JEE Main 2020 — Differentiation Question with Solution

Given ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1

= (sin(tan^–1x) + sin($$\frac{\pi }{2}$$ - tan^–1x))² – 1

= (sin(tan^–1x) + cos(tan^–1x))² – 1

= sin²(tan^–1x) + cos²(tan^–1x) + 2sin(tan^–1x)cos(tan^–1x) + 1

= 1 + sin(2tan^–1x) - 1

= sin(2tan^–1x)

Also given $$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left({\sin }^{-1}\left(f\left(x\right)\right)\right)$$

Integrating both sides we get

y = $$\frac{1}{2}$$ sin^-1 (f(x)) + C

= $$\frac{1}{2}$$ sin^-1 (sin(2tan^–1x)) + C

Given $$y\left(\sqrt{3}\right)=\frac{\pi }{6}$$ mean x = $$\sqrt{3}$$ and y = $$\frac{\pi }{6}$$

$$\therefore$$ $$\frac{\pi }{6}$$ = $$\frac{1}{2}$$ sin^-1 (sin(2tan^–1$$\sqrt{3}$$)) + C

$$\Rightarrow$$ $$\frac{\pi }{6}$$ = $$\frac{1}{2}$$ sin^-1 (sin(2$$\times$$$$\frac{\pi }{3}$$)) + C

$$\Rightarrow$$ $$\frac{\pi }{6}$$ = $$\frac{1}{2}$$ sin^-1 ($$\frac{\sqrt{3}}{2}$$) + C

$$\Rightarrow$$ $$\frac{\pi }{6}$$ = $$\frac{1}{2}$$ $$\times$$ $$\frac{\pi }{3}$$ + C

$$\Rightarrow$$ C = 0

Now y($$-\sqrt{3}$$) means when x = $$-\sqrt{3}$$ then find y.

y = $$\frac{1}{2}$$ sin^-1 (sin(2tan^–1x))

= $$\frac{1}{2}$$ sin^-1 (sin(2tan^–1($$-\sqrt{3}$$)))

= $$\frac{1}{2}$$ sin^-1 (sin(-2tan^–1($$\sqrt{3}$$)))

= $$\frac{1}{2}$$ sin^-1 (sin(-2$$\times$$$$\frac{\pi }{3}$$))

= $$\frac{1}{2}$$ sin^-1 (-sin(2$$\times$$$$\frac{\pi }{3}$$))

= $$\frac{1}{2}$$ sin^-1 (-$$\frac{\sqrt{3}}{2}$$)

= $$\frac{1}{2}$$ $$\times$$ -sin^-1 ($$\frac{\sqrt{3}}{2}$$)

= $$\frac{1}{2}$$ $$\times$$ -$$\frac{\pi }{3}$$

= -$$\frac{\pi }{6}$$