Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f(x)=x^3+x^2f'(1)+xf''(2)+f'''(3)$, then $f'''(3) =$

• A. 3
• B. 6
• C. 9
• D. -2
• E. $f''(2)$

Hint:

The $n$-th derivative of $x^n$ is $n!$, and all subsequent derivatives are zero.

Answer 1

The Correct Option is B

Step 1: Concept
$f'(1), f''(2),$ and $f'''(3)$ are constants.

Step 2: Analysis
Let $f(x) = x^3 + Ax^2 + Bx + C$. $f'(x) = 3x^2 + 2Ax + B$. $f''(x) = 6x + 2A$. $f'''(x) = 6$.

Step 3: Conclusion
Since the third derivative of any cubic function $x^3 + \dots$ is constant and equal to $3!$, we have $f'''(x) = 6$ for all $x$. Thus, $f'''(3) = 6$. Final Answer: (B)