# if

Question:

If $e^{y}+x y=e$, the ordered pair $\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)$ at $x=0$ is

equal to :

1. (1) $\left(\frac{1}{e},-\frac{1}{e^{2}}\right)$

2. (2) $\left(-\frac{1}{e}, \frac{1}{e^{2}}\right)$

3. (3) $\left(\frac{1}{e}, \frac{1}{e^{2}}\right)$

4. (4) $\left(-\frac{1}{e},-\frac{1}{e^{2}}\right)$

Correct Option: , 2

Solution:

Given, $e^{y}+x y=e$ ........$\ldots$ (i)

Putting $x=0$ in (i), $\Rightarrow e^{y}=e \Rightarrow y=1$

On differentiating (i) w. r. to $x$

$e^{y} \frac{d y}{d x}+x \frac{d y}{d x}+y=0$.....(ii)

Putting $y=1$ and $x=0$ in (ii),

$e \frac{d y}{d x}+0+1=0 \Rightarrow \frac{d y}{d x}=-\frac{1}{e}$

On differentiating (ii) w. r. to $x$,

$e^{y} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \cdot e^{y} \cdot \frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{d y}{d x}=0$..(iii)

Putting $y=1, x=0$ and $\frac{d y}{d x}=-\frac{1}{e}$ in (iii),

$e \frac{d^{2} y}{d x^{2}}+\frac{1}{e}-\frac{2}{e}=0 \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{1}{e^{2}}$

Hence, $\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right) \equiv\left(-\frac{1}{e}, \frac{1}{e^{2}}\right)$