Class 11 RD Sharma Solutions – Chapter 15 Linear Inequations – Exercise 15.1 | Set 2
Question 15. Solve:<− 5 in R.
Solution:
Given:<− 5
⇒ <
⇒ 6(5−2x) < 3(x−30)
⇒ 30 − 12x < 3x − 90
⇒ 15x > 120
⇒ x > 8
Thus, the solution set is (8, ∞).
Question 16. Solve:≥− 3.
Solution:
Given:≥− 3.
⇒≥
⇒ 2(4+2x) ≥ 3(x−60)
⇒ 8 + 4x ≥ 3x − 180
⇒ x ≥ −26
Thus, the solution set is [−26, ∞).
Question 17. Solve:− 2 <.
Solution:
Given:− 2 <
⇒<
⇒ 2x + 3 − 10 < 3x − 6
⇒ x > −1
Thus, the solution set is (−1, ∞).
Question 18. Solve: x−2 ≤
Solution:
Given: x−2 ≤
⇒ 3(x−2) ≤ 5x+8
⇒ 3x − 6 ≤ 5x + 8
⇒ 2x ≥ −14
⇒ x ≥ −7
Thus, the solution set is [−7, ∞).
Question 19. Solve:< 0.
Solution:
Given:< 0.
Case I: When 6x − 5 > 0 and 4x +1 < 0
⇒ x > 5/6 and x < −1/4, which is clearly impossible.
Case II: When 6x − 5 < 0 and 4x +1 > 0
⇒ x < 5/6 and x > −1/4
Thus, the solution set is (−1/4, 5/6).
Question 20. Solve:> 0.
Solution:
Given:> 0.
Case I: When 2x−3 > 0 and 3x−7 > 0
⇒ x > 3/2 and x > 7/3
⇒ x > 7/3 ….(a)
Case II: When 2x−3 < 0 and 3x−7 < 0
⇒ x < 3/2 and x < 7/3
⇒ x < 3/2 ….(b)
From (a) and (b), we get:
The solution set is (− ∞, 3/2)∪ (7/3, ∞).
Question 21. Solve:< 1.
Solution:
Given:< 1
⇒−1 < 0
⇒< 0
⇒> 0
Case I: When x−5 > 0 and x−2 > 0
⇒ x > 5 and x > 2
⇒ x > 5 ….(a)
Case II: When x−5 < 0 and x−2 < 0
⇒ x < 5 and x < 2
⇒ x < 2 ….(b)
From (a) and (b), we get:
The solution set is (− ∞, 2)∪ (5, ∞).
Question 22. Solve:≤ 2.
Solution:
Given:≤ 2
⇒− 2 ≤ 0
⇒≤ 0
⇒≤ 0
Case I: When 3−2x ≥ 0 and x−1 < 0
⇒ x ≥ 3/2 and x < 1
⇒ x < 1 …..(a)
Case II: 3−2x ≤ 0 and x−1 > 0
⇒ x ≥ 3/2 and x > 1
⇒ x ≥ 3/2 ….(b)
From (a) and (b), we get:
The solution set is (− ∞, 1)∪ (3/2, ∞).
Question 23. Solve:< 6
Solution:
Given:< 6
⇒−6 < 0
⇒< 0
⇒< 0
Case I: When 8x−33 > 0 and 2x−5 > 0
⇒ x > 33/8 and x > 5/2
⇒ x > 33/8 ….(a)
Case II: When 8x−33 < 0 and 2x−5 < 0
⇒ x < 33/8 and x <5/2
⇒ x < 5/2 ….(b)
From (a) and (b), we get:
The solution set is (− ∞, 5/2)∪ (33/8, ∞).
Question 24. Solve:< 1.
Solution:
Given:< 1
⇒− 1 < 0
⇒< 0
⇒< 0
Case I: When 4x−12 > 0 and x+6 < 0
⇒ x > −3 and x < −6, which is clearly not possible.
Case II: When 4x−12 < 0 and x+6 > 0
⇒ x < −3 and x > −6
The solution set is (− 3, 6).
Question 25. Solve:< 2.
Solution:
Given:< 2
⇒− 2 < 0
⇒< 0
⇒< 0
Case I: When 7x > 0 and 4−x < 0
⇒ x > 0 and x > 4
⇒ x > 4 ….(a)
Case II: When 7x < 0 and 4−x > 0
⇒ x < 0 and x > 4
⇒ x < 0 ….(b)
From (a) and (b), we get:
The solution set is (− ∞, 0)∪ (4, ∞).
Question 26. Solve:> 2.
Solution:
Given:> 2.
⇒− 2 > 0
⇒> 0
⇒< 0
Case I: When x+7 > 0 and x+3 < 0
⇒ x > −7 and x < −3
Case II: When x+7 < 0 and x+3 > 0
⇒ x < −7 and x > −3, which is clearly not possible.
The solution set is (−7, −3).
Question 27. Solve:> 4.
Solution:
Given:> 4
⇒− 4 > 0
⇒> 0
⇒> 0
⇒< 0
Case I: When 25x+17 > 0 and 8x+3 < 0
⇒ x > −17/25 and x < −3/8
Case II: When 25x+17 < 0 and 8x+3 > 0
⇒ x < −17/25 and x > −3/8, which is not clearly possible.
Hence the solution set is (−17/25, −3/8).
Question 28. Solve:> 1/2.
Solution:
Given:> 1/2.
⇒− 1/2 > 0
⇒> 0
Case I: When x+5 > 0 and 2x−10 > 0
⇒ x > −5 and x > 5
⇒ x > 5 ….(a)
Case II: When x+5 < 0 and 2x−10 < 0
⇒ x < −5 and x < 5
⇒ x < −5 ….(b)
From (a) and (b), we get:
The solution set is (− ∞, −5)∪ (5, ∞).
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