NCERT Solutions for Maths Class 12 Exercise 1.2

# NCERT Solutions for Maths Class 12 Exercise 1.2

Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 1.2.

## 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 x1, 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 x1, 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 x1, 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.

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)

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.