ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.2

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.2

Question 1.
A function is defined by f(x) = \(\frac{3 x^2+2 x-1}{x+1}\), x ∈ R, x ≠ – 1. Find the value of f(- 3) + 1.
Solution:
Given f(x) = \(\frac{3 x^2+2 x-1}{x+1}\), x ∈ R, x ≠ – 1
∴ f (- 3) = \(\frac{3(-3)^2+2(-3)-1}{-3+1}\)
= \(\frac{27-7}{-2}\)
= \(\frac{+20}{-2}\) = – 10
Thus, f(- 3)+ 1 = – 10 + 1 = – 9

Question 2.
(i) Is the function f defined by f(x) = \(\begin{cases}x^2, & 0 \leq x \leq 3 \\ 2 x & , 3<x \leq 5\end{cases}\)
Solution:
Given relation f(x) = \(\begin{cases}x^2, & 0 \leq x \leq 3 \\ 2 x & , 3<x \leq 5\end{cases}\)
When x = 3, f(3) = 3² = 9
and/(3) = 2 × 3 = 6
Thus, element 3 has two images 9 and 6.
but for relation to be a function, every elements has a unique image. So given relation is not a function.

(ii) A relation R in the set of real number R is defined as R = {(a, b): , √a = b) is a function or not. Justify.
Solution:
Given relation R = {(a, b) : √a = b}
Clearly √a is not defined for all a ∈ (- ∞, 0).
Thus every negative real number has no image in R. Thus, given relation is not a function.

Question 3.
A real function f is defined by f(x) = 3x – 2, x ∈ R, find the value of f(- 3) + f(0) × f(7).
Solution:
Given f(x) = 3x – 2, x ∈ R
f(- 3) = – 9 – 2 = – 11 ;
f(0) = 3 × 0 – 2 = – 2 ;
f(7) = 21 – 2 = 19
Thus, f(- 3) + f(0) × f(7) = – 11 – 2 × 19
= – 11 – 38 = – 49.

Question 4.
If a real function f is defined by f(x) = 2x² – 3, find x such that f(x) = 15.
Solution:
Given, f(x) = 2x² – 3 also, f(x) = 15
∴ 15 = 2x² – 3
⇒ 2x² = 18
⇒ x² = 9
⇒ x = ± 3

Question 5.
Given f(x) = x³ – 1, find x if f (x) is 215.
Solution:
Given f(x) = x³ – 1 also, f(x) = 215
x³ – 1 = 215
⇒ x³ = 216
⇒ x³ = 6³
⇒ x = 6
other values of x are imaginary.

Question 6.
A function is given by the formula f(x) = 144 – 16x². Calculate f(2). Also find the values of x when f(x) = 0.
Solution:
Given f(x) = 144 – 16x²
∴ f(2) = 144 – 16 x 4
= 144 – 64 = 80
Also, f(x) = 0
⇒ 144 – 16x² = 0
⇒ 16x² = 144
⇒ x² = 9
⇒ x = ± 3

Question 7.
A function ‘f‘ is defined by f(x) = 2x² + 3, for all x ∈ R. Find
(i) image of – 1 under f
(ii) element (elements) of the domain which has image 35.
Solution:
Given, f(x) = 2x² + 3 ∀ x ∈ R
(i) f(- 1) = 2 (- 1)² + 3
= 2 + 3 = 5

(ii) Now f(x) = 35
⇒ 2x² + 3 = 35
2x² = 32
x² = 16
∴ x = ± 4

Question 8.
If f(x) = 2x² + 1 and domain of f = {- 2, – 1, 0, 1, 3}, find
(i) range of f
(ii) f(3) × f(- 2)
(iii) x if f(x) = 9.
Solution:
(i) Given, f(x) = 2x² + 1 ;
D_f = {- 2, – 1, 0, 1, 3}
∴ f(- 2) = 2 (- 2)² + 1 = 9
f(- 1) = 2 (- 1)² + 1 = 3
f(0) = 2 (0)² + 1 = 1
f(1) = 2(1)² + 1 = 3
f(3) = 2 × 3² + 1 = 19
∴ Range of f = R_f = {9, 3, 1, 19}

(ii) f(3)= 19; f(- 2) = 9
∴ f(3) × f(- 2) = 19 × 9 = 171

(iii) Given f (x) = 9
⇒ 2x² + 1 = 9
⇒ 2x² = 8
⇒ x² = 4
i.e. x = ± 2.

Question 9.
A function ‘f ‘ is defined by f(x) = x² + 3, x ∈ N and x ≤ 5.
(i) Find the range of f.
(ii) Find f(2) × f(5).
(iii) Does f(- 3) exist ?
(iv) Find x when f(x) = 7.
Solution:
Given f(x) = x² + 3, x ∈ N and x ≤ 5
∴ x = 1, 2, 3, 4, 5
When x = 1;
f(1) = 1² + 3 = 4

When x = 2;
f(2) = 2² + 3 = 7

When x = 3 ;
f(3) = 3² + 3 = 12

When x = 4;
f(4) = 4² + 3 = 19

When x = 5;
f(5) = 5² + 3 = 28
∴ Rf = {4, 7, 12, 19, 28}

(ii) f(2) x f(5) = 7×28 = 196

(iii) since – 3 ∉ N
∴ f(- 3) does not exists.

(iv) Given f(x) = 7
⇒ x² + 3 = 7
⇒ x² = 4
⇒ x = ±2
but x ∈ N
∴ x = 2 [∵ – 2 ∉ N]

Question 10.
Find the domain of the following real functions :
(i) f(x) = |x + 2|
(ii) f(x) = – | x – 2 |
(iii) f(x) = 2 – | x – 1 |
(iv) f(x) = \(\frac{|x|-x}{2 x}\)
Solution:
(i) Given f(x) = |x + 2|
For D_f : f(x) must be a real number
⇒ | x + 2 | must be a real number, which is a real number ∀ x ∈ R
∴ D_f = R

(ii) Given f(x) = – | x – 2 |
For D_f ; f(x) must be a real number.
⇒ – | x – 2 | must be a real number, which is a real number ∀ x ∈ R
∴ D_f = R

(iii) Given f(x) = 2 – | x – 1 |
For D_f ; f(x) must be a real number
⇒ 2 – | x – 1 | must be a real number, which is a real number ∀ x ∈ R
∴ D_f = R

(iv) Given f(x) = \(\frac{|x|-x}{2 x}\)
For D_f ; f(x) must be a real number
⇒ \(\frac{|x|-x}{2 x}\) must be a real number
∴ x ≠ 0
Thus, D_f = R – {0}

Question 10 (old).
Find the domain of the following real functions :
(i) f(x) = \(\frac{x^2-4}{x+2}\)
(ii) f(x) = \(\frac{3 x}{28-x}\); (NCERT Exemplar)
(iii) f(x) = \(\frac{x}{x^2+3 x+2}\) (NCERT Exemplar)
(iv) f(x) = \(\frac{x^3-x+3}{x^2-1}\) (NCERT Exemplar)
Solution:
(i) Givern f(x) = \(\frac{x^2-4}{x+2}\)
For D_f, f(x) must be a real number
⇒ \(\frac{x^2-4}{x+2}\) must be a real number.
i.e., x + 2 ≠ 0
⇒ x ≠ – 2
∴ D_f = R – {- 2}

(ii) Given f(x) = \(\frac{3 x}{28-x}\)
For D_f, f(x) must be a real number
⇒ \(\frac{3 x}{28-x}\) must be a real number
∴ 28 – x ≠ 0
⇒ x ≠ 28
Thus, D_f = R – {28}

(iii) Given f(x) = \(\frac{x}{x^2+3 x+2}\)
For D_f, f(x) must be a real number.
\(\frac{x}{x^2+3 x+2}\), must be a real number I
∴ x² + 3x + 2 ≠ 0
⇒ (x + 1) (x + 2) ≠ 0
⇒ x ≠ – 1, – 2
∴ D_f = set of all real numbers except – 1 and – 2.
i.e. D_f = R – {- 1, – 2}

(iv) Given f(x) = \(\frac{x^3-x+3}{x^2-1}\)
For D_f, f (x) must be a real number x3 – x + 3
⇒ \(\frac{x^3-x+3}{x^2-1}\) must be a real number.
∴ x² – 1 ≠ 0
⇒ x ≠ ± 1
Thus, D_f = R – {- 1, 1}

Question 11.
Find the range of the following functions:
(i) f(x) = x² + 2, x ∈ R (NCERT)
(ii) f(x) = 3 – 2x, x ∈ R, x ≥ 1
Solution:
(i) Given f(x) = x² + 2, x ∈ R
For R_f ;
let y =f(x) = x² + 2
since x² ≥ 0 ∀ x ∈ R
⇒ x² + 2 ≥ 2
⇒ y ≥ 2
∴ R_f = [2, ∞).

(ii) Given f(x) = 3 – 2x, x ∈ R, x ≥ 1
For R_f;
let y = f(x) = 3 – 2x
since x ≥ 1
⇒ – 2x ≤ – 2
⇒ 3 – 2x ≤ 3 – 2 = 1
i.e. 3 – 2x ≤ 1
⇒ y ≤ 1
∴ R_f = (- ∞, 1].

Question 11 (old).
Find the domain of the following real functions :
(i) f(x) = \(\sqrt{x-2}\)
(ii) f(x) = \(\sqrt{2-3 x}\)
(iii) f(x) = \(\frac{1}{\sqrt{x-1}}\)
(iv) f(x) = \(\frac{1}{\sqrt{3-x}}\)
Solution:
(i) Given f(x) = \(\sqrt{x-2}\)
For D_f; f(x) must be a real number
∴ \(\sqrt{x-2}\) must be a real number
⇒ x – 2 ≥ 0
⇒ x ≥ 2
∴ D_f = [2, ∞)

(ii) Given f(x) = \(\sqrt{2-3 x}\)
For D_f : f(x) must be a real number
∴ \(\sqrt{2-3 x}\) must be a real number.
⇒ 2 – 3x ≥ 0
⇒ 2 ≥ 3x
⇒ x ≤ \(\frac{2}{3}\)
∴ D_f = (- ∞, \(\frac{2}{3}\)]

(iii) Given f(x) = \(\frac{1}{\sqrt{x-1}}\)
For D_f : f(x) must be a real number
⇒ \(\frac{1}{\sqrt{x-1}}\) must be a real number.
∴ x – 1 > 0
⇒ x > 1
Thus, D_f = (1, ∞)

(iv) Given f(x) = \(\frac{1}{\sqrt{3-x}}\)
For D_f : f(x) must be a real number
⇒ \(\frac{1}{\sqrt{3-x}}\) must be a real number.
∴ 3 – x > 0
⇒ x < 3
Thus, D_f = (- ∞, 3).

Question 12.
Find the domain and the range of the sy’ function f(x) = 2x² + 1. Also find f(- 2) and the numbers which are associated with the number 51 in its range.
Solution:
Given f(x) = 2x² + 1
For D_f; f(x) must be a real number
⇒ 2x² + 1 must be a real number, which is clearly a real number ∀ x ∈ R
∴ D_f = R
For R_f;
let y = f(x) = 2x² + 1
since x² > 0 ∀ x ∈ R
⇒ 2x² + 1 ≥ 1
⇒ y ≥ 1
∴ R_f = [1, ∞)
∴ f(- 2) = 2 (- 2)² + 1 = 9
Now let x ∈ D_f s.t f(x) = 51
⇒ 2x+ 1 = 51
⇒ 2x² = 50
⇒ x² = 25
⇒ x = ± 5

Question 13.
Find the domain and the range of the following functions :
(i) f(x) = \(\sqrt{x+2}\)
(ii) f(x) = \(\sqrt{3-2 x}\)
(iii) f(x) = \(\frac{1}{\sqrt{x-5}}\) (NCERT Exemplar)
Solution:
(i) Given f(x) = \(\sqrt{x+2}\)
For D_f; f(x) must be a real number
⇒ \(\sqrt{x+2}\) must be a real number
⇒ x + 2 ≥ 0
⇒ x ≥ – 2
∴ D_f = [- 2, ∞)
For R_f;
let y = f(x) = \(\sqrt{x+2}\)
since x ≥ – 2
⇒ x + 2 ≥ 0
⇒ \(\sqrt{x+2}\) ≥ 0
⇒ y ≥ 0
∴ R_f = [0, ∞)

(ii) Given f(x) = \(\sqrt{3-2 x}\)
For D_f; f(x) must be a real number
⇒ \(\sqrt{3-2 x}\) must be a real number
⇒ 3 – 2x ≥ 0
⇒ 2x ≤ 3
⇒ x ≤ \(\frac{3}{2}\)
∴ D_f = (- ∞, \(\frac{3}{2}\)]
For R_f
since x ≤ \(\frac{3}{2}\)
⇒ 2x ≤ 3
⇒ 3 – 2x ≥ 0
⇒ \(\sqrt{3-2 x}\) ≥ 0
⇒ y ≥ 0
∴ R_f = [0, ∞)

(iii) Given f(x) = \(\frac{1}{\sqrt{x-5}}\)
For D_f; f(x) must be a real number
⇒ \(\frac{1}{\sqrt{x-5}}\), must be a real number.
⇒ x – 5 > 0
⇒ x > 5
∴ D_f = (5, ∞)
For R_f;
Let y = f(x) = \(\frac{1}{\sqrt{x-5}}\)
since x > 5
⇒ x – 5 > 0
⇒ \(\sqrt{x-5}\) > 0
⇒ \(\frac{1}{\sqrt{x-5}}\) > 0
⇒ y > 0
∴ R_f = (0, ∞)

Question 14.
Find the domain and the range of the following functions :
(i) f(x) = \(\frac{4-x}{x-4}\) (NCERT Exemplar)
(ii) f(x) = \(\frac{x^2-9}{x-3}\)
(iii) f(x) = \(\frac{x+1}{2 x+1}\)
(iv) f(x) = \(\frac{4+x}{4-x}\)
Solution:
(i) Given f(x) = \(\frac{4-x}{x-4}\)
For D_f; f(x) must be a real number
∴ \(\frac{4-x}{x-4}\) must be a real number
∴ x – 4 ≠ 0
⇒ x ≠ 4
Thus, D_f = R – {4}
For R_f:
let y = f(x) = \(\frac{4-x}{x-4}\), x ≠ 4
⇒ y = – \(\frac{x-4}{x-4}\) = – 1
∴ R_f = {- 1}

(ii) Given f(x) = \(\frac{x^2-9}{x-3}\)
For D_f; f(x) must be a real number
⇒ \(\frac{x^2-9}{x-3}\) must be a real number.
⇒ x – 3 ≠ 0
⇒ x ≠ 3
∴ D_f = R – {3}
For R_f;
let y = f(x) = \(\frac{x^2-9}{x-3}\), x ≠ 3
⇒ y = \(\frac{(x-3)(x+3)}{x-3}\) = x + 3 (∵ x ≠ 3)
⇒ x = y – 3,
Since x ≠ 3
⇒ y ≠ 6, also x ∈ R
⇒ x + 3 ∈ R
⇒ y ∈ R
∴ R_f = R – {6}

(iii) Given f(x) = \(\frac{x+1}{2 x+1}\)
For D_f; f(x) must be a real number
⇒ \(\frac{x+1}{2 x+1}\) must be a real number.
⇒ 2x + 1 ≠ 0
⇒ x ≠ \(\frac{1}{2}\)
∴ D_f = R – {- \(\frac{1}{2}\)}
For R_f;
Let y = f(x) = \(\frac{x+1}{2 x+1}\)
⇒ 2xy + y = x + 1
⇒ 2xy – x = 1 – y
⇒ x (2y – 1) = 1 – y
⇒ x = \(\frac{1-y}{2 y-1}\)
but x ∈ R
⇒ \(\frac{1-y}{2 y-1}\), must be a real number
∴ 2y – 1 ≠ 0
⇒ y ≠ \(\frac{1}{2}\)
Thus, R_f = set of all real numbers except \(\frac{1}{2}\)
i.e., R_f = R – {\(\frac{1}{2}\)}

(iv) Given f(x) = \(\frac{4+x}{4-x}\)
For D_f;
f(x) must be a real number 4 + x
⇒ \(\frac{4+x}{4-x}\) must be a real number
⇒ 4 – x ≠ 0
⇒ x ≠ 4
∴ D = R – {4}
For R_f;
let y = f(x) = \(\frac{4+x}{4-x}\)
⇒ 4y – xy = 4 + x
⇒ 4y – 4 = x (y + 1)
⇒ x = \(\frac{4 y-4}{y+1}\)
But x ∈ R
⇒ \(\frac{4 y-4}{y+1}\) must be a real number
∴ y + 1 ≠ 0
y ≠ – 1
∴ R_f = R – {- 1}

Question 15.
Find the domain and the range of the following functions :
(i) f(x) = \(\sqrt{16-x^2}\) (NCERT Exemplar)
(ii) f(x) = \(\sqrt{x^2-9}\)
(iii) f(x) = \(\frac{1}{\sqrt{4-x^2}}\)
Solution:
(i) Given f(x) = \(\sqrt{16-x^2}\)
For D_f : f(x) must be a real number.
∴ \(\sqrt{16-x^2}\) must t>e a real number.
⇒ 16 – x² > 0
=> x² < 16
⇒ | x | ≤ 4
⇒ – 4 ≤ x ≤ 4
⇒ x ∈ [- 4, 4]
∴ D_f = [- 4, 4]
For R_f;
Let y = f(x) = \(\sqrt{16-x^2}\) ……………(1)
Since the square root of a real number is always non-negative.
∴ y ≥ 0
From eqn. (1) ;
y² = 16 – x²
⇒ x² = 16 – y²
Since x² ≥ 0
⇒ 16 – y² ≥ 0
⇒ y² ≤ 16
⇒ |y| ≤ 4
⇒ – 4 ≤ y ≤ 4 but y ≥ 0
⇒ 0 ≤ y ≤ 4
R_f = [0, 4]

(ii) Given f(x) = \(\sqrt{x^2-9}\)
For D_f; f(x) must be a real numebr
\(\sqrt{x^2-9}\) must be a real number
⇒ \(\sqrt{x^2-9}\) ≥ 0
∴ x² – 9 ≥ 0
⇒ |x| ≥ 3
⇒ x ≥ 3 or x ≤ – 3
∴ D_f = (- ∞, – 3] ∪ [3, ∞)
For R_f
Let y = f(x) = \(\sqrt{x^2-9}\) …………..(1)
since square root of a real number is always non-negative
∴ y ≥ 0
∴ from (1); y² = x² – 9
⇒ x² = y² + 9
since x² ≥ 0 ∀x ∈ R
⇒ y² ± 9 ≥ 0
⇒ y² ≥ – 9 which is true ∀ y ∈ R,also y ≥ 0
∴ R_f = [0, ∞)

(iii) Given f(x) = \(\frac{1}{\sqrt{4-x^2}}\)
For D_f; f(x) must be a real number.
∴ \(\frac{1}{\sqrt{4-x^2}}\) must be a real number.
⇒ 4 – x² > 0
⇒ x² < 4
⇒ |x| < 2
⇒ – 2 < x < 2
Thus D_f = (- 2, 2)
For R_{f ;}
let y = f(x) = \(\frac{1}{\sqrt{4-x^2}}\)
Since the square root of a real number is always non-negative.
∴ y > 0
On squaring eqn. (1) ; we have
y² = \(\frac{1}{4-x^2}\)
⇒ 4 – x² = \(\frac{1}{y^2}\)
⇒ x² = 4 – \(\frac{1}{y^2}\)
Since x² ≥ 0 ∀ x ∈ R
⇒ 4 – \(\frac{1}{y^2}\) ≥ 0, since y² > 0
⇒ 4y² – 1 ≥ 0
⇒ y² – \(\frac{1}{4}\) ≥ 0
(y – \(\frac{1}{2}\)) (y + \(\frac{1}{2}\)) ≥ 0
y ≥ \(\frac{1}{2}\) or y ≤ \(\frac{-1}{2}\), but y > 0
Thus y ≥ \(\frac{1}{2}\)
∴ R_f = [\(\frac{1}{2}\), ∞).

Question 16.
Find the domain and the range of the following functions :
(i) f(x) = |x – 3| (NCERT Exemplar)
(ii) f(x) = 3 – |x – 2|
Solution:
(i) Given f(x) = | x – 3 |
For D_f ; f(x) must be a real number
∴ x – 3 must be a real number, which is clearly a real number for all x ∈ R
∴ D_f = R
For R_f;
let y = f(x) = |x – 3|
Since |x| ≥ 0 ∀ x ∈ R
∴ x – 3 ≥ 0 ∀ x ∈ R
⇒ y ≥ 0
∴ R_f = [0, ∞)

(ii) Given f(x) = 3 – | x – 2|
For D_f; f(x) must be a real number
∴ 3 – |x – 2| must be a real number which is clearly a real number for all x ∈ R.
∴ D_f = R
For R_f; Let y = f(x) = |x – 3|
Since |x| ≥ 0 ∀ x ∈ R
∴ |x – 2| ≤ 0 ∀ x ∈ R
⇒ y ≤ 3
∴ R_f = [- ∞, 3)

Question 17.
If a real function fis defined by f(x) = \(\frac{|x|-x}{2 x}\), then find its domain and range.
Solution:
Given f(x) = \(\frac{|x|-x}{2 x}\)
For D_f ; f(x) must be a real number
∴ \(\frac{|x|-x}{2 x}\) must be a real number
⇒ 2x ≠ 0
⇒ x ≠ 0;
∴ D_f = R – {0}
For R_f;
Let y = f(x) = \(=\frac{|x|-x}{2 x}\), x ≠ 0
When x > 0, | x | = x
∴ f(x) = \(\frac{x-x}{2 x}\) = 0
When x < 0, |x| = – x
∴ f(- x) = \(\frac{-x-x}{2 x}=-\frac{2 x}{2 x}\) = – 1 [∵ x ≠ 0]
∴ R_f = {0, – 1}

Question 18.
Find the domain and the range of the function f defined by f(x) = \(\frac{|x-4|}{x-4}\). (NCERT Exemplar)
Solution:
Given f(x) = \(\frac{|x-4|}{x-4}\)
For domain of f (x), f (x) must be a real number
⇒ \(\frac{|x-4|}{x-4}\) must be a real number
⇒ x – 4 ≠ 0
⇒ x ≠ 4
∴ D_f set of all reals except 4 = R – {4}

For R_f:
let y = f(x) = x – 4

Case – I:
When x – 4 > 0
∴ y = \(-\frac{(x-4)}{x-4}\) = – 1

Case-II: When x – 4 < 0
⇒ | x – 4 | = – (x – 4)
∴ y = \(-\frac{(x-4)}{x-4}\) = – 1
Thus, R_f = {- 1, 1).

Question 19.
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 \(\left(\frac{1+x}{1-x}\right)\);
and f(y) = log \(\left(\frac{1+y}{1-y}\right)\)
L.H.S = f(x) + f(y)
= log \(\left(\frac{1+x}{1-x}\right)\) + log \(\left(\frac{1+y}{1-y}\right)\)
= log \(\left[\frac{(1+x)(1+y)}{(1-x)(1-y)}\right]\)
[ ∵ log a + log b = log ab]
= log \(\left[\frac{1+x+y+x y}{1-x-y+x y}\right]\) …………(1)

R.H.S = f\(\left(\frac{x+y}{1+x y}\right)\)
= log \(\left[\frac{1+\frac{x+y}{1+x y}}{1-\frac{x+y}{1+x y}}\right]\)
= log \(\left[\frac{1+x+y+x y}{1+x y-x-y}\right]\) ………………(2)

From eqn. (1) and eqn. (2); we have
L.H.S. = R.H.S.
⇒ f(x) + f(y) = f\(\left(\frac{x+y}{1+x y}\right)\)

Question 20.
Find the domain of the following functions :
(i) log (4x – 3)
(ii) \(\frac{1}{\log \left(9-x^2\right)}\)
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, 2√2}

Question 21.
Find the domain of the following functions :
(i) f(x) = \(\frac{1}{3-2 \sin x}\)
(ii) \(\frac{1}{\sqrt{1-\cos x}}\).
Solution:
(i) Given, f(x) = \(\frac{1}{3-2 \sin x}\)
For D_f; f(x) must be a real number
∴ \(\frac{1}{\sqrt{1-\cos x}}\) must be a real number.
i.e. 3 – 2 sin x ≠ 0
⇒ sin x ≠ \(\frac{3}{2}\), which is true for all x.
∴ D_f = R

(ii) Given f(x) = \(\frac{1}{\sqrt{1-\cos x}}\)
For D_f; f(x) must be a real number.
⇒ \(\frac{1}{\sqrt{1-\cos x}}\) must be a real number
⇒ 1 – cos x > 0
⇒ 1 > cos x
⇒ cos x < 1, which is true ∀ x ∈ R
Also 1 – cos x ≠ 0 and cos x ≠ 1 = cos 0
⇒ x ≠ 2nπ ∀ n ∈ I
∴ D_f = R – {2nπ, n ∈ I}

Question 22.
Find the range of the following functions :
(i) f(x) = 2 – 3 cos x
(ii) f(x) = 2 + 5 sin 3x.
Solution:
(i) Let y = f(x) = 2 – 3 cos x
since – 1 ≤ cos x < 3
⇒ – 3 ≤ 3 cos x ≤ 3
⇒ 3 ≥ – 3 cos x ≥ – 3
⇒ 5 ≥ 2 – 3 cos x ≥ – 1
⇒ – 1 ≤ 2 – 3 cos x ≤ 5
⇒ – 1 ≤ f(x) ≤ 5
⇒ – 1 ≤ y ≤ 5
∴ R_f = [- 1, 5]

(ii) Let y – f(x) = 2 + 5 sin 3x
Since – 1 < sin 3x < 1 ∀ x ∈ R
⇒ – 5 ≤ 5 sin 3x ≤ 5
⇒ 2 – 5 ≤ 2 + 5 sin 3x ≤ 5 + 2
⇒ – 3 ≤ y ≤ 7
∴ R_f= [- 3, 7]