Let P(x) be a polynomial of degree 2,with P(2 | vedclass

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

Question

(B) Using the Taylor series expansion of $P(x)$ about $x=2$:$P(x) = P(2) + P^{\prime}(2)(x-2) + \frac{P^{\prime \prime}(2)}{2!}(x-2)^2$Given $P(2)=-1,

Vedclass · Accepted Answer

$-0.002$

Vedclass · Answer

(B) Using the Taylor series expansion of $P(x)$ about $x=2$:$P(x) = P(2) + P^{\prime}(2)(x-2) + \frac{P^{\prime \prime}(2)}{2!}(x-2)^2$Given $P(2)=-1, P^{\prime}(2)=0, P^{\prime \prime}(2)=2$:$P(x) = -1 + 0(x-2) + \frac{2}{2}(x-2)^2$$P(x) = -1 + (x-2)^2$Now,substitute $x=1.001$:$P(1.001) = -1 + (1.001-2)^2$$P(1.001) = -1 + (-0.999)^2$$P(1.001) = -1 + 0.998001$$P(1.001) = -0.001999$Rounding to the nearest provided option,we get $-0.002$.

Let $P(x)$ be a polynomial of degree $2$,with $P(2)=-1, P^{\prime}(2)=0, P^{\prime \prime}(2)=2$. Then $P(1.001)$ is

Similar Questions

If $y = \sin(\sin x)$ and $y'' + f(x) \cdot y' + g(x) \cdot y = 0$,then $f(x) \cdot g(x) =$

If $y=44 x^{45}+45 x^{-44}$, then $y^{\prime \prime}=$

If $m \sin ^{-1} x = \log _{e} y$,then $(1 - x^{2}) y'' - x y'$ is equal to

If $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ and $f(0) = 1$,then $f(x)$ is equal to

If $x^3 + y^3 = 6$ and $\frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = \frac{m}{(xy)^n}$, where $m, n \in R$, then find the value of $\frac{m}{n}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module