JEE Main 2026 — Differentiation Question with Solution

Let the degree of the polynomial $f(x)$ be $n$.

The degree of $f^{\prime }(x)$ is $n-1$ and the degree of $f^{\prime \prime }(x)$ is $n-2$.

Since $f(x)=f^{\prime }(x)f^{\prime \prime }(x)$, equating the degrees on both sides gives:

$n=(n-1)+(n-2)\Rightarrow n=3$

Let $f(x)=a{x}^3+b{x}^2+cx+d$.

Given $f(0)=0$, we get $d=0$. Thus, $f(x)=a{x}^3+b{x}^2+cx$.

Differentiating $f(x)$ with respect to $x$:

$f^{\prime }(x)=3a{x}^2+2bx+c$

$f^{\prime \prime }(x)=6ax+2b$

Substituting these into the given equation $f(x)=f^{\prime }(x)f^{\prime \prime }(x)$:

$a{x}^3+b{x}^2+cx=(3a{x}^2+2bx+c)(6ax+2b)$

$a{x}^3+b{x}^2+cx=18a^2{x}^3+18ab{x}^2+(4b^2+6ac)x+2bc$

Comparing the coefficients of corresponding powers of $x$:

For $x^3$: $a=18a^2\Rightarrow a=\frac{1}{18}$ (since $a\ne 0$ for a cubic polynomial)

For the constant term: $0=2bc$

For $x$: $c=4b^2+6ac$

Substituting $a=\frac{1}{18}$ into the $x$ coefficient equation:

$c=4b^2+\frac{c}{3}\Rightarrow \frac{2c}{3}=4b^2\Rightarrow c=6b^2$

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 $f(x)=\frac{1}{18}{x}^3$.

Now, finding the required values:

$f^{\prime }(x)=\frac{1}{6}{x}^2\Rightarrow f^{\prime }(2)=\frac{4}{6}=\frac{2}{3}$

$f^{\prime \prime }(x)=\frac{1}{3}x\Rightarrow f^{\prime \prime }(2)=\frac{2}{3}$

${\int }_0^{2}f(x)dx={\int }_0^2\frac{1}{18}{x}^{3}dx=\frac{1}{18}{\left[\frac{x^4}{4}\right]}_0^2=\frac{16}{72}=\frac{2}{9}$

Substituting these into the given expression:

$36\left(f^{\prime }(2)+f^{\prime \prime }(2)+{\int }_0^{2}f(x)dx\right)=36\left(\frac{2}{3}+\frac{2}{3}+\frac{2}{9}\right)$

$=36\left(\frac{4}{3}+\frac{2}{9}\right)=36\left(\frac{14}{9}\right)=4\times 14=56$

Answer: $56$