Question

Let $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 a function differentiable at $x = \pi/2,$. Then \lambda equals

• A. $f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x};(x \neq 0)$
• B. $x = 0$
• C. $f(0)$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

Since $f ( x )$ is differentiable at $x = c$ therefore it is

continuous at $x = c$. Hence, $\lim _ { x \rightarrow c } f ( x ) = f ( c )$.