Find the value of k, for which f(x) = 1 + kx - 1 - kx x , if - 1 < 0 2x + 1 x - 1 , if0 < 1 is continuous at x = 0.

* lim _ x 0^ - f(x) = * lim _ h 0 f(0 - h) [ Let x = 0 - h,x 0^ - 0] = * lim _ h 0 f(-h) = * lim _ h 0 1 + k(-h) - 1 - k(-h) -h = * lim _ h 0 1 - kh - 1 + kh -h 1 - kh + 1 + kh 1 - kh + 1 + kh = * lim _ h 0 (1 - kh) - (1 + kh) -h 1 - kh + 1 + kh = * lim _ h 0 2k 1 - kh + 1 + kh = 2k 2 * lim _ x 0^ - f(x) = k - - - - - - - (i) Again * lim _ x 0^ + f(x) = * lim _ h 0 f(0 + h) [ Let x = 0 h,x 0^ + 0] * lim _ h 0 f( - h) = * lim _ h 0 2h + 1 h - 1 = 1 -1 * lim _ x 0^ + f(x) = - 1 - - - - - - - - - (ii) Also f(0) = 2 0 + 1 0 - 1 = - 1 f is continuous at x = 0 * lim _ x 0^ - f(x) = * lim _ x 0^ + f(x) = f(0) = - 1.

Find the value of k, for which f(x) = 1 + kx - 1 - kx x , if - 1 …

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Question

Find the value of k, for which $\text{f}(\text{x}) = $ $ \begin{matrix} \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}< 1 \end{matrix} $ is continuous at x = 0.

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Answer

$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(0 - \text{h}) [ \text{ Let x} = 0 - \text{h},\text{x}\rightarrow0^{-}\Rightarrow\text{h}\rightarrow0]$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(-\text{h}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\sqrt{1 + \text{k}(-\text{h})} - \sqrt{1 - \text{k}(-\text{h})}}{-\text{h}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{\sqrt{1 - \text{kh}} - \sqrt{1 + \text{kh}}}{-\text{h}}\times\frac{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}}{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}}$
$ = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{(1 - \text{kh)} - (1 + \text{kh)}}{-\text{h}\left\{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}\right\}} = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\frac{2\text{k}}{\left\{\sqrt{1 - \text{kh}} + \sqrt{1 + \text{kh}}\right\}}$
$ =\frac{2\text{k}}{2}$
$ \Rightarrow \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\text{f}(\text{x}) = \text{k}$ - - - - - - - (i)
Again $ \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{+}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}(0 + \text{h}) [ \text{Let x } = 0 \text{ h},\text{x}\rightarrow0^{+}\Rightarrow\text{h}\rightarrow0]$
$ \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{f}( - \text{h}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0} \frac{2\text{h} + 1 }{\text{h} - 1 } = \frac{1}{-1}$
$ \Rightarrow\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{+}}\text{f}(\text{x}) = - 1 $ - - - - - - - - - (ii)
Also $\text{f}(0) = \frac{2\times0 + 1 }{0 - 1 } = - 1 $
$\because\text{ f is continuous at x } = 0$
$ \therefore\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{-}}\text{f}(\text{x}) = \DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0^{+}}\text{f}(\text{x}) = \text{f}(0)\Rightarrow\text{k} = - 1.$