200 days streak special 4

Let $f(x) = \dfrac{4}{\sin x} + \dfrac{1}{1-\sin x}$ . We note that $f(x) > 0$ for $x \in \left( 0, \frac{\pi}{2} \right)$ . For $f(x)$ to have at least one root, $a = \min (f(x))$ , the minimum of $f(x)$ .

$\begin{aligned} f'(x) & = -\frac{4\cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2} \\ & = \frac{\cos x(-4+8\sin x - 4\sin^2 + \sin^2 x)}{\sin^2 x(1-\sin x)^2} \\ & = - \frac{\cos x(3\sin^2 x - 8 \sin x + 4)}{\sin^2 x(1-\sin x)^2} \\ & = - \frac{\cos x(3\sin x - 2)(\sin x -2)}{\sin^2 x(1-\sin x)^2} \\ \text{For } f'(x) & = 0 \\ & \Rightarrow \begin{cases} \cos x = 0 & \Rightarrow \sin x = 1 \quad \Rightarrow f(x) \to \infty \\ \sin x = \frac{2}{3} & \Rightarrow \dfrac{4}{\frac{2}{3}} + \dfrac{1}{1-\frac{2}{3}} = 6+3 = 9 \\ \sin x = \frac{1}{2} & \Rightarrow \dfrac{4}{\frac{1}{2}} + \dfrac{1}{1-\frac{1}{2}} =8 + 2 = 10 \end{cases} \\ & \\ \Rightarrow a & = \min (f(x)) =\boxed{9} \end{aligned}$

Let $\color{#3D99F6}{\sin x=k}$ . The quadratic can be simplified to $\color{#3D99F6}{a{k}^2-(3+a)k=4=0}$ . To find the minimum value of $\color{#3D99F6}{a}$ , let the roots of the equation be equal(why?). Hence we get the condition $\color{#3D99F6}{{(3+a)}^2=16a}$ , which gives $\color{#3D99F6}{a=1,9}$ . Since we have considered roots to be equal, the vertex of the parabola represented by the quadratic expression should lie between $\color{#3D99F6}{0}$ and $\color{#3D99F6}{\frac{\pi}{2}}$ . Hence $\color{#3D99F6}{0<\frac{3+a}{2a}<\frac{\pi}{2}}$ or $\color{#3D99F6}{a>\frac{3}{\frac{\pi}{2}-1}}$ . Hence the only value satisfying the condition is $\huge\color{#3D99F6}{\boxed{9}}$ .