OP Malhotra Class 12 Maths Solutions Chapter 2 Functions Ex 2(c)

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S Chand Class 12 ICSE Maths Solutions Chapter 2 Functions Ex 2(c)

Question 1.
(i) If f : R → R is defined by f (x) = 2x + 3, then find f^-1(x).
(ii) If the function f : R → R, defined by f(x) = 3x – 4 is invertible, then find f^-1.
(iii) If f : R → R defined by f(x) = \(\frac { 3x+5 }{ 2 }\) is an invertible function, then find f^-1.
Solution:
(i) Given f : R → R defined by
f(x) = 2x + 3 ∀x∈R
Injectivity : ∀x,y∈R s.t f(x) = f(y)
⇒ 2x + 3 = 2y + 3
⇒ 2x = 2y ⇒ x = y
∴ f is one – one.
Surjectivity : Let y∈R be any arbitrary element Then f(x) = y
⇒ 2x + 3 = y ⇒ x = \(\frac{y-3}{2}\)
Since y∈R ⇒ \(\frac{y-3}{2}\) ∈ R ⇒ x ∈ R
So ∀x∈R ∃ x ∈R s.t
f(x) = \(f\left(\frac{y-3}{2}\right)=2\left(\frac{y-3}{2}\right)\) + 3 = y
∴ f is onto
Thus f is bijective and f^-1 exists.
Let x∈R (codomain) and y∈R (domain)
s.t f(x) = y
⇒ 2x + 3 = y ⇒ x = \(\frac{y-3}{2}\)
⇒ f^-1(y) = \(\frac{y-3}{2}\)
Thus f^-1 : R → R defined by
f^-1(x) = \(\frac{y-3}{2}\) ∀x∈R

(ii) Given f : R → R defined by
f(x) = 3x – 4 ∀x∈R
one – one : ∀x, y∈R s.t f(x) = f(y)
⇒ 3x – 4 = 3y – 4
⇒ 3x = 3y ⇒ x = y
∴ f is one – one.
onto : Let y∈R be any arbitrary element Then f(x) = y
⇒ 3x – 4 = y ⇒ x = \(\frac{y+3}{3}\)
as y∈R ⇒ \(\frac{y+3}{2}\) ∈ R ⇒ x ∈ R
So ∀x∈R ∃ x ∈R s.t
f(x) = \(f\left(\frac{y+4}{3}\right)=3\left(\frac{y+4}{3}\right)\) – 4 = y
∴ f is onto
Thus f is one-one, onto and hence f is bijective and thus f^-1 exists.
Now, Let x∈R and y∈R (codomain)
s.t f(x) = y
⇒ 3x – 4= y ⇒ x = \(\frac{y+4}{2}\)
⇒ f-1 = \(\frac{y+4}{3}\)
Thus, f-1(x) = \(\frac{y+4}{3}\) ∀x∈R

(iii) Given f: R → R defined by 3x + 5
f(x) = \(\frac{3 x+5}{2}\) is mvertible
∴ f^-1 exists and f is one-one onto.
∀y∈R (codomain) ∃ x ∈R (domain)
s.t y = f(x)
⇒ y = \(\frac{3 x+5}{2}\) ⇒ 3x = 2y – 5
⇒ x = \(\frac{2 y-5}{3}\)
⇒ f^-1(y) = \(\frac{2 y-5}{3}\)
∴ f^-1(x) = \(\frac{2 y-5}{3}\) ∀x∈R

Question 2.
Let the function f which is invertible be defined by f(x) = \(\frac{2 x+1}{1-3 x}\), then show that f^-1(x) = \(\frac{x-1}{3 x+2}\).
Solution:
Given f is invertible and be defined by f(x) = \(\frac{2 x+1}{1-3 x}\)
∴ f^-1 exists and hence f is 1 – 1 and onto.
∀y∈R (codomain) ∃ x∈R (domain)
s.t y = f(x)

Question 3.
If f : R → R is invertible and is defined as f(x) = (1 – x)^1/3, then find f^-1(x).
Solution:
Given f : R → R is invertible and defined as f(x) = (1 – x)^1/3
Thus, f^-1 exists ⇒ f is 1-1 and onto.
∀y∈R (codomain) ∃ x∈R (domain)
s.t y = f(x)
⇒ y = (1 – x)^1/3
⇒ y³ = 1 – x ⇒ x = 1 – y³
⇒ f^-1(y) = 1 – y³
⇒ f^-1(x) = 1 – x³ ∀x ∈R

Question 4.
If f(x) = [4 – (x – 7)³]^1/5, is a real invertible function, then find f^-1(x).
Solution:
Given f(x) = [4 – (x – 7)³]^1/5 andfisa real invertible function
∴ f^-1 exists and f is one-one and onto.
∀y∈R (codomain) ∃ x∈R (domain)
s.t y = f(x)
⇒ y = f(x) = [4 – (x – 7)³]^1/5
⇒ y⁵ = 4 – (x – 7)³
⇒ (x – 7)³ = 4 – y⁵
⇒ x – 7 = \(\sqrt[3]{4-y^5}\)
⇒ x = 7 + \(\sqrt{4-y^5}\)
⇒ f^-1(y) = 7 + \(\sqrt[3]{4-x^5}\)
⇒ f^-1(x) = 7 + \(\sqrt[3]{4-x^5}\)

Question 5.
If f : N → Y is a functoin defined by f(x) = 4x + 3 where
Y = {y ∈ N : y = 4x + 3 for some x∈N}, then, find the inverse of f.
Solution:
Given f : N → Y be a function defined by f(x) = 4x + 3 ∀x∈R
where Y = {y ∈ N : y = 4x + 3 for some x∈N}
= {7, 11, 15, 19, …. ∞}
one – one : ∀x y ∈N s.t f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ x = y
∴ f is one – one.
onto : Let y∈4 be any arbitrary element then f(x) = y
4x + 3 = y ⇒ x = \(\frac{y-3}{4}\)
as y ∈ Y ⇒ \(\frac{y-3}{4}\) ∈ N ⇒ x∈N
So ∀y∈Y ∃ x∈N s.t
f(x) = \(f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)\) + 3 = y
∴ f is onto.
H ence, f is one – one, onto and f^-1 exists.
Let x∈N and y∈Y s.t. f(x) = y
⇒ 4x + 3 = y
⇒ x = \(\frac{y-3}{4} \Rightarrow f^{-1}(y)=\frac{y-3}{4}\)
∴ f^-1(x) = \(\frac{x-3}{4}\)

Question 6.
If f : R → R is defined by f(x) = x³, then find f^-1(8).
Solution:
Given f : R → R defined by f (x) = x³ ∀x∈R
one – one : ∀x y∈R  s.t f(x) = f(y)
⇒ x³ = y³
⇒ (x – y) (x² + xy + y²) = 0
⇒ x – y = 0
[∵ x² + xy + y² 0 ∀x, y∈R
∴ f is one – one
onto : Let y∈R be any arbitrary element then f(x) = y
x³ = y ⇒ x = 3\(\sqrt{y}\)
as y∈R ⇒ 3\(\sqrt{y}\) ∈R ⇒ x∈R
∀y∈R ∃ x∈R
s.t f(x) = f (3\(\sqrt{y}\)) = (3\(\sqrt{y}\))³ = y
∴ f is onto.
Thus, f is one – one, onto
∴ f^-1 exists.
Let x∈R (domain) and y∈R (codomain
s.t. f(x) = y
⇒ x³ = y ⇒ x = y^1/3
⇒ f^-1(y) = y^1/3
⇒ f^-1(8) = 8^1/3 = 2

Question 7.
If f : R → R is defined by f(x) = |x|, then
(a) f^-1(x) = – x
(b) f^-1(x) = \(\frac{1}{|x|}\)
(c) f^-1(x) does not exist
(d) f^-1(x) = \(\frac{1}{x}\)
Solution:
Given f: R → R is defined by
f(x) = |x| ∀x∈R
Since f(1) = 1 ; f(-1) = |- 1| = 1
so elements 1 and – 1 have same image 1 under f so different elements in R (domain) has same image in R (codomain)
∴ f is many one function
∴ f is not one – one. Thus f^-1 does not exists.

Question 8.
Consider f : R → R given by
f(x) = 4x + 3. Show thatfis invertible. Find the inverse of f.
Solution:
Given f : R → R defined by f(x) = 4x + 3 ∀x∈R
one – one : ∀x y∈R s.t f(x) = f(y)
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y ⇒ x = y
∴ f is one – one
onto : Let y∈R be any arbitraiy element then f(x) = y
4x + 3 = y ⇒ x = \(\frac{y-3}{4}\)
as y∈R ⇒ x = \(\frac{y-3}{4}\) ∈R ⇒ x ∈ R
So ∀y∈R ∃ x∈R s.t
f(x) = f\(\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)\) + 3 = y
∴ f is onto.
Thus, f is one – one, onto and hence, f is invertible.
Since f(x) = y
⇒ 4x + 3 = y ⇒ x = \(\frac{y-3}{4}\)
⇒ f^-1(y) =\(\frac{y-3}{4}\)
⇒ f^-1(x) = \(\frac{x-3}{4}\)

Question 9.
Show that f : [- 1, 1] → R, given by f(x) = \(\frac{x}{x+2}\) is one-one. Find the inverse of the function f : [- 1, 1] → Range f.
Solution:
Given f : [- 1, 1] → R, defined by
f(x) = \(\frac{x}{x+2}\), x ≠ – 2
∀x, y∈[-1, 1] s.t f(x) = f(y)
⇒ \(\frac{x}{x+2}=\frac{y}{y+2}\)
⇒ xy + 2x = xy + 2y ⇒ x = y
∴ f is one – one.
Now f : [- 1, 1] → Range f
Here codomain of f = Range of f
∴ f is onto
∴ f is one – one and onto
∴ f^-1 exists
Let x ∈ [-1, 1] and y∈R_f s.t. f(x) = y
⇒ \(\frac{x}{x+2}\) = y ⇒ x = xy + 2y
⇒ x(1 – y)= 2y ⇒ x = \(\frac{2 y}{1-y}\), y ≠ 1
⇒ f^-1(y) = \(\frac{2 y}{1-y}\) , y ≠ 1
⇒ f^-1(x) = \(\frac{2 x}{1-x}\), x ≠ 1

Question 10.
Let A = R – {2} and B = R – {1}. If A → B is a function defined by f(x) = \(\frac{x-1}{x-2}\), then show that f is one-one and onto. Hence, find f^-1.
Solution:
Given A = R – {2} and B = R – {1} and f be a function from A to B defined by f(x) = \(\frac{x-1}{x-2}\), x∈R
one – one : ∀x, y∈A = R – {2} s.t f(x) = f(y)
⇒ \(\frac{x-1}{x-2}=\frac{y-1}{y-2}\)
⇒ (x – 1) (y – 2) = (y – 1) (x – 2)
⇒ xy – 2x – y + 2 = xy – 2y – x + 2
⇒ – x = – y
⇒ x = y
∴f is one – one.
onto : Let y ∈ B = R – {1} be any arbitrary element
i.e. y ≠ 1 Then f(x) = y

∴ f is onto.
Thus, f is one – one, onto
∴ f is invertible and f^-1 exists.
Let f(x) = y
⇒ \(\frac{x-1}{x-2}=y \Rightarrow x=\frac{1-2 y}{1-y}\)
⇒ \(f^{-1}(y)=\frac{1-2 y}{1-y}=\frac{2 y-1}{y-1}\)
Thus, f^-1 : B → A is defined by
\(f^{-1}(x)=\frac{2 x-1}{x-1} \quad \forall x \in \mathrm{B}\)

Question 11.
Show that the function f in
A = R – \(\left\{\frac{2}{3}\right\}\) defined as f(x) = \(\frac { ∆L }{ L }\) is one-one and onto. Hence, find f^-1..
Solution:
Given f : A → R defined by
f(x) = \(\frac{4 x+3}{6 x-4} \quad \forall x \in A\)
where A = R – \(\left\{\frac{2}{3}\right\}\)
one – one : ∀x, y∈A s.t f(x) = f(y)
⇒ \(\frac{4 x+3}{6 x-4}=\frac{4 y+3}{6 y-4}\)
⇒ (4x + 3) (6y – 4) = (4y + 3)(6x – 4)
⇒ 24xy – 16x + 18y – 12 = 24xy – 16y + 18x – 12
⇒ 34y = 34x ⇒ y = x
⇒ x = y
∴ f is one – one.
onto : Let y∈R be any arbitrary element Then f(x) = y

∴ f is onto.
Thus, f is one – one and onto and hence f^-1 exists.
Let f(x) = y
⇒ \(\frac{4 x+3}{6 x-4}=y \Rightarrow x=\frac{4 y+3}{6 y-4}\)
⇒ f^-1 = \(\frac{4 y+3}{6 y-4}\)
Thus f^-1 : R → A is defined
f^-1 = \(\frac{4 y+3}{6 y-4}\) ∀ x∈R

Question 12.
Consider f : R⁺ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f^-1 of f given by f^-1(y) = \(\sqrt{y-4}\), where R, is the set of all non-negative numbers.
Solution:
f : R⁺ → [4, ∞] defined by
f(x) = x² + 4 ∀x∈R⁺
∀x, y∈R such that f (x) = f(y)
⇒ x² + 4 = y² + 4
⇒ x² = y²
⇒ x = y [∵ x, y∈R⁺]
∴ f is one-one function.
Let y ∈ [4, ∞) and y = f(x₀)
then y = x₀² + 4 ⇒ x₀ = \(\sqrt{y-4}\) (∵ x ∈ R⁺; x₀ ≥ 0)
as y ∈ [4, ∞) ⇒ \(\sqrt{y-4}\) ∈ R⁺ ⇒ x₀ ∈ R⁺ [∵ y ≥ 4 ⇒ y – 4 ≥ 0]
Now f(x₀) = x₀² + 4 = (\(\sqrt{y-4}\))² + 4 = y
∴ ∀y ∈ [4, ∞) ∃ 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 + 4
⇒ x = (\(\sqrt{y-4}\)) (∵x ∈ R⁺)
⇒ f^-1(y) = \(\sqrt{y-4}\)
⇒ f^-1(y) = \(\sqrt{y-4}\), ∀y ∈ [4, ∞)

Question 13.
Let f : R → R : f(x) = 10x + 7. Find the function g : R → R such that gof = fog = I_g.
Solution:
Given f : R → R defined by f(x) = 10x + 7.
Since gof = fog = I_R
∴ f and g are inverses of each other.
∴ f^-1 = f and g^-1 = f
one – one : ∀ x y∈R s.t f(x) = f(y)
⇒ 10x + 7 = 10y + 7
⇒ 10x = 10y ⇒ x = y
∴ f is one – one.
onto : Le y∈R be any arbitrary element Then
f(x) = y
⇒ 10x + 7 = y ⇒ x = \(\frac{y-7}{10}\)
Since y ∈ R ⇒ \(\frac{y-7}{10}\) ∈R ⇒ ∈R
So ∀y∈R ∃ X∈R s.t
f(x) = \(f\left(\frac{y-7}{10}\right)=10\left(\frac{y-7}{10}\right)\) + 7 = y
∴ f is onto.
Thus, f is one – one and onto and hence f^-1 exists.
Since f(x) = y ⇒ 10x + 7 = y
⇒ x = \(\frac{y-7}{10}\)
⇒ f^-1(y) = \(\frac{y-7}{10} \Rightarrow g(y)=\frac{y-7}{10}\)
∴ g : R → R is defined by
g(x) = \(\frac{y-7}{10}\) ∀x∈R

Question 14.
Show that the inverse function of the function f whose rule is f(x) = \(\frac{2 x+1}{3 x-2}, x \neq \frac{2}{3}\) is f it self.
Solution:

Thus, the given function f is invertible and it is the inverse of itself.
Since f(x) = y ⇒ 10x + 7 = y

Question 15.
Use composition to show that f and g are inverse of each other.
(i) f(x) = 2x – 6, g(x) = \(\frac { x }{ 2 }\) + 3
(ii) f(x) = \(\frac { 1 }{ x+1 }\), g(x) = \(\frac { 1-x }{ x }\)
(iii) f(x) = \(\frac{-3}{2 x+5}\), g(x) = \(\frac{-3-5 x}{2 x}\)
(iv) f(x) = x⁵, g(x) = 5\(\sqrt{x}\).
Solution:
(i) Given f(x) = 2x – 6, g(x) = \(\frac { x }{ 2 }\) + 3
Now (fog) (x) = f(g(x)) = f(\(\frac { x }{ 2 }\) + 3)
= 2(\(\frac { x }{ 2 }\) + 3) – 6 = x
gof(x) = g(f(x)) = g(2x – 6) = \(\frac{2 x-6}{2}\) + 3
= x – 3 + 3 = x
(fog)(x) = gof(x) = x
Hence, f and g are inverses of each other.