Miscellaneous Exercise (Revised) - Chapter 2 - Relations & Functions - Ncert Solutions class 11 - Maths

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Chapter 2: Relations & Functions - NCERT Solutions for Class 11 Maths

Miscellaneous Exercise Question 1.

The relation $f$ is defined by $f(x)=\left\{\begin{array}{ll}x^2 & 0 \leq x \leq 3 \\ 3 x & 3 \leq x \leq 10\end{array}\right.$. The relation $g$ is defined by $g(x)=\left\{\begin{array}{ll}x^2 & 0 \leq x \leq 2 \\ 3 x & 2 \leq x \leq 10\end{array}\right.$. Show that $f$ is a function and $g$ is not a function. Ans. Given: $f(x)=x^2 \quad 0 \leq x \leq 3$ and
$
f(x)=3 x \quad 3 \leq x \leq 10
$

At $x=3, f(3)=(3)^2=9$ and
$
f(3)=3 \times 3=9
$

It is observed that $f(x)$ takes unique value at each point in its domain $[0,10]$. Therefore, $f$ is a function.

Now, $g(x)=x^2 \quad 0 \leq x \leq 2$ and

$
g(x)=3 x \quad 2 \leq x \leq 10
$

At $x=2, g(2)=(2)^2=4$ and
$
g(2)=3 \times 2=6
$

Therefore, $g(x)$ does not have unique value at $x=2$.
Hence, $g(x)$ is not a function.

Miscellaneous Exercise Question 2.

If $f(x)=x^2$, find $\frac{f(1.1)-f(1)}{(1.1-1)}$.

Answer.

Given: $f(x)=x^2$
At $x=1.1 \quad f(1.1)=(1.1)^2=1.21$
and $f(1)=(1)^2=1$
$
\therefore \frac{f(1.1)-f(1)}{(1.1-1)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1
$
Miscellaneous Exercise Question 3.

Find the domain of the function $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.

Answer.

Given: $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$
$f(x)$ is a rational function of $x$
$f(x)$ assumes real values of all $x$ except for those values of $x$ for which
$
x^2-8 x+12=0
$

$
\begin{aligned}
& \Rightarrow(x-6)(x-2)=0 \\
& \Rightarrow x=2,6
\end{aligned}
$
$\therefore$ Domain of function $=\mathrm{R}-\{2,6\}$
Miscellaneous Exercise Question 4.

Find the domain and range of the real function $f$ defined by $f(x)=\sqrt{x-1}$.

Answer.

Given: $f(x)=\sqrt{x-1}, f(x)$ assumes real values if $x-1 \geq 0 \Rightarrow x \geq 1$
$
\Rightarrow x \in[1, \infty)
$

$\therefore$ Domain of $f(x)=[1, \infty)$
For $x \geq 1, f(x) \geq 0$
$\therefore$ Range of $f(x)=$ all real numbers $\geq 0=[0, \infty)$
Miscellaneous Exercise Question 5.

Find the domain and range of the real function $f$ defined by $f(x)=|x-1|$.

Answer.

Given: $f(x)=|x-1|$
The function $f(x)$ is defined for all values of $x$
$\therefore$ Domain of $f(x)=\mathrm{R}$
When $x>1,|x-1|=x-1>0$
When $x=1,|x-1|=0$
When $x<1,|x-1|=-x+1>0$
Miscellaneous Exercise Question 6.

Let $f=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}$ be a function from R into R. Determine the range of $f$.

Answer.

Here $f(x)=\frac{x^2}{1+x^2}$

Putting $y=\frac{x^2}{1+x^2}$
$
\begin{aligned}
& \Rightarrow y+y x^2=x^2 \\
& \Rightarrow x^2(1-y)=y \\
& \Rightarrow x^2=\frac{y}{1-y} \\
& \Rightarrow x= \pm \sqrt{\frac{y}{1-y}}
\end{aligned}
$

Now, $x$ will be real if $\frac{y}{1-y} \geq 0$

$
\begin{aligned}
& \Rightarrow \frac{y}{y-1} \leq 0 \\
& \Rightarrow-0 \leq y<1 \\
& \Rightarrow-y \in[0,1) \\
& \therefore \text { Range } f(x)=[0,1)
\end{aligned}
$
Miscellaneous Exercise Question 7.

Let $f: g: \mathrm{R} \rightarrow \mathrm{R}$ be defined respectively by $f(x)=x+1, g(x)=2 x-3$. Find $f+g \cdot f-g$ and $\frac{f}{g}$.

Answer.

Given : $f(x)=x+1$ and $g(x)=2 x-3$
Now, $(f+g)(x)=f(x)+g(x)=x+1+2 x-3=3 x-2$
And $(f-g)(x)=f(x)-g(x)=x+1-2 x+3=-x+4$
And $\frac{(f)}{(g)}(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2 x-3}, x \neq \frac{3}{2}$

Miscellaneous Exercise Question 8.

Let $f=\{(1,1),(2,3),(0,-1),(-1,-3)\}$ be a function from $\mathrm{Z}$ to $\mathrm{Z}$ defined by $f(x)=a x+b$ for some integers $a, b$. Determine $a, b$.

Answer.

Given: $f(x)=a x+b$ and
$
\begin{aligned}
& f=\{(1,1),(2,3),(0,-1),(1,-3)\} \\
& \Rightarrow f(1)=1, f(2)=3, f(0)=-1, f(-1)=-3
\end{aligned}
$

Now $f(1)=1 \Rightarrow a \times 1+b=1$
$
\Rightarrow a+b=1
$

And $f(2)=3 \Rightarrow a \times 2+b=3$
$
\Rightarrow 2 a+b=3
$

Solving eq. (i) and (ii), we get $a=2$ and $b=-1$
Miscellaneous Exercise Question 9.

Let $\mathrm{R}$ be a relation from $\mathrm{N}$ to $\mathrm{N}$ defined by $\mathrm{R}=$ $\left\{(a, b): a, b \in \mathrm{N}\right.$ and $\left.a=b^2\right\}$. Are the following true:

(i) $(a, a) \in \mathbf{R}$ for all $a \in \mathbf{N}$
(ii) $(a, b) \in \mathbf{R}$ implies $(b, a) \in \mathbf{R}$
(iii) $(a, b) \in \mathbf{R},(b, c) \in \mathbf{R}$ implies $(a, c) \in \mathbf{R}$

Answer.

Given: $\mathrm{R}=\left\{(a, b): a, b \in \mathrm{N}\right.$ and $\left.a=b^2\right\}$
(i) No, $(3,3) \notin \mathrm{R}$ because $3 \neq 3^2$
(ii)No, $(9,3) \in R$ but $(3,9) \notin R$
(iii)No, $(81,9) \in \mathrm{R}$ and $(9,3) \in R$ but $(81,3) \notin \mathrm{R}$
Miscellaneous Exercise Question 10.

Let $\mathbf{A}=\{1,2,3,4\}, \mathrm{B}=\{1,59,11,15,16\}$ and $f=\{(1,5),(2,9),(3$, $1),(4,5),(2,11)\}$.-Are the following true:
(i) $f$ is a relation from $\mathbf{A}$ to $\mathbf{B}$.
(ii) $f$ is a function from $\mathbf{A}$ to $\mathbf{B}$.

Justify your answer in each case.
Answer.

(i) Here $A=\{1,2,3,4\}$ and $B=\{1,5,9,11,15,16\}$
$
\begin{aligned}
& \therefore-A \times B=\{(1,1),(1,5),(1,9),(1,11),(1,15),(1,16),(2,1),(2,5),(2,9),(2, \\
& 11), \\
& (2,15),(2,16),(3,1),(3,5),(3,9),(3,11),(3,15),(3,16),(4,1),(4,5), \\
& (4,9),(4,11),(4,15),(4,16)\} \\
& f=\{(1,5),(2,9),(3,1),(4,5),(2,11)\}
\end{aligned}
$

Now, $(1,5),(2,9),(3,1),(4,5),(2,11) \in A \times B$
$\therefore-f$ is a relation from $\mathrm{A}$ to $\mathrm{B}$.
(ii) $\mathrm{f}$ is not a function because $(2,9) \in f$ and $(2,11) \in f$
Miscellaneous Exercise Question 11.

Let $f$ be a subset of $Z \times Z$ defined by $f=\{(a b, a+b): a, b \in Z\}$. Is $f$ a function from $\mathrm{Z}$ to $\mathrm{Z}$ ? Justify your answer.
Answer.

-We observed that $1 \times 4=4$ and $2 \times 2=4$
$
\begin{aligned}
& \Rightarrow(1 \times 4,1+4) \in f \text { and }(2 \times 2,2+2) \in f \\
& \Rightarrow(4,5) \in f \text { and }(4,4) \in f
\end{aligned}
$

It shows that $f$ is not a function from $\mathrm{Z}$ to $\mathrm{Z}$.
Miscellaneous Exercise Question 12.

Let $\mathrm{A}=\{9,10,11,1213\}$ and let $f: \mathrm{A} \rightarrow \mathrm{N}$ be defined by $f(n)=$ the highest prime factor of $n$. Find the range of $f$.

Answer.

Here $A=\{9,10,11,12,13\}$
For $n=9, f(9)=3$
[ $\because 9=3 \times 3$ and 3 is highest prime factor of 9 ]

For $n=10, f(10)=5$
[ $\because 10=2 \times 5$ and 5 is highest prime factor of 10]
For $n=11, f(11)=11$
[ $\because 11=1 \times 11$ and 11 is highest prime factor of 11]
For $n=12, f(12)=3$
[ $\because 12=3 \times 3 \times 2$ and 3 is highest prime factor of 12]
For $n=13, f(13)=13$
[ $\because 13=1 \times 13$ and 13 is highest prime factor of 13 ]
$\therefore$ Range of $f=\{5,11,3,13\}$
$=\{3,5,11,13\}$