JEE Main 2026 Apr 8 Shift 2, Mathematics Q18: Area as Definite Integral

JEE Main2026Apr 8, Shift 2Mathematics

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 $f^{1} (x) = f (x)$ . 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:

A
$\frac{1}{8} + lo g_{e} 2$
B
$\frac{1}{4} + lo g_{e} 2$
C
$\frac{5}{6} + 3 lo g_{e} 2$
D
$\frac{5}{6} + lo g_{e} 2$

Solution

Iterate $f$ : Direct computation gives

$f^{2} (x) = \frac{( x - 1 ) / ( x + 1 ) - 1}{( x - 1 ) / ( x + 1 ) + 1} = - \frac{1}{x}$ , $f^{3} (x) = \frac{- 1/ x - 1}{- 1/ x + 1} = \frac{1 + x}{1 - x}$ , $f^{4} (x) = x$ .

So $f$ has period $4$ under composition. Hence $f^{26} (x) = f^{2} (x) = - \frac{1}{x}$ , and $g (x) = - f^{26} (x) = \frac{1}{x}$ .

Set up the region. The line $2 y = 2 x - 3$ is $y = x - \frac{3}{2}$ (crosses $y = 0$ at $x = 3/2$ ). The curves $y = 1/ x$ and $y = x - 3/2$ meet where $\frac{1}{x} = x - \frac{3}{2}$ , i.e. $2 x^{2} - 3 x - 2 = 0 ⟹ x = 2$ (taking positive root). At $x = 2$ , $y = 1/2$ .

The bounded region splits into a triangle (under the line from $x = 3/2$ to $x = 2$ ) plus the area under $y = 1/ x$ from $x = 2$ to $x = 4$ .

Triangle area: base $= 2 - \frac{3}{2} = \frac{1}{2}$ , height $= \frac{1}{2}$ , area $= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .

Logarithmic strip: $\int_{2}^{4} \frac{1}{x} d x = ln 4 - ln 2 = ln 2$ .

$∴$ Total area $= \frac{1}{8} + lo g_{e} 2$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q18: Area as Definite Integral

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