JEE Main 2020 (Online) 2nd September Evening ShiftQuestion 1
The equation of the normal to the curve $y=(1+x)^{2y}+\cos^2(\sin^{-1}x)$ at $x=0$ is :
Differentiation and its Application
Solution
Concept. Simplify $\cos^2\left(\sin^{-1}x\right)=1-x^2$, find $y(0)$, then differentiate $(1+x)^{2y}$ with logarithms. Formulas used. (1) $\dfrac{d}{dx}(1+x)^{2y}=(1+x)^{2y}\left[2y'\ln(1+x)+\dfrac{2y}{1+x}\right]$Step 1 - Point. At $x=0$: $y=1+1=2$. Step 2 - Slope. $y=(1+x)^{2y}+1-x^2$. By (1), at $x=0$: $$y'(0)=1\cdot(0+4)-0=4.$$ Step 3 - Normal. Slope $-\frac14$: $y-2=-\frac x4$, i.e. $x+4y=8$. Answer: (A)