JEE Main 2021 (Online) 4th August Morning ShiftQuestion 1
The equation of the normal to the curve $y=e^{-|x|}$ at a point, where the curve cuts the line $x=-1$ is
Differentiation and its Application
Solution
Concept. For $x<0$, $y=e^{x}$; find the slope there and use the normal slope $-\frac{1}{y'}$. Formulas used. (1) Normal at $(x_1,y_1)$: $y-y_1=-\dfrac{1}{y'(x_1)}(x-x_1)$Step 1 - Point and slope. At $x=-1$: $y=e^{-1}$, $y'=e^{-1}$. So the normal slope is $-e$. Step 2 - Normal. By (1), $$y-\frac1e=-e(x+1).$$Step 3 - Rearrange. Multiply by $e$: $ey-1=-e^2x-e^2$, i.e. $e^2x+ey+(e^2-1)=0$. Answer: (C)