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

If $f$ is twice differentiable such that $f''(x) = -f(x)$,$f'(x) = g(x)$,$h'(x) = [f(x)]^2 + [g(x)]^2$,and $h(0) = 2$,$h(1) = 4$,then the equation $y = h(x)$ represents:

If $e^y(x+1)=1$,then $\frac{d^2 y}{d x^2} = $ . . . . . .

Let $f(x)$ be a polynomial in $x$. Then the second derivative of $f(e^x)$ is:

If the function $y = \sin^{-1} x$,then $\left(1-x^2\right) \frac{d^2 y}{d x^2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module