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AP EAPCET 05-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 62 of 160

Marks: +1, -0

If

f(x)=\begin{cases} \frac{1-\sin^3 x}{3\cos^2 x}, & x < \frac{\pi}{2} \\ \alpha, & x=\frac{\pi}{2} \\ \frac{\beta(1-\sin x)}{(\pi-2x)^2}, & x>\frac{\pi}{2} \end{cases}

is continuous at

x=\frac{\pi}{2}

\alpha \beta =

$1$
$-1$
2
$-2$

Solution: 👈: Video Solution

f(x)

is continuous at

\frac{\pi}{2}

\lim\limits_{x \rightarrow \frac{\pi^{-}}{2}} f(x) = \lim\limits_{x \rightarrow \frac{\pi^{-}}{2}} f(x) = f\left(\frac{\pi}{2}\right)

= \lim\limits_{x \rightarrow \frac{\pi^{-}}{2}} f(x) = \lim\limits_{x \rightarrow \frac{\pi^{-}}{2}} \left(\frac{1-\sin^3 x}{3\cos^2 x}\right)

= \lim\limits_{x \rightarrow \frac{\pi^{-}}{2}} \left(\frac{(1-\sin x)(1+\sin^2 x+\sin x)}{3(1-\sin x)(1+\sin x)}\right)

= \lim\limits_{x \rightarrow \frac{\pi^{-}}{2}} \left[\frac{1+\sin^2 x+\sin x}{3(1+\sin x)}\right] = \frac{3}{3(2)}

= \frac{1}{2}

and

\lim\limits_{x \rightarrow \frac{\pi^{+}}{2}} f(x) = \lim\limits_{x \rightarrow \frac{\pi^{+}}{2}} \frac{\beta(1-\sin x)}{(\pi-2x)^2}

= \lim\limits_{h \to 0^{+}} \frac{\beta\left(1-\sin\left(\frac{\pi}{2}+h\right)\right)}{\left(\pi-2\left(\frac{\pi}{2}+h\right)\right)^2}, \quad h = x-\frac{\pi}{2}

= \lim\limits_{h \to 0^{+}} \beta\left(\frac{1-\cos h}{4h^2}\right)

= \lim\limits_{h \to 0^{+}} \frac{\beta\left(2\sin^2\left(\frac{h}{2}\right)\right)}{4\times 4\left(\frac{h}{2}\right)^2} = \frac{\beta}{8}

\therefore \frac{1}{2} = \frac{\beta}{8} = \alpha

\Rightarrow \alpha = \frac{1}{2} \text{ and } \beta = 4

\therefore \alpha\beta = 2

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