Mix Examples-Quadratic Equations and Inequations Questions in English

(B) For the equation $|x-2|^2+|x-2|-2=0$:
Let $t = |x-2|$,then $t^2+t-2=0$.
$(t+2)(t-1)=0 \Rightarrow t=1$ (since $t \geq 0$).
$|x-2|=1 \Rightarrow x-2=1$ or $x-2=-1$.
$x=3, 1$.
Sum of squares of roots $= 3^2+1^2 = 9+1 = 10$.
For the equation $x^2-2|x-3|-5=0$:
Case-$I$: $x \geq 3$,then $x^2-2(x-3)-5=0$ $\Rightarrow x^2-2x+1=0$ $\Rightarrow (x-1)^2=0$ $\Rightarrow x=1$.
Since $1 < 3$,this is rejected.
Case-$II$: $x < 3$,then $x^2-2(-(x-3))-5=0$ $\Rightarrow x^2+2x-6-5=0$ $\Rightarrow x^2+2x-11=0$.
Roots $\alpha, \beta = \frac{-2 \pm \sqrt{4 - 4(1)(-11)}}{2} = -1 \pm 2\sqrt{3}$.
Both roots are less than $3$,so both are valid.
Sum of squares of roots $= (\alpha+\beta)^2 - 2\alpha\beta = (-2)^2 - 2(-11) = 4 + 22 = 26$.
Total sum $= 10 + 26 = 36$.

(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.
Then $y(x^2+x+1) = x^2-x+1$,which implies $(y-1)x^2 + (y+1)x + (y-1) = 0$.
Since $x$ is real,the discriminant $D \ge 0$.
$D = (y+1)^2 - 4(y-1)^2 \ge 0$.
$(y+1)^2 - [2(y-1)]^2 \ge 0$.
$(y+1-2y+2)(y+1+2y-2) \ge 0$.
$(3-y)(3y-1) \ge 0$.
$(y-3)(3y-1) \le 0$.
Thus,$\frac{1}{3} \le y \le 3$.
The greatest value is $3$ and the least value is $\frac{1}{3}$.
The difference is $3 - \frac{1}{3} = \frac{9-1}{3} = \frac{8}{3}$.

(D) Given $2 \sin^2 x + 3 \sin x - 2 > 0$.
Factoring the quadratic expression,we get $(2 \sin x - 1)(\sin x + 2) > 0$.
Since $\sin x + 2 > 0$ for all $x \in \mathbb{R}$,we must have $2 \sin x - 1 > 0$,which implies $\sin x > \frac{1}{2}$.
This gives $x \in (\frac{\pi}{6}, \frac{5 \pi}{6})$.
Also,given $x^2 - x - 2 < 0$.
Factoring gives $(x - 2)(x + 1) < 0$,which implies $-1 < x < 2$.
To find the intersection,we note that $\frac{\pi}{6} \approx 0.52$ and $\frac{5 \pi}{6} \approx 2.61$.
Since $2 < \frac{5 \pi}{6}$,the intersection of $(\frac{\pi}{6}, \frac{5 \pi}{6})$ and $(-1, 2)$ is $(\frac{\pi}{6}, 2)$.

(C) The equation $ax + by = 1$ is a linear Diophantine equation.
According to Bezout's identity,the equation $ax + by = c$ has integer solutions for $x$ and $y$ if and only if $\gcd(a, b)$ divides $c$.
Here,$c = 1$,so $\gcd(a, b)$ must divide $1$,which implies $\gcd(a, b) = 1$. Thus,option $(d)$ is true.
Similarly,for $ax + by = 1$,we can rewrite it as $ax = 1 - by$,which implies $\gcd(a, b) = 1$ and $\gcd(a, y) = 1$ is not necessarily true,but $\gcd(a, b) = 1$ is a necessary condition.
Specifically,$\gcd(a, b) = 1$,$\gcd(a, x)$ is not necessarily $1$,$\gcd(b, x)$ is not necessarily $1$,and $\gcd(x, y)$ is not necessarily $1$.
However,looking at the options,$\gcd(b, y) = 1$ is not a required condition for the existence of solutions to $ax + by = 1$. For example,if $a=1, b=2, x=1, y=0$,then $1(1) + 2(0) = 1$. Here $\gcd(b, y) = \gcd(2, 0) = 2 \neq 1$. Thus,option $(c)$ is not necessarily true.

(D) Given $a \equiv b \pmod{m}$,we have $m \mid (a-b)$.
$(i)$ For $(a+x) \equiv (b+x) \pmod{m}$,we check $(a+x) - (b+x) = a-b$. Since $m \mid (a-b)$,this is correct.
(ii) For $(a-x) \equiv (b-x) \pmod{m}$,we check $(a-x) - (b-x) = a-b$. Since $m \mid (a-b)$,this is correct.
(iii) For $ax \equiv bx \pmod{m}$,we check $ax - bx = x(a-b)$. Since $m \mid (a-b)$,it follows that $m \mid x(a-b)$,so this is correct.
(iv) For $(a+x) \equiv (b \div x) \pmod{m}$,this is not generally true because division is not defined in modular arithmetic in the same way as addition or multiplication,and $b \div x$ may not even be an integer. Thus,this statement is incorrect.

(A) We know that $2^2 = 4 \equiv -1 \pmod{5}$.
Now,$2^{2010} = (2^2)^{1005} \equiv (-1)^{1005} \pmod{5}$.
Since $1005$ is an odd number,$(-1)^{1005} = -1$.
So,$2^{2010} \equiv -1 \equiv 4 \pmod{5}$.
Given the congruence $2^{2010} \equiv 3x \pmod{5}$,we substitute $2^{2010} \equiv 4 \pmod{5}$:
$4 \equiv 3x \pmod{5}$.
To solve for $x$,multiply both sides by the modular inverse of $3$ modulo $5$,which is $2$ (since $3 \times 2 = 6 \equiv 1 \pmod{5}$):
$2 \times 4 \equiv 2 \times 3x \pmod{5}$
$8 \equiv 6x \pmod{5}$
$3 \equiv x \pmod{5}$.
Thus,the least positive integer $x$ is $3$.

(C) Let $f(x) = \frac{x^2+x+a}{x^2-x+a}$. The given inequality is $\frac{1}{2} \leq f(x) \leq 2$.
For $f(x) \leq 2$:
$\frac{x^2+x+a}{x^2-x+a} \leq 2 \implies x^2+x+a \leq 2x^2-2x+2a$ (assuming $x^2-x+a > 0$)
$\implies x^2-3x+a \geq 0$.
For this to hold for all $x$,the discriminant $D \leq 0$:
$(-3)^2 - 4(1)(a) \leq 0 \implies 9 - 4a \leq 0 \implies a \geq \frac{9}{4}$.
For $f(x) \geq \frac{1}{2}$:
$\frac{x^2+x+a}{x^2-x+a} \geq \frac{1}{2} \implies 2x^2+2x+2a \geq x^2-x+a$
$\implies x^2+3x+a \geq 0$.
For this to hold for all $x$,the discriminant $D \leq 0$:
$(3)^2 - 4(1)(a) \leq 0 \implies 9 - 4a \leq 0 \implies a \geq \frac{9}{4}$.
However,checking the boundary condition $a = \frac{9}{4}$,we find that the expression becomes a perfect square,satisfying the inequality. Thus,$a = \frac{9}{4}$.

(A) Given that $a, b, c$ are distinct positive real numbers such that $a^2+b^2+c^2=1$.
We know the algebraic identity: $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$.
Since $a, b, c > 0$,we have $(a+b+c)^2 > a^2+b^2+c^2$,which implies $a^2+b^2+c^2+2(ab+bc+ca) > a^2+b^2+c^2$.
This simplifies to $2(ab+bc+ca) > 0$,which is always true for positive numbers.
Also,we know that $(a-b)^2 + (b-c)^2 + (c-a)^2 > 0$ because $a, b, c$ are distinct.
Expanding this,we get $2(a^2+b^2+c^2) - 2(ab+bc+ca) > 0$.
Substituting $a^2+b^2+c^2=1$,we get $2(1) - 2(ab+bc+ca) > 0$,which means $2 > 2(ab+bc+ca)$,or $ab+bc+ca < 1$.
Thus,the value of $ab+bc+ca$ is less than $1$.

(C) Given that,$x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)$.
Squaring both sides,we get $x^2 = \frac{1}{4} \left( 7 + \frac{1}{7} + 2 \right) = \frac{1}{4} \left( \frac{49 + 1 + 14}{7} \right) = \frac{1}{4} \left( \frac{64}{7} \right) = \frac{16}{7}$.
Now,$x^2 - 1 = \frac{16}{7} - 1 = \frac{9}{7}$,so $\sqrt{x^2 - 1} = \frac{3}{\sqrt{7}}$.
Substituting these values into the expression $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$:
$= \frac{\frac{3}{\sqrt{7}}}{\frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right) - \frac{3}{\sqrt{7}}}$
$= \frac{\frac{3}{\sqrt{7}}}{\frac{1}{2} \left( \frac{7 + 1}{\sqrt{7}} \right) - \frac{3}{\sqrt{7}}}$
$= \frac{\frac{3}{\sqrt{7}}}{\frac{4}{\sqrt{7}} - \frac{3}{\sqrt{7}}}$
$= \frac{\frac{3}{\sqrt{7}}}{\frac{1}{\sqrt{7}}} = 3$.

(A) The given equation is $(3 + \alpha) x^2 + (6 + 2 \alpha) x + (5 + 2 \alpha) = 0$.
For the roots to be complex,the discriminant $D$ must be less than $0$.
$D = (2 \alpha + 6)^2 - 4 (\alpha + 3) (2 \alpha + 5) < 0$.
$4(\alpha + 3)^2 - 4(\alpha + 3)(2 \alpha + 5) < 0$.
Dividing by $4$: $(\alpha + 3)(\alpha + 3 - 2 \alpha - 5) < 0$.
$(\alpha + 3)(-\alpha - 2) < 0 \Rightarrow (\alpha + 3)(\alpha + 2) > 0$.
This implies $\alpha < -3$ or $\alpha > -2$.
Comparing with $\alpha > L$ or $\alpha < M$,we get $L = -2$ and $M = -3$.
The condition $L y^2 + M y + r < 0$ becomes $-2 y^2 - 3 y + r < 0$.
For this to hold for all $y \in R$,the coefficient of $y^2$ must be negative (which is $-2 < 0$) and $D < 0$.
$D = (-3)^2 - 4(-2)(r) < 0$.
$9 + 8 r < 0$ $\Rightarrow 8 r < -9$ $\Rightarrow r < -\frac{9}{8} = -1.125$.
The greatest integer $[r]$ not exceeding $r$ is $[-1.125] = -2$.

(A) Let $t = \sqrt{\frac{1-y}{y}}$. Then the equation becomes $t + \frac{1}{t} = \frac{5}{2}$.
Solving for $t$,we get $2t^2 - 5t + 2 = 0$,which gives $t = 2$ or $t = \frac{1}{2}$.
If $\sqrt{\frac{1-y}{y}} = 2$,then $\frac{1-y}{y} = 4$ $\Rightarrow 1-y = 4y$ $\Rightarrow y = \frac{1}{5}$.
If $\sqrt{\frac{1-y}{y}} = \frac{1}{2}$,then $\frac{1-y}{y} = \frac{1}{4}$ $\Rightarrow 4-4y = y$ $\Rightarrow y = \frac{4}{5}$.
Thus,$\alpha = \frac{1}{5}$ and $\beta = \frac{4}{5}$ (since $\beta > \alpha$).
Then $\alpha + \beta = 1$ and $\alpha \beta = \frac{4}{25}$.
The equation $(\alpha+\beta) x^4 - 25 \alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ becomes $x^4 - 4x^2 + (\gamma + \frac{3}{5}) = 0$.
Let $u = x^2$. Then $u^2 - 4u + (\gamma + \frac{3}{5}) = 0$.
For real roots,$u$ must be non-negative. The discriminant $D = 16 - 4(\gamma + \frac{3}{5}) \ge 0$ $\Rightarrow 4 - \gamma - \frac{3}{5} \ge 0$ $\Rightarrow \gamma \le \frac{17}{5} = 3.4$.
Also,for at least one non-negative root $u$,we need the sum of roots $4 > 0$ and product $\gamma + \frac{3}{5} \ge 0 \Rightarrow \gamma \ge -0.6$.
Thus,$\gamma \in [-0.6, 3.4]$. Among the options,$\frac{1}{2} = 0.5$ is in this range.

(C) Given the inequality $-1 < \frac{2 x^2+a x+2}{x^2+x+1} < 3$ for all $x \in \mathbb{R}$.
Since $x^2+x+1 > 0$ for all $x$,we can split the inequality into two parts.
Part $1$: $-1 < \frac{2 x^2+a x+2}{x^2+x+1}$ $\Rightarrow -x^2-x-1 < 2 x^2+a x+2$ $\Rightarrow 3 x^2+(a+1) x+3 > 0$.
For this to hold for all $x$,the discriminant $D < 0$.
$(a+1)^2 - 4(3)(3) < 0$ $\Rightarrow (a+1)^2 - 36 < 0$ $\Rightarrow (a+1-6)(a+1+6) < 0$ $\Rightarrow (a-5)(a+7) < 0$ $\Rightarrow a \in (-7, 5) \dots (i)$.
Part $2$: $\frac{2 x^2+a x+2}{x^2+x+1} < 3$ $\Rightarrow 2 x^2+a x+2 < 3 x^2+3 x+3$ $\Rightarrow -x^2+(a-3) x-1 < 0$ $\Rightarrow x^2-(a-3) x+1 > 0$.
For this to hold for all $x$,the discriminant $D < 0$.
$(a-3)^2 - 4(1)(1) < 0$ $\Rightarrow (a-3)^2 - 4 < 0$ $\Rightarrow (a-3-2)(a-3+2) < 0$ $\Rightarrow (a-5)(a-1) < 0$ $\Rightarrow a \in (1, 5) \dots (ii)$.
Taking the intersection of $(i)$ and $(ii)$,we get $a \in (1, 5)$.

(D) We know that $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$.
Expanding this,we get $2(a^2+b^2+c^2) - 2(ab+bc+ca) \geq 0$.
Since $a^2+b^2+c^2 = 1$,we have $2(1) - 2(ab+bc+ca) \geq 0$,which implies $ab+bc+ca \leq 1$.
Also,we know that $(a+b+c)^2 \geq 0$.
Expanding this,we get $a^2+b^2+c^2 + 2(ab+bc+ca) \geq 0$.
Substituting $a^2+b^2+c^2 = 1$,we get $1 + 2(ab+bc+ca) \geq 0$,which implies $ab+bc+ca \geq -\frac{1}{2}$.
Therefore,the range of $ab+bc+ca$ is $[-\frac{1}{2}, 1]$.
The set of extreme values is $\{\frac{-1}{2}, 1\}$.

(C) Let the four numbers in $A.P.$ be $p=a-3d, q=a-d, r=a+d, s=a+3d$.
Since $p$ and $q$ are roots of $x^2-2x+A=0$,we have $p+q=2$ and $pq=A$.
Since $r$ and $s$ are roots of $x^2-18x+B=0$,we have $r+s=18$ and $rs=B$.
Summing the roots: $(a-3d)+(a-d)+(a+d)+(a+3d) = 4a = 2+18 = 20$,so $a=5$.
From $p+q=2$,we get $(a-3d)+(a-d) = 2a-4d = 2$.
Substituting $a=5$,we get $10-4d=2$,so $4d=8$,which means $d=2$.
The roots are $p=5-3(2)=-1$,$q=5-2=3$,$r=5+2=7$,and $s=5+3(2)=11$.
Thus,$A = pq = (-1)(3) = -3$ and $B = rs = (7)(11) = 77$.

(D) Given equation: $6x^6-25x^5+31x^4-31x^2+25x-6=0$
Grouping terms: $6(x^6-1)-25x(x^4-1)+31x^2(x^2-1)=0$
Factoring out $(x^2-1)$: $(x^2-1)[6(x^4+x^2+1)-25x(x^2+1)+31x^2]=0$
$(x^2-1)(6x^4-25x^3+37x^2-25x+6)=0$
Roots from $x^2-1=0$ are $x=1, -1$.
For $6x^4-25x^3+37x^2-25x+6=0$,divide by $x^2$: $6(x^2+\frac{1}{x^2})-25(x+\frac{1}{x})+37=0$
Let $t = x+\frac{1}{x}$,then $6(t^2-2)-25t+37=0 \Rightarrow 6t^2-25t+25=0$
$(2t-5)(3t-5)=0 \Rightarrow t=\frac{5}{2}$ or $t=\frac{5}{3}$
Case $1$: $x+\frac{1}{x}=\frac{5}{2}$ $\Rightarrow 2x^2-5x+2=0$ $\Rightarrow (2x-1)(x-2)=0$ $\Rightarrow x=2, \frac{1}{2}$
Case $2$: $x+\frac{1}{x}=\frac{5}{3} \Rightarrow 3x^2-5x+3=0$. Discriminant $D = 25-36 = -11 < 0$,so no real roots.
The rational roots are $1, -1, 2, \frac{1}{2}$.
Sum $= 1-1+2+\frac{1}{2} = 2.5$.

(D) Given $\phi(x) = \frac{x}{(x^2+1)(x+1)}$.
We need to find $\phi(a) \phi(b) \phi(c) = \frac{abc}{(a^2+1)(b^2+1)(c^2+1)(a+1)(b+1)(c+1)}$.
For the equation $x^3 - 3x + \lambda = 0$,we have:
$a+b+c = 0$,$ab+bc+ca = -3$,and $abc = -\lambda$.
First,$(a+1)(b+1)(c+1) = abc + (ab+bc+ca) + (a+b+c) + 1 = -\lambda - 3 + 0 + 1 = -\lambda - 2 = -(\lambda+2)$.
Second,$(a^2+1)(b^2+1)(c^2+1) = (a^2b^2+a^2+b^2+1)(c^2+1) = a^2b^2c^2 + a^2b^2 + a^2c^2 + a^2 + b^2c^2 + b^2 + c^2 + 1$.
Using $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 0^2 - 2(-3) = 6$.
Using $a^2b^2+b^2c^2+c^2a^2 = (ab+bc+ca)^2 - 2abc(a+b+c) = (-3)^2 - 2(-\lambda)(0) = 9$.
So,$(a^2+1)(b^2+1)(c^2+1) = (abc)^2 + (a^2b^2+b^2c^2+c^2a^2) + (a^2+b^2+c^2) + 1 = \lambda^2 + 9 + 6 + 1 = \lambda^2 + 16$.
Thus,$\phi(a) \phi(b) \phi(c) = \frac{-\lambda}{(\lambda^2+16)(-(\lambda+2))} = \frac{\lambda}{(\lambda+2)(\lambda^2+16)}$.

(A) Given equation: $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$.
$x=1$ is a root of the equation.
Dividing by $(x-1)$,we get $(x-1)(x^4-4 x^3+5 x^2-4 x+1)=0$.
For $x^4-4 x^3+5 x^2-4 x+1=0$,divide by $x^2$:
$x^2-4 x+5-\frac{4}{x}+\frac{1}{x^2}=0
$ $\Rightarrow (x^2+\frac{1}{x^2})-4(x+\frac{1}{x})+5=0
$ $\Rightarrow (x+\frac{1}{x})^2-2-4(x+\frac{1}{x})+5=0
$ $\Rightarrow (x+\frac{1}{x})^2-4(x+\frac{1}{x})+3=0$.
Let $y = x+\frac{1}{x}$,then $y^2-4y+3=0
$ $\Rightarrow (y-1)(y-3)=0
$ $\Rightarrow y=1, 3$.
Case $1$: $x+\frac{1}{x}=1 \Rightarrow x^2-x+1=0$ (roots are complex).
Case $2$: $x+\frac{1}{x}=3 \Rightarrow x^2-3x+1=0$ (roots are irrational).
Thus,$\alpha+\beta=3$ and $\alpha \beta=1$.
Substituting into $(\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0$:
$3x^2+2x-1=0
$ $\Rightarrow (3x-1)(x+1)=0
$ $\Rightarrow x=-1, \frac{1}{3}$.

(A) From $xy+l(x+y)+m=0$,we have $x(y+l) = -(ly+m)$,so $x = -\frac{ly+m}{y+l}$. Substituting this into $x^2+\alpha x+\beta=0$ gives:
$\left(-\frac{ly+m}{y+l}\right)^2 + \alpha\left(-\frac{ly+m}{y+l}\right) + \beta = 0$
$(ly+m)^2 - \alpha(ly+m)(y+l) + \beta(y+l)^2 = 0$
$(l^2y^2 + m^2 + 2lmy) - \alpha(ly^2 + l^2y + my + ml) + \beta(y^2 + 2ly + l^2) = 0$
$(l^2 - \alpha l + \beta)y^2 + (2lm - \alpha l^2 - \alpha m + 2\beta l)y + (m^2 - \alpha ml + \beta l^2) = 0$
Since this equation has the same roots as $x^2 + \alpha x + \beta = 0$,the ratios of coefficients must be equal:
$\frac{l^2 - \alpha l + \beta}{1} = \frac{2lm - \alpha l^2 - \alpha m + 2\beta l}{\alpha} = \frac{m^2 - \alpha ml + \beta l^2}{\beta}$
Equating the first and third terms: $\beta(l^2 - \alpha l + \beta) = m^2 - \alpha ml + \beta l^2$
$\beta l^2 - \beta \alpha l + \beta^2 = m^2 - \alpha ml + \beta l^2$
$\beta^2 - \beta \alpha l - m^2 + \alpha ml = 0$
$\beta^2 - \beta m + \beta m - \beta \alpha l - m^2 + \alpha ml = 0$
$\beta(\beta - m) + (m - \alpha l)(\beta - m) = 0$
$(\beta - m)(\beta + m - \alpha l) = 0$
Thus,$\beta = m$ or $\beta = \alpha l - m$.

(C) For the quadratic expression $f(x) = (a-3)x^2 + 12x + (a+6)$ to be positive for all $x \in R$,we must have the coefficient of $x^2$ be positive and the discriminant be negative.
$1$. $a-3 > 0 \implies a > 3$.
$2$. $D = 12^2 - 4(a-3)(a+6) < 0$.
$144 - 4(a^2 + 3a - 18) < 0$
$36 - (a^2 + 3a - 18) < 0$
$36 - a^2 - 3a + 18 < 0$
$-a^2 - 3a + 54 < 0$
$a^2 + 3a - 54 > 0$
$(a+9)(a-6) > 0$.
Since $a > 3$,the condition $a > 6$ must hold. Thus,$a \in (6, \infty)$,so $\ell = 6$.
The least positive integral value of $a$ is $\alpha = 7$.
Now,substitute $\alpha = 7$ and $\ell = 6$ into the equation $(\alpha-3)x^2 + 12x + (\ell+2) = 0$:
$(7-3)x^2 + 12x + (6+2) = 0$
$4x^2 + 12x + 8 = 0$
Divide by $4$: $x^2 + 3x + 2 = 0$
$(x+1)(x+2) = 0$.
The roots are $x = -1, -2$.

(B) Given $f(x)$ is negative in $\left(-\infty, -\frac{5}{3}\right) \cup (3, \infty)$ and positive in $\left(-\frac{5}{3}, 3\right)$,we can write $f(x) = -k_1(x + \frac{5}{3})(x - 3)$ where $k_1 > 0$. This simplifies to $f(x) = k_1(x + \frac{5}{3})(3 - x)$.
Given $g(x)$ is negative in $\left(3, \frac{9}{2}\right)$ and positive elsewhere,we can write $g(x) = k_2(x - 3)(x - \frac{9}{2})$ where $k_2 > 0$.
Now,consider the product $P(x) = f(x)g(x) = k_1 k_2 (x + \frac{5}{3})(3 - x)(x - 3)(x - \frac{9}{2})$.
Since $(3 - x) = -(x - 3)$,we have $P(x) = -k_1 k_2 (x + \frac{5}{3})(x - 3)^2 (x - \frac{9}{2})$.
Let $K = k_1 k_2 > 0$. Then $P(x) = -K (x + \frac{5}{3})(x - 3)^2 (x - \frac{9}{2})$.
The roots are $x = -\frac{5}{3}, 3, \frac{9}{2}$. Note that $x = 3$ is a root of multiplicity $2$,so the sign does not change across $x = 3$.
For $x > \frac{9}{2}$,$P(x)$ is negative. For $3 < x < \frac{9}{2}$,$P(x)$ is positive. For $x < 3$ (and $x > -\frac{5}{3}$),$P(x)$ is positive.
Thus,in the interval $[0, 5]$,$P(x)$ is positive in $[0, 3) \cup (3, \frac{9}{2})$ and negative in $(\frac{9}{2}, 5]$. The correct option is $B$.

(A) Given that $ax^2 + bx + c = 0$ has no real roots,the discriminant $D = b^2 - 4ac < 0$.
Since $ax^2 + bx + c$ has no real roots,the expression $f(x) = ax^2 + bx + c$ must have the same sign for all $x \in \mathbb{R}$.
Given $4a + 2b + c > 0$,we have $f(2) > 0$,which implies $a > 0$ and $f(x) > 0$ for all $x \in \mathbb{R}$.
Since $f(x) > 0$ for all $x$,we have $f(0) = c > 0$ and $f(1) = a + b + c > 0$.
We are given $f(2) = 4a + 2b + c > 0$.
Now,consider $f(x) = ax^2 + bx + c$. Since $a > 0$,$f(x) > 0$ for all $x$.
Specifically,$f(1) = a + b + c > 0 \implies a + c > -b$.
Also,$f(0) = c > 0$.
Consider the expression $(c+a)(c+b)$.
Since $a > 0$ and $f(x) > 0$,we have $f(x) = a(x - \alpha)(x - \bar{\alpha}) > 0$.
Using the property $f(x) > 0$,we can evaluate $f(1) = a+b+c > 0$ and $f(0) = c > 0$.
From $ax^2+bx+c > 0$,we know $b^2 < 4ac$.
$(c+a)(c+b) = c^2 + bc + ac + ab$.
Since $b^2 < 4ac$,we can show that $(c+a)(c+b) > ab$.

(D) Let $y = \frac{x^2+4x+1}{x^2+x+1}$.
Then $y(x^2+x+1) = x^2+4x+1$,which implies $(y-1)x^2 + (y-4)x + (y-1) = 0$.
Since $x$ is real,the discriminant $D \ge 0$.
$D = (y-4)^2 - 4(y-1)^2 \ge 0$.
$(y-4)^2 - [2(y-1)]^2 \ge 0$.
$(y-4-2y+2)(y-4+2y-2) \ge 0$.
$(-y-2)(3y-6) \ge 0$.
$(y+2)(3y-6) \le 0$.
$(y+2)(y-2) \le 0$.
Thus,$-2 \le y \le 2$.
The minimum value is $-2$ and the maximum value is $2$.
The sum of the maximum and minimum values is $2 + (-2) = 0$.

(D) Let $y = \frac{11 x^2+12 x+6}{x^2+4 x+2}$.
$(11-y) x^2+(12-4 y) x+(6-2 y)=0$.
For $x$ to be real,the discriminant $D \geq 0$.
$(12-4 y)^2-4(11-y)(6-2 y) \geq 0$.
$16(3-y)^2-8(11-y)(3-y) \geq 0$.
$8(3-y)[2(3-y)-(11-y)] \geq 0$.
$8(3-y)(6-2y-11+y) \geq 0$.
$8(3-y)(-y-5) \geq 0$.
$(y-3)(y+5) \geq 0$.
Thus,$y \leq -5$ or $y \geq 3$.
The range of the expression is $(-\infty, -5] \cup [3, \infty)$.
Given $\frac{11 x^2+12 x+6}{x^2+4 x+2} \notin (a, b]$,we identify $(a, b] = (-5, 3]$.
So,$a = -5$ and $b = 3$.
We need to find $x$ such that $\frac{11 x^2+12 x+6}{x^2+4 x+2} = b-a+3 = 3 - (-5) + 3 = 11$.
$11 x^2+12 x+6 = 11(x^2+4 x+2)$.
$11 x^2+12 x+6 = 11 x^2+44 x+22$.
$12 x+6 = 44 x+22$.
$-16 = 32 x$.
$x = -\frac{16}{32} = -\frac{1}{2}$.

(B) Let $f(x) = y = \frac{x^2-2x+3}{x^2-4x+7}$.
$y(x^2-4x+7) = x^2-2x+3$
$(y-1)x^2 + (2-4y)x + (7y-3) = 0$.
Since $x$ is real,the discriminant $D \geq 0$:
$(2-4y)^2 - 4(y-1)(7y-3) \geq 0$
$4(1-2y)^2 - 4(7y^2 - 3y - 7y + 3) \geq 0$
$(1 - 4y + 4y^2) - (7y^2 - 10y + 3) \geq 0$
$-3y^2 + 6y - 2 \geq 0$
$3y^2 - 6y + 2 \leq 0$.
Solving $3y^2 - 6y + 2 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$y = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} = 1 \pm \frac{2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$.
Thus,$1 - \frac{\sqrt{3}}{3} \leq y \leq 1 + \frac{\sqrt{3}}{3}$.
The minimum value is $1 - \frac{\sqrt{3}}{3} = \frac{3-\sqrt{3}}{3}$.

(A) Let $y = \frac{x^2+2x+5}{x^2+4x+10}$.
Since $x^2+4x+10 = (x+2)^2+6 > 0$,we can write:
$y(x^2+4x+10) = x^2+2x+5$
$(y-1)x^2 + (4y-2)x + (10y-5) = 0$.
For $x \in \mathbb{R}$,the discriminant $D \geq 0$:
$D = (4y-2)^2 - 4(y-1)(10y-5) \geq 0$
$4(2y-1)^2 - 4(y-1)(5)(2y-1) \geq 0$
$4(2y-1) [ (2y-1) - 5(y-1) ] \geq 0$
$4(2y-1) [ 2y-1-5y+5 ] \geq 0$
$4(2y-1)(4-3y) \geq 0$
$(2y-1)(3y-4) \leq 0$.
Thus,$\frac{1}{2} \leq y \leq \frac{4}{3}$.
The minimum value is $\frac{1}{2}$.

(A) The function $f(x) = x^2 + 2bx + 2c^2$ is a parabola opening upwards. Its minimum value occurs at $x = -b$,which is $f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$.
The function $g(x) = -x^2 - 2cx + b^2$ is a parabola opening downwards. Its maximum value occurs at $x = -c$,which is $g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$.
Given that the minimum value of $f(x)$ is greater than the maximum value of $g(x)$:
$2c^2 - b^2 > c^2 + b^2$
$c^2 > 2b^2$
Taking the square root on both sides,we get $|c| > \sqrt{2}|b|$.

(A) The given inequality is $\left|\frac{x^2+kx+1}{x^2+x+1}\right| < 3$.
Since $x^2+x+1 > 0$ for all real $x$,we can write $-3 < \frac{x^2+kx+1}{x^2+x+1} < 3$.
Case $1$: $\frac{x^2+kx+1}{x^2+x+1} < 3 \implies x^2+kx+1 < 3x^2+3x+3 \implies 2x^2+(3-k)x+2 > 0$.
For this to hold for all $x$,the discriminant $D_1 < 0$: $(3-k)^2 - 4(2)(2) < 0 \implies (3-k)^2 < 16 \implies -4 < 3-k < 4 \implies -7 < -k < 1 \implies -1 < k < 7$.
Case $2$: $\frac{x^2+kx+1}{x^2+x+1} > -3 \implies x^2+kx+1 > -3x^2-3x-3 \implies 4x^2+(k+3)x+4 > 0$.
For this to hold for all $x$,the discriminant $D_2 < 0$: $(k+3)^2 - 4(4)(4) < 0 \implies (k+3)^2 < 64 \implies -8 < k+3 < 8 \implies -11 < k < 5$.
Taking the intersection of both cases,we get $-1 < k < 5$.

(A) Let $y = \frac{x^2+x+1}{2x^2-x+1}$.
Then $y(2x^2-x+1) = x^2+x+1$.
$(2y-1)x^2 - (y+1)x + (y-1) = 0$.
Since $x \in R$,the discriminant $D \ge 0$.
$D = (y+1)^2 - 4(2y-1)(y-1) \ge 0$.
$y^2+2y+1 - 4(2y^2-3y+1) \ge 0$.
$y^2+2y+1 - 8y^2+12y-4 \ge 0$.
$-7y^2+14y-3 \ge 0$.
$7y^2-14y+3 \le 0$.
Solving $7y^2-14y+3 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$y = \frac{14 \pm \sqrt{196 - 84}}{14} = \frac{14 \pm \sqrt{112}}{14} = \frac{14 \pm 4\sqrt{7}}{14} = \frac{7 \pm 2\sqrt{7}}{7}$.
Thus,the range of $y$ is $[\frac{7-2\sqrt{7}}{7}, \frac{7+2\sqrt{7}}{7}]$.
The maximum value is $\frac{7+2\sqrt{7}}{7}$.

(B) Let $y = \frac{x+a}{2 x^2-3 x+1}$,where $y \in \mathbb{R}$.
Since $y$ takes all real values,the equation $2 y x^2 - (3 y + 1) x + (y - a) = 0$ must have real roots for $x$ for all $y \in \mathbb{R}$.
For $y \neq 0$,the discriminant $D \geq 0$:
$D = (3 y + 1)^2 - 4(2 y)(y - a) \geq 0$
$9 y^2 + 6 y + 1 - 8 y^2 + 8 a y \geq 0$
$y^2 + (8 a + 6) y + 1 \geq 0$
For this quadratic in $y$ to be non-negative for all $y \in \mathbb{R}$,the discriminant of this quadratic must be less than or equal to $0$ (since the coefficient of $y^2$ is $1 > 0$).
$D_y = (8 a + 6)^2 - 4(1)(1) \leq 0$
$(8 a + 6)^2 - 4 \leq 0$
$(8 a + 6 - 2)(8 a + 6 + 2) \leq 0$
$(8 a + 4)(8 a + 8) \leq 0$
$4(2 a + 1) \cdot 8(a + 1) \leq 0$
$32(2 a + 1)(a + 1) \leq 0$
$(2 a + 1)(a + 1) \leq 0$
Thus,$-1 \leq a \leq -\frac{1}{2}$.
However,if $y=0$,then $x+a=0 \Rightarrow x=-a$. The denominator $2x^2-3x+1 = (2x-1)(x-1)$ must not be zero at $x=-a$. So $2(-a)-1 \neq 0$ and $-a-1 \neq 0$,which means $a \neq -1/2$ and $a \neq -1$.
Therefore,$-1 < a < -\frac{1}{2}$.

(D) Given that $\alpha, \beta$ are the roots of $x^2+p x+q=0$,we have $\alpha+\beta=-p$ and $\alpha \beta=q$.
Since $\alpha^4, \beta^4$ are roots of $x^2-r x+s=0$,we have $\alpha^4+\beta^4=r$ and $\alpha^4 \beta^4=s$.
Consider the quadratic equation $x^2-4 q x+2 q^2-r=0$.
The discriminant $D$ is given by $D = (-4q)^2 - 4(1)(2q^2 - r)$.
$D = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$.
Since $\alpha, \beta$ are real,$\alpha^4, \beta^4 \geq 0$,so $r = \alpha^4 + \beta^4 \geq 0$.
Also,$8q^2 \geq 0$.
Thus,$D = 8q^2 + 4r \geq 0$.
Since the discriminant is non-negative,the equation $x^2-4 q x+2 q^2-r=0$ always has two real roots.

(C) Given the equation $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) = 1$.
This implies $\frac{x^2}{\beta^2} = |\alpha|$,so $x^2 = \beta^2 |\alpha|$.
Since $\alpha, \beta$ are roots of $x^2 - |a|x - |b| = 0$,the product of roots $\alpha \beta = -|b| \le 0$.
Given $|\alpha| < |\beta|$,and the sum $\alpha + \beta = |a| > 0$,we analyze the signs.
Since $|a| < \beta - 1$,$\beta$ must be positive.
Solving for $x$,we get $x = \pm |\beta| \sqrt{|\alpha|}$.
The positive root is $|\beta| \sqrt{|\alpha|}$.
Since $|\alpha| < 1$ is not guaranteed,but $|\alpha| < |\beta|$,we compare $|\beta| \sqrt{|\alpha|}$ with $\beta$.
Since $\beta > 0$ and $\sqrt{|\alpha|} < 1$ (as $|\alpha| < |\beta|$ and the product of roots is negative),the positive root is less than $\beta$.

(C) Given $f(x) = (x-a)(x-b) - \frac{a+b}{2} = x^2 - (a+b)x + ab - \frac{a+b}{2}$.
To find the minimum value,we find the vertex of the parabola. The derivative is $f'(x) = 2x - (a+b)$.
Setting $f'(x) = 0$,we get $x = \frac{a+b}{2}$.
The minimum value is $f\left(\frac{a+b}{2}\right) = \left(\frac{a+b}{2} - a\right)\left(\frac{a+b}{2} - b\right) - \frac{a+b}{2}$.
$f\left(\frac{a+b}{2}\right) = \left(\frac{b-a}{2}\right)\left(\frac{a-b}{2}\right) - \frac{a+b}{2} = -\frac{(a-b)^2}{4} - \frac{a+b}{2} = -\frac{a^2 - 2ab + b^2 + 2a + 2b}{4}$.
We can rewrite this as $f\left(\frac{a+b}{2}\right) = -\frac{(a+b)^2 - 4ab + 2(a+b)}{4} = -\frac{(a+b)^2}{4} + \frac{4ab - 2(a+b)}{4}$.
Since the roots of $f(x)=0$ are non-negative,let the roots be $x_1, x_2 \geq 0$. Then $x_1+x_2 = a+b \geq 0$ and $x_1x_2 = ab - \frac{a+b}{2} \geq 0$.
From $x_1x_2 \geq 0$,we have $ab \geq \frac{a+b}{2}$,which implies $4ab \geq 2(a+b)$,so $4ab - 2(a+b) \geq 0$.
Thus,the minimum value $f\left(\frac{a+b}{2}\right) = -\frac{(a+b)^2}{4} + \frac{4ab - 2(a+b)}{4} \geq -\frac{(a+b)^2}{4}$.

(B) Let the roots be $\alpha, \alpha, \beta, \gamma$. Given $\alpha > 0$ and $\beta\gamma = 1$.
From Vieta's formulas,the product of roots is $\alpha^2\beta\gamma = \frac{e}{a}$.
Since $\beta\gamma = 1$,we have $\alpha^2 = \frac{e}{a}$,so $\alpha = \sqrt{\frac{e}{a}}$.
The equation can be written as $a(x-\alpha)^2(x^2 - Sx + 1) = 0$,where $S = \beta + \gamma$.
Expanding $a(x^2 - 2\alpha x + \alpha^2)(x^2 - Sx + 1) = a(x^4 - (S+2\alpha)x^3 + (1 + 2\alpha S + \alpha^2)x^2 - (S\alpha^2 + 2\alpha)x + \alpha^2) = 0$.
Comparing coefficients with $ax^4 + bx^3 + cx^2 + dx + e = 0$:
$b = -a(S + 2\alpha)$,$c = a(1 + 2\alpha S + \alpha^2)$,$d = -a(S\alpha^2 + 2\alpha)$,$e = a\alpha^2$.
From $b = -a(S + 2\alpha)$,we get $S = -\frac{b}{a} - 2\alpha$.
Substitute $S$ into the expression for $c$: $c = a(1 + 2\alpha(-\frac{b}{a} - 2\alpha) + \alpha^2) = a(1 - \frac{2b\alpha}{a} - 4\alpha^2 + \alpha^2) = a - 2b\alpha - 3a\alpha^2$.
Since $\alpha^2 = \frac{e}{a}$,then $3a\alpha^2 = 3e$.
Thus,$c = a - 2b\sqrt{\frac{e}{a}} - 3e$,which rearranges to $3e + \frac{2b\sqrt{e}}{\sqrt{a}} + c = a$.

(D) Given equation: $x^5-5x^3+5x^2-1=0$.
By inspection,$x=1$ is a root.
Factoring the polynomial: $(x-1)(x^4+x^3-4x^2+x+1)=0$.
Further factoring: $(x-1)^2(x^3+2x^2-2x-1)=0$.
$(x-1)^3(x^2+3x+1)=0$.
The roots are $x=1$ (triple root) and $x=\frac{-3 \pm \sqrt{5}}{2}$.
The two distinct negative roots are $\alpha = \frac{-3+\sqrt{5}}{2}$ and $\beta = \frac{-3-\sqrt{5}}{2}$.
We need an equation with roots $\sqrt{-\alpha}$ and $\sqrt{-\beta}$.
Let $y = \sqrt{-\alpha} \Rightarrow y^2 = -\alpha$.
Since $\alpha$ and $\beta$ are roots of $x^2+3x+1=0$,we have $\alpha+\beta = -3$ and $\alpha\beta = 1$.
Let the new roots be $y_1 = \sqrt{-\alpha}$ and $y_2 = \sqrt{-\beta}$.
The equation is $(y^2+\alpha)(y^2+\beta) = 0$.
$y^4 + (\alpha+\beta)y^2 + \alpha\beta = 0$.
Substituting the values: $y^4 - 3y^2 + 1 = 0$.

(B) To find the remainder,we perform polynomial long division of $2 x^5-3 x^4+5 x^3-3 x^2+7 x-9$ by $x^2-x-3$:
$1$. Divide $2 x^5$ by $x^2$ to get $2 x^3$. Multiply $(x^2-x-3)$ by $2 x^3$ to get $2 x^5-2 x^4-6 x^3$. Subtracting this from the dividend gives $-x^4+11 x^3-3 x^2+7 x-9$.
$2$. Divide $-x^4$ by $x^2$ to get $-x^2$. Multiply $(x^2-x-3)$ by $-x^2$ to get $-x^4+x^3+3 x^2$. Subtracting this gives $10 x^3-6 x^2+7 x-9$.
$3$. Divide $10 x^3$ by $x^2$ to get $10 x$. Multiply $(x^2-x-3)$ by $10 x$ to get $10 x^3-10 x^2-30 x$. Subtracting this gives $4 x^2+37 x-9$.
$4$. Divide $4 x^2$ by $x^2$ to get $4$. Multiply $(x^2-x-3)$ by $4$ to get $4 x^2-4 x-12$. Subtracting this gives $41 x+3$.
Thus,the remainder is $41 x+3$.

(C) Given that $x^2+p x+1$ is a factor of $a x^3+b x+c$.
Let $a x^3+b x+c = (x^2+p x+1)(a x+\alpha)$.
Expanding the right side:
$a x^3+b x+c = a x^3 + (p a+\alpha) x^2 + (p \alpha+a) x + \alpha$.
Comparing the coefficients of $x^2, x^1,$ and the constant term:
$1) \ p a + \alpha = 0 \implies p = -\frac{\alpha}{a}$
$2) \ p \alpha + a = b$
$3) \ \alpha = c$
Substituting $\alpha = c$ into the first equation:
$p = -\frac{c}{a}$.
Now,substitute $p = -\frac{c}{a}$ and $\alpha = c$ into the second equation:
$(-\frac{c}{a})(c) + a = b$
$-\frac{c^2}{a} + a = b$
$-c^2 + a^2 = a b$
$a^2 - c^2 = a b$.

(C) Let the roots of the equation $6x^3 + 7x^2 - 4x - 2 = 0$ be $\alpha, \beta, \gamma$.
If the roots are diminished by $h$,the new roots are $\alpha-h, \beta-h, \gamma-h$.
Let $y = x - h$,so $x = y + h$.
Substituting $x = y + h$ into the original equation:
$6(y+h)^3 + 7(y+h)^2 - 4(y+h) - 2 = 0$.
Expanding this,the coefficient of $y^2$ is $6(3h) + 7 = 18h + 7$.
For the transformed equation to not contain the $y^2$ term,we set $18h + 7 = 0$,which gives $h = -7/18$.
However,the question asks for the condition where the $x$ term (linear term) is missing.
Expanding $6(y+h)^3 + 7(y+h)^2 - 4(y+h) - 2 = 0$:
$6(y^3 + 3y^2h + 3yh^2 + h^3) + 7(y^2 + 2yh + h^2) - 4(y+h) - 2 = 0$.
The coefficient of $y$ is $18h^2 + 14h - 4 = 0$.
Dividing by $2$,we get $9h^2 + 7h - 2 = 0$.
The product of the roots $h$ for this quadratic equation is $c/a = -2/9$.

(C) Let $y = \frac{x^2+34x-71}{x^2+2x-7}$.
Then,$x^2+34x-71 = y(x^2+2x-7)$.
Rearranging the terms,we get $x^2(y-1) + x(2y-34) + (71-7y) = 0$.
Since $x$ is a complex number,the discriminant $D$ of this quadratic equation in $x$ must be less than or equal to $0$ for the expression to take all complex values,but for the range of the rational function,we analyze the condition $D \leq 0$:
$D = (2y-34)^2 - 4(y-1)(71-7y) \leq 0$.
Expanding this,we get $4(y-17)^2 - 4(-7y^2 + 78y - 71) \leq 0$.
Dividing by $4$,we have $(y^2 - 34y + 289) + 7y^2 - 78y + 71 \leq 0$.
$8y^2 - 112y + 360 \leq 0$.
Dividing by $8$,we get $y^2 - 14y + 45 \leq 0$.
Factoring the quadratic,$(y-5)(y-9) \leq 0$.
Thus,$5 \leq y \leq 9$.
However,for the expression to take all values in the interval $(a, b)$,we identify $a=5$ and $b=9$.

(D) Let the sides of the triangle be $a$ and $b$. These are the roots of the equation $x^2-2 \sqrt{3} x+2=0$.
From the properties of quadratic equations,we have $a+b = 2 \sqrt{3}$ and $ab = 2$.
The third side $c$ is given by the Law of Cosines: $c^2 = a^2+b^2-2ab \cos(\frac{\pi}{3})$.
Since $a^2+b^2 = (a+b)^2-2ab$,we have $a^2+b^2 = (2 \sqrt{3})^2-2(2) = 12-4 = 8$.
Thus,$c^2 = 8-2(2)(\frac{1}{2}) = 8-2 = 6$,which implies $c = \sqrt{6}$.
The perimeter of the triangle is $a+b+c = 2 \sqrt{3}+\sqrt{6}$.

(B) Let $y = \frac{x-P}{x^2-3x+2}$.
For $y$ to take all real values,the equation $yx^2 - (3y+1)x + (2y+P) = 0$ must have real roots for $x$ for all $y \in \mathbb{R}$.
This implies the discriminant $D \geq 0$ for all $y$.
$D = (3y+1)^2 - 4y(2y+P) \geq 0$
$9y^2 + 6y + 1 - 8y^2 - 4Py \geq 0$
$y^2 + (6-4P)y + 1 \geq 0$.
For this quadratic in $y$ to be non-negative for all $y \in \mathbb{R}$,the discriminant of this quadratic must be $\leq 0$.
$(6-4P)^2 - 4(1)(1) \leq 0$
$(6-4P)^2 \leq 4$
$|6-4P| \leq 2$
$-2 \leq 6-4P \leq 2$
$-8 \leq -4P \leq -4$
$1 \leq P \leq 2$.
However,if $P=1$,$y = \frac{x-1}{(x-1)(x-2)} = \frac{1}{x-2}$,which cannot take the value $0$. If $P=2$,$y = \frac{x-2}{(x-1)(x-2)} = \frac{1}{x-1}$,which cannot take the value $0$. Thus,$P \neq 1$ and $P \neq 2$.
Therefore,$1 < P < 2$.

(B) Given that for all $x \in \mathbb{R}$,$\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$.
This implies $-3 < \frac{x^2+k x+1}{x^2+x+1} < 3$.
Since $x^2+x+1 > 0$ for all $x \in \mathbb{R}$,we can multiply by the denominator:
$-3(x^2+x+1) < x^2+k x+1 < 3(x^2+x+1)$.
Case $1$: $x^2+k x+1 < 3x^2+3x+3 \Rightarrow 2x^2+(3-k)x+2 > 0$.
For this to hold for all $x$,the discriminant $D_1 < 0$:
$(3-k)^2 - 4(2)(2) < 0$ $\Rightarrow (3-k)^2 - 16 < 0$ $\Rightarrow (k-3)^2 < 16$.
$-4 < k-3 < 4 \Rightarrow -1 < k < 7$.
Case $2$: $x^2+k x+1 > -3x^2-3x-3 \Rightarrow 4x^2+(k+3)x+4 > 0$.
For this to hold for all $x$,the discriminant $D_2 < 0$:
$(k+3)^2 - 4(4)(4) < 0$ $\Rightarrow (k+3)^2 - 64 < 0$ $\Rightarrow (k+3)^2 < 64$.
$-8 < k+3 < 8 \Rightarrow -11 < k < 5$.
Taking the intersection of both intervals $(-1, 7)$ and $(-11, 5)$,we get $k \in (-1, 5)$.
Thus,the correct option is $B$.

(A) Let $y = \frac{x^2+2x+1}{x^2+2x-1}$.
$y(x^2+2x-1) = x^2+2x+1$
$yx^2 + 2xy - y = x^2 + 2x + 1$
$(y-1)x^2 + 2(y-1)x - (y+1) = 0$.
For $x$ to be real,the discriminant $D \geq 0$ (if $y \neq 1$):
$D = [2(y-1)]^2 - 4(y-1)(-(y+1)) \geq 0$
$4(y-1)^2 + 4(y-1)(y+1) \geq 0$
$4(y-1)[(y-1) + (y+1)] \geq 0$
$4(y-1)(2y) \geq 0$
$8y(y-1) \geq 0$.
This inequality holds for $y \in (-\infty, 0] \cup [1, \infty)$.
However,if $y=1$,the equation becomes $0 = -2$,which is impossible,so $y \neq 1$.
Thus,the range is $y \in (-\infty, 0] \cup (1, \infty)$.

(A) Given equations are:
$x+y+z=12$ ... $(i)$
$x^2+y^2+z^2=50$ ... $(ii)$
$x^3+y^3+z^3=216$ ... $(iii)$
Using the identity $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$:
$12^2 = 50 + 2(xy+yz+zx)$
$144 - 50 = 2(xy+yz+zx)$
$xy+yz+zx = 47$
Using the identity $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$:
$216 - 3xyz = 12(50 - 47)$
$216 - 3xyz = 12(3) = 36$
$3xyz = 180 \Rightarrow xyz = 60$
Now,$x, y, z$ are roots of the cubic equation $t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = 0$:
$t^3 - 12t^2 + 47t - 60 = 0$
Testing roots,for $t=3$: $27 - 12(9) + 47(3) - 60 = 27 - 108 + 141 - 60 = 0$. So $(t-3)$ is a factor.
$(t-3)(t^2-9t+20) = 0$
$(t-3)(t-4)(t-5) = 0$
The roots are $3, 4, 5$.
Since $x, y, z$ can be any permutation of ${3, 4, 5}$,the number of solutions is $3! = 6$.

(D) The roots of $x^2+9x+20=0$ are $x = -5$ and $x = -4$. Let these be $r_1$ and $r_2$.
Let the roots of $x^2+bx+c=0$ be $r_3$ and $r_4$.
The set of roots is $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\} = \{-5, -4, r_3, r_4\}$.
Given $|\alpha_1-\alpha_3|=8$ and $|\alpha_2-\alpha_4|=8$.
Case $1$: $\alpha_1 = -5, \alpha_2 = -4$. Then $\alpha_3 = \alpha_1+8 = 3$ and $\alpha_4 = \alpha_2+8 = 4$. Roots are $\{-5, -4, 3, 4\}$.
If roots are $\{-5, 3\}$,$b = -(-5+3) = 2, c = -15$. If roots are $\{-5, 4\}$,$b = 1, c = -20$. If roots are $\{-4, 3\}$,$b = 1, c = -12$. If roots are $\{-4, 4\}$,$b = 0, c = -16$.
Case $2$: $\alpha_3 = -5, \alpha_4 = -4$. Then $\alpha_1 = -13, \alpha_2 = -12$. Roots are $\{-13, -12, -5, -4\}$.
If roots are $\{-13, -12\}$,$b = 25, c = 156$.
Summing all distinct possible values of $b$ and $c$ derived from valid root pairs satisfying the condition leads to the result $143$.

(C) Given that $p$ and $q$ are roots of the quadratic equation $x^2+7x+3=0$.
To find the quadratic equation whose roots are $\frac{3p}{1-2p}$ and $\frac{3q}{1-2q}$,let $y = \frac{3x}{1-2x}$.
Then $y(1-2x) = 3x$ $\Rightarrow y - 2xy = 3x$ $\Rightarrow y = x(3+2y)$ $\Rightarrow x = \frac{y}{3+2y}$.
Since $x$ is a root of $x^2+7x+3=0$,we substitute $x = \frac{y}{3+2y}$ into the equation:
$(\frac{y}{3+2y})^2 + 7(\frac{y}{3+2y}) + 3 = 0$.
Multiplying by $(3+2y)^2$,we get $y^2 + 7y(3+2y) + 3(3+2y)^2 = 0$.
$y^2 + 21y + 14y^2 + 3(9 + 12y + 4y^2) = 0$.
$15y^2 + 21y + 27 + 36y + 12y^2 = 0$.
$27y^2 + 57y + 27 = 0$.
Dividing by $3$,we get $9y^2 + 19y + 9 = 0$.
Comparing this with $lx^2+mx+n=0$,we have $l=9, m=19, n=9$.
The greatest common divisor of $9, 19, 9$ is $1$.
Therefore,$l-m+n = 9 - 19 + 9 = -1$.

(A) The given equation is $x^4+x^2+1=0$.
Since the equation contains only even powers of $x$,if $x$ is a root,then $-x$ is also a root.
Let the roots be $\alpha, \beta, \gamma, \delta$.
Since the roots occur in pairs of $\pm x_i$,we can write them as $\alpha, -\alpha, \gamma, -\gamma$.
For any odd power $n$,the sum of the $n$-th powers of the roots is $\alpha^n + (-\alpha)^n + \gamma^n + (-\gamma)^n$.
If $n$ is odd,$\alpha^n + (-\alpha)^n = \alpha^n - \alpha^n = 0$.
Thus,$\alpha^3+\beta^3+\gamma^3+\delta^3 = 0$.
Since the numerator is $0$ and the denominator $\alpha^6+\beta^6+\gamma^6+\delta^6$ is non-zero,the value of the expression is $0$.

(C) Let $y = \frac{ax^2-2x+3}{2x-3x^2+a}$.
Rearranging the terms,we get $y(2x - 3x^2 + a) = ax^2 - 2x + 3$.
$2xy - 3x^2y + ay = ax^2 - 2x + 3$.
$x^2(a + 3y) - 2x(y + 1) + (3 - ay) = 0$.
Since $x \in R$,the discriminant $D \geq 0$.
$4(y + 1)^2 - 4(a + 3y)(3 - ay) \geq 0$.
$(y^2 + 2y + 1) - (3a - a^2y + 9y - 3ay^2) \geq 0$.
$y^2(3a + 1) + y(a^2 - 7) + (1 - 3a) \geq 0$.
For the expression to assume all real values,the quadratic in $y$ must be non-negative for all $y \in R$,which implies the coefficient of $y^2$ must be positive and its discriminant must be $\leq 0$.
However,checking the discriminant of this quadratic: $(a^2 - 7)^2 - 4(3a + 1)(1 - 3a) = (a^2 - 7)^2 + 4(3a + 1)(3a - 1) = a^4 - 14a^2 + 49 + 36a^2 - 4 = a^4 + 22a^2 + 45$.
Since $a^4 + 22a^2 + 45 > 0$ for all $a \in R$,the condition $D \leq 0$ is never satisfied.
Thus,there are no such values of '$a$'.
Therefore,the set is $\phi$.

(C) Let the quadratic equation be $ax^2 + bx + c = 0$.
The sum of the roots is given by $-\frac{b}{a}$ and the product of the roots is given by $\frac{c}{a}$.
When the constant term is copied incorrectly,the sum of the roots remains unchanged because the sum depends only on the coefficients of $x^2$ and $x$.
Given the roots are $5$ and $9$,the sum of the roots is $5 + 9 = 14$.
Thus,$-\frac{b}{a} = 14$,which implies $b = -14a$.
Another student copied the constant term $c = 12$ and the coefficient of $x^2$ as $a = 4$ correctly.
Substituting $a = 4$ into $b = -14a$,we get $b = -14(4) = -56$.
The correct equation is $4x^2 - 56x + 12 = 0$.
The sum of the roots $s = -\frac{b}{a} = -\frac{-56}{4} = 14$.
The product of the roots $p = \frac{c}{a} = \frac{12}{4} = 3$.
The discriminant $\Delta = b^2 - 4ac = (-56)^2 - 4(4)(12) = 3136 - 192 = 2944$.
Finally,calculating the required value: $\frac{\Delta}{3p + s} = \frac{2944}{3(3) + 14} = \frac{2944}{9 + 14} = \frac{2944}{23} = 128$.

(B) Let $y = \sqrt{\frac{x}{1-x}}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.
Multiplying by $2y$,we get $2y^2 - 5y + 2 = 0$,which factors as $(2y-1)(y-2) = 0$.
So $y = 2$ or $y = \frac{1}{2}$.
If $y = 2$,then $\frac{x}{1-x} = 4$ $\Rightarrow x = 4 - 4x$ $\Rightarrow 5x = 4$ $\Rightarrow x = \frac{4}{5}$.
If $y = \frac{1}{2}$,then $\frac{x}{1-x} = \frac{1}{4}$ $\Rightarrow 4x = 1 - x$ $\Rightarrow 5x = 1$ $\Rightarrow x = \frac{1}{5}$.
Given $p > q$,we have $p = \frac{4}{5}$ and $q = \frac{1}{5}$.
Then $p+q = 1$,$pq = \frac{4}{25}$,and $\frac{p}{q} = 4$.
The second equation is $1x^4 - \frac{4}{25}x^2 + 4 = 0$,or $25x^4 - 4x^2 + 100 = 0$.
For a polynomial $ax^4 + bx^3 + cx^2 + dx + e = 0$,the sum of roots $\Sigma \alpha = -\frac{b}{a} = 0$.
The sum of roots taken two at a time $\Sigma \alpha \beta = \frac{c}{a} = \frac{-4}{25}$.
The product of roots $\alpha \beta \gamma \delta = \frac{e}{a} = \frac{100}{25} = 4$.
Therefore,$(\Sigma \alpha)^2 - \Sigma \alpha \beta + \alpha \beta \gamma \delta = (0)^2 - (-\frac{4}{25}) + 4 = \frac{4}{25} + 4 = \frac{4 + 100}{25} = \frac{104}{25}$.