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

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 ∗ ?

Answer:

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 $f(x) = x^2$

Answer:

$f : N\rightarrow N$

$f(x) = x^2$

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 $f(x) = x^2$

Answer:

$f : Z \rightarrow Z$

$f(x) = x^2$

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 $f(x) = x^2$

Answer:

$f: R \rightarrow R$

$f(x) = x^2$

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 $f(x) = x^3$

Answer:

$f : N\rightarrow N$

$f(x) = x^3$

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 $f(x) = x^3$

Answer:

$f : Z \rightarrow Z$

$f(x) = x^3$

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 $x$ .

Answer:

$f : R\longrightarrow R$

$f (x) = [x]$

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.

Answer:

$f : R \rightarrow R$

$f (x) = | x |$

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

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.

Answer:

$f : R \rightarrow R$ 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.

Answer:

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

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

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

$f : A \rightarrow B$

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

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 $f(x) = 3 -4x$

Answer:

$f: R\rightarrow R$

$f(x) = 3 -4x$

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

Answer:

$f : R\rightarrow R$

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

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.

Answer:

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

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

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.

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.

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.

Answer:

$f : R \rightarrow R$

$f(x) = x^4$

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.

Answer:

$f : R\rightarrow R$

$f(x) = 3x$

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 .