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, ∞), then the area of the region bounded by the curves y=g(x), 2y=2x-3, y=0 and x=4 is:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #270

Let

f : (1, \infty) \to R

be a function defined as

f (x) = \frac{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 \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:

Clue 1 of 3: Iterated Functions

First, determine the pattern of the iterated function $f^{i} (x)$ . Pay attention to the cycle length of the function composition.

🧠 Full Solution Path

Step 1: Determine the Iterated Function

f^{26} (x)

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First, let's find the pattern of the iterated function $f (x) = \frac{x - 1}{x + 1}$ :

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

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

f^{3} (x) = f (f^{2} (x)) = f (- \frac{1}{x}) = \frac{- \frac{1}{x} - 1}{- \frac{1}{x} + 1} = \frac{- 1 - x}{- 1 + x} = \frac{x + 1}{1 - x}

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

The function has a cycle of 4, i.e., $f^{n + 4} (x) = f^{n} (x)$ . Therefore, $f^{26} (x) = f^{4 \times 6 + 2} (x) = f^{2} (x) = - \frac{1}{x}$ . Given $g (x) + f^{26} (x) = 0$ , we have $g (x) - \frac{1}{x} = 0 ⟹ g (x) = \frac{1}{x}$ .

Step 2: Identify the Region of Integration○ Expand

The region is bounded by the curves $y = g (x) = \frac{1}{x}$ , $y = x - \frac{3}{2}$ (from $2 y = 2 x - 3$ ), $y = 0$ , and $x = 4$ . Let's find the intersection points: Equating $y = \frac{1}{x}$ and $y = x - \frac{3}{2}$ : $\frac{1}{x} = x - \frac{3}{2} ⟹ 2 = 2 x^{2} - 3 x ⟹ 2 x^{2} - 3 x - 2 = 0$ Factoring the quadratic equation: $(2 x + 1) (x - 2) = 0$ . The solutions are $x = - \frac{1}{2}$ or $x = 2$ . Since the domain is $x \in (1, \infty)$ , the relevant intersection point is $x = 2$ . The line $y = x - \frac{3}{2}$ intersects the x-axis ( $y = 0$ ) at $x = \frac{3}{2}$ . We need to integrate from $x = \frac{3}{2}$ to $x = 4$ . The upper boundary of the region is determined by the lower envelope of the two functions $y = \frac{1}{x}$ and $y = x - \frac{3}{2}$ because the region is bounded by $y = 0$ from below. For $x \in [\frac{3}{2}, 2]$ , $x - \frac{3}{2} \leq \frac{1}{x}$ . For $x \in [2, 4]$ , $\frac{1}{x} \leq x - \frac{3}{2}$ . Thus, the area $A$ is given by the sum of two integrals:

A = \int_{3 / 2}^{2} (x - \frac{3}{2}) d x + \int_{2}^{4} \frac{1}{x} d x

Step 3: Calculate the Area○ Expand

Now, we evaluate each integral:

\int_{3 / 2}^{2} (x - \frac{3}{2}) d x = {[\frac{x^{2}}{2} - \frac{3}{2} x]}_{3 / 2}^{2}

= (\frac{2^{2}}{2} - \frac{3}{2} (2)) - (\frac{(3 / 2)^{2}}{2} - \frac{3}{2} (3 / 2))

= (2 - 3) - (\frac{9 / 4}{2} - \frac{9}{4}) = - 1 - (\frac{9}{8} - \frac{18}{8})

= - 1 - (- \frac{9}{8}) = - 1 + \frac{9}{8} = \frac{1}{8}

And for the second integral:

\int_{2}^{4} \frac{1}{x} d x = [\log_{e} | x |]_{2}^{4} = \log_{e} 4 - \log_{e} 2 = \log_{e} (\frac{4}{2}) = \log_{e} 2

Adding the two parts, the total area is:

A = \frac{1}{8} + \log_{e} 2

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Maths Question 18 – JEE-MAIN 2026

Clue 1 of 3: Iterated Functions

🧠 Full Solution Path