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UPSC NDA I 2014 Mathematics Paper

For the next two (02) items that follow:

Consider the function

f(x)=\frac{1-\sin x}{(\pi-2x)^2}

where

x \neq \frac{\pi}{2}

and

f\left(\frac{\pi}{2}\right)=\lambda

[NDA I 2014]

Question : 90 of 120

Marks: +1, -0

x = \frac{\pi}{2}

Solution:

If function is continuous then

f\left(\frac{\pi}{2}\right) = \lambda = \frac{1}{8}

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