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JEE Main Mathematics Functions 2026 JEE Main 2026 (06 April Shift 1)

JEE Main Mathematics Question (2026) — Solution

Question

Let e be the base of natural logarithm and let f: \ 1, 2, 3, 4\ \ 1, e, e^2, e^3\ and g: \ 1, e, e^2, e^3\ \ 1, 1 2 , 1 3 , 1 4 \ be two bijective functions such that f is strictly decreasing and g is strictly increasing. If (x) = [f^ -1 \ g^ -1 ( 1 2 ) \ ]^x, then the area of the region R = \ (x, y): x^2 y (x), 0 x 1\ is:

Options

  1. A. 3 - _e(2) 3 _e(2)
  2. B. 1 3 _e(2)
  3. C. 3 + _e(2)
  4. D. 3 + _e(2) 2 + _e(3)

Answer

A. 3 - _e(2) 3 _e(2)

Step-by-step solution

Given f: \ 1, 2, 3, 4\ \ 1, e, e^2, e^3\ is a strictly decreasing bijective function. Arranging the domain and codomain in increasing order, we get f(1) = e^3, f(2) = e^2, f(3) = e, and f(4) = 1. Given g: \ 1, e, e^2, e^3\ \ 1, 1 2 , 1 3 , 1 4 \ is a strictly increasing bijective function. Arranging the domain and codomain in increasing order, we get g(1) = 1 4 , g(e) = 1 3 , g(e^2) = 1 2 , and g(e^3) = 1. Evaluating (x) = [f^ -1 \ g^ -1 ( 1 2 ) \ ]^x: From the mapping of g, g^ -1 ( 1 2 ) = e^2. From the mapping of f, f^ -1 (e^2) = 2. Substituting these values, we get (x) = 2^x. The region R is given by x^2 y 2^x for 0 x 1. The area of the region R is: _ 0 ^ 1 (2^x - x^2) dx = [ 2^x _e(2) - x^3 3 ]_ 0 ^ 1 = ( 2 _e(2) - 1 3 ) - ( 1 _e(2) - 0 ) = 1 _e(2) - 1 3 = 3 - _e(2) 3 _e(2) Answer: 3 - _e(2) 3 _e(2)

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