Solution of quadratic inequations and Newton Formula Questions in English

(C) Given $f(x) = x^2 + 4x + 1$.
For option $A$: The discriminant $D = b^2 - 4ac = 4^2 - 4(1)(1) = 16 - 4 = 12 > 0$. Since $D > 0$,the quadratic has two distinct real roots,so $f(x)$ takes both positive and negative values. Thus,$f(x) > 0$ is not true for all $x$.
For option $B$: If $x \ge 0$,then $x^2 \ge 0$ and $4x \ge 0$. Thus $f(x) = x^2 + 4x + 1 \ge 1$. Specifically,$f(x) > 1$ for $x > 0$. This is a correct statement.
For option $C$: $f(x) \ge 1 \implies x^2 + 4x + 1 \ge 1 \implies x^2 + 4x \ge 0 \implies x(x + 4) \ge 0$. This inequality holds when $x \le -4$ or $x \ge 0$. Thus,$f(x) \ge 1$ when $x \le -4$ is a correct statement.
Note: Both $B$ and $C$ are mathematically correct statements based on the inequality $x(x+4) \ge 0$.

(D) Given the inequation: $x^2 - 4x < 12$
Subtract $12$ from both sides: $x^2 - 4x - 12 < 0$
Factor the quadratic expression: $(x - 6)(x + 2) < 0$
For the product of two factors to be negative,the factors must have opposite signs.
This inequality holds when $x$ is between the roots $-2$ and $6$.
Thus,the solution is $-2 < x < 6$.

(A) For a quadratic expression $Ax^2 + Bx + C > 0$ to be true for all $x \in R$,the discriminant $D$ must be less than $0$ and $A > 0$.
Here,$A = 1$,which is $> 0$.
The discriminant $D = B^2 - 4AC = (2a)^2 - 4(1)(10 - 3a) < 0$.
$4a^2 - 40 + 12a < 0$.
Dividing by $4$,we get $a^2 + 3a - 10 < 0$.
Factoring the quadratic,we get $(a + 5)(a - 2) < 0$.
This inequality holds when $a$ lies between the roots,so $-5 < a < 2$.

(A) Given the quadratic inequality $ax^2 - 2x + 4 > 0$ with $a < 0$.
First,find the roots of the corresponding quadratic equation $ax^2 - 2x + 4 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here,$x = \frac{2 \pm \sqrt{(-2)^2 - 4(a)(4)}}{2a} = \frac{2 \pm \sqrt{4 - 16a}}{2a} = \frac{1 \pm \sqrt{1 - 4a}}{a}$.
Since $a < 0$,the parabola $y = ax^2 - 2x + 4$ opens downwards.
The inequality $ax^2 - 2x + 4 > 0$ holds between the two roots.
Let $\alpha = \frac{1 + \sqrt{1 - 4a}}{a}$ and $\beta = \frac{1 - \sqrt{1 - 4a}}{a}$.
Since $a < 0$,$\frac{1 + \sqrt{1 - 4a}}{a} < \frac{1 - \sqrt{1 - 4a}}{a}$.
Thus,the solution is $\frac{1 + \sqrt{1 - 4a}}{a} < x < \frac{1 - \sqrt{1 - 4a}}{a}$.

(B) To solve the inequality $3x^2 + 14x + 11 > 0$,we first find the roots of the quadratic equation $3x^2 + 14x + 11 = 0$.
Using the factorization method: $3x^2 + 3x + 11x + 11 = 0$.
$3x(x + 1) + 11(x + 1) = 0$.
$(3x + 11)(x + 1) = 0$.
The roots are $x = -\frac{11}{3}$ and $x = -1$.
Since the coefficient of $x^2$ is positive $(3 > 0)$,the parabola opens upwards.
The expression $3x^2 + 14x + 11$ is positive outside the interval between the roots.
Therefore,the inequality holds for $x < -\frac{11}{3}$ or $x > -1$.

(B) Let $|x| = t$,where $t \geq 0$.
The inequality becomes $t^2 + 2t - 15 \geq 0$.
Factoring the quadratic expression: $(t + 5)(t - 3) \geq 0$.
Since $t \geq 0$,$t + 5$ is always positive.
Thus,we must have $t - 3 \geq 0$,which implies $t \geq 3$.
Substituting back $|x| \geq 3$,we get $x \geq 3$ or $x \leq -3$.

(C) Given the expression $f(x) = mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.
Multiplying by $x$ (since $x > 0$),we get $mx^2 - x + 1 \geq 0$.
For a quadratic $ax^2 + bx + c \geq 0$ to hold for all $x > 0$,we analyze the conditions.
If $m \leq 0$,then for large $x$,$mx^2 - x + 1$ will eventually become negative,so we must have $m > 0$.
For $mx^2 - x + 1 \geq 0$ to be true for all $x > 0$,the discriminant $D$ must be less than or equal to $0$.
$D = (-1)^2 - 4(m)(1) = 1 - 4m \leq 0$.
$1 \leq 4m \implies m \geq \frac{1}{4}$.
Thus,the minimum value of $m$ is $\frac{1}{4}$.

(B) Given the quadratic inequality: $2x^2 + 3x - 9 \le 0$
Step $1$: Factorize the quadratic expression.
$2x^2 + 6x - 3x - 9 \le 0$
$2x(x + 3) - 3(x + 3) \le 0$
$(2x - 3)(x + 3) \le 0$
Step $2$: Find the critical points by setting the factors to zero.
$2x - 3 = 0 \implies x = 3/2$
$x + 3 = 0 \implies x = -3$
Step $3$: Determine the interval where the product is less than or equal to zero.
For a quadratic $(x - \alpha)(x - \beta) \le 0$ where $\alpha < \beta$,the solution is $\alpha \le x \le \beta$.
Here,$-3 \le x \le 3/2$.
Thus,the correct option is $B$.

(D) Given the inequality: $\frac{4x^2 + 1}{64x^2 - 96x \sin \alpha + 5} < \frac{1}{32}$.
Assuming the denominator $64x^2 - 96x \sin \alpha + 5 > 0$,we cross-multiply:
$128x^2 + 32 < 64x^2 - 96x \sin \alpha + 5$
$64x^2 + 96x \sin \alpha + 27 < 0$.
For this quadratic in $x$ to be negative for all $x$,the discriminant $D$ must be positive,but here we require the inequality to hold for some $x$. However,the condition "for all real values of $x$" implies the quadratic $64x^2 + 96x \sin \alpha + 27$ must be negative,which is impossible for a parabola opening upwards. Re-evaluating the condition,the inequality holds if the discriminant $D > 0$ and the quadratic is negative between its roots.
$D = (96 \sin \alpha)^2 - 4(64)(27) > 0$
$9216 \sin^2 \alpha - 6912 > 0$
$\sin^2 \alpha > \frac{6912}{9216} = \frac{3}{4}$
$|\sin \alpha| > \frac{\sqrt{3}}{2}$.
This implies $\sin \alpha > \frac{\sqrt{3}}{2}$ or $\sin \alpha < -\frac{\sqrt{3}}{2}$.
For $\sin \alpha > \frac{\sqrt{3}}{2}$,$\alpha \in (\pi/3, 2\pi/3)$.
For $\sin \alpha < -\frac{\sqrt{3}}{2}$,$\alpha \in (4\pi/3, 5\pi/3)$.

(B) Given the inequality $2^{\log_{\sqrt{2}}(x - 1)} > x + 5$.
First,consider the domain of the logarithmic function: $x - 1 > 0 \implies x > 1$.
Using the property $\log_{a^n}(b) = \frac{1}{n} \log_a(b)$,we have $\log_{\sqrt{2}}(x - 1) = \log_{2^{1/2}}(x - 1) = 2 \log_2(x - 1) = \log_2((x - 1)^2)$.
Substituting this into the inequality: $2^{\log_2((x - 1)^2)} > x + 5$.
Since $2^{\log_2(y)} = y$,the inequality becomes $(x - 1)^2 > x + 5$.
Expanding the left side: $x^2 - 2x + 1 > x + 5$.
Rearranging the terms: $x^2 - 3x - 4 > 0$.
Factoring the quadratic: $(x - 4)(x + 1) > 0$.
The roots are $x = 4$ and $x = -1$.
The inequality holds for $x \in (-\infty, -1) \cup (4, +\infty)$.
Considering the domain $x > 1$,the intersection of $(x > 1)$ and $(x < -1 \text{ or } x > 4)$ is $x > 4$.
Thus,the solution set is $(4, +\infty)$.

(D) The given inequality is $kx^2 - 2x + k \geq 0$.
Case $1$: If $k = 0$,the inequality becomes $-2x \geq 0$,which implies $x \leq 0$. This holds for at least one real $x$,so $k = 0$ is a solution.
Case $2$: If $k \neq 0$,the inequality $kx^2 - 2x + k \geq 0$ holds for at least one real $x$ if the maximum value of the expression $f(x) = kx^2 - 2x + k$ is $\geq 0$ (if $k < 0$) or if the parabola opens upwards and has real roots or is always non-negative (if $k > 0$).
Alternatively,we can write $k(x^2 + 1) \geq 2x$,which implies $k \geq \frac{2x}{x^2 + 1}$.
Let $f(x) = \frac{2x}{x^2 + 1}$. The range of $f(x)$ is $[-1, 1]$.
For the inequality $k \geq f(x)$ to hold for at least one $x$,$k$ must be greater than or equal to the minimum value of $f(x)$.
Thus,$k \geq -1$ and $k \neq 0$ (from Case $1$,$k=0$ is included).
Therefore,the set of values is $k \in [-1, \infty)$.

(B) Let $f(t) = t^2 + 5mt - 3m + 1$,where $t = \{x\} \in [0, 1)$.
We require $f(t) < 0$ for all $t \in [0, 1)$.
Since $f(t)$ is a quadratic in $t$ with a positive leading coefficient,the parabola opens upward.
For $f(t) < 0$ to hold for all $t \in [0, 1)$,the values at the endpoints must satisfy $f(0) < 0$ and $f(1) \leq 0$ (since the interval is $[0, 1)$).
$f(0) = -3m + 1 < 0 \implies 3m > 1 \implies m > \frac{1}{3}$.
$f(1) = 1 + 5m - 3m + 1 = 2m + 2 \leq 0 \implies 2m \leq -2 \implies m \leq -1$.
Combining $m > \frac{1}{3}$ and $m \leq -1$,we find no such $m$ exists.
Thus,the number of integral values of $m$ is $0$.

(A) To solve the inequality $x^{2} \leq 4$,we can rewrite it as $x^{2} - 4 \leq 0$.
This is a difference of squares,which factors as $(x - 2)(x + 2) \leq 0$.
The roots of the corresponding equation $(x - 2)(x + 2) = 0$ are $x = 2$ and $x = -2$.
These roots divide the number line into three intervals: $(-\infty, -2)$,$(-2, 2)$,and $(2, \infty)$.
Testing a value in the interval $(-2, 2)$,such as $x = 0$: $(0 - 2)(0 + 2) = -4$,which is $\leq 0$.
Since the inequality includes the equality sign,the endpoints are included.
Thus,the solution set is the closed interval $[-2, 2]$.

(A) Given the inequality $x^{2} \leq 9$.
This can be rewritten as $x^{2} - 9 \leq 0$.
Factoring the expression,we get $(x - 3)(x + 3) \leq 0$.
For the product of two factors to be less than or equal to zero,the values of $x$ must lie between the roots of the equation $(x - 3)(x + 3) = 0$.
The roots are $x = 3$ and $x = -3$.
Testing the intervals,we find that for $x \in [-3, 3]$,the expression $(x - 3)(x + 3) \leq 0$ holds true.
Therefore,the solution set is $[-3, 3]$.

(A) For the quadratic inequality $ax^{2}+bx+c > 0$ to be valid for all $x \in \mathbb{R}$,the discriminant $D$ must be less than $0$ and $a > 0$.
Here,$a = 1 > 0$,so we only need $D < 0$.
$D = [-2(3k-1)]^{2} - 4(1)(8k^{2}-7) < 0$
$4(9k^{2}-6k+1) - 4(8k^{2}-7) < 0$
Dividing by $4$:
$9k^{2}-6k+1 - 8k^{2}+7 < 0$
$k^{2}-6k+8 < 0$
$(k-2)(k-4) < 0$
This implies $k \in (2, 4)$.
Since $k$ is an integer,the only integer value in the interval $(2, 4)$ is $k = 3$.

(B) Let $S_n = a^n + b^n$. Since $a$ and $b$ are roots of $x^2-7x-1=0$,by Newton's Sums,we have $S_{n+2} - 7S_{n+1} - S_n = 0$,which implies $S_{n+2} = 7S_{n+1} + S_n$.
We want to evaluate $\frac{S_{21} + S_{17}}{S_{19}}$.
From the recurrence relation,$S_{21} = 7S_{20} + S_{19}$.
Also,$S_{19} = 7S_{18} + S_{17}$,which implies $S_{17} = S_{19} - 7S_{18}$.
Substituting these into the expression:
$\frac{S_{21} + S_{17}}{S_{19}} = \frac{7S_{20} + S_{19} + S_{19} - 7S_{18}}{S_{19}} = \frac{7S_{20} + 2S_{19} - 7S_{18}}{S_{19}}$.
Since $S_{20} = 7S_{19} + S_{18}$,we have $S_{20} - S_{18} = 7S_{19}$.
Substituting this into the numerator:
$\frac{7(S_{20} - S_{18}) + 2S_{19}}{S_{19}} = \frac{7(7S_{19}) + 2S_{19}}{S_{19}} = \frac{49S_{19} + 2S_{19}}{S_{19}} = \frac{51S_{19}}{S_{19}} = 51$.

(B) Given the inequality $\frac{ax^2+2(a+1)x+9a+4}{x^2-8x+32} < 0$ for all $x \in R$.
Since the denominator $x^2-8x+32 = (x-4)^2 + 16 > 0$ for all $x \in R$,the inequality holds if and only if the numerator $f(x) = ax^2+2(a+1)x+9a+4 < 0$ for all $x \in R$.
For $f(x) < 0$ for all $x \in R$,we must have $a < 0$ and the discriminant $D < 0$.
However,the question asks for the set $S$ of positive integral values of $a$.
Since $a$ must be negative for the inequality to hold for all $x$,there are no positive integral values of $a$ that satisfy the condition.
Therefore,the set $S$ is empty,and the number of elements in $S$ is $0$.

(C) By Newton's sum formula for the equation $x^2 - (t^2 - 5t + 6)x + 1 = 0$,we have:
$a_{n+2} - (t^2 - 5t + 6)a_{n+1} + a_n = 0$
Rearranging the terms,we get:
$a_{n+2} + a_n = (t^2 - 5t + 6)a_{n+1}$
For $n = 2023$,this becomes:
$a_{2025} + a_{2023} = (t^2 - 5t + 6)a_{2024}$
Therefore,the ratio is:
$\frac{a_{2025} + a_{2023}}{a_{2024}} = t^2 - 5t + 6$
To find the minimum value of the quadratic expression $f(t) = t^2 - 5t + 6$,we complete the square:
$f(t) = (t - 5/2)^2 - 25/4 + 6 = (t - 5/2)^2 - 1/4$
The minimum value occurs at $t = 5/2$,which is $-1/4$.

(A) Given inequality: $4+11x-3x^2 > 0$
Multiply by $-1$ on both sides,the inequality sign reverses:
$3x^2 - 11x - 4 < 0$
Factorize the quadratic expression:
$3x^2 - 12x + x - 4 < 0$
$3x(x - 4) + 1(x - 4) < 0$
$(3x + 1)(x - 4) < 0$
To find the interval where the product is negative,we find the roots: $x = -\frac{1}{3}$ and $x = 4$.
Testing the intervals:
For $x < -\frac{1}{3}$,the expression is positive.
For $-\frac{1}{3} < x < 4$,the expression is negative.
For $x > 4$,the expression is positive.
Therefore,the solution set is $x \in \left(-\frac{1}{3}, 4\right)$.

(C) For the square root to be defined,$x^2+6x+5 \ge 0$,which implies $(x+5)(x+1) \ge 0$,so $x \in (-\infty, -5] \cup [-1, \infty)$.
For the inequality $\sqrt{x^2+6x+5} > (8-x)$ to hold,we must have $8-x < 0$ or ($8-x \ge 0$ and $x^2+6x+5 > (8-x)^2$).
Case $1$: $8-x < 0 \implies x > 8$. Since $x > 8$ satisfies $x \in [-1, \infty)$,this is a valid solution.
Case $2$: $8-x \ge 0 \implies x \le 8$. Squaring both sides: $x^2+6x+5 > 64-16x+x^2$.
$22x > 59 \implies x > \frac{59}{22}$.
Combining $x > \frac{59}{22}$ with $x \le 8$,we get $x \in (\frac{59}{22}, 8]$.
Combining Case $1$ and Case $2$: $x \in (\frac{59}{22}, 8] \cup (8, \infty) = (\frac{59}{22}, \infty)$.

(D) For the quadratic expression $f(x) = ax^2 + bx + c$ to be positive for all $x \in R$,we must have $a > 0$ and the discriminant $D < 0$.
Here,$a = 2k - 1$,$b = -2(3k - 2)$,and $c = 4k$.
Condition $1$: $a > 0 \implies 2k - 1 > 0 \implies k > \frac{1}{2}$.
Condition $2$: $D < 0 \implies b^2 - 4ac < 0$.
$[-2(3k - 2)]^2 - 4(2k - 1)(4k) < 0$.
$4(9k^2 - 12k + 4) - 16k(2k - 1) < 0$.
Divide by $4$: $(9k^2 - 12k + 4) - 4k(2k - 1) < 0$.
$9k^2 - 12k + 4 - 8k^2 + 4k < 0$.
$k^2 - 8k + 4 < 0$.
Roots of $k^2 - 8k + 4 = 0$ are $k = \frac{8 \pm \sqrt{64 - 16}}{2} = \frac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3}$.
Since $2\sqrt{3} \approx 3.46$,the roots are $4 - 3.46 = 0.54$ and $4 + 3.46 = 7.46$.
So,$0.54 < k < 7.46$.
Combining with $k > 0.5$,we get $0.54 < k < 7.46$.
The integral values of $k$ are $1, 2, 3, 4, 5, 6, 7$.
The sum is $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$.

(B) For the quadratic expression $f(x) = x^2 - 4ax + (5+a)$ to be greater than $0$ for all $x \in R$,the discriminant $D$ must be less than $0$.
$D = (-4a)^2 - 4(1)(5+a) < 0$
$16a^2 - 20 - 4a < 0$
$4a^2 - a - 5 < 0$
Factoring the quadratic: $(4a - 5)(a + 1) < 0$
This inequality holds when $a \in (-1, 5/4)$.
Comparing this with $a \in (\alpha, \beta)$,we get $\alpha = -1$ and $\beta = 5/4$.
Therefore,$4\beta + \alpha = 4(5/4) + (-1) = 5 - 1 = 4$.

(C) For the quadratic expression $ax^2 + bx + c > 0$ to be true for all $x \in R$,we must have $a > 0$ and the discriminant $D = b^2 - 4ac < 0$.
Here,$a = 1 > 0$,which is always true.
The discriminant $D = \{-(3k + 1)\}^2 - 4(1)(4k^2 + 3k - 3) < 0$.
Expanding this,we get $(9k^2 + 6k + 1) - (16k^2 + 12k - 12) < 0$.
$-7k^2 - 6k + 13 < 0$.
Multiplying by $-1$ reverses the inequality: $7k^2 + 6k - 13 > 0$.
Factoring the quadratic: $(7k + 13)(k - 1) > 0$.
The roots are $k = -\frac{13}{7}$ and $k = 1$.
For the expression to be positive,$k$ must lie outside the interval $[-\frac{13}{7}, 1]$.
Thus,the solution set is $k \in (-\infty, -\frac{13}{7}) \cup (1, \infty)$.

(B) For a quadratic expression $ax^2 + bx + c > 0$ to be true for all $x \in R$,the conditions are $a > 0$ and the discriminant $D < 0$.
Here,$a = 1$,which is $> 0$.
The discriminant $D = b^2 - 4ac < 0$.
$D = [-2(4k - 1)]^2 - 4(1)(15k^2 - 2k - 7) < 0$.
$4(16k^2 - 8k + 1) - 4(15k^2 - 2k - 7) < 0$.
Dividing by $4$,we get $16k^2 - 8k + 1 - 15k^2 + 2k + 7 < 0$.
$k^2 - 6k + 8 < 0$.
$(k - 2)(k - 4) < 0$.
This inequality holds for $2 < k < 4$.
The only integer value of $k$ in this interval is $k = 3$.

(B) For the first inequality: $x^2+5x+6 \geq 0 \Rightarrow (x+2)(x+3) \geq 0$. This holds for $x \in (-\infty, -3] \cup [-2, \infty)$.
For the second inequality: $x^2+3x-4 < 0 \Rightarrow (x+4)(x-1) < 0$. This holds for $x \in (-4, 1)$.
The intersection of these two sets is the set of $x$ values satisfying both conditions.
Intersection: $(-\infty, -3] \cup [-2, \infty) \cap (-4, 1) = (-4, -3] \cup [-2, 1)$.

(D) Given that $\alpha_1, \alpha_2, \alpha_3$ are the roots of $x^3+3x+2=0$.
By Newton's Sums,let $S_n = \alpha_1^n + \alpha_2^n + \alpha_3^n$.
The equation is $x^3 + 0x^2 + 3x + 2 = 0$.
For $n=1$: $S_1 + 0 = 0 \Rightarrow S_1 = 0$.
For $n=2$: $S_2 + 0(S_1) + 3(2) = 0 \Rightarrow S_2 = -6$.
For $n=3$: $S_3 + 0(S_2) + 3(S_1) + 2(3) = 0 \Rightarrow S_3 = -6$.
For $n=4$: $S_4 + 0(S_3) + 3(S_2) + 2(S_1) = 0$ $\Rightarrow S_4 + 3(-6) + 2(0) = 0$ $\Rightarrow S_4 = 18$.
For $n=5$: $S_5 + 0(S_4) + 3(S_3) + 2(S_2) = 0 \Rightarrow S_5 + 3(-6) + 2(-6) = 0$.
$S_5 - 18 - 12 = 0 \Rightarrow S_5 = 30$.

(A) Let $3^x = y$. Since $x \in R^{+}$,$y > 1$.
The given inequation is $y + \frac{3}{y} - 4 < 0$.
Multiplying by $y$ (since $y > 0$),we get $y^2 - 4y + 3 < 0$.
Factoring the quadratic,we have $(y - 1)(y - 3) < 0$.
This implies $1 < y < 3$.
Substituting back $y = 3^x$,we get $1 < 3^x < 3$.
Taking $\log_3$ on all sides,we get $\log_3(1) < x < \log_3(3)$,which simplifies to $0 < x < 1$.
Thus,the solution set is $(0, 1)$.

(D) Let the roots of the equation $x^3+x+1=0$ be $\alpha, \beta, \gamma$. Let $S_n = \alpha^n + \beta^n + \gamma^n$.
By Newton's Sums for the equation $x^3+p_1x^2+p_2x+p_3=0$,where $p_1=0, p_2=1, p_3=1$:
$S_1 + p_1 = 0$ $\Rightarrow S_1 + 0 = 0$ $\Rightarrow S_1 = 0$.
$S_2 + p_1S_1 + 2p_2 = 0$ $\Rightarrow S_2 + 0(0) + 2(1) = 0$ $\Rightarrow S_2 = -2$.
$S_3 + p_1S_2 + p_2S_1 + 3p_3 = 0$ $\Rightarrow S_3 + 0(-2) + 1(0) + 3(1) = 0$ $\Rightarrow S_3 = -3$.
$S_4 + p_1S_3 + p_2S_2 + p_3S_1 = 0 \Rightarrow S_4 + 0(-3) + 1(-2) + 1(0) = 0$.
$S_4 - 2 = 0 \Rightarrow S_4 = 2$.

(C) Given inequation: $x^2 - |x+2| + x > 0$
Case $I$: If $x+2 \geq 0$ (i.e., $x \geq -2$), then $|x+2| = x+2$.
The inequation becomes:
$x^2 - (x+2) + x > 0$
$\Rightarrow x^2 - 2 > 0$
$\Rightarrow (x-\sqrt{2})(x+\sqrt{2}) > 0$
The solution for this is $x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$.
Considering the condition $x \geq -2$, the intersection is $x \in [-2, -\sqrt{2}) \cup (\sqrt{2}, \infty)$.
Case $II$: If $x+2 < 0$ (i.e., $x < -2$), then $|x+2| = -(x+2)$.
The inequation becomes:
$x^2 - (-(x+2)) + x > 0$
$\Rightarrow x^2 + x + 2 + x > 0$
$\Rightarrow x^2 + 2x + 2 > 0$
$\Rightarrow (x+1)^2 + 1 > 0$
Since $(x+1)^2 + 1$ is always positive for all real $x$, the condition $x < -2$ is satisfied.
Combining Case $I$ and Case $II$:
$(-\infty, -2) \cup [-2, -\sqrt{2}) \cup (\sqrt{2}, \infty) = (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$

(B) Given inequality is $[x]^2 - 7[x] + 12 \leq 0$.
Let $y = [x]$. Then the inequality becomes $y^2 - 7y + 12 \leq 0$.
Factoring the quadratic expression,we get $(y - 4)(y - 3) \leq 0$.
This implies that $3 \leq y \leq 4$.
Since $y = [x]$,we have $3 \leq [x] \leq 4$.
This means $[x]$ can be either $3$ or $4$.
If $[x] = 3$,then $3 \leq x < 4$.
If $[x] = 4$,then $4 \leq x < 5$.
Combining these two intervals,we get $3 \leq x < 5$.

(C) For the first inequality: $x^2-7x+10 \geq 0$
$(x-2)(x-5) \geq 0$
Thus,$x \in (-\infty, 2] \cup [5, \infty)$.
For the second inequality: $2x+3-x^2 > 0$
$x^2-2x-3 < 0$
$(x-3)(x+1) < 0$
Thus,$x \in (-1, 3)$.
Taking the intersection of both intervals:
$(-\infty, 2] \cup [5, \infty) \cap (-1, 3) = (-1, 2]$.
Therefore,the set of all values of $x$ is $(-1, 2]$.

(A) Given the inequality $3x^2 - 16x + 4 > -16$.
This simplifies to $3x^2 - 16x + 20 > 0$.
Let $f(x) = 3x^2 - 16x + 20$. Here $a = 3, b = -16, c = 20$.
The discriminant $D = b^2 - 4ac = (-16)^2 - 4(3)(20) = 256 - 240 = 16 > 0$.
The roots of $3x^2 - 16x + 20 = 0$ are $x = \frac{16 \pm \sqrt{16}}{6} = \frac{16 \pm 4}{6}$,which gives $x = 2$ and $x = \frac{10}{3}$.
Since $a > 0$,$f(x) > 0$ for $x \in (-\infty, 2) \cup (\frac{10}{3}, \infty)$.
The interval $(0, \frac{10}{3})$ contains values like $x=1$ where $f(1) = 3 - 16 + 20 = 7 > 0$. Thus,$(A)$ is true.
Reason $(R)$ states that $ax^2 + bx + c$ and $a$ have the same sign for some $x$ when $D > 0$. This is true because the parabola opens in the direction of $a$ outside the roots. Thus,$(R)$ is true and explains $(A)$.

(D) Given the quadratic expression $f(x) = x^2+2px-2p+8 > 0$ for all real values of $x$.
For a quadratic $ax^2+bx+c > 0$ to be true for all $x \in R$,the conditions are $a > 0$ and the discriminant $D < 0$.
Here,$a = 1 > 0$,which is satisfied.
Now,calculate the discriminant $D = b^2 - 4ac < 0$:
$D = (2p)^2 - 4(1)(-2p+8) < 0$
$4p^2 + 8p - 32 < 0$
Divide by $4$:
$p^2 + 2p - 8 < 0$
Factorize the quadratic:
$(p+4)(p-2) < 0$
Using the sign scheme method,the expression is negative between the roots $p = -4$ and $p = 2$.
Therefore,the set of all possible values of $p$ is $p \in (-4, 2)$.

(A) Given the inequality: $x^2-5x-14 > 0$
Factorizing the quadratic expression: $(x+2)(x-7) > 0$
For the product to be positive,$x$ must be less than the smaller root or greater than the larger root: $x \in (-\infty, -2) \cup (7, \infty)$
The problem states that $x$ lies outside the interval $[\alpha, \beta]$.
Comparing the intervals,we identify $\alpha = -2$ and $\beta = 7$.
Therefore,the value of $\frac{\alpha}{\beta} = \frac{-2}{7}$.

(D) Given inequations are $x^2-1 \leq 0$ and $x^2-x-2 \geq 0$.
For $x^2-1 \leq 0$:
$(x-1)(x+1) \leq 0$
This implies $x \in [-1, 1]$.
For $x^2-x-2 \geq 0$:
$(x-2)(x+1) \geq 0$
This implies $x \in (-\infty, -1] \cup [2, \infty)$.
The intersection of the two sets $x \in [-1, 1]$ and $x \in (-\infty, -1] \cup [2, \infty)$ is the single point $\{-1\}$.
Thus,the set of all values of $x$ is $\{-1\}$.

(A) Given inequation: $3^x+3^{1-x}-4 < 0$
Multiply by $3^x$ (since $3^x > 0$):
$(3^x)^2 - 4 \cdot 3^x + 3 < 0$
Let $y = 3^x$,then $y^2 - 4y + 3 < 0$
Factorizing the quadratic: $(y-1)(y-3) < 0$
This implies $1 < y < 3$
Substituting $y = 3^x$: $1 < 3^x < 3$
$3^0 < 3^x < 3^1$
Since the base $3 > 1$,we have $0 < x < 1$
Thus,the solution set is $(0,1)$.

(C) Given the inequality: $|x|^2-5|x|+4 < 0$
Let $|x| = y$. Since $|x| \ge 0$,we have $y \ge 0$.
The inequality becomes $y^2-5y+4 < 0$.
Factoring the quadratic expression: $(y-4)(y-1) < 0$.
This implies $1 < y < 4$.
Substituting back $y = |x|$,we get $1 < |x| < 4$.
This inequality is equivalent to $|x| > 1$ and $|x| < 4$.
For $|x| > 1$,$x \in (-\infty, -1) \cup (1, \infty)$.
For $|x| < 4$,$x \in (-4, 4)$.
Taking the intersection of these two sets,we get $x \in (-4, -1) \cup (1, 4)$.

(D) Let $y = \frac{(x+2)(x+5)}{x+6}$.
$xy + 6y = x^2 + 7x + 10$
$x^2 + (7-y)x + (10-6y) = 0$
For $x$ to be real,the discriminant $\Delta \geq 0$.
$(7-y)^2 - 4(10-6y) \geq 0$
$y^2 - 14y + 49 - 40 + 24y \geq 0$
$y^2 + 10y + 9 \geq 0$
$(y+1)(y+9) \geq 0$
Thus,$y \in (-\infty, -9] \cup [-1, \infty)$.
The values of $y$ that do not lie in the range are in the interval $(-9, -1)$.

(C) Given inequation is $x^2 - 2x + 5 \leq 0$.
We can rewrite the expression by completing the square:
$x^2 - 2x + 1 + 4 \leq 0$
$(x - 1)^2 + 4 \leq 0$
Since $(x - 1)^2 \geq 0$ for all real $x$,it follows that $(x - 1)^2 + 4 \geq 4$.
Therefore,the expression $(x - 1)^2 + 4$ is always positive and can never be less than or equal to $0$.
Thus,there are no real values of $x$ that satisfy the given inequation.
The set of all solutions is the empty set,denoted by $\phi$.

(B) For the first inequation: $x^2-4x \leq 12$
$x^2-4x-12 \leq 0$
$(x-6)(x+2) \leq 0$
Thus,$x \in [-2, 6]$ ... $(i)$
For the second inequation: $x^2-2x \geq 15$
$x^2-2x-15 \geq 0$
$(x-5)(x+3) \geq 0$
Thus,$x \in (-\infty, -3] \cup [5, \infty)$ ... $(ii)$
Taking the intersection of $(i)$ and $(ii)$:
$[-2, 6] \cap ((-\infty, -3] \cup [5, \infty)) = [5, 6]$
Therefore,the common solution set is $[5, 6]$.

(D) $3^x+3^{1-x}-4 < 0$
$\Rightarrow 3^x+\frac{3}{3^x}-4 < 0$
Let $3^x=t$,where $t > 0$.
$\Rightarrow t+\frac{3}{t}-4 < 0$
$\Rightarrow t^2-4t+3 < 0$
$\Rightarrow (t-1)(t-3) < 0$
This inequality holds for $1 < t < 3$.
Substituting $t=3^x$ back,we get $1 < 3^x < 3$.
$\Rightarrow 3^0 < 3^x < 3^1$
Since the base $3 > 1$,the inequality sign remains the same:
$0 < x < 1$
Thus,the solution set is $x \in (0,1)$.

(B) $I \rightarrow |x|^2 - 4|x| + 3 < 0$
Let $t = |x|$,where $t \geq 0$. The inequality becomes $t^2 - 4t + 3 < 0$.
$(t - 1)(t - 3) < 0$,which implies $1 < t < 3$.
Since $t = |x|$,we have $1 < |x| < 3$.
This means $x \in (-3, -1) \cup (1, 3)$.
Thus,Statement $I$ is false.
$II \rightarrow x^2 - 8x + 15 > 0$
$(x - 3)(x - 5) > 0$.
The roots are $x = 3$ and $x = 5$. The inequality holds for $x < 3$ or $x > 5$.
Thus,Statement $II$ is true.

(A) Given the inequalities $x^2 < 4x + 77$ and $x^2 > 4$.
First,solve $x^2 - 4x < 77$.
Adding $4$ to both sides,we get $x^2 - 4x + 4 < 77 + 4$,which simplifies to $(x - 2)^2 < 81$.
Taking the square root,we have $-9 < x - 2 < 9$,which gives $-7 < x < 11$.
Next,solve $x^2 > 4$,which implies $x^2 - 4 > 0$,so $(x - 2)(x + 2) > 0$.
This holds when $x > 2$ or $x < -2$.
Combining the two conditions,we need $x$ such that $(-7 < x < 11)$ $AND$ $(x > 2 \text{ or } x < -2)$.
The intersection is $(-7, -2) \cup (2, 11)$.
The negative integers in this set are $\{-6, -5, -4, -3\}$.
The smallest negative integer in this set is $-6$.

(B) We have the inequality: $9 x-2 < (x+2)^2 < 12 x-3$.
This can be split into two parts:
Part $I$: $9 x-2 < (x+2)^2$
$9 x-2 < x^2+4 x+4$
$x^2-5 x+6 > 0$
$(x-3)(x-2) > 0$
So,$x \in (-\infty, 2) \cup (3, \infty)$ ... $(i)$
Part $II$: $(x+2)^2 < 12 x-3$
$x^2+4 x+4 < 12 x-3$
$x^2-8 x+7 < 0$
$(x-7)(x-1) < 0$
So,$x \in (1, 7)$ ... $(ii)$
Taking the intersection of $(i)$ and $(ii)$,we get:
$x \in (1, 2) \cup (3, 7)$
The integral values of $x$ in this interval are $\{4, 5, 6\}$.
Thus,the number of integral values is $3$.

(D) Given inequalities are:
$1$) $x^2 - 3x - 10 < 0$
$(x - 5)(x + 2) < 0$
This implies $x \in (-2, 5)$.
$2$) $10x - x^2 - 16 > 0$
Multiply by $-1$ (inequality sign reverses):
$x^2 - 10x + 16 < 0$
$(x - 2)(x - 8) < 0$
This implies $x \in (2, 8)$.
To find the values of $x$ that satisfy both inequalities simultaneously,we take the intersection of the two sets:
$x \in (-2, 5) \cap (2, 8) = (2, 5)$.
Therefore,the correct option is $D$.