JEE MAIN - Mathematics (2021 - 24th February Morning Shift - No. 15)
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 .......
equation $${4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha $$ has at least one solution in $$\left( {0,{\pi \over 2}} \right)$$ is .......
Answer
9
Explanation
$$f(x) = {4 \over {\sin x}} + {1 \over {1 - \sin x}}$$
Let sinx = t $$ \because $$ $$x \in \left( {0,{\pi \over 2}} \right) \Rightarrow 0 < t < 1$$
$$f(t) = {4 \over t} + {1 \over {1 - t}}$$
$$f'(t) = {{ - 4} \over {{t^2}}} + {1 \over {{{(1 - t)}^2}}}$$
$$ = {{{t^2} - 4{{(1 - t)}^2}} \over {{t^2}{{(1 - t)}^2}}}$$
$$ = {{(t - 2(1 - t))(t + 2(1 - t))} \over {{t^2}{{(1 - t)}^2}}}$$
$$ = {{(3t - 2)(2 - t)} \over {{t^2}{{(1 - t)}^2}}}$$
$${f_{\min }}$$ at $$t = {2 \over 3}$$
$${\alpha _{\min }} = f\left( {{2 \over 3}} \right) = {4 \over {{2 \over 3}}} + {1 \over {1 - {2 \over 3}}}$$
$$ = 6 + 3$$
$$ = 9$$
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 $$t = {2 \over 3}$$
$${\alpha _{\min }} = f\left( {{2 \over 3}} \right) = {4 \over {{2 \over 3}}} + {1 \over {1 - {2 \over 3}}}$$
$$ = 6 + 3$$
$$ = 9$$
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