JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Continuity and Differentiability

Matching degrees in $f=f^{\prime }{f}^{\prime \prime }$ gives $n=(n-1)+(n-2)$, so $n=3$. With $f(0)=0$ write $f(x)=a{x}^3+b{x}^2+cx$. Expanding $f^{\prime }{f}^{\prime \prime }=(3a{x}^2+2bx+c)(6ax+2b)$ and comparing coefficients: the $x^3$ term gives $18a^2=a$, so $a=\frac{1}{18}$; the constant term gives $2bc=0$ and the $x$ term gives $c=4b^2+6ac$, which together force $b=0$ and $c=0$. Hence $f(x)=\frac{x^3}{18}$, $f^{\prime }(x)=\frac{x^2}{6}$, $f^{\prime \prime }(x)=\frac{x}{3}$. Then $f^{\prime }(2)+f^{\prime \prime }(2)+{\int }_0^{2}fdx=\frac{2}{3}+\frac{2}{3}+\frac{2}{9}=\frac{14}{9}$, and $36\cdot \frac{14}{9}=56$.