Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{f}(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x} & ,-1 \leq x < 0 \\ \frac{2 x+1}{x-2} & , 0 \leq x \leq 1\end{array}\right.$ is continuous in the interval $[-1,1]$, then $\mathrm{m}$ is equal to
Options:
Solution:
2260 Upvotes
Verified Answer
The correct answer is:
$-\frac{1}{2}$
Since $\mathrm{f}(x)$ is continuous in $[-1,1]$, it is continuous at $x=0$.
$$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2} \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{(1+\mathrm{m} x-1+\mathrm{m} x)}{x(\sqrt{1+\mathrm{m} x}+\sqrt{1-\mathrm{m} x})}=\frac{2(0)+1}{0-2} \\
& \Rightarrow \frac{2 \mathrm{~m}}{1+1}=\frac{1}{-2} \\
& \Rightarrow \mathrm{m}=\frac{-1}{2}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+\mathrm{m} x}-\sqrt{1-\mathrm{m} x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2} \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{(1+\mathrm{m} x-1+\mathrm{m} x)}{x(\sqrt{1+\mathrm{m} x}+\sqrt{1-\mathrm{m} x})}=\frac{2(0)+1}{0-2} \\
& \Rightarrow \frac{2 \mathrm{~m}}{1+1}=\frac{1}{-2} \\
& \Rightarrow \mathrm{m}=\frac{-1}{2}
\end{aligned}
$$
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.