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

NCERT Solutions for Class 12 Maths chapter 1 exercise 1.2 is the most important exercise of chapter Relations and Functions and it includes topics like types of relations, functions, binary operations etc. Exercise 1.2 Class 12 Maths exposes students to questions like proving one to one functions etc. Such questions are generally asked in Board examinations. Solving NCERT syllabus for Class 12 Maths chapter 1 exercise 1.2 is recommended to students to score well in CBSE Class 12 board exam. The contribution of chapter Relations and Functions is high in competitive exams also like JEE main and NEET. There are many questions asked in previous years which are based on concepts of Class 12 Maths chapter 1 exercise 1.2. The NCERT chapter Relations and Functions also has the following exercise for practice.

Relations and Functions Exercise 1.1

Relations and Functions Exercise 1.3

Relations and Functions Exercise 1.4

Relations and Functions Miscellaneous Exercise

## NCERT Solutions For Class 12 Maths Chapter 1 Relations And Functions: Exercise 1.2

Question:1 Show that the function $f: R_* \longrightarrow R_{*}$ 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 ?

Given, $f: R_* \longrightarrow R_{*}$ is defined by $f(x) = \frac{1}{x}$ .

One - One :

$f(x)=f(y)$

$\frac{1}{x}=\frac{1}{y}$

$x=y$

$\therefore$ f is one-one.

Onto:

We have $y \in R_*$ , then there exists $x=\frac{1}{y} \in R_*$ ( Here $y\neq 0$ ) such that

$f(x)= \frac{1}{(\frac{1}{y})} = y$

$\therefore f is \, \, onto$ .

Hence, the function is one-one and onto.

If the domain R is replaced by N with co-domain being same as R ∗ i.e. $g: N \longrightarrow R_{*}$ defined by

$g(x)=\frac{1}{x}$

$g(x_1)=g(x_2)$

$\frac{1}{x_1}=\frac{1}{x_2}$

$x_1=x_2$

$\therefore$ g is one-one.

For $1.5 \in R_*$ ,

$g(x) = \frac{1}{1.5}$ but there does not exists any x in N.

Hence, function g is one-one but not onto.

Question:2(i) Check the injectivity and surjectivity of the following functions:

(i) $f : N\rightarrow N$ given by $f(x) = x^2$

$f : N\rightarrow N$

$f(x) = x^2$

One- one:

$x,y \in N$ then $f(x)=f(y)$

$x^{2}=y^{2}$

$x=y$

$\therefore$ f is one- one i.e. injective.

For $3 \in N$ there is no x in N such that $f(x)=x^{2}=3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:2(ii) Check the injectivity and surjectivity of the following functions:

(ii) $f : Z \rightarrow Z$ given by $f(x) = x^2$

$f : Z \rightarrow Z$

$f(x) = x^2$

One- one:

For $-1,1 \in Z$ then $f(x) = x^2$

$f(-1)= (-1)^{2}$

$f(-1)= 1$ but $-1 \neq 1$

$\therefore$ f is not one- one i.e. not injective.

For $-3 \in Z$ there is no x in Z such that $f(x)=x^{2}= -3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is neither injective nor surjective.

Question:2(iii) Check the injectivity and surjectivity of the following functions:

(iii) $f: R \rightarrow R$ given by $f(x) = x^2$

$f: R \rightarrow R$

$f(x) = x^2$

One- one:

For $-1,1 \in R$ then $f(x) = x^2$

$f(-1)= (-1)^{2}$

$f(-1)= 1$ but $-1 \neq 1$

$\therefore$ f is not one- one i.e. not injective.

For $-3 \in R$ there is no x in R such that $f(x)=x^{2}= -3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is not injective and not surjective.

Question:2(iv) Check the injectivity and surjectivity of the following functions:

(iv) $f: N \rightarrow N$ given by $f(x) = x^3$

$f : N\rightarrow N$

$f(x) = x^3$

One- one:

$x,y \in N$ then $f(x)=f(y)$

$x^{3}=y^{3}$

$x=y$

$\therefore$ f is one- one i.e. injective.

For $3 \in N$ there is no x in N such that $f(x)=x^{3}=3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:2(v) Check the injectivity and surjectivity of the following functions:

(v) $f : Z \rightarrow Z$ given by $f(x) = x^3$

$f : Z \rightarrow Z$

$f(x) = x^3$

One- one:

For $(x,y) \in Z$ then $f(x) = f(y)$

$x^{3}=y^{3}$

$x=y$

$\therefore$ f is one- one i.e. injective.

For $3 \in Z$ there is no x in Z such that $f(x)=x^{3}= 3$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:3 Prove that the Greatest Integer Function $f : R\longrightarrow R$ , given by $f (x) = [x]$ , is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$ .

$f : R\longrightarrow R$

$f (x) = [x]$

One- one:

For $1.5,1.7 \in R$ then $f(1.5)=\left [ 1.5 \right ] = 1$ and $f(1.7)=\left [ 1.7 \right ] = 1$

but $1.5\neq 1.7$

$\therefore$ f is not one- one i.e. not injective.

For $0.6 \in R$ there is no x in R such that $f(x)=\left [ 0.6 \right ]$

$\therefore$ f is not onto i.e. not surjective.

Hence, f is not injective but not surjective.

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.

$f : R \rightarrow R$

$f (x) = | x |$

$f (x) = | x | = x \, if\, x\geq 0 \,\, and \, \, -x\, if\, x< 0$

One- one:

For $-1,1 \in R$ then $f (-1) = | -1 |= 1$

$f (1) = | 1 |= 1$

$-1\neq 1$

$\therefore$ f is not one- one i.e. not injective.

For $-2 \in R$ ,

We know $f (x) = | x |$ is always positive there is no x in R such that $f (x) = | x |=-2$

$\therefore$ f is not onto i.e. not surjective.

Hence, $f (x) = | x |$ , is neither one-one nor onto.

Question:5 Show that the Signum Function $f : R \rightarrow R$ , given by

$f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right.$ is neither one-one nor onto.

$f : R \rightarrow R$ is given by

$f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right.$

As we can see $f(1)=f(2)=1$ , but $1\neq 2$

So it is not one-one.

Now, f(x) takes only 3 values (1,0,-1) for the element -3 in codomain $R$ ,there does not exists x in domain $R$ such that $f(x)= -3$ .

So it is not onto.

Hence, signum function is neither one-one nor 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.

$A = \{1, 2, 3\}$

$B = \{4, 5, 6, 7\}$

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

$f : A \rightarrow B$

$\therefore$ $f(1)=4,f(2)=5,f(3)=6$

Every element of A has a distant value in f.

Hence, it is one-one.

Question:7(i) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) $f: R\rightarrow R$ defined by $f(x) = 3 -4x$

$f: R\rightarrow R$

$f(x) = 3 -4x$

Let there be $(a,b) \in R$ such that $f(a)=f(b)$

$3-4a = 3 -4b$

$-4a = -4b$

$a = b$

$\therefore$ f is one-one.

Let there be $y \in R$ , $y = 3 -4x$

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

$f(x) = 3 -4x$

Puting value of x, $f(\frac{3-y}{4}) = 3 - 4(\frac{3-y}{4})$

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

$\therefore$ f is onto.

f is both one-one and onto hence, f is bijective.

Question:7(ii) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(ii) $f : R\rightarrow R$ defined by $f(x) = 1 + x^2$

$f : R\rightarrow R$

$f(x) = 1 + x^2$

Let there be $(a,b) \in R$ such that $f(a)=f(b)$

$1+a^{2} = 1 +b^{2}$

$a^{2}=b^{2}$

$a = \pm b$

For $f(1)=f(-1)=2$ and $1\neq -1$

$\therefore$ f is not one-one.

Let there be $-2 \in R$ (-2 in codomain of R)

$f(x) = 1 + x^2 = -2$

There does not exists any x in domain R such that $f(x) = -2$

$\therefore$ f is not onto.

Hence, f is neither one-one nor onto.

Question:8 Let A and B be sets. Show that $f : A \times B \rightarrow B \times A$ such that $f (a, b) = (b, a)$ is
bijective function.

$f : A \times B \rightarrow B \times A$

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

Let $(a_1,b_1),(a_2,b_2) \in A\times B$

such that $f (a_1, b_1) = f(a_2, b_2)$

$(b_1,a_1)=(b_2,a_2)$

$\Rightarrow$ $b_1= b_2$ and $a_1= a_2$

$\Rightarrow$ $(a_1,b_1) = (a_2,b_2)$

$\therefore$ f is one- one

Let, $(b,a) \in B\times A$

then there exists $(a,b) \in A\times B$ such that $f (a, b) = (b, a)$

$\therefore$ f is onto.

Hence, it is bijective.

Question:9 Let $f : N \rightarrow N$ be defined by $f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;even \end{matrix}\right.$ for all $n\in N$ . State whether the function f is bijective. Justify your answer.

$f : N \rightarrow N$ , $n\in N$

$f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;evem \end{matrix}\right.$

Here we can observe,

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

As we can see $f(1)=f(2)=1$ but $1\neq 2$

$\therefore$ f is not one-one.

Let, $n\in N$ (N=co-domain)

case1 n be even

For $r \in N$ , $n=2r$

then there is $4r \in N$ such that $f(4r)=\frac{4r}{2}=2r$

case2 n be odd

For $r \in N$ , $n=2r+1$

then there is $4r+1 \in N$ such that $f(4r+1)=\frac{4r+1+1}{2}=2r +1$

$\therefore$ f is onto.

f is not one-one but onto

hence, the function f is not bijective.

Question:10 Let $A = R - \{3\}$ and $B = R - \{1\}$ . Consider the function $f : A\rightarrow B$ defined by $f(x) = \left (\frac{x-2}{x-3} \right )$ . Is f one-one and onto? Justify your answer.

$A = R - \{3\}$

$B = R - \{1\}$

$f : A\rightarrow B$

$f(x) = \left (\frac{x-2}{x-3} \right )$

Let $a,b \in A$ such that $f(a)=f(b)$

$\left (\frac{a-2}{a-3} \right ) = \left ( \frac{b-2}{b-3} \right )$

$(a-2)(b-3)=(b-2)(a-3)$

$ab-3a-2b+6=ab-2a-3b+6$

$-3a-2b=-2a-3b$

$3a+2b= 2a+3b$

$3a-2a= 3b-2b$

$a=b$

$\therefore$ f is one-one.

Let, $b \in B = R - \{1\}$ then $b\neq 1$

$a \in A$ such that $f(a)=b$

$\left (\frac{a-2}{a-3} \right ) =b$

$(a-2)=(a-3)b$

$a-2 = ab -3b$

$a-ab = 2 -3b$

$a(1-b) = 2 -3b$

$a= \frac{2-3b}{1-b}\, \, \, \, \in A$

For any $b \in B$ there exists $a= \frac{2-3b}{1-b}\, \, \, \, \in A$ such that

$f(\frac{2-3b}{1-b}) = \frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}$

$=\frac{2-3b-2+2b}{2-3b-3+3b}$

$=\frac{-3b+2b}{2-3}$

$= b$

$\therefore$ f is onto

Hence, the function is one-one and onto.

Question:11 Let $f : R \rightarrow R$ 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.

$f : R \rightarrow R$

$f(x) = x^4$

One- one:

For $a,b \in R$ then $f(a) = f(b)$

$a^{4}=b^{4}$

$a=\pm b$

$\therefore f(a)=f(b)$ does not imply that $a=b$

example: and $2\neq -2$

$\therefore$ f is not one- one

For $2\in R$ there is no x in R such that $f(x)=x^{4}= 2$

$\therefore$ f is not onto.

Hence, f is neither one-one nor onto.

Option D is correct.

Question:12 Let $f : R\rightarrow R$ 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.

$f : R\rightarrow R$

$f(x) = 3x$

One - One :

Let $\left ( x,y \right ) \in R$

$f(x)=f(y)$

$3x=3y$

$x=y$

$\therefore$ f is one-one.

Onto:

We have $y \in R$ , then there exists $x=\frac{y}{3} \in R$ such that

$f(\frac{y}{3})= 3\times \frac{y}{3} = y$

$\therefore f is \, \, onto$ .

Hence, the function is one-one and onto.

The correct answer is A .

More About NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2

The NCERT Class 12 Maths chapter Relations and Functions has a total of 5 exercises including miscellaneous. Exercise 1.2 Class 12 Maths covers solutions to 12 main questions and their sub-questions. Most of the questions are related to proving a function one to one. Hence NCERT Solutions for Class 12 Maths chapter 1 exercise 1.2 can be referred for learning the concepts related to proof etc.

Also Read| NCERT Notes For Class 12 Mathematics Chapter 1

Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2

• The Class 12th Maths chapter 1 exercise provided here is solved by subject matter experts having rich experience in the domain of competitive exam preparation.

• Students can practice Exercise 1.2 Class 12 Maths to prepare various concepts like signus functions, one to one functions etc, many direct questions are asked in Board exams from this chapter.

• These Class 12 Maths chapter 1 exercise 1.2 solutions can be referred by students to revise before the exam and clarify any doubt regarding solution of exercise questions.

• NCERT Solutions for Class 12 Maths chapter 1 exercise 1.2 provided here are most recommended solutions for students aspiring to score well in examinations.

### Also see-

• NCERT exemplar solutions class 12 maths chapter 1

• NCERT solutions for class 12 maths chapter 1

NCERT Solutions Subject Wise

• NCERT solutions class 12 chemistry

• NCERT solutions for class 12 physics

• NCERT solutions for class 12 biology

• NCERT solutions for class 12 mathematics

Subject wise NCERT Exemplar solutions

• NCERT Exemplar Class 12th Maths
• NCERT Exemplar Class 12th Physics

• NCERT Exemplar Class 12th Chemistry

• NCERT Exemplar Class 12th Biology

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