** ** ** ** **NCERT Solutions for Maths Class 12 Exercise 1.2**

Hello Students! Welcome to **maths-formula.com**. In this post, you will find the complete** ****NCERT Solutions for Maths Class 12 Exercise 1.2**.

You can download the **PDF of NCERT Books Maths Chapter 1** for your easy reference while studying **NCERT Solutions for Maths Class 12 Exercise 1.2**.

Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

**NCERT Solutions for Maths Class 12 Exercise 1.2** helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from **NCERT Syllabus for Mathematics Class 12**.

**NCERT Solutions for Maths Class 12 Exercise 1.2** are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

If you want to recall **All Maths Formulas for Class 12**, you can find it by clicking this link.

If you want to recall **All** **Maths Formulas for Class 11**, you can find it by clicking this link.

**NCERT Solutions for Maths Class 12 Exercise 1.1**

**NCERT Solutions for Maths Class 12 Exercise 1.3**

**NCERT Solutions for Maths Class 12 Exercise 1.4**

**NCERT Solutions for Maths Class 12 Exercise 1.2**

**Maths Class 12 Ex 1.2 ****Question 1.**

**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(x_{1}) = 1/x_{1} and f(x_{2}) = 1/x_{2}

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

1/x_{1 }= 1/x_{2}

⇒ x_{1} = x_{2 }

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 n_{1}, n_{2} ∈
N in domain.

If f(n_{1}) = f(n_{2})

⇒ 1/n_{1} = 1/n_{2}

⇒ n_{1} = n_{2}

⇒ 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.**

**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, }x_{2 }∈
N, then f(x_{1}) = x_{1}² and f(x_{2}) = x_{2}²

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

⇒ x_{1}^{2} =
x_{2}^{2}

⇒ x_{1} = x_{2}

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) = x^{3}

(a) Let x_{1, }x_{2 }∈
N, then f(x_{1}) = x_{1}^{3} and f(x_{2}) = x_{2}^{3}

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

⇒ x_{1}^{3} =
x_{2}^{3}

⇒ x_{1} = x_{2}

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) = x^{3}

(a) Let x_{1, }x_{2 }∈ Z,
then f(x_{1}) = x_{1}^{3} and f(x_{2}) = x_{2}^{3}

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

⇒ x_{1}^{3} =
x_{2}^{3}

⇒ x_{1} = x_{2}

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.**

**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.**

**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.**

**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.**

**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.**

**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 x_{1},_{ }x_{2} ∈ R
in domain, then f(x_{1}) = 3 – 4x_{1} and f(x_{2}) = 3
– 4x_{2}

(a) If f(x_{1}) = f(x_{2}), then 3 – 4x_{1} = 3 –
4x_{2}

⇒ x_{1} = x_{2}.

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.**

**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(a_{1}, b_{1}) = (b_{1}, a_{1}) and
f(a_{2}, b_{2}) = (b_{2}, a_{2})

If 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}

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.**

**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

(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.**

**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.**

**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)** 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.**

**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 x_{1}, x_{2 }∈ **R in domain, then** f(x_{1})
= 3x_{1} and f(x_{2}) = 3x_{2}

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

⇒ 3x_{1} = 3x_{2}

⇒ x_{1} = x_{2}

⇒ 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.

**NCERT Solutions for Maths Class 12 Exercise 1.1**