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Question:
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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
Options:
\[
f(x)=\left\{\begin{array}{cl}
\frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}} ; & \text { if }-1 \leq \mathrm{x} \leq 0 \\
\frac{2 \mathrm{x}+1}{\mathrm{x}-1} ; & \text { if } 0 \leq \mathrm{x} \leq 1
\end{array}\right.
\]
is continuous at \( x=0 \), then the value of \( k \) is
Solution:
2930 Upvotes
Verified Answer
The correct answer is:
\( k=-1 \)
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
\[
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)
\]
\[
\begin{array}{l}
\text { So, } \lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \\
=\lim _{x \rightarrow 0^{-}} \frac{2 k(\sqrt{1-k x}+\sqrt{1+k x})}{x(\sqrt{x})}=k \\
\text { and } \lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-1}=-1=k \\
\text { So, } k=-1
\end{array}
\]
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