JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Since (2^ 1-a + 2^ 1+a ), f(a), and (3^a + 3^ -a ) are in A.P., we have: 2f(a) = 2^ 1-a + 2^ 1+a + 3^a + 3^ -a f(a) = (2^ -a + 2^a) + 1 2 (3^a + 3^ -a ) Using the AM-GM inequality, x + 1 x 2 for x > 0. 2^a + 2^ -a 2 and 3^a + 3^ -a 2. The minimum value of f(a) occurs when a = 0. = f(0) = (2^0 + 2^0) + 1 2 (3^0 + 3^0) = 2 + 1 = 3. Now, we evaluate the integral: I = _ _e 2 ^ _e 3 dx e^ 2x - e^ -2x = _ _e 2 ^ _e 3 e^ 2x e^ 4x - 1 dx Let e^ 2x = t 2e^ 2x dx = dt. When x = _e 2, t = e^ 2 _e 2 = 4. When x = _e 3, t = e^ 2 _e 3 = 9. I = 1 2 _ 4 ^ 9 dt t^2 - 1 I = 1 2 [ 1 2 _e | t-1 t+1 | ]_ 4 ^ 9 I = 1 4 ( _e 8 10 - _e 3 5 ) I = 1 4 ( _e 4 5 - _e 3 5 ) = 1 4 _e ( 4/5 3/5 ) = 1 4 _e ( 4 3 ) Answer: 1 4 _e ( 4 3 )