JEE Main 2026 (Online) 4th April Morning ShiftQuestion 1
Let $f$ be a real polynomial of degree $n$ such that $f(x)=f'(x)f''(x)$, for all $x\in\mathbb{R}$. If $f(0)=0$, then $36\left(f'(2)+f''(2)+\int_{0}^{2}f(x)\,dx\right)$ is equal to :
Definite Integration
Solution
Concept. Comparing degrees on both sides fixes $n$; comparing coefficients then fixes $f$. Formulas used. (1) $\deg(gh)=\deg g+\deg h$Step 1 - Degree. By (1), $n=(n-1)+(n-2)$, so $n=3$. With $f(0)=0$, write $f=ax^3+bx^2+cx$. Step 2 - Compare. $$f'f''=(3ax^2+2bx+c)(6ax+2b)=18a^2x^3+18abx^2+(4b^2+6ac)x+2bc.$$ $x^3$: $a=18a^2$, so $a=\frac1{18}$. $x^1$: $c=4b^2+\frac c3$, so $c=6b^2$. $x^0$: $0=2bc=12b^3$, so $b=0$ and $c=0$. Hence $f(x)=\dfrac{x^3}{18}$. Step 3 - The values. $$f'(2)=\frac{3\cdot4}{18}=\frac23,\qquad f''(2)=\frac{12}{18}=\frac23,\qquad\int_0^2\frac{x^3}{18}\,dx=\frac{16}{72}=\frac29.$$ Step 4 - Value. $36\left(\frac23+\frac23+\frac29\right)=36\cdot\frac{14}{9}=56$. Answer: (C)