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RD Chapter 15- Linear Inequations Ex-15.3 Interview Questions Answers

Question 1 : Solve each of the following system of equations in R.
|x + 1/3| > 8/3

Answer 1 :

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, ∞)

Question 2 : |4 – x| + 1 < 3

Answer 2 :

|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)

Question 3 : | (3x – 4)/2 | ≤ 5/12

Answer 3 :

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]

Question 4 : |x – 2| / (x – 2) > 0

Answer 4 :

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)

Question 5 : 1 / (|x| – 3) < ½

Answer 5 :

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, ∞)

Question 6 : (|x + 2| – x) / x < 2

Answer 6 :

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, ∞)


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