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

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

Question:1 Show that the function $f:R_{\ast }⟶R_{\ast }$ 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, $f:R_{\ast }⟶R_{\ast }$ 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 $y\in R_{\ast }$ , then there exists $x=\frac{1}{y}\in R_{\ast }$ ( Here $y\ne 0$ ) such that

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

$\therefore fisonto$ .

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.} $g:N⟶R_{\ast }$ 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 $1.5\in R_{\ast }$ ,

$g(x)=\frac{1}{1.5}$ 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) $f:N\to N$ given by $f(x)=x^2$

Answer:

$f:N\to N$

$f(x)=x^2$

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

Answer:

$f:Z\to Z$

$f(x)=x^2$

One- one:

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

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

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

$\therefore$ 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) $f:R\to R$ given by $f(x)=x^2$

Answer:

$f:R\to R$

$f(x)=x^2$

One- one:

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

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

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

$\therefore$ 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) $f:N\to N$ given by $f(x)=x^3$

Answer:

$f:N\to N$

$f(x)=x^3$

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

Answer:

$f:Z\to Z$

$f(x)=x^3$

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 $f:R⟶R$ , given by $f(x)=[x]$ , is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$ .

Answer:

$f:R⟶R$

$f(x)=[x]$

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 $1.5\ne 1.7$

$\therefore$ 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:

$f:R\to R$

$f(x)=|x|$

$f(x)=|x|=xifx\ge 0and-xifx<0$

One- one:

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

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

$-1\ne 1$

$\therefore$ 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 $f:R\to R$ , given by

$f(x)=\{\begin{array}{cc}1& ifx>0\\ 0& ifx=0\\ -1& ifx<0\end{array}$ is neither one-one nor onto.

Answer:

$f:R\to R$ is given by

$f(x)=\{\begin{array}{cc}1& ifx>0\\ 0& ifx=0\\ -1& ifx<0\end{array}$

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

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:

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

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

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

$f:A\to B$

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

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

Answer:

$f:R\to R$

$f(x)=3-4x$

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

Answer:

$f:R\to R$

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

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 $1\ne -1$

$\therefore$ 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 $f:A\times B\to B\times A$ such that $f(a,b)=(b,a)$ is
bijective function.

Answer:

$f:A\times B\to B\times A$

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

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 $f:N\to N$ be defined by $f(n)=\{\begin{array}{cc}\frac{n+1}{2}& ifnisodd\\ \frac{n}{2}& ifniseven\end{array}$ for all $n\in N$ . State whether the function f is bijective. Justify your answer.

Answer:

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

$f(n)=\{\begin{array}{cc}\frac{n+1}{2}& ifnisodd\\ \frac{n}{2}& ifnisevem\end{array}$

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 $1\ne 2$

$\therefore$ 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 $f:A\to B$ defined by $f(x)=\left(\frac{x-2}{x-3}\right)$ . Is f one-one and onto? Justify your answer.

Answer:

$A=R-\{3\}$

$B=R-\{1\}$

$f:A\to B$

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

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 $b\ne 1$

$a\in A$ 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 $f:R\to R$ 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:

$f:R\to R$

$f(x)=x^4$

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 $2\ne -2$

$\therefore$ 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 $f:R\to R$ 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:

$f:R\to R$

$f(x)=3x$

One - One :

Let $(x,y)\in R$

$f(x)=f(y)$

$3x=3y$

$x=y$

$\therefore$ f is one-one.

Onto:

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

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

$\therefore fisonto$ .

Hence, the function is one-one and onto.

The correct answer is A .