Question

If  \[f\left( x \right) = \frac{1 - \sin x}{\left( \pi - 2x \right)^2},\] when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x= π/2, where λ =

पर्याय

• 1/8

• 1/4

• 1/2

• none of thes

Answer 1

\[\frac{1}{8}\]

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

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

\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \pi - 2x \right)^2} = f\left( \frac{\pi}{2} \right)\]     ...(1)

Suppose 

\[\left( \frac{\pi}{2} - x \right) = t\] then

\[\lim_{t \to 0} \left[ \frac{1 - \sin \left( \frac{\pi}{2} - t \right)}{\left( 2t \right)^2} \right] = f\left( \frac{\pi}{2} \right) \left[ \text{ From eq }. (1) \right]\]
\[ \Rightarrow \lim_{t \to 0} \left[ \frac{1 - \cos t}{4 t^2} \right] = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow \frac{1}{4} \lim_{t \to 0} \left[ \frac{2 \sin^2 \left( \frac{t}{2} \right)}{t^2} \right] = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow \frac{1}{4} \lim_{t \to 0} \left[ \frac{\frac{2}{4} \sin^2 \left( \frac{t}{2} \right)}{\frac{t^2}{4}} \right] = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow \frac{1}{8} \lim_{t \to 0} \left[ \frac{\sin^2 \left( \frac{t}{2} \right)}{\frac{t^2}{4}} \right] = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow \frac{1}{8} \lim_{t \to 0} \left[ \frac{\sin \left( \frac{t}{2} \right)}{\frac{t}{2}} \right]^2 = f\left( \frac{\pi}{2} \right)\]
\[ \Rightarrow f\left( \frac{\pi}{2} \right) = \lambda = \frac{1}{8}\]