JEE Main 2022 — Mathematics Calculus

Given,

$$f(x)={\mathrm{tan}}^{-1}(\mathrm{sin}x-\mathrm{cos}x)$$f(x)=tan−1(sinx−cosx)

Now differentiating the function we get,

$${f}^{'}(x)=\frac{\mathrm{cos}x+\mathrm{sin}x}{{(\mathrm{sin}x-\mathrm{cos}x)}^{2}+1}$$f′(x)=(sinx−cosx)2+1cosx+sinx​

Now equating ${f}^{'}(x)=0$f′(x)=0 to find critical points,

$$\frac{\mathrm{cos}x+\mathrm{sin}x}{{(\mathrm{sin}x-\mathrm{cos}x)}^{2}+1}=0$$(sinx−cosx)2+1cosx+sinx​=0

$$\Rightarrow \mathrm{cos}x+\mathrm{sin}x=0$$⇒cosx+sinx=0

$$\Rightarrow \mathrm{tan}x=-1$$⇒tanx=−1

$$\therefore x=\frac{3\pi }{4}$$∴x=43π​

Now checking the value of function at critical point and boundary point we get,

$x$x	$0$0	$\frac{3\pi }{4}$43π​	$\pi$π
$f(x)$f(x)	$-\frac{\pi }{4}$−4π​	${\mathrm{tan}}^{-1}\sqrt{2}$tan−12​	$\frac{\pi }{4}$4π​

So, we get,

$$\begin{matrix}{(f(x))}_{\mathrm{max}}={\mathrm{tan}}^{-1}\sqrt{2} \\ \therefore {(f(x))}_{\mathrm{min}}=-\frac{\pi }{4}\end{matrix}]$$(f(x))max​=tan−12​∴(f(x))min​=−4π​​]

So, the sum will be $={\mathrm{tan}}^{-1}\sqrt{2}-\frac{\pi }{4}$=tan−12​−4π​

$$={\mathrm{cos}}^{-1}\frac{1}{\sqrt{3}}-\frac{\pi }{4}$$=cos−13​1​−4π​