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

Parents can use ISC Maths Class 11 Solutions Chapter 2 Relations and Functions MCQs to provide additional support to their children.

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

Choose the correct answer from the given four options in questions (1 to 32) :

Question 1.
If A = {- 1, 2, 5, 8}, B = {0, 1, 3, 6, 7} and R be the relation “is one less than” from to B, then k as a set of ordered pairs is
(a) ((- 1, 0), (2, 3), (5, 6)}
(b) {(- 1, 0), (2. 1), (8, 7)}
(c) {(0, 1), (2. 3), (6, 7)}
(d) {(1, 2), (2, 3), (5, 6), (6, 7), (7, 8)}
Answer:
(a) ((- 1, 0), (2, 3), (5, 6)}

Given A = {- 1, 2, 5, 8}
and B = {0, 1, 3, 6, 7}
(a, b) ∈ R a is one less than b ∀ a ∈ A, b ∈ B
Since 0 – 1 = – 1
∴ (- 1, 0) ∈ R
[∵ a – b = – 1]
2 – 3 = – 1
∴ (2, 3) ∈ R
5 – 6 = – 1
∴ (5, 6) ∈ R
∴ R = {(- 1, 0), (2. 3), (5, 6)}

Question 2.
If R = {(x, y) : x, y ∈ 2x + y = 8}, then domain of R is
(a) {0, 1, 2, 3, 4, 5}
(b) {0, 1, 2, 3, 4, 5, 6}
(c) {0, 1, 2, 3, 4}
(d) {0, 1, 2, 3}
Answer:
(c) {0, 1, 2, 3, 4}

Given 2x + y = 8 ∀ x, y ∈ W
⇒ y = 8 – 2x
when x = 1
∴ y = 8 – 2 = 6 ∈ W
∴ (1, 6)R
when x = 2
∴ y = 8 – 4 = 4 ∈ W
∴ (2, 4) ∈ W
when x = 0
∴ y = 8 – 0 = 8 ∈ W
∴ (0, 8) ∈ W
when x = 3
∴ y = 8 – 6 = 2 ∈ W
∴ (3, 2) ∈ W
when x = 4
∴ y = 8 – 8 = 0 ∈ W
∴ (4, 0) ∈ W
For other values of x, y ∉ W
∴ D_R = {0, 1, 2, 3, 4)

Question 3.
If R = {(x, y) : x, y ∈ N, x + 2y = 21}, then domain of R is
(a) {1, 2, 3, ………… 7, 8}
(b) {1, 2, 3, …………. 9, 10}
(c) {1, 3, 5, 7 ………….. 19}
(d) {1, 3, 5, 7, ………….. 15}
Answer:
(b) {1, 2, 3, …………. 9, 10}

Given x + 2y = 21
⇒ y = \(\frac{21-x}{2}\), x, y ∈ N
when x = 1
∴ y = \(\frac{21-1}{2}\) = 10 ∈ N
∴ (1, 10) ∈ R
when x = 2
∴ y = \(\frac{21-2}{2}=\frac{19}{2}\) ∉ N
∴ (2, \(\frac{19}{2}\)) ∉ R
when x = 3
∴ y – \(\frac{18}{2}\) = R
∴ (3, 9) ∈ R
when x = 4, 6, 8, 10, 12, 14, 16, 18, 20,
∴ y ∉ N
when x = 5, 7, 9, 11,13,15,17,19
∴ y = 8, 7, 6, 5, 4, 3, 2, 1
Thus (5, 8), (7, 7), (9, 6) (11, 5), (13, 4), (15, 3), (17, 2), (19, 1) ∈ R
For other values of x ∈ N, we does not get y ∈ N.
∴ R = {(1, 10), (3, 9), (5, 8), (7, 7), (9, 6), (11, 5), (13, 4), (15, 3), (17, 2), (19, 1)}
∴ Range of R = {1, 2, 1, ……………….. 9, 10}

Question 4.
Let A = {1, 2, 3, 4, 5, 6} and R be the relation defined on A by R = {(x, y) : x, y ∈ A, x divides y}, then range of R is
(a) {2, 3, 4 5, 6}
(b) {1, 2, 3, 4, 5}
(c) {2, 4, 6}
(d) {1, 2, 3, 4, 5, 6}
Answer:
(d) {1, 2, 3, 4, 5, 6}

Given R = ((x, y) : x, y ∈ A, x divides y)
Since 1/1, 1/2, ¡13, 1/4, 1/5, 1/6
∴ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1,6) ∈ R
2/2, 2/4, 2/16
∴ (2, 2) (2, 4), (2, 6) ∈ R
3/3, 3/6
∴ (3, 3), (3, 6) ∈ R
4/4, 5/5, 6/6
∴ (4, 4), (5, 5), (6, 6) ∈ R
∴ Range of R = {1, 2, 3, 4, 5, 6}.

Question 5.
Let A = (3, 5) and B = (7, 11) and R be the relation from A to B defined as R = {(a, b) : a ∈ A,b ∈ B, a – b is odd}, then
(a) R = A × B
(b) R = Φ
(c) R ⊂ A × B
(d) R ⊂ B × A
Answer:
(b) R = Φ

Given A = {3, 5}
and B = {7, 11}
and R = {(a, b) : a ∈ A, b ∈B, a- b is odd}
Since difference of two odd integers is always an even integer.
Here a and b both are odd.
∴ a – b is always even.
∴ R = Φ

Question 6.
Given R = {(x, y) : x, y ∈ Z, y = x – 3}, then which ordered pair belongs to R}
(a) (1, 4)
(b) (0, 3)
(c) (5, 2)
(d) (- 4, 1)
Answer:
(c) (5, 2)

Given R = {(x, y) ; x, y ∈ Z, y = x – 3}
(5, 2) ∈ R
[∵ 2 = 5 – 3]

Question 7.
If R = {(x, y) : x, y ∈ W, x² + y² = 169}, then domain of R is
(a) {0, 5, 12, 13}
(b) {- 13, – 12, – 5, 0, 5, 12, 13}
(c) {0, 1, 2, 3, 13}
(d) {0, ± 1, ± 2, …………….. ± 3}
Answer:
(a) {0, 5, 12, 13}

Given x² + y² = 169, x, y ∈ W
y = ± \(\sqrt{169-x^2}\) but y ∈ W
y = \(\sqrt{169-x^2}\)
when x = 0 ; y = 13
and when x = 5
∴ y = \(\sqrt{169-25}\) = 12
when x = 1, 2,3, 4. 6, 7, 8. 9, 10, 11 we have y ∉ W
when x =12 ;
y = \(\sqrt{169-144}\)
when x = 13 ; y = 0
For other values of x ∈ W, we have y ∉ W
∴ R = {(0, 13), (5, 12), (12, 5), (13, 0)}
∴ D_R = {0, 5, 12, 13}.

Question 8.
If A = {a, b} and B = {x, y, z}, then the number of relations from B to A is
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(d) 64

A = {a, b}
∴ n (A) = 2
and B = {x, y, z}
∴ n (B) = 3
∴ No. of relations from B to A = 2^{n (A × B)}
= 2^{n (A) n (B)}
= 2^{2 × 3} = 64

Question 9.
Let n (A) = m and n (B) = n, then the number of non-empty relations from A to B is
(a) mⁿ
(b) n^m – 1
(c) 2^mn – 1
(d) 2^mn
Solution:
(c) 2^mn – 1

Since relation is a subset of A × B
∴ No. of non-empty relations from A to B = 2^{n (A × B)}
= 2^{n(A) × n (B)} = 2 m^mn – 1
[Since Φ ⊆ A x ß ∴ Φ is also a relation]

Question 10.
Let A be a finite set containing n elements, then (he number of relations on A is
(a) 2ⁿ
(b) 2ⁿ²
(c) n²
(d) nⁿ
Solution:
(b) 2ⁿ²

Given A be a finite set containing n element.
Then relation on A is a subset of A × A.
∴ No. of relations on A = 2^{n (A × A)}
= 2^{n (A) × n (A)}
= 2ⁿ²

Question 11.
If A = {2, 3, 4, 5, 6} and R is a relation on setA defined by R = {(x, y) : y = x + 2, x, y ∈ A}, then R is
(a) {(2, 4), (3, 5), (4, 6)}
(b) {(4, 2), (5, 3), (6, 4)}
(c) ((2, 4), (5, 3), (4, 6)}
(ð) ((4, 2), (3, 5), (4, 6)}
Solution:
(a) {(2, 4), (3, 5), (4, 6)}

Given A = {2, 3, 4, 5, 6}
and R = {(x, y) : y = x + 2, x, y ∈ A}
When x = 2
∴ y = 2 + 2 = 4
∴ (2, 4) ∈ R
When x = 3
∴ y = 3 + 2 = 5
∴ (3, 5) ∈ R
When x = 4
∴ y = 4 + 2 = 6
∴ (4, 6) ∈ R
Thus, R = {(2, 4), (3, 5), (4, 6)}

Question 12.
Which of the following relations is a function?
(a) R = ((4, 6), (3, 9), (- 11, 6), (3, 11)}
(b) R = ((2, 1), (4, 3), (6, 5), (8, 7), (10, 9)}
(c) R = ((1, 2), (2, 4), (2, 6), (3, 5)}
(d) R = ((0, 1), (1, 3), (2, 4), (3, 1), (3, 5)}
Answer:
(c) R = ((1, 2), (2, 4), (2, 6), (3, 5)}

For option (a) ;
The element 3 has two images 9 and 11.
∴ given relation is not a function.
For option (b) ;
The element 2 have two images 4 and 6
∴ every element does not
have unique image
∴ given relation is not a function.
For option (d) ;
The element 3 has two images 1 and 5 and hence not a function.
In option (c) ;
every element has unique image
∴ given relation represents a function.

Question 13.
Which of the following arrow diagrams represents a function from X to Y?
(a)

(b)

(c)

(d)

Answer:
(c)

In option (b) ;
An element a ∈ X have two images p and s.
So it does not represents a function. Since every element in domain X has unique image in codomain Y.
In option (c) ;
An element c ∈ X has no image in Y. Thus does not represents a function.
In option (d) ;
An element a ∈ X has two images, Thus arrow diagram does not represents a function.
In option (a) ;
every element in X has unique image in Y.
∴ it represents a function.

Question 14.
Let A and B be two finite sets, then the number of functions from A to B is
(a) n (A) . n (B)
(b) 2 n (A) . n (B)
(c) {n (A)}^{n (B)}
(d) {n (B)}^{n (A)}
Answer:
(d) {n (B)}^{n (A)}

The required no. of functions from A to B = {n (B)}^{n (A)}

Question 15.
Let A be a finite set containing 3 elements, then the number of functions from A to B is
(a) 512
(b) 511
(c) 27
(d) 26
Answer:
Given A has 3 elements
∴ n(A) = 3
Thus, the no. of functions from A to A = [n(A)]^{n (A)}
= 3³ = 27

Question 16.
The domain of the function f defined by f(x) = \(\sqrt{a^2-x^2}\), (a > 0) is
(a) (- a, a)
(b) [- a, a]
(c) [0, a]
(d) (- a, 0]
Answer:
(c) [0, a]

Given A has 3 elements
∴ n (A) = 3
Thus, the no. of functions from A to A = [n(A)]^{n (A)}
= 3³ = 27

Question 17.
The domain of the function f defined by f(x) = \(\sqrt{x^2-9}\) is
(a) [- 3, 3]
(b) (- 3, 3)
(c) (- ∞, – 3] ∪ [3, ∞)
(d) [0, 3]
Answer:
(c) (- ∞, – 3] ∪ [3, ∞)

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
⇒ |x| ≥ 3
⇒ x ≥ 3 or x ≤ – 3
⇒ x ∈ [3, ∞) ∪ (- ∞ , – 3]
∴ D_f = [3, ∞) ∪ (- ∞ , – 3].

Question 18.
The domain of the function f defined by f (x) = \(\frac{x^2+2 x+3}{x^2-x-6}\) is
(a) R – [3, – 2]
(b) R – {- 3, 2}
(c) R – {3, – 2}
(d) R – (- 3, 2)
Answer:
(c) R – (3, – 2)

For D_f : f (x) must be a real number
⇒ \(\frac{x^2+2 x+3}{x^2-x-6}\) must be a real number
⇒ x² – x -6 ≠ 0
⇒ (x + 2) (x – 3) ≠ 0
⇒ x ≠ – 2, 3
∴ D_f = R – {- 2, 3}.

Question 19.
The domain of the functionf defined by f(x) = \(\frac{1}{\sqrt{|x|-x}}\)
(a) (- ∞, 0)
(b) (0, ∞)
(c) (- ∞, 0)
(d) Φ
Answer:
(a) (- ∞, 0)

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

Question 20.
The domain and range of thc real function f defincd by f (x) = \(\) are
(a) Domain = \(\left\{-\frac{1}{2}, \frac{1}{2}\right\}\),
Range = (- ∞, – 1] ∪ (0, ∞)
(b) Domain = R – \(\left\{-\frac{1}{2}, \frac{1}{2}\right\}\),
Range = (- ∞, – 1] ∪ (0, ∞)
(c) Domain = \(\left[-\frac{1}{2}, \frac{1}{2}\right]\),
Range = (- ∞, 0) ∪ (0, ∞)
(d) Domain = R – \(\left\{-\frac{1}{2}, \frac{1}{2}\right\}\),
Range = (- ∞, – 1] ∪ (2, ∞)
Solution:
(b) Domain = R – \(\left\{-\frac{1}{2}, \frac{1}{2}\right\}\),
Range = (- ∞, – 1] ∪ (0, ∞)

For D_f : f (x) must be a real number
⇒ \(\frac{1}{4 x^2-1}\) must be a real number
⇒ 4x² – 1 ≠ 0
⇒ x ≠ ± \(\frac{1}{2}\)
∴ D_f = R – {± \(\frac{1}{2}\)}
For R_f :
f (x) = \(\frac{1}{4 x^2-1}\) ∀ x ∈D_f
x² ≥ 0
⇒ 4x² – 1 ≥ – 1
⇒ \(\frac{1}{4 x^2-1}\) ≤ – 1
Let y = \(\frac{1}{4 x^2-1}\)
⇒ 4x² – 1 = \(\frac{1}{y}\)
⇒ x² = \(\frac{1+y}{4 y}\)
⇒ \(\frac{1+y}{4 y}\) ≥ 0
⇒ y (1 + y) ≥ 0
[∵ y² > 0 ∀ y ≠ 0 ∈R]
⇒ y ≥ 0 or y ≤ – 1
But y ≠ 0
⇒ y > 0 or y ≤ – 1
⇒ y ∈ (- ∞, – 1] ∪ (0, ∞)
∴ R_f = (- ∞, – 1] ∪ (0, ∞)

Question 21.
The domain and range of thereal function f defined by f(x) = \(\sqrt{x-1}\) are
(a) Domain = (1, ∞), Range = (0, ∞)
(b) Domain = [1, ∞), Range = (0, ∞)
(c) Domain = [1, ∞), Range = [0, ∞)
(d) Domain = (1, ∞), Range = [0, ∞)
Answer:
(c) Domain = [1, ∞), Range = [0, ∞)

For D_f : f (x) must be a real number
⇒ \(\sqrt{x-1}\) must be a real number
⇒ x – 1 ≥ 0
⇒ x ≥ 1
⇒ x ∈ [1, ∞)
∴ D_f = [1, ∞)
For range of f (x) ;
Let y = f (x)
= \(\sqrt{x-1}\) ∀ x ∈ D_f
⇒ x ≥ 1
⇒ x – 1 ≥ 0
⇒ \(\sqrt{x-1}\) ≥ 0
⇒ f (x) ≥ 0
∴ R_f = [0, ∞)

Question 22.
The domain and range of the real function f defined by f (x) = \(\frac{x-2}{2-x}\) are
(a) Domain = R – {2}, Range = {- 1}
(b) Domain = R – {- 2}, Range = {- 1}
(c) Domain = R – {- 2}, Range = {1}
(d) Domain = R – {2}, Range = {1}
Answer:
(a) Domain = R – {2}, Range = {- 1}

For D_f : f (x) must be a real number
⇒ \(\frac{x-2}{2-x}\) must be a real number
⇒ 2 – x ≠ 0
⇒ x ≠ 2
∴ D_f = R – {2}
For R_f :
f (x) = \(\frac{x-2}{2-x}\) ∀ x ∈ D_f
∴ x ≠ 2
∴ f (x) = – 1
∴ R_f = {- 1}.

Question 23.
The domain and range of the real functionf defined by \(\) are
(a) Domain = R, Range = {- 1, 1}
(b) Domain = R – {0}, Range = {- 1, 0, 1}
(c) Domain = R – {0}, Range = {- 1, 1}
(d) Domain = R, Range = {- 1, 0, 1}
Answer:
(c) Domain = R – {0}, Range = {- 1, 1}

For D_f : f (x) must be a real number
⇒ \(\frac{x}{|x|}\) ∈ R
∴ D_f = R – {0}
For R_f:
Let y = f (x)
= \(\frac{x}{|x|}\) ∀ x ∈ D_f
∴ x ≠ 0
when x > 0
∴ y = \(\frac{x}{x}\) = 1
when x < 0
∴ y = \(\frac{x}{x}\) = – 1
∴ R_f = {- 1, 1}

Question 24.
The domain and range of the function f given by f (x) = 2 – |x – 5| are
(a) domain = R⁺, Range = (- ∞, 1]
(b) Domain = R, Range = (- ∞, 2]
(c) Domain = R⁺, Range = (- ∞, 2]
(d) Domain = R, Range = (- ∞, 2}
Answer:
(b) Domain = R, Range = (- ∞, 2]

For D_f : f (x) must be a real number
⇒ 2 – |x – 5| mustbeareal number.
which is always a real number.
∴ D_f = R
For R_f :
|x – 5| ≥ 0
⇒ – |x – 5| ≤ 0
⇒ 2 – |x – 5| ≤ 2
⇒ f (x) ≤ 2
∴ R_f = (- ∞, 2]

Question 25.
The domain of the function f defined by f (x) = \(\sqrt{a-x}+\frac{1}{\sqrt{x^2-a^2}}\) is
(a) (- ∞, a]
(b) (- ∞, – a)
(c) (- ∞, – a)
(d) (a, ∞)
Answer:
(c) (- ∞, – a)

Given
f (x) = \(\sqrt{a-x}+\frac{1}{\sqrt{x^2-a^2}}\)
= g (x) + h (x)
where g (x) = \(\sqrt{a-x}\)
and h (x) = \(\frac{1}{\sqrt{x^2-a^2}}\)
For D_g : g (x) must be a real number
⇒ \(\sqrt{a-x}\) must be a real number
⇒ a – x ≥ 0
⇒ x ≤ a
⇒ x ∈ (- ∞, a]
∴ D_f = (- ∞, – a]
For D_f : h (x) must be a rcal number.
⇒ \(\frac{1}{\sqrt{x^2-a^2}}\) must be a real number
⇒ x² – a² > 0
⇒ x² > a²
⇒ |x| > a
⇒ x > a or x < – a
⇒ x ∈ (- ∞, – a) i (ci. a)
∴ D_f = D_g ∩ D_h
= (- x, a] ∩ {(- ∞, – a) ∪ (a, ∞)}
= (- ∞, – a)

Question 26.
The domain of the function f defined by f (x) = log (5 – 6x) is
(a) (- ∞, \(\frac{5}{6}\))
(b) (\(\frac{5}{6}\), ∞)
(c) (- ∞, \(\frac{5}{6}\))
(d) [\(\frac{5}{6}\), ∞)
Answer:
(a) (- ∞, \(\frac{5}{6}\))

For D_f : f (x) must be a real number
⇒ log_e (5 – 6x) ∈ R
⇒ 5 – 6x > 0
⇒ 5 > 6x
⇒ 6x < 5
⇒ x < \(\frac{5}{6}\)
∴ x ∈ (- ∞, \(\frac{5}{6}\))
∴ D_f = (- ∞, \(\frac{5}{6}\))

Question 27.
The domain of the function f (x) = \(\frac{1}{4-x^2}\) + log₁₀ (x² – x) is
(a) (- ∞, 0) ∪ (1, ∞)
(b) (- ∞ , 0) ∪ (1, ∞) – {- 2, 2}
(c) R – {- 2, 2}
(d) (- ∞, 0] ∪ [1, ∞) – {- 2, 2}
Solution:
(b) (- ∞ , 0) ∪ (1, ∞) – {- 2, 2}

Given f (x) = \(\frac{1}{4-x^2}\) + log₁₀ (x² – x)
f (x) = g (x) + h (x)
where g (x) = \(\frac{1}{4-x^2}\)
and h (x) = log₁₀ (x² – x)
For D_g : g (x) must be a real number.
⇒ \(\frac{1}{4-x^2}\) must be a real number.
⇒ 4 – x² ≠ 0
⇒ x ≠ ± 2
∴ D_g = R – {- 2, 2}
For D_h : h (x) must be a real number.
⇒ log₁₀ (x² – x) must be a real numbcr.
⇒ x² – x > 0
⇒ x (x – 1) > 0
⇒ x > 1 or x < 0
[∵ if a > b, (x – a) (x – b) > 0
⇒ x > a or x < b]
∴ D_h = (- ∞, 0) ∪ (1, ∞)
Thus D_f = D_g ∩ D_h
= (- ∞, 0) ∩ (1, ∞) – {- 2, 2).

Question 28.
If [x]² – 3 [x] + 2 = 0, where [.] denotes the greatest integer function, then
(a) x = {2, 3}
(b) x ∈ (1, 2]
(c) x ∈ [1, 2]
(d) x ∈ [1, 3)
Answer:
(d) x ∈ [1, 3)

Given [x]² – 3 [x] + 2 = 0
⇒ ([x] – 1) ([x] – 2) = 0
⇒ [x] = 1 or [x] = 2
⇒ 1 ≤ x < 2 or 2 ≤ x < 3
⇒ 1 ≤ x < 3
⇒ x ∈ [1, 3)

Question 29.
If f (x) = px + q, where p and q are integers f (- 1) = 1 and f (2) = 13, then p and q are
(a) p = 4, q = 5
(b) p = – 4, q = 5
(c) p = – 4, q = – 5
(d) p = 4, q = – 5
Answer:
(a) p = 4, q = 5

Given f (x) = p x + q ……………..(1)
f (- 1) = 1
∴ from (1) ;
1 = – p + q ……………(2)
f (2) = 13
∴ from (1) ;
13 = 2p + q
eqn. (3) – eqn. (2) ; we have
12 = 3p
⇒ p = 4
∴ from (2) ;
q = 5.

Question 30.
Let f (x) = \(\), then
(a) f (xy) = f (x) f (y)
(b) f (xy) ≥ f (x) f (y)
(c) f (xy) = f (x) . f (y)
(d) None of these
Solution:
(c) f (xy) = f (x) . f (y)

Given f (x) = \(\sqrt{1+x^2}\)
∴ f (xy) = \(\sqrt{1+x^2 y^2}\)
and f (y) = \(\sqrt{1+y^2}\)
Since (1 + x²) (1 + y²) = 1 + x² + y² + x²y² ≥ 1 + x²y²
⇒ \(\sqrt{\left(1+x^2\right)\left(1+y^2\right)} \geq \sqrt{1+x^2 y^2}\)
⇒ f (x) f (y) ≥ f (xy)

Question 31.
The domain for which the functions defined by f(x) = 6x² + 1 and g (x) = 11 – 7x are equal is
(a) {- 1, \(\frac{2}{3}\)}
(b) {3, \(\frac{5}{6}\)}
(c) {- 2, \(\frac{5}{6}\)}
(d) {2, \(\frac{2}{3}\)}
Answer:
(c) {- 2, \(\frac{5}{6}\)}

f (x) = g (x)
⇒ 6x² + 1 = 11 – 7x
⇒ 6x² + 7x – 10 = 0
⇒ (x + 2) (6x – 5) = 0
⇒ x = – 2, \(\frac{5}{6}\)
∴ D_f = D_g = {- 2, \(\frac{5}{6}\)}

Question 32.
If f (x) – 3 f(\(\frac{1}{x}\)) = 2x + 3 (x ≠ 0) then f (3) is equal to
(a) – \(\frac{3}{2}\)
(b) – \(\frac{5}{2}\)
(c) \(\frac{7}{2}\)
(d) – 1
Answer:
(b) – \(\frac{5}{2}\)

Given f (x) – 3 f (\(\frac{1}{x}\)) = 2x + 3 ……………(1)
Changing x to \(\frac{1}{x}\) in eqn. (1) ; we have
f (\(\frac{1}{x}\)) – 3 f (x) = 2 × \(\frac{1}{x}\) + 3 …………….(2)
eqn. (1) + 3 eqn. (2) ; we have
– 8 f (x) = 2x + 3 + 3 (\(\frac{2}{x}\) + 3)
⇒ – 8 f (x) = 2x + \(\frac{6}{x}\) + 12
⇒ – 8 f (3) = 2 × 3 + \(\frac{6}{3}\) + 12 = 20
⇒ f (3) = \(-\frac{20}{8}=-\frac{5}{2}\)