Question

If the function f(x) = $\begin{cases}\frac{k\cos x}{\pi - 2x} & ; x \neq \frac{\pi}{2} \\ 3 & ; x = \frac{\pi}{2} \end{cases}$ is continuous at x = $\frac{\pi}{2}$, then k is equal to

• A. 6
• B. 5
• C. -6
• D. 4

Hint:

L'Hôpital's Rule is a powerful tool for evaluating limits of indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Alternatively, one could use a substitution like \(h = x - \frac{\pi}{2}\) to solve the limit using standard trigonometric limits.

Answer 1

The Correct Option is A

Step 1: Concept Definition:
A function is continuous at a point x = a if its limit as x approaches a exists and is equal to the function's value at a. Mathematically, this is expressed as \(\lim_{x \to a} f(x) = f(a)\).

Step 2: Methodology:
We will verify the continuity condition at x = $\frac{\pi}{2}$ using the formula:
\[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) \]
The limit on the left side results in an indeterminate form (0/0), necessitating the use of L'Hôpital's Rule.

Step 3: Calculation:
Given the function's continuity at x = $\frac{\pi}{2}$, we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = f\left(\frac{\pi}{2}\right) \]
We are provided with \(f(\frac{\pi}{2}) = 3\).
Evaluating the limit: as x approaches $\frac{\pi}{2}$, cos(x) approaches 0 and ($\pi$ - 2x) approaches 0, resulting in the \(\frac{0}{0}\) indeterminate form.
Applying L'Hôpital's Rule by differentiating the numerator and denominator with respect to x:
\[ \text{Numerator derivative:} \frac{d}{dx}(k\cos x) = -k\sin x \]
\[ \text{Denominator derivative:} \frac{d}{dx}(\pi - 2x) = -2 \]
The limit simplifies to:
\[ \lim_{x \to \frac{\pi}{2}} \frac{-k\sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{k\sin x}{2} \]
Substituting x = $\frac{\pi}{2}$ into the simplified limit expression:
\[ \frac{k\sin(\frac{\pi}{2})}{2} = \frac{k(1)}{2} = \frac{k}{2} \]
For continuity, this limit must equal \(f(\frac{\pi}{2})\):
\[ \frac{k}{2} = 3 \]
\[ k = 6 \]

Step 4: Conclusion:
The calculated value for k is 6.