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) = e^{x} and g(x) = log_{e}x, then show that f o g = g o f and find f^{-1} and g^{-1}.

Solution:

Given f(x) = e^{x} and g(x) = log_{e}x

Now (f o g) (x) = f(g(x))

= f(log_{e}x) [∵ 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(e^{x}) [∵ f(x) = e^{x}]

= log_{e} (e^{x}) [∵ g(x) = log_{e}x]

= x log_{e} (e)

= x(1)

= x

∴ (g o f) (x) = x …….(2)

From (1) and (2)

f o g = g o f

Given f(x) = e^{x}

Let y = f(x) = e^{x} ⇒ x = f^{-1}(y)

and y = e^{x} ⇒ x = log_{e} (y)

∴ f^{-1}(y) = log_{e} (y) ⇒ f^{-1}(x) = log_{e} (x)

Let y = g(x) = log_{e} (x)

∵ y = g(x) ⇒ x = g^{-1}(y)

∵ y = log_{e} (x) ⇒ x = e^{y}

∴ g^{-1}(y) = e^{y} ⇒ g^{-1}(x) = e^{x}

∴ f^{-1}(x) = log_{e} (x) and g^{-1}(x) = e^{x}

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) = 2x^{2} + 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) = 2x^{2} + 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) = 2x^{2} + 3]

= 2(9x^{2} – 12x + 4) + 3

= 18x^{2} – 24x + 8 + 3

= 18x^{2} – 24x + 11

(ii) (gof) (x) = g(f(x))

= g(2x^{2} + 3) [∵ f(x) = 2x^{2} + 3]

= 3(2x^{2} + 3) – 2 [∵ g(x) = 3x – 2]

= 6x^{2} + 9 – 2

= 6x^{2} + 7

(iii) (fof) (0) = f(f(0))

= f(2(0) + 3) [∵ f(x) = 2x^{2} + 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) = 2x^{2} + 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) = x^{2} + 1, then find

(i) (f o f) (x^{2} + 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) = x^{2} + 1

(i) (f o f) (x^{2} + 1)

= f(f(x^{2} + 1))

= f[3(x^{2} + 1) – 1] [∵ f(x) = 3x – 1]

= f(3x^{2} + 2)

= 3(3x^{2} + 2) – 1

= 9x^{2} + 5

(ii) (f o g) (2)

= f(g(2))

= f(2^{2} + 1) [∵ g(x) = x^{2} + 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) = x^{2} + 1]

= 36a^{2} – 120a + 100 + 1

= 36a^{2} – 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) = x^{2}, h(x) = 2x for all x ∈ R, then find (f o (g o h)) (x).

Solution:

f(x) = 2, g(x) = x^{2}, 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) = x^{2}]

= f(4x^{2}) = 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) = 5^{x}

Solution:

f : R → (0, ∞) and f(x) = 5^{x}

Let y = f (x) = 5^{x}

y = f(x) ⇒ x = f^{-1}(y) ……(i)

and y = 5^{x} ⇒ log_{5} (y) = x ……..(ii)

From (i) and (ii)

f^{-1}(y) = log_{5}(y) ⇒ f^{-1}(x) = log5 (x)

(iii) f : (0, ∞) → R defined by f(x) = log_{2} (x).

Solution:

f : (0, ∞) → R and f(x) = log_{2} (x)

Let y = f(x) = log_{2} (x)

∵ y = f(x) ⇒ x = f^{-1}(y) ……..(i)

and y = log_{2}(x) ⇒ x = 2y

From (i) and (ii)

f^{-1}(y) = 2y ⇒ f^{-1}(x) = 2x

Question 9.

If f(x) = 1 + x + x^{2} + …… for |x| < 1 then show that f^{-1}(x) = \(\frac{x-1}{x}\)

Solution:

f(x) = 1 + x + x^{2} + ……..

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) = x^{2} + 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) = x^{2} + 1]

= 9x^{2} – 30x + 26

∴ (g o f) (x – 1) = 9x^{2} – 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) = x^{3} + 5 then find (f o g)^{-1} (x).

Solution:

f : R → R, g : R → R and f(x) = 2x – 3 and g(x) = x^{3} + 5

Now (fog) (x) = f(g(x))

= f(x^{3} + 5) [∵ g(x) = x^{2} + 5]

= 2(x^{3} + 5) – 3 [∵ f(x) = 2x – 3]

= 2x^{3} + 7

∴ (f o g) (x) = 2x^{3} + 7

Let y = (f o g) (x) = 2x^{3} + 7

∵ y = (fog)(x)

⇒ x = (fog)^{-1} (y) …….(1)

and y = 2x^{3} + 7

⇒ x^{3} = \(\frac{y-7}{2}\)

⇒ 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) = x^{2}, g(x) = 2^{x}. Then solve the equation (f o g) (x) = (g o f) (x)

Solution:

Given f(x) = x^{2} and g(x) = 2^{x}

Now (f o g) (x) = f(g(x))

= f(2^{x}) [∵ g(x) = 2^{x}]

= (2^{x})^{2}

= 2^{2x} [∵ f(x) = x^{2}]

∴ (f o g) (x) = 2^{2x} ……(1)

and (g o f) (x) = g(f(x))

= g(x^{2}) [∵ f(x) = x^{2}]

= \((2)^{x^{2}}\) [∵ g(x) = 2^{x}]

∴ (g o f) (x) = \((2)^{x^{2}}\)

∵ (f o g) (x) = (g o f) (x)

⇒ 2^{2x} = \((2)^{x^{2}}\)

⇒ 2^{x} = x^{2}

⇒ x^{2} – 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).