# Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions – Exercise 1.2

• Last Updated : 28 Jan, 2021

### 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.

### (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:AB 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.

### (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, x2R 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.

### (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

Therefore, f is not one-one

Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x)=2

Therefore, f is not onto.

### (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)

⇒ 3x = 3y

⇒ x=y

Therefore, f is one-one.

Also, for any real number (y) in co-domain R, there exists y/3 in R such that f(y/3) = 3(y/3) = y

Therefore, f is onto.

Hence, the correct answer is A.

My Personal Notes arrow_drop_up