JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given region is bounded by the curves y = 1, y = x^2, and xy = 27 in the first quadrant. Let us find the points of intersection of these curves: Intersection of y = 1 and y = x^2 gives x = 1. Intersection of y = x^2 and xy = 27 gives x(x^2) = 27 x^3 = 27 x = 3, so y = 9. Intersection of y = 1 and xy = 27 gives x = 27. We can find the area by integrating with respect to y from y = 1 to y = 9. For a given y, the value of x ranges from the parabola x = y to the hyperbola x = 27 y . The area A is given by: A = _ 1 ^ 9 ( 27 y - y ) dy Integrating the terms: A = [ 27 y - 2 3 y^ 3/2 ]_ 1 ^ 9 Substituting the limits: A = ( 27 9 - 2 3 (9)^ 3/2 ) - ( 27 1 - 2 3 (1)^ 3/2 ) A = ( 27 (3^2) - 2 3 (27) ) - ( 0 - 2 3 ) A = 54 3 - 18 + 2 3 A = 54 3 - 54 3 + 2 3 A = 54 3 - 52 3 Answer: 54 _e 3 - 52 3