JEE Advanced 2025 Paper 2 Q02 Mathematics Integration & Differential Equations Area Under Curves Hard

JEE Advanced 2025 Paper 2 · Q02 · Area Under Curves

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

  1. A. $\frac{17}{16} - \log_e 4$
  2. B. $\frac{33}{8} - \log_e 4$
  3. C. $\frac{57}{8} - \log_e 4$
  4. D. $\frac{17}{2} - \log_e 4$
Reveal answer + step-by-step solution

Correct answer:B

Solution

The region is bounded below by $y = 1/x$ and above by $y = (5x-1)/4$ for one stretch, then by $y = (17-4x)/4$. Find $x$ where $(5x-1)/4 = 1/x$: $5x^2 - x - 4 = 0$ $\to$ $x = 1$. Where $(5x-1)/4 = (17-4x)/4$: $x = 2$. Where $1/x = (17-4x)/4$: $4x^2 - 17x + 4 = 0$ $\to$ $x = 4$ (or $1/4$). Hence area $A = \int_1^2 [(5x-1)/4 - 1/x]\, dx + \int_2^4 [(17-4x)/4 - 1/x]\, dx$. First integral: $[5x^2/8 - x/4 - \ln x]_1^2 = (5/2 - 1/2 - \ln 2) - (5/8 - 1/4) = 13/8 - \ln 2$. Second: $[(17x - 2x^2)/4 - \ln x]_2^4 = (9 - \ln 4) - (13/2 - \ln 2) = 5/2 - \ln 2$. Total $= 13/8 + 5/2 - 2 \ln 2 = 33/8 - \ln 4$. Answer (B).

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