Question

If \( f(x) = 3x^3 + 2x^2 f'(1) + x f''(2) + f'''(3) \) then \( f(x) = ....... \)}

• A. \( \frac{1}{7}(3x^3 - 90x^2 + 72x + 18) \)
• B. \( \frac{1}{7}(21x^3 - 90x^2 + 72x + 126) \)
• C. \( 3x^3 - 90x^2 + 72x + 18 \)
• D. \( 3x^3 - 45x^2 + 36x + 9 \)

Hint:

Constants like $f'(k)$ are fixed values. Solve for them by differentiating the general expression and substituting $x=k$.

Answer 1

The Correct Option is A

Step 1: Identify Constants
Let $f'(1) = a, f''(2) = b, f'''(3) = c$.
$f(x) = 3x^3 + 2ax^2 + bx + c$.
Step 2: Differentiate
$f'(x) = 9x^2 + 4ax + b$.
$f''(x) = 18x + 4a$.
$f'''(x) = 18$. Thus, $c = 18$.
Step 3: Solve System
$b = f''(2) = 18(2) + 4a \implies b = 36 + 4a$.
$a = f'(1) = 9(1) + 4a + b$.
Substitute $b$: $a = 9 + 4a + 36 + 4a \implies 7a = -45 \implies a = -45/7$.
$b = 36 + 4(-45/7) = (252 - 180)/7 = 72/7$.
Step 4: Assemble f(x)
$f(x) = 3x^3 + 2(-45/7)x^2 + (72/7)x + 18$.
$f(x) = \frac{1}{7}(21x^3 - 90x^2 + 72x + 126)$. (Correction based on options).
Final Answer:(B)