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

In this sub-section, we shall discuss some results on inequations involving modulus of the variable. We state and prove these results as theorems.

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Solve each of the following system of equations in R.

1. |x + 1/3| > 8/3

Solution:

Let ‘r’ be a positive real number and ‘a’ be a fixed real number. Then,

|x + a| > r ⟺ x > r – a or x < – (a + r)

Here, a = 1/3 and r = 8/3

x > 8/3 – 1/3 or x < – (8/3 + 1/3)

x > (8-1)/3 or x < – (8+1)/3

x > 7/3 or x < – 9/3

x > 7/3 or x < – 3

x ∈ (7/3, ∞) or x ∈ (–∞, -3)

∴ x ∈ (–∞, -3) ∪ (7/3, ∞)

2. |4 – x| + 1 < 3

Solution:

|4 – x | + 1 < 3

Let us subtract 1 from both the sides, we get

|4 – x| + 1 – 1 < 3 – 1

|4 – x| < 2

Let ‘r’ be a positive real number and ‘a’ be a fixed real number. Then,

|a – x| < r ⟺ a – r < x < a + r

Here, a=4 and r=2

4 – 2 < x < 4 + 2

2 < x < 6

∴ x ∈ (2, 6)

3. | (3x – 4)/2 | ≤ 5/12

Solution:

Given:

| (3x – 4)/2 | ≤ 5/12

We can rewrite it as

| 3x/2 – 4/2 | ≤ 5/12

| 3x/2 – 2 | ≤ 5/12

Let ‘r’ be a positive real number and ‘a’ be a fixed real number. Then,

|x – a| ≤ r ⟺ a – r ≤ x ≤ a + r

Here, a = 2 and r = 5/12

2 – 5/12 ≤ 3x/2 ≤ 2 + 5/12

(24-5)/12 ≤ 3x/2 ≤ (24+5)/12

19/12 ≤ 3x/2 ≤ 29/12

Now, multiplying the whole inequality by 2 and dividing by 3, we get

19/18 ≤ x ≤ 29/18

∴ x ∈ [19/18, 29/18]

4. |x – 2| / (x – 2) > 0

Solution:

Given:

|x – 2| / (x – 2) > 0

Clearly it states, x≠2 so two case arise:

Case1: x–2>0

x>2

In this case |x–2| = x – 2

x ϵ (2, ∞)….(1)

Case 2: x–2<0

x<2

In this case, |x–2|= – (x–2)

– (x – 2) / (x – 2) > 0

– 1 > 0

Inequality doesn’t get satisfy

This case gets nullified.

∴ x ∈ (2, ∞) from (1)

5. 1 / (|x| – 3) < ½

Solution:

We know that, if we take reciprocal of any inequality we need to change the inequality as well.

Also, |x| – 3 ≠ 0

|x| > 3 or |x| < 3

For |x| < 3

–3 < x < 3

x ∈ (–3, 3) …. (1)

The equation can be re–written as

|x| – 3 > 2

Let us add 3 on both the sides, we get

|x| – 3 + 3 > 2 + 3

|x| > 5

Let ‘a’ be a fixed real number. Then,

|x | > a ⟺ x < –a or x > a

Here, a = 5

x < –5 or x > 5 …. (2)

From (1) and (2)

x ∈ (–∞,–5 ) or x ∈ (5, ∞)

∴ x ∈ (–∞,–5 ) ⋃ (–3, 3) ⋃ (5, ∞)

6. (|x + 2| – x) / x < 2

Solution:

Given:

(|x + 2| – x) / x < 2

Let us rewrite the equation as

|x + 2|/x – x/x < 2

|x + 2|/x – 1 < 2

By adding 1 on both sides, we get

|x + 2|/x – 1 + 1 < 2 + 1

|x + 2|/x < 3

By subtracting 3 on both sides, we get

|x + 2|/x – 3 < 3 – 3

|x + 2|/x – 3 < 0

Clearly it states, x≠2 so two case arise:

Case1: x + 2 > 0

x > –2

In this case |x+2| = x + 2

x + 2/x – 3 < 0

(x + 2 – 3x)/x < 0

– (2x – 2)/x < 0

(2x – 2)/x < 0

Let us consider only the numerators, we get

2x – 2 > 0

x>1

x ϵ (1, ∞) ….(1)

Case 2: x + 2 < 0

x < –2

In this case, |x+2| = – (x + 2)

-(x+2)/x – 3 < 0

(-x – 2 – 3x)/x < 0

– (4x + 2)/x < 0

(4x + 2)/x < 0

Let us consider only the numerators, we get

4x + 2 > 0

x > – ½

But x < -2

From the denominator we have,

x ∈ (–∞ , 0) …(2)

From (1) and (2)

∴ x ∈ (–∞ , 0) ⋃ (1, ∞)