JEE Main 2020 — Application Of Derivatives Question with Solution

Given equation of curve

y = (1+x)^2y + cos ²(sin^–1x)

at x = 0

$$\Rightarrow$$ y = (1 + 0)^2y + cos²(sin^–10)

$$\Rightarrow$$ y = 1 + 1

$$\Rightarrow$$ y = 2

So we have to find the normal at (0, 2)

Now, y = $$e^{2y\ln \left(1+x\right)}+{\cos }^2\left({\cos }^{-1}\sqrt{1-x^2}\right)$$

$$\Rightarrow$$ y = $$e^{2y\ln \left(1+x\right)}+{\left(\sqrt{1-x^2}\right)}^2$$

$$\Rightarrow$$ y = $$e^{2y\ln \left(1+x\right)}+\left(1-x^2\right)$$

Now differentiate w.r.t. x

y' = $$e^{2y\ln \left(1+x\right)}\left[2y.\left(\frac{1}{1+x}\right)+\ln \left(1+x\right).2y^{\prime }\right]$$ - 2x

Put x = 0 & y = 2

$$\Rightarrow$$ y' = $$e^{2y\ln \left(1+0\right)}\left[2y.\left(\frac{1}{1+0}\right)+\ln \left(1+0\right).2y^{\prime }\right]-2\times 0$$

$$\Rightarrow$$ y' = e⁰ [4 + 0] – 0

$$\Rightarrow$$ y' = 4 = slope of tangent to the curve

so slope of normal to the curve = - $$\frac{1}{4}$$

Hence equation of normal at (0, 2) is

y - 2 = - $$\frac{1}{4}$$(x - 0)

$$\Rightarrow$$ 4y – 8 = –x

$$\Rightarrow$$ x + 4y = 8