Q85 / 125 · BITSAT 2014 · Mathematics
Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\frac{1-\sin ^{3} \mathrm{x}}{3 \cos ^{2} \mathrm{x}}, \mathrm{x}<\frac{\pi}{2} \\ \mathrm{p}, \quad \mathrm{x}=\frac{\pi}{2} \\ \frac{\mathrm{q}(1-\sin \mathrm{x})}{(\pi-2 \mathrm{x})^{2}}, \mathrm{x}>\frac{\pi}{2}\end{array}\right.$
If $f(x)$ is continuous at $x=\frac{\pi}{2},(p, q)=$
- A(1,4)
- B$\left(\frac{1}{2}, 2\right)$
- Correct answerC$\left(\frac{1}{2}, 4\right)$
- DNone of these
Solution
$\mathrm{f}\left[(\pi / 2)^{-}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\sin ^{3}[(\pi / 2)-\mathrm{h}]}{3 \cos ^{2}[(\pi / 2)-\mathrm{h}]}$
$\quad=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos ^{3} \mathrm{~h}}{3 \sin ^{2} \mathrm{~h}}=\frac{1}{2}$
$\mathrm{f}\left[(\pi / 2)^{+}\right]=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}[1-\sin \{(\pi / 2)+\mathrm{h}\}]}{[\pi-2\{(\pi / 2)+\mathrm{h}\}]^{2}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{q}(1-\cosh )}{4 \mathrm{~h}^{2}}=\frac{\mathrm{q}}{8}$
$\therefore \mathrm{p}=\frac{1}{2}=\frac{\mathrm{q}}{8} \Rightarrow \mathrm{p}=\frac{1}{2}, \mathrm{q}=4$
Hence, the correct answer is option: 'C'.