NCERT Solutions Class 12 maths Chapter-1(Relation And Functions)Exercise 1.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-1 (Relation And Functions) Exercise 1.2. This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 1.2

Question 1.

Show that the function f: R_*⇢ R_* defined by f(x)=(1/x) is one-one and 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 same as R_*?

Solution:

One-one:

f(x)=f(y)

⇒1/x =1/y

⇒x=y

Therefore, f is one-one.

Onto:

It is clear that for y∈ R_* there exists x=(1/y)∈ R_* (exists as y ≠ 0) such that f(x)=1/(1/y)=y

Therefore, f is onto.

Thus, consider function g: N⇢R_* defined by g(x)=1/x

We have, f(x₁)=g(x₂)⇒1/x₁=1/x₂⇒x₁=x₂

Therefore, g is one-one.

Further, it is clear that g is not onto as for 1.2∈ R_* there does not exist any x in N such that g(x)=1/(1.2)

Hence, function g is one-one but not onto.

Question 2.

Check the injectivity and surjectivity of the following functions:

(i) f: N⇢N given by f(x)=x²

Solution:

It is seen that for x, y ∈ N, f(x)=f(y) ⇒x²=y²⇒x=y

Therefore, f is injective.

Now, 2 ∈ N but there does not exist any x in N such that f(x)=x²=2.

Therefore, f is not surjective.

(ii) f: Z⇢Z given by f(x)=x²

Solution:

It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.

-2 ∈ Z. But, there does not exist any x in Z such that f(x)= x²=-2.Therefore, f is not surjective

(iii) f: R⇢ R given by f(x)=x²

Solution:

It is seen that f(-1)=f(1), but -1 ≠1. Therefore, f is not injective.

-2 ∈ R. But, there does not exist any x in R such that f(x)= x²=-2.Therefore, f is not surjective.

(iv)f: N⇢N given by f(x)=x³

Solution:

It is seen that for x, y ∈ N, f(x)=f(y)⇒x³=y³⇒x=y. Therefore, f is injective.

2∈ N. But, there does not exist any element x in domain N such that f(x)=x³=2. Therefore, f is not surjective.

(v) f: Z⇢Z given by f(x)=x³

Solution:

It is seen that for x, y ∈Z, f(x)=f(y)⇒x³=y³⇒x=y. Therefore, f is injective.

2∈Z. But, there does not exist any element x in domain Z such that f(x)=x³=2. Therefore, f is 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:

It is seen that f(1.2)=[1.2]=1, f(1.9)=[1.9]=1.

f(1.2)=f(1.9), but 1.2≠1.9. Therefore, f is not one-one.

Consider 0.7∈R. It is known that f(x)=[x] is always an integer. Thus, there does not exist any element x ∈R such that f(x)=0.7. Therefore, f is not onto.

Hence, the greatest integer function is neither one-one nor 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 x is negative.

Solution:

It is seen that f(-1)=|-1|=1, f(1)=|1|=1.

f(-1)=f(1), but -1≠1. Therefore, f is not one-one.

Consider, -1∈R. It is known that f(x)=|x| is always non-negative. Thus, there does not exist any element x in domain R such that f(x)=|x|=-1. Therefore, f is not onto.

Hence, the modulus function is neither one-one nor onto.

Question 5.

Show that the signum function f: R⇢R given by, f(x)={ (1, if x>0), (0, if x=0), (-1, if x<0)} is neither one-one nor onto.

Solution:

It is seen that f(1)=f(2)=1, but 1≠2. Therefore, f is not one-one.

As f(x) takes only 3 values (1, 0, or -1) for the element -2 in co-domain R, there does not exist any x in domain R such that f(x)=-2. Therefore, f is not onto.

Hence, the signum function is neither one-one 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:

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.

f:A⇢B is defined as f={(1,4), (2,5), (3,6)}

Therefore, f(1)=4, f(2)=5, f(3)=6

It is seen that the images of distinct elements of A under f are distinct.

Hence, function f is one-one.

Question 7.

In each of the 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

Solution:

Let x₁, x₂ ∈R such that f(x₁)=f(x₂)

⇒3-4x₁=3-4x₂

⇒-4x₁=-4x₂

⇒x₁=x₂

Therefore, f is one-one.

For any real number (y) in R, there exists {(3-y)/4} in R such that f((3-y)/4)=3-4((3-y)/4)=y.

Therefore, f is onto

Hence, f is bijective.

(ii) f:R⇢R defined b f(x)=1+x²

Solution:

Let x₁, x₂ ∈ R such that f(x₁)=f(x₂)

⇒1+x₁²=1+x₂²

⇒x₁²=x₂²

⇒x₁=±x₂

Therefore, f(x₁)=f(x₂) does not imply that x₁=x₂

For instance, f(1)=f(-1)=2

Therefore, f is not one-one.

Consider, an element -2 in co-domain R.

It is seen that f(x)=1+x² is positive for all x ∈ R.

Thus, there does not exist any x in domain R such that f(x)=-1.

Therefore, f is not onto.

Hence, 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 (a, b)=(b, a) is bijective function.

Solution:

Let (a₁, b₁), (a₂, b₂) ∈ A x b such that f(a₁, b₁)=f(a₂, b₂)

⇒(b₁, a₁)=(b₂, a₂)

⇒b₁=b₂ and a₁=a₂

⇒(a₁, b₁)=(a₂, b₂)

Therefore, f is one-one.

Let (b,a) ∈ B x A such that f(a, b)=(b,a).

Therefore, f is onto.

Hence, f is bijective.

Question 9.

Let f: N⇢ N defined by f(n)={((n+1)/2, if n is odd), (n/2, if n is even) for all n ∈ N. State whether the function f is bijective. Justify your answer.

Solution:

It can be observed that:

f(1)=(1+1)/2=1 and f(2)=2/2=1

So, f(1)=f(2), where, 1≠2

Therefore, f is not one-one.

Therefore, it is not bijective. (Since, it needs to be both one-one and onto to be bijective).

Question 10.

Let A=R-{1}. Consider the function f: A⇢B defined by f(x)=(x-2)/(x-3). Is f one-one and onto? Justify your answer.

Solution:

Let x, y ∈ A such that f(x)=f(y)

⇒ (x-2)/(x-3)=(y-2)/(y-3)

⇒(x-2)(y-3)=(y-2)(x-3)

⇒ xy-3x-2y+6=xy-3y-2x+6

⇒ -3x-2y=-3y-2x

⇒ 3x-2x=3y-2y

⇒ x=y

Therefore, f is one-one.

Let, y ∈ B= R-{1}. Then y≠1.

The function f is onto if there exists x ∈ A such that f(x)=y

Now,

f(x)=y

⇒ (x-2)/(x-3)=y

⇒ x-2=xy-3y

⇒ x(1-y)=-3y+2

⇒ x=(2-3y)/(1-y) ∈ A

Thus, for any y ∈ B, there exists (2-3y)/(1-y) ∈ A such that f((2-3y)/(1-y))={((2-3y)/(1-y))-2}/{((2-3y)/(1-y))-3}=(2-3y-2+2y)/(2-3y-3+3y)=(-y)/(-1)=y

Therefore, f is onto.

Hence, function f is one-one and onto.

Question 11.

Let f: R⇢R be defined as f(x)=x⁴. 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:

Let x, y ∈ R such that f(x)=f(y)

⇒ x4=y4

⇒ x=±y

Therefore, f(x1)=f(x²) does not imply that x1=x²

For instance, f(1)=f(-1)=1

Therefore, f(1)=f(-1)=1