JEE Main 2026 · 8 Apr · Shift 2 · Mathematics Q18

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MathematicsSingle correct+4 / -1Question 18 of 75

Let $f:(1,\infty)\to\mathbf R$ be a function defined as $f(x)=\dfrac{x-1}{x+1}$ . 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 :

\dfrac18+\log_e2

\dfrac14+\log_e2

\dfrac56+3\log_e2

\dfrac56+\log_e2

Answer and worked solution

Official answer: A

Iterates. $f^2(x)=f\left(\tfrac{x-1}{x+1}\right)=-\tfrac1x$ and $f^4(x)=x$ . So $f^{26}=f^2$ and $g(x)=\tfrac1x$ .

Region. The line $y=x-\tfrac32$ meets $y=0$ at $x=\tfrac32$ and $y=\tfrac1x$ at $x=2$ . The region lies under the lower of the two curves from $x=\tfrac32$ to $x=4$ :
$\int_{3/2}^2\left(x-\tfrac32\right)dx+\int_2^4\frac{dx}x=\frac18+\ln2.$

(A)

Topics:Area between curves Composition and inverse functions

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JEE Main 2026 · 8 Apr · Shift 2, Mathematics question 18