NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 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 1.2.

- Relations and Functions Class 12 Ex 1.1
- Relations and Functions Class 12 Ex 1.3
- 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 1.2 |

Number of Questions Solved |
12 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2

**Question 1.**

**Show that the function f: R —> R defined by f (x) = is one-one onto, where R is the set of all non-zero real numbers. Is the result true, if the domain R is replaced by N with co-domain being same as R ?**

**Solution:**

(a) We observe the following properties of f:

(i) f(x) = if f(x_{1}) = f(x_{2})

=> x_{1} = x_{2}

Each x ∈ R has a unique image in codomain

=> f is one-one.

(ii) For each y belonging codomain then

or there is a unique pre image of y.

=> f is onto.

(b) When domain R is replaced by N. codomain remaining the same, then f: N—> R If f(x_{1}) = f(x_{2})

=> => n1 = n2 where n1; n2 ∈ N

=> f is one-one.

But for every real number belonging to codomain may not have a pre-image in N.

eg:

∴ f is not onto.

**Question 2.**

**Check the injectivity and surjectivity of the following functions:**

**(i) f: N -> N given by f (x) = x²**

**(ii) f: Z -> Z given by f (x) = x²**

**(iii) f: R -> R given by f (x) = x²**

**(iv) f: N -> N given by f (x) = x³**

**(v) f: Z ->Z given by f (x) = x³**

**Solution:**

(i) f: N —> N given by f (x) = x²

(a) f(x_{1}) =>f(x_{2})

=>x_{1}^{2} = x_{2}^{2} =>x_{1} = x_{2}

∴ f is one-one i.e. it is injective.

(b) There are such member of codomain which have no image in domain N.

e.g. 3 ∈ codomain N. But there is no pre-image in domain of f.

=> f is not onto i.e. not surjective.

(ii) f: z —> z given by f(x) = x²

(a) f (-1) = f (1) = 1 => -1 and 1 have the same image.

∴ f is not one-one i.e. not injective.

(b) There are many such elements belonging to codomain have no pre-image in its codomain z.

e.g. 3 ∈ codomain z but √3 ∉ domain z of f,

∴ f is not onto i.e. not surjective

(iii) f: R->R, given by f(x) = x²

(a) f is not one-one since f(-1) = f(1) = 1

– 1 and 1 have the same image i.e., f is not injective

(b) – 2∈ codomain R off but √-2 does not belong to domain R of f.

=> f is not onto i.e. f is not surjective.

(iv) Injective but not surjective.

(v) Injective but not surjective.

**Question 3.**

**Prove that the Greatest Integer Function f: R->R given by f (x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.**

**Solution:**

f: R—> R given by f (x) = [x]

(a) f(1. 2) = 1, f(1. 5) = 1 => f is not one-one

(b) All the images of x e R belonging to its domain have integers as the images in codomain. But no fraction proper or improper belonging to codomain of f has any pre-image in its domain.

=> f is not onto.

**Question 4.**

**Show that the Modulus Function f: R -> R given by f (x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is- x, if xis negative.**

**Solution:**

f: R->R given by f(x) = |x|

(a) f(-1) = |-1| = 1,f(1) = |1| = 1

=> -1 and 1 have the same image

∴ f is not one-one

(b) No negative value belonging to codomain of f has any pre-image in its domain

∴ f is not onto. Hence, f is neither one-one nor onto.

**Question 5.**

**Show that the Signum Function f: R–>R given by**

**f(x) = 1, if x > 0**

**f(x) = 0, if x = 0**

**f(x) = – 1, if x < 0**

**is neither one-one nor onto.**

**Solution:**

f: R–>R given by

f(x) = 1, if x > 0

f(x) = 0, if x = 0

f(x) = -1, if x < 0

(a) f(x_{1}) = f(x_{2}) = 1

∴ 1 and 2 have the same image i.e.

f(x_{1}) = f(x_{2}) = 1 for x>0

=> x_{1}≠x_{2}

Similarly f(x_{1}) = f(x_{2}) = – 1, for x<0 where x_{1} ≠ x_{2} => f is not one-one.

(b) Except – 1,0,1 no other member of codomain of f has any pre-image in its domain.

∴ f is not onto.

=> f is neither into nor onto.

**Question 6.**

**Let A= {1,2,3}, B = {4,5,6,7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B. Show that f is one-one.**

**Solution:**

A= {1,2,3},B= {4,5,6,7} f= {(1,4), (2,5), (3,6) }.

Every member of A has a unique image in B

∴ f is one – one.

**Question 7.**

**In each of die following cases, state whether the function is one-one, onto or bijective. Justify your answer.**

**(i) f: R–>R defined by f(x) = 3 – 4x**

**(ii) f: R–>R defined by f(x) = 1 + x²**

**Solution:**

(i) f: R —> R defined by 3 – 4x,

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

(a) f(x_{1}) = f(x_{2}) =>3 – 4x_{1} = 3 – 4x_{2}

=> x_{1} = x_{2}. This shows that f is one-one

(b) f(x) = y = 3 – 4x

For every value of y belonging to its codomain. There is a pre-image in its domain => f is onto.

Hence, f is one-one onto

(ii) f: R—>R given by f(x)= 1 + x²

(a) f(1) = 1 + 1 = 2,f(-1) = 1 +1 = 2

∴ f (-1) = f (1) = 2 i.e.-1 and 1 have the same image 2.

=> f is not one-one.

(b) No negative number belonging to its codomain has its pre-image in its domain

=> f is not onto. Thus f is neither one- one nor onto.

**Question 8.**

**Let A and B be sets. Show that f:A x B –>B x A such that f (a, b) = (b, a) is bijective function.**

**Solution:**

We have f: (A x B) —> B x A such that f (a, b) = b, a

(a) ∴ f(a1, b1)= (b1, a1) f(a2, b2) = (b2, a2) f(a1, b1) = f(a2, b2)

=>(b1, a1)

= (b2, a2)

=> b1 = b2 and a1 = a2 f is one-one

(b) Every member (p, q) belonging to its codomain has its pre-image in its domain as (q, p) f is onto. Thus, f is one-one and onto i.e. it is bijective.

**Question 9.**

**Let f: N —> N be defined by
f (n) = **

**,if n is odd**

f (n) = ,if n is even

for all n∈N

f (n) = ,if n is even

for all n∈N

**State whether the function f is bijective. Justify your answer.**

**Solution:**

f: N —> N, defined by

The elements 1, 2 belonging to domain of f have the same image 1 in its codomain

=> f is not one-one.

∴ it is not injective,

(b) Every member of codomain has pre-image in its domain e.g. 1 has two pre-images 1 and 2

=> f is onto. Thus f is not one-one but it is onto

=> f is not bijective.

**Question 10.**

**Let A = R-{3} and B = R-{1}. consider the function f: A -> B defined by f (x) = **

**Solution:**

Is f one-one and onto? Justify your answer.

f: A –> B where A = R – {3}, B = R – {1} f is defined by

**Question 11.**

**Let f: R -> R be defined as f (x)=x ^{4}. Choose the correct answer.**

**(a) f is one-one onto**

**(b) f is many-one onto**

**(c) f is one-one but not onto**

**(d) f is neither one-one nor onto**

**Solution:**

f(-1) = (-1)

^{4}= 1,f(1) = 1

^{4}= 1

∴ – 1, 1 have the same image 1 => f is not one- one

Further – 2 in the codomain of f has no pre-image in its domain.

∴ f is not onto i.e. f is neither one-one nor onto Option (d) is correct.

**Question 12.**

**Let f: R –> R be defined as f (x)=3x. Choose the correct answer.**

**(a) f is one-one onto**

**(b) f is many-one onto**

**(c) f is one-one but not onto**

**(d) f is neither one-one nor onto**

**Solution:**

f: R –> R is defined by f (x) = 3x

(a) f(x_{1}) = 3x_{1}, f(x_{2}) = 3x_{2}

=> f(x_{1}) = f(x_{2})

=> 3x_{1} = 3x_{2}

=> x_{1} = x_{2}

=> f is one-one

(b) for every member y belonging to co-domain has pre-image x in domain of f.

∵ y = 3x

=>

f is onto

f is one-one and onto. Option (a) is correct.

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