JEE Main 2019 — Mathematics Calculus

$${e}^{y}+xy=e$$ey+xy=e

differentiate w. r. t. $x$x

$${e}^{y}\frac{dy}{dx}+x\frac{dy}{dx}+y=0$$eydxdy​+xdxdy​+y=0

$$\Rightarrow \frac{dy}{dx}(x+{e}^{y})=-y, {\frac{dy}{dx}|}_{(0,1)}=-\frac{1}{e}$$⇒dxdy​(x+ey)=−y,dxdy​∣(0,1)​=−e1​

Again differentiate w. r. t. $x$x

$${e}^{y}.\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}.{e}^{y}.\frac{dy}{dx}+x.\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}+\frac{dy}{dx}=0$$ey.dx2d2y​+dxdy​.ey.dxdy​+x.dx2d2y​+dxdy​+dxdy​=0

$$\Rightarrow (x+{e}^{y})\frac{{d}^{2}y}{d{x}^{2}}+{(\frac{dy}{dx})}^{2}.{e}^{y}+2\frac{dy}{dx}=0$$⇒(x+ey)dx2d2y​+(dxdy​)2.ey+2dxdy​=0

Now, at $(0,1)$(0,1)

$$e\frac{{d}^{2}y}{d{x}^{2}}+\frac{1}{{e}^{2}}e+2(-\frac{1}{e})=0$$edx2d2y​+e21​e+2(−e1​)=0

$$\therefore \frac{{d}^{2}y}{d{x}^{2}}=\frac{1}{{e}^{2}}$$∴dx2d2y​=e21​

Hence, ordered pair $(\frac{dy}{dx},\frac{{d}^{2}y}{d{x}^{2}})=(-\frac{1}{e},\frac{1}{{e}^{2}})$(dxdy​,dx2d2y​)=(−e1​,e21​)