If f(x) = 3x^3 + 2x^2 f'(1) + x f''(2) + f''' | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = 3x^3 + Ax^2 + Bx + C$,where $A = 2f'(1)$,$B = f''(2)$,and $C = f'''(3)$.First,find the derivatives:$f'(x) = 9x^2 + 2Ax + B$$f''(x) =

Vedclass · Accepted Answer

$\frac{1}{7}(21x^3 - 90x^2 + 72x + 126)$

Vedclass · Answer

(B) Given $f(x) = 3x^3 + Ax^2 + Bx + C$,where $A = 2f'(1)$,$B = f''(2)$,and $C = f'''(3)$.First,find the derivatives:$f'(x) = 9x^2 + 2Ax + B$$f''(x) = 18x + 2A$$f'''(x) = 18$Now,evaluate the constants:$f'''(3) = 18$,so $C = 18$.$f''(2) = 18(2) + 2A = 36 + 2A$. Since $B = f''(2)$,we have $B = 36 + 2A$.$f'(1) = 9(1)^2 + 2A(1) + B = 9 + 2A + B$. Since $A = 2f'(1)$,we have $A = 2(9 + 2A + B) = 18 + 4A + 2B$.Substitute $B = 36 + 2A$ into the equation for $A$:$A = 18 + 4A + 2(36 + 2A)$$A = 18 + 4A + 72 + 4A$$A = 90 + 8A$$-7A = 90 \implies A = -\frac{90}{7}$.Now find $B$:$B = 36 + 2(-\frac{90}{7}) = 36 - \frac{180}{7} = \frac{252 - 180}{7} = \frac{72}{7}$.Thus,$f(x) = 3x^3 - \frac{90}{7}x^2 + \frac{72}{7}x + 18 = \frac{1}{7}(21x^3 - 90x^2 + 72x + 126)$.Therefore,the correct option is $B$.

If $f(x) = 3x^3 + 2x^2 f'(1) + x f''(2) + f'''(3)$,then $f(x) = $

Similar Questions

If $f(x)$ and $g(x)$ are twice differentiable functions satisfying $f''(x) = g''(x)$, $f'(1) = 2g'(1) = 4$ and $f(2) = 3g(2) = 9$, then the value of $f(x) - g(x)$ at $x = 4$ is:

If $y = \tan(\log x)$,then $\frac{d^2 y}{d x^2} =$

If $y = \sin^2 \alpha + \cos^2(\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$,then find $\frac{d^3y}{d\alpha^3}$ (keeping $\beta$ as a constant).

If $x = \sqrt{e^{\sin^{-1} t}}$ and $y = \sqrt{e^{\cos^{-1} t}}$,then $\frac{d^2 y}{dx^2}$ is

If $y=500 e^{7 x}+600 e^{-7 x},$ show that $\frac{d^{2} y}{d x^{2}}=49 y$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module