JEE Main 2021 (Online) 24th February Morning ShiftQuestion 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 $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$9$
Solution
Concept. With $s=\sin x\in(0,1)$, $\alpha$ must be at least the minimum of $h(s)=\frac4s+\frac{1}{1-s}$. Formulas used. (1) $h'(s)=-\dfrac{4}{s^2}+\dfrac{1}{(1-s)^2}$Step 1 - Critical point. By (1), $h'=0$ gives $\frac{2}{s}=\frac{1}{1-s}$ (both positive), so $s=\frac23$. Step 2 - Minimum. $h\to\infty$ at both ends of $(0,1)$, so $s=\frac23$ is the minimum: $h=6+3=9$. Step 3 - Answer. Every $\alpha\ge9$ is attained, so the least $\alpha$ is $9$. Answer: 9