Class 11: Linear Inequations – Exercise 15.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Solve each of the following systems of equations in R.

$\displaystyle \text{Question 1: } \Big| x+ \frac{1}{3} \Big| > \frac{8}{3}$

Answer:

$\displaystyle \text{Given } \Big| x+ \frac{1}{3} \Big| > \frac{8}{3}$

$\displaystyle \text{Case I: } x + \frac{1}{3} > \frac{8}{3} \Rightarrow x > \frac{-1}{3} + \frac{8}{3} \Rightarrow x > \frac{7}{3}$

$\displaystyle \text{Case II: } x + \frac{1}{3} < \frac{-8}{3} \Rightarrow x < \frac{-1}{3} - \frac{8}{3} \Rightarrow x < - 3$

$\displaystyle \text{Therefore the solution set is } \Big( \frac{7}{3} , \infty \Big) \cup ( -\infty, - 3)$

$\displaystyle \\$

$\displaystyle \text{Question 2: } |4-x| +1 < 3$

Answer:

$\displaystyle \text{Given } |4-x| +1 < 3$

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

$\displaystyle \Rightarrow -6 < -x < - 2$

$\displaystyle \Rightarrow 2 < x < 6$

$\displaystyle \text{Therefore the solution set is } (2, 6)$

$\displaystyle \\$

$\displaystyle \text{Question 3: } \Big| \frac{3x-4}{2} \Big| \leq \frac{5}{12}$

Answer:

$\displaystyle \text{Given } \Big| \frac{3x-4}{2} \Big| \leq \frac{5}{12}$

$\displaystyle \Rightarrow - \frac{5}{12} \leq \frac{3x-4}{2} \leq \frac{5}{12}$

$\displaystyle \Rightarrow -5 \leq 18x-24 \leq 5$

$\displaystyle \Rightarrow 19 \leq 18x \leq 29$

$\displaystyle \Rightarrow \frac{19}{18} \leq x \leq \frac{29}{18}$

$\displaystyle \text{Therefore the solution set is } \Big[ \frac{19}{18} , \frac{29}{18} \Big ]$

$\displaystyle \\$

$\displaystyle \text{Question 4: } \frac{|x-2|}{x-2} > 0$

Answer:

$\displaystyle \text{Given } \frac{|x-2|}{x-2} > 0$

$\displaystyle \text{Case I: When } x-2 \geq 0 \Rightarrow x \geq 2$

$\displaystyle \therefore \frac{|x-2|}{x-2} > 0 \Rightarrow \frac{x-2}{x-2} > 0 \Rightarrow x >2$

$\displaystyle \therefore x \in ( 2, \infty)$

$\displaystyle \text{Case II: When } x-2 < 0 \Rightarrow x < 2$

$\displaystyle \therefore \frac{|x-2|}{x-2} > 0 \Rightarrow \frac{-(x-2)}{x-2} > 0 \Rightarrow \frac{x-2} {x-2} < 0$

$\displaystyle \text{There is no possible solution. Therefore } x \in \phi$

Combining Case I and Case II we get the solution set as $\displaystyle ( 2, \infty)$

$\displaystyle \\$

$\displaystyle \text{Question 5: } \frac{1}{|x|-3} < \frac{1}{2}$

Answer:

$\displaystyle \text{Given } \frac{1}{|x|-3} < \frac{1}{2} \Rightarrow \frac{1}{|x|-3} - \frac{1}{2} < 0 \Rightarrow \frac{2 - ( |x|-3)}{2 ( |x|-3)} < 0 \Rightarrow \frac{5 - |x|}{2 |x|-6} < 0$

$\displaystyle \text{Case I: When } x \geq 0 \Rightarrow |x| = x$

$\displaystyle \text{Therefore } \frac{5 - |x|}{2 |x|-6} < 0 \Rightarrow \frac{5 - x}{2 x-6} < 0$

$\displaystyle \text{Case I(a) } 5-x < 0 \text{ and } 2x - 6 > 0$

$\displaystyle \Rightarrow x > 5 \text{ and } x > 3 \Rightarrow x > 5$

$\displaystyle \text{Case I(b) } 5-x > 0 \text{ and } 2x - 6 < 0$

$\displaystyle \Rightarrow x < 5 \text{ and } x < 3 \Rightarrow x < 3$

Therefore the solution set in Case I is $\displaystyle [0, 3) \cup ( 5, \infty)$

$\displaystyle \text{Case II: When } x < 0 \Rightarrow |x| = -x$

$\displaystyle \text{Therefore } \frac{5 - |x|}{2 |x|-6} < 0 \Rightarrow \frac{5 - (-x)}{2 (-x)-6} < 0 \Rightarrow \frac{5 x}{-2x-6} < 0$

Case II(a): $\displaystyle 5+x < 0 \text{ and } -2x - 6 > 0$

$\displaystyle \Rightarrow x < -5 \text{ and } x <-3 \Rightarrow x < - 5$

Case II(b): $\displaystyle 5+x > 0 \text{ and } -2x - 6 < 0$

$\displaystyle \Rightarrow x > -5 \text{ and } x > -3 \Rightarrow x > - 3$

Therefore the solution set in Case II is $\displaystyle (- \infty, -5) \cup ( -3, \infty)$

Combining both Case I and Case II, we get the complete solution set as

$\displaystyle [0, 3) \cup ( 5, \infty) \ \cup (- \infty, -5) \cup ( -3, \infty)$

$\displaystyle \Rightarrow ( - \infty, -5) \cup ( -3, 3) \cup ( 5, \infty)$

$\displaystyle \\$

$\displaystyle \text{Question 6: } \frac{|x+2|-x}{x} < 2$

Answer:

$\displaystyle \text{Given } \frac{|x+2|-x}{x} < 2 \Rightarrow \frac{|x+2|-x}{x} - 2 <0 \Rightarrow \frac{|x+2|-3x}{x} < 0$

$\displaystyle \text{Case I: When } x + 2 \geq 0 \Rightarrow x \geq -2$

$\displaystyle \Rightarrow |x+2| = ( x+2)$

$\displaystyle \therefore \frac{|x+2|-3x}{x} < 0 \Rightarrow \frac{x+2-3x}{x} < 0 \Rightarrow \frac{2-2x}{x} < 0$

$\displaystyle \text{Case I(a) } 2-2x < 0 \text{ and } x > 0$

$\displaystyle \Rightarrow x > 1 \text{ and } x > 0 \Rightarrow x > 1$

$\displaystyle \text{Case I(b) } 2-2x > 0 \text{ and } x < 0$

$\displaystyle \Rightarrow x < 1 \text{ and } x < 0 \Rightarrow x < 0$

Therefore the solution set in Case I and $\displaystyle x \geq -2$ is $\displaystyle [-2, 0) \cup ( 1, \infty)$

$\displaystyle \text{Case II: When } x + 2 < 0 \Rightarrow x < -2$

$\displaystyle \Rightarrow |x+2| = -( x+2)$

$\displaystyle \therefore \frac{|x+2|-3x}{x} < 0 \Rightarrow \frac{-(x+2)-3x}{x} < 0 \Rightarrow \frac{-4x-2}{x} < 0 \Rightarrow \frac{4x+2}{x} > 0$

Case II(a): $\displaystyle 4x+2 > 0 \text{ and } x > 0$

$\displaystyle \Rightarrow x > \frac{-1}{2} \text{ and } x > 0 \Rightarrow x > 0$

Case II(b): $\displaystyle 4x+2 < 0 \text{ and } x < 0$

$\displaystyle \Rightarrow x < \frac{-1}{2} \text{ and } x < 0 \Rightarrow x > \frac{-1}{2}$

Therefore the solution set in Case I and $\displaystyle x < -2$ is $\displaystyle (- \infty, -2) \cup ( 0, \infty)$

Combining both Case I and Case II, and we get the complete solution set as $\displaystyle [-2, 0) \cup ( 1, \infty) \ \cup \ (- \infty, -2) \cup ( 0, \infty)$

$\displaystyle \Rightarrow ( - \infty, 0) \cup ( 1, \infty)$

$\displaystyle \\$

$\displaystyle \text{Question 7: } \Big| \frac{2x-1}{x-1} \Big| > 2$

Answer:

$\displaystyle \text{Given } \Big| \frac{2x-1}{x-1} \Big| > 2$

$\displaystyle \text{Case I: } \frac{2x-1}{x-1} >2 \Rightarrow \frac{2x-1}{x-1} -2 > 0 \Rightarrow \frac{2x-1-2x+2}{x-1} > 0 \Rightarrow \frac{1}{x-1} >0$

$\displaystyle \Rightarrow x - 1 > 0 \Rightarrow x > 0$

Therefore the solution set for Case I is $\displaystyle ( 1, \infty)$

$\displaystyle \text{Case II: } \frac{2x-1}{x-1} < -2 \Rightarrow \frac{2x-1}{x-1} +2 < 0 \Rightarrow \frac{2x-1+2x-2}{x-1} < 0 \Rightarrow \frac{4x-3}{x-1} < 0$

Case II(a): $\displaystyle 4x-3 > 0 \text{ and } x-1 < 0$

$\displaystyle \Rightarrow x > \frac{3}{4} \text{ and } x < 1 \Rightarrow \frac{3}{4} < x < 1$

Case II(b): $\displaystyle 4x-3 < 0 \text{ and } x-1 > 0$

$\displaystyle \Rightarrow x < \frac{3}{4} \text{ and } x > 1 \Rightarrow \text{ No possible solution }$

Combining both Case I and Case II, and we get the complete solution set as $\displaystyle (1, \infty) \cup ( \frac{3}{4} , 1)$

$\displaystyle \\$

$\displaystyle \text{Question 8: } |x-1|+|x-2|+|x-3| \geq 6$

Answer:

$\displaystyle \text{Given: } |x-1|+|x-2|+|x-3| \geq 6$

$\displaystyle \text{For } |x-1|,$

$\displaystyle \text{When } x-1 \geq 0 \Rightarrow x \geq 1$

$\displaystyle \text{When } (x-1) < 0 \Rightarrow x < 0$

$\displaystyle \text{For } |x-2|,$

$\displaystyle \text{When } x-2 \geq 0 \Rightarrow x \geq 2$

$\displaystyle \text{When } (x-2) < 0 \Rightarrow x < 2$

$\displaystyle \text{For } |x-3|,$

$\displaystyle \text{When } x-3 \geq 0 \Rightarrow x \geq 3$

$\displaystyle \text{When } (x-3) < 0 \Rightarrow x < 3$

$\displaystyle \text{Case I: When } x < 1$

$\displaystyle (1-x)+ ( 2- x) + ( 3-x ) \geq 6 \Rightarrow 6 - 3x \geq 6 \Rightarrow x \leq 0$

Hence the solutions set is $\displaystyle ( - \infty, 0]$

$\displaystyle \text{Case II: When } 1 \leq x < 2$

$\displaystyle (x-1)+ ( 2- x) + ( 3-x ) \geq 6 \Rightarrow 4 - x \geq 6 \Rightarrow x \leq -2$

Hence the solutions set is null.

$\displaystyle \text{Case III: } \text{When } 2 \leq x < 3$

$\displaystyle (x-1)+ ( x-2) + ( 3-x ) \geq 6 \Rightarrow x \geq 6$

Hence the solutions set is null.

Case IV: $\displaystyle \text{When } x \geq 3$

$\displaystyle (x-1)+ ( x-2) + ( x-3 ) \geq 6 \Rightarrow 3x-6 \geq 6 \Rightarrow x \geq 4$

Hence the solutions set is $\displaystyle [4, \infty )$

From all the four cases the solution set is $\displaystyle ( - \infty, 0] \cup [4, \infty )$

$\displaystyle \\$

$\displaystyle \text{Question 9: } \frac{|x-2|-1}{|x-2|-2} \leq 0$

Answer:

$\displaystyle \text{Given: } \frac{|x-2|-1}{|x-2|-2} \leq 0$

$\displaystyle \text{Case I: When } x - 2 \geq 0 \Rightarrow x \geq 2$

$\displaystyle \text{Therefore } \frac{x-2-1}{x-2-2} \leq 0 \Rightarrow \frac{x-3}{x-4} \leq 0$

$\displaystyle \text{Case I(a) } x-3 \leq 0 \text{ and } x-4 > 0$

$\displaystyle \Rightarrow x \leq 3 \text{ and } x > 4$

Therefore the solution set is null set.

$\displaystyle \text{Case I(b) } x-3 \geq 0 \text{ and } x-4 < 0$

$\displaystyle \Rightarrow x \geq 3 \text{ and } x < 4$

$\displaystyle \text{Therefore the solution set is } [3, 4)$ .

Therefore the solution set in Case I and $\displaystyle x \geq 2$ is $\displaystyle [3, 4)$

$\displaystyle \text{Case II: When } x - 2 \leq 0 \Rightarrow x \leq 2$

$\displaystyle \text{Therefore } \frac{2-x-1}{2-x-2} \leq 0 \Rightarrow \frac{1-x}{-x} \leq 0 \Rightarrow \frac{x-1}{x} \geq 0$

Case II(a): $\displaystyle x-1 \leq 0 \text{ and } x > 0$

$\displaystyle \Rightarrow x \leq 1 \text{ and } x > 0$

$\displaystyle \text{Therefore the solution set is } (0, 1]$ .

Case II(b): $\displaystyle x-1 \geq 0 \text{ and } x < 0$

$\displaystyle \Rightarrow x \geq 1 \text{ and } x < 0$

Therefore the solution set is null.

Therefore the solution set in Case I and $\displaystyle x \leq 2$ is $\displaystyle (0, 1]$

Combining both Case I and Case II, and we get the complete solution set as $\displaystyle (0, 1] \cup ([3, 4)$

$\displaystyle \\$

$\displaystyle \text{Question 10: } \frac{1}{|x|-3} \leq \frac{1}{2}$

Answer:

$\displaystyle \text{Given: } \frac{1}{|x|-3} \leq \frac{1}{2}$

$\displaystyle \Rightarrow \frac{1}{|x|-3} - \frac{1}{2} \leq 0 \Rightarrow \frac{2 - |x| + 3}{2 ( |x|-3)} \leq 0 \Rightarrow \frac{5 - |x|}{2 |x| - 6} \leq 0$

$\displaystyle \text{Case I: When } x \geq 0 \Rightarrow |x| = x$

$\displaystyle \therefore \frac{5 - |x|}{2 |x| - 6} \leq 0 \Rightarrow \frac{5-x}{2x-6} \leq 0$

$\displaystyle \text{Case I(a) } 5-x \leq 0 \text{ and } 2x-6 > 0$

$\displaystyle \Rightarrow x \geq 5 \text{ and } x > 3$

$\displaystyle \text{Therefore the solution set is } [5, \infty)$ .

$\displaystyle \text{Case I(b) } 5-x \geq 0 \text{ and } 2x-6 < 0$

$\displaystyle \Rightarrow x \leq 5 \text{ and } x < 3$

Therefore the solution set is along with $\displaystyle x \geq 0$ is $\displaystyle [0, 3)$ .

Therefore the solution set in Case I is $\displaystyle [0, 3) \cup [5, \infty)$

$\displaystyle \text{Case II: When } x < 0 \Rightarrow |x| = -x$

$\displaystyle \therefore \frac{5 - |x|}{2 |x| - 6} \leq 0 \Rightarrow \frac{5-(-x)}{2(-x)-6} \leq 0 \Rightarrow \frac{5+x}{-2x-6} \leq 0 \Rightarrow \frac{5+x}{2x+6} \geq 0$

Case II(a): $\displaystyle 5+x \geq 0 \text{ and } 2x+6 > 0$

$\displaystyle \Rightarrow x \geq -5 \text{ and } x > -3$

$\displaystyle \Rightarrow x > -3$

$\displaystyle \text{Therefore the solution set is } (-3, \infty)$ .

Case II(b): $\displaystyle 5+x \leq 0 \text{ and } 2x+6 < 0$

$\displaystyle \Rightarrow x \leq -5 \text{ and } x < -3$

$\displaystyle \Rightarrow x \leq -5$

$\displaystyle \text{Therefore the solution set is } (-\infty, -5)$ .

Therefore the solution set in Case I is $\displaystyle (-3, \infty) \ \cup \ (-\infty, -5)$

Combining both Case I and Case II, and we get the complete solution set as $\displaystyle [0, 3) \cup [5, \infty) \ \cup \ (-3, \infty) \ \cup \ (-\infty, -5) \Rightarrow ( - \infty, -5) \cup (-3, 3) \cup ([5, \infty)$

$\displaystyle \\$

$\displaystyle \text{Question 11: } |x+1|+|x| > 3$

Answer:

$\displaystyle \text{Given: } |x+1|+|x| > 3$

$\displaystyle \text{When } (x+1) \geq 0 \Rightarrow x \geq -1 \text{ then } |x+1| = (x+1)$

$\displaystyle \text{When } (x+1) < 0 \Rightarrow x < -1 \text{ then } |x+1| = -(x+1)$

Similarly,

$\displaystyle \text{When } x \geq 0 \text{ then } |x| = x$

$\displaystyle \text{When } x < 0 \text{ then } |x| = -x$

$\displaystyle \text{Case I: When } x < -1$

$\displaystyle \Rightarrow -(x+1) - x > 3 \Rightarrow -2x-1 > 3 \Rightarrow -2x > 4 \Rightarrow x<-2$

$\displaystyle \therefore x \in ( - \infty , -2)$

$\displaystyle \text{Case II: When } -1 \leq x < 0$

$\displaystyle \Rightarrow (x+1) - x > 3 \Rightarrow 1 > 3$

This is not possible. $\displaystyle \text{Therefore } x \in \phi$

$\displaystyle \text{Case III: } \text{When } x \geq 0$

$\displaystyle \Rightarrow (x+1) + x > 3 \Rightarrow 2x+1 > 3 \Rightarrow x>1$

$\displaystyle \text{Therefore } x \in (1, \infty)$

Combining all the three cases, the solution set is $\displaystyle ( - \infty, -2) \cup ( 1,\infty)$

$\displaystyle \\$

$\displaystyle \text{Question 12: } 1 \leq |x-2| \leq 3$

Answer:

$\displaystyle \text{Given } 1 \leq |x-2| \leq 3$

$\displaystyle \text{For } |x-2| \geq 1$

$\displaystyle \text{When } x - 2 \geq 0 \Rightarrow x \geq 2 \text{ then } |x-2|= (x-2)$

$\displaystyle \therefore x-2 \geq 1 \Rightarrow x \geq 3 \text{ i.e } x \in [ 3, \infty)$

$\displaystyle \text{When } x - 2 < 0 \Rightarrow x < 2 \text{ then } |x-2| = -(x-2)$

$\displaystyle \therefore -(x-2) \geq 1 \Rightarrow x < 1 \text{ i.e } x \in ( - \infty, 1)$

Hence the solution set for the first equation is $\displaystyle ( - \infty, 1) \cup [ 3, \infty)$

$\displaystyle \text{For } |x-2| \leq 3$

$\displaystyle \text{When } x - 2 \geq 0 \Rightarrow x \geq 2 \text{ then } |x-2|= (x-2)$

$\displaystyle \therefore x-2 \leq 3 \Rightarrow x \leq 5 \text{ i.e } x \in (- \infty, 5]$

$\displaystyle \text{When } x - 2 < 0 \Rightarrow x < 2 \text{ then } |x-2| = -(x-2)$

$\displaystyle \therefore -(x-2) \leq 3 \Rightarrow x \geq -1 \text{ i.e } x \in [- 1, \infty)$

Hence the solution set for the first equation is $\displaystyle (- \infty, 5] \cup [- 1, \infty) \Rightarrow [-1, 5]$

Combining the two cases, we get the solution set as $\displaystyle ( - \infty, 1) \cup [ 3, \infty) \cup [-1, 5] \Rightarrow [-1, 1] \cup [3, 5]$

$\displaystyle \\$

$\displaystyle \text{Question 13: } |3-4x| \geq 9$

Answer:

$\displaystyle \Rightarrow 3 - 4x \geq 9 \text{ or } 3 - 4x \leq -9$

$\displaystyle \Rightarrow 4x \leq -6 \text{ or } 12 \leq 4x$

$\displaystyle \Rightarrow x \leq \frac{-3}{2} \text{ or } x \geq 3$

Therefore the solution set will be $\displaystyle \text{Therefore the solution set will be } \Big(- \infty, \frac{-3}{2} \Big) \cup [3, \infty)$

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Class 11: Linear Inequations – Exercise 15.3

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