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Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsFunctions5 Marks
Question
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions:
$\frac{\text{g}}{\text{f}}$
✓

Answer

We have,
$\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$
We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$
So, domain $\text{f}=[-1,\infty]$
Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for
$9-\text{x}^2\geq0$
$\Rightarrow\text{x}^2-9\leq0$
$\Rightarrow\text{x}^2-3^2\leq0$
$\Rightarrow\text{x}\in[-3,3]$
$\therefore\ \text{domain(g)}=[-3,3]$
Now,
$\text{domain(f)}\cap\text{domain(g)}$
$=[-1,\infty]\cap[-3,3]$
$=[-1,3]$
We have,
$\text{f(x)}=\sqrt{\text{x}+1}$
$\therefore\ \sqrt{\text{x}+1}=0$
$\Rightarrow\text{x}+1=0$
$\Rightarrow\text{x}=-1$
So, domain $\Big(\frac{\text{g}}{\text{f}}\Big)=[-1,3]-\{-1\}=[-1,3]$
$\therefore\ \frac{\text{f}}{\text{g}}:[-1,3]\rightarrow\text{R}$ is given by $\frac{\text{g}}{\text{f}}(\text{x})=\frac{\text{g(x)}}{\text{f(x)}}=\frac{\sqrt{9-\text{x}^2}}{\sqrt{\text{x}+1}}$

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