Question

Let \[ f(x)= \begin{cases} \dfrac{1-\sin^3 x}{3\cos^2 x}, & x < \dfrac{\pi}{2} \\ [6pt] p, & x = \dfrac{\pi}{2} \\ [6pt] \dfrac{q(1-\sin x)}{(\pi-2x)^2}, & x > \dfrac{\pi}{2} \end{cases} \] If \(f(x)\) is continuous at \(x=\dfrac{\pi}{2}\), then \((p,q)=\)

• A. \((1,4)\)
• B. \(\left(\dfrac{1}{2},2\right)\)
• C. \(\left(\dfrac{1}{2},4\right)\)
• D. None of these

Hint:

For continuity, LHL = RHL = value of the function.

Answer 1

The Correct Option is C

Step 1: Left hand limit as \(x\to \frac{\pi}{2}^-\): \[ \lim_{x\to \pi/2}\frac{1-\sin^3 x}{3\cos^2 x} =\lim_{x\to \pi/2}\frac{(1-\sin x)(1+\sin x+\sin^2 x)}{3\cos^2 x} =\frac{1}{2} \] Hence \(p=\dfrac{1}{2}\).
Step 2: Right hand limit as \(x\to \frac{\pi}{2}^+\): \[ \lim_{x\to \pi/2}\frac{q(1-\sin x)}{(\pi-2x)^2} =\frac{q}{4} \]
Step 3: Continuity gives: \[ \frac{q}{4}=\frac{1}{2}\Rightarrow q=4 \]