JEE Main 2021 — Application of Derivatives Question with Solution

Let

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

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

$\Rightarrow f\left(x\right)=4\mathrm{cosec}x+{\sec }^{2}x+\tan x\sec x$

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

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

$\Rightarrow f^{\text{'}\text{'}}\left(x\right)=-4\left[\frac{-{\sin }^{3}x-2\sin x{\cos }^{2}x}{{\sin }^{4}x}\right]+\frac{2\left(\sin x+1\right){\cos }^{4}x+3{\cos }^{2}x\sin x{\left(\sin x+1\right)}^2}{{\cos }^{6}x}$

$\Rightarrow f^{\text{'}\text{'}}\left(x\right)=-4\left[\frac{-{\sin }^{3}x-2\sin x\left(1-{\sin }^{2}x\right)}{{\sin }^{4}x}\right]+\frac{2\left(\sin x+1\right){\left(1-{\sin }^{2}x\right)}^2+3\left(1-{\sin }^{2}x\right)\sin x{\left(\sin x+1\right)}^2}{{\left(1-{\sin }^{2}x\right)}^3}$

For critical points,

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

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

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

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

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

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

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

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

$\Rightarrow \sin x=\frac{2}{3}$

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

Now, at $\sin x=\frac{2}{3}$, we get

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

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

Hence, it is the point of minima.

$\therefore f{\left(x\right)}_{\min }=\frac{4}{{\displaystyle \frac{2}{3}}}+\frac{1}{1-{\displaystyle \frac{2}{3}}}=9$

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

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

$\therefore {\alpha }_{\min }=9$