JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let the degree of the polynomial f(x) be n. The degree of f'(x) is n-1 and the degree of f''(x) is n-2. Since f(x) = f'(x) f''(x), equating the degrees on both sides gives: n = (n-1) + (n-2) n = 3 Let f(x) = ax^3 + bx^2 + cx + d. Given f(0) = 0, we get d = 0. Thus, f(x) = ax^3 + bx^2 + cx. Differentiating f(x) with respect to x: f'(x) = 3ax^2 + 2bx + c f''(x) = 6ax + 2b Substituting these into the given equation f(x) = f'(x) f''(x): ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b) ax^3 + bx^2 + cx = 18a^2x^3 + 18abx^2 + (4b^2 + 6ac)x + 2bc Comparing the coefficients of corresponding powers of x: For x^3: a = 18a^2 a = 1 18 (since a 0 for a cubic polynomial) For the constant term: 0 = 2bc For x: c = 4b^2 + 6ac Substituting a = 1 18 into the x coefficient equation: c = 4b^2 + c 3 2c 3 = 4b^2 c = 6b^2 From 0 = 2bc, either b = 0 or c = 0. In either case, since c = 6b^2, we get b = 0 and c = 0. Thus, the polynomial is f(x) = 1 18 x^3. Now, finding the required values: f'(x) = 1 6 x^2 f'(2) = 4 6 = 2 3 f''(x) = 1 3 x f''(2) = 2 3 _0^2 f(x)\,dx = _0^2 1 18 x^3\,dx = 1 18 [ x^4 4 ]_0^2 = 16 72 = 2 9 Substituting these into the given expression: 36 (f'(2) + f''(2) + _0^2 f(x)\,dx ) = 36 ( 2 3 + 2 3 + 2 9 ) = 36 ( 4 3 + 2 9 ) = 36 ( 14 9 ) = 4 14 = 56 Answer: 56