Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 1 Relations and Functions Ex 1.3

Question 1.

Let f : {1. 3 4} → {1,2, 5} and g: {1, 2, 5} → {1,3} be given by f = {(1, 2), (3, 5), (4,1)} and g = {(1, 3), (2, 3), (5,1)}. Write down

Solution:

f= {(1,2), (3,5), (4,1)}

g = {(I, 3), (2, 3), (5, 1)}

(f) = 2, g(2) = 3 ⇒ gof(1) = 3

f(3) = 5, g(5) = 1 ⇒ gof(3) = 1

f(4) = 1, g(1) = 3 ⇒ gof(4) = 3

⇒ gof = {(1,3), (3,1), (4, 3)].

Question 2.

Let f, g and h be functions from R to R. Show that

(i) (f + g)oh = foh + goh.

(ii) (f. g)oh = (foh) . (goh).

Solution:

f : R → R, g : R → R, h : R → R

Question 3.

Find gof and fog, if \(x^{\frac{1}{3}}\)

(i) f(x) = |x|, and g(x) = |5x – 2|.

(ii) f(x) = 8x³ and g(x) = \(x^{\frac{1}{3}}\)

Solution:

(i) f(x) = |x|, and g(x) = |5x – 2|

- gof(x) = g[f(x)]=g|x| = |5|x|-2|.
- fog(x) = f[g(x)] = f(|5x- 2|) =||5x-2|| = |5x – 2|.

(ii) f(x) = 8x³ and g(x) = \(x^{\frac{1}{3}}\)

- gof(x) = g[f(x)] = g(8x³) = (8x³)\(x^{\frac{1}{3}}\) = 2x.
- fog(x) = f[(g(x)] = f(\(x^{\frac{1}{3}}\)) = 8.(\(x^{\frac{1}{3}}\))³ = 8x.

Question 4.

If f(x) = \(\frac { 4x+3 }{ (6x-4) }\), x ≠ \(\frac { 2 }{ 3 }\) show that fof(x) = x for all x ≠ \(\frac { 2 }{ 3 }\) What is the inverse of f?

Solution:

f(x) = \(\frac { 4x+3 }{ (6x-4) }\), x ≠ \(\frac { 2 }{ 3 }\)

Question 5.

State with reasons whether each of the following functions has a inverse or not:

Solution:

(i) f : {1, 2, 3, 4} → {10} with

f= {(1, 10), (2, 10), (3, 10), (4, 10)}.

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with

g = {(5, 4), (6, 3), (7, 4), (8, 2)}.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with

h = {(2, 7), (3, 9), (4, 11), (5, 13)}.

Solution:

(i) f : {1, 2, 3, 4} → {10} with

f= {(1, 10), (2, 10), (3, 10), (4, 10)}.

f is not one-one since 1, 2, 3, 4 have same page 10.

⇒ f has no inverse.

(ii) g: {5, 6, 7, 8} → {1,2, 3,4} with

g = {(5, 4), (6, 3), (7, 4), (8, 2)}.

Here also 5 and 7 have the same image 4.

∴ g is not one-one.

Therefore, g is not invertible, i.e., It has no inverse.

(iii) h : {2,3,4,5} → {7,9,11,13} with

h = {(2, 7), (3, 9), (4, 11), (5, 13)}.

Hence, f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2}.

i. e., each element of X = {2,3,4,5} has a unique image in Y = {7,9,11,13}.

Similarly, each member of Y has a unique pre-image in X.

⇒ f is one-one and onto.

∴ f is invertible, i.e., It has an inverse.

Question 6.

Show that f: [- 1, 1] → R, given by f(x) = \(\frac { x }{ x+2 }\), x ≠ – 2 is one-one. Find the inverse of the function f : [- 1, 1} → Range f.

Solution:

Question 7.

Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Solution:

f : R → R given by f(x) = 4x + 3.

f(x_{1}) = 4x_{1} + 3, f(x_{2}) = 4x_{2} + 3

If f(x_{1}) = f(x_{2}),

then 4x_{1} + 3 = 4x_{2} + 3

or 4x_{1} = 4x_{2}

⇒ x_{1} = x_{2}.

⇒ f is one-one.

Aslo, let y = 4x + 3

or 4x = y – 3

∴ x = \(\frac { y-3 }{ 4 }\).

For each value of y e R and belonging to codomain of y, there is a pre-image in its domain.

∴f is onto.

i. e., f is one-one and onto.

∴ f is invertible.

So, f^{-1}(y) = g(y) = \(\frac { y-3 }{ 4 }\)

Question 8.

Consider f : R_{+} → [4, ∝) given by f(x) = x² + 4. Show that f is invertible with the inverse of f^{-1} of f given by f^{-1}(y) = \(\sqrt{y-4}\), where R_{+} is the set of all non-negative real numbers.

Solution:

f(x_{1}) = x²_{1} + 4 and f(x_{2}) = x²_{2} + 4

So, f(x_{1}) = f(x_{2}) ⇒ x²_{1} + 4 = x²_{2} + 4

or x²_{1} = x²_{2} ⇒ x_{1} = x_{2}

As x ∈ R, ∴ x > 0,

x²_{1} = x²_{2} ⇒ x_{1} = x_{2}

⇒ f is one-one.

Let y = x² + 4 or x² = y – 4 or x = ± \(\sqrt{y-4}\).

x being > 0, -vc sign not to be taken.

∴ = \(\sqrt{y-4}\)

∴ f^{-1}(y) = g(y) = \(\sqrt{y-4}\), y ≥ 4.

For every y ≥ 4, g(y) has real positive value.

∴ The inverse of f is f^{-1}(y) = \(\sqrt{y-4}\).

Question 9.

Consider f : → R+ [- 5, ∝) given by f(x) = 9x² + 6x – 5. Show that f is invertible with f^{-1}(y) = \(\left(\frac{\sqrt{y+6}-1}{3}\right)\)

Solution:

Let y be an arbitrary element in the range of f.

Let y = 9x² + 6x – 5

= 9x² + 6x + 1 – 6

= (3x +1)² – 6

⇒ y + 6 = (3x + 1)²

or 3x + 1 = \(\sqrt{y+6}\)

∴ x = \(\left(\frac{\sqrt{y+6}-1}{3}\right)\) = g(y).

Question 10.

Let f : X → Y be an invertible function. Show that f has a unique inverse.

Solution:

If f is invertible, then gof(x) = I_{x} and fog(y) = I_{y}.

⇒ f is one-one and onto.

Let there be two inverses g_{1} and g_{2}

∴ fog_{1} (y)= I_{y}, f0g_{2}(y) = I_{y}

I_{y} being unique for a given function f, we get

g_{1}(y) = g_{2}(y) [ ∵ f is one-one and onto.]

∴ f has a unique inverse.

Question 11.

Consider f : {1, 2, 3} → {a, b, c] given by f(1) = a, f(2) – b, f(3) = c. Find f^{-1} and show that (f^{-1})^{-1} = f.

Solution:

f : {1, 2, 3} → {a, b, c}

so that f(1) = a, f(2) = b, f(3) = c.

Now, let X = {1, 2, 3}, Y = {a, b, c).

∴ f : X → Y

∴ f^{-1} : Y → X such that f^{-1}(a) = 1, f^{-1}(b) = 2, f^{-1}(c) = 3.

Inverse of this function may be written as

(f^{-1})^{-1} : X → Y such that (f^{-1})^{-1} = a, (f^{-1})^{-1}(2) = b and (f^{-1})^{-1}(3) = c.

We also have:

f: X Y, such that f(1) = a, f(2) = b, f(3) = c.

⇒ (f^{-1})^{-1} = f

Question 12.

Let f : X → Y be an invertible function. Show that the inverse of f^{-1} is f, i.e., (f^{-1})^{-1} = f.

Solution:

f : X → Y is an invertible function.

∴ f is one-one and onto ⇒g : Y → X, where g is also one-one and onto such that

gof(x) = I_{x} and fog(y) = I_{y} ⇒ g = f^{-1}.

Now, f^{-1}o(f^{-1})^{-1} = I

and fo[f^{-1}o(f^{-1})^{-1}] = foI

or (fof^{-1})o(f^{-1})^{-1} = f

Io(f^{-1})^{-1} = f

⇒ (f^{-1})^{-1} = f

Question 13.

If f : R → R be given by f(x) = (3-x³)\(\left(3-x^{3}\right)^{\frac{1}{3}}\), when fof(x) is

(A) \(x^{\frac{1}{3}}\)

(B) x³

(C) x

(D) 3-x³.

Solution:

Question 14.

Let f : R – (- \(\frac { 4 }{ 3 }\)) → R be a function, defined as f(x) = \(\frac { 4x }{ 3x+4 }\), x ≠ – \(\frac { 4 }{ 5 }\).

The inverse off is map g: Range f → R – is given by \(\frac { 4 }{ 3 }\) is given by

(A) g(y) = \(\frac { 3y }{ 3-4y }\)

(B) g(y) = \(\frac { 4y }{ 4-3y }\)

(C) g(y) = \(\frac { 4y }{ 3-4y }\)

(D) g(y) = \(\frac { 3y }{ 4-3y }\)

Solution:

f : R – {\(\frac { -4 }{ 3 }\)} → R is a function defined by f(x) = \(\frac { 4x }{ 3x+4 }\)

Let y = \(\frac { 4x }{ 3x+4 }\).

⇒ 3xy + 4y = 4x

or x(4 – 3y) = 4y

∴ x = \(\frac { 4y }{ 4-3y }\)

∴ f^{-1}(4y) = g(y) = \(\frac { 4y }{ 4-3y }\)

So, part (B) is the correct answer.