Regular engagement with Understanding ISC Mathematics Class 11 Solutions Chapter 2 Relations and Functions Ex 2.6 can boost students confidence in the subject.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.6

Question 1.

Solve the following equations for x :

(i) [2x – 3] = 5

(ii) [2x + 1] = – 2

(iii) [2 – 3x] = 5.

Solution:

(i) [2x – 3] = 5

⇒ 5 ≤ 2x – 3 < 6

[∵ if k ∈ I, [x] = k iff k ≤ x < k + 1]

⇒ 5 + 3 ≤ 2x < 6 + 3

⇒ 8 ≤ 2x < 9

⇒ 4 ≤ x < \(\frac{9}{2}\)

Thus, the sclution set is [4, \(\frac{1}{2}\))

(ii) Since [2x + 1] – 2

⇒ – 2 ≤ 2x + 1 < – 1

⇒ [∵ [x] = k (k ∈ I) ⇔ k ≤ x < k + 1]

⇒ – 2 – 1 ≤ 2x < – 1 – 1

⇒ – 3 ≤ 2x < – 2

⇒ – \(\frac{3}{2}\) ≤ x < – 1

Thus, solution set is [- \(\frac{3}{2}\), – 1).

(iii) Given [2 – 3x] = 5

⇒ 5 ≤ 2 – 3x < 5 + 1 = 6

[∵ [x] = k (k ∈ I) iff k ≤ x < k + 1]

⇒ 5 – 2 ≤ – 3x < 6 – 2

⇒ 3 ≤ – 3x < 4 ⇒ \(-\frac{3}{3} \geq x>-\frac{4}{3}\)

⇒ – \(\frac{4}{3}\) < x ≤ 1

Thus, solution set is (- \(\frac{4}{3}\), – 1].

Question 2.

If f (x) = log \(\left(\frac{1+x}{1-x}\right)\), then show that f (x) + f (y) = f (\(\left(\frac{x+y}{1+x y}\right)\)).

Solution:

Given f (x) = log \(\left(\frac{1+x}{1-x}\right)\)

Question 3.

If f (x) = \(\frac{9^x}{9^x+3}\), then show that f (x) + f (1 – x) = 1.

Solution:

Given f (x) = \(\frac{9^x}{9^x+3}\) ;

f (1 – x) = \(\frac{9^{1-x}}{9^{1-x}+3}\)

= \(\frac{9.9^{-x}}{9.9^{-x}+3}\)

⇒ f (1 – x) = \(\frac{\frac{9}{9^x}}{\frac{9}{9^x}+3}\)

= \(\frac{9}{9+3.9^x}\)

= \(\frac{3}{3+9^x}\)

f (x) + f (1 – x) = \(\frac{9^x}{9^x+3}+\frac{3}{3+9^x}\)

= \(\frac{9^x+3}{9^x+3}\) = 1.

Question 4.

Find the domain of the following functions :

(i) log (4x – 3)

(ii) log \(\frac{1}{\log \left(9-x^2\right)}\)

Solution:

(i) Let f (x) = log (4x – 3)

For D_{f} : f (x) must be a real number

⇒ log (4x – 3) must be a real number

⇒ 4x – 3 > 0

⇒ x > \(\frac{3}{4}\)

∴ D_{f} = (\(\frac{3}{4}\), ∞)

(ii) Let f (x) = \(\frac{1}{\log \left(9-x^2\right)}\)

For D_{f} : f (x) must be a real number

⇒ \(\frac{1}{\log \left(9-x^2\right)}\) must be a real number

⇒ 9 – x^{2} > 0 and 9 – x^{2} ≠ 1

⇒ x^{2} < 9 and x^{2} ≠ 8

⇒ |x| < 3 and x ≠ ± 2√2

⇒ – 3 < x < 3 and x ≠ ± 2√2

⇒ D_{f} = (- 3, 3) – {± 2√2}