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The minimum value of \alpha  for which the equation \frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha has least one solution in \left ( 0,\frac{\pi}{2} \right ) is ______
Option: 1 7
Option: 2 8
Option: 3 9
Option: 4 10

Answers (1)

best_answer

 

\begin{aligned} \text { Let } f(x) &=\frac{4}{\sin x}+\frac{1}{1-\sin x} \\ y &=\frac{4-3 \sin x}{\sin x(1-\sin x)} \end{aligned}

Let sin x = t when t \in(0,1).

y=\frac{4-3 t}{t-t^{2}}

\frac{d y}{d t}=\frac{-3\left(t-t^{2}\right)-(1-2 t)(4-3 t)}{\left(t-t^{2}\right)^{2}}=0

\\\Rightarrow 3 \mathrm{t}^{2}-3 \mathrm{t}-\left(4-11 \mathrm{t}+6 \mathrm{t}^{2}\right)=0 \\ \\\Rightarrow 3 \mathrm{t}^{2}-8 \mathrm{t}+4=0

\begin{aligned} &\Rightarrow 3 \mathrm{t}^{2}-6 \mathrm{t}-2 \mathrm{t}+4=0\\ &\Rightarrow \mathrm{t}=\frac{2}{3}\quad\text{ and }\quad\mathrm{t}\neq2\\ \end{aligned}

\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha

\frac{12}{2}+\frac{3}{3-2}=\alpha

\alpha=9

Posted by

Suraj Bhandari

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