Let f(x)=ax^3+bx^2+cx+41 be such that f(1)=40 | vedclass

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

Question

(D) Given $f(x)=ax^3+bx^2+cx+41$. First derivative: $f'(x)=3ax^2+2bx+c$. Given $f'(1)=2$,so $3a+2b+c=2$ ... $(1)$. Second derivative: $f''(x)=6ax+2b$.

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(D) Given $f(x)=ax^3+bx^2+cx+41$. First derivative: $f'(x)=3ax^2+2bx+c$. Given $f'(1)=2$,so $3a+2b+c=2$ ... $(1)$. Second derivative: $f''(x)=6ax+2b$. Given $f''(1)=4$,so $6a+2b=4$,which simplifies to $3a+b=2$ ... $(2)$. From $(1)$ and $(2)$,subtract $(2)$ from $(1)$: $(3a+2b+c) - (3a+b) = 2 - 2$ $b+c=0$ ... $(3)$. Given $f(1)=40$: $a(1)^3+b(1)^2+c(1)+41=40$ $a+b+c+41=40$ $a+(b+c)=-1$. Using $(3)$,$b+c=0$,so $a+0=-1$,which gives $a=-1$. Substitute $a=-1$ into $(2)$: $3(-1)+b=2$ $-3+b=2 \Rightarrow b=5$. Using $(3)$,$5+c=0 \Rightarrow c=-5$. Finally,$a^2+b^2+c^2 = (-1)^2 + (5)^2 + (-5)^2 = 1 + 25 + 25 = 51$.

Let $f(x)=ax^3+bx^2+cx+41$ be such that $f(1)=40, f'(1)=2$ and $f''(1)=4$. Then $a^2+b^2+c^2$ is equal to :

Similar Questions

If $x = \sin \theta$ and $y = \cos(p \theta)$,then $(1 - x^2) y_2 =$

If $x=3 \cos t$ and $y=4 \sin t$,then $\frac{d^2 y}{d x^2}$ at the point $(x_0, y_0)=(\frac{3}{2} \sqrt{2}, 2 \sqrt{2})$ is

If $x = \log p$ and $y = \frac{1}{p}$,then

The $n^{th}$ derivative of $x^{n + 1}$ is

If $y=f(x)$ is a thrice differentiable function and a bijection,then $\frac{d^2 x}{d y^2}\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module