Question:

Let $f(x)=x^3+x^2f'(1)+2xf''(2)+f'''(3)$, $x\in\mathbb{R}$. Then the value of $f'(5)$ is:

$\dfrac{117}{5}$

$\dfrac{657}{5}$

$\dfrac{2}{5}$

$\dfrac{62}{5}$

Step-by-Step Solution

Key Concept: Let $a=f'(1)$, $b=f''(2)$, $c=f'''(3)$. Then $f(x)=x^3+ax^2+2bx+c$. Differentiate: $f'(x)=3x^2+2ax+2b$, $f''(x)=6x+2a$, $f'''(x)=6$.

$f'(5)=\dfrac{117}{5}$.

Correct Answer: 1