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 the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$, the limit of $f(x)$ as $x$ approaches $\frac{\pi}{2}$ must equal the value of $f(x)$ at $x = \frac{\pi}{2}$. That is, we need to find $\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = k$.
Let us first calculate the limit of the function as $x \to \frac{\pi}{2}$ for $x \neq \frac{\pi}{2}$: \[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^3 x}{3 \cos^2 x} \] At $x = \frac{\pi}{2}$, $\sin \frac{\pi}{2} = 1$ and $\cos \frac{\pi}{2} = 0$. Substituting these values into the expression gives: \[ \frac{1 - 1^3}{3 \cdot 0^2} = \frac{0}{0}. \] This is an indeterminate form, so we need to apply L'Hopital's Rule. To apply L'Hopital's Rule, we differentiate the numerator and denominator separately. The numerator is: \[ \frac{d}{dx}\left(1 - \sin^3 x\right) = -3 \sin^2 x \cdot \cos x. \] The denominator is: \[ \frac{d}{dx}\left(3 \cos^2 x\right) = -6 \cos x \cdot \sin x. \] Thus, we have: \[ \lim_{x \to \frac{\pi}{2}} \frac{-3 \sin^2 x \cos x}{-6 \cos x \sin x} = \lim_{x \to \frac{\pi}{2}} \frac{3 \sin x}{6} = \frac{3}{6} = \frac{1}{2}. \] Now, for the function to be continuous at $x = \frac{\pi}{2}$, the value of $f\left(\frac{\pi}{2}\right)$, which is $k$, must equal the limit we just calculated: \[ k = \frac{1}{6}. \] Hence, the correct value of $k$ is $\frac{1}{6}$.