The availability of OP Malhotra Maths Class 12 Solutions Chapter 11 Inequalities Chapter Test encourages students to tackle difficult exercises.
S Chand Class 11 ICSE Maths Solutions Chapter 11 Inequalities Chapter Test
Question 1.
Solve : 24x < 100, when (i) x is a natural number (ii) x is an integer
Solution:
Given 24x < 100 ⇒ x < \(\frac { 100 }{ 24 }\) = \(\frac { 25 }{ 6 }\) = 4\(\frac { 1 }{ 6 }\)
(i) When x ∈ N Then x = 1, 2, 3, 4
Then solution set = {1, 2, 3, 4}
(ii) When x is an integer
∴ x = ….., -3, -2, -1, 0, 1, 2, 3, 4
Thus solution set = { …., – 3, – 2, – 1, 0, 1, 2, 3, 4}
Question 2.
Solve the inequality
\(\frac { 1 }{ 2 }\)[\(\frac { 3 }{ 5 }\)x + 4] ≥ \(\frac { 1 }{ 3 }\)(x – 6)
Solution:
Given \(\frac { 1 }{ 2 }\)[\(\frac { 3 }{ 5 }\)x + 4] ≥ \(\frac { 1 }{ 3 }\)(x – 6)
⇒ 3(3x + 20) ≥ 10 (x – 6)
⇒ 9x – 10x ≥ – 60 – 60
⇒ 3[\(\frac { 3 }{ 5 }\)x + 4] ≥ 2 (x – 6)
⇒ 9x + 60 ≥ 10x – 60
⇒ – x ≥ – 120
⇒ x ≤ – 120
Then solution set = (- ∞, 120]
Question 3.
Solve : – 12 ≤ 4 – \(\frac { 3x }{ -5 }\) < 2
Solution:
Given 12 ≤ 4 – \(\frac { 3x }{ -5 }\) < 2
⇒ – 12 ≤ \(\frac{20+3 x}{5}\) < 2
⇒ – 60 – 20 ≤ 20 + 3x – 20 < 10 – 20
⇒ –\(\frac{80}{3}\) ≤ x < – \(\frac { 10 }{ 3 }\)
⇒ – 12 ≤ 4 + \(\frac { 3x }{ 5 }\) < 2
⇒ – 60 ≤ 20 + 3x < 10
⇒ – 80 ≤ 3x < – 10
⇒ x ∈[-\(\frac { 80 }{ 3 }\),- \(\frac { 10 }{ 3 }\))
Then solution set = [- \(\frac { 80 }{ 3 }\), – \(\frac { 10 }{ 3 }\))
Question 4.
Solve the following inequalities graphically
(i) 3x + 4y < 12
(ii) x > – 3
(iii) 5y – 3 ≤ 12
Solution:
(i) Given inequality be 3x + 4y < 12 first of all we draw the graph of line 3x + 4y = 12 it meets coordinate axes at a (4, 0) and B (0, 3).
Clearly (0, 0) satisfies the inequality 3x + 4y < 12 (∵ 3 × 0 + 4 × 0 = 0 < 12) Thus region containing origin gives the solution set of given inequality.
(ii) First of all, we draw the graph of line x = – 3 The line x = – 3 passing through the point A (- 3, 0) and || toy-axis. Clearly (0, 0) satisfies the inequality x > – 3. Thus region containing (0, 0) gives the solution set of x > – 3
(iii) First of all, we draw the graph of
5y – 3 = 12
i. e. 5y = 15 ⇒ y = 3
which passes through (0, 3) and parallel to x-axis.
Clearly (0,0) satisfies y ≤ 3. Thus region containing (0, 0) gives the solution set of given inequality.
Question 5.
Solve graphically : 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
Given inequalities are
3x + 4y ≤ 60 …(1)
x + 3y ≤ 30 …(2)
and x ≥ 0, y ≥ 0 …(3)
For region 3x + 4y ≤ 60 ;
The line 3x + 4y = 60 meets coordinate axes at A (20, 0) and B (0, 15).
Clearly (0,0) satisfies inequality (1). Thus region containing (0, 0) gives the solution set of inequality (1).
For region x + 3y ≤ 30 ;
The line x + 3y = 30 meet coordinate axes at C (30, 0) and D (0, 10). Clearly (0, 0) satisfies the inequality (2). Thus region containing origin gives the solution set of inequality (2).
For region x ≥ 0, y ≥ 0 ; It clearly represents the first quadrant of XOY plane.
The lines x + 3y = 30 and 3x + 4y= 60 intersects at P (12, 6).
Thus all points of common shaded region gives the solution set of given system of inequalities.