Question

If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\]  , then

 

पर्याय

• m = 1, n = 0

•   \[m = \frac{n\pi}{2} + 1\] 

• \[n = \frac{m\pi}{2}\] 

• \[m = n = \frac{\pi}{2}\]

   

Answer 1

  \[n = \frac{m\pi}{2}\]

Here,

\[f\left( \frac{\pi}{2} \right) = \frac{m\pi}{2} + 1\]

We have 

(LHL at  \[x = \frac{\pi}{2}\] = \[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} - h \right) = \lim_{h \to 0} m\left( \frac{\pi}{2} - h \right) + 1 = \frac{m\pi}{2} + 1\]

(RHL at   \[x = \frac{\pi}{2}\] = \[\lim_{x \to \frac{\pi}{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \frac{\pi}{2} + h \right) = \lim_{h \to 0} \left[ \sin\left( \frac{\pi}{2} + h \right) + n \right] = n + 1\] 

Thus,
If  \[f\left( x \right)\] is continuous at  \[x = \frac{\pi}{2}\] then  

\[\lim_{x \to \frac{\pi}{2}^-} f\left( x \right) = \lim_{x \to \frac{\pi}{2}^+} f\left( x \right)\]

\[\Rightarrow \frac{m\pi}{2} + 1 = n + 1\]
\[ \Rightarrow \frac{m\pi}{2} = n\]