NCERT Solutions for Maths Class 12 Exercise 1.2

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NCERT Solutions for Maths Class 12 Exercise 1.1

NCERT Solutions for Maths Class 12 Ex 1.2

Maths Class 12 Ex 1.2 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 being same as R?

Solution:
(a) The function f : R → R defined as f(x) = 1/x

(i) We have f(x₁) = 1/x₁ and f(x₂) = 1/x₂

If f(x₁) = f(x₂)
1/x₁= 1/x₂
⇒ x₁ = x₂
It means for each x ∈ R in domain, there is a unique image in codomain R.
Therefore, f is one-one.
(ii) For every y ∈ R in codomain, there exists an element x ∈ R in domain such that f(x) = y.
Therefore, f is onto.

(b) If the domain R is replaced by N and codomain remains the same, then the function is f : N → R

Let n₁, n₂ ∈ N in domain.

If f(n₁) = f(n₂)
⇒ 1/n₁ = 1/n₂

⇒ n₁ = n₂
⇒ f is one-one.
But for every y ∈ R in codomain, it may not have an element x ∈ N in domain such that y is an image of x. For example, 3/2, 4/5, -1 ∈ R in codomain, but these are not an image of any element of domain belonging to N.
Therefore, f is not onto.

Maths Class 12 Ex 1.2 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) The function f : N → N given by f(x) = x²
(a) Let x_1,x₂∈ N, then f(x₁) = x₁² and f(x₂) = x₂²

If f(x₁) = f(x₂)
⇒ x₁² = x₂²

⇒ x₁ = x₂
Therefore, f is one-one, i.e., it is injective.
(b) We have 2, 3 ∈ N in codomain, which are not an image of any element x ∈ N in domain. It means there is no natural number in domain whose square is 2 or 3.
Therefore, f is not onto, i.e., it is not surjective.
Hence, the function f is injective but not surjective.

(ii) The function f : Z → Z given by f(x) = x²
(a) Let 1, -1 ∈ Z in domain, then f(1) = f(-1) = 1

Here, f(1) = f(-1) = 1 but -1 and 1 are not equal. It means the elements 1 and -1 have the same image 1.
Therefore, f is not one-one, i.e., it is not injective.
(b) We have -2, -3 ∈ Z in codomain, which are not an image of any element x ∈ Z in domain. It means there is no integer in domain whose square is -2 or -3.
Therefore, f is not onto, i.e., it is not surjective.

Hence, the function f is neither injective nor surjective.

(iii) The function f : R → R given by f(x) = x²
(a) Let 1, -1 ∈ R in domain, then f(1) = f(-1) = 1

Here, f(1) = f(-1) = 1 but -1 and 1 are not equal. It means the elements 1 and -1 have the same image 1.
Therefore, f is not one-one, i.e., it is not injective.
(b) We have -2, -3 ∈ R in codomain, which are not an image of any element x ∈ R in domain. It means there is no integer in domain whose square is -2 or -3.
Therefore, f is not onto, i.e., it is not surjective.

Hence, the function f is neither injective nor surjective.

(iv) The function f : N → N given by f(x) = x³
(a) Let x_1,x₂∈ N, then f(x₁) = x₁³ and f(x₂) = x₂³

If f(x₁) = f(x₂)
⇒ x₁³ = x₂³

⇒ x₁ = x₂
Therefore, f is one-one, i.e., it is injective.
(b) We have 2, 3 ∈ N in codomain, which are not an image of any element x ∈ N in domain. It means there is no natural number in domain whose cube is 2 or 3.
Therefore, f is not onto, i.e., it is not surjective.
Hence, the function f is injective but not surjective.

(v) The function f : Z → Z given by f(x) = x³
(a) Let x_1,x₂∈ Z, then f(x₁) = x₁³ and f(x₂) = x₂³

If f(x₁) = f(x₂)
⇒ x₁³ = x₂³

⇒ x₁ = x₂
Therefore, f is one-one, i.e., it is injective.
(b) We have 2, 3 ∈ Z in codomain, which are not an image of any element x ∈ Z in domain. It means there is no natural number in domain whose cube is 2 or 3.
Therefore, f is not onto, i.e., it is not surjective.
Hence, the function f is injective but not surjective.

Maths Class 12 Ex 1.2 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:
The function f : R → R given by f(x) = [x]
(a) Let 1.3, 1.4 ∈ R in domain, then f(1.3) = 1 and f(1.4) = 1

Here, f(1.3) = f(1.4) = 1 but 1.3 ≠ 1.4

Therefore, f is not one-one.
(b) We have 1.1, 1.2 ∈ R in codomain, which are not an image of any element x ∈ R in domain because the ranges are only integers here.

Therefore, f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

Maths Class 12 Ex 1.2 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:
The function f : R → R given by f(x) = |x|
(a) Let 2, -2 ∈ R in domain, then f(2) = f(-2) = 2

Here, f(2) = f(-2) = 2 but -2 and 2 are not equal. It means the elements 2 and -2 have the same image 2. Therefore, f is not one-one.
(b) We have -2, -3 ∈ R in codomain, which are not an image of any element x ∈ R in domain. Therefore, f is not onto.

Hence, the modulus function is neither one-one nor onto.

Maths Class 12 Ex 1.2 Question 5.

Show that the Signum Function f : R → R, given by
f(x) = 1, if x > 0
f(x) = 0, if x = 0
f(x) = –1, if x < 0
is neither one-one nor onto.

Solution:
The function f : R → R given by
f(x) = 1, if x > 0
f(x) = 0, if x = 0
f(x) = -1, if x < 0
(a) Let 1, 2 ∈ R in domain, then f(1) = f(2) = 1
Here, f(1) = f(2) = 1 but 1 ≠ 2
Similarly, for -1, -2 ∈ R in domain, f(-1) = f(-2) = –1, but -1 ≠ -2

Therefore, f is not one-one.
(b) Except –1, 0, 1, there is no element y ∈ R in codomain which is an image of an element x ∈ R in domain such that f(x) = y.
Therefore, f is not onto.
Hence, f is neither one-one nor onto.

Maths Class 12 Ex 1.2 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:
We have A = {1, 2, 3}, B = {4, 5, 6, 7} and the function f = {(1, 4), (2, 5), (3, 6)}

From the given function, we can see that 4 is the image of 1, 5 is the image of 2 and 6 is the image of 3. Every element of A has a unique image in B.

Therefore, f is one-one.

Maths Class 12 Ex 1.2 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 + x²

Solution:
(i) The function f : R → R defined by f(x) = 3 – 4x.
Let x₁,x₂ ∈ R in domain, then f(x₁) = 3 – 4x₁ and f(x₂) = 3 – 4x₂
(a) If f(x₁) = f(x₂), then 3 – 4x₁ = 3 – 4x₂
⇒ x₁ = x₂.

Therefore, f is one-one.
(b) Let f(x) = 3 – 4x = y
Then x = (3 – y)/4
It means for every value of y ∈ R in codomain, there exists an element y ∈ R in domain such that f(x) = y. Therefore, f is onto.

Hence, the function f is bijective.

(ii) The function f : R → R defined by f(x) = 1 + x²
(a) Let 1, -1 ∈ R in domain, then f(1) = 1 + 1 = 2 and f(-1) = 1 + 1 = 2
Here, f(1) = f(-1) = 2, i.e., 1 and -1 have the same image 2. But 1 ≠ -1.
Therefore, f is not one-one.
(b) We have -2, -3 ∈ R in codomain, which are not an image of any element x ∈ R in domain. Therefore, f is not onto.

Hence, the function f is not bijective.

Maths Class 12 Ex 1.2 Question 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:
We have the function f : (A × B) → B × A such that f(a, b) = (b, a)
(a) Since f(a₁, b₁) = (b₁, a₁) and f(a₂, b₂) = (b₂, a₂)

If f(a₁, b₁) = f(a₂, b₂)
⇒ (b₁, a₁) = (b₂, a₂)
⇒ b₁ = b₂ and a₁ = a₂

Therefore, f is one-one.
(b) For every element (q, p) ∈ B × A in codomain, there exists an element (p, q) ∈ A × B in domain such that f(p, q) = (q, p). Therefore, f is onto.

Thus, f is a bijective function.

Maths Class 12 Ex 1.2 Question 9.

Let f : N → N be defined by
for all n ∈ N.
State whether the function f is bijective. Justify your answer.

Solution:
The function f : N → N, defined by

The elements 1, 2 ∈ N in domain of f have the same image 1 in its codomain

Therefore, f is not one-one.
(b) Every element in codomain N is an image of an element in domain N. For example, 1 in codomain is the image of 1 and 2 in domain.

Therefore, f is onto.

Hence, f is not bijective.

Maths Class 12 Ex 1.2 Question 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.

Solution:

The function f : A → B where A = R – {3}, B = R – {1} is defined by

Maths Class 12 Ex 1.2 Question 11.

Let f : R → R be defined as f(x) = x⁴. 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) For 1, -1 ∈ R in domain, f(1) = (1)⁴ = 1 and f(-1) = (-1)⁴ = 1
We have f(1) = f(-1) = 1 but 1 ≠ -1. It means for 1 and -1, we have the same image 1.

Therefore, f is not one-one.
For -2 ∈ R in the codomain of f, there is no element in domain whose image is -2.
Therefore, f is not onto. Thus, f is neither one-one nor onto.

Hence, option (D) is correct.

Maths Class 12 Ex 1.2 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) The function f : R → R is defined by f(x) = 3x
(a) Let x₁, x₂∈ R in domain, then f(x₁) = 3x₁ and f(x₂) = 3x₂
If f(x₁) = f(x₂)
⇒ 3x₁ = 3x₂
⇒ x₁ = x₂
⇒ Therefore, f is one-one.
(b) For every element y ∈ R in codomain, there exists an element x ∈ R in domain such that y = 3x or x = y/3.

Therefore, f is onto. Thus, f is one-one and onto.

Hence, option (A) is correct.

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