Search any question & find its solution
Question:
Answered & Verified by Expert
If the function $f(x)$, defined below, is continuous everywhere, then $k$ equals $f(x)=\left\{\begin{array}{cc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x < \infty \\ k, & x=0\end{array}\right.$
Options:
Solution:
1433 Upvotes
Verified Answer
The correct answer is:
$\log _e 4$
Given,
$f(x)=\left\{\begin{array}{ccc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x < \infty \\ k, & x=0\end{array}\right.$ is continuous everywhere.
Since, $f(x)$ is continuous everywhere
$\Rightarrow \quad f(x)$ will be continuous at $x=0$
$\Rightarrow \quad \lim _{x \rightarrow 0} f(x)=f(0)$
$\lim _{x \rightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1}=k$
$\left[\frac{0}{0}\right.$ form, so by applying $L^{\prime}$ Hospital rule $]$
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{2^x \log 2}{\frac{1}{2 \sqrt{1+x}}}=k \\ & \lim _{x \rightarrow 0} 2^{x+1} \sqrt{1+x} \log 2=k\end{aligned}$
$\begin{array}{ll}\Rightarrow & 2 \sqrt{1+0} \log 2=k \Rightarrow k=2 \log _e 2 \\ \Rightarrow & k=\log _e 4\end{array}$
$f(x)=\left\{\begin{array}{ccc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x < \infty \\ k, & x=0\end{array}\right.$ is continuous everywhere.
Since, $f(x)$ is continuous everywhere
$\Rightarrow \quad f(x)$ will be continuous at $x=0$
$\Rightarrow \quad \lim _{x \rightarrow 0} f(x)=f(0)$
$\lim _{x \rightarrow 0} \frac{2^x-1}{\sqrt{1+x}-1}=k$
$\left[\frac{0}{0}\right.$ form, so by applying $L^{\prime}$ Hospital rule $]$
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{2^x \log 2}{\frac{1}{2 \sqrt{1+x}}}=k \\ & \lim _{x \rightarrow 0} 2^{x+1} \sqrt{1+x} \log 2=k\end{aligned}$
$\begin{array}{ll}\Rightarrow & 2 \sqrt{1+0} \log 2=k \Rightarrow k=2 \log _e 2 \\ \Rightarrow & k=\log _e 4\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.