Thinking Process

First solve the first two inequalities, then solve the last two inequality to get range of $x$

Solution

Consider first two inequalities,

$\dfrac{4}{x+1} \leq 3 $

$\Rightarrow 4 \leq 3(x+1) $

$\Rightarrow 4 \leq 3 x+3 $

$\Rightarrow 4-3 \leq 3 x $

$\Rightarrow 1 \leq 3 x $

$\therefore \ x \geq \dfrac{1}{3} \quad \ldots (i)$

Now, consider last two inequalities,

$ 3 \leq \dfrac{6}{x+1} $

$\Rightarrow \ 3(x+1) \leq 6 $

$\Rightarrow \ 3 x+3 \leq 6 $

$\Rightarrow \ 3 x \leq 6-3 $

$\Rightarrow \ 3 x \leq 3 $

$\therefore \ x \leq 1 \quad \ldots (ii)$

From Eqs. (i) and (ii),

$ \begin{aligned} & \dfrac{1}{3} \leq x \leq 1 \end{aligned} $

$\Rightarrow \ x \in \left[\dfrac{1}{3}, 1 \right]$

Solution

Let, $|x-2|=y$

$\dfrac{y-1}{y-2} \leq 0, \quad y-2 \neq 0$

Case (1): $ \ y-1 \leq 0, \quad y-2 > 0$

$\Rightarrow \ y \leq 1, \quad y >2$

Case (2): $ \ y-1 \geq 0, \quad y-2 < 0$

$\Rightarrow \ y \geq 1 , \quad y <2$

$\therefore \ 1 \leq y < 2 $

$\Rightarrow \ 1 \leq |x-2| < 2$

Case (1): $ \ |x-2| \geq 1$

$\Rightarrow \ x-2 \geq 1 \Rightarrow x \geq 3$

Or $ \ x-2 \leq -1 \Rightarrow x \leq 1$

Case (2): $ \ |x-2| < 2$

$x-2 < 2 \Rightarrow x <4$

Or $ \ x-2 > -2 \Rightarrow x >0$

From case (1) and case (2)

$\therefore \ x \in (0,1] \cup [3,4)$

Solution

Given, $ \ \dfrac{1}{|x|-3} \leq \dfrac{1}{2} $

$\begin{aligned} & \Rightarrow \quad|x|-3 \geq 2 \quad \left[\because \dfrac{1}{a}<\dfrac{1}{b} \Rightarrow a>b \right] \\ \\ & \Rightarrow \quad |x| \geq 5 \quad \text { [adding } 3 \text { to both sides] } \\ \\ & \Rightarrow \quad x \leq-5 \text { or } x \geq 5 \quad[\because|x| \geq a \Rightarrow|x| \leq-a \Rightarrow|x| \geq a] \ & \end{aligned}$

$\Rightarrow \ x \in(-\infty,-5] \cup[5, \infty)\quad \ldots (i)$

$\begin{array}{ll} \text { Here, } & |x|-3 \neq 0 \\ \\ \Rightarrow & |x|-3<0 \ \text { or } \ |x|-3>0 \\ \\ \Rightarrow & |x|<3 \ \text { or } \ |x|>3 \\ \\ \Rightarrow & -3<x<3 \ \text { or } \ x<-3 \ \text { or } \ x>3 \quad \ldots (ii) \\ \\ & [\because|x|<a \Rightarrow-a<x<a \ \text { and } \ |x|>a \Rightarrow x<-a \ \text { or } \ x>a] \end{array}$

On combining results of Eqs. (i) and (ii), we get

$ x \in(-\infty,-5] \cup(-3,3) \cup[5, \infty) $

Solution

$|x-1| \leq 5$

$\Rightarrow \ -5 \leq x-1 \leq 5$

$\Rightarrow \ -4 \leq x \leq 6$

$\Rightarrow \ x \in [-4,6] \quad \ldots (i)$

And $|x| \geq 2$

$\Rightarrow \ x \leq -2 \ \text{or} \ x \geq 2$

$\Rightarrow \ x \in (- \infty , -2] \cup[2, \infty) \quad \ldots (ii)$

On combining Eqs. (i) and (ii), we get

$ x \in[-4,-2] \cup[2,6] $

Solution

We have, $ \ -5 \leq \dfrac{2-3 x}{4} $

$ \Rightarrow \ -20 \leq 2-3 x $

$ \Rightarrow \ 3 x \leq 2+20 $

$ \Rightarrow \ 3 x \leq 22 $

$ \Rightarrow \ x \leq \dfrac{22}{3} \quad \ldots (i)$

Also, $ \ \dfrac{2-3 x}{4} \leq 9 $

$ \Rightarrow \ 2-3 x \leq 36 $

$ \Rightarrow \ -3 x \leq 36-2 $

$\Rightarrow \ -3 x \leq 34 $

$\Rightarrow \ 3 x \geq-34 $

$\Rightarrow \ x \geq-\dfrac{34}{3} \quad \ldots (ii)$

On combining Eqs. (i) and (ii), we get

$-\dfrac{34}{3} \leq x \leq \dfrac{22}{3} $

$\Rightarrow \ x \in \left[\dfrac{-34}{3}, \dfrac{22}{3}\right]$

Solution

We have, $ \ 4 x+3 \geq 2 x+17 $

$\Rightarrow \ 4 x-2 x \geq 17-3 $

$ \Rightarrow \ 2 x \geq 14 $

$\Rightarrow \ x \geq \dfrac{14}{2} $

$\Rightarrow \ x \geq 7 \quad \ldots (i)$

Also, $ \ 3 x-5 <-2 $

$\Rightarrow \ 3 x <-2+5 $

$\Rightarrow \ 3 x<3 $

$\Rightarrow \ x <1 \quad \ldots (ii)$

On combining Eqs. (i) and (ii), we see that solution is not possible because nothing is common between these two solutions. (i.e., $x<1 \ \text{and} \ x \geq 7$ ).

Solution

Cost function, $C(x) = 26000 + 30x$

and revenue function, $R(x) = 43x$

For profit, $R(x) > C(x)$

$\Rightarrow$ $ 26000 + 30x < 43x$

$\Rightarrow$ $ 30x - 43x < -26000$

$\Rightarrow$ $ -13x < -26000$

$\Rightarrow$ $ 13x > 26000$

$\Rightarrow$ $x > \dfrac{26000}{13}$

$\therefore \ x > 2000$

$\therefore \ $ Hence, more than $2000$ cassettes must be produced to get profit.

Solution

Given, first $pH$ value $=8.48$

and second $pH$ value $=8.35$

Let third $pH$ value be $x$.

Since, it is given that average $pH$ value lies between $8.2$ and $8.5$ .

$\therefore \ 8.2<\dfrac{8.48+8.35+x}{3}<8.5 $

$\Rightarrow \ 8.2<\dfrac{16.83+x}{3}<8.5 $

$\Rightarrow \ 3 \times 8.2<16.83+x<8.5 \times 3 $

$\Rightarrow \ 24.6<16.83+x<25.5 $

$\Rightarrow \ 24.6-16.83<x<25.5-16.83 $

$\Rightarrow \ 7.77<x<8.67$

Thus, third pH value lies between $7.77$ and $8.67$ .

Solution

$x L$ of $3 %$ solution in added

$460L$ of $9 %$ solution in added

Total Solution $= (460+x)L$

Mixture $% = \dfrac{\left(460 \times \dfrac{9}{100}+ x \times \dfrac{3}{100}\right)}{(460+x)}\times 100$

According to the question,

$5 < \text{mixture %} <7$

$\Rightarrow \ 5 < \dfrac{(460 \times 9+3x)}{(460+x)} <7$

$\Rightarrow \ 5 \times (460+x) < (460 \times 9 +3x) < 7 \times (460+x)$

$\Rightarrow \ 2300+5x < 4140+3x < 3220+7x$

Now, $ \ 2300+5x < 4140+3x $

$\Rightarrow \ 2x < 4140-2300$

$\Rightarrow \ 2x < 1840$

$\Rightarrow \ x < 920 \quad \ldots (i)$

And $ \ 4140+3x < 3220 +7x$

$\Rightarrow \ 4x > 4140-3220$

$\Rightarrow \ 4x > 920$

$\Rightarrow \ x > 230 \quad \ldots (ii)$

From Eqs. (i) and (ii)

$920 > x > 230$

Hence, the number of litres of the $3$ solution of acid must be more than $230 L$ and less than $920 L$.

Solution

Let the required temperature be $x^{\circ} F$.

Given : $ F= \dfrac{9}{5} C+32 $

$ \begin{array}{l} \Rightarrow \ 5 F=9 C+160 \\ \\ \Rightarrow \ C=\dfrac{(5 F-160)}{9} \end{array} $

$ \because \ $ The temperature in degrees Celsius lies between $ 40^{\circ} \mathrm{C} $ to $ 45^{\circ} \mathrm{C} $

$ \begin{array}{l} \therefore \ 40^{\circ}<(5 \mathrm{~F}-160) / 9<45^{\circ} \\ \\ \Rightarrow \ 360^{\circ}<5 \mathrm{~F}-160<405^{\circ} \\ \\ \Rightarrow \ 520^{\circ}<5 \mathrm{~F}<565^{\circ} \\ \\ \Rightarrow \ 104^{\circ}<\mathrm{F}<113^{\circ} \end{array} $

Hence, the range of temperature in degree fahrenheit is $104^{\circ} F$ to $113^{\circ} F$.

Solution

Let the length of shortest side be $x cm$.

According to the given information,

Longest side $=2 \times$ Shortest side $ =2 x cm $

and third side $=2+$ Shortest side $ =(2+x) cm $

Perimeter of triangle $=x+2 x+(x+2)=4 x+2$

According to the question,

Perimeter $>166 cm$

$ \begin{matrix} \Rightarrow & 4 x+2>166 \\ \\ \Rightarrow & 4 x>166-2 \\ \\ \Rightarrow & 4 x>164 \\ \\ \therefore & x>\dfrac{164}{4}=41 cm \end{matrix} $

Hence, the minimum length of shortest side be $41 cm$.

Solution

Given that, $ \ T=30+25(x-3), 3 \leq x \leq 15 $

According to the question,

$ \begin{aligned} & 155<T<205 \\ \\ & \begin{matrix} \Rightarrow & 155<30+25(x-3)<205 \\ \\ \Rightarrow & 155-30<25(x-3)<205-30 \end{matrix} \\ \\ & \Rightarrow \quad 155-30<25(x-3)<205-30 \quad \text {[ subtracting 30 in whole ]} \\ \\ & \Rightarrow \quad 125<25(x-3)<175 \\ \\ & \Rightarrow \quad \dfrac{125}{25}<x-3<\dfrac{175}{25} \quad \text { [dividing by 25 in whole ]} \\ \\ & \Rightarrow \quad 5<x-3<7 \\ \\ & \Rightarrow \quad 5+3<x<7+3 \quad \text {[ adding 3 in whole] } \\ \\ & \Rightarrow \quad 8<x<10 \end{aligned} $

Hence, at the depth $8 km$ to $10 km$ temperature lies between $155^{\circ}$ to $205^{\circ} C$.

Solution

The given system of inequations is

$\dfrac{2 x+1}{7 x-1}>5 $

And $\quad \dfrac{x+7}{x-8}>2 $

Now, $\quad \dfrac{2 x+1}{7 x-1}-5>0 $

$\Rightarrow \quad \dfrac{(2 x+1)-5(7 x-1)}{7 x-1}>0 $

$\Rightarrow \quad \dfrac{2 x+1-35 x+5}{7 x-1}>0 $

$\Rightarrow \quad \dfrac{-33 x+6}{7 x-1}>0 $

$\Rightarrow \quad \dfrac{33 x-6}{7 x-1}<0 $

$\Rightarrow \quad \dfrac{33 \left(x-\dfrac{6}{33}\right)}{7 \left(x-\dfrac{1}{7}\right)} < 0$

$\Rightarrow \quad \dfrac{x-\dfrac{6}{33}}{x- \dfrac{1}{7}} < 0$

$\Rightarrow \quad x \in \left(\dfrac{1}{7}, \dfrac{6}{33} \right)$

$\Rightarrow \quad x \in \left(\dfrac{1}{7}, \dfrac{2}{11} \right) \quad \ldots (i)$

$ \dfrac{x+7}{x-8}>2 $

$\Rightarrow \ \dfrac{x+7}{x-8}-2>0 $

$\Rightarrow \ \dfrac{x+7-2(x-8)}{x-8}>0 $

$\Rightarrow \ \dfrac{x+7-2 x+16}{x-8}>0 $

$\Rightarrow \ \dfrac{-x+23}{x-8}>0 $

$\Rightarrow \ \dfrac{x-23}{x-8}<0 \quad \ldots (ii)$

$x \in(8,23) $

Since, the intersection of Eqs. (i) and (ii) is the null set. Hence, the given system of equation has no solution.

Solution

In the given figure, shaded region is in the first quadrant. Hence $ x \geq 0, y \geq $ 0.

The shaded region is the common area of lines $ x+y \leq 20,3 x+2 y \leq 48 $ in the first quadrant.

So, the linear inequalities representing the given shaded region are, $ x+y \leq 20,3 x+2 y \leq 48, x \geq 0, y \geq 0 $

Solution

Consider the line $x+y=4$.

We observe that the shaded region and the origin lie on the opposite side of this line and $(0,0)$ satisfies $x+y \leq 4$. Therefore, we must have $x+y \geq 4$ as the linear inequation corresponding to the line $x+y=4$.

Consider the line $x+y=8$, clearly the shaded region and origin lie on the same side of this line and $(0,0)$ satisfies the constraints $x+y \leq 8$. Therefore, we must have $x+y \leq 8$, as the linear inequation corresponding to the line $x+y=8$.

Consider the line $x=5$. It is clear from the graph that the shaded region and origin are on the left of this line and $(0,0)$ satisfy the constraint $x \leq 5$.

Hence, $x \leq 5$ is the linear inequation corresponding to $x=5$.

Consider the line $y=5$, clearly the shaded region and origin are on the same side (below) of the line and $(0,0)$ satisfy the constrain $y \leq 5$.

Therefore, $y \leq 5$ is an inequation corresponding to the line $y=5$.

We also notice that the shaded region is above the $X$-axis and on the right of the $Y$-axis i.e., shaded region is in first quadrant. So, we must have $x \geq 0, y \geq 0$.

Thus, the linear inequations comprising the given solution set are

$ x+y \geq 4 ; x+y \leq 8 ; x \leq 5 ; y \leq 5 ; x \geq 0 \text { and } y \geq 0 $

Solution

Consider the inequation $x+2 y \leq 3$ as an equation, we have

$x+2y=3$

$\begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 3 \\ \\ \hline y & 1.5 & 1 & 0 \\ \\ \hline \end{array}$

Now, $(0,0)$ satisfy the inequation $x+2 y \leq 3$.

So, half plane contains $(0,0)$ as the solution and the line $x+2 y=3$ intersect the coordinate axis at $(3,0)$ and $(0,3 / 2)$.

Consider the inequation $3 x+4 y \geq 12$ as an equation, we have $ \ 3 x+4 y=12$

$\begin{array}{|c|c|c|c|} \hline x & 0 & 2 & 4 \\ \\ \hline y & 3 & \dfrac{3}{2} & 0 \\ \\ \hline \end{array}$

Thus, coordinate axis intersected by the line $3 x+4 y=12$ at points $(4,0)$ and $(0,3)$.

Now, $(0,0)$ does not satisfy the inequation $3 x+4 y=12$.

Therefore, half plane of the solution does not contained $(0,0)$.

Consider the inequation $y \geq 1$ as an equation, we have

$ y=1 $

It represents a straight line parallel to $X$-axis passing through point $(0,1)$.

Now, $(0,0)$ does not satisfy the inequation $y \geq 1$.

Therefore, half plane of the solution does not contains $(0,0)$.

Clearly $x \geq 0$ represents the region lying on the right side of $Y$-axis.

The solution set of the given linear constraints will be the intersection of the above region.

It is clear from the graph the shaded portions do not have common region.

So, solution set is null set.

Solution

Consider the inequation $3 x+2 y \geq 24$ as an equation, we have $3 x+2 y=24$.

$\begin{array}{|c|c|c|c|} \hline x & 0 & 8 & 4 \\ \\ \hline y & 12 & 0 & 6 \\ \\ \hline \end{array}$

Hence, line $3 x+y=24$ intersect coordinate axes at points $(8,0)$ and $(0,12)$.

Now, $(0,0)$ does not satisfy the inequation $3 x+2 y \geq 24$.

Therefore, half plane of the solution set does not contains $(0,0)$.

Consider the inequation $3 x+y \leq 15$ as an equation, we have

$ 3 x+y=15 $

$\begin{array}{|c|c|c|c|} \hline x & 0 & 5 & 3 \\ \\ \hline y & 15 & 0 & 6 \\ \\ \hline \end{array}$

Line $3 x+y=15$ intersects coordinate axes at points $(5,0)$ and $(0,15)$.

Now, point $(0,0)$ satisfy the inequation $3 x+y \leq 15$.

Therefore, the half plane of the solution contain origin.

Consider the inequality $x \geq 4$ as an equation, we have

$ x=4 $

It represents a straight line parallel to $Y$-axis passing through $(4,0)$. Now, point $(0,0)$ does not satisfy the inequation $x \geq 4$.

Therefore, half plane does not contains $(0,0)$,

The graph of the above inequations is given below.

It is clear from the graph that there is no common region corresponding to these inequality. Hence, the given system of inequalities have no solution.

Solution

Consider the inequation $2 x+y \geq 8$ as an equation, we have

$2x + y = 8$

$\begin{array}{|c|c|c|c|} \hline x & 0 & 4 & 3 \\ \\ \hline y & 8 & 0 & 2 \\ \\ \hline \end{array}$

The line $2 x+y=8$ intersects coordinate axes at $(4,0)$ and $(0,8)$. Now, point $(0,0)$ does not satisfy the inequation $2 x+y \geq 8$. Therefore, half plane does not contain origin.

Consider the inequation $x+2 y \geq 10$, as an equation, we have

$ x+2 y=10 $

$\begin{array}{|c|c|c|c|} \hline x & 0 & 10 & 8 \\ \\ \hline y & 5 & 0 & 1 \\ \\ \hline \end{array}$

The line $2 x+y=8$ intersects the coordinate axes at $(10,0)$ and $(0,5)$.

Now, point $(0,0)$ does not satisfy the inequation $x+2 y \geq 10$.

Therefore, half plane does not contain $(0,0)$.

Consider the inequation $x \geq 0$ and $y \geq 0$ clearly, it represents the region in first quadrant. The graph of the above inequations is given below

It is clear from the graph that common shaded portion is unbounded.

Solution

Option (c) If $x<5$, then $-x>-5$

[If we multiply by negative numbers, then inequality get reversed]

• Option (a) $-x < -5$: This is incorrect because if $x<5$, then multiplying both sides by $-1$ reverses the inequality, resulting in $-x>-5$, not $-x<-5$.

• Option (b) $-x \leq-5$: This is incorrect because if $x<5$, then multiplying both sides by $-1$ reverses the inequality, resulting in $-x>-5$, not $-x \leq-5$.

• Option (d) $-x \geq-5$: This is incorrect because if $x<5$, then multiplying both sides by $-1$ reverses the inequality, resulting in $-x>-5$, not $-x \geq-5$.

Solution

Option (c) It is given that,

$ x<y, b<0 $

$\Rightarrow \quad \dfrac{x}{b}>\dfrac{y}{b}$ $\quad [\because b<0]$

• Option (a) $\dfrac{x}{b}<\dfrac{y}{b}$: This option is incorrect because when $b < 0$, dividing by a negative number reverses the inequality. Since $x < y$, $\dfrac{x}{b}$ will be greater than $\dfrac{y}{b}$, not less.

• Option (b) $\dfrac{x}{b} \leq \dfrac{y}{b}$: This option is incorrect for the same reason as (a). Dividing by a negative number reverses the inequality, so $\dfrac{x}{b}$ will be strictly greater than $\dfrac{y}{b}$, not less than or equal to.

• Option (d) $\dfrac{x}{b} \geq \dfrac{y}{b}$: This option is incorrect because while it correctly accounts for the reversal of the inequality when dividing by a negative number, it includes the possibility of equality $(\geq)$. Since $x < y$ and $b < 0$, $\dfrac{x}{b}$ will be strictly greater than $\dfrac{y}{b}$, not greater than or equal to.

Solution

Option (a) Given that, $-3 x+17<-13$

$\Rightarrow$ $3 x-17>13 \quad $ $[$ multiplying by $-1$ on both sides $]$

$\Rightarrow$ $3 x>13+17 \quad $ $[$ adding $17$ on both sides $]$

$\Rightarrow$ $3 x>30$

$\Rightarrow x>10$

• Option (b) $x \in[10, \infty)$ is incorrect because the inequality $x > 10$ does not include the value $10$ itself, whereas the interval notation $[10, ∞)$ includes $10$.

• Option (c) $x \in(-\infty, 10]$ is incorrect because the inequality $x > 10$ indicates that $x$ must be greater than $10$, not less than or equal to $10$.

• Option (d) $x \in(-10, 10)$ is incorrect because the inequality $x > 10$ indicates that $x$ must be greater than $10$, whereas the interval notation $(-10, 10)$ includes values less than $10$ and does not include values greater than $10$.

Solution

Option (b) Given, $ \ |x|<3 $

$ \Rightarrow \ -3<x<3 \quad[\because|x|<a \Rightarrow-a<x<a] $

• Option (a) $x \geq 3$ is incorrect because the condition $|x|<3$ implies that $x$ must be within the range of $-3$ and $3$, not greater than or equal to $3$.

• Option (c) $x \leq -3$ is incorrect because the condition $|x|<3$ implies that $x$ must be within the range of $-3$ and $3$, not less than or equal to $-3$.

• Option (d) $-3 \leq x \leq 3$ is incorrect because the condition $|x|<3$ implies that $x$ must be strictly between $-3$ and $3$, not including the endpoints $-3$ and $3$.

Solution

Option (d) Given, $|x| >b \ \text { and } \ b>0 $

$\Rightarrow \ x <-b \ \text { or } \ x>b$

$\Rightarrow \ x \in(-\infty,-b) \cup(b, \infty)$

• Option (a) $x \in(-b, \infty)$ is incorrect because it includes values of $x$ in the interval $(-b, b)$, which do not satisfy $|x| > b$.

• Option (b) $x \in(-\infty, b)$ is incorrect because it includes values of $x$ in the interval $(-b, b)$, which do not satisfy $|x| > b$.

• Option (c) $x \in(-b, b)$ is incorrect because it includes values of $x$ where $|x| \leq b$, which do not satisfy $|x| > b$.

Solution

Option (c) Given, $ \ |x-1| >5 $

$\Rightarrow \ (x-1) <-5 \ \text { or } \ (x-1)>5 \quad[\because|x|>a \Rightarrow x<a \ \text { or } \ x>a] $

$\Rightarrow \ x <-4 \ \text { or } \ x>6 $

$\Rightarrow \ x \in(-\infty,-4) \cup(6, \infty)$

• Option (a) $x \in(-4,6)$ is incorrect because it represents the interval where $x$ is between $-4$ and $6$, which does not satisfy the condition $|x-1|>5$. For $|x-1|>5$, $x$ must be either less than $-4$ or greater than $6$.

• Option (b) $x \in[-4,6]$ is incorrect because this interval includes the endpoints $-4$ and $6$ and also the values in between. It still does not satisfy $|x - 1| > 5$ for the same reason as in (a).

• Option (d) $x \in(-\infty,-4) \cup[6, \infty)$ is incorrect because it includes the interval $[6, \infty)$, which means $x \geq 6$. However, the correct condition $|x-1|>5$ requires $x > 6$, not $x \geq 6$.

Solution

Option (b) Given, $ \ |x+2| \leq 9,$

$\Rightarrow \ -9 \leq x+2 \leq 9 \quad {[\because \ |x| \leq a \Rightarrow-a \leq x \leq a]}$

$\Rightarrow \ -9-2 \leq x \leq 9-2 \quad \text { [subtracting 2 througout] } $

$\Rightarrow \ -11 \leq x \leq 7 $

$x \in [-11,7]$

• Option (a) $x \in(-7,11)$: This option is incorrect because the interval $(-7, 11)$ does not include the endpoints $-11$ and $7$, which are part of the solution to the inequality $|x+2| \leq 9$. The correct interval should be $[-11, 7]$.

• Option (c) $x \in(-\infty,-7) \cup(11, \infty)$: This option is incorrect because it represents the values of $x$ that are outside the interval $[-11, 7]$. The inequality $|x+2| \leq 9$ includes values within the interval $[-11, 7]$, not outside of it.

• Option (d) $x \in(-\infty,-7) \cup[11, \infty)$: This option is incorrect for the same reason as option (c). It represents values of $x$ that are outside the interval $[-11, 7]$. The inequality $|x+2| \leq 9$ includes values within the interval $[-11, 7]$, not outside of it.

Solution

Option (a) The given graph represent $x>-5$ and $x<5$.

On combining these two result, we get

$ |x|<5 \quad [\because -a < x < -a \Rightarrow |x|<a ;a >0] $

• Option (b) $|x| \leq 5$: This option is incorrect because the graph does not include the points $x = -5$ and $x = 5$. The graph shows open circles at these points, indicating that these values are not part of the solution set.

• Option (c) $|x| > 5$: This option is incorrect because the graph shows the region between $-5$ and $5$, not outside of it. The graph does not include values of $x$ that are greater than $5$ or less than $-5$.

• Option (d) $|x| \geq 5$: This option is incorrect because the graph does not include the points $x = -5$ and $x = 5$, nor does it include any values outside the interval $-5 < x < 5$. The graph only shows the region strictly between $-5$ and $5$.

Solution

Option (d) The given graph represents all the values greater than 5 except $x=5$ on the real line So,

i.e. $ \ x \in(5, \infty)$.

• Option (a) $x \in(-\infty, 5)$ is incorrect because it represents all values less than 5, which does not match the given graph that shows values greater than 5.

• Option (b) $x \in(-\infty, 5]$ is incorrect because it includes all values less than or equal to 5, which does not match the given graph that shows values greater than 5 and excludes 5 itself.

• Option (c) $x \in [5, \infty)$ is incorrect because this interval represents $x$ greater than or equal to $5$ which does not match the given graph that shows values strictly greater than $5$.

Solution

Option (b) The given graph represents all the values greater than $\dfrac{9}{2}$ including $\dfrac{9}{2}$ on the real line.

i.e. $ \ x \in \left[\dfrac{9}{2}, \infty \right) $

• Option (a) $x \in \left(\dfrac{9}{2}, \infty \right)$ is incorrect because it uses the same notation as the correct answer but does not include the correct interval notation. The correct interval should be $\left[\dfrac{9}{2}, \infty \right)$, indicating that $\dfrac{9}{2}$ is included.

• Option (c) $x \in \left(-\infty, \dfrac{9}{2} \right)$ is incorrect because it represents the interval $\left(-\infty, \dfrac{9}{2} \right)$, which includes all values less than $\dfrac{9}{2}$, not greater.

• Option (d) $x \in \left(-\infty, \dfrac{9}{2} \right]$ is incorrect for the same reason as option (c); it represents the interval $\left(-\infty, \dfrac{9}{2} \right)$, which includes all values less than $\dfrac{9}{2}$, not greater.

Solution

Option (a) The given graph represents all the values less than $\dfrac{7}{2}$ on the real line.

i.e. $ \ x \in \left(-\infty, \dfrac{7}{2} \right) $

• Option (b) $x \in \left(-\infty, \dfrac{7}{2} \right]$ is incorrect because this interval includes all real numbers less than or equal to $\dfrac{7}{2}$. It includes $\dfrac{7}{2}$ itself, which does not align with the condition of being strictly less than $\dfrac{7}{2}$.

• Option (c) $x \in \left[\dfrac{7}{2}, \infty \right)$ is incorrect because this interval represents all real numbers greater than or equal to $\dfrac{7}{2}$. This is the opposite of what we are looking for, as it includes values that are not less than $\dfrac{7}{2}$.

• Option (d) $x \in \left(\dfrac{7}{2}, \infty \right)$ is incorrect because it suggests that $ x $ is in the interval $ \left(\dfrac{7}{2}, \infty\right) $, which represents all values greater than $ \dfrac{7}{2} $, contrary to the given graph that represents values less than $ \dfrac{7}{2} $.

Solution

Option (b) The given graph represents all values less than $-2$ including $-2$ .

i.e. $ \ x \in(-\infty,-2] $

• Option (a) $x \in(-\infty,-2)$ is incorrect because it does not include the value $-2$, whereas the graph includes $-2$.

• Option (c) $x \in(-2, \infty]$ is incorrect because it represents values greater than $-2$, while the graph represents values less than $-2$.

• Option (d) $x \in(-2, \infty)$ is incorrect because it represents values greater than $-2$ and does not include $-2$, while the graph represents values less than $-2$ including $-2$.

Solution

(i) If $x<y$ and $b<0$ $\Rightarrow$ $\dfrac{x}{b}>\dfrac{y}{b}$

Hence, statement (i) is false.

(ii) If $x y>0$, then, $x>0, y>0$ or $x<0, y<0$.

Hence, statement (ii) is false.

(iii) If $x y>0$, then $x<0$ and $y<0$.

Hence, statement (iii) is true.

(iv) If $x y<0 \Rightarrow x<0, y>0$ or $x>0, y<0$.

Hence, statement (iv) is false.

(v) If $x<-5$ and $x<-2$

$x<−5$ implies that $x$ is less than $−5$.

$x<−2$ implies that $x$ is less than $−2$.

Then $ x \in(-\infty,-5) $

Hence, statement (v) is true.

(vi) If $x<-5$ and $x>2$, then $x$ have no value.

Hence, statement (vi) is false.

(vii) If $x>-2$ and $x<9$, then $x \in(-2,9)$.

Hence, statement (vii) is true.

(viii) If $|x|>5$, then either $x<-5$ or $x>5$.

$ \Rightarrow \ x \in(-\infty,-5) \cup(5, \infty) $

Hence, statement (viii) is false.

(ix) If $|x| \leq 4$, then

$-4 \leq x \leq 4 $

$\Rightarrow x \in [-4,4] $

Hence, statement (ix) is true.

(x)

The given graph represents $x \leq 3$.

Hence, statement $(x)$ is false.

(xi)

The given graph represents $x \geq 0$.

Hence, statement (xi) is true.

(xii)

The given graph represent $y \geq 0$.

Hence, statement (xii) is false.

(xiii) The solution set of $ x \geq 0 $ and $ y \leq 0 $ is the first quadrant including the axes. The given graph does not represent this correctly.

Solution set of $x \geq 0$ and $y \leq 0$ is

Hence, statement (xiii) is false.

(xiv) The solution set of $ x \geq 0 $ and $ y \leq 1 $ is the region where $ x $ is non-negative and $ y $ is less than or equal to 1.

Solution set of $x \geq 0$ and $y \leq 1$ is

Hence, statement (xiv) is false.

(xv)

The given graph represents $x+y \geq 0$.

Hence, statement (xv) is correct.

Solution

(i) If $-4 x \geq 12 $

$\Rightarrow x \leq-3$

(ii) If $\dfrac{-3}{4} x \leq-3$

$ \Rightarrow x \geq(-3) \times \dfrac{4}{-3} $

$\Rightarrow x \geq 4 $

(iii) $\dfrac{2}{x+2}>0$

For the entire fraction $\dfrac{2}{x+2}$ to be positive, the denominator $x+2$ must also be positive.

i.e. $ \ x+2 >0$

$\Rightarrow \ x > -2$

(iv) If $x>-5 \Rightarrow 4 x>-20$

(v) If $x>y$ and $z<0$, then

$ \ -x z>-y z \quad [ \because z \ \text{is a negative number}] $

(vi) If $p>0$ and $q<0$, then

$ p-q>p $

e.g., consider $2>0$ and $-3<0$.

$ 2-(-3)=2+3=5>2 $

(vii) If $|x+2|>5$, then

$ x+2<-5 \text { or } x+2>5 $

$\Rightarrow x<-5-2 \text { or } x>5-2 $

$\Rightarrow x<-7 \text { or } x>3 $

(viii) If $-2 x+1 \geq 9$, then

$ \begin{aligned} &-2 x \geq 9-1 & \\ \\ & \Rightarrow \ -2 x \geq 8 \\ \\ &\Rightarrow \ 2 x \leq-8 & \\ \\ & \Rightarrow \ x \leq-4 \end{aligned} $