JEE Main 2020 (Online) 5th September Evening ShiftQuestion 1
Which of the following points lies on the tangent to the curve $x^{4}e^{y}+2\sqrt{y+1}=3$ at the point $(1,0)$?
Differentiation and its Application
Solution
Concept. Differentiate implicitly, put $(1,0)$, then test the options on the tangent. Formulas used. (1) $4x^3e^y+x^4e^yy'+\dfrac{y'}{\sqrt{y+1}}=0$Step 1 - Slope. At $(1,0)$, by (1): $4+y'+y'=0$, so $y'=-2$. Step 2 - Tangent. $y=-2(x-1)$. Step 3 - Test. At $x=-2$: $y=6$. So $(-2,6)$ lies on it. Answer: (D)