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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If
$f(x)=
\begin{cases}
\frac{\sqrt{1-\cos2x}}{x\sqrt2}, & x\ne0 \\
k, & x=0
\end{cases}$
then the value of $k$ will make function $f$ continuous at $x=0$ is:
Solution
For continuity at $x=0$,
$\lim_{x\to0}f(x)=f(0)=k$
So evaluate limit:
$\lim_{x\to0}\frac{\sqrt{1-\cos2x}}{x\sqrt2}$
Using identity:
$1-\cos2x=2\sin^2x$
So,
$\sqrt{1-\cos2x}=\sqrt{2\sin^2x}=\sqrt2|\sin x|$
Hence,
$\frac{\sqrt{1-\cos2x}}{x\sqrt2}=\frac{\sqrt2|\sin x|}{x\sqrt2}=\frac{|\sin x|}{x}$
As $x\to0$,
$\frac{|\sin x|}{x}\to1$
Therefore,
$k=1$
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