JEE Advanced 2016 · Paper 1, Mathematics question 46

Factorise. The coefficient is $x^2+xy+4x+2y+4=(x+2)(x+2+y)$. Put $X=x+2$:
$$\frac{dy}{dX}=\frac{y^2}{X(X+y)}.$$

Solve. With $y=vX$:
$$X\frac{dv}{dX}=-\frac{v}{1+v}\implies\ln v+v=-\ln X+C\implies\ln y+\frac yX=C.$$
Through $(1,3)$: $X=3$, $y=3$, so $C=\ln3+1$. The curve is
$$\ln y+\frac y{x+2}=1+\ln3.$$

• (A), (B) On $y=x+2$: $\ln y+1=1+\ln3$, so $y=3$ only. Exactly one point: (A) is true, (B) is false.
• (C) On $y=X^2$: $2\ln X+X$ is increasing and already exceeds $1+\ln3$ at $X=2$, so there is no intersection for $x>0$. False.
• (D) On $y=(X+1)^2$: $2\ln(X+1)+\frac{(X+1)^2}X$ also exceeds $1+\ln3$ for $X>2$. No intersection. True.

(A), (D)