Solve: - 3 ≤ 4 - 7 x 2 ≤ 18

According to question Given that - 3 ≤ 4 - 7 x 2 ≤ 18 Substracting each side by 4 ⇒ - 3 - 4 ≤ 4 - 7 x 2 - 4 ≤ 18 - 4 ⇒ - 7 ≤ - 7 x 2 ≤ 14 ⇒ - 14 ≤ - 7 x ≤ 28 Dividing each side by - 7 ⇒ - 4 ≤ x ≤ 2 Hence interval for x is - 4 , 2

Solve: 3 ( 2 - x ) ≥ 2 ( 1 - x )

Given that, 3 ( 2 - x ) ≥ 2 ( 1 - x ) ⇒ 6 - 3 x ⩾ 2 - 2 x ⇒ - 3 x + 2 x ≥ 2 - 6 ⇒ - x ≥ - 4 Multiplying both sides by - 1 , we get, x ≤ 4 Hence, x ∈ ( - ∞ , 4 ]

Solve: 5 x - 6 x + 6 < 1

Given that, 5 x - 6 x + 6 < 1 ⇒ 5 x - 6 x + 6 - 1 < 0 ⇒ 5 x - 6 - x - 6 x + 6 < 0 ⇒ 4 x - 12 x + 6 < 0 ⇒ 4 x - 3 x + 6 < 0 ⇒ x - 3 x + 6 < 0 From here we have two cases Case (i): x - 3 < 0 and x + 6 > 0 ⇒ x < 3 and x > - 6 ⇒ x ∈ - 6 , 3 Case (ii): x - 3 > 0 and x + 6 < 0 ⇒ x > 3 and x < - 6 Which is not possible as their intersection is an empty set. Hence, the solution of the given inequality is - 6 , 3

Solve: 2 x - 1 3 ≥ 3 x - 2 4 - 2 - x 5 , x ∈ R

Given that, 2 x - 1 3 ≥ 3 x - 2 4 - 2 - x 5 ⇒ 2 x - 1 3 ≥ 5 ( 3 x - 2 ) - 4 ( 2 - x ) 20 ⇒ 2 x - 1 3 ≥ 15 x - 10 - 8 + 4 x 20 ⇒ 2 x - 1 3 ⩾ 19 x - 18 20 ⇒ 40 x - 20 ⩾ 57 x - 54 ⇒ 40 x - 57 x ≥ - 54 + 20 - 17 x ⩾ - 34 Dividing both side by - 17 , we get, x ≤ 2 Hence, x ∈ ( - ∞ , 2 ]

Solve graphically: x - y ≤ 1 , x + 2 y ≤ 8 , 2 x + y ≥ 2 , x , y ≥ 0

Given that, x - y ≤ 1 x 0 1 y -1 0 Hence, point is ( 0 , - 1 ) , ( 1 , 0 ) x + 2 y ≤ 8 x 0 8 y 4 0 Hence, point is ( 0 , 4 ) , ( 8 , 0 ) 2 x + y ≥ 2 x 0 1 y 2 0 Hence, point is ( 0 , 2 ) , ( 1 , 0 ) Let us plot the points and draw the graph to find the region represented by graphs. Check for the points 0 , 0 in all the inequalities C a s e I : x - y ≤ 1 0 - 0 = 0 ≤ 1 , which is true ⇒ the points 0 , 0 lies in plane x - y ≤ 1 . So, for this graph the solution region is left side of line x - y = 1 . C a s e I I : x + 2 y ≤ 8 ⇒ 0 ≤ 8 , which is true. Hence, the solution region is left side of the line x + 2 y = 8 . C a s e I I I : 2 x + y ≥ 2 ⇒ 0 ≥ 2 , which is false. Hence, the solution region is right side of the line 2 x + y = 2 . and x , y ≥ 0 ⇒ the graph lies in the first quadrant. The region containing all the solutions of a system of an inequality is the solution region. It is represented by the shaded region and is the required solution.

E M B I B E

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST

A P Prabhakaran Solutions for Chapter: Linear Inequalities, Exercise 1: SELF EVALUATION TEST

Author:A P Prabhakaran

A P Prabhakaran Mathematics Solutions for Exercise - A P Prabhakaran Solutions for Chapter: Linear Inequalities, Exercise 1: SELF EVALUATION TEST

Attempt the free practice questions on Chapter 6: Linear Inequalities, Exercise 1: SELF EVALUATION TEST with hints and solutions to strengthen your understanding. Golden MATHEMATICS CLASS 11 solutions are prepared by Experienced Embibe Experts.

Questions from A P Prabhakaran Solutions for Chapter: Linear Inequalities, Exercise 1: SELF EVALUATION TEST with Hints & Solutions

EASY

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 2

Solve: $4 x + 3 < 6 x + 7$

EASY

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 3

If the solution set of inequality $3 (\frac{x - 2}{5}) \geq 5 (\frac{2 - x}{3})$ is $x \in [a, \infty), \forall a \in R$ , then the value of $a$ is

EASY

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 4

Solve: $- 3 \leq 4 - \frac{7 x}{2} \leq 18$

EASY

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 5

Solve: $3 (2 - x) \geq 2 (1 - x)$

EASY

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 6

Solve: $\frac{5 x - 6}{x + 6} < 1$

EASY

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 7

Solve: $\frac{2 x - 1}{3} \geq \frac{3 x - 2}{4} - \frac{2 - x}{5}, x \in R$

MEDIUM

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 8

A man wants to cut three lengths from a single piece of board of length $91$ cm. The second length is to be $3$ cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least $5$ cm longer than the second?

HARD

11th CBSE

IMPORTANT

Golden MATHEMATICS CLASS 11>Chapter 6 - Linear Inequalities>SELF EVALUATION TEST>Q 9

Solve graphically: $x - y \leq 1, x + 2 y \leq 8, 2 x + y \geq 2, x, y \geq 0$