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This is the second of our many quizzes on absolute value inequalities. Click on this to access more such competitive quizzes on inequalities.

Absolute Value Inequalities Quiz 2

mathematics

 What will be the solution set of the inequalities $\left| 7t+12 \right|+6\dfrac{3}{7} < \dfrac{7}{3}\text{ and }\left| 7t+12 \right|+6\dfrac{3}{7}\le \dfrac{7}{3}$ ? 

 Simplify both the inequalities and then check the range of $t$. Absolute values can never be less than or equal to a negative number.

  $\begin{align} & \left| 7t+12 \right|+6\dfrac{3}{7} < \dfrac{7}{3}\text{ and }\left| 7t+12 \right|+6\dfrac{3}{7}\le \dfrac{7}{3}\\ &\Rightarrow \left| 7t+12 \right|+\dfrac{45}{7} < \dfrac{7}{3}\text{ and }\left| 7t+12 \right|+\dfrac{45}{7}\le \dfrac{7}{3}\\ &\Rightarrow \left| 7t+12 \right| < \dfrac{49-135}{21}\text{ and }\left| 7t+12 \right|\le \dfrac{49-135}{21}\\ &\Rightarrow \left| 7t+12 \right| < -\dfrac{86}{21}\text{ and }\left| 7t+12 \right|\le -\dfrac{86}{21}\\ & \text{Absolute values can never be less than or equal to a negative number. Hence, there is no solution for either of these.}\end{align}$

 What will be the solution set of the given inequalities? $\left| 3x-17 \right|+16\times \dfrac{4}{13} > 24\times \dfrac{1}{5}\text{ and }\left| 3x-17 \right|+16\times \dfrac{4}{13}\ge 24\times \dfrac{1}{5}\text{ }$ 

 Simplify both the inequalities and then check the range of $x$. Absolute values can never be less than or equal to a negative number.

  $\begin{align} & \left| 3x-17 \right|+16\times \dfrac{4}{13} > 24\times \dfrac{1}{5}\text{ and }\left| 3x-17 \right|+16\times \dfrac{4}{13}\ge 24\times \dfrac{1}{5}\text{ }\\ &\Rightarrow \left| 3x-17 \right|+\dfrac{64}{13} > \dfrac{24}{5}\text{ and }\left| 3x-17 \right|+\dfrac{64}{13}\ge \dfrac{24}{5}\text{ }\\ &\Rightarrow \left| 3x-17 \right| > \dfrac{312-320}{65}\text{ and }\left| 3x-17 \right|\ge \dfrac{312-320}{65}\\ &\Rightarrow \left| 3x-17 \right| > -\dfrac{8}{65}\text{ and }\left| 3x-17 \right|\ge -\dfrac{8}{65}\\ & \text{Absolute values are always positive or zero. It means, it will always be greater than or equal to a negative number. Hence, the solution for each of these is all real numbers. }\end{align}$

 $\dfrac{1}{4}-\dfrac{\dfrac{1}{4}\left| x \right|}{4+\left| x \right|}\ge \dfrac{1}{12}$ , the solution set to this inequality is $m\le x\le n$ then what is $n-m$ ?

 Simplify the given inequality and use $\left| x \right|\le a,a > 0$ then $-a\le x\le a$ .

 $\begin{align} & \dfrac{1}{4}-\dfrac{\dfrac{1}{4}\left| x \right|}{4+\left| x \right|}\ge \dfrac{1}{12}\\ &\Rightarrow \dfrac{4+\left| x \right|-\left| x \right|}{4\left( 4+\left| x \right| \right)}\ge \dfrac{1}{12}\\ &\Rightarrow \dfrac{1}{4+\left| x \right|}\ge \dfrac{1}{12}\\ &\Rightarrow 4+\left| x \right|\le 12\\ &\Rightarrow \left| x \right|\le 8\Rightarrow -8\le x\le 8\\ &\therefore n-m=8-\left( -8 \right)=16\end{align}$

 What will be the solution set of the inequality $\dfrac{\left| x+6 \right|}{2}-15 < 4$ ? 

 Simplify the given inequality and use $\left| x \right| < a,a > 0$ then $-a\le x\le a$ .

 $\begin{align} & \dfrac{\left| x+6 \right|}{2}-15 < 4\\ &\Rightarrow \dfrac{\left| x+6 \right|-30}{2} < 4\\ &\Rightarrow \left| x+6 \right|-30 < 8\\ &\Rightarrow \left| x+6 \right| < 38\\ &\Rightarrow -38 < x+6 < 38\\ &\Rightarrow -44 < x < 32\\ &\therefore x\in \left( -44,32 \right)\end{align}$

 If ${{\left( x-5 \right)}^{2}}-9\left( x-8 \right) < \left| 4x-35 \right|$ , which of the following express the allowable range for $x$ ? 

 Find the range of $x$ by taking the mode value as positive and negative and taking the intersection of both the ranges. 

 $\begin{align} & {{\left( x-5 \right)}^{2}}-9\left( x-8 \right) < \left| 4x-35 \right|\Rightarrow {{x}^{2}}-19x+97 < \left| 4x-35 \right|\\ & \text{Case 1: }\\ &{{x}^{2}}-19x+97 < 4x-35\Rightarrow {{x}^{2}}-23x+132 < 0\\ & \text{Critical point of inequality: }\\ &{{x}^{2}}-23x+132=0\Leftrightarrow x=11\text{ or }x=12\\ & \text{So, the solution of case 1 is }11 < x < 12.\\ & \\ & \text{Case 2: }\\ & {{x}^{2}}-19x+97 < -4x+35\Rightarrow {{x}^{2}}-15x+62 < 0\\ & \text{Critical point of inequality: }\\ & {{x}^{2}}-15x+62=0\Leftrightarrow x=\dfrac{15\pm \sqrt{-23}}{2}\\ & \text{So, case 2 has no solution.}\\ & \therefore\text{ The solution of the given inequality is }11 < x < 12\end{align}$