Question

If $f(x) = \left\{ \begin{array}{ll} \frac{1 - \sin^3 x}{3 \cos^2 x} & \text{for} \, x \neq \frac{\pi}{2}, \\ k & \text{for} \, x = \frac{\pi}{2}, \end{array} \right. $ is continuous at $x = \frac{\pi}{2}$, then the value of $k$ is:

• A. $\frac{3}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{2}$
• D. $1$

Hint:

When dealing with limits that give indeterminate forms like $\frac{0}{0}$, apply L'Hopital's Rule to differentiate the numerator and denominator separately and then evaluate the limit.

Answer 1

The Correct Option is B

For $f(x)$ to be continuous at $x = \frac{\pi}{2}$, the limit of $f(x)$ as $x$ approaches $\frac{\pi}{2}$ must equal $f\left(\frac{\pi}{2}\right)$, which is given as $k$. Thus, $\lim_{x \to \frac{\pi}{2}} f(x) = k$. We first evaluate the limit: \[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^3 x}{3 \cos^2 x} \] Substituting $x = \frac{\pi}{2}$ yields $\frac{1 - \sin^3(\frac{\pi}{2})}{3 \cos^2(\frac{\pi}{2})} = \frac{1 - 1^3}{3 \cdot 0^2} = \frac{0}{0}$, an indeterminate form. Applying L'Hopital's Rule: Differentiate the numerator: $\frac{d}{dx}(1 - \sin^3 x) = -3 \sin^2 x \cos x$. Differentiate the denominator: $\frac{d}{dx}(3 \cos^2 x) = -6 \cos x \sin x$. The limit becomes: \[ \lim_{x \to \frac{\pi}{2}} \frac{-3 \sin^2 x \cos x}{-6 \cos x \sin x} = \lim_{x \to \frac{\pi}{2}} \frac{\sin x}{2} = \frac{\sin(\frac{\pi}{2})}{2} = \frac{1}{2} \] For continuity, $k$ must equal this limit. \[ k = \frac{1}{2} \] Therefore, the value of $k$ is $\frac{1}{2}$.