Exercise 1.3-Additional Problems - Chapter 1 Sets Relations and Functions 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Find the domain and range of the function $f(x)=\frac{1}{\sqrt{x-5}}$
Solution:
Given that : $\mathrm{f}(\mathrm{x}) f(x)=\frac{1}{\sqrt{x-5}}$
Here, it is clear that / (x) is real when $x-5>0 \Rightarrow x>5$
Hence, the domain $=(5, \infty)$
Now to find the range put
For $x \in(5, \infty), y \in \mathrm{R}^{+}$.
Hence, the range of $f=R^{+}$.
Question 2.
If $f(x)=\frac{x-1}{x+1}$, then show that

Solution:
Given that $f(x)=\frac{x-1}{x+1}$
(i) $f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{1-x}{1+x}=\frac{-(x-1)}{x+1}=-f(x)$
Hence $f\left(\frac{1}{x}\right)=-f(x)$
(ii) $f\left(\frac{-1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{-\left(\frac{1}{x}+1\right)}{-\left(\frac{1}{x}-1\right)}=\frac{1+x}{1-x}=\frac{1}{\frac{1-x}{1+x}}=\frac{1}{\left(\frac{x-1}{x+1}\right)}=\frac{-1}{f(x)}$
Hence, $f\left(\frac{-1}{x}\right)=\frac{-1}{f(x)}$
Question 3.
Find the domain of each of the following functions given by: $f(x)=\frac{x^3-x+3}{x^2-1}$
Solution:
Here, $\mathrm{f}(\mathrm{x})$ is not defined if $\mathrm{x}^2-1 \neq 0$
$
\begin{aligned}
& (x-1)(x+1) \neq 0 \\
& x \neq 1, x \neq-1
\end{aligned}
$
Hence, the domain of $f=R-\{-1,1\}$
Question 4.
Find the range of the following functions given by $f(x)=1+3 \cos 2 \mathrm{x}$
Solution:
Given that: $f(x)=1+3 \cos 2 x$
We know that $-1 \leq \cos 2 \mathrm{x} \leq 1$
$
\begin{aligned}
& \Rightarrow-3 \leq 3 \cos 2 \mathrm{x} \leq 3 \Rightarrow-3+1 \leq 1+3 \cos 2 \mathrm{x} \leq 3+1 \\
& \Rightarrow-2 \leq 1+3 \cos 2 \mathrm{x} \leq 4 \Rightarrow-2 \leq \mathrm{f}(\mathrm{x}) \leq 4
\end{aligned}
$
Hence the range of $f=[-2,4]$
Question 5.
Find the domain and range of the function $f(x)=\frac{x^2-9}{x-3}$
Solution:

Domain off: Clearly $f(x)$ is not defined for $x-3=0$ i.e. $x=3$.
Therefore, Domain $(\mathrm{f})=\mathrm{R}-\{3\}$
Range off: Let $f(x)=y$. Then,
It follows from the above relation that $y$ takes all real values except 6 when $x$ takes values in the set $R$ $\{3\}$. Therefore, Range $(f)=R\{6\}$.
Question 6.
Find the range of the following functions given by $f(x)=\frac{1}{2-\sin 3 x}$
Solution:
We have $f(x)=\frac{1}{2-\sin 3 x}$
$
\begin{array}{ll} 
& -1 \leq \sin 3 x \leq 1 \text { for all } x \in \mathrm{R} \\
\Rightarrow & -1 \leq-\sin 3 x \leq 1 \text { for all } x \in \mathrm{R} \\
\Rightarrow & 1 \leq 2-\sin 3 x \leq 3 \text { for all } x \in \mathrm{R} \\
\Rightarrow & 2-\sin 3 x \neq 0 \text { for all } x \in \mathrm{R} \\
\Rightarrow & f(x)=\frac{1}{2-\sin 3 x} \text { is defined for all } x \in \mathrm{R}
\end{array}
$
Hence, domain $(f)=\mathrm{R}$.
Range of $f$ : As discused above
$
\begin{aligned}
& 1 \leq 2-\sin 3 x \leq 3 \text { for all } x \in \mathrm{R} \\
\Rightarrow & \frac{1}{3} \leq \frac{1}{2-\sin 3 x}-\sin 3 x \neq 0 \text { for all } x \in \mathrm{R} \\
\Rightarrow & \frac{1}{3} \leq f(x) \leq 1 \quad \text { for all } x \in \mathrm{R} \\
\Rightarrow & f(x) \in \mathrm{R}[1 / 3,1]
\end{aligned}
$
Hence, range $(f)=[1 / 3,1]$