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If $f(x)=\left\{\begin{array}{ll}m x+1 & x \leq \frac{\pi}{2} \\ \sin x+n & x>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then
which one of the following is correct?
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which one of the following is correct?
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Verified Answer
The correct answer is:
$\mathrm{n}=\mathrm{m}\left(\frac{\pi}{2}\right)$
Given function is
$f(x)=\left\{\begin{array}{ll}m x+1, & x \leq \frac{\pi}{2} \\ \sin x+n, & x>\frac{\pi}{2}\end{array}\right.$
As given this function is continuous at $x=\frac{\pi}{2}$.
So, limit of function when $\mathrm{x} \rightarrow \frac{\pi}{2}=\mathrm{f}\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}+}(\sin x+n)=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\sin \left(\frac{\pi}{2}+h\right)+n\right)=\frac{m \pi}{2}+1$
$\Rightarrow \quad \sin \frac{\pi}{2}+n=\frac{m \pi}{2}+1$
$\Rightarrow 1+\mathrm{n}=\frac{\mathrm{m} \pi}{2}+1$
$\Rightarrow \mathrm{n}=\frac{\mathrm{m} \pi}{2}$
$f(x)=\left\{\begin{array}{ll}m x+1, & x \leq \frac{\pi}{2} \\ \sin x+n, & x>\frac{\pi}{2}\end{array}\right.$
As given this function is continuous at $x=\frac{\pi}{2}$.
So, limit of function when $\mathrm{x} \rightarrow \frac{\pi}{2}=\mathrm{f}\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}+}(\sin x+n)=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\sin \left(\frac{\pi}{2}+h\right)+n\right)=\frac{m \pi}{2}+1$
$\Rightarrow \quad \sin \frac{\pi}{2}+n=\frac{m \pi}{2}+1$
$\Rightarrow 1+\mathrm{n}=\frac{\mathrm{m} \pi}{2}+1$
$\Rightarrow \mathrm{n}=\frac{\mathrm{m} \pi}{2}$
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