The minimum value of α for which the equation (4)/( x) + (1)/(1 -...

Let

$f\left(x\right) = \dfrac{4}{\sin x} + \dfrac{1}{1 - \sin x}$

$⇒ f\left(x\right) = \dfrac{4}{\sin x} + \dfrac{1 + \sin x}{1 - \sin^{2}x}$

$⇒ f\left(x\right) = 4cosecx + \sec^{2}x + \tan x \sec x$

$⇒ {f}^{'}\left(x\right) = \dfrac{- 4\cos x}{\sin^{2}x} + \dfrac{2\sin x + 1}{\cos^{3}x} + \dfrac{\sin^{2}x}{\cos^{3}x}$

$⇒ {f}^{'}\left(x\right) = \dfrac{- 4\cos x}{\sin^{2}x} + \dfrac{{\left(\sin x + 1\right)}^{2}}{\cos^{3}x}$

⇒ {f}^{' '}\left(x\right) = - 4\left[\dfrac{- \sin^{3}x - 2\sin x\cos^{2}x}{\sin^{4}x}\right] + \dfrac{2\left(\sin x + 1\right)\cos^{4}x + 3\cos^{2}xsinx{\left(\sin x + 1\right)}^{2}}{\cos^{6}x}

⇒ {f}^{' '}\left(x\right) = - 4\left[\dfrac{- \sin^{3}x - 2\sin x\left(1 - \sin^{2}x\right)}{\sin^{4}x}\right] + \dfrac{2\left(\sin x + 1\right){\left(1 - \sin^{2}x\right)}^{2} + 3\left(1 - \sin^{2}x\right)\sin x{\left(\sin x + 1\right)}^{2}}{{\left(1 - \sin^{2}x\right)}^{3}}

For critical points,

${f}^{'}\left(x\right) = 0$

$⇒ \dfrac{{\left(\sin x + 1\right)}^{2}}{\cos^{3}x} = \dfrac{4\cos x}{\sin^{2}x}$

$⇒ \sin^{2}x{\left(\sin x + 1\right)}^{2} = 4\cos^{4}x$

$⇒ \sin x\left(\sin x + 1\right) = 2\cos^{2}x$

$⇒ \sin^{2}x + \sin x = 2 - 2\sin^{2}x$

$⇒ 3\sin^{2}x + \sin x - 2 = 0$

$⇒ 3\sin^{2}x + 3\sin x - 2\sin x - 2 = 0$

$⇒ \left(3\sin x - 2\right)\left(\sin x + 1\right) = 0$

$⇒ \sin x = \dfrac{2}{3}$

Since, $\sin x > 0 \forall x \in \left(0, \dfrac{\pi}{2}\right)$

Now, at $\sin x = \dfrac{2}{3}$, we get

⇒ {f}^{' '}\left(x\right) = - 4\left[\dfrac{- \dfrac{8}{27} - \dfrac{4}{3}\left(1 - \dfrac{4}{9}\right)}{\dfrac{16}{81}}\right] + \dfrac{2 \times \dfrac{5}{3}{\left(1 - \dfrac{4}{9}\right)}^{2} + 3\left(1 - \dfrac{4}{9}\right) \times \dfrac{2}{3} \times \dfrac{25}{9}}{{\left(1 - \dfrac{4}{9}\right)}^{3}}

⇒ {f}^{' '}\left(x\right) = \left[4\left[\dfrac{28}{27} \times \dfrac{81}{16}\right] + \dfrac{2 \times \dfrac{5}{3}{\left(\dfrac{5}{9}\right)}^{2} + 3\left(\dfrac{5}{9}\right) \times \dfrac{2}{3} \times \dfrac{25}{9}}{{\left(\dfrac{5}{9}\right)}^{3}}\right] > 0

Hence, it is the point of minima.

∴ f{\left(x\right)}_\min = \dfrac{4}{\dfrac{2}{3}} + \dfrac{1}{1 - \dfrac{2}{3}} = 9

f{\left(x\right)}_\max \to \infty

$f\left(x\right)$ is continuous function

∴ {\alpha}_\min = 9