JEE PYQ: Application Of Derivatives Question 12
Question 12 - 2021 (24 Feb Shift 1)
The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x} + \frac{1}{1 - \sin x} = \alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is
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Answer: 9
Solution
Let $\sin x = t$, $0 < t < 1$. $f(t) = \frac{4}{t} + \frac{1}{1-t}$. $f’(t) = \frac{-4}{t^2} + \frac{1}{(1-t)^2} = 0 \Rightarrow t = \frac{2}{3}$. $f_{\min} = f\left(\frac{2}{3}\right) = 6 + 3 = 9$.
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