JEE Advanced 2025 — Paper 2 — MATH-I solution

JEE Advanced2025Paper 2MATH-I

Let $R$ denote the set of all real numbers. Then the area of the region

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

A
$\frac{17}{16} - lo g_{e} 4$
B
$\frac{33}{8} - lo g_{e} 4$
C
$\frac{57}{8} - lo g_{e} 4$
D
$\frac{17}{2} - lo g_{e} 4$

Solution

Step 1: Find the corner points.

The hyperbola $y = 1/ x$ meets $5 x - 4 y = 1$ at $x = 1$ , $y = 1$ , and meets $4 x + 4 y = 17$ at $x = 4$ , $y = 1/4$ . The two lines meet at $P (2, \frac{9}{4})$ .

Step 2: Decompose the region.

The region is bounded above by the two straight lines (meeting at $P$ ) and below by $y = 1/ x$ , from $x = 1$ to $x = 4$ . Split it into two trapezia using the vertical through $P$ , then subtract the area under $y = 1/ x$ from $x = 1$ to $x = 4$ .

Step 3: Compute the trapezoidal pieces.

Trapezium on $[1, 2]$ (between $5 x - 4 y = 1$ and $4 x + 4 y = 17$ above the chord at $y = 1$ ): area $= \frac{1}{2} (1 + \frac{9}{4}) = \frac{13}{8}$ .

Trapezium on $[2, 4]$ (between $4 x + 4 y = 17$ above and $y = 1/4$ baseline): area $= \frac{1}{2} (\frac{9}{4} + \frac{1}{4}) \cdot 2 = \frac{20}{8}$ .

Step 4: Subtract the area under $y = 1/ x$ .

$\int_{1}^{4} \frac{d x}{x} = lo g_{e} 4$ .

Step 5: Combine.

Total area $= \frac{13}{8} + \frac{20}{8} - lo g_{e} 4 = \frac{33}{8} - lo g_{e} 4$ .

$∴$ The correct option is (B).

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