find-the-least-value-of-such-that-4-sin-x-1-1-sin-x-has-at-least-one-solution-in-0-2-

Question Number 144833 by gsk2684 last updated on 29/Jun/21

find the least value of α such that \frac{4}{\sin x} + \frac{1}{1 - \sin x} = α

has at least one solution in (0 \frac{Π}{2})

Answered by liberty last updated on 29/Jun/21

α = f (x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}

let u = \sin x \Rightarrow f (u) = {4 u}^{- 1} + {(1 - u)}^{- 1}

f^{'} (u) = - \frac{4}{u^{2}} + \frac{1}{{(1 - u)}^{2}} = 0

\Rightarrow u^{2} - 4 (u^{2} - 2 u + 1) = 0

\Rightarrow - {3 u}^{2} + 8 u - 4 = 0

\Rightarrow {3 u}^{2} - 8 u + 4 = 0

\Rightarrow (3 u - 2) (u - 2) = 0

\Rightarrow u = \frac{2}{3}; minimum value

of α = \frac{4}{\sin x} + \frac{1}{1 - \sin x} = 6 + 3 = 9

Commented by gsk2684 last updated on 29/Jun/21

thank you

kindly explain why did not

find second derivative for verifying

whether it is positive or negative ?

Commented by liberty last updated on 30/Jun/21

i used first test derivative

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