Q108 / 180 · KCET 2018 · Mathematics
If
\[
f(x)=\left\{\begin{array}{cl}
\frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}} ; & \text { if }-1 \leq \mathrm{x} \leq 0 \\
\frac{2 \mathrm{x}+1}{\mathrm{x}-1} ; & \text { if } 0 \leq \mathrm{x} \leq 1
\end{array}\right.
\]
is continuous at \( x=0 \), then the value of \( k \) is
\[ f(x)=\left\{\begin{array}{cl} \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}} ; & \text { if }-1 \leq \mathrm{x} \leq 0 \\ \frac{2 \mathrm{x}+1}{\mathrm{x}-1} ; & \text { if } 0 \leq \mathrm{x} \leq 1 \end{array}\right. \]
is continuous at \( x=0 \), then the value of \( k \) is
- A\( k=1 \)
- Correct answerB\( k=-1 \)
- C\( k=0 \)
- D\( k=2 \)
Solution
Given that
\[ f(x)=\mid \begin{array}{cc} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & ;-1 \leq x < 0 \\ \frac{2 x+1}{x-1} & ; 0 \leq x \leq 1 \end{array} \]
For continues at \( x=0 \), we have
\[ \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \]
\[ \begin{array}{l} \text { So, } \lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \\ =\lim _{x \rightarrow 0^{-}} \frac{2 k(\sqrt{1-k x}+\sqrt{1+k x})}{x(\sqrt{x})}=k \\ \text { and } \lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-1}=-1=k \\ \text { So, } k=-1 \end{array} \]
\[ f(x)=\mid \begin{array}{cc} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & ;-1 \leq x < 0 \\ \frac{2 x+1}{x-1} & ; 0 \leq x \leq 1 \end{array} \]
For continues at \( x=0 \), we have
\[ \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \]
\[ \begin{array}{l} \text { So, } \lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \\ =\lim _{x \rightarrow 0^{-}} \frac{2 k(\sqrt{1-k x}+\sqrt{1+k x})}{x(\sqrt{x})}=k \\ \text { and } \lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-1}=-1=k \\ \text { So, } k=-1 \end{array} \]
Hence, the correct answer is option: 'B'.