Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(b) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(b)

I.

Question 1.
If f(x) = ex and g(x) = logex, then show that f o g = g o f and find f-1 and g-1.
Solution:
Given f(x) = ex and g(x) = logex
Now (f o g) (x) = f(g(x))
= f(logex) [∵ g(x) = $$\log _{e} x$$]
= $$e^{\left(\log _{e} x\right)}$$
= x
∴ (fog) (x) = x ………(1)
and (g o f) (x) = g(f(x))
= g(ex) [∵ f(x) = ex]
= loge (ex) [∵ g(x) = logex]
= x loge (e)
= x(1)
= x
∴ (g o f) (x) = x …….(2)
From (1) and (2)
f o g = g o f
Given f(x) = ex
Let y = f(x) = ex ⇒ x = f-1(y)
and y = ex ⇒ x = loge (y)
∴ f-1(y) = loge (y) ⇒ f-1(x) = loge (x)
Let y = g(x) = loge (x)
∵ y = g(x) ⇒ x = g-1(y)
∵ y = loge (x) ⇒ x = ey
∴ g-1(y) = ey ⇒ g-1(x) = ex
∴ f-1(x) = loge (x) and g-1(x) = ex Question 2.
If f(y) = $$\frac{y}{\sqrt{1-y^{2}}}$$, g(y) = $$\frac{y}{\sqrt{1+y^{2}}}$$ then show that (fog) (y) = y
Solution:
f(y) = $$\frac{y}{\sqrt{1-y^{2}}}$$ and g(y) = $$\frac{y}{\sqrt{1+y^{2}}}$$
Now, (fog) (y) = f(g(y)) ∴ (fog) (y) = y

Question 3.
If f : R → R, g : R → R are defined by f(x) = 2x2 + 3 and g(x) = 3x – 2, then find
(i) (fog)(x)
(ii) (gof) (x)
(iii) (fof) (0)
(iv) go(fof) (3)
Solution:
f : R → R, g : R → R and f(x) = 2x2 + 3; g(x) = 3x – 2
(i) (f o g) (x) = f(g(x))
= f(3x – 2) [∵ g(x) = 3x – 2]
= 2(3x- 2)2 + 3 [∵ f(x) = 2x2 + 3]
= 2(9x2 – 12x + 4) + 3
= 18x2 – 24x + 8 + 3
= 18x2 – 24x + 11

(ii) (gof) (x) = g(f(x))
= g(2x2 + 3) [∵ f(x) = 2x2 + 3]
= 3(2x2 + 3) – 2 [∵ g(x) = 3x – 2]
= 6x2 + 9 – 2
= 6x2 + 7

(iii) (fof) (0) = f(f(0))
= f(2(0) + 3) [∵ f(x) = 2x2 + 3]
= f(3)
= 2(3)2 + 3
= 18 + 3
= 21

(iv) g o (f o f) (3)
= g o (f (f(3)))
= g o (f (2(3)2 + 3)) [∵ f(x) = 2x2 + 3]
= g o (f(21))
= g(f(21))
= g(2(21)2 + 3)
= g(885)
= 3(885) – 2 [∵ g(x) = 3x – 2]
= 2653 Question 4.
If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x2 + 1, then find
(i) (f o f) (x2 + 1)
(ii) f o g (2)
(iii) g o f (2a – 3)
Solution:
f : R → R, g : R → R and f(x) = 3x – 1 ; g(x) = x2 + 1
(i) (f o f) (x2 + 1)
= f(f(x2 + 1))
= f[3(x2 + 1) – 1] [∵ f(x) = 3x – 1]
= f(3x2 + 2)
= 3(3x2 + 2) – 1
= 9x2 + 5

(ii) (f o g) (2)
= f(g(2))
= f(22 + 1) [∵ g(x) = x2 + 1]
= f(5)
= 3(5) – 1
= 14 [∵ f(x) = 3x – 1]

(iii) (g o f) (2a – 3)
= g(f(2a – 3))
= g[3(2a – 3) – 1] [∵ f(x) = 3x – 1]
= g(6a – 10)
= (6a – 10)2 + 1 [∵ g(x) = x2 + 1]
= 36a2 – 120a + 100 + 1
= 36a2 – 120a + 101

Question 5.
If f(x) = $$\frac{1}{x}$$, g(x) = √x for all x ∈ (0, ∞) then find (g o f) (x).
Solution:
f(x) = $$\frac{1}{x}$$, g(x) = √x, ∀ x ∈ (0, ∞)
(g o f) (x) = g(f(x))
= g($$\frac{1}{x}$$) [∵ f(x) = $$\frac{1}{x}$$]
= $$\sqrt{\frac{1}{x}}$$
= $$\frac{1}{\sqrt{x}}$$ [∵ g(x) = √x]
∴ (gof) (x) = $$\frac{1}{\sqrt{x}}$$

Question 6.
f(x) = 2x – 1, g(x) = $$\frac{x+1}{2}$$ for all x ∈ R, find (g o f) (x).
Solution:
f(x) = 2x – 1, g(x) = $$\frac{x+1}{2}$$ ∀ x ∈ R
(g o f) (x) = g(f(x))
= g(2x – 1) [∵ f(x) = 2x – 1]
= $$\frac{(2 x-1)+1}{2}$$
= x [∵ g(x) = $$\frac{x+1}{2}$$]
∴ (g o f) (x) = x Question 7.
If f(x) = 2, g(x) = x2, h(x) = 2x for all x ∈ R, then find (f o (g o h)) (x).
Solution:
f(x) = 2, g(x) = x2, h(x) = 2x, ∀ x ∈ R
[f o (g o h) (x)]
= [f o g (h(x))]
= f o g (2x) [∵ h(x) = 2x]
= f[g(2x)]
= f((2x)2) [∵ g(x) = x2]
= f(4x2) = 2 [∵ f(x) = 2]
∴ [f o (g o h) (x)] = 2

Question 8.
Find the inverse of the following functions.
(i) a, b ∈ R, f : R → R defined by f(x) = ax + b, (a ≠ 0).
Solution:
a, b ∈ R, f : R → R and f(x) = ax + b, a ≠ 0
Let y = f(x) = ax + b
⇒ y = f(x)
⇒ x = f-1(y) ……..(i)
and y = ax + b
⇒ x = $$\frac{y-b}{a}$$ ……..(ii)
From (i) and (ii)
f-1(y) = $$\frac{y-b}{a}$$
⇒ f-1(x) = $$\frac{x-b}{a}$$

(ii) f : R → (0, ∞) defined by f(x) = 5x
Solution:
f : R → (0, ∞) and f(x) = 5x
Let y = f (x) = 5x
y = f(x) ⇒ x = f-1(y) ……(i)
and y = 5x ⇒ log5 (y) = x ……..(ii)
From (i) and (ii)
f-1(y) = log5(y) ⇒ f-1(x) = log5 (x)

(iii) f : (0, ∞) → R defined by f(x) = log2 (x).
Solution:
f : (0, ∞) → R and f(x) = log2 (x)
Let y = f(x) = log2 (x)
∵ y = f(x) ⇒ x = f-1(y) ……..(i)
and y = log2(x) ⇒ x = 2y
From (i) and (ii)
f-1(y) = 2y ⇒ f-1(x) = 2x Question 9.
If f(x) = 1 + x + x2 + …… for |x| < 1 then show that f-1(x) = $$\frac{x-1}{x}$$
Solution:
f(x) = 1 + x + x2 + …….. Question 10.
If f : [1, ∞) ⇒ [1, ∞) defined by f(x) = $$2^{x(x-1)}$$ then find f-1(x).
Solution:  II.

Question 1.
If f(x) = $$\frac{x-1}{x+1}$$, x ≠ ±1, then verify (f o f-1) (x) = x.
Solution:
Given f(x) = $$\frac{x-1}{x+1}$$, x ≠ ±1
Let y = f(x) = $$\frac{x-1}{x+1}$$
∵ y = f(x) ⇒ x = f-1(y) ……(i)
and y = $$\frac{x-1}{x+1}$$ Question 2.
If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f : A → B, g : B → C are defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)-1 = f-1 o g-1.
Solution:
A = {1, 2, 3}, B = {α, β, γ},
f : A → B and f = {(1, α), (2, γ), (3, β)}
⇒ f(1) = α, f(2) = γ, f(3) = β
∵ Distinct elements of A have distinct f – images in B, f: A → B is an injective function.
Range of f = {α, γ, β} = B(co-domain)
∴ f : A → B is a surjective function.
Hence f : A → B is a bijective function.
B = {α, β, γ}, C = {p, q, r}, g : B → C and g : {(α, q), (β, r), (γ, p)}
⇒ g(α) = q, g(β) = r, g(γ) = p
∴ Distinct elements of B have distinct g – images in C, g : B → C is an injective function.
Range of g = {q, r, p} = C, (co-domain)
∴ g : B → C is a surjective function
Hence g : B → C is a bijective function
Now f = {(1, α), (2, γ), (3, β)}
g = {(α, q), (β, r), (γ, p)}
g o f = {(1, q), (2, p), (3, r)}
∴ (g o f)-1 = {(q, 1), (r, 3), (p, 2)} ………(1)
g-1 = {(q, α), (r, β), (p, γ)}
f-1 = {(α, 1), (γ, 2),(β, 3)}
Now f-1 o g-1 = {(q, 1), (r, 3), (p, 2)} …….(2)
From eq’s (1) and (2)
(gof)-1 = f-1 o g-1 Question 3.
If f : R → R, g : R → R defined by f(x) = 3x – 2, g(x) = x2 + 1, then find
(i) (g o f-1) (2)
(ii) (g o f)(x – 1)
Solution:
f : R → R, g : R → R and f(x) = 3x – 2
f is a bijective function ⇒ its inverse exists
Let y = f(x) = 3x – 2
∵ y = f(x) ⇒ x = f-1(y) …….(i)
and y = 3x – 2
⇒ x = $$\frac{y+2}{3}$$ ……..(ii)
From (i) and (ii)
f-1(y) = $$\frac{y+2}{3}$$
⇒ f-1(x) = $$\frac{x+2}{3}$$
Now (gof-1) (2)
= g(f-1(2)) ∴ (g o f-1) (2) = $$\frac{25}{9}$$

(ii) (g o f) (x -1)
= g(f(x – 1))
= g(3(x – 1) – 2) [∵ f(x) = 3x – 2]
= g(3x – 5)
= (3x – 5)2 + 1 [∵ g(x) = x2 + 1]
= 9x2 – 30x + 26
∴ (g o f) (x – 1) = 9x2 – 30x + 26

Question 4.
Let f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a), (4, b), (1, c), (3, d)} then show that (gof)-1 = f-1 o g-1
Solution:
f = {(1, a), (2, c), (4, d), (3, b)}
∴ f-1 = {(a, 1), (c, 2), (d, 4), (b, 3)}
g-1 = {(2, a), (4, b), (1, c), (3, d)}
∴ g = {(a, 2), (b, 4), (c, 1), (d, 3)}
(g o f) = {(1, 2), (2, 1), (4, 3), (3, 4)}
∴ (gof)-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ……….(1)
f-1 o g-1 = {(2, 1), (4, 3), (1, 2), (3, 4)} ……..(2)
From eq’s (1) and (2), we observe (gof)-1 = f-1 o g-1

Question 5.
Let f : R → R, g : R → R be defined by f(x) = 2x – 3, g(x) = x3 + 5 then find (f o g)-1 (x).
Solution:
f : R → R, g : R → R and f(x) = 2x – 3 and g(x) = x3 + 5
Now (fog) (x) = f(g(x))
= f(x3 + 5) [∵ g(x) = x2 + 5]
= 2(x3 + 5) – 3 [∵ f(x) = 2x – 3]
= 2x3 + 7
∴ (f o g) (x) = 2x3 + 7
Let y = (f o g) (x) = 2x3 + 7
∵ y = (fog)(x)
⇒ x = (fog)-1 (y) …….(1)
and y = 2x3 + 7
⇒ x3 = $$\frac{y-7}{2}$$
⇒ x = $$\left(\frac{y-7}{2}\right)^{\frac{1}{3}}$$ …..(2)
From eq’s (1) and (2),
(f o g)-1 (y) = $$\left(\frac{y-7}{2}\right)^{\frac{1}{3}}$$
∴ (f o g)-1 (x) = $$\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$$ Question 6.
Let f(x) = x2, g(x) = 2x. Then solve the equation (f o g) (x) = (g o f) (x)
Solution:
Given f(x) = x2 and g(x) = 2x
Now (f o g) (x) = f(g(x))
= f(2x) [∵ g(x) = 2x]
= (2x)2
= 22x [∵ f(x) = x2]
∴ (f o g) (x) = 22x ……(1)
and (g o f) (x) = g(f(x))
= g(x2) [∵ f(x) = x2]
= $$(2)^{x^{2}}$$ [∵ g(x) = 2x]
∴ (g o f) (x) = $$(2)^{x^{2}}$$
∵ (f o g) (x) = (g o f) (x)
⇒ 22x = $$(2)^{x^{2}}$$
⇒ 2x = x2
⇒ x2 – 2x = 0
⇒ x(x – 2) = 0
⇒ x = 0, x = 2
∴ x = 0, 2

Question 7.
If f(x) = $$\frac{x+1}{x-1}$$, (x ≠ ±1) then find (fofof) (x) and (fofofof) (x).
Solution:
f(x) = $$\frac{x+1}{x-1}$$, (x ≠ ±1)
(i) (fofof) (x) = (fof) [f(x)]
= (fof) $$\left(\frac{x+1}{x-1}\right)$$ [∵ f(x) = $$\left(\frac{x+1}{x-1}\right)$$] (ii) (fofofof) (x) = f[(f o f o f) (x)]
= f [f(x)] {from (1)} In the above problem if a number of f is even its answer is x and if a number of f is odd its answer is f(x).