If f(x) = begin{cases} frac{sqrt{pi} - sqrt{c | vedclass

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(D) For $f(x)$ to be right continuous at $x = -1$,we must have $\lim_{x \rightarrow -1^+} f(x) = f(-1)$.Applying $L'H\hat{o}pital's$ rule to the limit

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$2$

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(D) For $f(x)$ to be right continuous at $x = -1$,we must have $\lim_{x \rightarrow -1^+} f(x) = f(-1)$.Applying $L'H\hat{o}pital's$ rule to the limit $\lim_{x \rightarrow -1^+} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}$:$\lim_{x \rightarrow -1^+} \frac{-\frac{1}{2\sqrt{\cos^{-1} x}} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)}{\frac{1}{2\sqrt{x+1}}} = \lim_{x \rightarrow -1^+} \frac{\sqrt{x+1}}{\sqrt{\cos^{-1} x} \sqrt{1-x} \sqrt{1+x}}$$= \lim_{x \rightarrow -1^+} \frac{1}{\sqrt{\cos^{-1} x} \sqrt{1-x}} = \frac{1}{\sqrt{\cos^{-1}(-1)} \sqrt{1-(-1)}} = \frac{1}{\sqrt{\pi} \sqrt{2}} = \frac{1}{\sqrt{2\pi}}$.Given $f(-1) = \frac{1}{\sqrt{\lambda \pi}}$,we equate: $\frac{1}{\sqrt{2\pi}} = \frac{1}{\sqrt{\lambda \pi}}$.Squaring both sides gives $2\pi = \lambda \pi$,so $\lambda = 2$.

If $f(x) = \begin{cases} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}, & x \neq -1 \\ \frac{1}{\sqrt{\lambda \pi}}, & x = -1 \end{cases}$ is right continuous at $x = -1$,then $\lambda = $

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