JEE Advanced 2021 Paper 1 Mathematics: Paragraph 1

The tangent at $P(x,y)$ has y-intercept $y-xy'$. Setting this equal to $-x^2$ gives $y'-\dfrac yx=x$. The integrating factor is $\dfrac1x$, so $\dfrac{d}{dx}\left(\dfrac yx\right)=1$ and $y=x^2+Cx$. Using $f(1)=3$ gives $C=2$, hence $f(x)=x^2+2x$. This is a parabola cutting the $x$-axis at $x=-2$ and $x=0$ and lying below the axis between them. The bounded area is $\displaystyle\int_{-2}^{0}-(x^2+2x)\,dx=\left[-\frac{x^3}{3}-x^2\right]_{-2}^{0}=-\left(\frac{8}{3}-4\right)=\frac43\approx1.33$.

From the differential equation, $f(x)=x^2+2x$ with $f(1)=3$ and $f'(x)=2x+2$, so $f'(1)=4$. The tangent at $(1,3)$ is $y=4x-1$. The tangent meets the $y$-axis at $(0,-1)$ while the curve meets it at the origin, so the region lies between $x=0$ and $x=1$ with the curve above the tangent (their difference is $x^2-2x+1=(x-1)^2\ge0$). The area is $\displaystyle\int_0^1(x-1)^2dx=\frac13\approx0.33$. As a check, the tangent to a convex curve always lies below it, which agrees with the sign of the integrand.