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The minimum value of alpha for which the equation 4 sin... | null (Mathematics) | DivineJEE

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Mathematics

Exam

Mathematics

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 .......

Correct Answer:

9

Explanation

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 }} \text{ 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

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