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Question: If \(f(x) = \left\{ \begin{matrix} \frac{1 - \sin x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \lambda, &...

If f(x)={1−sin⁡xπ−2x,x≠π2λ,x=π2 f(x) = \left\{ \begin{matrix} \frac{1 - \sin x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \lambda, & x = \frac{\pi}{2} \end{matrix} \right.\ be continuous at x=π/2,x = \pi/2, , then

value of λ\lambda is

A

–1

B

1

C

0

D

2

Answer

0

Explanation

Solution

f(x)f(x) is continuous at x=π2x = \frac{\pi}{2}, then lim⁡x→π/2f(x)=f(0)\lim_{x \rightarrow \pi/2}f(x) = f(0) or

λ=lim⁡x→π/21−sin⁡xπ−2x\lambda = \lim_{x \rightarrow \pi/2}\frac{1 - \sin x}{\pi - 2x} , (00 form)\left( \frac{0}{0}\text{ form} \right)

Applying L-Hospital’s rule, λ=lim⁡x→π/2−cos⁡x−2\lambda = \lim_{x \rightarrow \pi/2}\frac{- \cos x}{- 2}

⇒λ=lim⁡x→π/2cos⁡x2=0.\lambda = \lim_{x \rightarrow \pi/2}\frac{\cos x}{2} = 0.