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

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

Answer:

The function $f : R \rightarrow \{x \in R : - 1 < x < 1\}$ 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.

$x-y\neq 2xy$

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

Answer:

$f : R \rightarrow R$

$f (x) = x ^3$

One-one:

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

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

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

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

• It will contradict given condition of cubes being equal.

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

Question:3 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 $A =\left \{ 0,1 \right \}$ and $B =\left \{ 0,1,2 \right \}$

then we cannot say that B is related to A.

$\therefore$ R is not symmetric.

If $ARB \, \, \, and \, \, \, BRC, \, \, then \, \, A\subset B \, \, \, and \, \, B\subset C$

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

$\therefore$ R is transitive.

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

Question:4 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:5 Let $A = \{- 1, 0, 1, 2\}$ , $B = \{- 4, - 2, 0, 2\}$ and $f, g : A \rightarrow B$ be functions defined by $f(x) = x^2 -x, x\in A$ and $g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A$ . Are $f$ and $g$ equal? Justify your answer. (Hint: One may note that two functions $f : A \rightarrow B$ and $g : A \rightarrow B$ such that $f (a) = g (a) \;\forall a \in A$ , are called equal functions).

Answer:

Given :

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

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

It can be observed that

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

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

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

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

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

$f(0)=g(0)$

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

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

$f(1)=g(1)$

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

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

$f(2)=g(2)$

$\therefore \, \, \, f(a)=g(a) \forall a\in A$

Hence, f and g are equal functions.

Question:6 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

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

$(1,1),(2,2),(3,3) \in R$ , 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:7 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

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

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.