JEE Main 2019 — Mathematics Calculus

$$2y={({\mathrm{cot}}^{-1}\frac{\sqrt{3}\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x-\sqrt{3}\mathrm{sin}x})}^{2}$$2y=(cot−1cosx−3​sinx3​cosx+sinx​)2

$$={({\mathrm{cot}}^{-1}(\frac{\sqrt{3}+\mathrm{tan}x}{1-\sqrt{3}\mathrm{tan}x}))}^{2}$$=(cot−1(1−3​tanx3​+tanx​))2

$$={({\mathrm{cot}}^{-1}\mathrm{tan}(\frac{\pi }{3}+x))}^{2}$$=(cot−1tan(3π​+x))2

$$={({\mathrm{cot}}^{-1}\mathrm{cot}(\frac{\pi }{2}-(\frac{\pi }{3}+x)) )}^{2}$$=(cot−1cot(2π​−(3π​+x)))2

\Rightarrow 2y={\begin{matrix}{(\frac{\pi }{6}-x)}^{2}, x\in (0,\frac{\pi }{6}) \\ {(\pi +\frac{\pi }{6}-x)}^{2}, x\in (\frac{\pi }{6},\frac{\pi }{2})\end{matrix}

$$\Rightarrow \frac{2dy}{dx}=2(\frac{\pi }{6}-x).(-1)\Rightarrow \frac{dy}{dx}=x-\frac{\pi }{6}, x\in (0,\frac{\pi }{6})$$⇒dx2dy​=2(6π​−x).(−1)⇒dxdy​=x−6π​,x∈(0,6π​)

And $\frac{dy}{dx}=x-\frac{7\pi }{6}, x\in (\frac{\pi }{6},\frac{\pi }{2})$dxdy​=x−67π​,x∈(6π​,2π​)

Left Hand and Right Hand Derivatives are not same so function is non derivable at $x=\frac{\pi }{6}$x=6π​.

Hence, $\frac{dy}{dx}$dxdy​ does not exist for all values in the given interval.