Let $\mathrm{f}:(1, \infty) \rightarrow \mathbf{R}$ be function defined as $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-1}{\mathrm{x}+1}$. Let $\mathrm{f}^{\mathrm{i}+1}(\mathrm{x})=\mathrm{f}\left(\mathrm{f}^{\mathrm{i}}(\mathrm{x})\right)$, $\mathrm{i}=1,2, \ldots, 25$, where $\mathrm{f}^{1}(\mathrm{x})=(\mathrm{x})$. If $\mathrm{g}(\mathrm{x})+\mathrm{f}^{26}(\mathrm{x})=0$, $x \in(1, \infty)$, then the area of the region bounded by the curves $\mathrm{y}=\mathrm{g}(\mathrm{x}), 2 \mathrm{y}=2 \mathrm{x}-3, \mathrm{y}=0$ and $\mathrm{x}=4$ is:

(A) $\frac{1}{8}+\log _{\mathrm{e}} 2$

(B) $\frac{1}{4}+\log _{\mathrm{e}} 2$

(C) $\frac{5}{6}+3 \log _{e} 2$

(D) $\frac{5}{6}+\log _{e} 2$

Area Under Curves — JEE Main · 2026 · 08 Apr 2026 · Shift 2 Question with Solution

Let

f : (1, \infty) \to R

be function defined as

f (x) = \frac{x - 1}{x + 1}

. Let

f^{i + 1} (x) = f (f^{i} (x))

i = 1, 2, \dots, 25

, where

f^{1} (x) = (x)

. If

g (x) + f^{26} (x) = 0

x \in (1, \infty)

, then the area of the region bounded by the curves

y = g (x), 2 y = 2 x - 3, y = 0

and

x = 4

is:

Answer: (A)

Iterating

f (x) = \frac{x - 1}{x + 1}

gives

f^{2} (x) = - \frac{1}{x}

f^{3} (x) = \frac{1 + x}{1 - x}

and

f^{4} (x) = x

, so the iterates repeat with period

4

. Since

26 \equiv 2 (mod 4)

f^{26} (x) = f^{2} (x) = - \frac{1}{x}

, hence

g (x) = - f^{26} (x) = \frac{1}{x}

. The region is bounded by

y = \frac{1}{x}

, the line

y = x - \frac{3}{2}

, the

x

-axis and

x = 4

. The line meets the

x

-axis at

x = \frac{3}{2}

and meets

y = \frac{1}{x}

where

\frac{1}{x} = x - \frac{3}{2}

, i.e.

2 x^{2} - 3 x - 2 = 0

, giving

x = 2

. The area is the triangle under the line from

x = \frac{3}{2}

x = 2

plus the region under

y = \frac{1}{x}

from

x = 2

x = 4

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} + \int_{2}^{4} \frac{d x}{x} = \frac{1}{8} + ln 2.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option