JEE Main 2018 — Mathematics Calculus

Given, $x^2+y^2+\sin y=4$x2+y2+siny=4 After differentiating the above equation $w$w. r. t. $x$x we get $$\begin{aligned} &2 x+2 y \frac{d y}{d x}+\cos y \frac{d y}{d x}=0 \ldots(1) \\ &\Rightarrow 2 x+(2 y+\cos y) \frac{d y}{d x}=0 \\ &\Rightarrow \frac{d y}{d x}=\frac{-2 x}{2 y+\cos y} \\ &\text { At }(-2,0),\left(\frac{d y}{d x}\right)_{(-2,0)}=\frac{-2 \times-2}{2 \times 0+\cos 0} \\ &\Rightarrow\left(\frac{d y}{d x}\right)_{(-2,0)}=\frac{4}{0+1} \\ &\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=4 \ldots(2) \end{aligned}$$​2x+2ydxdy​+cosydxdy​=0…(1)⇒2x+(2y+cosy)dxdy​=0⇒dxdy​=2y+cosy−2x​ At (−2,0),(dxdy​)(−2,0)​=2×0+cos0−2×−2​⇒(dxdy​)(−2,0)​=0+14​⇒(dxdy​)(2,0)​=4…(2)​ Again differentiating equation (1) w. r. t to $x$x, we get $2+2\left(\frac{d y}{d x}\right)^2+2 y \frac{d^2 y}{d x^2}-\sin y\left(\frac{d y}{d x}\right)^2+\cos \mathrm{y} \frac{d^2 y}{d x^2}=0$2+2(dxdy​)2+2ydx2d2y​−siny(dxdy​)2+cosydx2d2y​=0 $\Rightarrow 2+(-2 \sin y)\left(\frac{d y}{d x}\right)^2+(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=0$⇒2+(−2siny)(dxdy​)2+(2y+cosy)dx2d2y​=0 $\Rightarrow(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=-2-(2-\sin y)\left(\frac{d y}{d x}\right)^2$⇒(2y+cosy)dx2d2y​=−2−(2−siny)(dxdy​)2 $\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-(2-\sin y)\left(\frac{d y}{d x}\right)^2}{2 y+\cos \mathrm{y}}$⇒dx2d2y​=2y+cosy−2−(2−siny)(dxdy​)2​ So, at $(-2,0)$(−2,0) $\frac{d^2 y}{d x^2}=\frac{-2-(2-0) \times 4^2}{2 \times 0+1}$dx2d2y​=2×0+1−2−(2−0)×42​ $\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-2 \times 16}{1}$⇒dx2d2y​=1−2−2×16​ $\Rightarrow \frac{d^2 y}{d x^2}=-34$⇒dx2d2y​=−34