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Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSFunctions4 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: $\frac{\text{f}}{\text{g}}$
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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, $\text{g(x)}=\sqrt{9-\text{x}^2}$ $\therefore\ 9-\text{x}^2=0$ $\Rightarrow\text{x}^2-9=0$ $\Rightarrow(\text{x}-3)(\text{x}+3)=0$ $\Rightarrow\text{x}=\pm3$ So, domain $\Big(\frac{\text{f}}{\text{g}}\Big)=[-1,3]-[-3,3]=[-1,3]$ $\therefore\ \frac{\text{f}}{\text{g}}:[-1,3]\rightarrow\text{R}$ is given by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\text{f(x)}}{\text{g(x)}}=\frac{\sqrt{\text{x}+1}}{\sqrt{9-\text{x}^2}}$

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