We are given a polynomial $f$ such that $f(3x)=f'(x)\,f''(x) \qquad \text{for all } x\in\mathbb R.$

We must determine which option is correct.

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

• Then $f(3x)$ is also a polynomial of degree $n$.
• $f'(x)$ has degree $n-1$.
• $f''(x)$ has degree $n-2$.
• So $f'(x)f''(x)$ has degree $(n-1)+(n-2)=2n-3.$

Since $f(3x)=f'(x)f''(x),$ the degrees must be equal: $n=2n-3 \implies n=3.$

Hence $f$ is a cubic polynomial.

Let $f(x)=ax^3+bx^2+cx+d, \qquad a\neq 0.$

Then $f'(x)=3ax^2+2bx+c,$ $f''(x)=6ax+2b.$

Also, $f(3x)=a(3x)^3+b(3x)^2+c(3x)+d=27ax^3+9bx^2+3cx+d.$

Now compute $f'(x)f''(x)$: \begin{align*} f'(x)f''(x) &= (3ax^2+2bx+c)(6ax+2b) \ &= 18a^2x^3 + 6abx^2 + 12abx^2 + 4b^2x + 6acx + 2bc \ &= 18a^2x^3 + 18abx^2 + (4b^2+6ac)x + 2bc. \end{align*}

Compare coefficients of $27ax^3+9bx^2+3cx+d = 18a^2x^3 + 18abx^2 + (4b^2+6ac)x + 2bc.$

Equating coefficients:

• Coefficient of $x^3$: $27a=18a^2.$ Since $a\neq 0$, $27=18a \implies a=\frac{3}{2}.$

• Coefficient of $x^2$: $9b=18ab.$ Substitute $a=\frac32$: $9b=18\cdot \frac32\, b = 27b.$ Hence $18b=0 \implies b=0.$

• Coefficient of $x$: $3c=4b^2+6ac.$ With $b=0$ and $a=\frac32$, $3c=0+6\cdot\frac32\,c=9c.$ So $6c=0 \implies c=0.$

• Constant term: $d=2bc=0.$ Hence $d=0.$

Therefore, $f(x)=\frac32 x^3.$

Then

Evaluate at $x=2$: $f(2)=\frac32(8)=12,$ $f'(2)=\frac92(4)=18,$ $f''(2)=9(2)=18.$

Check the options:

• A: $f(2)+f'(2)=12+18=30\neq 28$ → False
• B: $f''(2)-f'(2)=18-18=0$ → True
• C: $f''(2)-f(2)=18-12=6\neq 4$ → False
• D: $f(2)-f'(2)+f''(2)=12-18+18=12\neq 10$ → False

Hence the correct option is $\boxed{\text{B}}.$