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

Concept: A function is continuous at a point \( x = a \) if: \[ \lim_{x \to a} f(x) = f(a) \] If the function gives an indeterminate form at that point, we evaluate the limit using standard techniques like L'Hôpital's Rule.
Step 1: {Substitute \( x = \frac{\pi}{2} \) into the function.}
\[ f\left(\frac{\pi}{2}\right) = \frac{k \cos \frac{\pi}{2}}{\pi - 2\cdot \frac{\pi}{2}} = \frac{k \cdot 0}{\pi - \pi} = \frac{0}{0} \] This is an indeterminate form.
Step 2: {Apply L'Hôpital's Rule.}
\[ \lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x} = \lim_{x \to \frac{\pi}{2}} \frac{-k \sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{k \sin x}{2} \]
Step 3: {Evaluate the limit.}
\[ = \frac{k \sin \frac{\pi}{2}}{2} = \frac{k}{2} \]
Step 4: {For continuity, limit must equal function value.}
Since the function approaches \( \frac{k}{2} \), it must be finite and well-defined. Hence, \[ \frac{k}{2} = 1 \quad \Rightarrow \quad k = 2 \]