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UPSC NDA I 2015 Mathematics Solved Paper

For the next two (2) items that follow :

Consider the function

f(x)=\begin{cases} \frac{\alpha \cos x}{\pi-2x} & \text{if } x \neq \frac{\pi}{2} \\ 3 & \text{if } x = \frac{\pi}{2} \end{cases}

which is continuous at

x=\frac{\pi}{2}

,where

\alpha

is a constant.

[NDA I 2015]

Question : 74 of 120

Marks: +1, -0

What is the value of

\alpha

Solution:

For a function to be continuous, the limit of the function at x should beequal to the value of the function at x.

Hence

\Rightarrow \lim\limits_{x\to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)

\Rightarrow \lim\limits_{x\to \frac{\pi}{2}} \frac{\alpha \cos(x)}{\pi - 2x} = 3

\Rightarrow \lim\limits_{x\to \frac{\pi}{2}} \frac{-\alpha \sin(x)}{-2} = 3

(L Hopital's rule)

\Rightarrow \frac{\alpha}{2} = 3

\therefore \alpha = 6

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