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

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 – x² > 0 and 9 – x² ≠ 1
⇒ x² < 9 and x² ≠ 8
⇒ |x| < 3 and x ≠ ± 2√2
⇒ – 3 < x < 3 and x ≠ ± 2√2
⇒ D_f = (- 3, 3) – {± 2√2}