Question

If \( f(x) = \begin{cases} \sqrt{\pi - \cos^{-1} x}, & x = -1 \\ \frac{\sqrt{2(1 + x)}}{\pi + \cos^{-1} x}, & x \neq -1 \end{cases} \) is right continuous at \( x = -1 \), then \( \lambda = \)

• A. \( 1 \)
• B. \( \pi \)
• C. \( 2\pi \)
• D. \( 2 \)

Hint:

For a function \( f(x) \) to be right continuous at \( x = a \), the limit of the function as \( x \) approaches \( a \) from the right must be equal to the value of the function at \( a \), i.e., \( \lim_{x \to a^+} f(x) = f(a) \). Remember the range of the inverse cosine function is \( [0, \pi] \).

Answer 1

The Correct Option is D

For \( f(x) \) to be right continuous at \( x = -1 \), we must have \( \lim_{x \to -1^+} f(x) = f(-1) \).
Given \( f(-1) = \lambda \).
We need to find \( \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{\sqrt{2(1 + x)}}{\pi + \cos^{-1} x} \).
As \( x \to -1^+ \), \( \sqrt{2(1 + x)} \to 0 \) and \( \pi + \cos^{-1} x \to \pi + \cos^{-1}(-1) = \pi + \pi = 2\pi \).
So, \( \lim_{x \to -1^+} f(x) = \frac{0}{2\pi} = 0 \).
For right continuity, \( \lambda = 0 \).
There is a discrepancy with the provided answer.
Let's re-examine the function definition at \( x = -1 \).
If \( f(-1) \) was meant to be related to the first expression, there might be a misunderstanding.
Assuming the question is stated correctly and \( f(-1) = \lambda \), then \( \lambda = 0 \).
If there's a typo and \( f(-1) \) should be evaluated from the first expression at \( x = -1 \), then \( f(-1) = \sqrt{\pi - \cos^{-1}(-1)} = 0 \), leading to \( \lambda = 0 \).
The provided correct answer suggests a different interpretation or a possible error in the question.