"The minimum value of $\alpha$ for which the equation ${4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha$ has at least one solution in $\left( {0,{\pi \over 2}} \right)$ is ......."

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Accepted Answer

"$f(x) = {4 \over {\sin x}} + {1 \over {1 - \sin x}}$

Let sinx = t $\because$ $x \in \left( {0,{\pi \over 2}} \right) \Rightarrow 0 < t < 1$

$f(t) = {4 \over t} + {1 \over {1 - t}}$

$f'(t) = {{ - 4} \over {{t^2}}} + {1 \over {{{(1 - t)}^2}}}$

$= {{{t^2} - 4{{(1 - t)}^2}} \over {{t^2}{{(1 - t)}^2}}}$

$= {{(t - 2(1 - t))(t + 2(1 - t))} \over {{t^2}{{(1 - t)}^2}}}$

$= {{(3t - 2)(2 - t)} \over {{t^2}{{(1 - t)}^2}}}$

${f_{\min }}$ at $t = {2 \over 3}$

$${\alpha _{\min }} = f\left( {{2 \over 3}} \right) = {4 \over {{2 \over 3}}} + {1 \over {1 - {2 \over 3}}}$$

$= 6 + 3$

$= 9$"

The minimum value of $\alpha$ for which the
equation ${4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha$ has at least one solution in $\left( {0,{\pi \over 2}} \right)$ is .......

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The minimum value of $\alpha$ for which the equation ${4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha$ has at least one solution in $\left( {0,{\pi \over 2}} \right)$ is .......

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The minimum value of $\alpha$ for which the
equation ${4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha$ has at least one solution in $\left( {0,{\pi \over 2}} \right)$ is .......