Exercise 1.2 (Revised) - Chapter 1 - Relations And Functions - Ncert Solutions class 12 - Maths

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NCERT Solutions Class 12 Maths -  Exercise 1.2 (Revised Syllabus)

Ex 1.2 Question 1:

Show that the function $f . \mathbf{R}_* \rightarrow \mathbf{R}_*$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $\mathbf{R}_*$ is the set of all non-zero real numbers. Is the result true, if the domain $R_*$ is replaced by $\mathbf{N}$ with co-domain being same as $R_*$ ?

Answer:
It is given that $f: \mathbf{R}_* \rightarrow \mathbf{R}_*$ is defined by $f(x)=\frac{1}{x}$.
One-one:
$
\begin{aligned}
& f(x)=f(y) \\
& \Rightarrow \frac{1}{x}=\frac{1}{y} \\
& \Rightarrow x=y
\end{aligned}
$
$\therefore f$ is one-one.
Onto:

$
x=\frac{1}{y} \in \mathrm{R} .(\text { Exists as } y \neq 0)
$

It is clear that for $y \in \mathbf{R}_*$, there exists
such that
$
f(x)=\frac{1}{\left(\frac{1}{y}\right)}=y \text {. }
$
$\therefore f$ is onto.

Thus, the given function (f) is one-one and onto.
Now, consider function $g: \mathbf{N} \rightarrow \mathbf{R}$ defined by
$
g(x)=\frac{1}{x}
$

We have,
$
g\left(x_1\right)=g\left(x_2\right) \Rightarrow \frac{1}{x_1}=\frac{1}{x_2} \Rightarrow x_1=x_2
$
$\therefore g$ is one-one.
Further, it is clear that $g$ is not onto as for $1.2 \in \mathbf{R}_*$ there does not exit any $x$ in $\mathbf{N}$ such that $g(x)=\frac{1}{1.2}$ Hence, function $g$ is one-one but not onto.

Ex 1.2 Question 2:

Check the injectivity and surjectivity of the following functions:
(i) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^2$
(ii) $f: Z \rightarrow Z$ given by $f(x)=x^2$
(iii) $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^2$
(iv) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^3$
(v) $f: Z \rightarrow Z$ given by $f(x)=x^3$

Answer:
(i) $f: \mathbf{N} \rightarrow \mathbf{N}$ is given by,
$
f(x)=x^2
$

It is seen that for $x, y \in \mathbf{N}, f(x)=f(y) \Rightarrow x^2=y^2 \Rightarrow x=y$.
$\therefore f$ is injective.
Now, $2 \in \mathbf{N}$. But, there does not exist any $x$ in $\mathbf{N}$ such that $f(x)=x^2=2$.
$\therefore f$ is not surjective.
Hence, function $f$ is injective but not surjective.
(ii) $f: Z \rightarrow Z$ is given by,
$
f(x)=x^2
$

It is seen that $f(-1)=f(1)=1$, but $-1 \neq 1$.
$\therefore f$ is not injective.
Now, $-2 \in Z$. But, there does not exist any element $x \in Z$ such that $f(x)=x^2=-2$.
$\therefore f$ is not surjective.

Hence, function $f$ is neither injective nor surjective.
(iii) $f \cdot \mathbf{R} \rightarrow \mathbf{R}$ is given by,
$
f(x)=x^2
$

It is seen that $f(-1)=f(1)=1$, but $-1 \neq 1$.
$\therefore f$ is not injective.
Now, $-2 \in \mathbf{R}$. But, there does not exist any element $x \in \mathbf{R}$ such that $f(x)=x^2=-2$.
$\therefore f$ is not surjective.
Hence, function $f$ is neither injective nor surjective.
(iv) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by,
$
f(x)=x^3
$

It is seen that for $x, y \in \mathbf{N}, f(x)=f(y) \Rightarrow x^3=y^3 \Rightarrow x=y$.
$\therefore$ fis injective.
Now, $2 \in \mathbf{N}$. But, there does not exist any element $x$ in domain $\mathbf{N}$ such that $f(x)=x^3=2$.

$\therefore f$ is not surjective
Hence, function $f$ is injective but not surjective.
(v) $f: Z \rightarrow Z$ is given by,
$
f(x)=x^3
$

It is seen that for $x, y \in \mathbf{Z}, f(x)=f(y) \Rightarrow x^3=y^3 \Rightarrow x=y$.
$\therefore f$ is injective.
Now, $2 \in Z$. But, there does not exist any element $x$ in domain $Z$ such that $f(x)=x^3=2$.
$\therefore f$ is not surjective.
Hence, function $f$ is injective but not surjective.

Ex 1.2 Question 3:

Prove that the Greatest Integer Function $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=[x]$, is neither one-once nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$.

Answer:
$f: R \rightarrow R$ is given by,
$
f(x)=[x]
$

It is seen that $f(1.2)=[1.2]=1, f(1.9)=[1.9]=1$.
$\therefore f(1.2)=f(1.9)$, but $1.2 \neq 1.9$.
$\therefore f$ is not one-one.
Now, consider $0.7 \in \mathbf{R}$.
It is known that $f(x)=[x]$ is always an integer. Thus, there does not exist any element $x \in \mathbf{R}$ such that $f(x)=0.7$.
$\therefore f$ is not onto.
Hence, the greatest integer function is neither one-one nor onto.

Question 4:

Show that the Modulus Function $f: R \rightarrow R$, given by $f(x)=|x|$, is neither oneone nor onto, where is $x$, if $x$ is positive or 0 and $|x|$ is -1, if $x$ is negative.

Answer:

 Modulus Function $f: R \rightarrow R$, given by $f(x)=|x|$

Now
$
|x|= \begin{cases}-x, & \text { if } x<0 \\ 0, & \text { if } x=0 \\ x, & \text { if } x \geq 0\end{cases}
$
$\Rightarrow f$ contains $(-1,1),(1,1),(-2,2),(2,2)$
Thus negative integers are not images of any element. $\quad \therefore f$ is not one-one.
Also second set $\mathrm{R}$ contains some negative numbers which are not images of any real number.
$\therefore f$ is not onto.

Ex 1.2 Question 5:

Show that the Signum Function $f: R \rightarrow \mathbb{R}$, given by $f(x)=\left\{\begin{array}{l}1, \text { if } x>0 \\ 0, \text { if } x=0 \\ -1, \text { if } x<0\end{array}\right.$ is neither one-one nor onto.

Answer.

Signum Function $f: \mathrm{R} \rightarrow \mathrm{R}$, given by $f(x)=\left\{\begin{array}{l}1, \text { if } x>0 \\ 0, \text { if } x=0 \\ -1, \text { if } x<0\end{array}\right.$
$
\begin{aligned}
& f(1)=f(2)=1 \\
& \Rightarrow f\left(x_1\right)=f\left(x_2\right)=1 \text { for } n>0 \\
& \Rightarrow x_1 \neq x_2 \quad \therefore f \text { is not one-one. }
\end{aligned}
$

Except $-1,0,1$ no other members of co-domain of $f$ has any pre-image its domain.
$\therefore f$ is not onto.
Therefore, $f$ is neither one-one nor onto.
Ex 1.2 Question 6.

Let $\mathbf{A}=\{1,2,3\}, \mathbf{B}=\{\mathbf{4}, 5,6,7\}$ and let $f=\{(1,4),(2,5),(3,6)\}$ be a function from $\mathrm{A}$ to
B. Show that $f$ is one-one.

Answer.

$\mathrm{A}=\{1,2,3\}, \mathrm{B}=\{4,5,6,7\}$ and $f=\{(1,4),(2,5),(3,6)\}$
Here, $f(1)=4, f(2)=5$ and $f(3)=6$
Here, also distinct elements of A have distinct images in B.
Therefore, $f$ is a one-one function.

Ex 1.2 Question 7.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) $f: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=3-4 x$
(ii) $f: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=1+x^2$

Answer.

(i) $f: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=3-4 x$
Now, if $x_1, x_2 \in \mathrm{R}$, then $f\left(x_1\right)=3-4 x_1$ and $f\left(x_2\right)=3-4 x_2$
And if $f\left(x_1\right)=f\left(x_2\right)$, then $x_1=x_2 \quad \therefore f$ is one-one.
Again, if every element of $\mathrm{Y}(-\mathrm{R})$ is image of some element of $\mathrm{X}(\mathrm{R})$ under $f$, i.e., for every $y \in \mathrm{Y}$, there exists an element $x$ in $\mathrm{X}$ such that $f(x)=y$.

Now $y=3-4 x$
$
\begin{aligned}
& \Rightarrow x=\frac{3-y}{4} \\
& \therefore f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right) \\
& \Rightarrow f(x)=3-3+y=y
\end{aligned}
$
$\therefore f$ is onto or bijective function.

(ii) $f: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=1+x^2$

Now, if $x_1, x_2 \in R$, then $f\left(x_1\right)=1+x_1^2$ and $f\left(x_2\right)=1+x_2^2$
And if $f\left(x_1\right)=f\left(x_2\right)$, then $x_1^2=x_2^2$
$\Rightarrow x_1= \pm x_2 \therefore f$ is not one-one.
Again, if every element of $\mathrm{Y}(-\mathrm{R})$ is image of some element of $\mathrm{X}(\mathrm{R})$ under $f$, i.e., for every $y \in \mathrm{Y}$, there exists an element $x$ in $\mathrm{X}$ such that $f(x)=y$.

Now, $y=1+x^2 \Rightarrow x= \pm \sqrt{y-1}$
$\therefore f(\sqrt{y-1})=1+y-1=-y \neq y$
$\therefore f$ is not onto.
Therefore, $f$ is not bijective.

Ex 1.2 Question 8.

Let $\mathbf{A}$ and B be sets. Show that $f: \mathbf{A} \times \mathbf{B} \rightarrow \mathbf{B} \times$ A such that $f(a, b)=(b, a)$ is a bijective function.

Answer.

Injectivity: Let $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right) \in$ A x B such that $f\left(a_1, b_1\right)=f\left(a_2, b_2\right)$
$
\begin{aligned}
& \Rightarrow\left(b_1, a_1\right)=\left(b_2, a_2\right) \\
& \Rightarrow b_1=b_2 \text { and } a_1=a_2 \\
& \Rightarrow\left(a_1, b_1\right)=\left(a_2, b_2\right) \\
& \Rightarrow\left(a_1, b_1\right)=\left(a_2, b_2\right) \text { for all }\left(a_1, b_1\right),\left(a_2, b_2\right) \in \text { Ax B }
\end{aligned}
$

So, $f$ is injective.
Surjectivity: Let $(b, a)$ be an arbitrary element of $\mathrm{B} \times \mathrm{A}$. Then $b \in \mathrm{B}$ and $a \in \mathrm{A}$.

$
\Rightarrow(a, b) \in \mathrm{AxB}
$

Thus, for all $(b, a) \in \mathrm{B} \times \mathrm{A}$, their exists $(a, b) \in \mathrm{A} \times \mathrm{B}$ such that $f(a, b)=(b, a)$
So, $f: \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{B} \times \mathrm{A}$ is an onto function, therefore $f$ is bijective.

Ex 1.2 Question 9.

Let $f: \mathrm{N} \rightarrow \mathrm{N}$ be defined by $f(n)=\left\{\begin{array}{l}\frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { if } n \text { is even }\end{array}\right.$ for all $n \in \mathrm{N}$.

State whether the function $f$ is bijective. Justify your answer.
Answer.

$f: \mathrm{N} \rightarrow \mathrm{N}$ be defined by $f(n)=\left\{\begin{array}{l}\frac{n+1}{2}, \text { if } n \text { is odd } \\ \frac{n}{2}, \text { if } n \text { is even }\end{array}\right.$
(a) $f(1)=\frac{n+1}{2}=\frac{1+1}{2}=\frac{2}{2}=1$ and $f(2)=\frac{n}{2}=\frac{2}{2}=1$

The elements 1,2 , belonging to domain of $f$ have the same image 1 in its co-domain.
So, $f$ is not one-one, therefore, $f$ is not injective.
(b) Every number of co-domain has pre-image in its domain e.g., 1 has two pre-images 1 and 2.

So, $f$ is onto, therefore, $f$ is not bijective.

Ex 1.2 Question 10.

Let $\mathbf{A}=\mathbf{R}-\{3\}$ and $\mathbf{B}=\mathbf{R}-\{1\}$. Consider the function $f: \mathbf{A} \rightarrow \mathbf{B}$ defined by $f(x)=\left(\frac{x-2}{x-3}\right)$. Is $f$ one-one and onto? Justify your answer.

Answer.

$\mathrm{A}=\mathrm{R}-\{3\}$ and $\mathrm{B}=\mathrm{R}-\{1\}$ and $f(x)=\frac{x-2}{x-3}$

Let $x_1, x_2 \in$ A, then $f\left(x_1\right)=\frac{x_1-2}{x_1-3}$ and $f\left(x_2\right)=\frac{x_2-2}{x_2-3}$
Now, for $f\left(x_1\right)=f\left(x_2\right)$
$
\begin{aligned}
& \Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \\
& \Rightarrow\left(x_1-2\right)\left(x_2-3\right)=\left(x_2-2\right)\left(x_1-3\right) \\
& \Rightarrow x_1 x_2-3 x_1-2 x_2+6=x_1 x_2-2 x_1-3 x_2+6 \\
& \Rightarrow-3 x_1-2 x_2=-2 x_1-3 x_2
\end{aligned}
$
$\Rightarrow x_1=x_2 \therefore f$ is one-one function.
Now $y=\frac{x-2}{x-3}$
$
\begin{aligned}
& \Rightarrow y(x-3)=x-2 \\
& \Rightarrow x y-3 y=x-2 \\
& \Rightarrow x(y-1)=3 y-2 \\
& \Rightarrow x=\frac{3 y-2}{y-1}
\end{aligned}
$
$
\begin{aligned}
& \therefore f\left(\frac{3 y-2}{y-1}\right)=\frac{\frac{3 y-2}{y-1}-2}{\frac{3 y-2}{y-1}-3}=\frac{3 y-2-2 y+2}{3 y-2-3 y+3}=y \\
& \Rightarrow f(x)=y
\end{aligned}
$

$
\Rightarrow f(x)=y
$

Therefore, $f$ is an onto function.

Ex 1.2 Question 11.

Let $f: \mathrm{R} \rightarrow \mathrm{R}$ 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.

$f(x)=x^4$ and $\underline{\mathrm{R}} \rightarrow \underline{\mathrm{R}}$
Let $x_1, x_2 \in \underline{\mathrm{R}}$, then $f\left(x_1\right)=x_1^4$ and $f\left(x_2\right)=x_2^4$
$
\begin{aligned}
& \therefore x_1^4=x_2^4 \\
& \Rightarrow \pm x_1= \pm x_2
\end{aligned}
$

Therefore, $f$ is not one-one function.
Now, $y=x^4$
$
\begin{aligned}
& \Rightarrow x= \pm y^{\frac{1}{4}} \\
& \therefore f\left(y^{\frac{1}{4}}\right)=y^{\frac{1}{4}}=y \text { and } f\left(-y^{\frac{1}{4}}\right)=-y^{\frac{1}{4}}=y
\end{aligned}
$

Therefore, $f$ is not onto function.
Therefore, option (D) is correct.

Ex 1.2 Question 12.

Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $f(x)=3 x$. 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.

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

Therefore, $f$ is one-one function.
Now, consider $y \in \mathrm{R}$ (co-domain of $f$ ) certainly $x=\frac{y}{3} \in \mathrm{R}$ (domain of $f$ )
Thus for all $y \in \mathrm{R}$ (co-domain of $f$ ) there exists $x=\frac{y}{3} \in \mathrm{R}$ (domain of $f$ ) such that
$
f(x)=f\left(\frac{y}{3}\right)=3 \cdot \frac{y}{3}=y
$

Therefore, $f$ is onto function.
Therefore, option (A) is correct.