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: Understanding the Concept:
For a function to be continuous at a point x = a, the limit of the function as x approaches a must exist and be equal to the value of the function at that point. That is, \(\lim_{x \to a} f(x) = f(a)\).
Step 2: Key Formula or Approach:
We will apply the condition for continuity at x = $\frac{\pi}{2}$.
\[ \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) \] The limit on the left side is an indeterminate form (0/0), so we can evaluate it using L'Hôpital's Rule.
Step 3: Detailed Explanation:
Given the function is continuous 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 given that \(f(\frac{\pi}{2}) = 3\).
Now we evaluate the limit. As x $\to \frac{\pi}{2}$, cos(x) $\to$ 0 and ($\pi$ - 2x) $\to$ 0. This is a \(\frac{0}{0}\) indeterminate form.
Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to x:
\[ \text{Derivative of numerator:} \frac{d}{dx}(k\cos x) = -k\sin x \] \[ \text{Derivative of denominator:} \frac{d}{dx}(\pi - 2x) = -2 \] So, the limit becomes:
\[ \lim_{x \to \frac{\pi}{2}} \frac{-k\sin x}{-2} = \lim_{x \to \frac{\pi}{2}} \frac{k\sin x}{2} \] Now, substitute x = $\frac{\pi}{2}$ into the simplified 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: Final Answer:
The value of k is 6.