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.
For \(x\lt \dfrac{\pi}{2}\), \[ f(x)=\frac{1-\sin^3x}{3\cos^2x} \] Now, \[ 1-\sin^3x=(1-\sin x)(1+\sin x+\sin^2x) \] Also, \[ \cos^2x=1-\sin^2x=(1-\sin x)(1+\sin x) \] Therefore, \[ \frac{1-\sin^3x}{3\cos^2x} = \frac{(1-\sin x)(1+\sin x+\sin^2x)} {3(1-\sin x)(1+\sin x)} \] \[ = \frac{1+\sin x+\sin^2x}{3(1+\sin x)} \] Taking limit as \(x\to \dfrac{\pi}{2}^{-}\), \[ \sin x\to 1 \] Hence, \[ \lim_{x\to \frac{\pi}{2}^{-}}f(x) = \frac{1+1+1}{3(1+1)} \] \[ =\frac{3}{6} \] \[ =\frac{1}{2} \]

Step 2: Use continuity to find \(\alpha\).
Since \(f(x)\) is continuous at \[ x=\frac{\pi}{2}, \] we must have \[ \alpha=\frac{1}{2} \]

Step 3: Find the right hand limit.
For \(x\gt \dfrac{\pi}{2}\), \[ f(x)=\frac{\beta(1-\sin x)}{(\pi-2x)^2} \] Let \[ h=x-\frac{\pi}{2} \] Then, \[ \sin x=\sin\left(\frac{\pi}{2}+h\right)=\cos h \] Also, \[ \pi-2x=\pi-2\left(\frac{\pi}{2}+h\right)=-2h \] So, \[ (\pi-2x)^2=4h^2 \] Therefore, \[ \frac{\beta(1-\sin x)}{(\pi-2x)^2} = \frac{\beta(1-\cos h)}{4h^2} \] Using \[ 1-\cos h \sim \frac{h^2}{2}, \] we get \[ \lim_{h\to 0^+}\frac{\beta(1-\cos h)}{4h^2} = \frac{\beta}{8} \]

Step 4: Apply continuity condition.
For continuity, \[ \frac{\beta}{8}=\alpha \] Since \[ \alpha=\frac{1}{2}, \] we get \[ \frac{\beta}{8}=\frac{1}{2} \] \[ \beta=4 \]

Step 5: Find \(\alpha\beta\).
\[ \alpha\beta=\frac{1}{2}\times 4 \] \[ =2 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{2} \]