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

NCERT solutions for Class 12 Maths Chapter 1 Relations and Functions: Miscellaneous Exercise

Question:1 Let $f:R\to R$ be defined as $f(x)=10x+7$ . Find the function $g:R\to R$ such that $gof=fog=I_R$ .

Answer:

$f:R\to R$

$f(x)=10x+7$

$g:R\to R$ and $gof=fog=I_R$

For one-one:

$f(x)=f(y)$

$10x+7=10y+7$

$10x=10y$

$x=y$

Thus, f is one-one.

For onto:

For $y\in R$ , $y=10x+7$

$x=\frac{y-7}{10}\in R$

Thus, for $y\in R$ , there exists $x=\frac{y-7}{10}\in R$ such that

$f(x)=f(\frac{y-7}{10})=10(\frac{y-7}{10})+7=y-7+7=y$

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let $g:R\to R$ as $f(y)=\frac{y-7}{10}$

$gof(x)=g(f(x))=g(10x+7)=\frac{(10x+7)-7}{10}=\frac{10x}{10}=x$

$fog(x)=f(g(x))=f(\frac{y-7}{10})=10\frac{y-7}{10}+7=y-7+7=y$

$\therefore$ $gof(x)=I_R$ and $fog(x)=I_R$

Hence, $g:R\to R$ defined as $g(y)=\frac{y-7}{10}$

Question:2 Let $f:W\to W$ be defined as $f(n)=n-1$ , if n is odd and $f(n)=n+1$ , if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer:

$f:W\to W$

$f(n)=n-1$ if n is odd

$f(n)=n+1$ if n is even.

For one-one:

Taking x as odd number and y as even number.

$f(x)=f(y)$

$x-1=y+1$

$x-y=2$

Now, Taking y as odd number and x as even number.

$f(x)=f(y)$

$x+1=y-1$

$x-y=-2$

This is also impossible.

If both x and y are odd :

$f(x)=f(y)$

$x-1=y-1$

$x=y$

If both x and y are even :

$f(x)=f(y)$

$x+1=y+1$

$x=y$

$\therefore$ f is one-one.

Onto:

Any odd number 2r+1 in codomain of N is an image of 2r in domain N and any even number 2r in codomain N is the image of 2r+1 in domain N.

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Sice, f is invertible.

Let $g:W\to W$ as $m+1$ if m is even and $m-1$ if m is odd.

When x is odd.

$gof(x)=g(f(x))=g(n-1)=n-1+1=n$

When x is even

$gof(x)=g(f(x))=g(n+1)=n+1-1=n$

Similarly, m is odd

$fog(x)=f(g(x))=f(m-1)=m-1+1=m$

m is even ,

$fog(x)=f(g(x))=f(m+1)=m-1+1=m$

$\therefore$ $gof(x)=I_W$ and $fog(x)=I_W$

Hence, f is invertible and the inverse of f is g i.e. $f^{-1}=g$ , which is the same as f.

Hence, inverse of f is f itself.

Question:3 If f : R → R is defined by f(x) = x ² – 3x + 2, find f (f (x)).

Answer:

This can be solved as following

f : R → R

$f(x)=x^2-3x+2$

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

$=(x^4+9x^2+4-6x^3-12x+4x^2)-3x^2+9x-6+2$

$=x^4-6x^3+10x^2-3x$

Question:4 Show that the function $f:R\to \{x\in R:-1<x<1\}$ defined by $f(x)=\frac{x}{1+|x|}$ $x\in R$ is one one and onto function.

Answer:

The function $f:R\to \{x\in R:-1<x<1\}$ defined by

$f(x)=\frac{x}{1+|x|}$ , $x\in R$

One- one:

Let $f(x)=f(y)$ , $x,y\in R$

$\frac{x}{1+\left|x\right|}=\frac{y}{1+\left|y\right|}$

It is observed that if x is positive and y is negative.

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

Since x is positive and y is negative.

$x>y\Rightarrow x-y>0$ but 2xy is negative.

$x-y\ne 2xy$

Thus, the case of x is positive and y is negative is removed.

Same happens in the case of y is positive and x is negative so this case is also removed.

When x and y both are positive:

$f(x)=f(y)$

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

$x(1+y)=y(1+x)$

$x+xy=y+xy$

$x=y$

When x and y both are negative : $f(x)=f(y)$

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

$x(1-y)=y(1-x)$

$x-xy=y-xy$

$x=y$

$\therefore$ f is one-one.

Onto:

Let $y\in R$ such that $-1<y<1$

If y is negative, then $x=\frac{y}{y+1}\in R$

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

If y is positive, then $x=\frac{y}{1-y}\in R$

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

Thus, f is onto.

Hence, f is one-one and onto.

Question:5 Show that the function $f:R\to R$ given by $f(x)=x^3$ is injective.

Answer:

$f:R\to R$

$f(x)=x^3$

One-one:

Let $f(x)=f(y)x,y\in R$

$x^3=y^3$

We need to prove $x=y$ .So,

• Let $x\ne y$ then there cubes will not be equal i.e. $x^3\ne y^3$ .

• It will contradict given condition of cubes being equal.

• Hence, $x=y$ and it is one -one which means it is injective

Question:6 Give examples of two functions $f:N\to Z$ and $g:Z\to Z$ such that $gof$ is injective but g is not injective. (Hint : Consider $f(x)=x$ and $g(x)=|x|$ ).

Answer:

$f:N\to Z$

$g:Z\to Z$

$f(x)=x$

$g(x)=|x|$

One - one:

Since $g(x)=|x|$

$f(1)=\left|1\right|=1$

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

As we can see $f(1)=f(-1)$ but $1\ne -1$ so $g(x)$ is not one-one.

Thus , g(x) is not injective.

$gof:N\to Z$

$gof(x)=g(f(x))=g(x)=$ $\left|x\right|$

Let $gof(x)=gof(y)x,y\in N$

$\left|x\right|=\left|y\right|$

Since, $x,y\in N$ so x and y are both positive.

$\therefore x=y$

Hence, gof is injective.

Question:7 Give examples of two functions $f:N\to N$ and $g:N\to N$ such that $gof$ is onto but $f$ is not onto.

(Hint : Consider $f(x)=x+1$ and $g(x)=\{\begin{array}{cc}x-1& ifx>1\\ 1& ifx=1\end{array}$

Answer:

$f:N\to N$ and $g:N\to N$

$f(x)=x+1$ and $g(x)=\{\begin{array}{cc}x-1& ifx>1\\ 1& ifx=1\end{array}$

Onto :

$f(x)=x+1$

Consider element in codomain N . It is clear that this element is not an image of any of element in domain N .

$\therefore$ f is not onto.

$gof:N\to N$

$gof(x)=g(f(x))=g(x+1)=x+1-1=xsincex\in N\Rightarrow x+1>1$

Now, it is clear that $y\in N$ , there exists $x=y\in N$ such that $gof(x)=y$ .

Hence, $gof$ is onto.

Question:8 Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if $A\subset B$ . Is R an equivalence relation on P(X)? Justify your answer.

Answer:

Given a non empty set X, consider P(X) which is the set of all subsets of X.

Since, every set is subset of itself , ARA for all $A\in P(x)$

$\therefore$ R is reflexive.

Let $ARB\Rightarrow A\subset B$

This is not same as $B\subset A$

If $A=\{0,1\}$ and $B=\{0,1,2\}$

then we cannot say that B is related to A.

$\therefore$ R is not symmetric.

If $ARBandBRC,thenA\subset BandB\subset C$

this implies $A\subset C$ $=ARC$

$\therefore$ R is transitive.

Thus, R is not an equivalence relation because it is not symmetric.

Question:9 Given a non-empty set X, consider the binary operation $\ast :P(X)\times P(X)\to P(X)$ given by $A\ast B=A\cap B\mathrm{\forall }A,B\in P(X)$ , where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Answer:

Given $\ast :P(X)\times P(X)\to P(X)$ is defined as $A\ast B=A\cap B\mathrm{\forall }A,B\in P(X)$ .

As we know that $A\cap X=A=X\cap A\mathrm{\forall }A\in P(X)$

$\Rightarrow$ $A\ast X=A=X\ast A\mathrm{\forall }AinP(X)$

Hence, X is the identity element of binary operation *.

Now, an element $A\in P(X)$ is invertible if there exists a $B\in P(X)$ ,

such that $A\ast B=X=B\ast A$ (X is identity element)

i.e. $A\cap B=X=B\cap A$

This is possible only if $A=X=B$ .

Hence, X is only invertible element in $P(X)$ with respect to operation *

Question:10 Find the number of all onto functions from the set $\{1,2,3,...,n\}$ to itself.

Answer:

The number of all onto functions from the set $\{1,2,3,...,n\}$ to itself is permutations on n symbols 1,2,3,4,5...............n.

Hence, permutations on n symbols 1,2,3,4,5...............n = n

Thus, total number of all onto maps from the set $\{1,2,3,...,n\}$ to itself is same as permutations on n symbols 1,2,3,4,5...............n which is n.

Question:11(i) Let $S=\{a,b,c\}$ and $T=\{1,2,3\}$ . Find $F^{-1}$ of the following functions F from S to T, if it exists.

(i) $F=\{(a,3),(b,2),(c,1)\}$

Answer:

$F:S\to T$

$F:\{a,b,c\}\to \{1,2,3\}$ is defined as $F=\{(a,3),(b,2),(c,1)\}$

$F(a)=3,F(b)=2,F(c)=1$

$\therefore F^{-1}:T\to S$ is given by

$F^{-1}=\{(3,a),(2,b),(1,c)\}$

Question:11(ii) Let $S=\{a,b,c\}$ and $T=\{1,2,3\}$ . Find $F^{-1}$ of the following functions F from S to T, if it exists.

(ii) $F=\{(a,2),(b,1),(c,1)\}$

Answer:

$F:S\to T$

$F:\{a,b,c\}\to \{1,2,3\}$ is defined as $F=\{(a,2),(b,1),(c,1)\}$

$F(a)=2,F(b)=1,F(c)=1$ , F is not one-one.

So inverse of F does not exists.

Hence, F is not invertible i.e. $F^{-1}$ does not exists.

Question:12 Consider the binary operations $\ast :R\times R\to R$ and $\circ :R\times R\to R$ defined as $a\ast b=|a-b|$ and $a\circ b=a\mathrm{\forall }a\in R$ . Show that ∗ is commutative but not associative, $\circ$ is associative but not commutative. Further, show that $\mathrm{\forall }a,b,c\in R$ , $a\ast (b\circ c)=(a\ast b)\circ (a\ast c)$ . [If it is so, we say that the operation ∗ distributes over the operation $\circ$ ]. Does $\circ$ distribute over ∗? Justify your answer.

Answer:

Given $\ast :R\times R\to R$ and $\circ :R\times R\to R$ is defined as
$a\ast b=|a-b|$ and $a\circ b=a\mathrm{\forall }a,b\in R$

For $a,b\in R$ , we have

$a\ast b=|a-b|$

$b\ast a=|b-a|=|-(a-b)|=|a-b|$

$\therefore a\ast b=b\ast a$

$\therefore$ the operation is commutative.

$(1\ast 2)\ast 3=(|1-2|)\ast 3=1\ast 3=|1-3|=2$

$1\ast (2\ast 3)=1\ast (|2-3|)=1\ast 1=|1-1|=0$

$\therefore (1\ast 2)\ast 3\ne 1\ast (2\ast 3)$ where $1,2,3\in R$

$\therefore$ the operation is not associative

Let $a,b,c\in R$ . Then we have :

$a\ast (b\circ c)=a\ast b=|a-b|$

$(a\ast b)\circ (a\ast c)=|a-b|\circ |a-c|=|a-b|$

Hence, $a\ast (b\circ c)=(a\ast b)\circ (a\ast c)$

Now,

$1\circ (2\ast 3)=1\circ (|2-3|)=1\circ 1=1$

$(1\circ 2)\ast (1\circ 3)=1\ast 1=|1-1|=0$

$\therefore 1\circ (2\ast 3)\ne (1\circ 2)\ast (1\circ 3)$ for $1,2,3\in R$

Hence, operation o does not distribute over operation *.

Question:13 Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

Answer:

It is given that

$\ast :P(X)\times P(X)\to P(X)$ be defined as

$A\ast B=(A-B)\cup (B-A),A,BϵP(X).$

Now, let $AϵP(X)$ .
Then,

$A\ast \varphi =(A-\varphi )\cup (\varphi -A)=A\cup \varphi =A$
And

$\varphi \ast A=(\varphi -A)\cup (A-\varphi )=\varphi \cup A=A$

Therefore,

$A\ast \varphi =\varphi \ast A=A,AϵP(X)$

Therefore, we can say that $\varphi$ is the identity element for the given operation *.

Now, an element A $ϵ$ P(X) will be invertible if there exists B $ϵ$ P(X) such that

$A\ast B=\varphi =B\ast A(\because \varphi isanidentityelement)$

Now, We can see that

$???????A\ast A=(A-A)\cup (A-A)=\varphi \cup \varphi =\varphi ,AϵP(X).$ $A\ast A=(A-A)\cup (A-A)=\varphi \cup \varphi =\varphi ,$ such that $AϵP(X)$

Therefore, by this we can say that all the element A of P(X) are invertible with $A^{-1}=A$

Question:14 Define a binary operation ∗ on the set $\{0,1,2,3,4,5\}$ as $a\ast b=\{\begin{array}{cc}a+b& ifa+b<6\\ a+b-6& ifa+b\ge 6\end{array}$ Show that zero is the identity for this operation and each element $a\ne 0$ of the set is invertible with $6-a$ being the inverse of $a$ .

Answer:

X = $\{0,1,2,3,4,5\}$ as

$a\ast b=\{\begin{array}{cc}a+b& ifa+b<6\\ a+b-6& ifa+b\ge 6\end{array}$

An element $c\in X$ is identity element for operation *, if $a\ast c=a=c\ast a\mathrm{\forall }a\in X$

For $a\in X$ ,

$a\ast 0=a+0=a[a\in X\Rightarrow a+0<6]$

$0\ast a=0+a=a[a\in X\Rightarrow a+0<6]$

$\therefore a\ast 0=a=0\ast a$ $\mathrm{\forall }a\in X$

Hence, 0 is identity element of operation *.

An element $a\in X$ is invertible if there exists $b\in X$ ,

such that $a\ast b=0=b\ast a$ i.e. $\{\begin{array}{cc}a+b=0=b+a& ifa+b<6\\ a+b-6=0=b+a-6& ifa+b\ge 6\end{array}$

means $a=-b$ or $b=6-a$

But since we have X = $\{0,1,2,3,4,5\}$ and $a,b\in X$ . Then $a\ne -b$ .

$\therefore b=a-x$ is inverse of a for $a\in X$ .

Hence, inverse of element $a\in X$ , $a\ne 0$ is 6-a i.e. , $a^{-1}=6-a$

Question:15 Let $A=\{-1,0,1,2\}$ , $B=\{-4,-2,0,2\}$ and $f,g:A\to B$ be functions defined by $f(x)=x^2-x,x\in A$ and $g(x)=2|x-\frac{1}{2}|-1,x\in A$ . Are $f$ and $g$ equal? Justify your answer. (Hint: One may note that two functions $f:A\to B$ and $g:A\to B$ such that $f(a)=g(a)\mathrm{\forall }a\in A$ , are called equal functions).

Answer:

Given :

$A=\{-1,0,1,2\}$ , $B=\{-4,-2,0,2\}$

$f,g:A\to B$ are defined by $f(x)=x^2-x,x\in A$ and $g(x)=2|x-\frac{1}{2}|-1,x\in A$ .

It can be observed that

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

$g(-1)=2|-1-\frac{1}{2}|-1=2\left|\frac{-3}{2}\right|-1=3-1=2$

$f(-1)=g(-1)$

$f(0)=(0{)}^2-(0)=0+0=0$

$g(0)=2|0-\frac{1}{2}|-1=2\left|\frac{-1}{2}\right|-1=1-1=0$

$f(0)=g(0)$

$f(1)=(1{)}^2-(1)=1-1=0$

$g(1)=2|1-\frac{1}{2}|-1=2\left|\frac{1}{2}\right|-1=1-1=0$

$f(1)=g(1)$

$f(2)=(2{)}^2-(2)=4-2=2$

$g(2)=2|2-\frac{1}{2}|-1=2\left|\frac{3}{2}\right|-1=3-1=2$

$f(2)=g(2)$

$\therefore f(a)=g(a)\mathrm{\forall }a\in A$

Hence, f and g are equal functions.

Question:16 Let $A=\{1,2,3\}$ . Then number of relations containing $(1,2)$ and $(1,3)$ which are reflexive and symmetric but not transitive is

(A) 1
(B) 2
(C) 3
(D) 4

Answer:

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

The smallest relations containing $(1,2)$ and $(1,3)$ which are
reflexive and symmetric but not transitive is given by

$R=\{(1,1),(2,2),(3,3),(1,2),(1,3),(2,1),(3,1)\}$

$(1,1),(2,2),(3,3)\in R$ , so relation R is reflexive.

$(1,2),(2,1)\in R$ and $(1,3),(3,1)\in R$ , so relation R is symmetric.

$(2,1),(1,3)\in R$ but $(2,3)\notin R$ , so realation R is not transitive.

Now, if we add any two pairs $(2,3)$ and $(3,2)$ to relation R, then relation R will become transitive.

Hence, the total number of the desired relation is one.

Thus, option A is correct.

Question:17 Let $A=\{1,2,3\}$ . Then number of equivalence relations containing $(1,2)$ is
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

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

The number of equivalence relations containing $(1,2)$ is given by

$R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$

We are left with four pairs $(2,3)$ , $(3,2)$ , $(1,3),(3,1)$ .

$(1,1),(2,2),(3,3)\in R$ , so relation R is reflexive.

$(1,2),(2,1)\in R$ and $(2,3),(3,2)\notin R$ , so relation R is not symmetric.

$(1,3),(3,1)\notin R$ , so realation R is not transitive.

Hence , equivalence relation is bigger than R is universal relation.

Thus the total number of equivalence relations cotaining $(1,2)$ is two.

Thus, option B is correct.

Question:18 Let $f:R\to R$ be the Signum Function defined as $f(x)=\{\begin{array}{cc}1& x>0\\ 0& x=0\\ -1& x<0\end{array}$ and $g:R\to R$ be the Greatest Integer Function given by $g(x)=[x]$ , where $[x]$ is greatest integer less than or equal to $x$ . Then, does $fog$ and $gof$ coincide in $(0,1]$ ?

Answer:

$f:R\to R$ is defined as $f(x)=\{\begin{array}{cc}1& x>0\\ 0& x=0\\ -1& x<0\end{array}$

$g:R\to R$ is defined as $g(x)=[x]$

Let $x\in (0,1]$

Then we have , $[1]=1$ if x=1 and $[x]=0if0<x<1$

$\therefore gof(x)=g(f(x))=g(1)=\left[1\right]=1(sincex>0)$

Hence,for $x\in (0,1]$ , $fog(x)=0$ and $gof(x)=1$ .

Hence , gof and fog do not coincide with $(0,1]$ .

Question:19 Number of binary operations on the set are(A) 10(B) 16(C) 20(D ) 8

Answer:

Binary operations on the set $\{a,b\}$ are is a function from $\{a,b\}\times \{a,b\}\to \{a,b\}$

i.e. * is a function from $\{(a,b),(b,a),(a,a),(b,b)\}\to \{a,b\}$

Hence, the total number of binary operations on set $\{a,b\}$ is $2^4=16.$

Hence, option B is correct