Question

If \[ f(x)= \begin{cases} \dfrac{1-\sin^3 x}{3\cos^2 x}, & x\lt \dfrac{\pi}{2} \\[6pt] \alpha, & x=\dfrac{\pi}{2} \\[6pt] \dfrac{\beta(1-\sin x)}{(\pi-2x)^2}, & x\gt \dfrac{\pi}{2} \end{cases} \] is continuous at \(x=\dfrac{\pi}{2}\), then \(\alpha\beta=\)

• A. \(1\)
• B. \(-1\)
• C. \(2\)
• D. \(-2\)

Hint:

For continuity of a piecewise function at a point, always equate: \[ \text{LHL}=\text{value of function}=\text{RHL}. \] Also remember: \[ 1-\cos h\sim \frac{h^2}{2} \quad \text{as} \quad h\to 0. \]

Answer 1

The Correct Option is C

Step 1: Find the left-hand limit at $x = \pi/2$.
For $x < \pi/2$, $f(x) = \dfrac{1 - \sin^3 x}{3\cos^2 x}$. As $x \to \pi/2^-$, let $h = \pi/2 - x \to 0^+$. Then $\sin x = \cos h \to 1$ and $\cos x = \sin h \to 0$.

Step 2: Use substitution to evaluate the left limit.
$1 - \sin^3 x = 1 - \cos^3 h$ and $3\cos^2 x = 3\sin^2 h$. So: \[ \lim_{h\to 0^+} \frac{1 - \cos^3 h}{3\sin^2 h} = \lim_{h\to 0^+} \frac{(1-\cos h)(1+\cos h+\cos^2 h)}{3\sin^2 h}. \] Use $1-\cos h \approx h^2/2$ and $\sin^2 h \approx h^2$: \[ = \frac{(h^2/2)(3)}{3h^2} = \frac{1}{2}. \] So LHL $= 1/2$.

Step 3: For continuity, set $\alpha = \text{LHL}$.
$f(\pi/2) = \alpha = 1/2$.

Step 4: Find the right-hand limit at $x = \pi/2$.
For $x > \pi/2$, $f(x) = \dfrac{\beta(1-\sin x)}{(\pi - 2x)^2}$. Let $h = x - \pi/2 \to 0^+$. Then $1 - \sin x = 1 - \cos h \approx h^2/2$ and $\pi - 2x = -2h$, so $(\pi-2x)^2 = 4h^2$.

Step 5: Evaluate the right limit.
\[ \lim_{h\to 0^+} \frac{\beta \cdot h^2/2}{4h^2} = \frac{\beta}{8}. \] For continuity: $\dfrac{\beta}{8} = \dfrac{1}{2}$, so $\beta = 4$.

Step 6: Compute $\alpha\beta$.
$\alpha\beta = \dfrac{1}{2} \times 4 = 2$.
\[ \boxed{2} \]