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If $f(x)=\left\{\begin{array}{cl}\frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}, & x \neq-1 \\ \frac{1}{\sqrt{\lambda \pi}}, & x=-1\end{array}\right.$ is right
continuous at $x=-1$, then $\lambda=$
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continuous at $x=-1$, then $\lambda=$
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Verified Answer
The correct answer is:
2
$\lim _{x \rightarrow-1} \frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}=\lim _{x \rightarrow-1} \frac{-\frac{1}{2 \sqrt{\cos ^{-1} x}} \cdot \frac{-1}{\sqrt{1-x^2}}}{\frac{1}{2 \sqrt{x+1}}}$
$\begin{aligned} & =\lim _{x \rightarrow-1} \frac{\sqrt{x+1}}{\sqrt{\cos ^{-1} x} \sqrt{1-x^2}} \\ & =\lim _{x \rightarrow-1} \frac{1}{\sqrt{\cos ^{-1} x} \sqrt{1-x}}=\frac{1}{\sqrt{\pi} \sqrt{1+1}} \\ & \Rightarrow \lim _{x \rightarrow-1} \frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}=\frac{1}{\sqrt{2 \pi}}\end{aligned}$
$\because f(x)$ is right continuous on $x=-1$
$\begin{aligned} & \therefore \lim _{x \rightarrow-1^{+}} \frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}=f(-1) \\ & \Rightarrow \frac{1}{\sqrt{2 \pi}}=\frac{1}{\sqrt{\lambda \pi}} \Rightarrow \lambda=2 .\end{aligned}$
option (4) is correct.
$\begin{aligned} & =\lim _{x \rightarrow-1} \frac{\sqrt{x+1}}{\sqrt{\cos ^{-1} x} \sqrt{1-x^2}} \\ & =\lim _{x \rightarrow-1} \frac{1}{\sqrt{\cos ^{-1} x} \sqrt{1-x}}=\frac{1}{\sqrt{\pi} \sqrt{1+1}} \\ & \Rightarrow \lim _{x \rightarrow-1} \frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}=\frac{1}{\sqrt{2 \pi}}\end{aligned}$
$\because f(x)$ is right continuous on $x=-1$
$\begin{aligned} & \therefore \lim _{x \rightarrow-1^{+}} \frac{\sqrt{\pi}-\sqrt{\cos ^{-1} x}}{\sqrt{x+1}}=f(-1) \\ & \Rightarrow \frac{1}{\sqrt{2 \pi}}=\frac{1}{\sqrt{\lambda \pi}} \Rightarrow \lambda=2 .\end{aligned}$
option (4) is correct.
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