Word problem of Linear inequalities Questions in English

(C) Let the length of the shortest side be $x \text{ cm}$.
The longest side is $2x \text{ cm}$ and the third side is $(x + 2) \text{ cm}$.
The perimeter of the triangle is the sum of all sides: $P = x + 2x + (x + 2) = 4x + 2$.
According to the problem, the perimeter is greater than $166 \text{ cm}$:
$4x + 2 > 166$
Subtract $2$ from both sides:
$4x > 164$
Divide by $4$:
$x > 41$.
Since $x$ must be greater than $41$, the minimum integer length of the shortest side is $42 \text{ cm}$.

(N/A) Given the temperature formula: $T = 30 + 25(x - 3)$,for $3 \leq x \leq 15$.
We need to find the depth $x$ such that $155 < T < 205$.
Substituting the expression for $T$:
$155 < 30 + 25(x - 3) < 205$
Subtract $30$ from all parts:
$155 - 30 < 25(x - 3) < 205 - 30$
$125 < 25(x - 3) < 175$
Divide by $25$:
$5 < x - 3 < 7$
Add $3$ to all parts:
$5 + 3 < x < 7 + 3$
$8 < x < 10$
Thus,the temperature will be between $155^{\circ}C$ and $205^{\circ}C$ at a depth between $8 \text{ km}$ and $10 \text{ km}$.

(NONE) For the first inequality: $\frac{2x+1}{7x-1} > 5$
$\Rightarrow \frac{2x+1}{7x-1} - 5 > 0$
$\Rightarrow \frac{2x+1 - 35x + 5}{7x-1} > 0$
$\Rightarrow \frac{-33x + 6}{7x-1} > 0$
$\Rightarrow \frac{33x - 6}{7x-1} < 0$
$\Rightarrow \frac{x - 2/11}{x - 1/7} < 0$
Thus,$x \in (1/7, 2/11)$.
For the second inequality: $\frac{x+7}{x-8} > 2$
$\Rightarrow \frac{x+7}{x-8} - 2 > 0$
$\Rightarrow \frac{x+7 - 2x + 16}{x-8} > 0$
$\Rightarrow \frac{-x + 23}{x-8} > 0$
$\Rightarrow \frac{x - 23}{x-8} < 0$
Thus,$x \in (8, 23)$.
The intersection of the intervals $(1/7, 2/11)$ and $(8, 23)$ is empty.
Therefore,there is no solution for the given system of inequalities.

(A) Given the inequality $\frac{x^{2}-1}{x^{2}+1} \geq 0$.
Since $x^{2} + 1 > 0$ for all real $x$,we can multiply both sides by $(x^{2} + 1)$ without changing the inequality sign.
This simplifies to $x^{2} - 1 \geq 0$.
Factoring the expression,we get $(x - 1)(x + 1) \geq 0$.
Using the wavy curve method (sign scheme),the expression is non-negative when $x \leq -1$ or $x \geq 1$.
Thus,$x \in (-\infty, -1] \cup [1, \infty)$.

(C) The given inequalities are $x \geq 0, y \geq 0, y \leq 6$,and $x+y \leq 3$.
$1$. The conditions $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.
$2$. The line $x+y = 3$ passes through $(3, 0)$ and $(0, 3)$. Since $(0, 0)$ satisfies $x+y \leq 3$,the region lies towards the origin.
$3$. The line $y = 6$ is a horizontal line. Since $x+y \leq 3$ already implies $y \leq 3$ (given $x \geq 0$),the condition $y \leq 6$ is redundant as the region is already bounded by $x+y \leq 3$ and the axes.
$4$. The intersection of these regions forms a triangle with vertices $(0, 0), (3, 0)$,and $(0, 3)$.
Since this region has finite boundaries,it is a bounded region in the first quadrant.

(C) The given system of inequalities is:
$x - y \leq -1$ $(1)$
$x - y \geq 0$ $(2)$
$x \geq 0, y \geq 0$ $(3)$
From inequality $(1)$,we have $x - y \leq -1$,which implies $y \geq x + 1$.
From inequality $(2)$,we have $x - y \geq 0$,which implies $y \leq x$.
Combining these,we get $x + 1 \leq y \leq x$.
This implies $x + 1 \leq x$,which simplifies to $1 \leq 0$.
Since $1 \leq 0$ is a false statement,there is no value of $x$ and $y$ that can satisfy both inequalities simultaneously.
Therefore,the region represented by these inequalities does not exist.

(D) To determine the feasible region,we analyze the given constraints:
$1$. $x + y \leq 4$
$2$. $3x + 3y \geq 18 \implies x + y \geq 6$
$3$. $x \geq 0, y \geq 0$
From the first constraint,the region must lie on or below the line $x + y = 4$.
From the second constraint,the region must lie on or above the line $x + y = 6$.
Since these two lines are parallel and represent disjoint half-planes ($x+y \leq 4$ and $x+y \geq 6$),there is no point $(x, y)$ that can satisfy both inequalities simultaneously.
Therefore,the feasible region does not exist.

(B) The feasible region is defined by the inequalities $3x + 4y \leq 12$,$x \geq 0$,and $y \geq 1$.
We need to find integer pairs $(x, y)$ that satisfy these conditions.
Since $y \geq 1$,we test integer values for $y$:
$1$. If $y = 1$: The inequality becomes $3x + 4(1) \leq 12$,which simplifies to $3x \leq 8$,so $x \leq 2.66$. Since $x \geq 0$,the possible integer values for $x$ are $0, 1, 2$. The points are $(0, 1), (1, 1), (2, 1)$.
$2$. If $y = 2$: The inequality becomes $3x + 4(2) \leq 12$,which simplifies to $3x \leq 4$,so $x \leq 1.33$. Since $x \geq 0$,the possible integer values for $x$ are $0, 1$. The points are $(0, 2), (1, 2)$.
$3$. If $y = 3$: The inequality becomes $3x + 4(3) \leq 12$,which simplifies to $3x \leq 0$,so $x \leq 0$. Since $x \geq 0$,the only integer value for $x$ is $0$. The point is $(0, 3)$.
$4$. If $y > 3$: For $y = 4$,$3x + 16 \leq 12$ implies $3x \leq -4$,which has no non-negative integer solution for $x$.
Summing these up,the points are $(0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (0, 3)$.
There are $6$ such points.

(B) To determine the region,we analyze the given inequalities:
$1$. $2x + 3y \leq 5$: This represents a region on or below the line $2x + 3y = 5$.
$2$. $4x - 3y \leq -2$ or $3y \geq 4x + 2$: This represents a region on or above the line $3y = 4x + 2$.
$3$. $x \geq 0$: This restricts the region to the right of the $y$-axis (first and fourth quadrants).
Intersection points:
- Intersection of $2x + 3y = 5$ and $3y = 4x + 2$: Substituting $3y$,we get $2x + (4x + 2) = 5 \implies 6x = 3 \implies x = 0.5$. Then $3y = 4(0.5) + 2 = 4 \implies y = 4/3$.
- When $x = 0$,$2x + 3y = 5 \implies y = 5/3$ and $3y = 4x + 2 \implies y = 2/3$.
Since the region includes $x = 0$ and values of $y$ where $x$ can be negative for the second inequality,but the constraint $x \geq 0$ limits it,we observe the region spans across the $y$-axis boundary. Specifically,for $x \geq 0$,the region exists in the first quadrant,but the boundary $4x - 3y + 2 \leq 0$ allows $x$ to be $0$ while $y \geq 2/3$,and $2x + 3y - 5 \leq 0$ allows $x=0$ while $y \leq 5/3$. Thus,the region lies in the first quadrant.

(B) Given $a, b > 0$ and $a+2b \leq 1$.
Radius of circle $C_1 = ab^3$ and radius of circle $C_2 = b^2$.
Area $A_1 = \pi(ab^3)^2 = \pi a^2b^6$.
Area $A_2 = \pi(b^2)^2 = \pi b^4$.
Thus,$\frac{A_1}{A_2} = \frac{\pi a^2b^6}{\pi b^4} = a^2b^2 = (ab)^2$.
Using the $AM$-$GM$ inequality for $a$ and $2b$:
$\frac{a+2b}{2} \geq \sqrt{a \cdot 2b} = \sqrt{2ab}$.
Since $a+2b \leq 1$,we have $\frac{1}{2} \geq \sqrt{2ab}$.
Squaring both sides,$\frac{1}{4} \geq 2ab$,which implies $ab \leq \frac{1}{8}$.
Therefore,$(ab)^2 \leq (\frac{1}{8})^2 = \frac{1}{64}$.
The maximum value of $\frac{A_1}{A_2}$ is $\frac{1}{64}$.

(C) Given the inequality: $\sqrt{n+1}-\sqrt{n-1} < 0.2$
Rationalizing the left side:
$\frac{(\sqrt{n+1}-\sqrt{n-1})(\sqrt{n+1}+\sqrt{n-1})}{\sqrt{n+1}+\sqrt{n-1}} < 0.2$
$\frac{(n+1)-(n-1)}{\sqrt{n+1}+\sqrt{n-1}} < 0.2$
$\frac{2}{\sqrt{n+1}+\sqrt{n-1}} < 0.2$
$\frac{2}{0.2} < \sqrt{n+1}+\sqrt{n-1}$
$10 < \sqrt{n+1}+\sqrt{n-1}$
For $n=25$: $\sqrt{26}+\sqrt{24} \approx 5.099 + 4.899 = 9.998 < 10$ (False)
For $n=26$: $\sqrt{27}+\sqrt{25} \approx 5.196 + 5 = 10.196 > 10$ (True)
Thus,the least positive integer $n$ is $26$.

(C) Let $x$ be the number of correct answers,$y$ be the number of wrong answers,and $z$ be the number of unattempted questions.
Given that the total number of questions is $30$,we have $x + y + z = 30$.
The total score is given by $4x + 0y + 1z = 60$,which simplifies to $4x + z = 60$.
From the second equation,$z = 60 - 4x$.
Since $z \ge 0$,we have $60 - 4x \ge 0$,which implies $x \le 15$.
Also,since $x + y + z = 30$,we substitute $z = 60 - 4x$ to get $x + y + (60 - 4x) = 30$,which simplifies to $y = 3x - 30$.
Since $y \ge 0$,we have $3x - 30 \ge 0$,which implies $x \ge 10$.
Thus,$10 \le x \le 15$.
The possible integer values for $x$ are $10, 11, 12, 13, 14, 15$.
Counting these,there are $6$ possible values for $x$.

(C) The region is bounded by the lines $y = mx + 3$,$y = nx + 3$,and the $x$-axis $(y = 0)$.
For $y = 0$,the $x$-intercepts are $x_1 = -\frac{3}{m}$ (for $m > 0$) and $x_2 = -\frac{3}{n}$ (for $n < 0$).
The intersection point of the two lines $y = mx + 3$ and $y = nx + 3$ is $(0, 3)$.
The base of the triangle on the $x$-axis has length $b = |x_2 - x_1| = |-\frac{3}{n} - (-\frac{3}{m})| = 3 |\frac{1}{m} - \frac{1}{n}|$.
The height of the triangle is $h = 3$.
Area $A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 3 |\frac{1}{m} - \frac{1}{n}| \times 3 = \frac{9}{2} (\frac{1}{m} - \frac{1}{n})$.
To minimize the area,we maximize $m$ and $n$. Given $m \leq \sqrt{3}$ and $n \geq -\sqrt{3}$,we take $m = \sqrt{3}$ and $n = -\sqrt{3}$.
Area $= \frac{9}{2} (\frac{1}{\sqrt{3}} - \frac{1}{-\sqrt{3}}) = \frac{9}{2} (\frac{2}{\sqrt{3}}) = \frac{9}{\sqrt{3}} = 3 \sqrt{3}$.

(A) The triangle is bounded by $x > 0$,$y > 0$,and $3x + 4y < 60$. Since $b$ is a multiple of $a$,let $b = ka$ for some integer $k \ge 1$.
For a fixed $x = a$,we have $3a + 4y < 60$,so $y < \frac{60 - 3a}{4} = 15 - 0.75a$.
Since $y = ka$,we have $ka < 15 - 0.75a$,which implies $k < \frac{15}{a} - 0.75$.
For $a=1$: $k < 15 - 0.75 = 14.25 \Rightarrow k \in \{1, 2, \dots, 14\}$ ($14$ points).
For $a=2$: $k < 7.5 - 0.75 = 6.75 \Rightarrow k \in \{1, 2, \dots, 6\}$ ($6$ points).
For $a=3$: $k < 5 - 0.75 = 4.25 \Rightarrow k \in \{1, 2, 3, 4\}$ ($4$ points).
For $a=4$: $k < 3.75 - 0.75 = 3 \Rightarrow k \in \{1, 2\}$ ($2$ points).
For $a=5$: $k < 3 - 0.75 = 2.25 \Rightarrow k \in \{1, 2\}$ ($2$ points).
For $a=6$: $k < 2.5 - 0.75 = 1.75 \Rightarrow k \in \{1\}$ ($1$ point).
For $a=7$: $k < 2.14 - 0.75 = 1.39 \Rightarrow k \in \{1\}$ ($1$ point).
For $a=8$: $k < 1.875 - 0.75 = 1.125 \Rightarrow k \in \{1\}$ ($1$ point).
For $a \ge 9$: $k < \frac{15}{a} - 0.75 \le \frac{15}{9} - 0.75 = 1.66 - 0.75 = 0.91$,no positive integer $k$ exists.
Total points $= 14 + 6 + 4 + 2 + 2 + 1 + 1 + 1 = 31$.

(D) The given inequality is $|n^2 - 10n + 19| < 6$.
This is equivalent to $-6 < n^2 - 10n + 19 < 6$.
Case $1$: $n^2 - 10n + 19 < 6 \Rightarrow n^2 - 10n + 13 < 0$.
The roots of $n^2 - 10n + 13 = 0$ are $n = \frac{10 \pm \sqrt{100 - 52}}{2} = 5 \pm \sqrt{12} = 5 \pm 2\sqrt{3}$.
Since $2\sqrt{3} \approx 3.46$,the range is $n \in (5 - 3.46, 5 + 3.46) = (1.54, 8.46)$.
Case $2$: $n^2 - 10n + 19 > -6$ $\Rightarrow n^2 - 10n + 25 > 0$ $\Rightarrow (n - 5)^2 > 0$.
This is true for all $n \in \mathbb{Z}$ except $n = 5$.
Combining these,$n \in \{2, 3, 4, 6, 7, 8\}$.
The number of such elements is $6$.

(B) The given constraints are $|x-y| \leqslant 1$ and $x, y \geqslant 0$.
The inequality $|x-y| \leqslant 1$ is equivalent to $-1 \leqslant x-y \leqslant 1$,which can be split into two inequalities: $y \leqslant x+1$ and $y \geqslant x-1$.
Combining these with $x \geqslant 0$ and $y \geqslant 0$,we observe the region in the first quadrant.
For any $x \geqslant 0$,we can choose $y$ such that $x-1 \leqslant y \leqslant x+1$.
As $x$ increases indefinitely,$y$ can also increase indefinitely (e.g.,by choosing $y=x$).
Since the region extends infinitely in the first quadrant,the solution set is an unbounded set.

(D) To find the feasible region,we analyze the given constraints:
$1$. $2x + 3y \leqslant 18$: This represents the region on or below the line $2x + 3y = 18$. The intercepts are $(9, 0)$ and $(0, 6)$.
$2$. $x + y \geqslant 10$: This represents the region on or above the line $x + y = 10$. The intercepts are $(10, 0)$ and $(0, 10)$.
$3$. $x \geqslant 0, y \geqslant 0$: This restricts the region to the first quadrant.
Comparing the two lines:
For $2x + 3y = 18$,the maximum $x$-value is $9$ and the maximum $y$-value is $6$.
For $x + y = 10$,the minimum $x$-value is $10$ and the minimum $y$-value is $10$.
Since the region defined by $2x + 3y \leqslant 18$ lies entirely below the line $x + y = 10$ in the first quadrant,there is no point $(x, y)$ that satisfies both $2x + 3y \leqslant 18$ and $x + y \geqslant 10$ simultaneously.
Therefore,the feasible region is a null set.

(D) Given the system of linear inequations:
$1) 2x + 3y \leq 18$
$2) x + y \geq 10$
$3) x \geq 0, y \geq 0$
For the first inequation $2x + 3y \leq 18$,the boundary line is $2x + 3y = 18$. The intercepts are $(9, 0)$ and $(0, 6)$. Since the origin $(0, 0)$ satisfies $2(0) + 3(0) \leq 18$,the feasible region lies towards the origin.
For the second inequation $x + y \geq 10$,the boundary line is $x + y = 10$. The intercepts are $(10, 0)$ and $(0, 10)$. Since the origin $(0, 0)$ does not satisfy $0 + 0 \geq 10$,the feasible region lies away from the origin.
Comparing the two regions: the first region is below the line passing through $(9, 0)$ and $(0, 6)$,while the second region is above the line passing through $(10, 0)$ and $(0, 10)$. These two regions do not intersect in the first quadrant $(x \geq 0, y \geq 0)$.
Therefore,there is no common region that satisfies all the given inequations. Thus,the feasible region is an empty set.

(A) The system of linear inequalities is given by:
$1) x+y \geq 1$
$2) 7x+9y \leq 63$
$3) y \leq 5$
$4) x \leq 6$
$5) x \geq 0, y \geq 0$
To find the feasible region,we analyze the boundaries:
- The line $x+y=1$ passes through $(1,0)$ and $(0,1)$. The region $x+y \geq 1$ is away from the origin.
- The line $7x+9y=63$ passes through $(9,0)$ and $(0,7)$. The region $7x+9y \leq 63$ is towards the origin.
- The lines $x=6$ and $y=5$ are vertical and horizontal lines respectively,bounding the region.
- The conditions $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.
By plotting these lines and considering the inequalities,the feasible region is the bounded polygon in the first quadrant that satisfies all conditions. Comparing this with the provided options,the figure in the solution image (which matches option $A$) correctly represents the intersection of all these half-planes.

(D) To find the common region,we analyze the given inequations:
$1$. $x+y \geq 5$: This represents the region on or above the line $x+y=5$. Since $(0,0)$ does not satisfy this ($0 \geq 5$ is false),the region is on the non-origin side.
$2$. $y \leq 4$: This represents the region on or below the line $y=4$.
$3$. $x \geq 2$: This represents the region on or to the right of the line $x=2$.
$4$. $x, y \geq 0$: This restricts the region to the first quadrant.
By plotting these lines,we observe the shaded region in the figure. The region is bounded by the lines $x=2$,$y=4$,and $x+y=5$. Since the region is enclosed by these lines,it is a bounded region. Furthermore,the origin $(0,0)$ does not lie in the shaded region,so it is on the non-origin side.
Therefore,the correct option is $D$.

(A) Given,$C(x) = 20x + 4000$ and $R(x) = 60x + 2000$.
To earn a profit,the revenue must be greater than the cost,i.e.,$R(x) - C(x) > 0$.
Substituting the given functions:
$(60x + 2000) - (20x + 4000) > 0$
$60x + 2000 - 20x - 4000 > 0$
$40x - 2000 > 0$
$40x > 2000$
$x > \frac{2000}{40}$
$x > 50$.
Thus,the number of items $x$ must be greater than $50$ to earn a profit.

(B) Let the breadth of the rectangle be $x \ cm$.
Then,the length of the rectangle is $5x \ cm$.
The perimeter of a rectangle is given by $P = 2(\text{length} + \text{breadth})$.
Substituting the values,$P = 2(5x + x) = 2(6x) = 12x$.
Given that the minimum perimeter is $180 \ cm$,we have $P \geq 180$.
Therefore,$12x \geq 180$.
Dividing both sides by $12$,we get $x \geq 15$.
Since $x$ represents the breadth,the breadth must be at least $15 \ cm$.

(C) Given the inequation $\frac{x^{2}+6x-7}{|x+4|} < 0$.
Since $|x+4|$ is always positive for $x \neq -4$,the expression is negative only when the numerator is negative.
Thus,$x^{2}+6x-7 < 0$ and $x \neq -4$.
Factoring the quadratic expression,we get $(x+7)(x-1) < 0$.
This inequality holds for $x \in (-7, 1)$.
Excluding the point where the denominator is zero $(x = -4)$,the solution set is $(-7, -4) \cup (-4, 1)$.

(D) Given the inequality: $\frac{8x^2-14x-9}{3x^2-7x-6} > 2$ \\ Subtract $2$ from both sides: $\frac{8x^2-14x-9}{3x^2-7x-6} - 2 > 0$ \\ $\frac{8x^2-14x-9 - 2(3x^2-7x-6)}{3x^2-7x-6} > 0$ \\ $\frac{8x^2-14x-9 - 6x^2+14x+12}{3x^2-7x-6} > 0$ \\ $\frac{2x^2+3}{3x^2-7x-6} > 0$ \\ Since $2x^2+3$ is always positive for all real $x$,the inequality holds if $3x^2-7x-6 > 0$ \\ Factor the denominator: $3x^2-9x+2x-6 = 3x(x-3)+2(x-3) = (3x+2)(x-3) > 0$ \\ The critical points are $x = -2/3$ and $x = 3$ \\ Testing the intervals $(-\infty, -2/3)$,$(-2/3, 3)$,and $(3, \infty)$,the expression is positive in $(-\infty, -2/3) \cup (3, \infty)$.

(B) Given the inequality: $\frac{x^2-1}{(x-4)(x-3)} \geq 1$.
Subtract $1$ from both sides: $\frac{x^2-1}{(x-4)(x-3)} - 1 \geq 0$.
Simplify the expression: $\frac{x^2-1 - (x^2-7x+12)}{(x-4)(x-3)} \geq 0$.
$\frac{x^2-1 - x^2+7x-12}{(x-4)(x-3)} \geq 0$.
$\frac{7x-13}{(x-4)(x-3)} \geq 0$.
Find the critical points: $x = \frac{13}{7}, x = 3, x = 4$.
Using the wavy curve method on the number line,we test the intervals:
For $x > 4$,the expression is positive.
For $3 < x < 4$,the expression is negative.
For $\frac{13}{7} \leq x < 3$,the expression is positive.
For $x < \frac{13}{7}$,the expression is negative.
Thus,the solution set is $[\frac{13}{7}, 3) \cup (4, \infty)$.

(A) Given the inequality: $\frac{x-1}{3x+4} - \frac{x-3}{3x-2} < 0$
Taking the common denominator:
$\frac{(x-1)(3x-2) - (x-3)(3x+4)}{(3x+4)(3x-2)} < 0$
Expanding the numerators:
$\frac{(3x^2 - 2x - 3x + 2) - (3x^2 + 4x - 9x - 12)}{(3x+4)(3x-2)} < 0$
Simplifying the numerator:
$\frac{(3x^2 - 5x + 2) - (3x^2 - 5x - 12)}{(3x+4)(3x-2)} < 0$
$\frac{3x^2 - 5x + 2 - 3x^2 + 5x + 12}{(3x+4)(3x-2)} < 0$
$\frac{14}{(3x+4)(3x-2)} < 0$
Since the numerator $14$ is positive,the expression is negative only when the denominator is negative:
$(3x+4)(3x-2) < 0$
The roots of the denominator are $x = -\frac{4}{3}$ and $x = \frac{2}{3}$.
Using the wavy curve method (sign scheme),the expression $(3x+4)(3x-2)$ is negative between the roots.
Thus,$x \in \left(-\frac{4}{3}, \frac{2}{3}\right)$.

(D) Given inequality: $\frac{14 x}{x+1}-\frac{9 x-30}{x-4} < 0$
Simplify the expression:
$\frac{14 x(x-4)-(9 x-30)(x+1)}{(x+1)(x-4)} < 0$
$\frac{14 x^2-56 x-(9 x^2+9 x-30 x-30)}{(x+1)(x-4)} < 0$
$\frac{14 x^2-56 x-(9 x^2-21 x-30)}{(x+1)(x-4)} < 0$
$\frac{5 x^2-35 x+30}{(x+1)(x-4)} < 0$
$\frac{5(x^2-7 x+6)}{(x+1)(x-4)} < 0$
$\frac{5(x-1)(x-6)}{(x+1)(x-4)} < 0$
Using the wavy curve method (sign scheme) on the number line with critical points $-1, 1, 4, 6$:
The expression is negative in the intervals $(-1, 1)$ and $(4, 6)$.
Thus,$x \in (-1, 1) \cup (4, 6)$.

(B) Given the inequality: $\frac{7 x^2-5 x-18}{2 x^2+x-6} < 2$
Subtract $2$ from both sides: $\frac{7 x^2-5 x-18}{2 x^2+x-6} - 2 < 0$
$\Rightarrow \frac{7 x^2-5 x-18 - 2(2 x^2+x-6)}{2 x^2+x-6} < 0$
$\Rightarrow \frac{7 x^2-5 x-18 - 4 x^2-2 x+12}{(2 x-3)(x+2)} < 0$
$\Rightarrow \frac{3 x^2-7 x-6}{(2 x-3)(x+2)} < 0$
Factor the numerator: $3 x^2-7 x-6 = 3 x^2-9 x+2 x-6 = 3 x(x-3)+2(x-3) = (3 x+2)(x-3)$
So,the inequality becomes: $\frac{(3 x+2)(x-3)}{(2 x-3)(x+2)} < 0$
The critical points are $x = -2, -\frac{2}{3}, \frac{3}{2}, 3$.
Using the wavy curve method,we test the intervals:
For $x > 3$,the expression is positive.
For $\frac{3}{2} < x < 3$,the expression is negative.
For $-\frac{2}{3} < x < \frac{3}{2}$,the expression is positive.
For $-2 < x < -\frac{2}{3}$,the expression is negative.
For $x < -2$,the expression is positive.
Since we need the expression to be less than $0$,the solution is $x \in \left(-2, -\frac{2}{3}\right) \cup \left(\frac{3}{2}, 3 \right)$.

(D) Given,$1 \leq \frac{3x^2-7x+8}{x^2+1} \leq 2$. Since $x^2+1 > 0$ for all $x \in R$,we can multiply throughout by $(x^2+1)$.
First,consider $1 \leq \frac{3x^2-7x+8}{x^2+1} \implies x^2+1 \leq 3x^2-7x+8 \implies 2x^2-7x+7 \geq 0$.
The discriminant of $2x^2-7x+7$ is $D = (-7)^2 - 4(2)(7) = 49 - 56 = -7 < 0$. Since the leading coefficient is positive,$2x^2-7x+7 > 0$ for all $x \in R$.
Next,consider $\frac{3x^2-7x+8}{x^2+1} \leq 2 \implies 3x^2-7x+8 \leq 2x^2+2 \implies x^2-7x+6 \leq 0$.
Factoring the quadratic,we get $(x-1)(x-6) \leq 0$.
This inequality holds for $x \in [1, 6]$.
Thus,the minimum value is $1$ and the maximum value is $6$.

(D) Given inequality: $\frac{8x^2+16x-51}{(2x-3)(x+4)} > 3$
Subtract $3$ from both sides: $\frac{8x^2+16x-51 - 3(2x^2+5x-12)}{(2x-3)(x+4)} > 0$
Simplify the numerator: $8x^2+16x-51 - 6x^2-15x+36 = 2x^2+x-15$
Factor the numerator: $2x^2+6x-5x-15 = 2x(x+3)-5(x+3) = (2x-5)(x+3)$
So,the inequality becomes: $\frac{(2x-5)(x+3)}{(2x-3)(x+4)} > 0$
The critical points are $x = -4, -3, \frac{3}{2}, \frac{5}{2}$.
Using the wavy curve method (sign scheme) for the intervals $(-\infty, -4), (-4, -3), (-3, \frac{3}{2}), (\frac{3}{2}, \frac{5}{2}), (\frac{5}{2}, \infty)$:
The expression is positive in the intervals $(-\infty, -4) \cup (-3, \frac{3}{2}) \cup (\frac{5}{2}, \infty)$.

(C) The given inequality is $\frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}$.
First,for the expression to be defined,we must have $6+x-x^2 \geq 0$,which implies $x^2-x-6 \leq 0$,so $(x-3)(x+2) \leq 0$. Thus,$x \in [-2, 3]$.
If $6+x-x^2 = 0$,then $x = -2$ or $x = 3$. Both satisfy the inequality as $0 \geq 0$.
If $6+x-x^2 > 0$,we can divide by $\sqrt{6+x-x^2}$:
$\frac{1}{2x+5} \geq \frac{1}{x+4} \Rightarrow \frac{1}{2x+5} - \frac{1}{x+4} \geq 0$
$\Rightarrow \frac{x+4-(2x+5)}{(2x+5)(x+4)} \geq 0$ $\Rightarrow \frac{-x-1}{(2x+5)(x+4)} \geq 0$
$\Rightarrow \frac{x+1}{(2x+5)(x+4)} \leq 0$.
Using the sign scheme for critical points $-4, -2.5, -1$,the expression is $\leq 0$ for $x \in (-\infty, -4) \cup (-2.5, -1]$.
Combining this with the domain $x \in [-2, 3]$,we get $x \in [-2, -1] \cup \{3\}$.

(B) We have two inequalities to solve:
$5x - 1 < (x + 1)^2$ and $(x + 1)^2 < 7x - 3$.
For the first inequality:
$5x - 1 < x^2 + 2x + 1$
$x^2 - 3x + 2 > 0$
$(x - 1)(x - 2) > 0$
This implies $x \in (-\infty, 1) \cup (2, \infty)$.
For the second inequality:
$(x + 1)^2 < 7x - 3$
$x^2 + 2x + 1 < 7x - 3$
$x^2 - 5x + 4 < 0$
$(x - 1)(x - 4) < 0$
This implies $x \in (1, 4)$.
Taking the intersection of both intervals:
$x \in ((-\infty, 1) \cup (2, \infty)) \cap (1, 4) = (2, 4)$.
The integers in the interval $(2, 4)$ is only $x = 3$.
Thus,there is only $1$ integral value.

(A) Given the inequality: $\sqrt{9x^2+6x+1} < (2-x)$
Since $9x^2+6x+1 = (3x+1)^2$,the expression becomes $\sqrt{(3x+1)^2} < 2-x$.
This simplifies to $|3x+1| < 2-x$.
For the square root to be defined,we must have $2-x > 0$,which implies $x < 2$.
Now,solving the absolute value inequality $|3x+1| < 2-x$:
$-(2-x) < 3x+1 < 2-x$
Case $1$: $3x+1 < 2-x$
$4x < 1 \Rightarrow x < \frac{1}{4}$
Case $2$: $3x+1 > -(2-x)$
$3x+1 > -2+x$
$2x > -3 \Rightarrow x > -\frac{3}{2}$
Combining these,we get $-\frac{3}{2} < x < \frac{1}{4}$.
Thus,$x \in \left(-\frac{3}{2}, \frac{1}{4}\right)$.

(C) The quadrilateral is bounded by the lines $x = 0$ ($y$-axis),$y = 0$ ($x$-axis),$2x + y = 2$,and $x + y = 5$.
For a point $(x, y)$ to be inside the quadrilateral with natural number coordinates $(x, y \in \{1, 2, 3, \dots\})$,it must satisfy the inequalities:
$1) \ x > 0$
$2) \ y > 0$
$3) \ 2x + y > 2$
$4) \ x + y < 5$
Testing integer values for $x$:
If $x = 1$: $2(1) + y > 2 \implies 2 + y > 2 \implies y > 0$. Also $1 + y < 5 \implies y < 4$. So $y \in \{1, 2, 3\}$. Points: $(1, 1), (1, 2), (1, 3)$.
If $x = 2$: $2(2) + y > 2 \implies 4 + y > 2 \implies y > -2$. Also $2 + y < 5 \implies y < 3$. So $y \in \{1, 2\}$. Points: $(2, 1), (2, 2)$.
If $x = 3$: $2(3) + y > 2 \implies 6 + y > 2 \implies y > -4$. Also $3 + y < 5 \implies y < 2$. So $y = 1$. Point: $(3, 1)$.
If $x \ge 4$: $x + y < 5$ cannot be satisfied for natural numbers $y \ge 1$.
The total number of such points is $3 + 2 + 1 = 6$.

(C) Given $S = \{x \in \mathbb{Z} : x^2 - 7x + 6 \leq 0 \text{ and } x^2 - 3x > 0\}$ ...$(i)$
First,solve the inequality $x^2 - 7x + 6 \leq 0$:
$(x - 6)(x - 1) \leq 0$
This implies $x \in [1, 6]$.
Next,solve the inequality $x^2 - 3x > 0$:
$x(x - 3) > 0$
This implies $x \in (-\infty, 0) \cup (3, \infty)$.
Now,find the intersection of these two intervals:
$S = \{x \in \mathbb{Z} : x \in [1, 6] \cap ((-\infty, 0) \cup (3, \infty))\}$
$S = \{x \in \mathbb{Z} : x \in (3, 6] \cap \mathbb{Z}\}$
$S = \{4, 5, 6\}$.
The number of elements in set $S$ is $3$.

(D) Given equation: $|x|+|y|=|x-3|+|y-2|$.
Case $I$: $0 \leq x < 3$ and $0 \leq y < 2$.
In this region,$|x|=x$,$|y|=y$,$|x-3|=-(x-3)$,and $|y-2|=-(y-2)$.
Substituting these into the equation: $x+y = -(x-3) - (y-2)$.
$x+y = -x+3-y+2$.
$2x+2y = 5$.
$x+y = 5/2$.
Case $II$: $x \geq 3$ and $y \geq 2$.
In this region,$|x|=x$,$|y|=y$,$|x-3|=x-3$,and $|y-2|=y-2$.
Substituting these into the equation: $x+y = (x-3) + (y-2)$.
$x+y = x+y-5$.
$0 = -5$,which is impossible.
Therefore,the condition $x+y = 5/2$ holds for $0 \leq x < 3$ and $0 \leq y < 2$.

(A) We have the system of inequalities:
$x^2-4x+3 > 0$ and $x^2-2x-8 \leq 0$
First,solve $x^2-4x+3 > 0$:
$(x-3)(x-1) > 0$
This implies $x \in (-\infty, 1) \cup (3, \infty)$.
Next,solve $x^2-2x-8 \leq 0$:
$(x-4)(x+2) \leq 0$
This implies $x \in [-2, 4]$.
Finally,find the intersection of the two sets:
$x \in ((-\infty, 1) \cup (3, \infty)) \cap [-2, 4]$
$x \in [-2, 1) \cup (3, 4]$

(C) The given inequality is $\frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2x+9}$.
First,for the square root to be defined,we must have $12-x-x^2 \geq 0$,which implies $x^2+x-12 \leq 0$,so $(x+4)(x-3) \leq 0$,giving $x \in [-4, 3]$.
Also,the denominators must be non-zero: $x+10 \neq 0 \implies x \neq -10$ and $2x+9 \neq 0 \implies x \neq -4.5$.
Case $1$: If $12-x-x^2 = 0$,then $x = -4$ or $x = 3$. Both satisfy the inequality $0 \leq 0$.
Case $2$: If $12-x-x^2 > 0$,we can divide by $\sqrt{12-x-x^2}$:
$\frac{1}{x+10} \leq \frac{1}{2x+9} \implies \frac{1}{x+10} - \frac{1}{2x+9} \leq 0 \implies \frac{2x+9-x-10}{(x+10)(2x+9)} \leq 0 \implies \frac{x-1}{(x+10)(2x+9)} \leq 0$.
Using the sign scheme for $\frac{x-1}{(x+10)(2x+9)} \leq 0$ with $x \in (-4, 3)$:
The critical points are $-10, -4.5, 1$. Within $(-4, 3)$,the relevant interval is $(-4, 1]$.
Combining Case $1$ and Case $2$,we get $x \in [-4, 1] \cup \{3\}$.

(B) Let $m = n+4$. Since $n \in \mathbb{N}$,$n \geq 1$,so $m \geq 5$.
The given inequality becomes $2^m + 12 \geq km$,which implies $k \leq \frac{2^m + 12}{m}$ for all $m \geq 5$.
To find the greatest integer $k$,we need to find the minimum value of the function $f(m) = \frac{2^m + 12}{m}$ for $m \in \{5, 6, 7, \dots\}$.
For $m = 5$: $f(5) = \frac{2^5 + 12}{5} = \frac{32 + 12}{5} = \frac{44}{5} = 8.8$.
For $m = 6$: $f(6) = \frac{2^6 + 12}{6} = \frac{64 + 12}{6} = \frac{76}{6} \approx 12.66$.
For $m = 7$: $f(7) = \frac{2^7 + 12}{7} = \frac{128 + 12}{7} = \frac{140}{7} = 20$.
As $m$ increases,$f(m)$ increases for $m \geq 5$.
Thus,the minimum value is $8.8$ at $m = 5$.
Since $k \leq f(m)$ for all $m$,$k$ must be less than or equal to the minimum value of $f(m)$.
Therefore,$k \leq 8.8$.
The greatest integer $k$ is $8$.

(B) Let $L_1(x, y) = x+2y-5$ and $L_2(x, y) = 3x-4y+5$. The origin $(0,0)$ gives $L_1(0,0) = -5$ and $L_2(0,0) = 5$. Since they have opposite signs,the origin lies between the lines.
For a point $P = ((\alpha-1)^2, \alpha)$ to lie in the same region as the origin,$L_1(P)$ must have the same sign as $L_1(0,0)$,i.e.,$L_1(P) < 0$,and $L_2(P)$ must have the same sign as $L_2(0,0)$,i.e.,$L_2(P) > 0$.
Condition $1$: $(\alpha-1)^2 + 2\alpha - 5 < 0$ $\Rightarrow \alpha^2 - 2\alpha + 1 + 2\alpha - 5 < 0$ $\Rightarrow \alpha^2 - 4 < 0$ $\Rightarrow -2 < \alpha < 2$.
Condition $2$: $3(\alpha-1)^2 - 4\alpha + 5 > 0$ $\Rightarrow 3(\alpha^2 - 2\alpha + 1) - 4\alpha + 5 > 0$ $\Rightarrow 3\alpha^2 - 10\alpha + 8 > 0$ $\Rightarrow (3\alpha-4)(\alpha-2) > 0$ $\Rightarrow \alpha < \frac{4}{3}$ or $\alpha > 2$.
Combining the conditions: $\alpha \in (-2, 2) \cap (\alpha < \frac{4}{3} \cup \alpha > 2) \Rightarrow \alpha \in (-2, \frac{4}{3})$.
Since $\alpha \in \mathbb{Z}$,the possible values are $\alpha \in \{-1, 0, 1\}$.
Thus,there are $3$ such points.

(D) Given the inequality $\sqrt{x+2} > \sqrt{8-x^2}$.
First,we determine the domain of the functions:
$x+2 \geq 0 \Rightarrow x \geq -2$ $(i)$
$8-x^2 \geq 0$ $\Rightarrow x^2 \leq 8$ $\Rightarrow x \in [-2\sqrt{2}, 2\sqrt{2}]$ $(ii)$
Combining $(i)$ and $(ii)$,the domain is $x \in [-2, 2\sqrt{2}]$.
Now,squaring both sides of the inequality:
$x+2 > 8-x^2$
$x^2 + x - 6 > 0$
$(x+3)(x-2) > 0$
This inequality holds for $x \in (-\infty, -3) \cup (2, \infty)$ $(iii)$.
Taking the intersection of the domain $[-2, 2\sqrt{2}]$ and the solution $(iii)$:
$x \in (2, 2\sqrt{2}]$.
The maximum value of $x$ in this interval is $2\sqrt{2}$.

(B) Given the inequation: $3^x + 3^{1-x} - 4 < 0$.
Substitute $3^{1-x} = \frac{3}{3^x}$ into the expression: $3^x + \frac{3}{3^x} - 4 < 0$.
Multiply the entire inequality by $3^x$ (since $3^x > 0$ for all $x \in R$): $(3^x)^2 - 4(3^x) + 3 < 0$.
Let $y = 3^x$. The inequality becomes $y^2 - 4y + 3 < 0$.
Factor the quadratic expression: $(y - 1)(y - 3) < 0$.
This implies $1 < y < 3$.
Substituting back $y = 3^x$,we get $1 < 3^x < 3$.
Since $3^0 = 1$ and $3^1 = 3$,we have $3^0 < 3^x < 3^1$.
Comparing the exponents,we get $0 < x < 1$.
Thus,the solution set is $(0, 1)$.

(A) The given inequality is $|x-1|+|x+1| < 4$.
We define the function $f(x) = |x-1|+|x+1|$.
Case $1$: $x < -1$. Then $f(x) = -(x-1) - (x+1) = -2x$.
$-2x < 4 \Rightarrow x > -2$. So,$x \in (-2, -1)$.
Case $2$: $-1 \leq x \leq 1$. Then $f(x) = -(x-1) + (x+1) = 2$.
$2 < 4$ is always true for $x \in [-1, 1]$.
Case $3$: $x > 1$. Then $f(x) = (x-1) + (x+1) = 2x$.
$2x < 4 \Rightarrow x < 2$. So,$x \in (1, 2)$.
Combining all cases,the solution set is $(-2, -1) \cup [-1, 1] \cup (1, 2) = (-2, 2)$.

(C) For the square root to be defined,we must have $x^2+x-2 \ge 0$,which implies $(x+2)(x-1) \ge 0$. Thus,$x \in (-\infty, -2] \cup [1, \infty)$.
Case $1$: If $1-x < 0$,i.e.,$x > 1$,the inequality $\sqrt{x^2+x-2} > 1-x$ is always true because the left side is non-negative and the right side is negative.
Case $2$: If $1-x \ge 0$,i.e.,$x \le 1$,we square both sides: $x^2+x-2 > (1-x)^2$.
$x^2+x-2 > 1-2x+x^2$.
$x-2 > 1-2x$ $\Rightarrow 3x > 3$ $\Rightarrow x > 1$.
Combining this with the condition $x \le 1$,we get no solution from this case.
However,considering the domain $x \in (-\infty, -2] \cup [1, \infty)$,for $x > 1$,the inequality holds.
Checking $x \le -2$: If $x = -2$,$\sqrt{4-2-2} = 0$ and $1-(-2) = 3$. $0 > 3$ is false.
Thus,the solution set is $(1, \infty)$.

(D) Let $g$ be the number of apples the girl receives and $b$ be the number of apples the boy receives. We have $g + b = 11$.
By the Pigeonhole Principle,if we distribute $11$ items into $2$ groups,at least one group must contain at least $\lceil 11/2 \rceil = 6$ items.
However,we check the given options using the condition $g + b = 11$.
If $g < 4$ and $b < 8$,then $g \leq 3$ and $b \leq 7$. Adding these gives $g + b \leq 10$,which contradicts $g + b = 11$.
Therefore,it must be that $g \geq 4$ or $b \geq 8$.

(B) We know that $|A + B| = |A| + |B|$ if and only if $AB \geq 0$.
Given the equation $|x^2 + x - 9| = |x| + |x^2 - 9|$.
Let $A = x$ and $B = x^2 - 9$. Then $A + B = x^2 + x - 9$.
The condition for the equality to hold is $x(x^2 - 9) \geq 0$.
Factoring the expression,we get $x(x - 3)(x + 3) \geq 0$.
Using the wavy curve method (sign scheme),the solution set is $x \in [-3, 0] \cup [3, \infty)$.
Comparing this with the given form $[\alpha, \beta] \cup [\gamma, \infty)$,we identify $\alpha = -3$,$\beta = 0$,and $\gamma = 3$.
Therefore,the value of $(\alpha^2 + \beta^2 + \gamma^2) = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$.