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# 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* â‡¢ 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)

â‡’1/x =1/y

â‡’x=y

Therefore, f is one-one.

Onto:

It is clear that for yâˆˆ R* there exists x=(1/y)âˆˆ R* (exists as y â‰  0) such that f(x)=1/(1/y)=y

Therefore, f is onto.

Thus, consider function g: Nâ‡¢R* defined by g(x)=1/x

We have, f(x1)=g(x2)â‡’1/x1=1/x2â‡’x1=x2

Therefore, g is one-one.

Further, it is clear that g is not onto as for 1.2âˆˆ 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: Nâ‡¢N given by f(x)=x2

Solution:

It is seen that for x, y âˆˆ N, f(x)=f(y) â‡’x2=y2â‡’x=y

Therefore, f is injective.

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

Therefore, f is not surjective.

### (ii) f: Zâ‡¢Z given by f(x)=x2

Solution:

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

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

### (iii) f: Râ‡¢ R given by f(x)=x2

Solution:

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

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

#### (iv)f: Nâ‡¢N given by f(x)=x3

Solution:

It is seen that for x, y âˆˆ N, f(x)=f(y)â‡’x3=y3â‡’x=y. Therefore, f is injective.

2âˆˆ 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: Zâ‡¢Z given by f(x)=x3

Solution:

It is seen that for x, y âˆˆZ, f(x)=f(y)â‡’x3=y3â‡’x=y. Therefore, f is injective.

2âˆˆZ. 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: Râ‡¢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 1.2â‰ 1.9. Therefore, f is not one-one.

Consider 0.7âˆˆR. It is known that f(x)=[x] is always an integer. Thus, there does not exist any element x âˆˆR 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:Râ‡¢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 -1â‰ 1. Therefore, f is not one-one.

Consider, -1âˆˆR. 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: Râ‡¢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 1â‰ 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:Aâ‡¢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.

### (i) f:Râ‡¢R defined by f(x)=3-4x

Solution:

Let x1, x2 âˆˆR such that f(x1)=f(x2)

â‡’3-4x1=3-4x2

â‡’-4x1=-4x2

â‡’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:Râ‡¢R defined b f(x)=1+x2

Solution:

Let x1, x2 âˆˆ R such that f(x1)=f(x2)

â‡’1+x12=1+x22

â‡’x12=x22

â‡’x1=Â±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 âˆˆ 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 â‡¢B x A such that (a, b)=(b, a) is bijective function.

Solution:

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

â‡’(b1, a1)=(b2, a2)

â‡’b1=b2 and a1=a2

â‡’(a1, b1)=(a2, b2)

Therefore, f is one-one.

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

Therefore, f is onto.

Hence, f is bijective.

### Question 9. Let f: Nâ‡¢ N defined by f(n)={((n+1)/2, if n is odd), (n/2, if n is even) for all n âˆˆ 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, 1â‰ 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: Aâ‡¢B defined by f(x)=(x-2)/(x-3). Is f one-one and onto? Justify your answer.

Solution:

Let x, y âˆˆ A such that f(x)=f(y)

â‡’ (x-2)/(x-3)=(y-2)/(y-3)

â‡’(x-2)(y-3)=(y-2)(x-3)

â‡’ xy-3x-2y+6=xy-3y-2x+6

â‡’ -3x-2y=-3y-2x

â‡’ 3x-2x=3y-2y

â‡’ x=y

Therefore, f is one-one.

Let, y âˆˆ B= R-{1}. Then yâ‰ 1.

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

Now,

f(x)=y

â‡’ (x-2)/(x-3)=y

â‡’ x-2=xy-3y

â‡’ x(1-y)=-3y+2

â‡’ x=(2-3y)/(1-y) âˆˆ A

Thus, for any y âˆˆ B, there exists (2-3y)/(1-y) âˆˆ 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 âˆˆ R such that f(x)=f(y)

â‡’ x4=y4

â‡’ x=Â±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 âˆˆ R such that f(x)=f(y)

â‡’ 3x = 3y

â‡’ 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.

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