Inequalities Class 11 OP Malhotra Exe-11B ISC Maths Solutions Ch-11 Solutions. In this article you would learn about Inequations Involving Absolute Values. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Inequalities Class 11 OP Malhotra Exe-11B ISC Maths Solutions Ch-11
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-11 | Inequalities |
| Writer | O.P. Malhotra |
| Exe-11(B) | Inequations Involving Absolute Values. |
Exercise- 11B
Inequalities Class 11 OP Malhotra Exe-11B Solution.
Que-1: Solve and graph the solution set.
(i) |x-3| < 4 (ii) |x-3| > 4 (iii) |x-3| ≥ 4 (iv) |1-3x| < 4 (v) 2+3|2y-1| > 8
Sol: (i) |x-3| < 4
⇒ −4 < x−3 < 4
−4+3 < x < 4+3
⇒ −1 < x < 7
{x : −1 < x < 7}
(ii) |x-3| > 4
⇒ x−3 < −4 or x−3 > 4
x < −1 or x > 7
{x : x < −1 or x > 7}
(iii) |x-3| ≥ 4
x−3 ≤ −4 or x−3 ≥ 4
⇒ x ≤ −1 or x ≥ 7
{x : x ≤ −1 or x ≥ 7}
(iv) |1-3x| < 4
−4 < 1−3x < 4
−5 < −3x < 3
5/3 > x > −1
⇒ −1 < x < 5/3
{x : −1 < x < 5/3}
(v) 2+3|2y-1| > 8
3 ∣2y−1∣ > 6
⇒ ∣2y−1∣ > 2
= 2y-1 < -2 or 2y-1 > 2
= y < -1/2 or y > 3/2
{y : y > 3/2 or y < -1/2}
Que-2: 3|x-6|-4 ≤ 11
Sol: 3|x-6|-4 ≤ 11
3 ∣x−6∣ ≤ 11+4 = 15
|x-6| ≤ 15/3 = 5
∣x−6∣ ≤ 5
⇒ −5 ≤ x−6 ≤ 5
−5+6 ≤ x ≤ 5+6
⇒ 1 ≤ x ≤ 11
Que-3: 2|3p-5|+1 > 7
Sol: 2∣3p−5∣+1 > 7
2 ∣3p−5∣ > 7−1 = 6
∣3p−5∣ > 6/2 = 3
∣A∣ > B ⇒ A < −B or A > B
3p−5 < −3 or 3p−5 > 3
3p−5 < −3
⇒ 3p < 2 ⇒ p < 2/3
3p−5 > 3
⇒ 3p > 8 ⇒ p > 8/3
{p : p < 2/3 or p > 8/3}
Que-4: |5-(m+3)|+8 < 15
Sol: Given : ∣5−(m+3)∣ + 8 < 15
∣5−m−3∣ + 8 < 15
⇒ ∣2−m∣ + 8 < 15
∣2−m∣ < 15−8 = 7
∣2−m∣ < 7
⇒ −7 < 2−m < 7
−9 < −m < 5
9 > m > −5
⇒ −5 < m < 9
1 < m < 15.
Que-5: 1/[|x|-3] ≤ 1/2
Sol: Given : 1/[|x|-3] ≤ 1/2
∣x∣−3 ≠ 0 ⇒ ∣x∣ ≠ 3 ⇒ x ≠ 3, −3
|x|-3 ≠ 0 ⇒ |x| ≠ 3
∣x∣ > 3 ⇒ x < −3 or x > 3
1/[|x|-3] ≤ 1/2
1 ≤ (1/2) (∣x∣−3)
2 ≤ ∣x∣−3 ⇒ ∣x∣ ≥ 5
∣x∣ ≥ 5 ⇒ x ≤ −5 or x ≥ 5
(−∞,−5) ∪ (-3,3) ∪ [5,∞)
Que-6: 1/[|x|-1] ≥ 1/2
Sol: Given : 1/[|x|-1] ≥ 1/2
|x| – 1 ≠ 0 = |x| ≠ 1
1 ≥ {(∣x∣−1)/2}
⇒ 2 ≥ ∣x∣−1 ⇒ ∣x∣ ≤ 3
1 < ∣x∣ ≤ 3
[−3,−1) ∪ (1,3]
Que-7: [|x-2|-1]/[|x-2|-2] ≤ 0
Sol: [|x-2|-1]/[|x-2|-2] ≤ 0
Let y = |x-2|
{(y-1)/(y-2)} ≤ 0
y ∈ [1,2)
Substitute back y = |x-2|
1 ≤ ∣x−2∣ < 2
1 ≤ ∣x−2∣ < 2 ⇒ This is a compound absolute value inequality.
⇒ −2 < x−2 ≤ −1 or 1 ≤ x−2 < 2
Solving both
−2 < x−2 ≤ −1 ⇒ 0 < x ≤ 1
1 ≤ x−2 < 2 ⇒ 3 ≤ x < 4
(0,1] ∪ [3,4)
Que-8: |x-1| ≤ 5, |x| ≥ 2
Sol: Given : Inequality 1 : |x-1| ≤ 5
−5 ≤ x−1 ≤ 5
−4 ≤ x ≤ 6
x ∈ [−4,6]
Inequality 2 : ∣x∣ ≥ 2
x ≤ −2 or x ≥ 2
x ∈ (−∞,−2] ∪ [2,∞)
From Inequality 1 : x ∈ [−4,6]
From Inequality 2 : x ∈ (−∞,−2] ∪ [2,∞)
[−4,6] ∩ (−∞,−2] = [−4,−2]
[−4,6] ∩ [2,∞) = [2,6]
(−4,−2] ∪ [2,6]
–: End of Inequalities Class 11 OP Malhotra Exe-11B ISC Maths Ch-11 Solutions. :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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