JEE Advanced 2021 Paper 2 Mathematics: Paragraph 2

The tangent condition: at $T(a,\ln a)$ the slope is $\frac1a$ and the line through the origin requires $\ln a=a\cdot\frac1a=1$, so $a=e$ and $T=(e,1)$, consistent with the line $y=x/e$. Region $R$ lies between the line (above) and the curve from $x=1$ to $x=e$, together with the triangle under the line from $x=0$ to $x=1$ bounded below by the axis. Equivalently, area $=$ (triangle under the line on $[0,e]$) $-\int_1^e\ln x\,dx=\frac12\cdot e\cdot1-\left[x\ln x-x\right]_1^e=\frac e2-1\approx0.359$. The other options come from omitting the subtraction or mis-evaluating the integral.

Rotating about the $x$-axis, use washers/discs: the line sweeps a cone from $x=0$ to $x=e$ of volume $\pi\int_0^e\frac{x^2}{e^2}dx=\frac{\pi e}{3}$, and the portion of the curve from $x=1$ to $x=e$ removes $\pi\int_1^e(\ln x)^2dx=\pi\left[x\ln^2x-2x\ln x+2x\right]_1^e=\pi(e-2)$. Hence $V=\frac{\pi e}{3}-\pi(e-2)=\pi\left(2-\frac{2e}{3}\right)=\frac{2\pi}{3}(3-e)\approx0.590$. The first option forgets to subtract the removed solid, and the second subtracts but forgets the cone.