JEE PYQ: Differentiation Question 15
Question 15 - 2019 (12 Apr Shift 1)
If $e^y + xy = e$, the ordered pair $\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)$ at $x = 0$ is equal to:
(1) $\left(\frac{1}{e}, -\frac{1}{e^2}\right)$ (2) $\left(-\frac{1}{e}, \frac{1}{e^2}\right)$ (3) $\left(\frac{1}{e}, \frac{1}{e^2}\right)$ (4) $\left(-\frac{1}{e}, -e^2\right)$
Show Answer
Answer: (2) $\left(-\frac{1}{e}, \frac{1}{e^2}\right)$
Solution
At $x = 0$: $y = 1$. $\frac{dy}{dx} = -\frac{1}{e}$, $\frac{d^2y}{dx^2} = \frac{1}{e^2}$.
BETA
