Let $\mathbb{R}$ denote the set of all real numbers. Then the area of the region \[ \left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x>0, y>\frac{1}{x}, 5 x-4 y-1>0,4 x+4 y-17<0\right\} \] is

(B) $\frac{33}{8}-\log _{e} 4$

(A) $\frac{17}{16}-\log _{e} 4$

(C) $\frac{57}{8}-\log _{e} 4$

(D) $\frac{17}{2}-\log _{e} 4$

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Area Under Curves

Let

R

denote the set of all real numbers. Then the area of the region

{(x, y) \in R \times R : x > 0, y > \frac{1}{x}, 5 x - 4 y - 1 > 0, 4 x + 4 y - 17 < 0}

Answer: (B)

The region is bounded by

y = \frac{1}{x}

y = \frac{5 x - 1}{4}

, and

y = \frac{17 - 4 x}{4}

, with

x > 0

.

Find intersection points:

$\frac{1}{x} = \frac{5 x - 1}{4}$ gives $5 x^{2} - x - 4 = 0$ , solving yields $x = 1$ (taking positive root).
$\frac{5 x - 1}{4} = \frac{17 - 4 x}{4}$ gives $x = 2$ .
$\frac{1}{x} = \frac{17 - 4 x}{4}$ gives $4 x^{2} - 17 x + 4 = 0$ , solving yields $x = 4$ .

The area consists of two parts:

From $x = 1$ to $x = 2$ : area between $y = \frac{5 x - 1}{4}$ and $y = \frac{1}{x}$ .
From $x = 2$ to $x = 4$ : area between $y = \frac{17 - 4 x}{4}$ and $y = \frac{1}{x}$ .

Computing:

Area = \int_{1}^{2} (\frac{5 x - 1}{4} - \frac{1}{x}) d x + \int_{2}^{4} (\frac{17 - 4 x}{4} - \frac{1}{x}) d x

Evaluating these integrals gives

\frac{13}{8} + \frac{20}{8} - ln 4 = \frac{33}{8} - lo g_{e} 4

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option