Mathematics · 2026
JEE Main · 8 April 2026, Shift 2 · Q18
Let f:(1, ∞) → R be a function defined as f(x)=(x-1)/(x+1). Let f^i+1(x)=f(f^i(x)), i=1,2, …, 25, where f^1(x)=f(x). If g(x)+f^26(x)=0, x ∈(1, ∞),…
Let $\displaystyle f:(1, \infty) \rightarrow \mathbf{R}$ be a function defined as $\displaystyle f(x)=\frac{x-1}{x+1}$. Let $\displaystyle f^{i+1}(x)=f\left(f^i(x)\right), i=1,2, \ldots, 25$, where $\displaystyle f^1(x)=f(x)$. If $\displaystyle g(x)+f^{26}(x)=0, x \in(1, \infty)$, then the area of the region bounded by the curves $\displaystyle y=\mathrm{g}(x), 2 y=2 x-3, y=0$ and $\displaystyle x=4$ is :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{1}{8}+\log _{\mathrm{e}} 2$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.