In this chapter, we provide NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 pdf, free NCERT solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 book pdf download.

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) = .****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) =

(a) gof(x) = g(f(x)) = g(8x³) = = 2x

(b) fog (x) = f(g (x))=f() = 8.()³ = 8x

**Ex 1.3 Class 12 Maths Question 4.****If , show that fof (x) = x, for all . What is the inverse of f?****Solution:**

(a) fof (x) = f(f(x)) =

**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) = is one-one. Find the inverse of the function f: [-1,1] –> Range f.****Hint – For y ∈ Range f, y = f (x) = for some x in [- 1,1], i.e., x =****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

∴

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) =

**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**

**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) = , then fof (x) is****(a)****(b) x³****(c) x****(d) (3 – x³)****Solution:**

f: R-> R defined by f(x) =

fof (x) = f[f(x)] =

=

=

=

= x

**Ex 1.3 Class 12 Maths Question 14.****Let f: be a function defined as f (x) = . The inverse of f is the map g: Range f–> given by****(a)****(b)****(c)****(d)****Solution:**

(b)

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