Q102 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Mathematics
$f(x)=\left\{\begin{array}{cl}\frac{1-\sin ^3 x}{3 \cos ^2 x}, & x < \frac{\pi}{2} \\ \alpha, & x=\frac{\pi}{2} \text { is continuous at } x=\frac{\pi}{2}, \\ \frac{\beta(1-\sin x)}{(\pi-2 x)^2}, & x>\frac{\pi}{2}\end{array}\right.$ then $\alpha \beta=$
- A$1$
- B$-1$
- Correct answerC$2$
- D$-2$
Solution
If $f(x)$ is contineous at $x=\pi / 2$
$\begin{aligned} & \left.L H L\right|_{x=\pi / 2}=\left.R H L\right|_{x=\pi / 2}=f\left(\frac{\pi}{2}\right) \\ & L H L=\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{1-\sin ^3 x}{3 \cos ^2 x}\end{aligned}$
$\because \frac{0}{0}$ form, using L-Hospital Rule
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{-3 \sin ^2 x \cos x}{-3(2 \cos x) \sin x}=\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin x}{2}$
$=\frac{1}{2}$
$\because f\left(\frac{\pi}{2}\right)=L H L$
$\therefore \alpha=\frac{1}{2}$ ...(i)
$R H L=\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{\beta(1-\sin x)}{(\pi-2 x)^2}=\frac{1}{2}$
Using L-Hospital Rule twice
$\begin{aligned} & \lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{\beta(-\cos x)}{2(\pi-2 x)(-2)}=\frac{1}{2} \\ & \lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{\beta \sin x}{(-4)(-2)}=\frac{1}{2}\end{aligned}$
$\begin{aligned} \frac{\beta}{8} & =\frac{1}{2} \\ \beta & =4\end{aligned}$
$\therefore \alpha \beta=\frac{1}{2} \times 4$
$=2$
Hence, the correct answer is option: 'C'.