Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(a) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(a)

Question 1.

Determine whether the solution set is finite or infinite or empty:

(i) x < 1000, x ∈ N

Solution:

Finite

(ii) x < 1, x ∈ Z (set of integers)

Solution:

Infinite

(iii) x < 2, x is a positive integer.

Solution:

Finite

(iv) x < 1, x is a positive integer.

Solution:

Empty

Question 2.

Solve as directed:

(i) 5x ≤ 20 in positive integers, in integers.

Solution:

5x ≤ 20

⇒ \(\frac{5 x}{5} \leq \frac{20}{5}\)

⇒ x ≤ 4

If x is a positive integer, then the solution set is {1, 2, 3, 4}

If x is an integer, then the solution set is:

S = {x : x ∈ Z and x ≤ 4}

= { ….. -3, -2, -1, 0, 1, 2, 3, 4}

(ii) 2x + 3 > 15 in integers, in natural numbers.

Do you mark any difference in the solution sets?

Solution:

2x + 3 > 15

⇒ 2x + 3 – 3 > 15 – 3

⇒ 2x > 12

⇒ \(\frac{2 x}{2}>\frac{12}{2}\)

⇒ x > 6

If x ∈ Z, then the solution set is S = (x : x ∈ Z and x > 6}

= {7, 8, 9…… }

If x ∈ N. then the solution set is S = {x : x ∈ N and x > 6}

= {7, 8, 9…… }

Two solution sets are the same.

(iii) 5x + 7 < 32 in integers, in non-negative integers.

Solution:

5x + 7 < 32

⇒ 5x + 7 – 7 < 32 – 7

⇒ 5x < 25

⇒ \(\frac{5 x}{5}<\frac{25}{5}\)

⇒ x < 5

If x ∈ Z, then the solution set is S = { x : x ∈ Z and x < 5 }

= {…..-3, -2, -1, 0, 1, 2, 3, 4}

If x is a non-negative solution then the solution set is S = {x : x is a non-negative integer < 5}

= (0, 1,2, 3,4}

(iv) -3x – 8 > 19, in integers, in real numbers.

Solution:

– 3x – 8 > 19

⇒ – 3 x – 8 + 8 > 19 + 8

⇒ – 3x > 27

⇒ \(\frac{-3 x}{-3}<\frac{27}{-3}\)

⇒ x < – 9

If x ∈ Z, then the solution set is S = (x : x ∈ Z and x < – 9}

= { ……..- 11, – 10}

If x ∈ R then the solution set is S = {x : x ∈ R and x < – 9}

= (∞, – 9)

(v) |x – 3| < 11, in N and in R.

Solution:

|x – 3| < 11

⇒ – 1 < x – 3 < 11

⇒ – 11 + 3 < x – 3 + 3 < 11+3

⇒ – 8 < x < 14

If x ∈ N the solution set is S = {1, 2, 3, 4, 5……..12, 13}

If x ∈ R then the solution set is: S = {x : x ∈ R and – 8 < x < 14}

= (- 8, 14)

Question 3.

Solve as directed:

(i) 2x + 3 > x – 7 in R

Solution:

2x + 3 > x – 7

⇒ 2x – x > – 7 – 3

⇒ x > – 10

x ∈ R, the solution set is S = (x : x ∈ R and x > – 10} = (-10, ∞)

(ii) \(\frac{x}{2}+\frac{7}{3}\) < 3x – 1 in R

Solution:

\(\frac{x}{2}+\frac{7}{3}\) < 3x – 1

\(\frac{3 x+14}{6}\) < 3x – 1

⇒ 3x + 14 < 18x – 6

⇒ 3x – 18x < – 6 – 14

⇒ – 15x < – 20

⇒ \(\frac{-15 x}{-15}>\frac{-20}{-15}\)

⇒ x > \(\frac{4}{3}\)

If x ∈ R, the solution set is S = \(\left(\frac{4}{3}, \infty\right)=\left\{x: x \in R \text { and } x>\frac{4}{3}\right\}\)

(iii) \(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\) for non-negative real numbers.

Solution:

\(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\)

⇒ \(\frac{15 x-10 x+6 x}{30}\) ≤ \(\frac{11}{3}\)

⇒ 11x ≤ \(\frac{11}{3}\) × 30

⇒ 11x ≤ 110

⇒ x ≤ 10

If x is a non-negative real number then the solution set is S = {x : x ∈ R and 0 ≤ x ≤ 10}

= {0, 10}

(iv) 2(3x – 1) < 7x + 1 < 3 (2x + 1) for real values.

Solution:

2(3x – 1) < 7x + 1 < 3(2x + 1)

⇒ 6x – 2 < 7x + 1< 6x + 3

⇒ – 2 < x + 1 < 3

⇒ – 3 < x < 2

If x ∈ R, the solution set is S = (x : x ∈ R and -3 < x < 2}

= {-3, 2}

(v) 7(x – 3) ≤ 4 (x + 6), for non-negative integral values.

Solution:

7(x – 3) ≤ 4(x + 6)

⇒ 7x – 21 ≤ 4x + 24

⇒ 7x – 4x ≤ 24 + 21

⇒ 3x ≤ 45

⇒ x ≤ 9

If x is a non-negative integer the solution set is S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(vi) Convert to linear inequality and solve for natural numbers: (x – 2) (x – 3) < (x + 3) (x – 1)

Solution:

(x – 2) (x – 3) < (x + 3) (x – 1)

⇒ x^{2} – 5x + 6 < x^{2} + 2x – 3

⇒ – 5x + 6 < 2x – 3

⇒ – 5x – 2x < – 3 – 6

⇒ – 7x < – 9

⇒ x > \(\frac{9}{7}\)

If x ∈ N, the solution set is S = {2, 3, 4 }

(vii) Solve in R, \(\frac{x}{2}\) + 1 ≤ 2x – 5 < x. Also, find its solution in N.

Solution:

\(\frac{x}{2}\) + 1 ≤ 2x – 5 < x

⇒ \(\frac{x}{2}\) +1 ≤ 2x – 5 and 2x – 5 < x

⇒ \(\frac{x}{2}\) – 2x ≤ – 5 – 1 and x < 5

⇒ \(\frac{-3x}{2}\) ≤ – 6 and x < 5

⇒ – 3x ≤ – 12 and x < 5

⇒ x ≥ 4 and x < 5

⇒ 4 ≤ x < 5

If x ∈ R, the solution set is S = {x : x ∈ R and 4 < x < 5}

= {4, 5}

If x ∈ N, the solution set is S = { 4 }

(viii) Solve in R and also in Z: \(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)

Solution:

\(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)

⇒ \(\frac{3 x+1}{5} \geq \frac{5 x+10-15+9 x}{15}\)

⇒ 3x + 1 ≥ \(\frac{14 x-5}{3}\)

⇒ 9x + 3 ≥ 14x – 5

⇒ 9x – 14x ≥ – 5 – 3

⇒ – 5x ≥ – 8

⇒ x ≤ \(\frac{8}{5}\)

If x ∈ R, then the solution set is S = (x : x ∈ R and x ≤ \(\frac{8}{5}\)}

= (- ∞, \(\frac{8}{5}\))

If x ∈ Z, then the solution set is S = { x : x ∈ Z and x ≤ \(\frac{8}{5}\)}

= {……. -3, -2, -1, 0, 1}

Question 4.

Solve |x – 1| >1 and represent the solution on the number line.

[Exhaustive hints: By definition of modulus function

For x – 1 ≥ 0 or x ≥ 1, |x – 1| > 1

⇔ x – 1 > 1 ⇔ x > 2 ⇔ x ∈ (2, ∞)

For x- 1 < 0 or x < 1, |x – 1| > 1

⇔ – (x – 1) > 1

⇔ x – 1 < -1 (multiplication by -1 reverses the inequality)

⇔ x < 0 ⇔ x ∈ ( -∞, 0)

∴ The solution set is the Union,

(-∞, 0) ∪ (2, ∞) Show this as two disjoint open intervals on the number line, i.e., real line.]

Solution:

|x – 1| > 1

⇒ – 1 > x – 1 > 1

⇒ 0 > x > 2

⇒ x < 0 and x > 2

∴ The solution set is S = {x : x ∈ R, x < 0 and x > 2}

= (-∞, 0) ∪ (2, ∞)

We can show this solution in the number line as

Question 5.

Solve in R and represent the solution on the number line.

(i) |x – 5| < 1

Solution:

|x – 5| < 1

⇒ – 1< x – 5 < 1

⇒ 4 < x < 6

If x ∈ R, then the solution set is S = (4, 6)

We can represent the solution on the number line as

(ii) \(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)

Solution:

\(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)

⇒ \(\frac{x}{5}<\frac{4 x+2+1-3 x}{6}\)

⇒ \(\frac{x}{5}<\frac{x+3}{6}\)

⇒ 6x < 5x + 15

⇒ x < 15

If x ∈ R, the solution set is S = (-∞, 5)

We can represent the solution on the number line as

(iii) 2x + 1 ≥ 0

Solution:

2x + 1 ≥ 0

⇒ 2x ≥ -1

⇒ x ≥ -1/2

If x ∈ R, then the solution set is S = [\(-\frac{1}{2}\), ∞]

We can represent the solution on the number line as

(iv) \(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)

Solution:

\(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)

⇒ 3x – 3 ≤ 2x + 2 < 3x – 1

⇒ 3x – 3 ≤ 2x + 2 and 2x + 2 < 3x – 1

⇒ x ≤ 5 and – x < – 3

⇒ x ≤ 5 and x > 3

⇒ 3 < x ≤ 5

If x ∈ R, the solution set is S = {3, 5}

We can represent the solution on the number line as

Question 6.

In a triangle, ABC; AB, BC, and CA are x, 3x + 2, and x + 4 units respectively where x ∈ N. Find the length of its sides. (Hint: Apply triangle-inequality).

Solution:

Given AB = x

BC = 3x + 2

and CA = x + 4

Now AB + AC > BC (Triangle inequality)

⇒ x + x + 4 > 3x + 2

⇒ 2x + 4 > 3x + 2

⇒ – x > – 2

⇒ x < 2

As x ∈ N we have x = 1

The sides of triangle ABC are

AB = 1 unit

BC = 5 units

and CA = 5 units

Question 7.

The length of one side of a parallelogram is 1 cm. shorter than that of its adjacent side. If its perimeter is at least 26 c.m., find the minimum possible lengths of its sides.

Solution:

Let the longer side = x cm

∴ The smaller side = (x – 1) cm

Perimeter = 2(x + (x – 1)) = 4x – 2 cm

According to the question

4x – 2 ≥ 26

⇒ 4x ≥ 28

⇒ x > 7

The minimum value of x = 7.

∴ The minimum length of the sides is 7cm and 6 cm.

Question 8.

The length of the largest side of a quadrilateral is three times that of its smallest side. Out of the other two sides, the length of one is twice that of the smallest and the other is 1 cm. longer than the smallest. If the perimeter of the quadrilateral is at most 36 c.m., then find the maximum possible lengths of its sides.

Solution:

Let the smallest side = x cm.

Largest side = 3 times x = 3x cm.

The other two sides are 2x cm and x + 1 cm.

⇒ The perimeter = x + 3x + 2x + x + 1

= 7x + 1 cm

According to the question:

7x + 1 ≤ 36

⇒ 7x ≤ 35

⇒ x ≤ 5

Maximum value of x = 5

∴ The maximum possible length of sides are x = 5 cm, 3x = 15 cm, 2x = 10 cm, and x + 1 = 6 cm.

Question 9.

Find all pairs of consecutive odd numbers each greater than 20, such that their sum is less than 60.

Solution:

Let two consecutive odd numbers are

2n – 1 and 2n + 1

Now 2n – 1 > 20 and 2n + 1 > 20

But their sum = 2n – 1 + 2n + 1

= 4n < 60

⇒ n < 15

for n = 14 two numbers are 27, 29

for n = 13 two numbers are 25, 27

for n = 12 two numbers are 23, 25

for n = 11 two numbers are 21, 23

∴ All pairs are 21, 23; 23, 25; 25, 27 and 27, 29

Question 10.

Find all pairs of even numbers each less than 35, such that their sum is at least 50.

Solution:

Let two even numbers be x and y.

According to the question

x < 35, y < 35 and x + y ≥ 50

⇒ x ≤ 34, y ≤ 34 and x + y ≥ 50

⇒ x + y ≤ 70, x + y ≥ 50

⇒ 50 ≤ x + y ≤ 70

If x + y = 50 the numbers are {34, 16}, {32, 18}, {30, 20}, {28, 22}, {26, 24}

If x + y = 52 the numbers are {34, 18}, {32, 20}, {30, 22}, {28, 24}, {26, 26}

If x + y = 34 the numbers are {34, 20}, {32, 22}, {30, 24}

If x + y = 56 the numbers are {34, 22}, {32, 24}, {30, 26}, {28, 28}

If x + y = 58 the numbers are {34, 24}, {32, 26}, {30, 28}

If x + y = 60 the numbers are {34, 26}, {32, 28}, {30, 30}

If x + y = 62 the numbers are {34, 28}, {32, 30}

If x + y = 64 the numbers are {34, 30}, {32, 32}

If x + y = 68 the numbers are {34, 34}