Let f be a real polynomial of degree n such that f(x)=f^'(x)f^''(x), for all x R.

Question

Let $f$ be a real polynomial of degree $n$ such that $f(x)={f}^{'}(x){f}^{''}(x)$, for all $x\in R$. If $f(0)=0$, then $36({f}^{'}(2)+{f}^{''}(2)+{\int }_{2}^{0}f(x)dx)$ is equal to:

Accepted Answer

Let the degree of the real polynomial $f$ be $n$.

Since $f(x)={f}^{'}(x){f}^{''}(x)$ for all $x$, we compare degrees:

$\begin{gathered} deg(f)=deg({f}^{'})+deg({f}^{''}) \ n=(n-1)+(n-2)=2n-3 \ \Rightarrow n=3 \end{gathered}$

So $f$ must be a cubic polynomial. Write

$\begin{gathered} f(x)=a{x}^{3}+b{x}^{2}+cx+d \end{gathered}$

Given $f(0)=0$, we get $d=0$. Hence

$\begin{gathered} f(x)=a{x}^{3}+b{x}^{2}+cx \end{gathered}$

Now find derivatives:

$\begin{gathered} {f}^{'}(x)=3a{x}^{2}+2bx+c \ {f}^{''}(x)=6ax+2b \end{gathered}$

Use the given condition $f(x)={f}^{'}(x){f}^{''}(x)$:

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

Now compare coefficients of like powers of $x$.

For ${x}^{3}$:

$\begin{gathered} a=18{a}^{2} \ \Rightarrow a=\frac{1}{18}(a
e 0) \end{gathered}$

For ${x}^{2}$:

$\begin{gathered} b=6ab+12ab \ b=18ab \end{gathered}$

With $a=\frac{1}{18}$, this becomes $b=b$, so it gives no new condition on $b$.

For $x$:

$\begin{gathered} c=4{b}^{2}+6ac \ c=4{b}^{2}+\frac{c}{3}\, (\because 6a=\frac{1}{3}) \ \Rightarrow \frac{2c}{3}=4{b}^{2}\Rightarrow c=6{b}^{2} \end{gathered}$

For the constant term:

$\begin{gathered} 0=2bc\Rightarrow bc=0 \end{gathered}$

But $c=6{b}^{2}$, so $bc=b(6{b}^{2})=6{b}^{3}=0\Rightarrow b=0$, and then $c=0$.

So the polynomial is

$\begin{gathered} f(x)=\frac{1}{18}{x}^{3} \end{gathered}$

Now compute the required value:

$\begin{gathered} {f}^{'}(x)=\frac{1}{6}{x}^{2}\Rightarrow {f}^{'}(2)=\frac{1}{6}\cdot 4=\frac{2}{3} \ {f}^{''}(x)=\frac{1}{3}x\Rightarrow {f}^{''}(2)=\frac{1}{3}\cdot 2=\frac{2}{3} \ {\int }_{2}^{0}f(x)\, dx={\int }_{2}^{0}\frac{1}{18}{x}^{3}\, dx=\frac{1}{18}{[\frac{{x}^{4}}{4}]}_{2}^{0}=\frac{1}{18}\cdot \frac{16}{4}=\frac{2}{9} \end{gathered}$

Add them:

$\begin{gathered} {f}^{'}(2)+{f}^{''}(2)+{\int }_{2}^{0}f(x)\, dx \ =\frac{2}{3}+\frac{2}{3}+\frac{2}{9}=\frac{14}{9} \end{gathered}$

Finally,

$\begin{gathered} 36({f}^{'}(2)+{f}^{''}(2)+{\int }_{2}^{0}f(x)\, dx)=36\cdot \frac{14}{9}=56 \end{gathered}$

Solutions · 1