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If $f(x)=\left \{\begin{array}{1 1}mx+1, & if\;x\leq\large\frac{\pi}{2}\\\sin x+n, & if\;x>\large\frac{\pi}{2}\end{array}\right.$,is continuous at $x=\large\frac{\pi}{2}$,then

\[\begin{array}{1 1}(A)\;m=1,n=0 & (B)\;m=\large\frac{n\pi}{2}+1\\(C)\;n=\large\frac{m\pi}{2} & (D)\;m=n=\large\frac{\pi}{2}\end{array}\]

1 Answer

Toolbox:
  • A real valued function $f(x)$ is continuous at a point $'a'$ in its domain if $\lim\limits_{\large x\to a}f(x)=f(a)$
  • If a given function is continuous ,then its LHL = RHL.
Step 1:
$f(x)=\left \{\begin{array}{1 1}mx+1, & if\;x\leq\large\frac{\pi}{2}\\\sin x+n, & if\;x>\large\frac{\pi}{2}\end{array}\right.$
Since the given function is continuous at $x=\large\frac{\pi}{2}$ the LHD = RHD.
(i.e) $\lim\limits_{\large x\to {\large\frac{\pi}{2}}^-}f(x)=\lim\limits_{\large x\to {\large\frac{\pi}{2}}^+}f(x)$
$\Rightarrow \lim\limits_{\large x\to \large\frac{\pi}{2}}mx+1=\lim\limits_{\large x\to \large\frac{\pi}{2}}\sin x+n$
Step 2:
Applying the limits we get,
$m\big(\large\frac{\pi}{2}\big)$$+1=\sin\large\frac{\pi}{2}$$+n$
But $\sin\large\frac{\pi}{2}$$=1$
Therefore $m\large\frac{\pi}{2}$$+1=1+n$
$\Rightarrow n=\large\frac{m\pi}{2}$
Hence the correct option is $C$
answered Jul 4, 2013 by sreemathi.v
 

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