NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3.

- Relations and Functions Class 12 Ex 1.1
- Relations and Functions Class 12 Ex 1.2
- Relations and Functions Class 12 Ex 1.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Relations and Functions |

Exercise |
Ex 1.3 |

Number of Questions Solved |
14 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3

**Ex 1.3 Class 12 Maths 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 g of.**

**Solution:**

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

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

f(1) = 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)}

**Ex 1.3 Class 12 Maths Question 2.**

**Let f, g and h be functions from R to R. Show that (f + g) oh = foh + goh, (f • g) oh = (foh) • (goh)**

**Solution:**

f + R –> R, g: R –> R, h: R –> R

(i) (f+g)oh(x)=(f+g)[h(x)]

= f[h(x)]+g[h(x)]

={foh} (x)+ {goh} (x)

=>(f + g) oh = foh + goh

(ii) (f • g) oh (x) = (f • g) [h (x)]

= f[h (x)] • g [h (x)]

= {foh} (x) • {goh} (x)

=> (f • g) oh = (foh) • (goh)

**Ex 1.3 Class 12 Maths Question 3.**

**Find gof and fog, if**

**(i) f (x) = |x| and g (x) = |5x – 2|**

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

**Solution:**

(i) f(x) = |x|, g(x) = |5x – 2|

(a) gof(x) = g(f(x)) = g|x|= |5| x | – 2|

(b) fog(x) = f(g (x)) = f(|5x – 2|) = ||5 x – 2|| = |5x-2|

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

(a) gof(x) = g(f(x)) = g(8x³) = \({ { (8x }^{ 3 }) }^{ 1/3 }\) = 2x

(b) fog (x) = f(g (x))=f(\({ x }^{ 1/3 }\)) = 8.(\({ x }^{ 1/3 }\))³ = 8x

**Ex 1.3 Class 12 Maths Question 4.**

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

**Solution:**

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

(a) fof (x) = f(f(x)) = \(f\frac { 4x+3 }{ 6x-4 } \)

**Ex 1.3 Class 12 Maths Question 5.**

**State with reason whether following functions have inverse**

**(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: {1,2,3,4,5}–>{7,9,11,13} with h = {(2,7), (3,9), (4,11), (5,13)}**

**Solution:**

f: {1,2,3,4} –> {10} with f = {(1,10), (2,10), (3,10), (4,10)}

(i) f is not one-one since 1,2,3,4 have the same image 4.

=> 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

∴ g is not one-one. Therefore g is not invertible.

(iii) f has an inverse

**Ex 1.3 Class 12 Maths Question 6.**

**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.**

**Hint – For y ∈ Range f, y = f (x) = \(\frac { x }{ (x+2) } \) for some x in [- 1,1], i.e., x = \(\frac { 2y }{ (1-y) } \)**

**Solution:**

**Ex 1.3 Class 12 Maths 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} or x_{1} = x_{2}

f is one-one

Also let y = 4x + 3, or 4x = y – 3

∴ \(x=\frac { y-3 }{ 4 } \)

For each value of y ∈ R and belonging to co-domain of y has a pre-image in its domain.

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

f is invertible and f^{-1} (y) = g (y) = \(\frac { y-3 }{ 4 } \)

**Ex 1.3 Class 12 Maths Question 8.**

**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) = √y-4 , where R_{+} is the set of all non-negative real numbers.**

**Solution:**

f(x

_{1}) = x

_{1}

^{2}+ 4 and f(x

_{2}) = x

_{2}

^{2}+ 4

f(x

_{1}) = f(x

_{2}) => x

_{1}

^{2}+ 4 = x

_{2}

^{2}+ 4

or x

_{1}

^{2}= x

_{2}

^{2}=> x

_{1}= x

_{2}As x ∈ R

∴ x>0, x

_{1}

^{2}= x

_{2}

^{2}=> x

_{1}= x

_{2}=>f is one-one

Let y = x² + 4 or x² = y – 4 or x = ±√y-4

x being > 0, -ve sign not to be taken

x = √y – 4

∴ f

^{-1}(y) = g(y) = √y-4 ,y ≥ 4

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

∴ The inverse of f is f

^{-1}(y) = √y-4

**Ex 1.3 Class 12 Maths Question 9.**

**Consider f: R _{+} –> [- 5, ∞) given by f (x) = 9x² + 6x – 5. Show that f is invertible with**

**\({ f }^{ -1 }(y)=\left( \frac { \left( \sqrt { y+6 } \right) -1 }{ 3 } \right) \)**

**Solution:**

Let y be an arbitrary element in range of f.

Let y = 9x² + 6x – 5 = 9x² + 6x + 1 – 6

=> y = (3x + 1)² – 6

=> y + 6 = (3x + 1)²

=> 3x + 1 = √y + 6

**Ex 1.3 Class 12 Maths Question 10.**

**Let f: X –> Y be an invertible function. Show that f has unique inverse.**

**Hint – suppose g _{1} and g_{2} are two inverses of f. Then for all y∈Y, fog_{1}(y)=I_{y}(y)=fog_{2}(y).Use one-one ness of f.**

**Solution:**

If f is invertible gof (x) = I

_{x}and fog (y) = I

_{y}

∴ f is one-one and onto.

Let there be two inverse g

_{1}and g

_{2}

fog

_{1}(y) = I

_{y}, fog

_{2}(y) = I

_{y}

I

_{y}being unique for a given function f

=> g

_{1}(y) = g

_{2}(y)

f is one-one and onto

f has a unique inverse.

**Ex 1.3 Class 12 Maths Question 11.**

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

^{-1})

^{-1}(2) = b, (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

**Ex 1.3 Class 12 Maths 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}] =fol

or (fof

^{-1})

^{-1}o (f

^{-1})

^{-1}= f

=> Io (f

^{-1})

^{-1}= f

=> (f

^{-1})

^{-1}= f

**Ex 1.3 Class 12 Maths Question 13.**

**If f: R –> R be given by f(x) = \({ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \), then fof (x) is**

**(a) \({ x }^{ \frac { 1 }{ 3 } } \)**

**(b) x³**

**(c) x**

**(d) (3 – x³)**

**Solution:**

f: R-> R defined by f(x) = \({ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \)

fof (x) = f[f(x)] = \({f{ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }} \)

= \({ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} }^{ 3 } \right] }^{ \frac { 1 }{ 3 } } \)

= \({ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} } \right] }\)

= \({ \left( { x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }\)

= x

**Ex 1.3 Class 12 Maths Question 14.**

**Let f: \(R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R\) be a function defined as f (x) = \(\frac { 4x }{ 3x+4 } \) . The inverse of f is the map g: Range f–> \(R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R\) 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:**

(b)

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