These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.2

Question 1.

Show that the function f : R_{*} → R_{*}, defined by f(x) = \(\frac { 1 }{ x }\) 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:

a. One-one

Let x_{1}, x_{1} ∈ R, such that f(x_{1}) = f(x_{2})

For each y ∈ R_{*} there exists x = \(\frac { 1 }{ y }\) ∈ R,

such that f(x) = y.

∴ f is onto

b. One-one

Let x_{1}, x_{2} ∈ N such that f(x_{1}) = f(x_{2})

⇒ \(\frac{1}{x_{1}}=\frac{1}{x_{2}}\) ⇒ x_{1} = x_{2}

∴ f is one-one

Onto

The domain of f is N = {1, 2, 3, ……. }

The range of f = {1, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\), \(\frac { 1 }{ 4 }\), ……. } ≠ R

∴ f is not onto

Another method

2 ∈ R_{*}.

The pre-image of 2 is \(\frac { 1 }{ 2 }\) ∉ N .

Hence f is not onto.

Question 2.

Check the injectivity and surjectivity of the following functions

i. f : N → N given by f (x) = x²

ii. f : Z → Z given by f (x) = x²

iii. f : R → R given by f (x) = x²

iv. f : N → N given by f (x) = x³

v. f : Z → Z given by f (x) = x³

Solution:

i. Injectivity

Let x_{1}, x_{1} ∈ N such that f(x_{1}) = f(x_{2})

⇒ x²_{1} = x²_{2}

⇒ x²_{1} – x²_{2} = 0

⇒ (x_{1} – x_{2})(x_{1} + x_{2}) = 0

⇒ x_{1} – x_{2} = 0 since x_{1} + x_{2} ≠ 0 as x_{1}, x_{1} ∈ N

⇒ x_{1} = x_{2}

∴ f is injective

Surjectivity

Let y = 3 ∈ N, the co-domain of f

f(x) = 3 ⇒ x² = 3

⇒ x = ±\(\sqrt{3}\) ∉ N , the domain of f

∴f is not surjective

ii. Injectivity

Let x_{1} = 2, x_{2} = – 2

f(x_{1}) = 2² = 4, f(x_{2}) = (- 2)² = 4

i.e., f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴ f is not injective

Surjectivity

Let y = 2 ∈ Z, the co-domain of f

∴ f(x) = 2 ⇒ x² = 2 ⇒ x = ± \(\sqrt{2}\) ∉ Z, the domain of f

∴ f is not surjective.

iii. Injectivity

Let x_{1} = 2, x_{2} = – 2

f(x_{1}) = 2² = 4, f(x_{2}) = (- 2)² = 4

i.e., f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴ f is not injective (one-one)

Surjectivity

Let y = – 1 ∈ R, the co-domain of f

f(x) = – 1 ⇒ x² = -1

⇒ x = ± \(\sqrt{1}\) ∉ R, the domain of f

∴ f is not surjective (onto)

iv. Injectivity

Let x_{1}, x_{2} ∈ N such that f(x_{1}) = f(x_{1}) = f(x_{2})

⇒ x³_{1} = x³_{2} ⇒ x³_{1} – x³_{2} = 0

⇒ (x_{1} – x_{2}) (x²_{1} + x_{1}x_{2} + x²_{2}2) = 0

⇒ x_{1} – x_{2} = 0 since x²_{1} + x_{1}x_{2} + x²_{2} ≠ 0

⇒ x_{1} = x_{2}

∴f is injective (one-one)

Surjectivity

Let y = 4 ∈ N, the co-domain of f

⇒ f(x) = 4 ⇒ x_{3} = 4

⇒ x = 4^{1/3} ∉ N, the domain of f

∴f is not surjective (onto)

v. Injectivity

Let x_{1}, x_{2} ∈ Z such that f(x_{1}) = f(x_{2})

⇒ x³_{1} = x³_{2} ⇒ x_{1} = x_{2}

∴ f is injective (one-one)

Surjectivity

Let y = 4 ∈ Z, the co-domain of f such that

f(x) = 4

⇒ x_{3} = 4 ⇒ x = 4^{1/3} ∉ Z, the domain of f

∴ f is not surjective (onto)

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:

Refer Example 26

One-one

Let x_{1} = 2.1, x_{2} = 2.5

f(x_{1}) = f(2.1) = [2.1] = 2

f(x_{2}) = f(2.5) = [2.5] = 2

i.e. f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴ f is not one-one.

Onto

Let y = 2.5 ∈ R, the co-domain of f

f(x) = 2.5 ⇒ [x] = 2.5, which is not possible.

∴ f is not 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:

One-one

Let x_{1} = 1 and x_{2} = – 1 ∈ R

f(x_{1}) = f(1) = |1| = 1

f(x_{2}) = f(- 1) = |- 1|= 1

i.e., f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴ f is not one-one Onto

Let y = – 1 ∈ R, the co-domain of f

f(x) = – 1 ⇒ |x|= – 1 which is not possible

∴ f is not onto.

Question 5.

Show that the Signum Function f : R → R, given by

\(f(x)=\left\{\begin{array}{c}

1, \text { if } x>0 \\

0, \text { if } x=0 \\

-1, \text { if } x<0

\end{array}\right.\)

is neither one-one nor onto.

Solution:

One-one

Let x_{1} = 1, x_{2} = 2

f(x_{1}) = f(1) = 1

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

i.e. f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴ f is not one-one

Onto

The co-domain of f is R

The range of f is {-1, 0, 1} ≠ R

∴ f is not 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:

Different elements in A have different images in B. Hence 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

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

Solution:

i. One-one

Let x_{1}, x_{2} 6 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

Onto

Let y ∈ R, such that y = f(x)

⇒ y = 3 – 4x

⇒ 4x = 3 – y

⇒ x = \(\frac { 3-y }{ 4 }\) ∈ R

f(x) = f\(\frac { 3-y }{ 4 }\) = 3 – 4(\(\frac { 3-y }{ 4 }\))

= 3 – 3 + y = y

For each y ∈ R there exists x = \(\frac { 3-y }{ 4 }\) ∈ R

such that f(x) = y

∴f is onto

Since f is one-one and onto, f is bijective.

Another Method

f(x) = 3 – 4x is a linear function from R to R.

∴ f is one-one and onto.

Hence f is bijective.

ii. One-one

Let x_{1} = 1, x_{2} = – 1 ∈ R

f(x_{1}) = f(1) = 1 + 1² = 2

f(x_{2})= f(-1) = 1 + (- 1)² = 2

i.e., f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴f is not one-one

Onto

Let y = – 1 ∈ R, the co-domain of f

f(x) = y ⇒ 1 + x^{2} = – 1

⇒ x^{2} = – 2

⇒ x = ± \(\sqrt{-2}\) ∉ R

∴f is not onto.

Thus 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 f(a, b) = (b, a) is bijective function.

Solution:

Let (a_{1}, b_{1}) and (a_{2}, b_{2}) ∈ A x 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.

Let (b, a) ∈ B x A ⇒ b ∈ B and a ∈ A

⇒ (a, b) ∈ A x B

f(a, b) = (b, a)

i.e., corresponding to each (b, a) ∈ B x A, there exists (a, b) ∈ A x B such that f(a, b) = (b, a)

∴ f is onto

Hence f is a bijection.

Question 9.

Let f : N → N be defined by

f(n) = \(\left\{\begin{array}{l}

\frac{n+1}{2}, \text { if } n \text { is odd } \\

\frac{n}{2}, \text { if } n \text { is even }

\end{array} \text { for all } n \in \mathbf{N}\right.\)

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

Solution:

Let x_{1} = 1 and x_{2} = 2 ∈ N

f(x_{1}) = f(1) = \(\frac { 1+1 }{ 2 }\) = 1 and f(x_{2}) = f(2) = \(\frac { 2 }{ 2 }\) = 1

∴ f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

Hence f is not one-one and hence not bijective.

Question 10.

Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = \(\left(\frac{x-2}{x-3}\right)\). Is f one-one and onto? Justify your answer.

Solution:

One-one

One-one

Let y ∈ B such that f(x) = y

For each y ∈ B, there exists x = \(\left(\frac{2-2y}{1-y}\right)\) ∈ A

such that f(x) = y

∴ f is onto.

Question 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:

d. f is neither one-one nor onto.

One-one

Let x_{1} = 1, x_{2} = – 1 ∈ R

f(x_{1}) = f(1) = (1)^{4} = 1

f(x_{2}) = f(-1) = (-1)^{4}= 1

f(x_{1}) = f(x_{2}) for x_{1} ≠ x_{2}

∴ f is not one-one

Onto

The co-domain of f is R.

The range of f is [0, ∞), the non-negative real numbers.

Since range of f ≠ co-domain of f is not onto.

Hence f is neither one-one nor onto.

Question 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:

a. f is one-one onto

f(x_{1}) = f(x_{2}) ⇒ 3x_{1} = 3x_{2} ⇒ x_{1} = x_{2}

∴ f is one-one

For y ∈ co-domain off there exists \(\frac { y }{ 3 }\) in the domain of f such that f(\(\frac { y }{ 3 }\)) – 3(\(\frac { y }{ 3 }\)) = y

Hence f is onto.