JEE Advanced 2019 · Paper 1, Mathematics question 11

The differential equation. The tangent at $P(x,y)$ meets the $y$-axis at $(0,\,y-xy')$. So
$$PY_P^2=x^2+x^2y'^2=1\implies y'=\pm\frac{\sqrt{1-x^2}}x.$$
The curve is in the first quadrant and falls to $(1,0)$, so $y'<0$:
$$xy'+\sqrt{1-x^2}=0.$$
(B) is true and (D) is false.

Integrate, using $\displaystyle\int\frac{\sqrt{1-x^2}}x\,dx=\sqrt{1-x^2}-\ln\frac{1+\sqrt{1-x^2}}x$:
$$y=\ln\frac{1+\sqrt{1-x^2}}x-\sqrt{1-x^2}+C.$$
$y(1)=0$ gives $C=0$, so (A) is true and (C) is false.

(A), (B)