The minimum values of a for which the equation 4 s i n x + 1 1 − s i n x = α has at least one solution in ( 0 , π 2 ) i s _ _ _ _ _ .

4 s i n x + 1 1 − s i n x = a ∀ x ∈ ( 0 , π 2 ) Let sin x = t, t ∈ (0, 1) g ( t ) = 4 t + 1 1 − t g ' ( t ) = 0 ⇒ t = 2 3 g ' ' ( 2 3 ) > 0 ∴ g ( t ) m i n i m u m = 4 2 / 3 + 1 1 − 2 / 3 = 9 Minimum value of a for which solution exist = 9

The minimum values of a for which the equation $\frac{4}{s i n x} + \frac{1}{1 - s i n x} = α$ has at least one solution in $(0, \frac{π}{2}) i s_____.$

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Alok Kumar Singh | Contributor-Level 10

7 months ago

$\frac{4}{s i n x} + \frac{1}{1 - s i n x} = a \forall x \in (0, \frac{π}{2})$

Let sin x = t, t $\in$ (0, 1)

$g (t) = \frac{4}{t} + \frac{1}{1 - t}$

$g' (t) = 0 \Rightarrow t = \frac{2}{3}$

$g'' (\frac{2}{3}) > 0$

$∴ g {(t)}_{m i n i m u m} = \frac{4}{2 / 3} + \frac{1}{1 - 2 / 3} = 9$

Minimum value of a for which solution exist = 9

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