Miscellaneous Exercise (Revised) - Chapter 1 - Relations And Functions - Ncert Solutions class 12 - Maths

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NCERT Solutions Class 12 Maths: Relations and Functions Revised Syllabus

Miscellaneous Exercise Question 1:

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

Answer.

It is given that $\mathrm{f}: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1<x<1\}$ is defined as $\mathrm{f}(\mathrm{x})=\frac{x}{1+|x|}, x \in \mathbf{R}$
For one- one
Suppose $f(x)=f(y)$, where $x, y \in R$
$
\Rightarrow \frac{x}{1+|x|}=\frac{y}{1+|y|}
$

It can be observed that if $\mathrm{x}$ is positive and $\mathrm{y}$ is negative.
Then, we have
$
\frac{x}{1+x}=\frac{y}{1-y} \Rightarrow 2 x y=x-y
$

Since, $x$ is positive and $y$ is negative,
$\mathrm{x}>\mathrm{y} \Rightarrow x-y>0$, but, $2 \mathrm{xy}$ is negative
Then, $2 x y \neq x-y$.
Thus, the case of $x$ being positive and $y$ being negative can be ruled out.
Under a similar argument, $\mathrm{x}$ being negative and $\mathrm{y}$ being positive can also be ruled out.
$\therefore \mathrm{x}$ and $\mathrm{y}$ have to be either positive or negative.
When $x$ and $y$ are both positive,we have
$
\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y}) \Rightarrow \frac{x}{1+x}=\frac{y}{1+y} \Rightarrow x+x y=y+x y \Rightarrow x=y
$

When $x$ and $y$ are both negative,we have
$
\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y}) \Rightarrow \frac{x}{1-x}=\frac{y}{1-y} \Rightarrow x-x y=y-x y \Rightarrow x=y
$
$\therefore \mathrm{f}$ is one - one
For Onto

Now, let $\mathrm{y} \in \mathbf{R}$ such that $-1<\mathrm{y}<1$.
If $\mathrm{y}$ is negative then, there exists $\mathrm{x}=\frac{y}{1+y} \in \mathbf{R}$ such that
$
\mathrm{f}(\mathrm{x})=f\left(\frac{y}{1+y}\right)=\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}=\frac{y}{1+y-y}=y
$

If $\mathrm{y}$ is positive, then, there exists $\mathrm{x}=\frac{y}{1-y} \in \mathbf{R}$ such that
$
\mathrm{f}(\mathrm{x})=f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left|\frac{y}{1-y}\right|}=\frac{\frac{y}{1-y}}{1+\left(\frac{y}{1-y}\right)}=\frac{y}{1-y+y}=y
$

Therefore, $\mathrm{f}$ is onto. Hence, $\mathrm{f}$ is one -one and onto.

Miscellaneous Exercise 2.

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

Answer.

Let $x_1, x_2 \in \mathrm{R}$ be such that $f\left(x_1\right)=f\left(x_2\right)$
$
\begin{aligned}
& \Rightarrow x_1^3=x_2^3 \\
& \Rightarrow x_1=x_2
\end{aligned}
$

Therefore, $f$ is one-one function, hence $f(x)=x^3$ is injective.

Miscellaneous Exercise 3. Given a non empty set $X$, consider $P$ (X) which is the set of all subsets of $X$.

Define the relation $\mathrm{R}$ in $\mathrm{P}(\mathrm{X})$ as follows:
For subsets $A, B$ in $P(X), A R B$ if and only if $A \subset B$. Is $R$ an equivalence relation on $P(X)$ ? Justify your answer.

Answer.

(i) $\mathrm{A} \subset \mathrm{A} \therefore \mathrm{R}$ is reflexive.
(ii) $\mathrm{A} \subset \mathrm{B} \neq \mathrm{B} \subset \mathrm{A} \therefore \mathrm{R}$ is not commutative.
(iii) If $\mathrm{A} \subset \mathrm{B}, \mathrm{B} \subset \mathrm{C}$, then $\mathrm{A} \subset \mathrm{C} \therefore \mathrm{R}$ is transitive.

Therefore, $\mathrm{R}$ is not equivalent relation.

Miscellaneous Exercise 4.

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

Answer.

The number of onto functions that can be defined from a finite set A containing $n$ elements onto a finite set $\mathrm{B}$ containing $n$ elements $=2^n-n$.

Miscellaneous Exercise 5.

Let $\mathbf{A}=\{-1,0,1,2\}, \mathbf{B}=\{-4,-2,0,2\}$ and $f, g: \mathrm{A} \rightarrow \mathrm{B}$ be the 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 $f: \mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{g}: \mathrm{A} \rightarrow \mathrm{B}$ such that $f(a)=g(a) \forall a \in \mathbf{A}$, are called equal functions).

Answer.

When $x=-1$ then $f(x)=1^2+1=2$ and $g(x)=2\left|-1-\frac{1}{2}\right|-1=2$
At $x=0, f(0)=0$ and $g(0)=2\left|-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0$
At $x=1, f(1)=1^2-1=0$ and $g(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0$

At $x=2, f(2)=2^2-2=2$ and $g(2)=2\left|2-\frac{1}{2}\right|-1=3-1=2$
Thus for each $a \in$ A, $f(a)=g(a)$
Therefore, $f$ and $g$ are equal function.

Miscellaneous Exercise 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.

It is clear that 1 is reflexive and symmetric but not transitive.
Therefore, option (A) is correct.
Miscellaneous Exercise 7.

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

Answer.

2
Therefore, option (B) is correct.