Relation and Function - NCERT Exemplar Problems and Solution (Short Answer Type)

Short Answer Type NCERT Exemplar Problems and Solution

Question – 1 Let$ A = (a, b, c)$ and the relation $R$  be defined on $A$  as follows:
$R = ((a, a), (b, c), (a, b)).$
Then, write minimum number of ordered pairs to be added in $R $ to make $ R$ reflexive and transitive.

Solution:

In order to make R reflexive,$ (b, b) $ and $(c, c) $ will be added to $R$

And in order to make $ R$  transitive,$ (a, c)$ will be added to $R.$

Therefore, The minimum number of order pair to be added to $R$ will be $(b, b), (c, c) $ and $ (a, c) $ ♦

Question – 2  Let $D$ be the domain of real-valued function f defined by $f(x)=\sqrt{25-x^2} $ then, write $D.$

Solution:

Here given D is the domain of $f(x)=\sqrt{25-x^2} $

Therefore, $ 25-x^2 \geq 0$

$\Rightarrow 25\geq x^2$

$ \Rightarrow -5\leq x\leq 5$

Therefore, $D = [- 5, 5]$ ♦

Question - 3   Let $ f,g: R\rightarrow R$  be defined by  $f(x)=2x+1$ and $g(x)=x^2-2, \hspace{5pt} \forall x\in R,$  respectively. Then find $g o f.$

Solution: Given $ f(x)=2x+1 $ and $g(x)=x^2-2,$  $\forall x \in R$

Therefore, $ gof(x)=g(f(x))$

$=g(2x+1)$

$={(2x+1)}^2 -2 $ $= 4x^2+4x-1$ ♦

Question -4  Let $ f,g:R\rightarrow R$ be the function defined by $f(x)=2x-3$ $\forall x \in R $. Write $f^{-1}$

Solution: 

Given, $y=2x-3$ $\Rightarrow 2x=y-3$ $\Rightarrow x=\frac{y+3}{2}$ Therefore $f^{-1}(x)=\frac{x+3}{2}$ ♦

Question – 5 If $A = {a, b, c, d}$ and the function $f = {(a, b), (b, d), (c, a), (d, c)}$ write $f^{-1}$

Solution:

Given, $f = {(a, b), (b, d), (c, a), (d, c)}$

Therefore, $f^{-1}={(b, a), (d, b), (c, a), (c, d)}$ ♦

 Question – 6  If $ f:R\rightarrow R$ is defined by $f(x)=x^2-3x+2$ write $f(f(x))$

Solution:

Given, $ f(x)=x^2-3x+2$ therefore $f(f(x))=f(x^2-3x+2)$

$ = {(x^2-3x+2)}^2 -3(x^2-3x+2)+2$

$=x^4 -6x^3+10x^2-3x$  ♦

Question – 7   Is $g = ((1, 1), (2, 3, (3, 5), (4, 7))$ a function? If $g$ is described by $g(x) = \alpha x + \beta$, then what value should be assigned to $\alpha$ and $\beta$?

Solution:

Given, $g = ((1, 1), (2, 3, (3, 5), (4, 7))$

Therefore, each of the element of domain will be have unique image.

Consequently, g is a function. Since $ g(x)=\alpha x+\beta$ therefore

$g(1)=\alpha \times 1 +\beta$ thus $1=\alpha+\beta$ Also $g(2)=\alpha \times 2+\beta $ therefore solving these two equations we get $\alpha=2 $ and $\beta =-1$ ♦

Question-8: Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.

(i) {$(x,y)$:$ x$ is a person and $y$ is mother of$ x$}

(ii) {$(a,b)$: $a$ is a person , $b$ is an acestor of $a$}.

Solution:

(i) Let's check whether the relation defined in (i) is a function or not. For this, check whether it maps every element of domain uniquely or not. Which is easy to verify since $x$ cannot have two mothers. So if $(x,y)\in R$ then its not possible for any $z\in codomain $ such that $(x,z)\in R$ So clearly it's a function. Next to check, R is injective or not. For this Let mother $y$ has two children $x_1, x_2$ then for $x_1\neq x_2$ $f(x_1)=f(x_2)$ that is mother is same. so $R$ is not one-one.(Injective)♦

(ii) Similarly, check whether it is a function or not so look here it is possible that one element is mapped to 2 or more elements. For this let's take an example  $a$ is ancestor of $b$ and $b$ is ancestor of $c$.

$(a,b)\in R$ clearly also $ c$ is ancestor of both $a$ and $b$.So $(a,c)\in R$. So it's not a Function.♦

Question -9: If the mapping $f$ and $g$ are given by

$f$={$(1,2),(3,5)(4,1)$} and $g$={$(2,3),(5,1),(1,3)$}, Write $fog$.

Domain of $fog$ is domain $g$. So $fog(2)=f(g(2))=f(3))=5$

$fog(5)=f(g(5))=f(1)=2$ $fog(1)=f(g(1))=f(3))=5$

so $fog$={$(2,5),(5,2),(1,5)$}♦

Question-10: Let $C$ be the set of complex numbers. Prove that the mapping $f : C\rightarrow R $ given by
$f (z) = |z|, \forall z \in C$, is neither one-one nor onto.

Solution:  For $-1, i\in$ C, clearly $-1\neq i$  $|-1|=|i|$ So it's not one-one.  Now there exsist no preimage of negative real numbers. Since, there exist no complex number whose modulus is negative. So it's not Onto. ♦
Question-11: Let the function $f : R \rightarrow R$ be defined by $f (x) = cosx, \forall x \in R.$ Show that$ f$ is
neither one-one nor onto.

Solution:  For $0\neq 2\pi$ $cos0=cos2\pi$ so it's not one-one. Also Preimage of $2$ does not exist. We know that Range of $cos(x)$ is $[-1,1]$. Alternatively, since Range$\neq$ codomain. Therefore $cos(x)$ is not onto.
Question-12: Let $X$ = {$1, 2, 3$}and $Y$ = {$4, 5$}. Find whether the following subsets of $X \times Y$ are
functions from $X$ to$ Y$ or not.
(i) $f$ = {$(1, 4), (1, 5), (2, 4), (3, 5)$} (ii) $g $= {$(1, 4), (2, 4), (3, 4)$}
(iii)$ h$ = {$(1,4), (2, 5), (3, 5)$} (iv) $k$ = {$(1,4), (2, 5)$}.

Solution: (i) is not Function. Since, $1$ is mapped to $4$ and $5$. So image of $1$ is not unique.

(ii) It is a Function, Since every element of Domain Mapped Uniquely.

(iii) Image of $1,2,3 $are unique.So it is a function.

(iv) Similarly, It is also a function.
Question-13: If functions $f : A \rightarrow B$ and $g : B \rightarrow A$ satisfy $g o f = I$ $A$, then show that $f$ is one-one
and $g$ is onto.

Solution:  Given $gof=I$ Let $ f(x_1)=f(x_2)$$\Rightarrow g(f(x_1))=g(f(x_2))$ [Since $g$ is a function, $g$ is Well-defined] $\Rightarrow x_1=x_2$ [Since $gof(x)=x$]

Next , To prove onto, Let there exist $x\in A$  Now $g(f(x)) =x$ For $f(x)\in B$ So,$ g$ is Onto.

" This question is important for Board Examination"

Question-14: Let $f : R \rightarrow R$ be the function defined by $ f (x) =\frac{1}{2-cos(x)}$ $x\in R$.Then, find
the range of$ f$.

Solution: $-1\leq cos(X) \leq 1$ $\Rightarrow 1\geq -cos(x) \geq -1 $ , $\Rightarrow 3\geq 2-cos(x) \geq 1 $ So $\frac{1}{3} \leq \frac{1}{1-2cos(x)} \leq 1$. So Range= $[1/3,1]$. ♦
Question-15: Let $n$ be a fixed positive integer. Define a relation $R$ in $Z$ as follows: $a, b\in Z$,
$aRb$ if and only if $ a - b$ is divisible by $n$ . Show that $R$ is an equivalence relation.

Solution:

(i) Since $a-a=0$ is divisible by $n$ For every $a\in Z$ Therefore $(a,a)\in R$ for every $a\in Z$. So $R$ is Reflexive.

(ii) Let $(a,b)\in R$ that is $a-b$ is divisible by $n$ Then $-(a-b)$ is also divisible by $n$. i.e., $b-a$ is divisible by $n$. So $(b,a)\in R$. So, $R$ is Symmetric.

(iii) Let $(a,b), (b,c)\in R$ then  $a-b=r\times n$ .....(1)

$a-b=r\times n$                                       .....(1)

and   $b-c=r\times m$                              ....(2)

for some $n,m\in Z$. So , Substracting (2) from (1) we get, $a-c= r \times(n-m) $. and $n-m \in Z$ [Clousure law of addition of Z]

Therefore $a-c$ is divisible by $n$ i.e. $(a,c) \in R$ Therefore $R$ is transitive.

So. $R$ is an Equivalence Relation.  ♦