Well-structured OP Malhotra Maths Class 12 Solutions Chapter 11 Inequalities Ex 11(a) facilitate a deeper understanding of mathematical principles.
S Chand Class 11 ICSE Maths Solutions Chapter 11 Inequalities Ex 11(a)
Question 1.
You are given the following numbers :
-2.6 5.1 -3 0.4 1.2 -3.1 4.7
Fill in the blanks.
(i) A = {x : x ≥ – 3} = {…}
(ii) B = {x : x ≤ 1} = {…}
Solution:
(i) A = {x : x ≥ – 3}
= {-3, -2.6, 0.4, 1.2, 4.7, 5.1}
(ii) B = {x : x ≤ 1} = {-2.6, -3, 0.4, -3.1}
Question 2.
If the replacement set is {- 2, -1, +1, + 2, + 4, + 5, + 9}, what is the solution set of each of the following mathematical sentences?
(i) x + \(\frac { 3 }{ 2 }\) > \(\frac { 5 }{ 2 }\)
(ii) 2x – 5 ≥ 10
(iii) 3y + 2 ≤ \(\frac { 5 }{ 2 }\)
Solution:
Given replacement set is {-2, -1, 1, 2, 4, 5, 9}
(i) x + \(\frac { 3 }{ 2 }\) > \(\frac { 5 }{ 2 }\) ⇒ x + \(\frac { 3 }{ 2 }\) – \(\frac { 3 }{ 2 }\) > \(\frac { 5 }{ 2 }\) – \(\frac { 3 }{ 2 }\) ⇒ x > 1
∴ x = 2, 4, 5, 9
∴ Solution set = {2, 4, 5, 9}
(ii) 2x – 5 ≥ 10 ⇒ 2x ≥ 15 ⇒ x ≥ \(\frac { 15 }{ 2 }\) = 7\(\frac { 1 }{ 2 }\)
but replacement set is {- 2, -1, + 1, 2, 4, 5, 9}
∴ x = 9
Hence solution set = {9}
(iii) Given 3y + 2 ⇒ 3y ≤ \(\frac { 5 }{ 2 }\) – 2
⇒ 3y ≤ \(\frac { 1 }{ 2 }\) ⇒ y ≤ \(\frac { 1 }{ 6 }\)
But replacement set is {- 2, – 1, 1, 2, 4, 5, 9}
∴ x = -2, -1
Thus solution set = {- 2, – 1}
Question 3.
List the solution set of 30 – 4 (2x – 1) < 30, given that x is a positive integer.
Solution:
Given 30 – 4 (2x – 1) < 30
⇒ – 4 (2x – 1) < 0 ⇒ (2x – 1) > 0
[if a < b => – a > – b]
⇒ 2x > 1 ⇒ x > \(\frac { 1 }{ 2 }\) but x ∈ N
∴ x = 1, 2, 3
Thus solution set = (1, 2, 3, ……}
Question 4.
(i) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}.
Form all ordered pairs (x,y) such that x is a factor of y and x < y.
(ii) Find the truth set of the inequality x > y + 2 where (x, y) ∈ {(1, 2), (2,3), (5,1), (7,3), (5, 6), (6, 5)}.
Solution:
Given x ∈ {2, 4, 6, 9}
and y ∈ {4, 6, 18, 27, 54}
Now 2 be a factor of 4, 6, 18, 54
We want to find those ordered pair (x, y)
s.t x is a factor of y and x < y and Let A be its solution set
∴ (2, 4), (2, 6), (2, 18), (2, 54) ∈ A. 4 be a factor of 4 but 4 ≮ 4
∴ (4, 4) ∉ A 6 be a factor of 18, 54
∴ (6, 18), (6,54) ∈ A 9 be a factor of 18, 27, 54
∴ (9, 18), (9, 27), (9,54) ∈ A
Thus solution set A = {(2,4), (2,6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)}
(ii) Here x > y + 2, Let A be the truth set
Now 5 > 3 = 1 + 2 ∴ (5, 1) ∈ A
7 > 5 = 3 + 2 ∴ (7, 3) ∈ A
∴ Truth set = {(5, 1), (7, 3)}
Question 5.
P is the solution set of 8x – 1 > 5x + 2 and Q is the solution set of 7x – 2 ≥ 3 (x + 6), where x ∈ N. Find the set P ∩ Q.
Solution:
Given 8x – 1 > 5x + 2
⇒ 8x – 5x > 1 + 2
⇒ 3x > 3 ⇒ x > 1
Also x ∈ N
∴ Solution set P = {2, 3, 4, 5, …..}
Given 7x – 2 ≥ 3 (x + 6)
⇒ 7x – 2 ≥ 3x + 18
⇒ 7x – 3x ≥ 18 + 2
⇒ 4x ≥ 20 ⇒ x ≥ 5, x ∈ N
Solution set Q = {x ≥ 5, x ∈ N}
= {5, 6, 7, 8,….}
Thus P ∩ Q = {5, 6, 7, 8, }
Question 6.
Find the solution of the inequality
2 ≤ 2p – 3 ≤ 5, p ∈ R Hence, graph the solution set on the number line.
Solution:
Given 2 ≤ 2p – 3 ≤ 5
⇒ 2 + 3 ≤ 2p – 3 + 3 ≤ 5 + 3
⇒ 5 ≤ 2p ≤ 8 ⇒ \(\frac { 5 }{ 2 }\) ≤ p ≤ 4
∴ solution set = {\(\frac { 5 }{ 2 }\) ≤ p ≤ 4, p ∈ R}
= [\(\frac { 5 }{ 2 }\), 4]
Solution on real line is represented by dark line between \(\frac { 5 }{ 2 }\) and 4.
Further, there are dark dots at \(\frac { 5 }{ 2 }\) and 4 indicate that both points are included in solution set.
Question 7.
If x is a negative integer, find the solution set of \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) (x + 1) > 0.
Solution:
Given, \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 3 }\) (x +1) > 0
⇒ \(\frac { 2 }{ 3 }\) + \(\frac { x }{ 3 }\) + \(\frac { 1 }{ 3 }\) > 0
⇒ \(\frac { x }{ 3 }\) + 1 > 0 ⇒ \(\frac { x }{ 3 }\) > – 1 ⇒ x > – 3
but x is a negative integer
∴ x = – 2, – 1
Thus solution set = {- 1, – 2}
Question 8.
Solve the inequality: 3 – 2x ≥ x – 12, given that x ∈ N.
Solution:
Given 3 – 2x ≥ x – 12
⇒ 3 + 12 ≥ 2x + x ⇒ 15 ≥ 3x ⇒ 3x ≤ 15
⇒ x ≤ 5, Also x ∈ N
∴ x = 1, 2, 3, 4, 5
Thus solution set = {1, 2, 3, 4, 5}
Question 9.
Find the range of values of x which satisfies – 2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < 9\(\frac { 1 }{ 3 }\), x ∈ R. Graph these values of x on the number line.
Solution:
Given -2\(\frac { 2 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < 9\(\frac { 1 }{ 3 }\)
⇒ –\(\frac { 8 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) < \(\frac { 28 }{ 3 }\)
⇒ –\(\frac { 8 }{ 3 }\) – \(\frac { 1 }{ 3 }\) ≤ x + \(\frac { 1 }{ 3 }\) – \(\frac { 1 }{ 3 }\) < \(\frac { 28 }{ 3 }\) – \(\frac { 1 }{ 3 }\)
⇒ -3 ≤ x < 9, but x ∈ R
∴ Solution set = [- 3, 9)
The solution set represented on real line is given as under. There is a dark dot at x = – 3 and hollow dot at 9 indicate that – 3 included in solution set and 9 is not included in solution set.
Question 10.
A = {x: – 1 < x ≤ 5, x ∈ R},
B = {x : -4 ≤ x < 3, x ∈ R}.
Represent (i) A ∩ B (ii) A’ ∩ B, on different number lines.
Solution:
Given A = {x : – 1 < x ≤ 5, x ∈ R} = (- 1, 5]
and B = {x : – 4 ≤ x < 3, x ∈ R} = [- 4, 3)
∴ A ∩ B = {x : – 1 < x < 3, x ∈ R}
Now A’ = {x : x ≤ – 1 or x > 5, x ∈ R}
= (- ∞, – 1) ∪ (5, ∞)
Thus A’ ∩ B = {x : – 4 ≤ x ≤ -1, X ∈ R} = [-4, -1]
Question 11.
Find the range of values of x, which satisfy the inequality :
\(\frac{-1}{5}\) ≤ \(\frac{3 x}{10}\) + 1 < \(\frac{2}{5}\); x ∈ R Graph the solution set on the number line.
Solution:
Given –\(\frac{1}{5}\) ≤ \(\frac{3 x}{10}\) + 1 < \(\frac{2}{5}\)
⇒ –\(\frac{1}{5}\) – 1 ≤ \(\frac{3 x}{10}\) + 1 < \(\frac{2}{5}\) – 1
⇒ –\(\frac{6}{5}\) ≤ \(\frac{3 x}{10}\) < – \(\frac{3}{5}\)
⇒ –\(\frac{6}{5}\) × \(\frac{10}{3}\) ≤ x < – \(\frac{3}{5}\) × \(\frac{10}{3}\)
⇒ -4 ≤ x < – 2, x ∈ R
Thus solution set= {x : – 4 ≤ x < – 2, x ∈ R} = [- 4, – 2)
Solution set of given inequality is represented on real line is as under. There is a dark dot at – 4 which indicate that – 4 included in solution set while hollow dot at – 2 included in solution set.
Question 12.
Find the range of values of x which satisfy \(\frac{-1}{3}\) ≤ \(\frac{x}{2}\) – 1 \(\frac{1}{3}\) < \(\frac{1}{6}\); x ∈ R Graph these values of x on the real number line.
Solution:
Given \(\frac{-1}{3}\) ≤ \(\frac{x}{2}\) – 1 \(\frac{1}{3}\) < \(\frac{1}{6}\)
⇒ \(\frac{-1}{3}\) ≤ \(\frac{x}{2}\) – \(\frac{4}{3}\) < \(\frac{1}{6}\)
⇒ \(\frac{-1}{3}\) + \(\frac{4}{3}\) ≤ \(\frac{x}{2}\) – \(\frac{4}{3}\) + \(\frac{4}{3}\) < \(\frac{1}{6}\) + \(\frac{4}{3}\)
⇒ 1 ≤ \(\frac{x}{2}\) < \(\frac{3}{2}\) ⇒ 2 ≤ x < 3
∴ Solution set = {x : 2 ≤ x < 3, x ∈ R} = [2, 3)
Solution set represented on real line is given as under. There is dark dot at 2 and hollow dot at 3 indicates that 2 is included while 3 is not included in solution set.
Question 13.
Solve the following inequation, and graph the solution set, on the number line.
2x – 3 < x + 2 ≤ 3x + 5, x ∈ R.
Solution:
Given 2x – 3 < x + 2 ≤ 3x + 5
Now, 2x – 3 < x + 2 ⇒ 2x – x < 3 + 2
⇒ x < 5 …(1)
and x + 2 ≤ 3x + 5 ⇒ 2 – 5 ≤ 3x – x
⇒ -3 ≤ 2x ⇒ \(\frac{-3}{2}\) ≤ x …(2)
∴ from (1) and (2) ; we have
Solution set = {x ; \(\frac{-3}{2}\) ≤ x < 5, x ∈ R}
= [\(\frac{-3}{2}\) , 5)
Solution set represented on real line is given as under:
There is a dark dot at \(\frac{-3}{2}\) which indicates that \(\frac{-3}{2}\) included in solution set whole hollow dot at 5 indicate that 5 is not included in solution set.
Question 14.
Solve the inequation on :
– 3 ≤ 3 – 2x < 9, x ∈ R.
Solution:
Given -3 ≤ 3 – 2x < 9,x ∈ R
Now -3 ≤ 3 – 2x ⇒ 2x ≤ 6 ⇒ x ≤ 3
and 3 – 2x < 9 ⇒ 3 – 9 < 2x ⇒ – 3 < x
∴ Solution set = {x : – 3 < x ≤ 3, x ∈ R} = (-3, 3]
Thus solution set is represented on real line is given as under:
There is a dark dot at 3 indicate that 3 is included in solution set and hollow dot at – 3 indicate that – 3 is not included in solution set.
Question 15.
Find the value of x which satisfies the inequation
-2 ≤ \(\frac{1}{2}\) – \(\frac{2x}{3}\) ≤ 1\(\frac{5}{6}\), x ∈ N Graph the solution on a number line.
Solution:
Given -2 ≤ \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ 1\(\frac { 5 }{ 6 }\)
Now -2 ≤ \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ⇒ \(\frac { 2x }{ 3 }\) ≤ \(\frac { 1 }{ 2 }\) + 2
⇒ \(\frac { 2x }{ 3 }\) ≤ \(\frac { 5 }{ 2 }\) ⇒ x ≤ \(\frac { 5 }{ 2 }\) × \(\frac { 3 }{ 2 }\) ⇒ x ≤ \(\frac { 15 }{ 4 }\)
Also \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ 1\(\frac { 5 }{ 6 }\) ⇒ \(\frac { 1 }{ 2 }\) – \(\frac { 2x }{ 3 }\) ≤ \(\frac { 11 }{ 6 }\)
⇒ \(\frac { 1 }{ 2 }\) – \(\frac { 11 }{ 6 }\) ≤ \(\frac { 2x }{ 3 }\) ⇒ – \(\frac { 4 }{ 3 }\) ≤ \(\frac { 2x }{ 3 }\)
⇒ x ≥ – \(\frac { 4 }{ 3 }\) × \(\frac { 3 }{ 2 }\) = -2
∴ Solution set = {x : -2 ≤ x ≤ \(\frac { 15 }{ 4 }\), x ∈ N} = {1, 2, 3}
Solution set represented on real line is given as under :
Question 16.
Write down the range of real values of x for which the inequation x > 3 and -2 ≤ x < 5 are both true.
Solution:
The solution set of given inequality x > 3, x ∈ R is given as under :
The solution set of given inequality
-2 ≤ x < 5, x ∈ R represented on real line is given as under :
Clearly the common portion of two graphs is the set {x : 3 < x < 5, x ∈ R}
Hence the required range set = {x : x ∈ R, 3 < x < 5} = {3, 5}
Question 17.
Solve and graph the solution set of 3x – 4 > 11 or 5 – 2x ≥ 7.
Solution:
Given 3x – 4 > 11 ⇒ 3x > 15 ⇒ x > 5
or 5 – 2x ≥ 7 ⇒ -2x ≥ 2 ⇒ x ≤ – 1
∴ Solution set = {x > 5 or x ≤ – 1, x ∈ R}
= (-∞, – 1] ∪ (5, ∞)
This solution set is represented on real line is given as under :
Clearly there is a dark dot at -1 and hence included in solution set while hollow dot at 5 indicates that 5 is not included in solution set.
Question 18.
The following diagram represents two inequations A and B on number lines.
(i) Write down A, B and B’ in set builder form.
(ii) Represent A ∩ B and A ∩ B’ on two different number lines.
Solution:
In A, there is a dark line between – 1 and 5. Here, dark dot at 1 and hollow dot at 5 which shows that – 1 included while 5 is not included in solution set.
∴ A = {x : – 1 ≤ x < 5, x ∈ R} = [- 1, 5)
In B, there is a dark line between 2 and 7. Here dark dot at 2 indicates that 2 is included in solution set while hollow dot at 7 indicates that 7 is not included in solution set.
∴ B = {x : 2 ≤ x < 7, x ∈ R} = [2, 7)
∴ A ∩ B = {x ; 2 ≤ x < 5 ; x ∈ R}
This solution set represented on real line is given as under :
Further B’ = {x ; x < 2 and x ≥ 7}
∴ A ∩ B’ = {x ; – 1 ≤ x < 2, x ∈ R}
Question 19.
P is the solution set of 7x – 2 > 4x + 1 and Q is the solution set of (9x – 45) ≥ 5 (x – 5); where x ∈ R. Represent
(i) P ∩ Q
(ii) P – Q,
(iii) P ∩ Q’ on different lines.
Solution:
Given inequality be
7x – 2 > 4x + 1 …(1)
⇒ 7x – 4x > 2 + 1 ⇒ 3x > 3 ⇒ x > 1
∴ Solution set of eqn. (1) = P
= {x : x > 1, x ∈ R} = (1, ∞)
Also given another inequality be
9x – 45 ≥ 5 (x – 5) …(2)
⇒ 9x – 45 ≥ 5x – 25
⇒ 9x – 5x ≥ 45 – 25
⇒ 4x ≥ 20 ⇒ x ≥ 5
∴ Solution set Q = {x ; x ≥ 5, x ∈ R}
(i) P ∩ Q = {x : x ≥ 5, x ∈ R}
This solution set represented on real line is given as under:
There is a dark dot at 5 indicates that 5 is included in solution set.
(ii) P – Q = P ∩ Q’
= {x : x > 1, x ∈ R) ∩ {x : x < 5, x ∈ R} = {x : 1 < x < 5, x ∈ R}
The solution set is represented on real line is given as under. There are hollow dots at 1 and 5 shows that 1 and 5 both are not included in solution set.
(iii) Same as (ii) [∵ P ∩ Q’ = P – Q]
Question 20.
Solve the following system of inequalities (x ∈ R).
(i) 2x – 7 > 5 – x
11 – 5x ≤ 1
(ii) 2x + 5 ≤ 0
x – 3 ≤ 0
(iii) 4x + 3 ≥ 2x + 17
3x – 5 < – 2
(iv) 5x – 7 < 3 (x + 3) 1 – \(\frac { 3x }{ 2 }\) ≥ x – 4
(v) \(\frac { 5x }{ 4 }\) + \(\frac { 3x }{ 8 }\) > \(\frac { 39 }{ 8 }\)
\(\frac { 2x – 1}{ 12 }\) – \(\frac { x – 1 }{ 3 }\) < \(\frac { 3x + 1 }{ 4 }\)
(vi) 2(2x + 3) – 10 < 6 (x – 2) \(\frac{2 x-3}{4}\) + 6 ≥ 2 + \(\frac{4x}{3}\)
(vii) -11 ≤ 4x – 3 ≤ 13
Solution:
(i) Given 2x – 7 > 5 – x
⇒ 2x + x > 5 + 7
⇒ 3x > 12 ⇒ x > 4 …(i)
and 11 – 5x ≤ 1
⇒ -5x ≤ 1 – 11
⇒ -5x ≤ – 10 ⇒ x ≥ 2 …(ii)
Clearly the value of x which is common to both (i) and (ii) be x > 4.
Thus solution set = { x : x > 4, x ∈ R} = (4, ∞)
(ii) Given 2x + 5 ≤ 0 ⇒ x ≤ – \(\frac{5}{2}\) ….(1)
and x – 3 ≤ 0 ⇒ x ≤ 3 …(2)
Clearly the values of x which is common to (1) and (2) be x ≤ \(\frac{-5}{2}\)
∴ Solution set = {x : x ≤ \(\frac{-5}{2}\), x ∈ R}
= (- ∞, \(\frac{-5}{2}\)]
(iii) Given 4x + 3 ≥ 2x + 17
⇒ 4x – 2x ≥ 17 – 3 ⇒ 2x ≥ 14
⇒ x ≥ 7 …(1)
Also, 3x – 5 < – 2 ⇒ 3x < 3
⇒ x < 1 …(2)
Clearly there is no value of x which is common to (1) and (2). Hence given system of inequalities has no solution.
∴ Solution set = Φ
(iv) Given 5x – 7 < 3 (x + 3)
⇒ 5x – 7 < 3x + 9
⇒ 5x – 3x < 9 + 7
⇒ 2x < 16 ⇒ x < 8 …(1)
Also, 1 – \(\frac { 3x }{ 2 }\) ≥ x – 4
⇒ 1 + 4 ≥ x + \(\frac { 3x }{ 2 }\)
⇒ 5 ≥ \(\frac { 5x }{ 2 }\) ⇒ \(\frac { 5x }{ 2 }\) ≤ 5
⇒ x ≤ 2 …(2)
The values of x which is common to (1) and (2) is given by x ≤ 2
∴ Solution set = {x : x ≤ 2, x ∈ R} = ( – ∞, 2]
(v) Given \(\frac { 5x }{ 4 }\) + \(\frac { 3x }{ 8 }\) > \(\frac { 39 }{ 8 }\)
⇒ \(\frac{10 x+3 x}{8}\) > \(\frac{39}{8}\)
⇒ 13x > 39 ⇒ x > 3 …(1)
and \(\frac{2 x-1}{12}\) – \(\frac{x-1}{3}\) < \(\frac{3 x+1}{4}\)
⇒ \(\frac{2 x-1-4(x-1)}{12}\) < \(\frac{3 x+1}{4}\)
⇒ \(\frac{2 x-1-4 x+4}{12}\) < \(\frac{3 x+1}{4}\)
The values of x which is common to (1) and (2) be given by x > 3, x ∈ R
∴ Solution set = {x : x > 3, x ∈ R} = (3, ∞)
(vi) Given
2 (2x + 3) – 10 < 6 (x – 2)
⇒ 4x + 6 – 10 < 6x – 12
⇒ 4x – 4 < 6x – 12 ⇒ 6x – 4x > 12 – 4
⇒ 2x > 8 ⇒ x > 4 …(1)
Also, \(\frac{2 x-3}{4}\) + 6 ≥ 2 + \(\frac{4 x}{3}\);
Multiply both sides by 12 ; we have
⇒ 3 (2x – 3) + 72 ≥ 24 + 16x
⇒ 6x – 9 + 72 ≥ 24 + 16x
⇒ 63 – 24 ≥ 16x – 6x
⇒ 10x ≤ 39
⇒ x ≤ \(\frac{39}{10}\) = 3\(\frac{9}{10}\) …(2)
Clearly there is no value of x which is common to (1) and (2).
Hence given system of equalities has no solution.
∴ Solution set = Φ
(vii) Given – 11 ≤ 4x – 3 ≤ 13
⇒ -11 + 3 ≤ 4x – 3 + 3 ≤ 13 + 3
⇒ – 8 ≤ 4x ≤ 16
⇒ -2 ≤ x ≤ 4
⇒ x ∈ [- 2, 4]
∴ Solution set = {x ; – 2 ≤ x ≤, 4, x ∈ R} = [-2,4]
Question 21.
Solve and graphs the solution set
(i) | x – 3 | < 4
(ii) | x – 3 | > 4
(iii) | x + 3 | > 4
(iv) 11 – 3x | < 4 (v) 2 + 3 | 2y – l | > 8
Solution:
(i) Given | x – 3 | < 4
⇒ -4 < x – 3 < 4
[∵| x | < l ⇒ – l < x < l]
⇒ -4 + 3 < x – 3 + 3 < 4 + 3
= – 1 < x < 7
∴ Solution set = {x : – 1 < x < 7, x ∈ R} = (-1, 7)
This solution set represented on real line is given as under :
There are hollow dots at – 1 and 7
∴ both points – 1 and 7 are not included in solution set.
(ii) Given | x – 3 | > 4 ⇒ x – 3 > 4
or x – 3 < – 4
[∵ | x | > l ⇒ x >l or x <- l]
⇒ x > 7 or x < – l
∴ Solution set = {x : x > 7 or x < -1, x ∈ R} = (- ∞, – 1) ∪ (7, ∞)
Thus solution set is represented on real line is given as under :
Now hollow dots at x = – 1 and x = 7 indicates both points are not included in solution set.
(iii) Given | x + 3 | ≥ 4
⇒ x + 3 ≥ 4 or x + 3 ≤ -4
i.e. x ≥ l or x ≥ – 7
∴ Solution set = {x : x ≥ 1 or x ≤ -7, x ∈ R} = (- ∞,- 7] ∪ [1, ∞)
Hence solution set is represented on real line is given as under :
Clearly dark dots at x = – 7 and x = 1 indicates that – 7 and 1 both are included in solution set.
(iv) Given | 1 – 3x | < 4 ⇒ – 4 < 1 – 3x < 4
⇒ -4 – 1 < -3x <4 – 1
⇒ – 5 < – 3x < 3
⇒ \(\frac { 5 }{ 3 }\) > x > 1 [if a < b ⇒ – a > – b]
⇒ -1 < x < \(\frac { 5 }{ 3 }\)
∴ Solution set = {x : -1 < x < \(\frac { 5 }{ 3 }\), x ∈ R} = (-1, \(\frac { 5 }{ 3 }\))
Hence solution set is represented on real line is given as under :
Here hollow dots at -1, \(\frac { 5 }{ 3 }\) indicates that both points -1,\(\frac { 5 }{ 3 }\) are not included in solution set.
(v) Given 2 + 3 | 2y – 1 | > 8
⇒ 3 | 2y – 1| > 6 ⇒ | 2y – 1 | > 2
⇒ 2y – 1 > 2 or 2y – 1 < – 2
[∵ | x | > a ⇒ x > a or x < – a]
⇒ 2y >3 or 2y < – 1
⇒ y > \(\frac { 3 }{ 2 }\) or y < – \(\frac { 1 }{ 2 }\)
Thus solution set = { y : y > \(\frac { 3 }{ 2 }\) or y < – \(\frac { 1 }{ 2 }\), t ∈ R}
= ( – ∞, – \(\frac { 1 }{ 2 }\)) ∪ (\(\frac { 3 }{ 2 }\), ∞)
This solution set is represented on real line is given as under :
There are hollow dots at –\(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 2 }\) shows that both points are not included in solution set.
Question 22.
3 | x – 6 | – 4 ≤ 11
Solution:
Given 3 | x – 6 | – 4 ≤ 11
⇒ 3 | x – 6 | ≤ 15
⇒ | x -6 | ≤ 5
⇒ – 5 ≤ x – 6 ≤ 5
[∵ | x | ≤ l ⇒ – l ≤ x ≤ l]
⇒ ≤ 5 + 6 ≤ x – 6 + 6 ≤ 5 + 6
⇒ 1 ≤ x ≤ 11
∴ Solution set = {x, 1 ≤ x ≤ 11, x ∈ R] = [1, 11]
Question 23.
2 | 3p – 5 | + 1 > 7
Solution:
Given 2 | 3p – 5 | + 1 > 7
⇒ | 3p – 5 | > 3
⇒ 3p – 5 > 3 or 3p – 5 > – 3
[∵ | x | > l ⇒ x > l or x < -1]
⇒ 3p > 8 or 3p < 2
⇒ p > \(\frac { 8 }{ 3 }\) or p < \(\frac { 2 }{ 3 }\)
∴ Solution set = {p : p > \(\frac { 8 }{ 3 }\) or p < \(\frac { 2 }{ 3 }\)}
Question 24.
| 5 – (m – 3) | + 8 < 15
Solution:
Given | 5 – (m – 3) | + 8 < 15
⇒ | 8 – m | < 7 ⇒ – 7 < 8 – m < 7
⇒ – 7 – 8 < – m < 7 – 8
⇒ – 15 < – m < – 1 ⇒ 15 > m > 1 ⇒ 1 < m < 15
∴ Solution set = {m ; 1 < m < 15}