#### NCERT Solutions for Class 12th: Ch 1 Relations and Functions Exercise 1.2 Math

Page No: 10

**Exercise 1.2**

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 being same as R∗?**Answer**

2. Check the injectivity and surjectivity of the following functions:

(i)

*f*: N → N given by f(x) =*x*^{2}
(ii)

*f*: Z → Z given by f(x) =*x*^{2}
(iii)

*f*: R → R given by f(x) =*x*^{2}
(iv)

*f*: N → N given by f(x) = x^{3}
(v)

*f*: Z → Z given by f(x) = x^{3}**Answer**

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.**Answer**

*f*: R → R is given by, f(x) = [x]

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
∴

*f*is not one-one.
Now, 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.
∴

*f*is not onto.
Hence, the greatest integer function is neither one-one nor onto.

Page No. 11

4. Show that the Modulus Function

*f*: R → R, given by*f*(x) = |x|, is neither one-ne nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.**Answer**

It is clear that

*f*(-1) = I-1| = 1
and

*f*(1) = I1l = 1
∴

*f*(-1) =*f*(1), but -1 ≠ 1
∴

*f*is not one-one.
Now, consider -1 ∈ R.

It is known that

*f*(x) = Ix| is always non-negative.
Thus, there does not exist any element x in domain R such that

*f*(x) = Ix| = -1.
∴

*f*is not onto.
Hence, the modulus function is neither one-one nor onto.

5. Show that the Signum Function

*f*: R → R, given by**Answer**

It is seen that

*f*(1) =*f*(2) = 1, but 1 ≠ 2.
∴

*f*is not one-one.
Now, 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.
∴

*f*is not onto.
Hence, the Signum function is neither one-one nor onto.

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.**Answer**

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

A → B is defined as

*f*= {(1, 4), (2, 5), (3, 6)}*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.
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
(ii)

*f*: R → R defined by*f*(x) = 1+x^{2}**Answer**

(i)

*f*: R → 12 is defined as*f*(x) = 3 - 4x.
Let x

_{1}, x_{2}∈ R such that f(x_{1}) = f(x_{2})
⇒ 3-4x

_{1}= 3-4x_{2}
⇒ -4x

_{1}= -4x_{2 }
⇒ x

_{1}= x_{2}
∴ f is one-one.

∴ f is onto.

Hence, f is bijective.

(ii) f: R → R is defined as f(x) = 1 + x2

Let x

_{1}, x_{2}∈ R such that f(x_{1}) = f(x_{2})
⇒ 1+ x

_{1}^{2}= 1 + x_{2}^{2}
⇒ x

_{1}^{2}= x_{2}^{2}
⇒ x

_{1}= ± x_{2}
∴ f(x

_{1}) = f(x_{2}) does not imply that x_{1}= x_{2}
8. Let A and B be sets. Show that f : A×B → B×A such that f(a, b) = (b, a) is
bijective function.

**Answer**

f: A × B → B × A is defined as f(a, b) = (b, a).

Let (a

_{1}, b_{1}), (a_{2}, b_{2}) ∈ A × B such that f(a_{1}, b_{1}) = f(a_{2}, b_{2})
⇒ (b

_{1}, a_{1}) = (b_{2}, a_{2})
⇒ b

_{1}= b_{2}and a_{1}= a_{2}
⇒ (a

_{1}, b_{1}) = (a_{2}, b_{2})
∴ f is one – one.

Now, let (b, a) ∈ B × A be any element.

Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a). [By definition of f]

∴ f is onto.

Hence, f is bijective.

9.

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

**Answer**

10. Let A = R – {3} and B = 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.

**Answer**

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.

**Answer**

f: R → R is defined as f(x) = x

^{4}
Let x, y ∈ R such that f(x) = f(y).

⇒ x

^{4}= y^{4}
⇒ x = ± y

∴ f(x) = f(y) does not imply that x = y.

For example f(1) = f(–1) = 1

∴ 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.

∴ f is not onto.

Hence, function f is neither one – one nor onto.

The correct answer is D.

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.

**Answer**

f: R → R is defined as f(x) = 3x.

Let x, y ∈ R such that f(x) = f(y).

⇒ 3x = 3y

⇒ x = y

∴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

∴ f is onto.

Hence, function f is one – one and onto.

The correct answer is A.

Exercise 1.3

Exercise 1.4

Miscellaneous Exercise

Class 12 Math NCERT Solutions