1$.
We know $(1-\alpha)(1-\beta) = \frac{1}{3}$,so $(\alpha-1)(\beta-1) = \frac{1}{3}$.
Substituting this into the equation,we get $\frac{1}{3} \beta = 1$,which implies $\beta = 3$.
Then $(1-\alpha)(1-3) = \frac{1}{3} \Rightarrow -2(1-\alpha) = \frac{1}{3} \Rightarrow 1-\alpha = -\frac{1}{6} \Rightarrow \alpha = \frac{7}{6}$.
Thus,$f(x) = (x-\frac{7}{6})(x-3)$.
Finally,$f(-3) = (-3 - \frac{7}{6})(-3 - 3) = (-\frac{25}{6})(-6) = 25$.

(D) For set $S$: The inequality is $\frac{|x+3|-1}{|x|-2} \geq 0$.
Critical points are $x = -4, -2, 2, -2$ (from $|x+3|=1 \implies x+3 = \pm 1$ and $|x|=2 \implies x = \pm 2$).
Testing intervals in $[-6, 3] \setminus \{-2, 2\}$:
$[-6, -4] \cup (-2, 2) \cup (2, 3]$.
For set $T$: $x^2 - 7|x| + 9 \leq 0$. Let $y = |x|$,then $y^2 - 7y + 9 \leq 0$.
Roots are $y = \frac{7 \pm \sqrt{49-36}}{2} = \frac{7 \pm \sqrt{13}}{2} \approx \frac{7 \pm 3.6}{2}$.
So $y \in [1.7, 5.3]$. Since $x \in \mathbb{Z}$,$|x| \in \{2, 3, 4, 5\}$,so $x \in \{-5, -4, -3, -2, 2, 3, 4, 5\}$.
Intersecting $S$ and $T$: $S \cap T = \{-5, -4, 3\}$.
The number of elements is $3$.

(D) Given $P_{n} = \alpha^{n} - \beta^{n}$ and the quadratic equation $x^{2} - x - 4 = 0$.
Since $\alpha$ and $\beta$ are roots,we have $\alpha^{2} = \alpha + 4$ and $\beta^{2} = \beta + 4$.
Consider $P_{n} - P_{n-1} = (\alpha^{n} - \beta^{n}) - (\alpha^{n-1} - \beta^{n-1}) = \alpha^{n-1}(\alpha - 1) - \beta^{n-1}(\beta - 1)$.
Since $\alpha^{2} - \alpha = 4$ and $\beta^{2} - \beta = 4$,we have $\alpha - 1 = \frac{4}{\alpha}$ and $\beta - 1 = \frac{4}{\beta}$.
Thus,$P_{n} - P_{n-1} = \alpha^{n-1} \cdot \frac{4}{\alpha} - \beta^{n-1} \cdot \frac{4}{\beta} = 4(\alpha^{n-2} - \beta^{n-2}) = 4P_{n-2}$.
Now,simplify the expression: $\frac{P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14})}{P_{13} P_{14}} = \frac{(P_{15} - P_{14})(P_{16} - P_{15})}{P_{13} P_{14}}$.
Using the relation $P_{n} - P_{n-1} = 4P_{n-2}$,we get $P_{15} - P_{14} = 4P_{13}$ and $P_{16} - P_{15} = 4P_{14}$.
Substituting these values: $\frac{(4P_{13})(4P_{14})}{P_{13} P_{14}} = 16$.

(C) Let $\frac{2a-1}{b} = \alpha$ and $\frac{2b-1}{a} = \beta$,where $\alpha, \beta \in \mathbb{Z}^+$.
Since $a, b \ge 1$,we have $2a-1 \ge 1$ and $2b-1 \ge 1$,so $\alpha, \beta \ge 1$.
From the equations,$2a-1 = \alpha b$ and $2b-1 = \beta a$.
Substituting $b = \frac{2a-1}{\alpha}$ into the second equation: $2(\frac{2a-1}{\alpha}) - 1 = \beta a$.
$4a - 2 - \alpha = \alpha \beta a$,which implies $a(4 - \alpha \beta) = \alpha + 2$.
Since $a > 0$ and $\alpha + 2 > 0$,we must have $4 - \alpha \beta > 0$,so $\alpha \beta < 4$.
Possible pairs $(\alpha, \beta)$ are $(1, 1), (1, 2), (1, 3), (2, 1), (3, 1)$.
Case $1$: $(\alpha, \beta) = (1, 1) \implies a(4-1) = 1+2 \implies 3a = 3 \implies a = 1$. Then $b = \frac{2(1)-1}{1} = 1$. Pair: $(1, 1)$.
Case $2$: $(\alpha, \beta) = (1, 2) \implies a(4-2) = 1+2 \implies 2a = 3$,no integer solution.
Case $3$: $(\alpha, \beta) = (1, 3) \implies a(4-3) = 1+2 \implies a = 3$. Then $b = \frac{2(3)-1}{1} = 5$. Pair: $(3, 5)$.
Case $4$: $(\alpha, \beta) = (2, 1) \implies a(4-2) = 2+2 \implies 2a = 4 \implies a = 2$. Then $b = \frac{2(2)-1}{2} = 1.5$,not an integer.
Case $5$: $(\alpha, \beta) = (3, 1) \implies a(4-3) = 3+2 \implies a = 5$. Then $b = \frac{2(5)-1}{3} = 3$. Pair: $(5, 3)$.
The ordered pairs are $(1, 1), (3, 5), (5, 3)$. Total $3$ pairs.

(C) Given the equation $x^4+4y^4+16z^4+64=32xyz$.
Applying the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality to the four positive terms $x^4, 4y^4, 16z^4, 64$:
$\frac{x^4+4y^4+16z^4+64}{4} \geq \sqrt[4]{x^4 \cdot 4y^4 \cdot 16z^4 \cdot 64} = \sqrt[4]{4096 x^4 y^4 z^4} = 8|xyz|$.
Thus,$x^4+4y^4+16z^4+64 \geq 32|xyz|$.
For the equality to hold,we must have $x^4 = 4y^4 = 16z^4 = 64$,which implies $|x|=2\sqrt{2}, |y|=2, |z|=\sqrt{2}$.
Also,the product $xyz$ must be positive for the equality $32xyz = 32|xyz|$ to hold.
The possible combinations for $(x, y, z)$ are:
$1$. $(2\sqrt{2}, 2, \sqrt{2}) \implies x+y+z = 2\sqrt{2}+2+\sqrt{2} = 3\sqrt{2}+2$.
$2$. $(2\sqrt{2}, -2, -\sqrt{2}) \implies x+y+z = 2\sqrt{2}-2-\sqrt{2} = \sqrt{2}-2$.
$3$. $(-2\sqrt{2}, 2, -\sqrt{2}) \implies x+y+z = -2\sqrt{2}+2-\sqrt{2} = 2-3\sqrt{2}$.
$4$. $(-2\sqrt{2}, -2, \sqrt{2}) \implies x+y+z = -2\sqrt{2}-2+\sqrt{2} = -\sqrt{2}-2$.
There are $4$ distinct possible values for the sum $x+y+z$.

(D) Let $f(a, b, c) = \frac{a^2+b^2+c^2}{ab+bc+ca}$.
Case $1$: If $ab+bc+ca > 0$,we know that $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$,which implies $2(a^2+b^2+c^2) - 2(ab+bc+ca) \geq 0$,so $a^2+b^2+c^2 \geq ab+bc+ca$.
Thus,$\frac{a^2+b^2+c^2}{ab+bc+ca} \geq 1$.
Case $2$: If $ab+bc+ca < 0$,we use the identity $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) \geq 0$.
This implies $a^2+b^2+c^2 \geq -2(ab+bc+ca)$.
Since $ab+bc+ca < 0$,dividing by it reverses the inequality sign:
$\frac{a^2+b^2+c^2}{ab+bc+ca} \leq -2$.
Combining both cases,the range of $S$ is $(-\infty, -2] \cup [1, \infty)$.

(C) Given equations are:
$q = p(4-p) \dots (i)$
$r = q(4-q) \dots (ii)$
$p = r(4-r) \dots (iii)$
Adding equations $(i)$,$(ii)$,and $(iii)$,we get:
$p+q+r = 4(p+q+r) - (p^2+q^2+r^2)$
$p^2+q^2+r^2 = 3(p+q+r)$
Consider the case where $p=q=r$. Then $p = p(4-p) \Rightarrow p = 4p - p^2 \Rightarrow p^2 - 3p = 0$,which gives $p=0$ or $p=3$.
If $p=q=r=0$,then $p+q+r = 0$.
If $p=q=r=3$,then $p+q+r = 3+3+3 = 9$.
Since $p^2+q^2+r^2 = 3(p+q+r)$,and by Cauchy-Schwarz inequality $(p+q+r)^2 \le 3(p^2+q^2+r^2)$,we have $(p+q+r)^2 \le 3(3(p+q+r)) = 9(p+q+r)$.
Thus,$p+q+r \le 9$.
The maximum value is $9$.

(B) Given $x^2+y^2=2z^2$ with $HCF(x, y, z)=1$.
Check $I$: Let $x=1, y=7, z=5$. Here $1^2+7^2=50=2(5^2)$. $HCF(1, 7, 5)=1$. $4$ does not divide $1$ or $7$. Thus,$I$ is false.
Check $II$: $x^2+y^2=2z^2 \implies x^2+y^2 \equiv 2z^2 \pmod 3$. Squares modulo $3$ are $0, 1$. If $z^2 \equiv 0$,then $x^2+y^2 \equiv 0 \implies x, y$ are multiples of $3$,contradicting $HCF=1$. If $z^2 \equiv 1$,then $x^2+y^2 \equiv 2 \implies x^2 \equiv 1, y^2 \equiv 1$. Then $x \equiv \pm 1, y \equiv \pm 1$. Thus $x+y \equiv 0$ or $x-y \equiv 0 \pmod 3$. $II$ is true.
Check $III$: $x^2+y^2=2z^2 \implies x^2-z^2 = z^2-y^2$. Modulo $5$,squares are $0, 1, 4$. If $z^2 \equiv 0$,then $x^2+y^2 \equiv 0 \implies x, y \equiv 0$,contradiction. If $z^2 \equiv 1$,$x^2+y^2 \equiv 2$. Possible pairs $(x^2, y^2)$ are $(1, 1)$. Then $x^2-y^2 \equiv 0 \pmod 5$. If $z^2 \equiv 4$,$x^2+y^2 \equiv 8 \equiv 3$. Possible pairs $(x^2, y^2)$ are $(4, 4)$. Then $x^2-y^2 \equiv 0 \pmod 5$. In both cases,$5$ divides $x^2-y^2$,so $5$ divides $z(x^2-y^2)$. $III$ is true.

(B) Given the equation $x^4+y^4+z^4+1=4xyz$.
By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for four positive numbers $x^4, y^4, z^4, 1$:
$\frac{x^4+y^4+z^4+1}{4} \geq \sqrt[4]{x^4 y^4 z^4 \cdot 1}$
Substituting the given equation $x^4+y^4+z^4+1 = 4xyz$:
$\frac{4xyz}{4} \geq |xyz|$
$xyz \geq |xyz|$
This inequality holds if and only if $xyz \geq 0$. For equality to hold in $AM \geq GM$,all terms must be equal:
$x^4 = y^4 = z^4 = 1$
This implies $|x| = 1, |y| = 1, |z| = 1$. Thus,$x, y, z \in \{1, -1\}$.
Substituting these into the original equation $x^4+y^4+z^4+1=4xyz$:
$1+1+1+1 = 4xyz$ $\Rightarrow 4 = 4xyz$ $\Rightarrow xyz = 1$.
The possible triples $(x, y, z)$ such that their product is $1$ are:
$(1, 1, 1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1)$.
There are $4$ such triples.

(D) Given the system of equations:
$x+y^2=12 \quad \dots(i)$
$x^2+y=12 \quad \dots(ii)$
Subtracting $(i)$ from $(ii)$ gives:
$(x^2-x) + (y-y^2) = 0$
$(x^2-y^2) - (x-y) = 0$
$(x-y)(x+y) - (x-y) = 0$
$(x-y)(x+y-1) = 0$
This implies $x=y$ or $x+y=1$.
Case $1$: If $x=y$,then $x^2+x=12$ $\Rightarrow x^2+x-12=0$ $\Rightarrow (x+4)(x-3)=0$. Thus,$x=3, y=3$ and $x=-4, y=-4$. This gives $2$ solutions.
Case $2$: If $x+y=1$,then $y=1-x$. Substituting into $(ii)$: $x^2+(1-x)=12 \Rightarrow x^2-x-11=0$. The discriminant $D = (-1)^2 - 4(1)(-11) = 1+44 = 45 > 0$. This quadratic equation has $2$ distinct real roots for $x$,each giving a corresponding real value for $y$. This gives $2$ more solutions.
Total number of ordered pairs $(x, y)$ is $2+2=4$.

(D) Given that $x$ and $y$ are positive real numbers such that $x+y=1$.
We want to find the minimum value of $f(x, y) = \frac{1}{x} + \frac{1}{y}$.
Using the Arithmetic Mean-Harmonic Mean $(AM \geq HM)$ inequality for two positive numbers $x$ and $y$:
$\frac{x+y}{2} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}}$
Substitute $x+y=1$ into the inequality:
$\frac{1}{2} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}}$
Multiplying both sides by $2(\frac{1}{x} + \frac{1}{y})$,we get:
$\frac{1}{x} + \frac{1}{y} \geq 4$.
Thus,the minimum value is $4$.

(A) Given the equation $2^x + 3^y = 5^{xy}$ for positive integers $x, y \in \mathbb{Z}^+$.
If $x = 1$ and $y = 1$,then $2^1 + 3^1 = 2 + 3 = 5$ and $5^{1 \times 1} = 5^1 = 5$. Thus,$(1, 1)$ is a solution.
If $x > 1$ or $y > 1$,we can divide the equation by $5^{xy}$ to get $\frac{2^x}{5^{xy}} + \frac{3^y}{5^{xy}} = 1$.
This can be rewritten as $\left(\frac{2}{5^y}\right)^x + \left(\frac{3}{5^x}\right)^y = 1$.
For $x, y \ge 1$,the terms $\frac{2^x}{5^{xy}}$ and $\frac{3^y}{5^{xy}}$ decrease rapidly as $x$ or $y$ increase.
Specifically,for $x, y \ge 1$,$2^x + 3^y < 5^{xy}$ for all pairs except $(1, 1)$.
Therefore,the only ordered pair $(x, y)$ that satisfies the equation is $(1, 1)$.

(B) Given the equation $x^3+y^3=65$.
We can factorize the expression as $(x+y)(x^2-xy+y^2)=65$.
Since $x$ and $y$ are integers,$(x+y)$ must be a divisor of $65$.
The divisors of $65$ are $\pm 1, \pm 5, \pm 13, \pm 65$.
Testing integer values for $x$ and $y$:
If $x=1$,then $1+y^3=65 \implies y^3=64 \implies y=4$.
If $y=1$,then $x^3+1=65 \implies x^3=64 \implies x=4$.
Thus,$(1, 4)$ and $(4, 1)$ are solutions.
For other values,if $x$ or $y$ are negative or larger,the sum of cubes does not yield $65$ for integer pairs.
Therefore,there are exactly $2$ ordered pairs.

(C) Given that $a, b$ are roots of $x^2-5cx-6d=0$,we have:
$a+b=5c$ $(1)$
$ab=-6d$ $(2)$
Given that $c, d$ are roots of $x^2-5ax-6b=0$,we have:
$c+d=5a$ $(3)$
$cd=-6b$ $(4)$
Subtracting $(3)$ from $(1)$:
$(a+b)-(c+d) = 5c-5a$
$(a-c)+(b-d) = -5(a-c)$
$b-d = -6(a-c) = 6(c-a)$ $(5)$
Adding $(1)$ and $(3)$:
$(a+b)+(c+d) = 5c+5a$
$(b+d) = 4(a+c)$ $(6)$
From $(2)$ and $(4)$,$ab-cd = -6d+6b = 6(b-d)$.
Substituting $(5)$ into this: $ab-cd = 6(6(c-a)) = 36(c-a)$.
Also,$ab-cd = a(5c-a) - c(5a-c) = 5ac - a^2 - 5ac + c^2 = c^2-a^2 = (c-a)(c+a)$.
Since $a, b, c, d$ are distinct,$c-a \neq 0$,so $c+a = 36$.
Substituting this into $(6)$: $b+d = 4(36) = 144$.

(C) Given $a+b+c=0$ and $a^2+b^2+c^2=1$.
Let $S = (3a+5b-8c)^2+(-8a+3b+5c)^2+(5a-8b+3c)^2$.
Expanding each term:
$(3a+5b-8c)^2 = 9a^2+25b^2+64c^2+30ab-48ac-80bc$
$(-8a+3b+5c)^2 = 64a^2+9b^2+25c^2-48ab-80ac+30bc$
$(5a-8b+3c)^2 = 25a^2+64b^2+9c^2-80ab+30ac-48bc$
Summing these expressions:
$S = (9+64+25)a^2 + (25+9+64)b^2 + (64+25+9)c^2 + (30-48-80)ab + (-48-80+30)ac + (-80+30-48)bc$
$S = 98a^2 + 98b^2 + 98c^2 - 98ab - 98ac - 98bc$
$S = 98(a^2+b^2+c^2) - 98(ab+bc+ca)$
Since $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = 0$,we have $ab+bc+ca = -\frac{1}{2}(a^2+b^2+c^2) = -\frac{1}{2}(1) = -\frac{1}{2}$.
Substituting these values:
$S = 98(1) - 98(-\frac{1}{2}) = 98 + 49 = 147$.

(D) For statement $I$: We check if $n^2+3 \equiv 0 \pmod{17}$.
This implies $n^2 \equiv -3 \equiv 14 \pmod{17}$.
The quadratic residues modulo $17$ are $0^2=0, 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25 \equiv 8, 6^2=36 \equiv 2, 7^2=49 \equiv 15, 8^2=64 \equiv 13$.
Since $14$ is not in the set of quadratic residues ${0, 1, 2, 4, 8, 9, 13, 15, 16}$,$n^2+3$ is never divisible by $17$. Thus,$I$ is true.
For statement $II$: We check if $n^2+4 \equiv 0 \pmod{17}$.
This implies $n^2 \equiv -4 \equiv 13 \pmod{17}$.
From the list of quadratic residues above,$8^2 = 64 = 3 \times 17 + 13$,so $8^2 \equiv 13 \pmod{17}$.
For $n=8$,$n^2+4 = 64+4 = 68 = 4 \times 17$,which is divisible by $17$.
Thus,$II$ is false.
Therefore,$I$ is true and $II$ is false.

(A) Given equations are:
$|a|=\sqrt{4-\sqrt{5-a}}$
$|b|=\sqrt{4+\sqrt{5-b}}$
$|c|=\sqrt{4-\sqrt{5+c}}$
$|d|=\sqrt{4+\sqrt{5+d}}$
Squaring the first equation: $a^2 = 4 - \sqrt{5-a} \implies a^2 - 4 = -\sqrt{5-a}$.
Squaring again: $(a^2 - 4)^2 = 5 - a \implies a^4 - 8a^2 + 16 = 5 - a \implies a^4 - 8a^2 + a + 11 = 0$.
Similarly,for $b, c, d$,we observe that $a, b, -c, -d$ are the roots of the polynomial equation $x^4 - 8x^2 + x + 11 = 0$.
By Vieta's formulas,the product of the roots of $x^4 + 0x^3 - 8x^2 + x + 11 = 0$ is given by the constant term,which is $11$.
Thus,$a \cdot b \cdot (-c) \cdot (-d) = 11 \implies abcd = 11$.

(B) Let $S = (a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$.
Expanding the squares,we get $S = 2(a_1^2+a_2^2+a_3^2+a_4^2) - 2(a_1a_2+a_2a_3+a_3a_4+a_4a_1)$.
Given $a_1^2+a_2^2+a_3^2+a_4^2=1$,so $S = 2 - 2(a_1+a_3)(a_2+a_4)$.
Since $a_1+a_2+a_3+a_4=0$,we have $a_2+a_4 = -(a_1+a_3)$.
Let $x = a_1+a_3$,then $a_2+a_4 = -x$.
Thus,$S = 2 - 2(x)(-x) = 2 + 2x^2$.
To minimize $S$,we need to minimize $x^2$.
Since $a_1^2+a_2^2+a_3^2+a_4^2=1$,by Cauchy-Schwarz inequality,$(a_1+a_3)^2 \leq 2(a_1^2+a_3^2)$ and $(a_2+a_4)^2 \leq 2(a_2^2+a_4^2)$.
Summing these,$x^2 + (-x)^2 \leq 2(a_1^2+a_2^2+a_3^2+a_4^2) = 2$,so $2x^2 \leq 2$,which means $x^2 \leq 1$.
The minimum value of $x^2$ is $0$ (when $a_1+a_3=0$ and $a_2+a_4=0$).
If $x=0$,$S = 2 + 2(0) = 2$.
Since $2$ lies in the interval $(1.5, 2.5)$,the correct option is $B$.

(B) The given equation is $nx^2 + 7\sqrt{n}x + n = 0$.
The discriminant $D$ is given by $D = (7\sqrt{n})^2 - 4(n)(n) = 49n - 4n^2 = n(49 - 4n)$.
For the roots to be distinct,$D \neq 0$. Since $n$ is a positive integer,$49 - 4n = 0$ implies $n = 12.25$,which is not an integer. Thus,$D \neq 0$ for all $n \in \mathbb{Z}^+$. Hence,statement $I$ is correct.
For the roots to be real,$D \geq 0$,which implies $n(49 - 4n) \geq 0$. Since $n > 0$,we have $49 - 4n \geq 0$,so $n \leq 12.25$. The possible values for $n$ are $\{1, 2, 3, \dots, 12\}$. This is a finite set,so statement $II$ is incorrect.
The product of the roots of a quadratic equation $ax^2 + bx + c = 0$ is given by $\frac{c}{a}$. Here,the product is $\frac{n}{n} = 1$,which is an integer. Hence,statement $III$ is correct.
Therefore,statements $I$ and $III$ are correct.

(B) For $x, y, z > 0$,we analyze each condition:
$I.$ $x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) = 0$. Since $x+y+z > 0$,this implies $(x-y)^2+(y-z)^2+(z-x)^2 = 0$,so $x=y=z$.
$II.$ $x^3+y^2z+yz^2=3xyz$. Dividing by $xyz$,we get $\frac{x^2}{yz} + \frac{y}{x} + \frac{z}{x} = 3$. By $AM$-$GM$ inequality,$\frac{x^2}{yz} + \frac{y}{x} + \frac{z}{x} \ge 3 \sqrt[3]{\frac{x^2}{yz} \cdot \frac{y}{x} \cdot \frac{z}{x}} = 3(1) = 3$. Equality holds if $\frac{x^2}{yz} = \frac{y}{x} = \frac{z}{x}$,which implies $x=y=z$.
$III.$ $x^3+y^2z+z^2x=3xyz$. Let $x=1, y=2, z=0.5$. Then $1 + (4)(0.5) + (0.25)(1) = 1 + 2 + 0.25 = 3.25 \neq 3(1)(2)(0.5) = 3$. However,testing values like $x=1, y=1, z=1$ gives $1+1+1=3$. This condition does not force $x=y=z$ for all positive reals.
$IV.$ By $AM$-$GM$,$\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$. Thus $(x+y+z)^3 \ge 27xyz$. Equality holds if and only if $x=y=z$.
Thus,$I, II,$ and $IV$ imply $x=y=z$.

(C) Given that $r$ is a root of the equation $x^2+2x+6=0$,we have $r^2+2r+6=0$,which implies $r^2+2r = -6$.
We need to evaluate the expression $E = (r+2)(r+3)(r+4)(r+5)$.
Grouping the terms as follows:
$E = [(r+2)(r+5)] \times [(r+3)(r+4)]$
$E = (r^2+7r+10)(r^2+7r+12)$
Let $y = r^2+7r$. Then $E = (y+10)(y+12) = y^2+22y+120$.
Alternatively,using the relation $r^2 = -2r-6$:
$E = (r^2+5r+6)(r^2+9r+20)$
Substituting $r^2 = -2r-6$:
$E = (-2r-6+5r+6)(-2r-6+9r+20)$
$E = (3r)(7r+14)$
$E = 21(r^2+2r)$
Since $r^2+2r = -6$,we get:
$E = 21 \times (-6) = -126$.

(B) Let $P_n$ be the population in year $n$. According to the problem,$P_{n+2} - P_n = k P_{n+1}$,where $k$ is a constant.
Given populations:
$P_{2010} = 39$
$P_{2011} = 60$
$P_{2013} = 123$
Let $P_{2012} = x$.
For $n = 2010$:
$P_{2012} - P_{2010} = k P_{2011}$ $\Rightarrow x - 39 = k(60)$ $\Rightarrow k = \frac{x - 39}{60} \quad (i)$
For $n = 2011$:
$P_{2013} - P_{2011} = k P_{2012}$ $\Rightarrow 123 - 60 = k(x)$ $\Rightarrow 63 = kx$ $\Rightarrow k = \frac{63}{x} \quad (ii)$
Equating $(i)$ and $(ii)$:
$\frac{x - 39}{60} = \frac{63}{x}$
$x(x - 39) = 60 \times 63$
$x^2 - 39x - 3780 = 0$
Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{39 \pm \sqrt{(-39)^2 - 4(1)(-3780)}}{2}$
$x = \frac{39 \pm \sqrt{1521 + 15120}}{2}$
$x = \frac{39 \pm \sqrt{16641}}{2}$
$x = \frac{39 \pm 129}{2}$
Since population must be positive,$x = \frac{39 + 129}{2} = \frac{168}{2} = 84$.
Thus,the population in $2012$ was $84$.

(B) Given $a+b+c=0$.
We know that $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) = 0$.
Thus,$q + 2(ab+bc+ca) = 0$,which implies $ab+bc+ca = -\frac{q}{2}$.
Squaring both sides,$(ab+bc+ca)^2 = \frac{q^2}{4}$.
Also,$(ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2 + 2abc(a+b+c) = a^2b^2+b^2c^2+c^2a^2 + 0 = \frac{q^2}{4}$.
Now,consider $q^2 = (a^2+b^2+c^2)^2 = a^4+b^4+c^4 + 2(a^2b^2+b^2c^2+c^2a^2)$.
Substituting $r = a^4+b^4+c^4$ and $a^2b^2+b^2c^2+c^2a^2 = \frac{q^2}{4}$,we get:
$q^2 = r + 2(\frac{q^2}{4}) = r + \frac{q^2}{2}$.
Rearranging the terms,$q^2 - \frac{q^2}{2} = r$,which simplifies to $\frac{q^2}{2} = r$,or $q^2 = 2r$.

(C) Given the system of equations:
$x+y=a$ $(1)$
$\frac{x^2}{x-1}+\frac{y^2}{y-1}=4$ $(2)$
From $(2)$,we have $\frac{x^2(y-1)+y^2(x-1)}{(x-1)(y-1)}=4$.
Substituting $y=a-x$,the denominator becomes $(x-1)(a-x-1) = -(x-1)^2 + (a-2)(x-1)$.
The numerator simplifies as $x^2(a-x-1) + (a-x)^2(x-1) = x^2a - x^3 - x^2 + (a^2 - 2ax + x^2)(x-1) = x^2a - x^3 - x^2 + a^2x - a^2 - 2ax^2 + 2ax + x^3 - x^2 = a^2x - a^2 + 2ax - 2x^2$.
Setting the equation equal to $4(x-1)(y-1)$,we simplify to $(a-2)(xy - (a-2)) = 0$.
If $a=2$,the equation becomes $\frac{x^2}{x-1} + \frac{(2-x)^2}{1-x} = 4$,which simplifies to $\frac{x^2 - (4-4x+x^2)}{x-1} = 4$,so $\frac{4x-4}{x-1} = 4$,which is $4=4$. This holds for all $x \neq 1$,meaning infinitely many solutions.
If $a \neq 2$,then $xy = a-2$. Since $x+y=a$,$x$ and $y$ are roots of $t^2 - at + (a-2) = 0$.
The discriminant $D = a^2 - 4(a-2) = a^2 - 4a + 8 = (a-2)^2 + 4 > 0$,so there are always two distinct real roots for any $a$.
However,we must ensure $x \neq 1$ and $y \neq 1$. If $x=1$,then $1+y=a \Rightarrow y=a-1$. Substituting into $xy=a-2$ gives $1(a-1) = a-2$,which is $a-1=a-2$,impossible.
Thus,for any $a \neq 2$,the system has exactly two solutions. Since $a \in [1, 2014]$,there are $2014$ integers total,and excluding $a=2$,there are $2013$ values.

(C) Let $a = \frac{x}{y}$,$b = \frac{y}{z}$,and $c = \frac{z}{x}$.
Given that $a+b+c = 7$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 9$.
Note that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{y}{x}+\frac{z}{y}+\frac{x}{z} = 9$.
We know the algebraic identity $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
This can be rewritten as $a^3+b^3+c^3-3abc = (a+b+c)((a+b+c)^2-3(ab+bc+ca))$.
Here,$abc = (\frac{x}{y})(\frac{y}{z})(\frac{z}{x}) = 1$.
Also,$ab+bc+ca = (\frac{x}{y})(\frac{y}{z}) + (\frac{y}{z})(\frac{z}{x}) + (\frac{z}{x})(\frac{x}{y}) = \frac{x}{z} + \frac{y}{x} + \frac{z}{y} = 9$.
Substituting these values into the identity:
$a^3+b^3+c^3-3(1) = (7)((7)^2 - 3(9))$.
$a^3+b^3+c^3-3 = 7(49-27)$.
$a^3+b^3+c^3-3 = 7(22) = 154$.

(D) Let the dimensions of the cuboid be $x, y,$ and $z$. The faces have dimensions $(x, y), (y, z),$ and $(x, z)$.
The sum of the perimeter and area for each face is given by:
$2(x+y) + xy = 16 \quad (i)$
$2(y+z) + yz = 24 \quad (ii)$
$2(x+z) + xz = 31 \quad (iii)$
Adding $4$ to both sides of each equation to factorize:
$xy + 2x + 2y + 4 = 16 + 4 \implies (x+2)(y+2) = 20 \quad (iv)$
$yz + 2y + 2z + 4 = 24 + 4 \implies (y+2)(z+2) = 28 \quad (v)$
$xz + 2x + 2z + 4 = 31 + 4 \implies (x+2)(z+2) = 35 \quad (vi)$
Let $X = x+2, Y = y+2, Z = z+2$. Then $XY=20, YZ=28, XZ=35$.
Multiplying these: $(XYZ)^2 = 20 \times 28 \times 35 = 19600 \implies XYZ = 140$.
$Z = \frac{XYZ}{XY} = \frac{140}{20} = 7 \implies z+2 = 7 \implies z = 5$.
$X = \frac{XYZ}{YZ} = \frac{140}{28} = 5 \implies x+2 = 5 \implies x = 3$.
$Y = \frac{XYZ}{XZ} = \frac{140}{35} = 4 \implies y+2 = 4 \implies y = 2$.
The volume $V = xyz = 3 \times 2 \times 5 = 30$.
Since $30$ lies between $28$ and $35$,option $(d)$ is correct.

(A) Given the expression $P(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$.
Evaluate $P(x)$ at $x=a$:
$P(a) = \frac{(a-b)(a-c)}{(a-b)(a-c)} + 0 + 0 = 1$.
Evaluate $P(x)$ at $x=b$:
$P(b) = 0 + \frac{(b-c)(b-a)}{(b-c)(b-a)} + 0 = 1$.
Evaluate $P(x)$ at $x=c$:
$P(c) = 0 + 0 + \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1$.
Since $P(x)$ is a polynomial of degree at most $2$ and it takes the value $1$ for three distinct values $x = a, b, c$,it must be the constant polynomial $P(x) = 1$ for all real $x$.
Therefore,the simplified value is $1$.

(A) Given,$x+\frac{1}{x}=a$ and $x^2+\frac{1}{x^3}=b$.
Squaring $x+\frac{1}{x}=a$,we get:
$x^2+\frac{1}{x^2}+2=a^2 \Rightarrow x^2+\frac{1}{x^2}=a^2-2$.
Cubing $x+\frac{1}{x}=a$,we get:
$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=a^3 \Rightarrow x^3+\frac{1}{x^3}=a^3-3a$.
Now,consider the sum:
$(x^2+\frac{1}{x^2}) + (x^3+\frac{1}{x^3}) = (a^2-2) + (a^3-3a)$.
Rearranging the terms:
$(x^2+\frac{1}{x^3}) + (x^3+\frac{1}{x^2}) = a^3+a^2-3a-2$.
Substituting $x^2+\frac{1}{x^3}=b$:
$b + (x^3+\frac{1}{x^2}) = a^3+a^2-3a-2$.
Therefore,$x^3+\frac{1}{x^2} = a^3+a^2-3a-2-b$.

(A) Given equations are:
$(i) x = x^2 + y^2$
$(ii) y = 2xy$
From equation $(ii)$:
$y - 2xy = 0$
$y(1 - 2x) = 0$
This implies $y = 0$ or $x = \frac{1}{2}$.
Case $1$: If $y = 0$,substitute into $(i)$:
$x = x^2 + 0^2$
$x^2 - x = 0$
$x(x - 1) = 0$
So,$x = 0$ or $x = 1$.
This gives pairs $(0, 0)$ and $(1, 0)$.
Case $2$: If $x = \frac{1}{2}$,substitute into $(i)$:
$\frac{1}{2} = (\frac{1}{2})^2 + y^2$
$\frac{1}{2} = \frac{1}{4} + y^2$
$y^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$
$y = \pm \frac{1}{2}$.
This gives pairs $(\frac{1}{2}, \frac{1}{2})$ and $(\frac{1}{2}, -\frac{1}{2})$.
Thus,the total number of pairs $(x, y)$ is $4$.

(A) Consider the equation $x^2 + y^2 = a^2 + b^2 + c^2$ where $x, y, a, b, c$ are primes.
Case $1$: If all primes are odd,then $x^2, y^2, a^2, b^2, c^2 \equiv 1 \pmod{4}$ or $1 \pmod{8}$.
Specifically,for any odd prime $p$,$p^2 \equiv 1 \pmod{8}$.
Then $x^2 + y^2 \equiv 1 + 1 = 2 \pmod{8}$ and $a^2 + b^2 + c^2 \equiv 1 + 1 + 1 = 3 \pmod{8}$.
Since $2 \not\equiv 3 \pmod{8}$,there are no solutions where all variables are odd primes.
Case $2$: If at least one variable is $2$.
If $x=2$,then $4 + y^2 = a^2 + b^2 + c^2$. If $y$ is also $2$,then $8 = a^2 + b^2 + c^2$. The only primes whose squares are less than $8$ are $2$. If $a=b=c=2$,then $a^2 + b^2 + c^2 = 4 + 4 + 4 = 12 \neq 8$. If any are $2$,we cannot satisfy the sum.
Testing small primes,we find no combination of prime numbers satisfies the equation.
Thus,the number of solutions is $0$.

(B) Given the quadratic equation $x^2+ax+b=0$. Since $a-b$ is a root,it must satisfy the equation:
$(a-b)^2 + a(a-b) + b = 0$
$a^2 - 2ab + b^2 + a^2 - ab + b = 0$
$b^2 - 3ab + b + 2a^2 = 0$
$b^2 + b(1-3a) + 2a^2 = 0$
Solving for $b$ using the quadratic formula:
$b = \frac{-(1-3a) \pm \sqrt{(3a-1)^2 - 8a^2}}{2} = \frac{(3a-1) \pm \sqrt{9a^2 - 6a + 1 - 8a^2}}{2} = \frac{(3a-1) \pm \sqrt{a^2 - 6a + 1}}{2}$
For $b$ to be an integer,the discriminant $D = a^2 - 6a + 1$ must be a perfect square,say $k^2$.
$a^2 - 6a + 1 = k^2$
$(a-3)^2 - 8 = k^2$
$(a-3)^2 - k^2 = 8$
$(a-3-k)(a-3+k) = 8$
Since the product is $8$ (even),both factors must be even. Let $X = a-3-k$ and $Y = a-3+k$. Then $XY = 8$ and $X+Y = 2(a-3)$.
Possible pairs $(X, Y)$ such that $XY=8$ and $X, Y$ are even:
$1$) $X=2, Y=4 \implies 2(a-3)=6 \implies a-3=3 \implies a=6$. Then $k=1$,$b = \frac{(3(6)-1) \pm 1}{2} = \frac{17 \pm 1}{2} \implies b=9, 8$.
$2$) $X=-4, Y=-2 \implies 2(a-3)=-6 \implies a-3=-3 \implies a=0$. Then $k=1$,$b = \frac{(3(0)-1) \pm 1}{2} = \frac{-1 \pm 1}{2} \implies b=0, -1$.
Thus,the pairs $(a, b)$ are $(6, 9), (6, 8), (0, 0), (0, -1)$.
There are $4$ such ordered pairs.

(C) The given equation is $|x|^2 - 2|x| + |\lambda - 3| = 0$.
This can be rewritten as $(|x| - 1)^2 - 1 + |\lambda - 3| = 0$,or $(|x| - 1)^2 = 1 - |\lambda - 3|$.
For $x$ to be a real solution,we must have $1 - |\lambda - 3| \ge 0$,which implies $|\lambda - 3| \le 1$,so $2 \le \lambda \le 4$.
Since $x$ must be an integer,$(|x| - 1)^2$ must be a perfect square,i.e.,$0$ or $1$.
Case $1$: $(|x| - 1)^2 = 0 \implies |x| = 1 \implies x = \pm 1$. Then $1 - |\lambda - 3| = 0 \implies |\lambda - 3| = 1 \implies \lambda = 4$ or $\lambda = 2$.
If $\lambda = 4, x = 1, -1$,then $x + \lambda$ can be $1+4=5$ or $-1+4=3$.
If $\lambda = 2, x = 1, -1$,then $x + \lambda$ can be $1+2=3$ or $-1+2=1$.
Case $2$: $(|x| - 1)^2 = 1 \implies |x| - 1 = \pm 1 \implies |x| = 2$ or $0 \implies x = \pm 2, 0$. Then $1 - |\lambda - 3| = 1 \implies |\lambda - 3| = 0 \implies \lambda = 3$.
If $\lambda = 3, x = 2, -2, 0$,then $x + \lambda$ can be $2+3=5, -2+3=1, 0+3=3$.
The set $S = \{5, 3, 1\}$. The largest element is $5$.

(A) Given the quadratic equation $x^2+60^{\frac{1}{4}} x+a=0$,we have the sum of roots $\alpha+\beta = -60^{\frac{1}{4}}$ and the product of roots $\alpha \beta = a$.
We are given $\alpha^4+\beta^4 = -30$.
Using the identity $\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha \beta)^2$,we substitute the values:
$(\alpha^2+\beta^2)^2 - 2a^2 = -30$.
Since $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha \beta = (-60^{\frac{1}{4}})^2 - 2a = 60^{\frac{1}{2}} - 2a$,we have:
$(60^{\frac{1}{2}} - 2a)^2 - 2a^2 = -30$.
Expanding the square:
$60 + 4a^2 - 4a(60^{\frac{1}{2}}) - 2a^2 = -30$.
Simplifying the equation:
$2a^2 - 4\sqrt{60}a + 90 = 0$.
This is a quadratic equation in $a$. The product of the roots $a_1 a_2$ is given by the constant term divided by the leading coefficient:
$\text{Product} = \frac{90}{2} = 45$.

(A) The given equation is $|x^2-8x+15|-2x+7=0$.
Case $1$: $x^2-8x+15 \geq 0$,which implies $x \leq 3$ or $x \geq 5$.
The equation becomes $x^2-8x+15-2x+7=0$,so $x^2-10x+22=0$.
The roots are $x = \frac{10 \pm \sqrt{100-88}}{2} = 5 \pm \sqrt{3}$.
Since $5+\sqrt{3} \geq 5$ and $5-\sqrt{3} \approx 3.268$ (which is not $\leq 3$ or $\geq 5$),only $x = 5+\sqrt{3}$ is a valid root.
Case $2$: $x^2-8x+15 < 0$,which implies $3 < x < 5$.
The equation becomes $-(x^2-8x+15)-2x+7=0$,so $-x^2+8x-15-2x+7=0$,which simplifies to $-x^2+6x-8=0$ or $x^2-6x+8=0$.
Factoring gives $(x-2)(x-4)=0$,so $x=2$ or $x=4$.
Since $3 < x < 5$,only $x=4$ is a valid root.
The sum of all roots is $(5+\sqrt{3}) + 4 = 9+\sqrt{3}$.

(A) For the first equation: $x^2-12x+[x]+31=0$.
Since $x = [x] + \{x\}$,we have $x^2-12([x]+\{x\})+[x]+31=0$.
Rearranging gives $\{x\} = \frac{x^2-11[x]+31}{12}$.
Alternatively,$x^2-12x+31 = -[x]$.
Let $[x] = k$,then $x^2-12x+31+k=0$.
Solving for $x$,$x = \frac{12 \pm \sqrt{144-4(31+k)}}{2} = 6 \pm \sqrt{36-31-k} = 6 \pm \sqrt{5-k}$.
For $x$ to be real,$5-k \geq 0$,so $k \leq 5$.
Also,$k \leq 6 \pm \sqrt{5-k} < k+1$.
Case $k=5$: $x = 6 \pm 0 = 6$. But $[6]=6 \neq 5$.
Case $k=4$: $x = 6 \pm 1 = 7, 5$. $[7]=7 \neq 4$,$[5]=5 \neq 4$.
Case $k=3$: $x = 6 \pm \sqrt{2} \approx 7.41, 4.58$. $[4.58]=4 \neq 3$.
Case $k=2$: $x = 6 \pm \sqrt{3} \approx 7.73, 4.26$. $[4.26]=4 \neq 2$.
Case $k=1$: $x = 6 \pm 2 = 8, 4$. $[4]=4 \neq 1$.
Case $k=0$: $x = 6 \pm \sqrt{5} \approx 8.23, 3.76$. $[3.76]=3 \neq 0$.
Checking $x^2-12x+[x]+31=0$ again: if $x \in [k, k+1)$,then $x^2-12x+k+31=0$.
For $k=5$,$x^2-12x+36=0 \implies x=6$,not in $[5,6)$.
For $k=4$,$x^2-12x+35=0 \implies (x-5)(x-7)=0 \implies x=5$,which is in $[4,5)$. So $x=5$ is a root.
For $k=3$,$x^2-12x+34=0 \implies x = 6 \pm \sqrt{2}$,neither in $[3,4)$.
Thus,$m=1$.
For the second equation: $x^2-5|x+2|-4=0$.
If $x \geq -2$,$x^2-5(x+2)-4=0 \implies x^2-5x-14=0 \implies (x-7)(x+2)=0 \implies x=7, -2$.
If $x < -2$,$x^2-5(-x-2)-4=0 \implies x^2+5x+6=0 \implies (x+3)(x+2)=0 \implies x=-3, -2$.
The set of roots is $\{7, -2, -3\}$,so $n=3$.
Then $m^2+mn+n^2 = 1^2 + (1)(3) + 3^2 = 1+3+9 = 13$.
Wait,checking the options,if $m=0, n=3$,$m^2+mn+n^2=9$. Let's re-verify $m$.
$x^2-12x+[x]+31=0$. If $x=5$,$25-60+5+31=1
eq 0$.
Actually,there are no real roots for the first equation,$m=0$.
Then $m^2+mn+n^2 = 0^2+0(3)+3^2 = 9$.

(D) Given the quadratic equation $x^2-70x+\lambda=0$ with roots $\alpha, \beta \in \mathbb{N}$.
By Vieta's formulas,$\alpha+\beta=70$ and $\alpha\beta=\lambda$.
We are given that $\frac{\lambda}{2} \notin \mathbb{N}$ and $\frac{\lambda}{3} \notin \mathbb{N}$,which implies $\lambda$ is not divisible by $2$ or $3$. Thus,$\lambda$ is not divisible by $6$.
Since $\lambda = \alpha(70-\alpha)$,we test values of $\alpha$ such that $\lambda$ is not divisible by $2$ or $3$.
If $\alpha=1$,$\lambda=69$ (divisible by $3$,reject).
If $\alpha=2$,$\lambda=136$ (divisible by $2$,reject).
If $\alpha=3$,$\lambda=201$ (divisible by $3$,reject).
If $\alpha=4$,$\lambda=264$ (divisible by $2$ and $3$,reject).
If $\alpha=5$,$\lambda=5 \times 65 = 325$. $325$ is not divisible by $2$ or $3$. This is the minimum value.
For $\alpha=5, \beta=65, \lambda=325$,we calculate the expression:
$|\alpha-\beta| = |5-65| = 60$.
$\sqrt{\alpha-1} = \sqrt{4} = 2$.
$\sqrt{\beta-1} = \sqrt{64} = 8$.
Expression $= \frac{(2+8)(325+35)}{60} = \frac{10 \times 360}{60} = 10 \times 6 = 60$.

(A) Let $f(x) = (a+b-2c)x^2 + (b+c-2a)x + (c+a-2b)$.
Sum of coefficients: $f(1) = (a+b-2c) + (b+c-2a) + (c+a-2b) = 0$.
Since $f(1) = 0$,$x = 1$ is one root of the equation.
Let the roots be $1$ and $\alpha$. From the product of roots formula,$1 \cdot \alpha = \frac{c+a-2b}{a+b-2c}$.
Thus,$\alpha = \frac{c+a-2b}{a+b-2c}$.
Case $(I)$: If $-1 < \alpha < 0$,then $-1 < \frac{c+a-2b}{a+b-2c} < 0$. Solving this inequality given $0 < c < b < a$ leads to $b > \frac{a+c}{2}$. Since the geometric mean of $a$ and $c$ is $\sqrt{ac}$ and $\sqrt{ac} < \frac{a+c}{2}$,$b$ cannot be the geometric mean.
Case $(II)$: If $0 < \alpha < 1$,then $0 < \frac{c+a-2b}{a+b-2c} < 1$. Solving this leads to $b < \frac{a+c}{2}$. Since it is possible for $\sqrt{ac} < b < \frac{a+c}{2}$,$b$ may be the geometric mean of $a$ and $c$.
Therefore,both statements are true.

(B) The given equation is $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$.
This can be rewritten as $a^2x^2-2abx+b^2+b^2x^2-2bcx+c^2=0$.
This simplifies to $(ax-b)^2+(bx-c)^2=0$.
Since $a, b, c$ are real,we must have $ax-b=0$ and $bx-c=0$,which implies $x = b/a = c/b$.
Thus,$b^2 = ac$,meaning $a, b, c$ are in geometric progression.
For $a, b, c$ to be sides of a triangle,they must satisfy the triangle inequality: $a+b > c$,$b+c > a$,and $c+a > b$.
Substituting $b = ax$ and $c = bx = ax^2$:
$a + ax > ax^2$ $\Rightarrow x^2 - x - 1 < 0$ $\Rightarrow \frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}$.
$ax + ax^2 > a$ $\Rightarrow x^2 + x - 1 > 0$ $\Rightarrow x > \frac{-1+\sqrt{5}}{2}$ (since $x > 0$).
$a + ax^2 > ax \Rightarrow x^2 - x + 1 > 0$ (always true).
Combining these,we get $\frac{\sqrt{5}-1}{2} < x < \frac{\sqrt{5}+1}{2}$.
So,$\alpha = \frac{\sqrt{5}-1}{2}$ and $\beta = \frac{\sqrt{5}+1}{2}$.
Then $\alpha^2 + \beta^2 = \frac{5+1-2\sqrt{5}}{4} + \frac{5+1+2\sqrt{5}}{4} = \frac{12}{4} = 3$.
Therefore,$12(\alpha^2+\beta^2) = 12(3) = 36$.

(D) Given $x^2+20x-2020=0$ has roots $a, b$. By Vieta's formulas,$a+b = -20$ and $ab = -2020$.
Given $x^2-20x+2020=0$ has roots $c, d$. By Vieta's formulas,$c+d = 20$ and $cd = 2020$.
We need to evaluate $E = ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$.
Expanding the expression: $E = a^2c - ac^2 + a^2d - ad^2 + b^2c - bc^2 + b^2d - bd^2$.
Grouping terms: $E = a^2(c+d) + b^2(c+d) - c^2(a+b) - d^2(a+b)$.
$E = (c+d)(a^2+b^2) - (a+b)(c^2+d^2)$.
Using $a^2+b^2 = (a+b)^2 - 2ab$ and $c^2+d^2 = (c+d)^2 - 2cd$:
$a^2+b^2 = (-20)^2 - 2(-2020) = 400 + 4040 = 4440$.
$c^2+d^2 = (20)^2 - 2(2020) = 400 - 4040 = -3640$.
Substituting these values: $E = (20)(4440) - (-20)(-3640)$.
$E = 88800 - 72800 = 16000$.

(A) The given equation is $3x^2 + x - 1 = 4|x^2 - 1|$.
Case $1$: If $x \in [-1, 1]$,then $|x^2 - 1| = -(x^2 - 1) = 1 - x^2$.
The equation becomes $3x^2 + x - 1 = 4(1 - x^2)$ $\Rightarrow 3x^2 + x - 1 = 4 - 4x^2$ $\Rightarrow 7x^2 + x - 5 = 0$.
Let $f(x) = 7x^2 + x - 5$. Here $f(-1) = 7 - 1 - 5 = 1 > 0$ and $f(1) = 7 + 1 - 5 = 3 > 0$. Since $f(0) = -5 < 0$,the quadratic $f(x)$ has two real roots in the interval $(-1, 1)$.
Case $2$: If $x \in (-\infty, -1] \cup [1, \infty)$,then $|x^2 - 1| = x^2 - 1$.
The equation becomes $3x^2 + x - 1 = 4(x^2 - 1)$ $\Rightarrow 3x^2 + x - 1 = 4x^2 - 4$ $\Rightarrow x^2 - x - 3 = 0$.
Let $g(x) = x^2 - x - 3$. The roots are $x = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}$.
Since $\sqrt{13} \approx 3.6$,the roots are $x_1 \approx \frac{4.6}{2} = 2.3$ (which is $\ge 1$) and $x_2 \approx \frac{-2.6}{2} = -1.3$ (which is $\le -1$). Both roots are valid.
Thus,there are $2 + 2 = 4$ real roots in total.

(B) Given the quadratic equation $2x^2 + (\cos \theta)x - 1 = 0$.
Sum of roots $\alpha_\theta + \beta_\theta = -\frac{\cos \theta}{2}$ and product of roots $\alpha_\theta \beta_\theta = -\frac{1}{2}$.
We need to find $\alpha_\theta^4 + \beta_\theta^4 = (\alpha_\theta^2 + \beta_\theta^2)^2 - 2(\alpha_\theta \beta_\theta)^2$.
Since $\alpha_\theta^2 + \beta_\theta^2 = (\alpha_\theta + \beta_\theta)^2 - 2\alpha_\theta \beta_\theta = \frac{\cos^2 \theta}{4} + 1$.
Thus,$\alpha_\theta^4 + \beta_\theta^4 = (\frac{\cos^2 \theta}{4} + 1)^2 - 2(-\frac{1}{2})^2 = (\frac{\cos^2 \theta}{4} + 1)^2 - \frac{1}{2}$.
Let $u = \cos^2 \theta$,where $u \in [0, 1]$.
Let $f(u) = (\frac{u}{4} + 1)^2 - \frac{1}{2}$.
For $u = 0$,$f(0) = (1)^2 - \frac{1}{2} = \frac{1}{2} = m$.
For $u = 1$,$f(1) = (\frac{1}{4} + 1)^2 - \frac{1}{2} = (\frac{5}{4})^2 - \frac{1}{2} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16} = M$.
Therefore,$16(M + m) = 16(\frac{17}{16} + \frac{1}{2}) = 16(\frac{17 + 8}{16}) = 25$.

(C) Let $f(x) = x^2+3x+2 = (x+1)(x+2)$.
Let $g(x) = \min \{|x-3|, |x+2|\}$.
We need to find the number of intersection points of $y = f(x)$ and $y = g(x)$.
For $x < -2$,$f(x) > 0$ and $g(x) = |x-3| = 3-x$. Solving $x^2+3x+2 = 3-x$ gives $x^2+4x-1 = 0$,so $x = -2 \pm \sqrt{5}$. Since $x < -2$,$x = -2-\sqrt{5}$ is a solution.
For $-2 \le x < 0.5$,$g(x) = |x+2| = x+2$. Solving $x^2+3x+2 = x+2$ gives $x^2+2x = 0$,so $x=0$ or $x=-2$. Both are in the interval.
For $x \ge 0.5$,$g(x) = |x-3|$. The parabola $f(x)$ is increasing and $g(x)$ is decreasing or increasing,but $f(x)$ grows much faster,so there are no further solutions.
The solutions are $x = -2-\sqrt{5}$,$x = -2$,and $x = 0$. Thus,there are $3$ real solutions.

(C) Given equation: $x^2+|2x-3|-4=0$.
Case $I$: $x \geq \frac{3}{2}$.
The equation becomes $x^2 + 2x - 3 - 4 = 0$,which is $x^2 + 2x - 7 = 0$.
Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{-2 \pm \sqrt{32}}{2} = -1 \pm 2\sqrt{2}$.
Since $x \geq \frac{3}{2}$,we accept $x = 2\sqrt{2} - 1$.
Case $II$: $x < \frac{3}{2}$.
The equation becomes $x^2 - (2x - 3) - 4 = 0$,which is $x^2 - 2x - 1 = 0$.
Using the quadratic formula,$x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.
Since $x < \frac{3}{2}$,we accept both $x = 1 + \sqrt{2}$ and $x = 1 - \sqrt{2}$.
The roots are $2\sqrt{2}-1$,$1+\sqrt{2}$,and $1-\sqrt{2}$.
Sum of squares $= (2\sqrt{2}-1)^2 + (1+\sqrt{2})^2 + (1-\sqrt{2})^2$.
$= (8 - 4\sqrt{2} + 1) + (1 + 2\sqrt{2} + 2) + (1 - 2\sqrt{2} + 2)$.
$= 9 - 4\sqrt{2} + 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 15 - 4\sqrt{2}$.