JEE Main 2026 Apr 4 Shift 1, Mathematics Q17: Integration as an Inverse Process of Differentiation

Degree count: $\mathrm{deg}f=\mathrm{deg}{f}^{\prime }+\mathrm{deg}{f}^{\prime \prime }=(n-1)+(n-2)$, forcing $n=3$.

Since $f(0)=0$, write $f(x)=a{x}^3+b{x}^2+cx$. Then $f^{\prime }(x)=3a{x}^2+2bx+c$ and $f^{\prime \prime }(x)=6ax+2b$.

Compare $f=f^{\prime }{f}^{\prime \prime }$ coefficient by coefficient.

$x^3$: $a=18a^2\Rightarrow a=\frac{1}{18}$ (taking $a\ne 0$).

The remaining coefficient comparisons yield $b=0$ and $c=0$. So $f(x)=\frac{x^3}{18}$.

Then $f^{\prime }(x)=\frac{x^2}{6},f^{\prime \prime }(x)=\frac{x}{3}$:

$f^{\prime }(2)=\frac{2}{3},f^{\prime \prime }(2)=\frac{2}{3},{\int }_0^{2}f(x)dx=\frac{1}{18}\cdot \frac{16}{4}=\frac{2}{9}.$

Sum $=\frac{2}{3}+\frac{2}{3}+\frac{2}{9}=\frac{12+2}{9}=\frac{14}{9}.$

$36\cdot \frac{14}{9}=56.$