ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.5

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.5

Question 1.
Let f : {1, 2, 3} → {a, b, c} be a function defined by f (1) = a, f (2) = b and f (3) = c. Show that f is invertible and find f^-1. Also show that (f^-1)^-1 = f. (NCERT)
Solution:
Given, f : {1, 2, 3} → [a, b, c} defined by f (1) = a ; f (2) = b ; f (3) = c
∴ Different elements have different images
∴ f is one-one.
∀ y ∈ {a, b, c} ∃ x ∈ {1, 2, 3} s.t f (x) = y
∴ f is onto.
∴ f is one-one, onto
∴ f^-1 exists
and f^-1 = {(a, 1), (6, 2), (c, 3)}, also f^-1 is 1 – 1, onto therefore its inverse exists.
i.e. (f^-1)^-1 exists.
Thus (f^-1)^-1 = {(1, a), (2, b), (3, c)} = f.

Question 2.
Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as below have inverse. Find f^-1 if it exists.
(i) f = {(1, 1), (2, 2), (3, 3)}
(ii) f = {(1, 2), (2, 1), (3, 1)}
(iii) f = {(1, 3), (3, 2), (2, 1)}
(iv) f = {(1, 3), (2, 2), (3, 1)}.
Solution:
Given, f : S → S defined by
f = {(1, 1), (2, 2), (3, 3)} and S = {1, 2, 3}
∴ f (1) = 1 ; f (2) = 2 ; f (3) = 3
As different elements in domain of S. has different images in S (codomain off)
Also for any y ∈ S ∃ x ∈ S s.t.f (x) = y
∴ f is 1 – 1, onto
Thus f^-1 exists and f^-1 (1) = 1 ;
f^-1 (2) = 2 ; f^-1 (3) = 3
i.e., f^-1 = {(1, 1), (2, 2), (3, 3)} =f.

Question 3.
Explain why the following functions f : X → Y do not have inverses :
(i) X = R – {0}, Y = R and f(x) = \(\frac{1}{x}\) for all x ∈ X.
(ii) X = {- 1, 0, 1, 2}, Y = {2, 3} and f (x) = 2 for x = – 1, 0, 1 and f (2) = 3.
(iii) X = R = Y and f (x) = x² + 1 for all x ∈ R.
Solution:
(i) Given f : X → Y defined by
f(x) = \(\frac{1}{x}\) ∀ x ∈ X
where X = R – {0} and Y = R Since 0 e Y = R
Then f(x) = 0 ⇒ – = 0 does not gives any
real solution in X = R.
Hence 0 has no pre-image in X
∴ f is not onto.
Thus f is not invertible.
∴ f : X → Y do not have inverse.

(ii) Given X = {- 1, 0, 1, 2} ; Y = {2, 3} and
f(x) = 2 for X = – 1, 0, 1
So different elements – 1, 0, 1 of X have same image 2 in Y.
∴ f is not 1 – 1.
Hence f is not invertible.
∴ f : X → Y do not have inverse.

(iii) Given X = R = Y
and f (x) = x² + 1 ∀ x ∈ R
Clearly, 1, – 1 ∈ R s.t f(1) = 1² + 1 = 2
and f(- 1) = (- 1)² + 1 = 2
∴ different elements of X have same images in Y.
∴ f is not one-one.
Thus f is not invertible.
Thus f → Y donot have inverse.

Question 4.
Let S = {a, b, c} and T = {1, 2, 3}. Find f^{– 1} of the following functions f from S to T, if it exists.
(i) f ={(a, 3),(b, 2),(c, 1)}
(ii) f = {(a, 2), (b, 1 ),(c, 1)}
Solution:
(i) Given, S = {a, b, c} and T = {1, 2, 3} and f : S → T
defined by f(a) = 3, f(b) = 2, f(c) = 1
Since different elements in S have different images in T
∴ f is one-one.
Also ∀ y ∈ T ∃ x ∈ S s.t. f(x) = y
∴ f is onto.
∴ f is 1 – 1, onto.
Thus f^{– 1} exists s.t. f^{– 1} : T → S
and f^{– 1} = {(3, a), (2, b), (1, c)}

(ii) Given, f(a) = 2, f(b) = 1, f(c) = 1
∴ The elements b, c have images 1 under f.
∴ f is not 1 – 1.
∴ f^{– 1} does not exists.

Question 5.
Let R be the set of all real numbers. Show that the functions f : R → R defined by f(x) = 4x + 5 is invertible. Also find inverse of f. (NCERT Exemplar)
Solution:
Given f : R → R is defined as f (x) = 4x + 5
Injectivity : ∀ x, y ∈ R such that f (x) = f(y)
⇒ 4x + 5 = 4y + 5
⇒ x = y
∴ f is one-one.

Surjectivity:
Let y ∈ R be any arbitrary element.
Then f (x) = y
⇒ 4x + 5 = y
⇒ x = \(\frac{y-5}{4}\)
Since y ∈ R
⇒ \(\frac{y-5}{4}\) ∈ R i.e. x ∈ R
Thus, ∀ y ∈ R ∃ x ∈ R s.t.
⇒ f(x) = f (\(\frac{y-5}{4}\))
= 4 × \(\frac{y-5}{4}\) + 5 = y
Thus f is onto.
Hence f is one – one and onto.
∴ f is bijective.
Thus f^{– 1} exists.
Let x ∈ R (domain) and y ∈ R (codomain).
Then f(x) = y
⇒ 4x + 5 = y
⇒ x = \(\frac{y-5}{4}\)
⇒ f^{– 1} (y) = \(\frac{y-5}{4}\)
Hence f^{– 1} : R → R is defined by
f^{– 1}(x) = \(\frac{x-5}{4}\) ∀ x ∈ R.

Question 6.
Show that the function f : R → R defined by f(x) = \(\frac{2 x-1}{3}\) is one-one and onto. Also find the inverse of the function f.
Solution:
Given a function f : R → R defined by
f(x) = \(\frac{2 x-1}{3}\) ∀ x ∈ R

Injectivity :
∀ x, y ∈ R s.t f (x) = f(y)
⇒ \(\frac{2 x-1}{3}\) = \(\frac{2 y-1}{3}\)
⇒ 2x – 1 = 2y – 1
⇒ x = y
∴ f is injective i.e. one-one

Subjectivity :
Let y ∈ R be any arbitrary element then f (x) = y
⇒ \(\frac{2 x-1}{3}\) = y
⇒ x = \(\frac{3 y+1}{2}\)
Since y ∈ R
⇒ \(\frac{3 y+1}{2}\) ∈ R
⇒ x ∈ R
Thus, ∀ y ∈ R ∃ 3x ∈ R
s.t f(x) = f (\(\frac{3 y+1}{2}\))
= \(\frac{2\left(\frac{3 y+1}{2}\right)-1}{3}\) = y
∴ f is surjective i.e. onto.
Hence f is 1 – 1 and onto i.e.f is bijective.
Thus f is invertible and its inverse exists.

To find f^{– 1} : Let f(x) = y
⇒ \(\frac{2 x-1}{3}\) = y
Since y = f(x) and f is invertible
f^{– 1} (y) = x = \(\frac{3 y+1}{2}\)
Thus, inverse fiinction f^{– 1} : R → R defined by
f^{– 1} (y) = \(\frac{3 y+1}{2}\) ∀ y ∈ R.

Question 7.
Consider a bijective function
f : R⁺ ₀ → (7, ∞) given by f(x) = 16x² + 24x + 7, where R is the set of positive real numbers. Find the inverse function of f.
Solution:
Given bijective function f : R⁺ ₀ → b (7, ∞) given by
f(x) = 16x² + 24x + 7

To find f^-1 :
Let f(x) = y
⇒ 16x² + 24x + 7y
⇒ (4x + 3)² – 2 = y
⇒ (4x + 3)² = y + 2
4x + 3 = \(\sqrt{y+2}\)
[∵ X ∈ R⁺ ₀ ⇒ 4x + 3 > 0]
⇒ x = \(\frac{\sqrt{y+2}-3}{4}\)
Now f(x) = y and f is bijective i.e. one-one and onto and hence invertible.
f^-1 (y) = \(\frac{\sqrt{y+2}-3}{4}\)

Question 8.
If A = R – {- 3) and B = R – {2} and a function f : A → B be given by f (x) = \(\frac{2 x+1}{x+3}\), then show that the function f is one-one and onto. Hence, find f^-1.
Solution:
Given a functionf: A → B be defined by
f (x) = \(\frac{2 x+1}{x+3}\)
where A = R – {- 3}and B = R – {2}

Injectivity :
∀ x, y ∈ A s.t f(x) = f(y)
⇒ \(\frac{2 x+1}{x+3}=\frac{2 y+1}{y+3}\)
⇒ (2x + 1) (y + 3) = (2y + 1) (x + 3)
⇒ 2xy + 6x + y + 3 = 2xy + 6y + x + 3
⇒ 5x = 5y
⇒ x = y
Thus f is injective i.e. one-one.

Subjectivity :
Let y ∈ B be any arbitrary element.
Then y = f(x) and y ≠ 2, y ∈ R
⇒ \(\frac{2 x+1}{x+3}\) = y
⇒ 2x + 1 = xy + 3y
⇒ x (2 – y) = 3y – 1
⇒ x = \(\frac{3 y-1}{2-y}\)
Since y ∈ R
\(\frac{3 y-1}{2-y}\) ∈ R as y ≠ 2
Also, \(\frac{3 y-1}{2-y}\) ≠ – 3
[If \(\frac{3 y-1}{2-y}\) = – 3
⇒ 3y – 1 = – 6 + 3y
⇒ I = 6, which is false]
∴ x = \(\frac{3 y-1}{2-y}\) ∈ R – {- 3} = A
f(x) = f \(\left(\frac{3 y-1}{2-y}\right)\)
= \(\frac{2\left(\frac{3 y-1}{2-y}\right)+1}{\frac{3 y-1}{2-y}+3}\)
= \(\frac{6 y-2+2-y}{3 y-1+6-3 y}\) = y
Thus f is onto.
Hrnce f is one-one, onto.
∴ f is invertible.
Thus f^-1.
⇒ x = \(\frac{1-3 y}{y-2}\)
Since y = f(x) and f be invertible.
∴ f^-1 (y) = x
⇒ f^-1 (y) = x = \(\frac{1-3 y}{y-2}\)
Thus inverse function f^-1; B → A defined by
f^-1 (y) = \(\frac{1-3 y}{y-2}\) ∀ y ∈ B.

Question 9.
If f : A → A and A = R – {\(\frac{8}{5}\)}, show that the function f (x) = \(\frac{8 x+3}{5 x-8}\) is one-one onto. Hence, find f^-1. (ISC 2019)
Solution:
Given f : A → A and A = R – {\(\frac{8}{5}\)} defined 8x + 3 by f(x) = \(\frac{8 x+3}{5 x-8}\)

one-one:
∀ x, y ∈ A s.t. f(x) = f(y)
⇒ \(\)
⇒ (8x + 3) (5y – 8) = (8y + 3) (5x – 8)
⇒ 40xy – 64x + 15y – 24 = 40xy – 64y + 15x – 24
⇒ 79x = 79y
⇒ x = y .
Thus f is one-one.

onto:
Let us find the range of f i.e. R_f,
Let y = f(x) = \(\frac{8 x+3}{5 x-8}\)
⇒ y (5x – 8) = 8x + 3
⇒ 5xy – 8y = 8x + 3
⇒ x (5y – 8) = 8y + 3
⇒ x = \(\frac{8 y+3}{5 y-8}\)

Now x ∈ R if 5y – 8 ≠ 0
⇒ x ≠ \(\frac{8}{5}\)
Thus R_f = R – {\(\frac{8}{5}\)} = A
∴ R_f = Codomain of f = A
⇒ f is onto.

To find f^-1 :
Let y = f(x) = \(\frac{8 x+3}{5 x-8}\)
⇒ 5xy – 8y = 8x+3
⇒ x (5y – 8) = 8y + 3
⇒ x = \(\frac{8 y+3}{5 y-8}\)
Now y = f (x) and f is inversible.
Then x = f^-1(y) = \(\frac{8 y+3}{5 y-8}\)
∴ f^-1 : R → R defined by f^-1(x) = \(\frac{8 x+3}{5 x-8}\)

Question 10.
Consider the function
f : R – {- \(\frac{4}{3}\)} → R – {\(\frac{4}{3}\)} detined by f(x) = \(\frac{4 x+3}{3 x+4}\). Show that f is bijective. Find the inverse of f and hence, find f^-1(0) and x such that f^-1 (x) = 2.
Solution:
Given f : R – {- \(\frac{4}{3}\)} → R – {\(\frac{4}{3}\)} definef by
f(x) = \(\frac{4 x+3}{3 x+4}\)

One-one :
∀ x, y ∈ R – {- \(\frac{4}{3}\)} s.t. f(x) = f(y)
⇒ \(\frac{4 x+3}{3 x+4}=\frac{4 y+3}{3 y+4}\)
⇒ (4x + 3) (3y + 4) = (3x + 4) (4y + 3)
⇒ 12xy + 16x + 9y + 12 = 12xy + 9x + 16y + 12
⇒ 7x = 7y
⇒ x = y
Thus f is one-one.

Onto:
Let y ∈ R – {\(\frac{4}{3}\)} be any arbitrary element i.e. y ≠ \(\frac{4}{3}\) and y ∈ R
Let y = f(x) = \(\frac{4 x+3}{3 x+4}\)
⇒ y (3x + ) = 4x + 3
⇒ 3xy + 4y = 4x + 3
⇒ x (3y – 4) = 3 – 4y
⇒ x = \(\frac{3-4 y}{3 y-4}\)
Since y ∈ R and y ≠ \(\frac{4}{3}\)
∴ \(\frac{3-4 y}{3 y-4}\) ∈ R
Also, \(\frac{3-4 y}{3 y-4} \neq-\frac{4}{3}\)
[If \(\frac{3-4 y}{3 y-4}\) = \(\frac{4}{3}\)
⇒ 9 – 12y = – 12y + 16
⇒ 9 = 16, which is false]
∴ \(\frac{3-4 y}{3 y-4}\) ∈ R – {- \(\frac{4}{3}\)}
Thus ∀ y ∈ R – {- \(\frac{4}{3}\)} ∃
x = \(\frac{3-4 y}{3 y-4}\) ∈ R – {- \(\frac{4}{3}\)}
s.t f(x) = f (\(\frac{3-4 y}{3 y-4}\))
= \(\frac{4\left(\frac{3-4 y}{3 y-4}\right)+3}{3\left(\frac{3-4 y}{3 y-4}\right)+4}\)
= \(\frac{12-16 y+9 y-12}{9-12 y+12 y-16}\)
= \(\frac{-7 y}{-7}\) = y
Thus f is Onto.
∴ f is 1 – 1 and onto.
Hence f is bijective.
Thus, f^-1 exists.

To find f^-1 :
Let y = f(x) = \(\frac{4 x+3}{3 x+4}\)
⇒ x = \(\frac{3-4 y}{3 y-4}\),
Now y = f(x) and f is invertible.
⇒ f^-1(y) = \(\frac{3-4 y}{3 y-4}\)
Thus the function f^-1 : R → R defined by
f^-1 (y) = \(\frac{3-4 y}{3 y-4}\)
f^-1 (0) = \(\frac{3-0}{0-4}\)
= – \(\frac{3}{4}\)

Now given f^-1 (x) = 2
⇒ \(\frac{3-4 x}{3 x-4}\) = 2
⇒ 3 – 4x = 6x – 8
⇒ 11 = 10x
⇒ x = \(\frac{11}{10}\).

Question 11.
Consider the function f : R₊ → [9, oo) defined by f (x) = x² + 9, where R₊ is the set of all non-negative real numbers. Show that f is invertible. Also find the inverse of f.
Solution:
f : R⁺ → [9, ∞] defined by
f (x) = x² + 9 ∀ x ∈ R⁺
∀ x, y ∈ R such that f (x) = f (y)
⇒ x² + 9 = y² + 9
⇒ x² = y²
⇒ x = y [∵ x, y ∈ R⁺]
∴ f is one-one function.
Let y ∈ [9, ∞) and y = f(x₀)
then y = x₀² + 9
⇒ x₀ = \(\sqrt{y-9}\)
(∵ x₀ ∈ R⁺; x₀ ≥ 0)
as y ∈ [9, ∞)
⇒ \(\sqrt{y-9}\) ∈ R⁺
⇒ x₀ ∈ R⁺ ;
⇒ x₀ ≥ 0
[∵ y ≥ 9 ⇒ y – 9 ≥ 0]
Now f (x₀) = x₀² + 9
= (\(\sqrt{y-9}\))² + 9 = y
∴ ∀ y ∈ [9, ∞) ∃ x₀ ∈ R⁺ s.t. f (x₀) = y
∴ f is onto.
Thus f is 1 – 1, onto.
∴ f^-1 exists.
Now x = f^-1,(y) then
y = f (x) = x² + 9
⇒ x = (\(\sqrt{y-9}\)) (∵ x ∈ R⁺)
⇒ f^-1 (y) = \(\sqrt{y-9}\)
⇒ f^-1 (x) = \(\sqrt{y-9}\) ∀ x ∈ [9, ∞)

Question 12.
If R₀ is the set of all non-zero real numbers, show that the function f : R₀ → R₀ defined by f(x) = \(\frac{3}{x}\) is invertible and it is the inverse of itself.
Solution:
Given a function f : R₀ → R₀ defined by
f(x) = \(\frac{3}{x}\).
where R₀ set of all non-zero real numbers
i.e. R₀ = R – {0}

Injectivity :
∀ x, y ∈ R₀ s.t f(x) = f(y)
⇒ \(\frac{3}{x}\) = \(\frac{3}{y}\)
⇒ x = y [∵ x, y ≠ 0]
∴ f is injective i.e. one-one.

Surjectivity :
Let y ∈ R₀ be any arbitrary element
then y = f(x) = \(\frac{3}{x}\)
⇒ x = \(\frac{3}{y}\)
Since y ∈ R₀
= \(\frac{3}{y}\) ∈ R₀ as y ≠ 0
∴ x ∈ R₀
Thus, ∀ y ∈ R₀ ∃ x ∈ R₀
s.t f(x) = f (\(\frac{3}{y}\)) = \(\frac{3}{3}\) × y = y
∴ f is onto.
Thus, f is one-one and onto and hence bijective.
∴ f is invertible and f^-1 exists.

To find f^-1 :
Let f (x) = y
⇒ \(\frac{3}{x}\) = y
⇒ x = \(\frac{3}{y}\)
Since y = f(x) and f is invertible
∴ f^-1 (y) = x = \(\frac{3}{y}\)
Thus inverse function f^-1 : R₀ —+ R₀
defined by f^-1 (y) = \(\frac{3}{y}\)
f^-1 (x) = \(\frac{3}{x}\) ∀ x ∈ R₀
defined by f^-1 (y) = \(\frac{3}{y}\)
⇒ f^-1 (x) = \(\frac{3}{x}\) ∀ x ∈ R₀
also f(x) = \(\frac{3}{x}\) ∀ x ∈ R₀
∴ f^-1 (x) = f(x) ∀ x ∈ R₀
⇒ f^-1 = f
Hence, f is the inverse of itself.

Question 13.
Let f : N → 5 be a function defined by f(x) = 9x² + 6x – 5, where S is the range of f. Show thatf is inversible. Find the
inverse of f and hence find f^-1 (43) and f^-1 (163).
Solution:
Given f: R⁺ – [- 5, ∞) is defined by
f(x) = 9x² + 6x – 5

Injective:
∀ x, y ∈ R such that f(x) = f(y)
⇒ 9x² + 6x – 5 = 9y + 6y – 5
⇒ 9 (x – y) + 6(x – y) = 0
⇒ (x – y) [9 (x + y) + 6] = 0
⇒ x – y = 0
[∵ 9 (x + y) + 6 ≠ 0 ∀ x, y ∈ R]
⇒ x = y
Thus f is injective.

Surjective:
Let y ∈ [- 5, ∞) be any arbitrary element
Then f(x) = y
⇒ y = 9x + 6x – 5
⇒ y = (3x + 1)² – 6
⇒ y + 6 = (3x + 1)²
⇒ 3x + 1 = ± \(\sqrt{y+6}\)
⇒ x = \(\frac{-1 \pm \sqrt{y+6}}{3}\)
Since y ∈ [- 5, ∞)
⇒ y ≥ – 5
⇒ y + 6 ≥ 1
⇒ \(\sqrt{y+6}\) ≥ 1
⇒ \(\frac{-1+\sqrt{y+6}}{3}\) ≥ 0
⇒ x = \(\frac{-1+\sqrt{y+6}}{3}\) ≥ 0 i.e., x ∈ R⁺
Thus ∀ y ∈ [- 5, ∞) ∃ x ∈ R such that
f(x) = y
∴ f is onto.
Hence f is 1-1 and onto, thus f is bijective
∴ f^-1 exists and let f^-1 denotes the inverse of f.
Then (fof^-1)(x) = x ∀ x ∈ [- 5, ∞)
⇒ f [f^-1(x)] = x ∀ x ∈ [- 5, ∞)
⇒ 9 {f^-1 (x))² + 6 {f^-1(x)} – 5 = x ∀ x ∈ [- 5, ∞)
⇒ {3 f^-1(x) + 1}² = x + 6 ∀ x ∈ [- 5, ∞)
3 f^-1(x) + 1 = \(\sqrt{x+6}\) ∀ x ∈ [- 5, ∞)
f^-1(x) = \(\frac{-1+\sqrt{x+6}}{3}\), ∀ x ∈ [- 5, ∞)
f^-1(43) = \(\frac{-1+\sqrt{43+6}}{3}=\frac{-1+\sqrt{49}}{3}\)
= \(\frac{-1+7}{3}\) = 2
f^-1 (163) = \(=\frac{-1+\sqrt{163+6}}{3}\)
= \(\frac{-1+\sqrt{169}}{3}\)
= \(\frac{-1+13}{3}\) = 4.

Question 14.
If function f : R → R is defined by f(x) = 5x – 3,thcn find function g : R → R such that gof = I_R = fog.
Solution:
Given a function f: R → R defined by
f(x) = 5x – 3 ∀ x ∈ R
Let y ∈ R be any arbitrary element, as y ∈ R
⇒ \(\frac{y+3}{5}\) ∈ R
Let us define g : R → R by
g(y) = \(\frac{y+3}{5}\) ∀ y ∈ R
As, f : R → R and g : R → R,
∴ gof and fog both composite functions exists.
(gof) (x) = g (f{x)) = g (5x – 3)
= \(\frac{5 x-3+3}{5}\) = x ∀ x ∈ R
and (fog) (y) = f(g (y))
= f \(\left(\frac{y+3}{5}\right)\) = 5 \(\left(\frac{y+3}{5}\right)\) – 3
= y + 3 – 3 = y ∀ y ∈ R
∴ gof = fog = I_R
Thus, the required function g : R → R defined by
g(y) = \(\frac{y+3}{5}\) ∀ y ∈ R

Question 15.
If f = {(1, 5), (2, 3), (3, 0), (4, – 2)}, find f^-1.
Solution:
Given f = {(1, 5), (2, 3), (3, 0), (4, – 2)}
Clearly f(1) = 5 ; f(2) = 3 ; f(3) = 0 ; f(4) = – 2
∴ different elements have different images underf.
f is one-one.
Also corresponding to every element (in R_f) has pre-image in domain of f.
Thus f is onto.
Hence f is 1 – 1 and onto
∴ f^-1 exists.
∴ f^-1 = {(5, 1), 0, 2), (0, 3), (- 2, 4)}.

Question 16.
If f : R → R is a function defined by f(x) = 3x + 5, then find f^-1.
Solution:
Given f : R → R is defined by f(x) = 3x + 5 ∀ x ∈ R

Injectivity :
∀ x, y ∈ R s.t f (x) = f(y)
⇒ 3x + 5 = 3y + 5
⇒ x = y
∴ f is one-one

Surjectivity :
Let y ∈ R be any arbitrary element
Then f(x) = y
⇒ y = 3x + 5
⇒ x = \(\frac{y-5}{3}\)
Since y ∈ R
⇒ \(\frac{y-5}{3}\) ∈ R
⇒ x ∈ R
Thus ∀ y ∈ R ∃ x ∈ R
s.t. f(x) = f (\(\frac{y-5}{3}\)) = 3 (\(\frac{y-5}{3}\)) + 5 = y
∴ f is onto.
Thus, f is 1-1 and onto.
∴ f^-1 exists.
Let f(x) = y
⇒ 3x + 5 = y
⇒ x = \(\frac{y-5}{3}\)
Since y = f(x) and f is invertible.
∴ f^-1 (y) = x = \(\frac{y-5}{3}\)
Thus inversible function f^-1 : R → R defined by
f^-1 (y) = \(\frac{y-5}{3}\) ∀ y ∈ R

Question 17.
If f : R → R defined by f (x) = \(\frac{3 x-2}{5}\) is inversible, find the inverse of f.
Solution:
Given f : R → R defined by
f(x) = \(\frac{3 x-2}{5}\) is inversible
∴ f is one-onto and onto.
∴ inverse f i.e. f^-1 exists.
Let f(x) = y
⇒ y = \(\frac{3 x-2}{5}\)
⇒ x = \(\frac{5 y+2}{3}\)
Since y = f (x) and f is invertible.
∴ f^-1 (y) = x = \(\frac{5 y+2}{3}\)
Thus, inverse function f^-1 : R → R defined by
f^-1 (y) = \(\frac{5 y+2}{3}\) ∀ y ∈ R.

Question 18.
If f : R → R defined by f(x) = \(\frac{2 x-7}{4}\) is inversible, find f^-1.
Solution:
Given f : R → R defined by
f(x) = \(\frac{2 x-7}{4}\) is inversible
∴ f is one-one and onto.
∀ y ∈ R ∃ x ∈ R s.t. f(x) = y
⇒ y = f(x) = \(\frac{2 x-7}{4}\)
⇒ 4y = 2x – 7
⇒ x = \(\frac{4 y+7}{2}\)
Sincey = f(x) and f is invertible
∴ f^-1 (y) = x = \(\frac{4 y+7}{2}\)
Thus, inverse function f^-1 : R → R defined by
f^-1 (y) = \(\frac{4 y+7}{2}\) ∀ y ∈ R.

Question 19.
If f is an invertible function defined as f(x) = \(\frac{3 x-4}{5}\), write f^-1 (x).
Solution:
Given f^-1 (x) = \(\frac{3 x-4}{5}\) and f is invertible.
∴ f is one-one and onto.
Since f is onto.
∴ ∀ y ∈ R ∃ x ∈ R s.t. f(x) = y
⇒ \(\frac{3 x-4}{5}\) = y
⇒ x = \(\frac{5 y+4}{3}\)
Since, y = f(x) and f is invertible
∴ f^-1 (y) = x = \(\frac{5 y+4}{3}\)
f^-1 (x) = \(\frac{5 x+4}{3}\)
Thus inverse function f^-1 : R → R defined by
f^-1 (x) = \(\frac{5 x+4}{3}\) ∀ x ∈ R

Question 20.
If f : R → R is a function defined by f^-1 (x) = ax – b, a, b ∈ R, a ≠ 0, then find f^-1.
Solution:
Given f : R → R defined by
f(x) = ax – b, a, b ∈ R, a ≠ 0

Injectivity:
∀ x, y ∈ R s.t f(x) = f(y)
⇒ ax – b = ay – b
⇒ ax = ay (∵ a ≠ 0)
∴ f is one-one.

Subjectivity :
Let y ∈ R be any arbitrary element
Then f(x) = y
⇒ ax – b = y
⇒ x = \(\frac{y+b}{a}\)
since y ∈ R ⇒ \(\frac{y+b}{a}\) ∈ R (∵ a ≠ 0)
∴ x ∈ R
∀ y ∈ R ∃ x ∈ R
S.t f(x) = f (\(\frac{y+b}{a}\))
= a (\(\frac{y+b}{a}\)) – b
= y + b – b = y
Thus f is onto.
∴ f is one-one and onto.
∴ f is invertible and f^-1 exists.

To find f^-1 :
Let f(x) = y
y = ax – b
∴ \(\frac{y+b}{a}\) = x
since y = f(x) and f is invertible
∴ f^-1 (y) = x = \(\frac{y+b}{a}\)
Thus, inverse function f^-1 : R → R defined by
f^-1 (y) = \(\frac{y+b}{a}\) ∀ y ∈ R.

Question 21.
If f : R – {- \(\frac{3}{5}\)} → R is a function defined by f(x) = \(\frac{2 x}{5 x+3}\), then find f^-1 : range of f → R – {- \(\frac{3}{5}\)}.
Solution:
Given f : R – {- \(\frac{3}{5}\)} → R defined as
f(x) = \(\frac{2 x}{5 x+3}\)
∀ y ∈ R s.t. f(x) = f(y)
⇒ \(\frac{2 x}{5 x+3}=\frac{2 y}{5 y+3}\)
⇒ 10xy + 6x = 10 xy + 6y
⇒ x = y
∴ f is 1-1.
Obviously f : R – {- \(\frac{3}{5}\)} → R Range (f) is onto
∴ f is 1-1 and onto
∴ f^-1 exists.
Let f(x) = y
⇒ y = \(\frac{2 x}{5 x+3}\)
⇒ 5xy + 3y = 2x
⇒ x (5y – 2) = – 3y
⇒ x = \(\frac{3 y}{2-5 y}\) ∈ R – {- \(\frac{3}{5}\)}
[at x = \(-\frac{3}{5}\); \(-\frac{3}{5}\) = \(\frac{3 y}{2-5 y}\)
⇒ – 6 + 15y = 15y
⇒ – 6 = 0 which is wrong]
∴ f^-1 (y) = \(\frac{3 y}{2-5 y}\)
∴ f^-1 : Range (f) → R – (- \(\frac{3}{5}\)} is defined as
f^-1 (x) = \(\frac{3 x}{2-5 x}\)

Question 22.
If f : R – {- \(\frac{4}{3}\)} → R is a function defined by f(x) = \(\frac{4 x}{3 x+4}\), then find f^-1. (NCERT)
Solution:
Here we want to prove that
f : R – {- \(\frac{4}{3}\)} → Range (f) is bijective
∀ y ∈ R – {- \(\frac{4}{3}\)} such that f (x) = f (y)
∴ \(\frac{4 x}{3 x+4}=\frac{4 y}{3 y+4}\)
⇒ 4x (3y + 4) = 4y (3x + 4)
⇒ 12xy + 16x = 12xy + 16y
∴ f is one-one.

Obviously f : R – {- \(\frac{4}{3}\)} → range (f) is onto.
∴ f is 1 – 1 and onto
∴ f is bijective.
Hence f^-1 exists.
Then f (f^-1(x)) = x ∀ x ∈ range (f)
⇒ \(\frac{4 f^{-1}(x)}{3 f^{-1}(x)+4}\) = x ∀ x ∈ range (f)
⇒ 4 f^-1(x) = 3x f^-1(x) + 4x ∀ x ∈ range (f)
⇒ (4 – 3x) f^-1 (x) = 4x ∀ x ∈ range (f)
⇒ f^-1 (x) = \(\frac{4 x}{4-3 x}\) ∀ x ∈ range (f).