JEE Advanced 2025 — Paper 2 — Question 18

The region is defined by $x>0$, $y>1/x$, $y<(5x-1)/4$, and $y<(17-4x)/4$.

The two lines intersect at $x=2$.

The curve $y=1/x$ meets the first line at $x=1$ and the second line at $x=4$.

Thus the area is computed from $x=1$ to $x=4$ with upper bound being $(5x-1)/4$ for $1\le x\le2$ and $(17-4x)/4$ for $2\le x\le4$.

Lower bound is $y=1/x$. Area $= \int_{1}^{2}\left(\frac{5x-1}{4} - \frac{1}{x}\right)dx + \int_{2}^{4}\left(\frac{17-4x}{4} - \frac{1}{x}\right)dx$. First integral: $\frac{1}{4}\left[\frac{5x^2}{2} - x\right]_{1}^{2} - [\ln x]_{1}^{2} = \frac{13}{8} - \ln 2$. Second integral: $\frac{1}{4}\left[17x - 2x^2\right]_{2}^{4} - [\ln x]_{2}^{4} = \frac{5}{2} - \ln 2 = \frac{20}{8} - \ln 2$. Total area $\frac{13}{8} + \frac{20}{8} - 2\ln 2 = \frac{33}{8} - \ln 4$.