JEE Main 2026 — Mathematics Algebra

Given $f(x) = \dfrac{x-1}{x+1}$f(x)=x+1x−1​. Let us find the first few compositions of $f(x)$f(x):

$$f^2(x) = f(f(x)) = \dfrac{\dfrac{x-1}{x+1}-1}{\dfrac{x-1}{x+1}+1} = \dfrac{x-1-x-1}{x-1+x+1} = \dfrac{-2}{2x} = -\dfrac{1}{x}$$f2(x)=f(f(x))=x+1x−1​+1x+1x−1​−1​=x−1+x+1x−1−x−1​=2x−2​=−x1​

$$f^3(x) = f(f^2(x)) = f\left(-\dfrac{1}{x}\right) = \dfrac{-\dfrac{1}{x}-1}{-\dfrac{1}{x}+1} = \dfrac{-1-x}{-1+x} = \dfrac{x+1}{1-x}$$f3(x)=f(f2(x))=f(−x1​)=−x1​+1−x1​−1​=−1+x−1−x​=1−xx+1​

$$f^4(x) = f(f^3(x)) = f\left(\dfrac{x+1}{1-x}\right) = \dfrac{\dfrac{x+1}{1-x}-1}{\dfrac{x+1}{1-x}+1} = \dfrac{x+1-1+x}{x+1+1-x} = \dfrac{2x}{2} = x$$f4(x)=f(f3(x))=f(1−xx+1​)=1−xx+1​+11−xx+1​−1​=x+1+1−xx+1−1+x​=22x​=x

Since $f^4(x) = x$f4(x)=x, the sequence of functions is periodic with a period of $4$4.

Therefore, $f^{26}(x) = f^{4 \times 6 + 2}(x) = f^2(x) = -\dfrac{1}{x}$f26(x)=f4×6+2(x)=f2(x)=−x1​.

We are given $g(x) + f^{26}(x) = 0$g(x)+f26(x)=0, which implies:

$$g(x) - \dfrac{1}{x} = 0 \Rightarrow g(x) = \dfrac{1}{x}$$g(x)−x1​=0⇒g(x)=x1​

We need to find the area of the region bounded by the curves $y = \dfrac{1}{x}$y=x1​, $y = x - \dfrac{3}{2}$y=x−23​ (from $2y = 2x - 3$2y=2x−3), $y = 0$y=0, and $x = 4$x=4.

First, find the point of intersection of $y = \dfrac{1}{x}$y=x1​ and $y = x - \dfrac{3}{2}$y=x−23​:

$$\dfrac{1}{x} = x - \dfrac{3}{2} \Rightarrow 2x^2 - 3x - 2 = 0$$x1​=x−23​⇒2x2−3x−2=0

$$(2x+1)(x-2) = 0$$(2x+1)(x−2)=0

Since $x \in (1, \infty)$x∈(1,∞), we get $x = 2$x=2.

The line $y = x - \dfrac{3}{2}$y=x−23​ intersects the x-axis ($y = 0$y=0) at $x = \dfrac{3}{2}$x=23​.

The required area $A$A is bounded by $y = x - \dfrac{3}{2}$y=x−23​ from $x = \dfrac{3}{2}$x=23​ to $x = 2$x=2, and by $y = \dfrac{1}{x}$y=x1​ from $x = 2$x=2 to $x = 4$x=4.

$$A = \int_{3/2}^{2} \left(x - \dfrac{3}{2}\right) dx + \int_{2}^{4} \dfrac{1}{x} dx$$A=∫3/22​(x−23​)dx+∫24​x1​dx

Evaluating the first integral:

$$\int_{3/2}^{2} \left(x - \dfrac{3}{2}\right) dx = \left[ \dfrac{1}{2}\left(x - \dfrac{3}{2}\right)^2 \right]_{3/2}^{2} = \dfrac{1}{2} \left(2 - \dfrac{3}{2}\right)^2 - 0 = \dfrac{1}{2} \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{8}$$∫3/22​(x−23​)dx=[21​(x−23​)2]3/22​=21​(2−23​)2−0=21​(21​)2=81​

Evaluating the second integral:

$$\int_{2}^{4} \dfrac{1}{x} dx = [\log_e x]_{2}^{4} = \log_e 4 - \log_e 2 = \log_e 2$$∫24​x1​dx=[loge​x]24​=loge​4−loge​2=loge​2

Total Area = $\dfrac{1}{8} + \log_e 2$81​+loge​2