Question

Find the value of \( k \) if the function \( f(x) = \frac{k \cos x}{\pi - 2x} \) is continuous at \( x = \frac{\pi}{2} \).

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

Hint:

When substitution gives \( \frac{0}{0} \), use L'Hôpital's Rule by differentiating numerator and denominator separately.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the value of a constant \( k \) that ensures the function \( f(x) \) is continuous at a specific point \( x = \frac{\pi}{2} \).
Continuity at a point requires that the limit of the function as it approaches that point equals the function's value at that point.
Step 2: Key Formula or Approach:
For a function to be continuous at \( x = a \), the condition is:
\[ \lim_{x \to a} f(x) = f(a) \]
If direct substitution results in an indeterminate form like \( \frac{0}{0} \), we apply L'Hôpital's Rule:
\[ \lim_{x \to a} \frac{g(x)}{h(x)} = \lim_{x \to a} \frac{g'(x)}{h'(x)} \]
Step 3: Detailed Explanation:
We evaluate the limit of \( f(x) \) as \( x \to \frac{\pi}{2} \):
\[ L = \lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x} \]
Substituting \( x = \frac{\pi}{2} \) gives \( \frac{k \cos(\pi/2)}{\pi - 2(\pi/2)} = \frac{0}{0} \).
Applying L'Hôpital's Rule by differentiating the numerator and denominator with respect to \( x \):
\[ L = \lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}(k \cos x)}{\frac{d}{dx}(\pi - 2x)} \]
\[ L = \lim_{x \to \frac{\pi}{2}} \frac{-k \sin x}{-2} = \frac{k}{2} \lim_{x \to \frac{\pi}{2}} \sin x \]
Since \( \sin(\pi/2) = 1 \), we get:
\[ L = \frac{k}{2}(1) = \frac{k}{2} \]
Assuming the function is defined such that \( f(\frac{\pi}{2}) = 1 \) (standard for these types of MHT CET problems to find \( k \)) or simply finding the limit that makes it finite:
Setting \( \frac{k}{2} = 1 \):
\[ k = 2 \]
Step 4: Final Answer:
The value of \( k \) that satisfies the limit condition for continuity is \( 2 \).