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

# Exercise 1.2

**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∗?**

**Solution:**

**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}**Solution:**

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

For x, y ∈ N => f(x) = f(y) which implies x^{2} = y^{2}

x = y

Therefore f is injective.

There are such numbers of co-domain which have no image in domain N.

Say, 3 ∈ N, but there is no pre-image in domain of f. such that f(x) = x^{2} = 3.

f is not surjective.

Therefore, f is injective but not surjective.

**(ii)** Given, f : Z → Z given by f(x) = x^{2}

Here, Z = {0, ±1, ±2, ±3, ±4, …..}

f(-1) = f(1) = 1

But -1 not equal to 1.

f is not injective.

There are many numbers of co-domain which have no image in domain Z.

For example, -3 ∈ co-domain Z, but -3 ∉ domain Z

f is not surjective.

Therefore, f is neither injective nor surjective.

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

f(-1) = f(1) = 1

But -1 not equal to 1.

f is not injective.

There are many numbers of co-domain which have no image in domain R.

For example, -3 ∈ co-domain R, but there does not exist any x in domain R where x

2

= -3

f is not surjective.

Therefore, f is neither injective nor surjective.

(iv) f : N → N given by f(x) = x^{3}

For x, y ∈ N => f(x) = f(y) which implies x

3 = y^{3}

x = y

Therefore f is injective.

There are many numbers of co-domain which have no image in domain N.

For example, 4 ∈ co-domain N, but there does not exist any x in domain N where x^{3}= 4.

f is not surjective.

Therefore, f is injective but not surjective.

(v) f : Z → Z given by f(x) = x^{3}

For x, y ∈ Z => f(x) = f(y) which implies x

3 = y^{3}

x = y

Therefore f is injective.

There are many numbers of co-domain which have no image in domain Z.

For example, 4 ∈ co-domain N, but there does not exist any x in domain Z where x^{3}= 4.

f is not surjective.

Therefore, f is injective but not surjective.

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

**Solution:**

Function f : R → R, given by f(x) = [x]

f(x) = 1, because 1 ≤ x ≤ 2

f(1.2) = [1.2] = 1

f(1.9) = [1.9] = 1

But 1.2 ≠ 1.9

f is not one-one.

There is no fraction proper or improper belonging to co-domain of f has any pre-image in its

domain.

For example, f(x) = [x] is always an integer

for 0.7 belongs to R there does not exist any x in domain R where f(x) = 0.7

f is not onto.

Hence proved, the Greatest Integer Function is neither one-one nor onto.

**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:**

f : R → R, given by f(x) = | x |, defined as

**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} and

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

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

Here, also distinct elements of A have distinct images in B.

Therefore, 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}**

**Solution:**

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

If x_{1}, x_{2} ∈ R then

f(x_{1}) = 3 – 4x_{1} and

f(x_{2}) = 3 – 4x_{2}

If f(x_{1}) = f(x_{2}) then x_{1} = x_{2}

Therefore, f is one-one.

Again,

f(x) = 3 – 4x

or y = 3 – 4x

or x = (3-y)/4 in R

f((3-y)/4) = 3 – 4((3-y)/4) = y

f is onto.

Hence f is onto or bijective.

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

If x

_{1}, x

_{2}∈ R then

f(x

_{1}) = 1 + 𝑥

_{1}

^{2}and

f(x2) = 1 + 𝑥2

2

If f(x1) = f(x2) then 𝑥

_{1}

^{2}= 𝑥

_{2}

^{2}

This implies x1 ≠ x2

Therefore, f is not one-one

Again, if every element of co-domain is image of some element of Domain under f, such that

f(x) = y

f(x) = 1 + x

^{2}

y = f(x) = 1 + x^{2}

or x = ±√1 − 𝑦

Therefore, f(√1 − 𝑦 ) = 2 – y ≠ y

Therefore, f is not onto or bijective.

**8. Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective**

**function.**

**Solution: **

**Step 2: Check for Surjectivity:**

Let (b, a) be any element of B x A. Then a ∈ A and b ∈ B

This implies (a, b) ∈ A x B

For all (b, a) ∈ B x A, their exists (a, b) ∈ A x B

Therefore, f: A x B -> B x A is bijective function.

**9. Let f : N → N be defined by**

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

**Solution:**

For n = 1, 2

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

f(2) = (n)/2 = (2)/2 = 1

f(1) = f(2), but 1 ≠ 2

f is not one-one.

For a natural number, “a” in co-domain N

If n is odd

n = 2k + 1 for k ∈ N , then 4k + 1 ∈ N such that

f(4k+1) = (4k+1+1)/2 = 2k + 1

If n is even

n= 2k for some k ∈ N such that

f(4k) = 4k/2 = 2k

f is onto

Therefore, f is onto but not bijective function.

**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.**

**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 : R → R be defined as f(x) = x

^{4}

let x and y belongs to R such that, f(x) = f(y)

x

4 = y

^{4}or x = ± y

f is not one-one function.

Now, y = f(x) = x

4

Or x = ± y

^{1/4}

f(y

^{1/4}) = y and f(-y

^{1/4}) = -y

Therefore, f is not onto function.

**Option D is correct.**

**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 be defined as f(x) = 3x

let x and y belongs to R such that f(x) = f(y)

3x = 3y or x = y

f is one-one function.

Now, y = f(x) = 3x

Or x = y/3

f(x) = f(y/3) = y

Therefore, f is onto function.

**Option (A) is correct.**