Let … be a function defined as …. Let …,…

Single correct+4 / -1Answer matches NTA’s final key

Let $f : (1, \infty) \to \mathbb{R}$ be a function defined as $f(x) = \frac{x - 1}{x + 1}$ . Let $f^{i+1}(x) = f\left(f^{i}(x)\right)$ , $i = 1, 2, \dots, 25$ , 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 $y = g(x)$ , $2y = 2x - 3$ , $y = 0$ and $x = 4$ is:

$\frac18 + \log_e 2$
$\frac14 + \log_e 2$
$\frac56 + 3\log_e 2$
$\frac56 + \log_e 2$

Solution

Find the cycle. $f^2(x) = \dfrac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \dfrac{-2}{2x} = -\dfrac1x$ .

Then $f^4(x) = -\dfrac{1}{-1/x} = x$ : the iterates repeat with period 4. Since $26 = 4 \cdot 6 + 2$ , $f^{26}(x) = f^2(x) = -\frac1x$ , and $g(x) = \frac1x$ .

Set up the region. The line $y = x - \frac32$ meets $y = 0$ at $x = \frac32$ and meets $y = \frac1x$ where $x^2 - \frac32 x - 1 = 0$ , that is $x = 2$ . Between $\frac32$ and $2$ the line is the lower of the two curves; from $2$ to $4$ the hyperbola is. The region under the lower curve, above $y = 0$ , up to $x = 4$ :

$\text{Area} = \int_{3/2}^{2}\left(x - \tfrac32\right)dx + \int_2^4 \frac{dx}{x} = \frac12\left(\frac12\right)^2 + \ln 2 = \frac18 + \ln 2.$

Answer: $\frac18 + \log_e 2$ .

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Integral Calculus

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