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What should be the value of \mathrm{\lambda} such that the function defined below is continuous at

\mathrm{x=\pi / 2 ? \quad f(x)= \begin{cases}\frac{\lambda \cdot \cos x}{\frac{\pi}{2}-x} & \text { if } x \neq \pi / 2 \\ 1 & \text { if } x=\pi / 2\end{cases}}

Option: 1

Zero


Option: 2

\mathrm{2 \pi}


Option: 3

1


Option: 4

\mathrm{\pi / 2}


Answers (1)

best_answer

Given,             \mathrm{f(x)=\left\{\begin{array}{cc} \frac{\lambda \cos x}{\frac{\pi}{2}-x} & ; \text { if } x \neq \frac{\pi}{2} \\ 1 & ; \quad x=\frac{\pi}{2} \end{array}\right.}

           \mathrm{\lim _{x \rightarrow \pi / 2} f(x)=\lim _{x \rightarrow \pi / 2} \frac{\lambda \cos x}{\left(\frac{\pi}{2}-x\right)} ; \quad\{\mathrm{LHL}=\mathrm{RHL}\}}

                                \mathrm{=\lambda \lim _{x \rightarrow \pi / 2} \frac{(-\sin x)}{(-1)}}{ By L' Hospital rule}

                                \mathrm{=\lambda \times \sin \left(\frac{\pi}{2}\right)=\lambda}

For the function to be continuous,

                \mathrm{\lim _{x \rightarrow \pi / 2} f(x)=f\left(\frac{\pi}{2}\right)}

\Rightarrow                           \mathrm{\lambda=1}

Posted by

Pankaj Sanodiya

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