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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find the value of k in this question, so that the function f is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}},&\text{if}-1\leq0\\\frac{2\text{x}+1}{\text{x}-1},&\text{if }0\leq\text{x}\leq1\end{cases}$ at x = 0.
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Answer

We have, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}},&\text{if}-1\leq\text{x<0}\\\frac{2\text{x}+1}{\text{x}-1},&\text{if }0\leq\text{x}\leq1\end{cases}$ at x = 0
$\text{L.H.L}=\lim\limits_{\text{h}\rightarrow0^-}\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0^-}\bigg(\frac{\sqrt{1+\text{kx}}-\sqrt{1-\text{kx}}}{\text{x}}\bigg)\cdot\bigg(\frac{\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}{\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0^-}\frac{1+\text{kx}-1+\text{kx}}{\text{x}\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}$
$=\lim\limits_{\text{h}\rightarrow0^-}\frac{2\text{kx}}{\text{x}\sqrt{1+\text{kx}}+\sqrt{1-\text{kx}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{k}}{\sqrt{1+\text{k}(0-\text{h})}+\sqrt{1-\text{k}(0-\text{h})}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{k}}{\sqrt{1-\text{kh}}+\sqrt{1+\text{kh}}}$
$=\frac{2\text{k}}{2}=\text{k}$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow0^+}\frac{2\text{x}+1}{\text{x}-1}=\lim\limits_{\text{h}\rightarrow0}\frac{2(0+\text{h})+1}{(0+\text{h})-1}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}+1}{\text{h}-1}=-1$
Also $\text{f}(0)=\frac{2\times0+1}{0-1}=-1$
We must have L.H.L = R.H.L = f(0)
⇒ k = -1

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