JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given integral is f(x) = ( 16x + 24 x^2 + 2x - 15 ) dx. Factorizing the denominator, we get x^2 + 2x - 15 = (x + 5)(x - 3). Using partial fractions, we can write: 16x + 24 (x + 5)(x - 3) = A x + 5 + B x - 3 16x + 24 = A(x - 3) + B(x + 5) Substituting x = 3, we get 72 = 8B B = 9. Substituting x = -5, we get -56 = -8A A = 7. Therefore, the integral becomes: f(x) = ( 7 x + 5 + 9 x - 3 ) dx f(x) = 7 _e|x + 5| + 9 _e|x - 3| + C Given f(4) = 14 _e(3), we substitute x = 4: f(4) = 7 _e(9) + 9 _e(1) + C = 14 _e(3) + C Since f(4) = 14 _e(3), we get C = 0. Thus, f(x) = 7 _e|x + 5| + 9 _e|x - 3|. Now, substituting x = 7: f(7) = 7 _e(12) + 9 _e(4) f(7) = 7 _e(2^2 3) + 9 _e(2^2) f(7) = 7(2 _e(2) + _e(3)) + 18 _e(2) f(7) = 14 _e(2) + 7 _e(3) + 18 _e(2) f(7) = 32 _e(2) + 7 _e(3) f(7) = _e(2^ 32 3^7) Comparing this with f(7) = _e(2^ 3^ ), we get = 32 and = 7. Therefore, + = 32 + 7 = 39. Answer: 39