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

### Question 1. Solve

Solution:

Using the property of modulus operator, we know

|x| > a â‡’ x < -a or x > a

Therefore,  or

â‡’ x <  or x >

â‡’ x < -3 or x >

Hence, we can conclude x lies in the range ( -âˆž, -3) âˆª ( ,âˆž )

### Question 2. Solve | 4 – x | + 1 < 3

Solution:

We have,  | 4 – x | + 1 < 3

â‡’ | 4 – x | < 2

Using the property of modulus operator, we know

|x| < a â‡’ -a < x < a

â‡’ -2 < 4 – x <  2

â‡’ -6 < -x < -2

â‡’ 6 > x > 2

â‡’ 2 < x < 6

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

### Question 3. Solve  â‰¤

Solution:

Using the property of modulus operator, we know

|x| â‰¤ a â‡’ -a â‰¤ x â‰¤ a

â‡’ –  â‰¤  â‰¤

â‡’ –  â‰¤ 3x – 4 â‰¤

â‡’ –  + 4 â‰¤ 3x â‰¤  + 4

â‡’  â‰¤ 3x â‰¤

â‡’  â‰¤ x â‰¤

Hence, we can conclude x lies in the range [  ,  ]

### Question 4. Solve  > 0

Solution:

Using the property of modulus operator, we have

| x -2 | = x-2 when x â‰¥ 2 or 2-x when x < 2

since,  > 0 for x > 2

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

### Question 5. Solve

Solution:

We are given,

â‡’  < 0

â‡’  < 0

â‡’  < 0

â‡’  < 0

Now, we have two cases:

Case 1: When x â‰¥ 0, then |x| = x

< 0

â‡’ (5 – x < 0 and x – 3 > 0) or ( 5 – x > 0 and x – 3 < 0)

â‡’ (x >5 and x > 3) or ( x < 5 and x < 3)

â‡’ x > 5 and x < 3

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

Case 2: When x â‰¤ 0 then |x| = -x

< 0

â‡’  > 0

â‡’ (x + 5 > 0 and x + 3 > 0) or ( x + 5 <0 and x + 3 < 0)

â‡’ ( x > -5 and x > -3 ) or ( x < -5 and x < -3 )

â‡’ x > -3 or x < -5

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

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

### Question 6.  Solve  < 2

Solution:

We have  < 2

â‡’  – 2 < 0

â‡’  < 0

â‡’  < 0

Now, we have two cases:

Case 1: When x â‰¥ -2, then |x+2| = x+2,

< 0

â‡’  < 0

â‡’  < 0

â‡’  > 0

â‡’ (x – 1 > 0 and x > 0) or ( x – 1 < 0 and x < 0)

â‡’ (x > 1 and x > 0) or ( x < 1 and x < 0)

â‡’ x > 1 or x < 0

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

Case 1: When x â‰¤ -2, then |x+2| = -(x+2),

< 0

â‡’  < 0

â‡’  < 0

â‡’  > 0

â‡’ (2x -+ 1 > 0 and x > 0) or ( 2x + 1 < 0 and x < 0)

â‡’ (x > -1/2 and x > 0) or ( x < -1/2 and x < 0)

â‡’ x > 0 or x < -1/2

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

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

### Question 7. Solve  > 2

Solution:

We have  > 2

â‡’  – 2 > 0

â‡’  > 0 or  +2 < 0

â‡’  > 0 or  < 0

â‡’ x-1 >0 or  < 0

â‡’ x-1 > 0 or [ (4x-3 > 0 and x-1 < 0) or ( 4x-3 < 0 and x-1 > 0) ]

â‡’ x > 1 or [ (x > 3/4  and x < 1) or ( x < 3/4 and x > 1) ]

â‡’ x > 1 or [ 3/4 < x < 1 or âˆ…]

â‡’ 3/4 < x < 1 or x > 1

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

### Question 8. Solve |x-1| + |x-2| + |x-3| â‰¥ 6

Solution:

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

As,  | x-1 | = ( x-1, when x â‰¥ 1 and 1-x when x < 1 )

similarly, | x-2 | = ( x-2, when x â‰¥ 2 and 2-x when x < 2 )

and | x-3 | = ( x-3, when x â‰¥ 3 and 3-x when x < 3 )

Now, we have four cases:

Case 1: When x < 1

1 – x + 2 – x + 3 – x â‰¥ 6

â‡’ 6 -3x â‰¥ 6

â‡’ x â‰¤ 0

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

Case 2: when 1 â‰¤ x < 2

x – 1 + 2 – x + 3 – x â‰¥ 6

â‡’ 4 – x â‰¥ 6

â‡’ x â‰¤ -2

using case 2 , we see x has no values so x âˆˆ âˆ…

Case 3: when 2 â‰¤ x < 3

x – 1 + x – 2 + 3 – x â‰¥ 6

â‡’ x  â‰¥ 6

using case 3 , we see x has no values so x âˆˆ âˆ…

Case 4: when x â‰¥ 3

x – 1 + x – 2 + x – 3 â‰¥ 6

â‡’ 3x – 6 â‰¥ 6

â‡’ x  â‰¥ 4

using case 4 , we see x has no values so x âˆˆ [ 4, âˆž )

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

### Question 9. Solve  â‰¤ 0

Solution:

We have,  â‰¤ 0

Case 1: when x â‰¥ 2, then | x -2 | = x – 2

â‰¤ 0

â‡’  â‰¤ 0

â‡’ ( x – 3 â‰¤ 0 and x – 4 > 0 ) or ( x – 3 â‰¥ 0 and x – 4 < 0)

â‡’ ( x â‰¤ 3 and x > 4 ) or ( x â‰¥ 3 and x < 4)

â‡’ âˆ… or ( 3 â‰¤ x < 4)

â‡’ 3 â‰¤ x < 4

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

Case 2: when x â‰¤ 2, then | x – 2| = 2 – x,

â‰¤ 0

â‡’  â‰¤ 0

â‡’  â‰¤ 0

â‡’  ( x – 1 â‰¤ 0 and x > 0 ) or ( x – 1 â‰¥ 0 and x< 0)

â‡’  ( x â‰¤ 1 and x > 0 ) or ( x â‰¥ 1 and x< 0)

â‡’ ( 0 < x â‰¤ 1) or âˆ…

â‡’ 0 < x â‰¤ 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

Solution:

We have,

â‡’  â‰¤ 0

â‡’  â‰¤ 0

â‡’  â‰¤ 0

Case 1: when x â‰¥ 0 then |x| =x

â‡’  â‰¤ 0

â‡’ ( 5 – x â‰¤ 0 and x – 3 > 0 ) or ( 5 – x â‰¥ 0 and x – 3 < 0)

â‡’  ( x â‰¥ 5 and x > 3 ) or ( x â‰¤ 5 and x < 3)

â‡’ x â‰¥ 5 or x < 3

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

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

â‡’  â‰¤ 0

â‡’  â‰¥ 0

â‡’ ( x + 5 > 0 and x + 3 > 0 ) or ( x + 5 < 0 and x + 3 < 0)

â‡’  ( x > -5 and x > -3 ) or ( x < -5 and x < -3)

â‡’ x > -3 or x < -5

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

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

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

Solution:

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

â‡’ |x + 1| = ( x + 1 when x â‰¥ -1 and -(x + 1) when x < -1 )

similarly, |x| = (x when x â‰¥ 0 and -x when x < 0)

Case 1: When x < -1

|x + 1| + |x| > 3

â‡’ – (x+1) -x > 3

â‡’ -2x -1 > 3

â‡’ x < -2

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

Case 2: When -1 â‰¤ x < 0

|x + 1| + |x| > 3

â‡’  (x+1) + x > 3

â‡’ 2x > 2

â‡’ x > 1

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

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

### Question 12. Solve 1 â‰¤ |x – 2| â‰¤ 3

Solution:

We have, 1 â‰¤ |x – 2| â‰¤ 3

Case 1: |x – 2| â‰¥ 1

â‡’ ((x – 2) â‰¤ -1 or (x-2) â‰¥ 1)

â‡’ ( x â‰¤ 1 or x â‰¥ 3)

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

Case 2:  |x – 2| â‰¤ 3

â‡’  ( -3 â‰¤ (x-2) â‰¤ 3)

â‡’ (-1 â‰¤ x â‰¤ 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| â‰¥ 9

Solution:

We have, |3-4x| â‰¥ 9

therefore, by using property of modulus we know, |x| â‰¥ a â‡’ x â‰¤ -a or x â‰¥ a

â‡’ (3-4x) â‰¤ -9 or (3-4x) â‰¥ 9

â‡’ -4x â‰¤ -12 or  -4x â‰¥ 6

â‡’ x â‰¥ 3 or  x â‰¤ -3/2

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

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