If $f(x) = \left\{ \begin{array}{r} mx + 1,\ \ \ \ x \leq \frac{\pi}{2} \ \sin{x + n,\ \ \ \ x \frac{\pi}{2}} \end{array} ight.\$ is continuous t $x = \frac{\pi}{2}$, then

Question

If $f(x) = \left\{ \begin{array}{r} mx + 1,\ \ \ \ x \leq \frac{\pi}{2} \ \sin{x + n,\ \ \ \ x > \frac{\pi}{2}} \end{array} ight.\$ is continuous t $x = \frac{\pi}{2}$, then

Dr. Abubacker · Accepted Answer

(c)
Since, $f(x)$ is continuous at $x = \frac{\pi}{2}$
$\lim_{x \rightarrow \frac{\pi^{- 1}}{2}}\ \ (mx + 1) = \lim_{x \rightarrow \frac{\pi^{+}}{2}}\ \ \ \ \left( \sin{x + n} \right)$
$\Rightarrow \ \ \ \ m\frac{\pi}{2} + 1 = \sin{\frac{\pi}{2} + n}$
$\Rightarrow \ \ \ \ \frac{m\pi}{2} = n$

Q. If $f(x) = \left\{ \begin{array}{r} mx + 1,\ \ \ \ x \leq \frac{\pi}{2} \\ \sin{x + n,\ \ \ \ x > \frac{\pi}{2}} \end{array} \right.\$ is continuous t $x = \frac{\pi}{2}$, then

Explanation

Reviewed by Dr. Abubacker

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