Skip to content
Related Articles

Related Articles

Improve Article

Class 11 RD Sharma Solutions – Chapter 15 Linear Inequations – Exercise 15.3

  • Last Updated : 25 Jan, 2021
Geek Week

Question 1. Solve \left | x + \frac{1}{3} \right | > \frac{8}{3}

Solution:

Using the property of modulus operator, we know

|x| > a ⇒ x < -a or x > a

Therefore, \left ( x + \frac{1}{3} \right ) < \frac{-8}{3}  or \left ( x + \frac{1}{3} \right ) > \frac{8}{3}

⇒ x < \frac{-8}{3} - \frac{1}{3}  or x > \frac{8}{3} - \frac{1}{3}



⇒ x < -3 or x > \frac{7}{3}

Hence, we can conclude x lies in the range ( -∞, -3) ∪ ( \frac{7}{3} ,∞ )

Question 2. Solve | 4 – x | + 1 < 3

Solution:

We have,  | 4 – x | + 1 < 3

⇒ | 4 – x | < 2

Using the property of modulus operator, we know

|x| < a ⇒ -a < x < a

⇒ -2 < 4 – x <  2 



⇒ -6 < -x < -2

⇒ 6 > x > 2

⇒ 2 < x < 6 

Hence, we can conclude x lies in the range (2, 6)

Question 3. Solve \left | \frac{3x - 4}{2} \right |  ≤ \frac{5}{12}

Solution:

Using the property of modulus operator, we know

|x| ≤ a ⇒ -a ≤ x ≤ a

⇒ – \frac{5}{12}  ≤ \frac{3x - 4}{2}  ≤ \frac{5}{12}  

⇒ – \frac{5}{6}  ≤ 3x – 4 ≤ \frac{5}{6}

⇒ – \frac{5}{6}  + 4 ≤ 3x ≤ \frac{5}{6}  + 4



⇒ \frac{19}{6}  ≤ 3x ≤ \frac{29}{6}

⇒ \frac{19}{18}  ≤ x ≤ \frac{29}{18}

Hence, we can conclude x lies in the range [ \frac{19}{18}  , \frac{29}{18}  ]

Question 4. Solve \frac{|x - 2|}{x - 2}  > 0

Solution:

Using the property of modulus operator, we have

| x -2 | = x-2 when x ≥ 2 or 2-x when x < 2

since, \frac{|x-2|}{x-2}  > 0 for x > 2

Hence, we can conclude x lies in the range ( 2, ∞)

Question 5. Solve \frac{1}{|x|-3} < \frac{1}{2}

Solution:

We are given, \frac{1}{|x|-3} < \frac{1}{2}



⇒ \frac{1}{|x|-3} - \frac{1}{2}  < 0

⇒ \frac{2 - (|x|-3)}{2(|x|-3)}  < 0

⇒ \frac{2 - |x|+ 3}{|x|-3}  < 0

⇒ \frac{5 - |x|}{|x|-3}  < 0

Now, we have two cases:

Case 1: When x ≥ 0, then |x| = x

\frac{5-x}{x-3}  < 0

⇒ (5 – x < 0 and x – 3 > 0) or ( 5 – x > 0 and x – 3 < 0)

⇒ (x >5 and x > 3) or ( x < 5 and x < 3)

⇒ x > 5 and x < 3



Therefore, we can conclude, from case 1 that x lies in the range [ 0,3) U (5,∞)  

Case 2: When x ≤ 0 then |x| = -x

\frac{5+x}{-x-3}  < 0

⇒ \frac{x+5}{x+3}  > 0

⇒ (x + 5 > 0 and x + 3 > 0) or ( x + 5 <0 and x + 3 < 0)

⇒ ( x > -5 and x > -3 ) or ( x < -5 and x < -3 )

⇒ x > -3 or x < -5

Therefore, we can conclude, from case 2 that x lies in the range [ -∞, -5) U (-3,0)

Now, taking the union of above two cases we can conclude x lies in the range (-∞, -5) U ( -3,3) U (5, ∞)

Question 6.  Solve \frac{|x + 2| - x}{x}  < 2

Solution:

We have \frac{|x + 2| - x}{x}  < 2

⇒ \frac{|x + 2| - x}{x}  – 2 < 0

⇒ \frac{|x + 2| - x -2x}{x}  < 0

⇒ \frac{|x + 2| - 3x}{x}  < 0

Now, we have two cases:

Case 1: When x ≥ -2, then |x+2| = x+2,

\frac{x + 2 - 3x}{x}  < 0

⇒ \frac{2-2x}{x}  < 0

⇒ \frac{-2(x-1)}{x}  < 0

⇒ \frac{x-1}{x}  > 0



⇒ (x – 1 > 0 and x > 0) or ( x – 1 < 0 and x < 0)

⇒ (x > 1 and x > 0) or ( x < 1 and x < 0)

⇒ x > 1 or x < 0

Therefore, we can conclude, from case 1 that x lies in the range [ -2,0) U (1,∞)

Case 1: When x ≤ -2, then |x+2| = -(x+2),

\frac{-(x + 2) - 3x}{x}  < 0

⇒ \frac{-x-2-3x}{x}  < 0

⇒ \frac{-2(2x+1)}{x}  < 0

⇒ \frac{2x+1}{x}  > 0

⇒ (2x -+ 1 > 0 and x > 0) or ( 2x + 1 < 0 and x < 0)

⇒ (x > -1/2 and x > 0) or ( x < -1/2 and x < 0)

⇒ x > 0 or x < -1/2

Therefore, we can conclude, from case 2 that x lies in the range ( -∞,-2 ] U (0,∞)

Now, taking the union of above two cases we can conclude x lies in the range [ -2,0 ) U (1,∞) U (-∞, -2] U ( 0, ∞) i.e., x belongs to (-∞,0) U (1,∞)

Question 7. Solve |\frac{2x-1}{x-1}|  > 2

Solution:

We have |\frac{2x-1}{x-1}|  > 2

⇒ |\frac{2x-1}{x-1}|  – 2 > 0

⇒ \frac{2x-1}{x-1}  > 0 or \frac{2x-1}{x-1}  +2 < 0

⇒ \frac{1}{x-1}  > 0 or \frac{4x-3}{x-1}  < 0

⇒ x-1 >0 or \frac{4x-3}{x-1}  < 0



⇒ x-1 > 0 or [ (4x-3 > 0 and x-1 < 0) or ( 4x-3 < 0 and x-1 > 0) ]

⇒ x > 1 or [ (x > 3/4  and x < 1) or ( x < 3/4 and x > 1) ]

⇒ x > 1 or [ 3/4 < x < 1 or ∅]

⇒ 3/4 < x < 1 or x > 1

Hence, we can conclude x lies in the range ( 3/4, 1) U (1, ∞ )

Question 8. Solve |x-1| + |x-2| + |x-3| ≥ 6

Solution:

We have, |x-1| + |x-2| + |x-3| ≥ 6 ———————let this be equation (1)

As,  | x-1 | = ( x-1, when x ≥ 1 and 1-x when x < 1 )

similarly, | x-2 | = ( x-2, when x ≥ 2 and 2-x when x < 2 )

and | x-3 | = ( x-3, when x ≥ 3 and 3-x when x < 3 )

Now, we have four cases:

Case 1: When x < 1

1 – x + 2 – x + 3 – x ≥ 6

⇒ 6 -3x ≥ 6

⇒ x ≤ 0

So, we see x lies in the range (-∞ ,0]

Case 2: when 1 ≤ x < 2

x – 1 + 2 – x + 3 – x ≥ 6

⇒ 4 – x ≥ 6

⇒ x ≤ -2

using case 2 , we see x has no values so x ∈ ∅

Case 3: when 2 ≤ x < 3

x – 1 + x – 2 + 3 – x ≥ 6

⇒ x  ≥ 6

using case 3 , we see x has no values so x ∈ ∅

Case 4: when x ≥ 3

x – 1 + x – 2 + x – 3 ≥ 6

⇒ 3x – 6 ≥ 6

⇒ x  ≥ 4

using case 4 , we see x has no values so x ∈ [ 4, ∞ )

Combining all the cases, we get to know x lies in the range ( -∞, 0 ] U [ 4, ∞)

Question 9. Solve \frac{|x-2|-1}{|x-2|-2}  ≤ 0 

Solution:

We have, \frac{|x-2|-1}{|x-2|-2}  ≤ 0 

Case 1: when x ≥ 2, then | x -2 | = x – 2

\frac{x-2-1}{x-2-2}  ≤ 0

⇒ \frac{x-3}{x-4}  ≤ 0

⇒ ( x – 3 ≤ 0 and x – 4 > 0 ) or ( x – 3 ≥ 0 and x – 4 < 0)

⇒ ( x ≤ 3 and x > 4 ) or ( x ≥ 3 and x < 4)

⇒ ∅ or ( 3 ≤ x < 4)

⇒ 3 ≤ x < 4



using case 1 , we see x lies in the range [ 3, 4 )

Case 2: when x ≤ 2, then | x – 2| = 2 – x,

\frac{2-x-1}{2-x-2}  ≤ 0

⇒ \frac{1-x}{-x}  ≤ 0

⇒ \frac{x-1}{x}  ≤ 0

⇒  ( x – 1 ≤ 0 and x > 0 ) or ( x – 1 ≥ 0 and x< 0)

⇒  ( x ≤ 1 and x > 0 ) or ( x ≥ 1 and x< 0)

⇒ ( 0 < x ≤ 1) or ∅

⇒ 0 < x ≤ 1

using case 2 , we see x lies in the range ( 0, 1 ]

Combining all the cases, we get to know x lies in the range ( 0, 1 ] U [ 3, 4)

Question 10. Solve \frac{1}{|x|-3} \leq \frac{1}{2}

Solution:

We have, \frac{1}{|x|-3} \leq \frac{1}{2}

⇒ \frac{1}{|x|-3} - \frac{1}{2}  ≤ 0

⇒ \frac{2 - (|x-3|)}{2(|x-3|)}  ≤ 0

⇒ \frac{5 - |x|}{|x|-3}  ≤ 0

Case 1: when x ≥ 0 then |x| =x

⇒ \frac{5-x}{x-3}  ≤ 0

⇒ ( 5 – x ≤ 0 and x – 3 > 0 ) or ( 5 – x ≥ 0 and x – 3 < 0)

⇒  ( x ≥ 5 and x > 3 ) or ( x ≤ 5 and x < 3)



⇒ x ≥ 5 or x < 3

using case 1, we see x lies in the range ( 0, 3 ) U [5, ∞ )

Case 2: when x < 0 then |x| = -x

⇒ \frac{5+x}{-x-3}  ≤ 0

⇒ \frac{x+5}{x+3}  ≥ 0

⇒ ( x + 5 > 0 and x + 3 > 0 ) or ( x + 5 < 0 and x + 3 < 0)

⇒  ( x > -5 and x > -3 ) or ( x < -5 and x < -3)

⇒ x > -3 or x < -5

using case 2, we see x lies in the range ( -∞, -5 ) U (-3, ∞ )

Combining both the cases, we get to know x lies in the range ( -∞, -5 ) U (-3, ∞ ) U ( 0, 3 ) U [5, ∞ )

Question 11. Solve |x + 1| + |x| > 3

Solution:

We have, |x + 1| + |x| > 3

⇒ |x + 1| = ( x + 1 when x ≥ -1 and -(x + 1) when x < -1 )

similarly, |x| = (x when x ≥ 0 and -x when x < 0)

Case 1: When x < -1

|x + 1| + |x| > 3

⇒ – (x+1) -x > 3

⇒ -2x -1 > 3

⇒ x < -2

using case `1, we see x lies in the range ( -∞, -2 )

Case 2: When -1 ≤ x < 0

|x + 1| + |x| > 3

⇒  (x+1) + x > 3

⇒ 2x > 2

⇒ x > 1

using case `2, we see x lies in the range ( 1, ∞ )

Combining both the cases, we get to know x lies in the range ( -∞, -2 ) U ( 1, ∞ )

Question 12. Solve 1 ≤ |x – 2| ≤ 3

Solution: 

We have, 1 ≤ |x – 2| ≤ 3

Case 1: |x – 2| ≥ 1

⇒ ((x – 2) ≤ -1 or (x-2) ≥ 1) 

⇒ ( x ≤ 1 or x ≥ 3)

using case `1, we see x lies in the range ( -∞,1 ] U [3,∞) 

Case 2:  |x – 2| ≤ 3

⇒  ( -3 ≤ (x-2) ≤ 3)

⇒ (-1 ≤ x ≤ 5)

using case `2, we see x lies in the range [-1, 5]

Combining both the cases, we get to know x lies in the range [-1, 1 ] U [ 3, 5 ]

Question 13. Solve |3-4x| ≥ 9

Solution:

We have, |3-4x| ≥ 9

therefore, by using property of modulus we know, |x| ≥ a ⇒ x ≤ -a or x ≥ a

⇒ (3-4x) ≤ -9 or (3-4x) ≥ 9

⇒ -4x ≤ -12 or  -4x ≥ 6

⇒ x ≥ 3 or  x ≤ -3/2

Hence, we can conclude x lies in the range ( -∞, -3/2] U [ 3, ∞ )

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :