ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.4

Continuous practice using ML Aggarwal Class 11 ISC Solutions Chapter 2 Relations and Functions Ex 2.4 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.4

Question 1.
If f(x) = x², find \(\frac{f(1.1)-f(1)}{1.1-1}\)
Solution:
Given f(x) = x²
∴ f (1.1) = (1.1)² = 1.21
∴ \(\frac{f(1.1)-f(1)}{1.1-1}=\frac{1.21-1}{1.1-1}\)
= \(\frac{0.21}{0.1}\)
= \(\frac{21}{100} \times \frac{10}{1}\)
= \(\frac{21}{10}\) = 2.1.

Question 2.
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) = \(\frac{3(-3)^2+2(-3)-1}{-3+1}\)
= \(\frac{27-6-1}{-2}\)
= \(\frac{20}{-2}\) = – 10
∴ f (- 3) + 1 = – 10 + 1 = – 9.

Question 3.
Is the relation f defined by f(x) = \(\begin{cases}x^2, & 0 \leq x \leq 3 \\ 2 x, & 3 \leq x \leq 5\end{cases}\) a function ? Justify your answer.
Solution:
Relation f defined by
f(x) = \(\begin{cases}x^2, & 0 \leq x \leq 3 \\ 2 x, & 3 \leq x \leq 5\end{cases}\)
when x = 3, f(3) has two values 3² and 2 × 3 i.e. 9 and 6.
Hence 3 has two images 9 and 6.
Thus relation f is not a function.
Since every element has no unique image.

Question 4.
A real functionf 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
Now – 3 ∈ R
∴ f (- 3)= – 9 – 2 = – 11
0, 7 ∈ R
∴ f (0) = 0 – 2 = – 2
and f(7) = 21 – 2 = 19
Thus, f (- 3) + f (0) × f (7)
= – 11 + (- 2) × 19
=- 11 – 38 = – 49

Question 5.
Given f(x) = x³ – 1, find x if x is 215.
Solution:
Givenf(x) = x³ – 1,
since f(x) = 215
⇒ x³ – 1 = 215
⇒ x³ = 16
⇒ x² – 6² = 0
⇒ (x – 6) (x² – 6x + 36) = 0
⇒ x = 6 ∈ R
since x² + 6x + 36 = 0 does not gives real values of x.

Short answer questions (6 to 10) :

Question 6.
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:
Given f(x) = 2x² + 1
and domain of f = {- 2, – 1, 0, 1, 3}
(i) f (- 2) = 2 (- 2)² + 1 = 9 ;
f (1) = 2 (1)² + 1 = 3
f (- 1)² = (- 1)² + 1 = 3 ;
f (3) = 2 (3)² + 1 = 19
f (0) = 2 (0)² + 1 = 1
Thus R_f = {9, 3, 1, 3, 19}

(ii) f (3) × f (- 2) = 19 × 9 = 171

(iii) Since, f(x) = 9
⇒ 2x² + 1 = 9
⇒ 2x² = 8
⇒ x² = 4
⇒ x = ± 2

Question 7.
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 a functionf defined by
f(x) = x² + 3, x ∈ N and x ≤ 5
(i) Since x ∈ N, x ≤ 5
∴ x = 1, 2, 3, 4, 5
When x = 1,
f(x) = f(1)
= 1² + 3 = 4
When x = 2,
f(x) = f(2)
= 2² + 3 = 7
Wben x = 3,
f(x) = f(3)
= 3² + 3 = 12
When x = 4,
f(x) = f(4)
= 4² + 3 = 19
When x = 5,
f(x) = f(5)
= 5² + 3 = 28
∴ Range of f = {4, 7, 12, 19, 28}

(ii) f(2) × f(5) = 7 × 28 = 196

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

(iv) Since it is given that f(x) = 7
⇒ x² + 3 = 7
⇒ x² = 4
⇒ x = ± 2 but x = – 2N
∴ x = 2.

Question 8.
Find the domain of the following real functions :
(i) f(x) = \(\frac{x^2-4}{x+2}\)
(ii) f(x) = \(\frac{3 x}{28-x}\) (Exemplar)
(iii) f(x) = \(\frac{x}{x^2+3 x+2}\) (Exemplar)
(iv) f(x) = \(\frac{x^3-x+3}{x^2-1}\) (Exemplar)
Solution:
(i) Given 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
⇒ x + 2 ≠ 0
⇒ x ≠ – 2
Thus, D_f = set of all real number except – 2
= 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) 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 real number
⇒ x² + 3x + 2 ≠ 0
⇒ x = – 1, – 2
∴ D_f = set of all real numbers except – 1 and – 2.
i.e. R – {- 1, – 2}

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

Question 9.
Find the range of the following functions:
(i) f(x) = x² + 2, x ∈ R
(ii) f(x) = 3- 2x, x ∈ R, x ≥ 1.
Solution:
(i) Given f(x) = x² + 2, x ∈ R
For R: since x² ≥ 0 ∀ x ∈ R
⇒ x² + 2 ≥ 2
⇒ f (x) ≥ 2 ∀ x ∈ R
∴ R_f = [2, ∞)

(ii) f(x) = 3 – 2x, x ∈ R, x ≥ 1
since x ≥ 1
⇒ – 2x ≤ – 2
⇒ 3 – 2x ≤ 3 – 2 = 1
⇒ f(x) ≤ 1 ∀ x ∈ R, x ≥ 1
∴ R_f = (- ∞, 1].

Question 10.
Find the domain and the range of the 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
2x + 1 must be a real number, which is a real number for all x ∈ R
∴ D_f = R
For R_f : since x² ≥ 0 ∀ x ∈ R
2x² ≥ 0
⇒ 2x² + 1 ≥ 1
⇒ f (x) ≥ 1 ∀ x ∈ R
∴ R_f = [1, ∞)
Now f (- 2) = 2 (- 2)² + 1 = 9
[∵ – 2 ∈D_f]
Let x ∈ D_f s.t f (x) = 51
⇒ 2x² + 1 = 51
⇒ 2x² = 50
⇒ x² = 25
⇒ x = ± 5.

Long answer questions (11 to 18) :

Question 11.
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) = \(\frac{1}{\sqrt{x-5}}\)
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 square root of real number is always non -negative
∴ y ≥ 0
[As x ≥ – 2
⇒ x + 2 ≥ 0
⇒ \(\sqrt{x+2}\) ≥ 0
⇒ y ≥ 0]
On squaring ; we have
y² = x + 2
⇒ x = y² – 2
Since x ≥ – 2
⇒ y² – 2 ≥ – 2
⇒ y² ≥ 0 ∀ y ∈ R
∴ 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
⇒ 3 ≥ 2x
⇒ 2x ≤ 3
⇒ x ≤ \(\frac{3}{2}\)
∴ D_f = (- ∞, \(\frac{3}{2}\)]
For R_f :
Let y = f(x) = \(\sqrt{3-2 x}\)
as 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) = \(\frac{1}{\sqrt{x-5}}\) ∀ x ∈ D_f
since x ∈ D_f
⇒ x > 5
⇒ x – 5 > 0
⇒ \(\sqrt{x-5}\) > 0
⇒ \(\frac{1}{\sqrt{x-5}}\) > 0
⇒ y > 0
∴ R_f = (0, ∞).

Question 12.
Find the domain and the range of the following functions:
(i) f(x) = \(\frac{4-x}{x-4}\)
(ii) f(x) = \(\frac{x^2-9}{x-3}\)
(iii) f(x) = \(\frac{x+1}{2 x+1}\)
(iv) f(x) = \(\frac{4+x}{4-x}\)
(v) f(x) = \(\frac{1}{4-x^2}\)
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
∴ D_f = R – {4}
For R_f :
Let y = f(x) = \(\frac{4-x}{x-4}\) ∀ x ∈R
i.e. x ≠ 4
⇒ y = \(\frac{x-4}{-(4-x)}\) = – 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 ∈D_f
since x ∈ D_f
⇒ x ≠ 3
⇒ x – 3 ≠ 0
⇒ x = y – 3 and x ≠ 3
⇒ y – 3 ≠ 3
⇒ y ≠ 6
∴ 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)
= \(\frac{x+1}{2 x+1}\) ∀ x ∈ R – {- \(\frac{1}{2}\)}
⇒ 2xy + y = x + 1
⇒ 2xy – x = 1 – y
⇒ x = \(\frac{1-y}{2 y-1}\) but x ∈ R – {- \(\frac{1}{2}\)}
⇒ \(\frac{1-y}{2 y-1}\) must be a real number
[since \(\frac{1-y}{2 y-1} \neq-\frac{1}{2}\)]
⇒ 2y – 1 ≠ 0
⇒ y ≠ \(\frac{1}{2}\)
∴ 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
⇒ \(\frac{4+x}{4-x}\) must be a real number
⇒ 4 – x ≠ 0
⇒ x ≠ 4
∴ D_f = R – {4}
For R_f :
Let y = f (x)
= \(\frac{4+x}{4-x}\) ∀ x ∈ R – {4}
⇒ 4y -xy = 4 + x
⇒ 4y – 4 = x (1+y)
⇒ x = \(\frac{4(y-1)}{1+y}\) ∈ R
[since x ≠ 4 if \(\frac{4(y-1)}{1+y}\) = 4
⇒ 4y – 4 = 4 + 4y
⇒ – 4 = 4, which is false]
⇒ \(\frac{4(y-1)}{1+y}\) must be a real number
⇒ 1 + y ≠ 0
⇒ y ≠ – 1
∴ R_f = R – {- 1}

(v) Given f(x) = \(\frac{1}{4-x^2}\)
For D_f : f (x) must be a real number
⇒ \(\frac{1}{4-x^2}\) must be a real number
⇒ 4 – x² ≠ 0
⇒ x² ≠ 4
⇒ x ≠ ± 2
∴ D_f = R – {- 2, 2}
Let R_f :
Let y = f(x) ∀ x ∈ D_f
⇒ 4 – x² = \(\frac{1}{y}\), y ≠ 0
⇒ x² = 4 – \(\frac{1}{y}\), y ≠ 0
But x² ≥ 0 ∀ x ∈ D_f
⇒ 4 – \(\frac{1}{y}\) ≥ 0, y ≥ 0
[Multiplying both sides by positive real number y² ≥ 0]
⇒ y² (4 – \(\frac{1}{4}\)) ≥ 0
⇒ y (4y – 1) ≥ 0
⇒ y (y – \(\frac{1}{4}\)) ≥ 0
⇒ y ≥ \(\frac{1}{4}\) or y ≤ 0 but y ≠ 0
⇒ R_f = (- ∞, 0) ∪ [\(\frac{1}{4}\), ∞).

Question 13.
Find the domain and the range of the following functions :
(i) f (x) = \(\sqrt{16-x^2}\)
(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 be a real number
⇒ 16 – x²2 ≥ 0
⇒ x² ≤ 16
⇒ |x| ≤ 4
⇒ – 4 ≤ x ≤ 4
∴ D_f = [- 4, 4]
For R_f :
Let y = f (x) =\(\sqrt{16-x^2}\) ∀ x ∈ D_f
Since square root of a real number is always non-negative y ≥ 0
On squaring ; we have
y² = 16 – x²
⇒ x² = 16 – y²
but x² ≥ 0 ∀ x ∈ D_f
⇒ 16 – y² ≥ 0
⇒ – 4 ≤ y ≤ 4 but y ≥ 0
∴ 0 ≤ y ≤ 4
Thus, R_f = [0, 4].

(ii) Given f (x) = \(\sqrt{x^2-9}\)
For D_f : f (x) must be a real number
⇒ \(\sqrt{x^2-9}\) must be a real number
⇒ x² – 9 ≥ 0
⇒ x² ≥ 9
⇒ |x| ≥ 3
⇒ x ≥ 3 or x ≤ – 3
∴ D_f = (- ∞, – 3] ∪ [3, ∞)
For R_f :
Let y = f (x) = \(\sqrt{x^2-9}\) ∀ x ∈ D_f
Since square root of a real number is always non-negative
∴ y ≥ 0
On squaring ; we have
y² = x² – 9
⇒ x² = y² + 9
Since x² ≥ 0 ∀ x ∈ D_f
⇒ y² ≥ – 9 ∀ y ∈ R
but y ≥ 0
∴ R_f = [0, x).

(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
∴ D_f = (- 2, 2)
For R_f :
Let y = f (x) = \(\frac{1}{\sqrt{4-x^2}}\) y ≠ 0
Since square root of a real number is always non-negative
∴ y ≥ 0
On squaring ; we have
y² = \(\frac{1}{4-x^2}\)
⇒ 4 – x² = \(\frac{1}{y^2}\)
Since x² ≥ 0 ∀ x ∈ D_f
⇒ 4 – \(\frac{1}{y^2}\) ≥ 0
Multiplying both sides by a positive real number i.e. y²
∴ y² (4 – \(\frac{1}{y^2}\)) ≥ 0
⇒ (4y² – 1) ≥ 0
⇒ y² ≥ \(\frac{1}{4}\)
⇒ |y| ≥ \(\frac{1}{2}\)
⇒ y ≥ \(\frac{1}{2}\) or y ≤ – \(\frac{1}{2}\) but y > 0
⇒ y ≥ \(\frac{1}{2}\)
∴ R_f = [\(\frac{1}{2}\) , ∞).

Question 14.
Find the domain and the range of the following functions :
(i) f (x) = |x – 3|
(ii) f (x) = 3 – |x – 2|
Solution:
(i) f (x) = |x – 3|
For D_f : f(x) must be a real number
⇒ |x – 3| must be a real number
which is a real number for all x ∈ R
∴ D_f = R
For R_f:
Let y = f (x) = |x – 3| ∀ x ∈ R
since |x| ≥ 0 ∀ x ∈ R
⇒ |x – 3| ≥ 0
⇒ y ≥ 0.
∴ R_f = [0, ∞)

(ii) f (x) = 3 – |x – 2|
For D_f : f (x) must be a real number
⇒ 3 – |x – 2| must be a real number
⇒ \(\frac{|x|-x}{2 x}\) must be a real number.
which is a real number for all x ∈ R
∴ D_f = R
For R_f :
Let y = f (x) = 1 – |x – 2|
Since |x| ≥ 0 ∀ x ∈ R
∴ |x – 2| ≥ 0 ∀ x ∈ R
⇒ – |x – 2| ≤ 0
⇒ 3 – |x – 2| ≤ 3
⇒ y ≤ 3
∴ R_f = (- ∞, 3].

Question 15.
If a real functionf is 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
⇒ x ≠ 0
∴ D_f = R – {0}
Let R_f :
Let y = f (x)
= \(\frac{|x|-x}{2 x}\) ∀ x ∈ D_f
i.e. x ≠ 0
When x > 0, |x| = x
∴ y = \(\frac{x-x}{2 x}\) = 0
When x < 0, |x| = – x
∴ y = \(\frac{-x-x}{2 x}=\frac{-2 x}{2 x}\) = – 1
[∵ x ≠ 0]
∴ R_f = {0, – 1}

Question 16.
Find the domain and the range of the function f defined by f (x) = \(\frac{|x-4|}{x-4}\).
Solution:
Given f (x) = \(\frac{|x-4|}{x-4}\)
For D_f : f (x) must be a real number
⇒ \(\frac{|x-4|}{x-4}\) must be a real number
⇒ x – 4 ≠ 0
⇒ x ≠ 4
∴ D_f = R – {4}
For R_f :
Let y = \(\frac{|x-4|}{x-4}\) ∀ x ∈ D_f
∴ x ≠ 4
Case – I :
When x > 4
⇒ x – 4 > 0
⇒ |x – 4| = x – 4
∴ y = \(\frac{x-4}{x-4}\) = 1 [∵ x ≠ 4]
Case – II :
When x < 4
⇒ x – 4 < 0
∴ y = \(-\frac{(x-4)}{x-4}\) = – 1
[∵ x ≠ 4]
∴ R_f = {1, – 1}.

Question 17.
If f and g are two real functions defined by f(x) = 2x + 1 and g (x) = 4x – 7, then for what real numbers x
(i) f(x) = g(x) ?
(ii) f(x) < g(x) ?
Solution:
(i) Given f(x) = 2x – 1
and g (x) = 4x – 7
Since f(x) = g (x)
⇒ 2x + 1 = 4x – 7
⇒ 2x = 8
⇒ x = 4
(ii) Since f (x) < g (x)
⇒ 2x + 1 < 4x – 7
⇒ 8 < 2x
⇒ 2x > 8
⇒ x < 4.

Question 18.
Find the domain for which the functions f (x) = 3x² – 1 and g (x) = 3 + x are equal.
Solution:
Given f(x) = 3x² – 1
and g (x) = 3 + x
Since the functions f and g are equal
∴ f (x) = g (x)
⇒ 3x² – 1 = 3 + x
⇒ 3x² – x – 4 = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x = – 1, \(\frac{4}{3}\)
Thus, the domain for which functions f and g are equal be {- 1, \(\frac{4}{3}\)}.