Question

Let $f$ be a polynomial function such that $f\left(3x\right)=f^{\prime }\left(x\right)\cdot f^{\prime \prime }\left(x\right),$ for all $x\in R.$ Then:

• A. $f\left(2\right)+f^{\prime }\left(2\right)=28$
• B. $f^{\prime \prime }\left(2\right)-f^{\prime }\left(2\right)=0$
• C. $f^{\prime \prime }\left(2\right)-f\left(2\right)=4$
• D. $f\left(2\right)-f^{\prime }\left(2\right)+f^{\prime \prime }\left(2\right)=10$

Answer 1

The correct option is B $f^{\prime \prime }\left(2\right)-f^{\prime }\left(2\right)=0$

The correct answer is $B$

Now,

If $f\left(x\right)$ is of first degree its second derivatives is identically null, so also $f\left(x\right)$ would have to be identically null.

To satisfy the equation $f\left(3x\right)=f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)$

Let, then $f\left(x\right)$ be a generic polynomial of degree $n\ge 2$. Then $f^{\prime }\left(x\right)$ will have degree $(n-1)$ and $f^{\prime \prime }\left(x\right)$ degree $(n-2)$.

Now, The product $f^{\prime }\left(x\right).f"\left(x\right)$ is a polynomial can be equal for every $x$ only is they the same degree

$f\left(3x\right)=f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)$.