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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&\text{if }0\leq\text{ x}<1\\\text{a},&\text{if }1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\text{if }\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous on $(0,\infty),$ then find the most suitable value of a and b.
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Answer

Given, f is continuous on $(0,\infty)$
$\therefore$ f is continuous at x = 1 and $\sqrt{2}$
Ar x = 1, we have
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{(1-\text{h})^2}{\text{a}}\bigg]=\frac{1}{\text{a}}$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
Also,
At $\text{x}=\sqrt{2},$ we have
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{2\text{b}^2-4\text{b}}{(\sqrt{2}+\text{h})^2}\bigg]=\frac{2\text{b}^2-4\text{b}}{2}=\text{b}^2-2\text{b}$
f is continuous at x = 1 and $\sqrt{2}$
$\therefore\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}$
$\Rightarrow\frac{1}{\text{a}}=\text{a}$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}^2=1$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}=\pm1$ and $\text{b}^2-2\text{b}=\text{a}\ ...(\text{i})$
If a = 1, then
$\text{b}^2-2\text{b}=\text{a}$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}-1=0$
$\Rightarrow\text{b}=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm2\sqrt{2}}{2}\\=1\pm\sqrt{2}$
If a = -1, then
$\Rightarrow\text{b}^2-2\text{b}=-1$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}+1=0$
$\Rightarrow(\text{b}-1)^2=0$
$\Rightarrow\text{b}=1$
Hence, the most suitable value of a and b are
$\text{a}=-1,\text{ b}=1$ or $\text{a}=1,\text{ b}=1\pm\sqrt{2}$

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