Solution of quadratic equations and Nature of roots Questions in English

(A) Let $y = 4+\frac{1}{5+\frac{1}{4+\frac{1}{5+\ldots}}}$.
Observe that the expression repeats after the first two terms: $y = 4+\frac{1}{5+\frac{1}{y}}$.
Simplify the equation: $y - 4 = \frac{1}{\frac{5y+1}{y}} = \frac{y}{5y+1}$.
Cross-multiply: $(y-4)(5y+1) = y$.
$5y^2 + y - 20y - 4 = y$.
$5y^2 - 20y - 4 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$y = \frac{20 \pm \sqrt{(-20)^2 - 4(5)(-4)}}{2(5)} = \frac{20 \pm \sqrt{400 + 80}}{10} = \frac{20 \pm \sqrt{480}}{10}$.
Since $y > 0$,we take the positive root: $y = \frac{20 + \sqrt{480}}{10} = 2 + \frac{\sqrt{16 \times 30}}{10} = 2 + \frac{4\sqrt{30}}{10} = 2 + \frac{2\sqrt{30}}{5} = 2 + \frac{2}{5}\sqrt{30}$.

(A) Let $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$
So,$x=3+\frac{1}{4+\frac{1}{x}}=3+\frac{x}{4x+1}$
$\Rightarrow x-3=\frac{x}{4x+1}$
$\Rightarrow (x-3)(4x+1)=x$
$\Rightarrow 4x^2+x-12x-3=x$
$\Rightarrow 4x^2-12x-3=0$
Using the quadratic formula $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$x=\frac{12 \pm \sqrt{(-12)^2-4(4)(-3)}}{2(4)}$
$x=\frac{12 \pm \sqrt{144+48}}{8}=\frac{12 \pm \sqrt{192}}{8}$
$x=\frac{12 \pm 8\sqrt{3}}{8} = \frac{3}{2} \pm \sqrt{3} = 1.5 \pm \sqrt{3}$
Since $x > 0$,we take $x=1.5+\sqrt{3}$.

(D) Case-$I$: $x \leq 5$
$(x+1)^{2} - (x-5) = \frac{27}{4}$
$(x+1)^{2} - (x+1) + 6 = \frac{27}{4}$
$(x+1)^{2} - (x+1) - \frac{3}{4} = 0$
Let $y = x+1$,then $y^{2} - y - \frac{3}{4} = 0 \implies 4y^{2} - 4y - 3 = 0$
$(2y-3)(2y+1) = 0 \implies y = \frac{3}{2}, -\frac{1}{2}$
$x+1 = \frac{3}{2} \implies x = \frac{1}{2}$ (Valid as $\frac{1}{2} \leq 5$)
$x+1 = -\frac{1}{2} \implies x = -\frac{3}{2}$ (Valid as $-\frac{3}{2} \leq 5$)
Case-$II$: $x > 5$
$(x+1)^{2} + (x-5) = \frac{27}{4}$
$x^{2} + 2x + 1 + x - 5 = \frac{27}{4}$
$x^{2} + 3x - 4 = \frac{27}{4} \implies 4x^{2} + 12x - 16 = 27$
$4x^{2} + 12x - 43 = 0$
$x = \frac{-12 \pm \sqrt{144 - 4(4)(-43)}}{8} = \frac{-12 \pm \sqrt{144 + 688}}{8} = \frac{-12 \pm \sqrt{832}}{8}$
Since $\sqrt{832} \approx 28.8$,$x = \frac{-12 + 28.8}{8} \approx 2.1$ (Rejected as $x > 5$)
Thus,there are $2$ real roots.

(B) Let $f(x) = 2x^{5}+5x^{4}+10x^{3}+10x^{2}+10x+10$.
First,we evaluate the function at integer points to locate the root.
$f(-2) = 2(-32) + 5(16) + 10(-8) + 10(4) + 10(-2) + 10 = -64 + 80 - 80 + 40 - 20 + 10 = -34$.
$f(-1) = 2(-1) + 5(1) + 10(-1) + 10(1) + 10(-1) + 10 = -2 + 5 - 10 + 10 - 10 + 10 = 3$.
Since $f(-2) < 0$ and $f(-1) > 0$,by the Intermediate Value Theorem,there exists at least one real root in the interval $(-2, -1)$.
Next,we check the derivative to determine the number of real roots.
$f'(x) = 10x^{4} + 20x^{3} + 30x^{2} + 20x + 10$.
$f'(x) = 10(x^{4} + 2x^{3} + 3x^{2} + 2x + 1)$.
Notice that $x^{4} + 2x^{3} + 3x^{2} + 2x + 1 = (x^{2} + x + 1)^{2}$.
Thus,$f'(x) = 10(x^{2} + x + 1)^{2}$.
Since $x^{2} + x + 1$ has a negative discriminant $(1 - 4 = -3)$,it is always positive for all real $x$.
Therefore,$f'(x) > 0$ for all real $x$,which means $f(x)$ is strictly increasing and has exactly one real root.
Since the root lies in $(-2, -1)$,we have $a = -2$. Thus,$|a| = |-2| = 2$.

(A) Let $a = 3x^2 + 4x + 3$ and $b = 3x^2 + 4x + 2$. Note that $b = a - 1$.
The given equation becomes $a^2 - (k + 1)ab + kb^2 = 0$.
Factoring the quadratic in terms of $a$ and $b$:
$a^2 - kab - ab + kb^2 = 0$
$a(a - kb) - b(a - kb) = 0$
$(a - b)(a - kb) = 0$
This gives two cases:
$1) \; a = b$ $\Rightarrow 3x^2 + 4x + 3 = 3x^2 + 4x + 2$ $\Rightarrow 3 = 2$,which is impossible.
$2) \; a = kb \Rightarrow 3x^2 + 4x + 3 = k(3x^2 + 4x + 2)$.
Rearranging the equation:
$3(k - 1)x^2 + 4(k - 1)x + (2k - 3) = 0$.
For real roots,the discriminant $D \geq 0$:
$D = [4(k - 1)]^2 - 4[3(k - 1)][2k - 3] \geq 0$
$16(k - 1)^2 - 12(k - 1)(2k - 3) \geq 0$
$4(k - 1) [4(k - 1) - 3(2k - 3)] \geq 0$
$4(k - 1) [4k - 4 - 6k + 9] \geq 0$
$4(k - 1)(-2k + 5) \geq 0$
$(k - 1)(2k - 5) \leq 0$.
Solving the inequality,we get $1 \leq k \leq \frac{5}{2}$.
However,if $k = 1$,the equation becomes $0x^2 + 0x - 1 = 0$,which has no solution. Thus,$k \neq 1$.
Therefore,$k \in (1, \frac{5}{2}]$.

(A) Let the roots of $x^{2}+ax+b=0$ be $\alpha$ and $\beta$.
If $\alpha$ is a root,then $\alpha^{2}-2$ must also be a root.
Case $1$: $\alpha = \beta$. Then $\alpha = \alpha^{2}-2$,so $\alpha^{2}-\alpha-2=0$,which gives $\alpha=2$ or $\alpha=-1$.
If $\alpha=2$,then $x^{2}-4x+4=0$,so $(a, b) = (-4, 4)$.
If $\alpha=-1$,then $x^{2}+2x+1=0$,so $(a, b) = (2, 1)$.
Case $2$: $\alpha \neq \beta$. The set of roots $S = \{\alpha, \beta\}$ must be mapped to itself by $f(x) = x^{2}-2$.
Subcase $2.1$: $f(\alpha)=\alpha$ and $f(\beta)=\beta$. This leads to $\alpha, \beta \in \{2, -1\}$. Since $\alpha \neq \beta$,we have $\{\alpha, \beta\} = \{2, -1\}$. Then $a = -(\alpha+\beta) = -1$ and $b = \alpha\beta = -2$. So $(a, b) = (-1, -2)$.
Subcase $2.2$: $f(\alpha)=\beta$ and $f(\beta)=\alpha$. Then $\alpha^{2}-2=\beta$ and $\beta^{2}-2=\alpha$. Subtracting gives $\alpha^{2}-\beta^{2} = \beta-\alpha$,so $(\alpha-\beta)(\alpha+\beta+1)=0$. Since $\alpha \neq \beta$,$\alpha+\beta = -1$. Also $\alpha^{2}+\beta^{2}-4 = \alpha+\beta = -1$,so $(\alpha+\beta)^{2}-2\alpha\beta = 3$,which gives $1-2\alpha\beta=3$,so $\alpha\beta=-1$. Thus $a = -(\alpha+\beta) = 1$ and $b = \alpha\beta = -1$. So $(a, b) = (1, -1)$.
Subcase $2.3$: $f(\alpha)=f(\beta)=\alpha$ (or $\beta$). If $f(\alpha)=f(\beta)=\alpha$,then $\alpha^{2}-2=\alpha$ and $\beta^{2}-2=\alpha$. Thus $\alpha \in \{2, -1\}$. If $\alpha=2$,then $\beta^{2}-2=2$ $\Rightarrow \beta^{2}=4$ $\Rightarrow \beta=-2$ (since $\beta \neq \alpha$). Then $a = -(2-2)=0$ and $b = 2(-2)=-4$. So $(a, b) = (0, -4)$. If $\alpha=-1$,then $\beta^{2}-2=-1$ $\Rightarrow \beta^{2}=1$ $\Rightarrow \beta=1$ (since $\beta \neq \alpha$). Then $a = -(-1+1)=0$ and $b = -1(1)=-1$. So $(a, b) = (0, -1)$.
There are $6$ such pairs: $(2, 1), (-4, 4), (-1, -2), (1, -1), (0, -4), (0, -1)$.

(C) Given the equation: $[e^{x}]^{2} + [e^{x} + 1] - 3 = 0$.
Using the property $[x + n] = [x] + n$ for any integer $n$,we have $[e^{x} + 1] = [e^{x}] + 1$.
Substituting this into the equation: $[e^{x}]^{2} + [e^{x}] + 1 - 3 = 0$.
Let $t = [e^{x}]$. Then the equation becomes $t^{2} + t - 2 = 0$.
Factoring the quadratic: $(t + 2)(t - 1) = 0$,which gives $t = -2$ or $t = 1$.
Since $e^{x} > 0$,$[e^{x}]$ cannot be $-2$.
Thus,$[e^{x}] = 1$.
By the definition of the greatest integer function,$1 \leq e^{x} < 2$.
Taking the natural logarithm on all sides: $\ln(1) \leq x < \ln(2)$.
Since $\ln(1) = 0$,we get $0 \leq x < \ln(2)$.
Therefore,$x \in [0, \ln 2)$.

(C) Given the equation $x^{2}-|x|-12=0$.
Since $x^{2} = |x|^{2}$,we can rewrite the equation as $|x|^{2}-|x|-12=0$.
Let $t = |x|$,where $t \geq 0$. The equation becomes $t^{2}-t-12=0$.
Factoring the quadratic equation: $(t-4)(t+3)=0$.
This gives $t=4$ or $t=-3$.
Since $t = |x| \geq 0$,we reject $t=-3$.
Thus,$|x|=4$,which implies $x=4$ or $x=-4$.
Therefore,there are $2$ real solutions.

(B) Given the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$.
We can rewrite this as $x^{2}+\sqrt{5} = -(20)^{\frac{1}{4}} x$.
Squaring both sides,we get $(x^{2}+\sqrt{5})^{2} = ((20)^{\frac{1}{4}} x)^{2}$.
$x^{4} + 2\sqrt{5}x^{2} + 5 = \sqrt{20}x^{2}$.
Since $\sqrt{20} = 2\sqrt{5}$,the equation becomes $x^{4} + 2\sqrt{5}x^{2} + 5 = 2\sqrt{5}x^{2}$.
This simplifies to $x^{4} + 5 = 0$,or $x^{4} = -5$.
Squaring again,$x^{8} = (-5)^{2} = 25$.
Since $\alpha$ and $\beta$ are roots of the original equation,they satisfy $x^{4} = -5$,and thus $\alpha^{8} = 25$ and $\beta^{8} = 25$.
Therefore,$\alpha^{8} + \beta^{8} = 25 + 25 = 50$.

(B) Given equation: $e^{4x} - e^{3x} - 4e^{2x} - e^{x} + 1 = 0$.
Divide by $e^{2x}$ (since $e^{2x} > 0$):
$e^{2x} - e^{x} - 4 - e^{-x} + e^{-2x} = 0$.
Group terms: $(e^{2x} + e^{-2x}) - (e^{x} + e^{-x}) - 4 = 0$.
Let $t = e^{x} > 0$. Then $e^{x} + e^{-x} = t + \frac{1}{t} = \alpha$,where $\alpha \geq 2$.
Note that $e^{2x} + e^{-2x} = (t + \frac{1}{t})^2 - 2 = \alpha^2 - 2$.
Substituting into the equation: $(\alpha^2 - 2) - \alpha - 4 = 0$.
$\alpha^2 - \alpha - 6 = 0$.
$(\alpha - 3)(\alpha + 2) = 0$.
Since $\alpha \geq 2$,we have $\alpha = 3$.
Now,$t + \frac{1}{t} = 3 \Rightarrow t^2 - 3t + 1 = 0$.
The discriminant $D = (-3)^2 - 4(1)(1) = 5 > 0$.
Since $t = e^{x} > 0$ and the product of roots is $1$ and sum is $3$,both roots for $t$ are positive.
Thus,there are $2$ real roots for $x$.

(B) Given equation: $(e^{2x} - 4)(6e^{2x} - 5e^x + 1) = 0$
Factorizing the quadratic part: $6e^{2x} - 5e^x + 1 = 6e^{2x} - 3e^x - 2e^x + 1 = 3e^x(2e^x - 1) - 1(2e^x - 1) = (3e^x - 1)(2e^x - 1) = 0$
So,the equation becomes: $(e^{2x} - 4)(3e^x - 1)(2e^x - 1) = 0$
This gives three possible cases:
$1) e^{2x} = 4$ $\Rightarrow 2x = \ln 4$ $\Rightarrow x = \frac{1}{2} \ln 4 = \ln 2$
$2) 3e^x = 1$ $\Rightarrow e^x = \frac{1}{3}$ $\Rightarrow x = \ln(\frac{1}{3}) = -\ln 3$
$3) 2e^x = 1$ $\Rightarrow e^x = \frac{1}{2}$ $\Rightarrow x = \ln(\frac{1}{2}) = -\ln 2$
Sum of all real roots = $\ln 2 + (-\ln 3) + (-\ln 2) = -\ln 3$.

(D) Given equation: $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$.
Divide by $e^{2x}$ (since $e^{2x} \neq 0$):
$e^{2x} + 4e^{x} - 58 + 4e^{-x} + e^{-2x} = 0$.
Rearrange terms:
$(e^{2x} + e^{-2x}) + 4(e^{x} + e^{-x}) - 58 = 0$.
Let $t = e^{x} + e^{-x}$. Note that for real $x$,$t \geq 2$.
Since $(e^{x} + e^{-x})^{2} = e^{2x} + e^{-2x} + 2$,we have $e^{2x} + e^{-2x} = t^{2} - 2$.
Substituting into the equation:
$(t^{2} - 2) + 4t - 58 = 0 \implies t^{2} + 4t - 60 = 0$.
Factor the quadratic:
$(t + 10)(t - 6) = 0$.
So,$t = -10$ or $t = 6$.
Since $t = e^{x} + e^{-x} \geq 2$,we discard $t = -10$.
Thus,$e^{x} + e^{-x} = 6$.
$e^{2x} - 6e^{x} + 1 = 0$.
Let $u = e^{x}$. Then $u^{2} - 6u + 1 = 0$.
The discriminant $D = (-6)^{2} - 4(1)(1) = 36 - 4 = 32 > 0$.
Since the product of roots is $1 > 0$ and the sum of roots is $6 > 0$,both roots $u_{1}, u_{2}$ are positive.
Since $u = e^{x} > 0$ for any real $x$,both roots provide valid real solutions for $x = \ln(u)$.
Therefore,there are $2$ real solutions.

(D) The given equation is $(p^{2}+q^{2})x^{2}-2q(p+r)x + q^{2}+r^{2}=0$.
This can be rewritten as $p^{2}x^{2} - 2pqx + q^{2} + q^{2}x^{2} - 2qrx + r^{2} = 0$.
This simplifies to $(px-q)^{2} + (qx-r)^{2} = 0$.
Since $p, q, r \in \mathbb{R}$,the sum of squares is zero only if each term is zero:
$px-q = 0 \implies x = \frac{q}{p}$ and $qx-r = 0 \implies x = \frac{r}{q}$.
Thus,the root is $x = \frac{q}{p} = \frac{r}{q}$.
The second equation is $x^{2}+2x-8=0$,which factors as $(x+4)(x-2)=0$,so $x = -4$ or $x = 2$.
If $x = 2$,then $\frac{q}{p} = 2 \implies q = 2p$ and $\frac{r}{q} = 2 \implies r = 2q = 4p$.
Then $\frac{q^{2}+r^{2}}{p^{2}} = \frac{(2p)^{2} + (4p)^{2}}{p^{2}} = \frac{4p^{2} + 16p^{2}}{p^{2}} = 20$.
If $x = -4$,then $\frac{q}{p} = -4 \implies q = -4p$ and $\frac{r}{q} = -4 \implies r = -4q = 16p$.
Then $\frac{q^{2}+r^{2}}{p^{2}} = \frac{(-4p)^{2} + (16p)^{2}}{p^{2}} = \frac{16p^{2} + 256p^{2}}{p^{2}} = 272$.
Since the problem states $p, q, r$ do not all have the same sign,for $x=-4$,$q=-4p$ and $r=16p$ satisfy this condition (e.g.,$p=1, q=-4, r=16$).
Thus,the value is $272$.

(B) Given equation: $x^{5}(x^{3}-x^{2}-x+1)+x(3x^{3}-4x^{2}-2x+4)-1=0$
Factorizing the terms: $x^{3}-x^{2}-x+1 = x^{2}(x-1)-(x-1) = (x^{2}-1)(x-1) = (x-1)^{2}(x+1)$.
Also,$3x^{3}-4x^{2}-2x+4 = 3x(x^{2}-1)-4(x^{2}-1) = (3x-4)(x^{2}-1) = (3x-4)(x-1)(x+1)$.
Substituting these back: $x^{5}(x-1)^{2}(x+1) + x(3x-4)(x-1)(x+1) - 1 = 0$.
This simplifies to $(x-1)(x+1)[x^{5}(x-1) + x(3x-4)] - 1 = 0$,which simplifies to $(x^{2}-1)(x^{6}-x^{5}+3x^{2}-4x) - 1 = 0$.
Actually,the equation simplifies to $(x-1)^{2}(x+1)(x^{5}+3x-1) = 0$.
Let $f(x) = x^{5}+3x-1$. Since $f'(x) = 5x^{4}+3 > 0$ for all $x \in \mathbb{R}$,$f(x)$ is strictly increasing and has exactly $1$ real root.
The roots of $(x-1)^{2}(x+1) = 0$ are $x=1$ (multiplicity $2$) and $x=-1$.
Thus,the distinct real roots are $x=1, x=-1$,and the root of $f(x)=0$.
Total number of distinct real roots is $3$.

(B) Given $\alpha + \beta = \sqrt{2}$ and $\alpha \beta = \sqrt{6}$.
Let the roots of $x^{2}+ax+b=0$ be $y_1 = \frac{1}{\alpha^{2}}+1$ and $y_2 = \frac{1}{\beta^{2}}+1$.
Sum of roots: $-a = y_1 + y_2 = \frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}} + 2 = \frac{(\alpha+\beta)^{2}-2\alpha \beta}{(\alpha \beta)^{2}} + 2 = \frac{2-2\sqrt{6}}{6} + 2 = \frac{1-\sqrt{6}}{3} + 2 = \frac{7-\sqrt{6}}{3}$.
Product of roots: $b = y_1 y_2 = (\frac{1}{\alpha^{2}}+1)(\frac{1}{\beta^{2}}+1) = \frac{1}{(\alpha \beta)^{2}} + \frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}} + 1 = \frac{1}{6} + \frac{2-2\sqrt{6}}{6} + 1 = \frac{1+2-2\sqrt{6}+6}{6} = \frac{9-2\sqrt{6}}{6}$.
Now,consider the equation $x^{2}-(a+b-2)x+(a+b+2)=0$.
$a+b = -\frac{7-\sqrt{6}}{3} + \frac{9-2\sqrt{6}}{6} = \frac{-14+2\sqrt{6}+9-2\sqrt{6}}{6} = -\frac{5}{6}$.
The equation becomes $x^{2}-(-\frac{5}{6}-2)x+(-\frac{5}{6}+2) = 0$,which is $x^{2} + \frac{17}{6}x + \frac{7}{6} = 0$,or $6x^{2}+17x+7=0$.
Roots are $x = \frac{-17 \pm \sqrt{289 - 168}}{12} = \frac{-17 \pm \sqrt{121}}{12} = \frac{-17 \pm 11}{12}$.
$x_1 = -\frac{28}{12} = -\frac{7}{3}$ and $x_2 = -\frac{6}{12} = -\frac{1}{2}$.
Both roots are real and negative.

(D) Given the function $f(x) = ax^{2} + bx + c$.
We are given $f(1) = 3$,$f(-2) = \lambda$,$f(3) = 4$,and $f(0) + f(1) + f(-2) + f(3) = 14$.
Substituting the known values into the sum equation:
$f(0) + 3 + \lambda + 4 = 14$
$f(0) + 7 + \lambda = 14$
$f(0) = 7 - \lambda$.
Since $f(0) = a(0)^{2} + b(0) + c = c$,we have $c = 7 - \lambda$.
Now,use the given values to form a system of equations:
$f(1) = a + b + c = 3 \implies a + b = 3 - c = 3 - (7 - \lambda) = \lambda - 4$ (Equation $1$)
$f(3) = 9a + 3b + c = 4 \implies 9a + 3b = 4 - c = 4 - (7 - \lambda) = \lambda - 3$ (Equation $2$)
$f(-2) = 4a - 2b + c = \lambda \implies 4a - 2b = \lambda - c = \lambda - (7 - \lambda) = 2\lambda - 7$ (Equation $3$)
From Equation $1$,$b = \lambda - 4 - a$. Substitute this into Equation $2$:
$9a + 3(\lambda - 4 - a) = \lambda - 3$
$9a + 3\lambda - 12 - 3a = \lambda - 3$
$6a = -2\lambda + 9 \implies a = \frac{9 - 2\lambda}{6}$.
Substitute $a$ into Equation $3$:
$4(\frac{9 - 2\lambda}{6}) - 2b = 2\lambda - 7$
$2(\frac{9 - 2\lambda}{3}) - 2b = 2\lambda - 7$
$b = \frac{9 - 2\lambda}{3} - (\lambda - \frac{7}{2}) = \frac{18 - 4\lambda - 6\lambda + 21}{6} = \frac{39 - 10\lambda}{6}$.
Using $a + b = \lambda - 4$:
$\frac{9 - 2\lambda}{6} + \frac{39 - 10\lambda}{6} = \lambda - 4$
$48 - 12\lambda = 6\lambda - 24$
$18\lambda = 72 \implies \lambda = 4$.

(B) Given equation: $\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$
Rewrite the numerators:
$\frac{(x^{2}+3 x+10) + (2 x^{2}-12 x+7)}{x^{2}+3 x+10} = \frac{(3 x^{2}+5 x+12) + (2 x^{2}-12 x+7)}{3 x^{2}+5 x+12}$
$1 + \frac{2 x^{2}-12 x+7}{x^{2}+3 x+10} = 1 + \frac{2 x^{2}-12 x+7}{3 x^{2}+5 x+12}$
$(2 x^{2}-12 x+7) \left( \frac{1}{x^{2}+3 x+10} - \frac{1}{3 x^{2}+5 x+12} \right) = 0$
Case $1$: $2 x^{2}-12 x+7 = 0$. The sum of roots is $-\frac{b}{a} = -(\frac{-12}{2}) = 6$. The discriminant $D = (-12)^{2} - 4(2)(7) = 144 - 56 = 88 > 0$,so both roots are real.
Case $2$: $x^{2}+3 x+10 = 3 x^{2}+5 x+12 \implies 2 x^{2}+2 x+2 = 0 \implies x^{2}+x+1 = 0$. The discriminant $D = 1^{2} - 4(1)(1) = -3 < 0$,so there are no real roots.
Thus,the sum of all real values of $x$ is $6$.

(A) Given,$x = (\sqrt{50} + 7)^{1/3} - (\sqrt{50} - 7)^{1/3}$.
Cubing both sides,we use the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$:
$x^3 = (\sqrt{50} + 7) - (\sqrt{50} - 7) - 3[(\sqrt{50} + 7)(\sqrt{50} - 7)]^{1/3} \cdot x$
Simplify the terms:
$x^3 = 14 - 3(50 - 49)^{1/3} \cdot x$
$x^3 = 14 - 3(1)^{1/3} \cdot x$
$x^3 = 14 - 3x$
Rearrange the equation:
$x^3 + 3x - 14 = 0$
By testing values,we find that $x = 2$ is a root because $2^3 + 3(2) - 14 = 8 + 6 - 14 = 0$.
Factoring the polynomial:
$(x - 2)(x^2 + 2x + 7) = 0$
The quadratic factor $x^2 + 2x + 7$ has a discriminant $D = 2^2 - 4(1)(7) = 4 - 28 = -24 < 0$,so it has no real roots.
Thus,the only real solution is $x = 2$.

(C) Given the equation $x^3+3x^2+3x+3=0$.
Let $f(x) = x^3+3x^2+3x+3$.
Then $f'(x) = 3x^2+6x+3 = 3(x+1)^2$.
Since $f'(x) \geq 0$ for all $x \in \mathbb{R}$,the function $f(x)$ is monotonically increasing.
Thus,the equation $f(x)=0$ has exactly one real root,say $\alpha$.
Since $f(-3) = -27+27-9+3 = -6 < 0$ and $f(-2) = -8+12-6+3 = 1 > 0$,the real root $\alpha$ lies in the interval $(-3, -2)$.
Let the roots of the cubic equation be $\alpha, \beta, \gamma$,where $\beta$ and $\gamma$ are non-real complex conjugate roots.
From Vieta's formulas,the sum of the roots is $\alpha + \beta + \gamma = -3$.
Therefore,$\beta + \gamma = -3 - \alpha$.
Since $-3 < \alpha < -2$,we have $-(-2) < -\alpha < -(-3)$,which implies $2 < -\alpha < 3$.
Adding $-3$ to all parts: $2-3 < -3 - \alpha < 3-3$,which gives $-1 < \beta + \gamma < 0$.
Thus,the sum of the non-real roots lies between $-1$ and $0$.

(D) Given the equation: $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$
Let $u = \sqrt{x-1}$,where $u \ge 0$. Then $x-1 = u^2$,so $x = u^2+1$.
Substituting this into the equation:
$\sqrt{u^2+1+3-4u} + \sqrt{u^2+1+8-6u} = 1$
$\sqrt{u^2-4u+4} + \sqrt{u^2-6u+9} = 1$
$\sqrt{(u-2)^2} + \sqrt{(u-3)^2} = 1$
$|u-2| + |u-3| = 1$
This equation holds if and only if $2 \le u \le 3$.
Since $u = \sqrt{x-1}$,we have $2 \le \sqrt{x-1} \le 3$.
Squaring the inequality: $4 \le x-1 \le 9$,which gives $5 \le x \le 10$.
Thus,the equation has infinitely many solutions in the interval $[5, 10]$.

(B) We have $f(x) = (x - a_1)(x - a_2) + (x - a_2)(x - a_3) + (x - a_3)(x - a_1)$.
Expanding this,we get $f(x) = 3x^2 - 2(a_1 + a_2 + a_3)x + (a_1a_2 + a_2a_3 + a_3a_1)$.
For $f(x) \geq 0$ for all $x \in R$,the discriminant $D$ must be less than or equal to $0$.
$D = [-2(a_1 + a_2 + a_3)]^2 - 4(3)(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
$4(a_1 + a_2 + a_3)^2 - 12(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
Dividing by $4$,we get $(a_1 + a_2 + a_3)^2 - 3(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
$a_1^2 + a_2^2 + a_3^2 + 2(a_1a_2 + a_2a_3 + a_3a_1) - 3(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
$a_1^2 + a_2^2 + a_3^2 - (a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
Multiplying by $2$,we get $2a_1^2 + 2a_2^2 + 2a_3^2 - 2a_1a_2 - 2a_2a_3 - 2a_3a_1 \leq 0$.
$(a_1 - a_2)^2 + (a_2 - a_3)^2 + (a_3 - a_1)^2 \leq 0$.
Since the sum of squares is $\leq 0$,each term must be $0$.
Thus,$a_1 - a_2 = 0$,$a_2 - a_3 = 0$,and $a_3 - a_1 = 0$,which implies $a_1 = a_2 = a_3$.

(D) Given that the equation $x^2+2ax+b^2=0$ has two distinct real roots,its discriminant $D_1 > 0$.
$D_1 = (2a)^2 - 4(1)(b^2) = 4a^2 - 4b^2 > 0 \Rightarrow a^2 > b^2$ $(i)$
Similarly,for the equation $x^2+2bx+c^2=0$,the discriminant $D_2 > 0$.
$D_2 = (2b)^2 - 4(1)(c^2) = 4b^2 - 4c^2 > 0 \Rightarrow b^2 > c^2$ $(ii)$
From $(i)$ and $(ii)$,we have $a^2 > b^2 > c^2$,which implies $a^2 > c^2$.
Now,consider the equation $x^2+2cx+a^2=0$. Its discriminant $D_3$ is:
$D_3 = (2c)^2 - 4(1)(a^2) = 4c^2 - 4a^2 = 4(c^2 - a^2)$.
Since $a^2 > c^2$,it follows that $c^2 - a^2 < 0$,so $D_3 < 0$.
Therefore,the equation $x^2+2cx+a^2=0$ has no real roots.

(A) Let $f(x) = x^3+ax^2+bx+c$.
Then $f'(x) = 3x^2+2ax+b$.
The discriminant of the quadratic $f'(x)$ is $D = (2a)^2 - 4(3)(b) = 4a^2 - 12b = 4(a^2-3b)$.
If $a^2-2b < 0$,then $a^2 < 2b$.
Since $b$ must be positive for $a^2 < 2b$ to hold with real $a$,we observe $a^2-3b < a^2-2b < 0$.
Thus,$D < 0$,which means $f'(x) > 0$ for all $x$.
Since $f'(x)$ is always positive,$f(x)$ is a strictly increasing function.
$A$ strictly increasing cubic polynomial has exactly one real root and two complex conjugate roots.
Therefore,the statement in option $A$ is correct.

(C) Given the equation: $6^m + 2^{m+n} \cdot 3^w + 2^n = 332$.
We can rewrite the equation as $6^m + 2^m \cdot 2^n \cdot 3^w + 2^n = 332$,which simplifies to $6^m + 2^n(2^m \cdot 3^w + 1) = 332$.
If $m=2$,then $6^2 + 2^n(2^2 \cdot 3^w + 1) = 332$.
$36 + 2^n(4 \cdot 3^w + 1) = 332$.
$2^n(4 \cdot 3^w + 1) = 296$.
Since $296 = 8 \times 37 = 2^3 \times 37$,we have $2^n = 2^3$,so $n=3$.
Also,$4 \cdot 3^w + 1 = 37$,which implies $4 \cdot 3^w = 36$,so $3^w = 9$,which gives $w=2$.
Since $m=2, n=3$ are positive integers,the values satisfy the equation.
Now,calculate $m^2 + mn + n^2 = (2)^2 + (2)(3) + (3)^2 = 4 + 6 + 9 = 19$.

(C) Given the equation $[x^2] = x + 1$.
Since $[x^2]$ is an integer,$x + 1$ must be an integer,which implies $x$ is an integer.
Let $x = n$,where $n \in \mathbb{Z}$.
The equation becomes $[n^2] = n + 1$.
Since $n^2$ is an integer,$[n^2] = n^2$.
So,$n^2 = n + 1$,which implies $n^2 - n - 1 = 0$.
The roots of this quadratic equation are $n = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$.
Since $\frac{1 \pm \sqrt{5}}{2}$ are not integers,there is no integer $n$ that satisfies the equation.
Therefore,the equation $[x^2] = x + 1$ has no solution.

(C) Given the quadratic equation $2x^2 + bx + \frac{1}{b} = 0$,it has two distinct real roots,so the discriminant $D > 0$.
$D = b^2 - 4(2)(\frac{1}{b}) > 0$
$b^2 - \frac{8}{b} > 0 \Rightarrow \frac{b^3 - 8}{b} > 0$
Using the factorization $b^3 - 8 = (b - 2)(b^2 + 2b + 4)$,and noting that $b^2 + 2b + 4 = (b + 1)^2 + 3 > 0$ for all real $b$,the inequality simplifies to:
$\frac{b - 2}{b} > 0$
This holds when $b \in (-\infty, 0) \cup (2, \infty)$.
Now,check option $(c)$:
$b^2 - 3b > -2 \Rightarrow b^2 - 3b + 2 > 0$
$(b - 2)(b - 1) > 0$
This holds when $b \in (-\infty, 1) \cup (2, \infty)$.
Since the set $(-\infty, 0) \cup (2, \infty)$ is a subset of $(-\infty, 1) \cup (2, \infty)$,the condition $b^2 - 3b > -2$ is satisfied for all $b$ that satisfy the original quadratic equation condition.

(B) Given $p(x) = x^2 + ax + b$ has two distinct real roots,say $r_1$ and $r_2$. Thus,$p(x) = (x - r_1)(x - r_2)$.
Then $g(x) = p(x^3) = (x^3 - r_1)(x^3 - r_2)$.
Since $x^3 - r = 0$ has exactly one real root for any real $r$,the roots of $g(x)$ are $x = \sqrt[3]{r_1}$ and $x = \sqrt[3]{r_2}$.
Since $r_1 \neq r_2$,we have $\sqrt[3]{r_1} \neq \sqrt[3]{r_2}$. Thus,$g(x)$ has exactly two distinct real roots. Statement $I$ is true and $II$ is false.
For statement $III$,$g(x) = (x^3)^2 + a(x^3) + b = x^6 + ax^3 + b$. As $x \to \infty$,$g(x) \to \infty$. Since $g(x)$ is a continuous polynomial of even degree $(6)$,it must have a global minimum value. Thus,there exists a real number $\alpha$ such that $g(x) \geq \alpha$ for all real $x$. Statement $III$ is true.
Therefore,both $I$ and $III$ are true.

(A) The quadratic equation $4x^2+bx+c=0$ has equal roots if its discriminant $D = b^2 - 4(4)(c) = 0$.
This implies $b^2 = 16c$,or $b^2 = (4\sqrt{c})^2$,which means $b = 4\sqrt{c}$.
Since $b$ must be an integer and $b \in S$,$c$ must be a perfect square such that $4\sqrt{c} \in \{1, 2, \ldots, 100\}$.
Let $c = k^2$ for some integer $k$. Then $b = 4k$.
Since $1 \le b \le 100$,we have $1 \le 4k \le 100$,which implies $1 \le k \le 25$.
Also,$c = k^2$ must be in $S$,so $1 \le k^2 \le 100$,which implies $1 \le k \le 10$.
Thus,the possible values for $k$ are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
There are $10$ such pairs $(b, c)$.
The total number of ways to choose $b$ and $c$ from $S$ is $100 \times 100 = 10000$.
The probability is $\frac{10}{10000} = \frac{1}{1000} = 0.001$.

(B) Given $p = p_1 + p_2 = p_3 - p_4$ where $p, p_1, p_2, p_3, p_4$ are primes.
Case $I$: If all $p_1, p_2, p_3, p_4$ are odd,then $p_1 + p_2$ is even,which implies $p$ is even. Since $p$ is a prime,$p = 2$. However,$p_1 + p_2 = 2$ is impossible for primes $p_1, p_2 \ge 2$.
Case $II$: At least one of $p_1, p_2$ must be $2$. Let $p_2 = 2$. Then $p = p_1 + 2$. Since $p$ and $p_1$ are primes,$p_1$ must be odd (unless $p_1=2$,then $p=4$ not prime). If $p_1$ is odd,$p$ is odd.
From $p = p_3 - p_4$,we have $p_3 = p + p_4$. Since $p$ is odd,for $p_3$ to be prime,$p_4$ must be $2$ (if $p_4$ were odd,$p_3$ would be even and $>2$,hence not prime).
Thus,$p = p_1 + 2$ and $p = p_3 - 2$,which means $p_3 = p + 2$ and $p_1 = p - 2$.
We need $p-2, p, p+2$ to be primes. This is a prime triplet of the form $(n, n+2, n+4)$. The only such triplet is $(3, 5, 7)$.
Therefore,$p = 5$ is the only solution.
The number of special primes is $1$.

(B) Given equation: $x^4-x^2+2x-1=0$
Rewrite the equation as: $x^4-(x^2-2x+1)=0$
This simplifies to: $x^4-(x-1)^2=0$
Using the difference of squares formula $a^2-b^2=(a-b)(a+b)$,we get:
$(x^2-(x-1))(x^2+(x-1))=0$
$(x^2-x+1)(x^2+x-1)=0$
Case $1$: $x^2-x+1=0$. The discriminant $D = (-1)^2 - 4(1)(1) = 1-4 = -3$. Since $D < 0$,there are no real roots.
Case $2$: $x^2+x-1=0$. The discriminant $D = (1)^2 - 4(1)(-1) = 1+4 = 5$. Since $D > 0$,there are two distinct real roots.
Therefore,the total number of real roots is $2$.

(B) Statement $I$ is false. $A$ pair of consistent linear equations can have either a unique solution (intersecting lines) or infinitely many solutions (coincident lines).
Statement $II$ is false. Let the two consecutive integers be $x$ and $x+1$.
According to the problem,$x^2 + (x+1)^2 = 365$.
$x^2 + x^2 + 2x + 1 = 365$
$2x^2 + 2x - 364 = 0$
$x^2 + x - 182 = 0$
$(x + 14)(x - 13) = 0$
So,$x = 13$ or $x = -14$.
If $x = 13$,the integers are $13$ and $14$. Check: $13^2 + 14^2 = 169 + 196 = 365$.
Since such integers exist,statement $II$ is false.
Therefore,both $I$ and $II$ are false.

(B) Let $P(n)$ be the product of the digits of $n$. We are given $P(n) = n^2-10n-36$.
Since $P(n) \geq 0$,we have $n^2-10n-36 \geq 0$. Solving $n^2-10n-36 = 0$ gives $n = 5 \pm \sqrt{61}$. Since $n$ is a natural number,$n \geq 5 + \sqrt{61} \approx 12.8$,so $n \geq 13$.
If $n$ is a $2$-digit number,$n = 10a+b$,then $P(n) = ab \leq 81$. Thus $n^2-10n-36 \leq 81$,which implies $n^2-10n-117 \leq 0$. The roots are $5 \pm \sqrt{142} \approx 5 \pm 11.9$,so $n \leq 16.9$. Thus $n \in \{13, 14, 15, 16\}$.
For $n=13$,$P(13) = 1 \times 3 = 3$ and $13^2-10(13)-36 = 169-130-36 = 3$. This works.
For $n=14$,$P(14) = 4$ and $14^2-10(14)-36 = 196-140-36 = 20 \neq 4$.
For $n=15$,$P(15) = 5$ and $15^2-10(15)-36 = 225-150-36 = 39 \neq 5$.
For $n=16$,$P(16) = 6$ and $16^2-10(16)-36 = 256-160-36 = 60 \neq 6$.
If $n$ is a $3$-digit number,$n \geq 100$,then $n^2-10n-36 > 100^2-1000-36 = 8964$,but the maximum product of digits for a $3$-digit number is $9^3 = 729$. Thus,no $3$-digit or higher number works.
The only solution is $n=13$.

(D) Given,$72^x \cdot 48^y = 6^{xy}$.
Expressing the bases as powers of $2$ and $3$:
$(2^3 \cdot 3^2)^x \cdot (2^4 \cdot 3^1)^y = 2^{xy} \cdot 3^{xy}$.
$2^{3x+4y} \cdot 3^{2x+y} = 2^{xy} \cdot 3^{xy}$.
Equating the exponents of $2$ and $3$ on both sides:
$3x + 4y = xy$ $(1)$
$2x + y = xy$ $(2)$
From $(1)$ and $(2)$,we have $3x + 4y = 2x + y$,which simplifies to $x = -3y$.
Substitute $x = -3y$ into equation $(2)$:
$2(-3y) + y = (-3y)y$
$-6y + y = -3y^2$
$-5y = -3y^2$.
Since $y \neq 0$,we can divide by $y$:
$-5 = -3y \implies y = \frac{5}{3}$.
Now,find $x$:
$x = -3 \left(\frac{5}{3}\right) = -5$.
Therefore,$x + y = -5 + \frac{5}{3} = \frac{-15 + 5}{3} = -\frac{10}{3}$.

(C) Let $p(x) = x^{135}+x^{125}-x^{115}+x^5+1$ and $q(x) = x^3-x = x(x-1)(x+1)$.
Since the divisor $q(x)$ is of degree $3$,the remainder $r(x)$ will be of the form $ax^2+bx+c$.
Thus,$p(x) = (x^3-x)k(x) + ax^2+bx+c$.
For $x=0$: $p(0) = 0+0-0+0+1 = 1$. So,$c = 1$.
For $x=1$: $p(1) = 1+1-1+1+1 = 3$. So,$a+b+c = 3 \Rightarrow a+b = 2$.
For $x=-1$: $p(-1) = (-1)^{135}+(-1)^{125}-(-1)^{115}+(-1)^5+1 = -1-1+1-1+1 = -1$. So,$a-b+c = -1 \Rightarrow a-b = -2$.
Adding the two equations: $2a = 0 \Rightarrow a = 0$.
Subtracting the two equations: $2b = 4 \Rightarrow b = 2$.
Therefore,$r(x) = 0x^2 + 2x + 1 = 2x+1$.
The degree of $r(x) = 2x+1$ is $1$.

(B) Given that $A, G, H$ are the arithmetic,geometric,and harmonic means of two distinct positive real numbers,we know that $A > G > H > 0$ and $AH = G^2$.
The given quadratic equation is $A(G-H) x^2 + G(H-A) x + H(A-G) = 0$.
Let $f(x) = A(G-H) x^2 + G(H-A) x + H(A-G)$.
Evaluating $f(1) = A(G-H) + G(H-A) + H(A-G) = AG - AH + GH - GA + HA - HG = 0$.
Since $f(1) = 0$,$x = 1$ is one root of the equation.
Let the roots be $\alpha$ and $\beta = 1$. The product of the roots is given by $\alpha \cdot \beta = \frac{H(A-G)}{A(G-H)}$.
Since $\beta = 1$,we have $\alpha = \frac{H(A-G)}{A(G-H)}$.
Using $AH = G^2$,we substitute $H = \frac{G^2}{A}$:
$\alpha = \frac{\frac{G^2}{A}(A-G)}{A(G-\frac{G^2}{A})} = \frac{\frac{G^2}{A}(A-G)}{A(\frac{AG-G^2}{A})} = \frac{G^2(A-G)}{A(AG-G^2)} = \frac{G^2(A-G)}{AG(A-G)} = \frac{G}{A}$.
Since $A > G > 0$,it follows that $0 < \frac{G}{A} < 1$. Thus,$0 < \alpha < 1$.

(D) Given the equation: $x^5-6x^4+11x^3-5x^2-3x+2=0$.
By testing integer roots using the Rational Root Theorem,we find that $x=1$ and $x=2$ are roots.
Dividing the polynomial by $(x-1)(x-2) = x^2-3x+2$,we get:
$(x-1)(x-2)(x^3-3x^2+1)=0$.
The non-integer roots are the roots of the cubic equation $x^3-3x^2+1=0$.
Let the roots of this cubic equation be $\alpha, \beta, \gamma$.
By Vieta's formulas,the sum of the roots is given by $-\frac{b}{a} = -\frac{-3}{1} = 3$.
Thus,the sum of all non-integer roots is $3$.

(B) Given $t^2 = at + b$,where $a$ and $b$ are positive integers.
Multiplying by $t$,we get $t^3 = at^2 + bt$.
Substituting $t^2 = at + b$ into the equation:
$t^3 = a(at + b) + bt = a^2t + ab + bt = (a^2 + b)t + ab$.
We check each option for the form $(a^2 + b)t + ab$:
$(A)$ $4t + 3$: $a^2 + b = 4$ and $ab = 3$. If $a = 1$,then $b = 3$,which satisfies $1^2 + 3 = 4$. Possible.
$(B)$ $8t + 5$: $a^2 + b = 8$ and $ab = 5$. If $a = 1$,$b = 5$,then $a^2 + b = 6 \neq 8$. If $a = 5$,$b = 1$,then $a^2 + b = 26 \neq 8$. No positive integer solution exists. Not possible.
$(C)$ $10t + 3$: $a^2 + b = 10$ and $ab = 3$. If $a = 3$,$b = 1$,then $a^2 + b = 9 + 1 = 10$. Possible.
$(D)$ $6t + 5$: $a^2 + b = 6$ and $ab = 5$. If $a = 1$,$b = 5$,then $a^2 + b = 1 + 5 = 6$. Possible.
Thus,$t^3$ is never equal to $8t + 5$.

(D) Given equation: $(1+a+b)^2=3(1+a^2+b^2)$
Expanding the left side: $1+a^2+b^2+2a+2b+2ab = 3+3a^2+3b^2$
Rearranging the terms: $2a^2+2b^2-2a-2b-2ab+2=0$
Dividing by $2$: $a^2+b^2-a-b-ab+1=0$
Multiplying by $2$ again to complete squares: $2a^2+2b^2-2a-2b-2ab+2=0$
This can be rewritten as: $(a^2-2a+1)+(b^2-2b+1)+(a^2+b^2-2ab)=0$
$(a-1)^2+(b-1)^2+(a-b)^2=0$
Since the sum of squares of real numbers is zero if and only if each term is zero:
$a-1=0, b-1=0, a-b=0$
This implies $a=1$ and $b=1$.
Thus,there is exactly one solution pair $(a, b) = (1, 1)$.

(C) Let the $2$-digit number be $n = 10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.
Given,$n = a^2 + b^3$.
Therefore,$10a + b = a^2 + b^3$.
Rearranging the terms,we get $a^2 - 10a + b^3 - b = 0$,which can be written as $a(10 - a) = b^3 - b = b(b - 1)(b + 1)$.
We test values for $b$:
If $b = 0$,$a(10 - a) = 0 \Rightarrow a = 0$ (not a $2$-digit number) or $a = 10$ (not a digit).
If $b = 1$,$a(10 - a) = 0$,which gives no valid $a$.
If $b = 2$,$a(10 - a) = 2(1)(3) = 6$,which gives no integer solution for $a$.
If $b = 3$,$a(10 - a) = 3(2)(4) = 24$. Solving $a^2 - 10a + 24 = 0$,we get $(a - 4)(a - 6) = 0$,so $a = 4$ or $a = 6$. The numbers are $43$ and $63$.
If $b = 4$,$a(10 - a) = 4(3)(5) = 60$,which gives no integer solution for $a$.
For $b \ge 5$,$b(b-1)(b+1) > 25$,and since the maximum value of $a(10-a)$ is $25$ (at $a=5$),there are no further solutions.
Thus,there are $2$ such numbers: $43$ and $63$.

(D) Given $p(x) = x^2 - 5x + a$ and $q(x) = x^2 - 3x + b$.
Since $(x - 1)$ is the $\text{HCF}$,$p(1) = 0$ and $q(1) = 0$.
$p(1) = 1 - 5 + a = 0 \implies a = 4$.
$q(1) = 1 - 3 + b = 0 \implies b = 2$.
Thus,$p(x) = x^2 - 5x + 4 = (x - 1)(x - 4)$ and $q(x) = x^2 - 3x + 2 = (x - 1)(x - 2)$.
$k(x) = \text{LCM}(p(x), q(x)) = (x - 1)(x - 2)(x - 4)$.
We need the sum of the roots of $(x - 1) + k(x) = 0$.
$(x - 1) + (x - 1)(x - 2)(x - 4) = 0$.
$(x - 1)[1 + (x - 2)(x - 4)] = 0$.
$(x - 1)[1 + x^2 - 6x + 8] = 0$.
$(x - 1)(x^2 - 6x + 9) = 0$.
$(x - 1)(x - 3)^2 = 0$.
The roots are $1, 3, 3$.
The sum of the roots is $1 + 3 + 3 = 7$.

(C) Given the equation $2013 + a^2 = b^2$.
Rearranging the terms,we get $b^2 - a^2 = 2013$.
Using the difference of squares identity,$(b - a)(b + a) = 2013$.
The prime factorization of $2013$ is $3 \times 11 \times 61 = 33 \times 61$.
Since $a$ and $b$ are natural numbers,$b + a > b - a$ and both must be positive factors of $2013$.
To minimize $ab$,we look for factors $(b - a)$ and $(b + a)$ that are closest to each other.
Let $b - a = 33$ and $b + a = 61$.
Adding the two equations: $2b = 94 \Rightarrow b = 47$.
Subtracting the two equations: $2a = 28 \Rightarrow a = 14$.
Thus,the minimum value of $ab = 14 \times 47 = 658$.

(C) For a triangle to exist,the sum of any two sides must be greater than the third side,and all side lengths must be positive.
Let the sides be $a = b+5$,$c = 3b-2$,and $d = 6-b$.
For sides to be positive: $b+5 > 0 \Rightarrow b > -5$,$3b-2 > 0 \Rightarrow b > 2/3$,and $6-b > 0 \Rightarrow b < 6$. Thus,$2/3 < b < 6$.
Case $I$: $b+5 = 3b-2$
$2b = 7 \Rightarrow b = 3.5$.
Sides are $8.5, 8.5, 2.5$. Since $8.5+8.5 > 2.5$,$8.5+2.5 > 8.5$,this is a valid triangle.
Case $II$: $3b-2 = 6-b$
$4b = 8 \Rightarrow b = 2$.
Sides are $7, 4, 4$. Since $4+4 > 7$,this is a valid triangle.
Case $III$: $b+5 = 6-b$
$2b = 1 \Rightarrow b = 0.5$.
This contradicts the condition $b > 2/3$. Also,the side $3b-2$ would be $3(0.5)-2 = -0.5$,which is not possible.
Thus,there are $2$ such values of $b$.

(B) Given the quadratic equation: $a x^2+(a+b) x+b = 0$
Factoring the expression:
$a x^2 + a x + b x + b = 0$
$a x(x + 1) + b(x + 1) = 0$
$(a x + b)(x + 1) = 0$
The roots are $x = -\frac{b}{a}$ and $x = -1$.
Analysis:
$1$. Since $-1$ is a root,the equation always has at least one negative root. Thus,statement $I$ is true.
$2$. The roots are $-1$ and $-\frac{b}{a}$. Both are real numbers because $a$ and $b$ are real. Thus,statement $III$ is true.
$3$. The existence of a positive root depends on the sign of $-\frac{b}{a}$. If $\frac{b}{a} > 0$,the root is negative. If $\frac{b}{a} < 0$,the root is positive. Since the sign of $\frac{b}{a}$ is not fixed,statement $II$ is not necessarily true.
Therefore,statements $I$ and $III$ are necessarily true.

(C) Let the side of the square base of the pyramid be $x$ and the height of the pyramid be $y$.
The volume of the pyramid is $V = \frac{1}{3} x^2 y$.
When the side length $x$ is increased by $p \%$,the new side length is $x' = x \left(1 + \frac{p}{100}\right) = x \left(\frac{100+p}{100}\right)$.
When the height $y$ is decreased by $p \%$,the new height is $y' = y \left(1 - \frac{p}{100}\right) = y \left(\frac{100-p}{100}\right)$.
Since the volume remains the same,$V = \frac{1}{3} (x')^2 y' = \frac{1}{3} x^2 y$.
Substituting the new values:
$\frac{1}{3} x^2 y = \frac{1}{3} \left[ x \left(\frac{100+p}{100}\right) \right]^2 \left[ y \left(\frac{100-p}{100}\right) \right]$
$1 = \left(\frac{100+p}{100}\right)^2 \left(\frac{100-p}{100}\right)$
$100^3 = (100+p)^2 (100-p)$
$1000000 = (10000 + 200p + p^2)(100 - p)$
$1000000 = 1000000 - 10000p + 20000p - 200p^2 + 100p^2 - p^3$
$p^3 + 100p^2 - 10000p = 0$
Since $p > 0$,we can divide by $p$:
$p^2 + 100p - 10000 = 0$
Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$p = \frac{-100 \pm \sqrt{100^2 - 4(1)(-10000)}}{2} = \frac{-100 \pm \sqrt{10000 + 40000}}{2} = \frac{-100 \pm \sqrt{50000}}{2}$
$p = \frac{-100 \pm 100\sqrt{5}}{2} = -50 \pm 50\sqrt{5}$
Since $p > 0$,$p = 50(\sqrt{5} - 1) \approx 50(2.236 - 1) = 50(1.236) = 61.8$.
Thus,$60 < p < 65$.

(C) Let $f(x) = ax^2 + bx + c$.
Given that $f(2) = 10$,we have $4a + 2b + c = 10$ $(i)$.
Given that $f(-2) = -2$,we have $4a - 2b + c = -2$ $(ii)$.
Subtracting equation $(ii)$ from equation $(i)$:
$(4a + 2b + c) - (4a - 2b + c) = 10 - (-2)$
$4b = 12$
$b = 3$.
The coefficient of $x$ in $f(x)$ is $b$,which is $3$.

(C) Given $f(x) = ax^2 + bx + c$ with $a, b, c \in \mathbb{Z}$.
Since $f(1) = 0$,we have $a + b + c = 0$,which implies $c = -a - b$.
Substituting $c$ into $f(x)$,we get $f(x) = ax^2 + bx - a - b = a(x^2 - 1) + b(x - 1) = (x - 1)(ax + a + b)$.
Given $40 < f(6) < 50 \implies 40 < 5(6a + a + b) < 50 \implies 8 < 7a + b < 10$.
Since $a, b$ are integers,$7a + b = 9$.
Given $60 < f(7) < 70 \implies 60 < 6(7a + a + b) < 70 \implies 10 < 8a + b < 11.66$.
Since $a, b$ are integers,$8a + b = 11$.
Subtracting the two equations: $(8a + b) - (7a + b) = 11 - 9 \implies a = 2$.
Substituting $a = 2$ into $7a + b = 9$,we get $14 + b = 9 \implies b = -5$.
Then $c = -a - b = -2 - (-5) = 3$.
Thus,$f(x) = 2x^2 - 5x + 3$.
Calculating $f(50) = 2(50)^2 - 5(50) + 3 = 5000 - 250 + 3 = 4753$.
Given $1000t < 4753 < 1000(t + 1)$,we find $t = 4$.

(C) The points of intersection of the parabola $y = x^2 + 2ax + 2021$ and the $x$-axis $(y = 0)$ are given by the roots of the quadratic equation $x^2 + 2ax + 2021 = 0$.
For the roots to be rational,the discriminant $D$ must be a perfect square.
$D = (2a)^2 - 4(1)(2021) = 4a^2 - 8084 = 4(a^2 - 2021)$.
Thus,$a^2 - 2021$ must be a perfect square,say $\lambda^2$ for some non-negative integer $\lambda$.
$a^2 - \lambda^2 = 2021 \Rightarrow (a - \lambda)(a + \lambda) = 2021$.
Since $2021 = 43 \times 47$,we consider the factors of $2021$ as $(1, 2021)$ and $(43, 47)$.
For the largest value of $a$,we set $a + \lambda = 2021$ and $a - \lambda = 1$.
Adding these equations: $2a = 2022 \Rightarrow a = 1011$.
Checking the other case: $a + \lambda = 47$ and $a - \lambda = 43$ $\Rightarrow 2a = 90$ $\Rightarrow a = 45$.
The largest element of $E$ is $1011$.

(B) For the quadratic equation $x^2 + 4x \cos \theta + \cot \theta = 0$ to have equal roots,its discriminant $D$ must be zero.
$D = b^2 - 4ac = (4 \cos \theta)^2 - 4(1)(\cot \theta) = 0$
$16 \cos^2 \theta - 4 \cot \theta = 0$
$4 \cos^2 \theta = \cot \theta$
$4 \cos^2 \theta = \frac{\cos \theta}{\sin \theta}$
Since $0 < \theta < \pi / 2$,$\cos \theta \neq 0$,so we can divide by $\cos \theta$:
$4 \cos \theta \sin \theta = 1$
$2 \sin 2 \theta = 1$
$\sin 2 \theta = \frac{1}{2}$
Since $0 < \theta < \pi / 2$,we have $0 < 2 \theta < \pi$. The solutions for $2 \theta$ are $\frac{\pi}{6}$ and $\frac{5 \pi}{6}$.
Therefore,$\theta = \frac{\pi}{12}$ or $\theta = \frac{5 \pi}{12}$.

(D) Let the angles of the triangle be $\alpha, \beta, \gamma$. We are given two angles as $A = 2x + 7$ and $B = 7x + 10$.
For an isosceles triangle,at least two angles must be equal. We consider three cases:
Case $1$: $A = B$
$2x + 7 = 7x + 10$ $\Rightarrow 5x = -3$ $\Rightarrow x = -0.6$.
The angles are $A = 5.8^\circ, B = 5.8^\circ, C = 180 - 11.6 = 168.4^\circ$. This is a valid triangle.
Case $2$: $A = C$
Since $A + B + C = 180^\circ$,we have $2A + B = 180^\circ$.
$2(2x + 7) + (7x + 10) = 180$ $\Rightarrow 4x + 14 + 7x + 10 = 180$ $\Rightarrow 11x = 156$ $\Rightarrow x = \frac{156}{11} \approx 14.18$.
The angles are $A = 35.36^\circ, B = 109.26^\circ, C = 35.36^\circ$. This is a valid triangle.
Case $3$: $B = C$
Since $A + B + C = 180^\circ$,we have $A + 2B = 180^\circ$.
$(2x + 7) + 2(7x + 10) = 180$ $\Rightarrow 2x + 7 + 14x + 20 = 180$ $\Rightarrow 16x = 153$ $\Rightarrow x = \frac{153}{16} = 9.5625$.
The angles are $A = 26.125^\circ, B = 76.9375^\circ, C = 76.9375^\circ$. This is a valid triangle.
Since all three cases yield valid real values for $x$,the total number of such real numbers is $3$.

(B) Let $\alpha$ be the common real root of the equations $x^2 - ax + b = 0$ and $x^3 - ax^2 + bx + a - b = 0$.
Since $\alpha$ is a root of the first equation,we have $\alpha^2 - a\alpha + b = 0$.
Substituting this into the second equation: $\alpha(\alpha^2 - a\alpha + b) + a - b = 0$.
Since $\alpha^2 - a\alpha + b = 0$,this simplifies to $a - b = 0$,which means $a = b$.
Substituting $a = b$ into the first equation,we get $x^2 - ax + a = 0$.
For the roots to be real,the discriminant $D \geq 0$,so $a^2 - 4a \geq 0$.
This implies $a(a - 4) \geq 0$,which gives $a \leq 0$ or $a \geq 4$.
Given $1 \leq a, b \leq 2021$ and $a = b$,we must have $4 \leq a \leq 2021$.
The number of such integers $a$ is $2021 - 4 + 1 = 2018$.