JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given f(x) = x-1 x+1 . Let us find the first few compositions of f(x): f^2(x) = f(f(x)) = x-1 x+1 -1 x-1 x+1 +1 = x-1-x-1 x-1+x+1 = -2 2x = - 1 x f^3(x) = f(f^2(x)) = f (- 1 x ) = - 1 x -1 - 1 x +1 = -1-x -1+x = x+1 1-x f^4(x) = f(f^3(x)) = f ( x+1 1-x ) = x+1 1-x -1 x+1 1-x +1 = x+1-1+x x+1+1-x = 2x 2 = x Since f^4(x) = x, the sequence of functions is periodic with a period of 4. Therefore, f^ 26 (x) = f^ 4 6 + 2 (x) = f^2(x) = - 1 x . We are given g(x) + f^ 26 (x) = 0, which implies: g(x) - 1 x = 0 g(x) = 1 x We need to find the area of the region bounded by the curves y = 1 x , y = x - 3 2 (from 2y = 2x - 3), y = 0, and x = 4. First, find the point of intersection of y = 1 x and y = x - 3 2 : 1 x = x - 3 2 2x^2 - 3x - 2 = 0 (2x+1)(x-2) = 0 Since x (1, ), we get x = 2. The line y = x - 3 2 intersects the x-axis (y = 0) at x = 3 2 . The required area A is bounded by y = x - 3 2 from x = 3 2 to x = 2, and by y = 1 x from x = 2 to x = 4. A = _ 3/2 ^ 2 (x - 3 2 ) dx + _ 2 ^ 4 1 x dx Evaluating the first integral: _ 3/2 ^ 2 (x - 3 2 ) dx = [ 1 2 (x - 3 2 )^2 ]_ 3/2 ^ 2 = 1 2 (2 - 3 2 )^2 - 0 = 1 2 ( 1 2 )^2 = 1 8 Evaluating the second integral: _ 2 ^ 4 1 x dx = [ _e x]_ 2 ^ 4 = _e 4 - _e 2 = _e 2 Total Area = 1 8 + _e 2