Answer
a) $ Q(x)=1+kx+\dfrac{k(k-1)}{2}x^2$
b) $0 \lt x \lt 0.21544$
Work Step by Step
a) The Taylor series for $ f(x)$ at $ x=0$ is as follows: $ Q(x)=f(0)+f^{,}(x-0)+\dfrac{f^{,}(0)}{2!}(x-0)^2=1+kx+\dfrac{k(k-1)}{2}x^2$
b) Consider the Remainder Estimation Theorem to find the Error.
So, $|R_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$
and $|R_2(x)| =|\dfrac{ 3! x^3}{3 !}|=|x^3|$
Since, $|x| \lt \dfrac{1}{100}$, thus,
$|x|^3 \lt \dfrac{1}{100}$
or, $\dfrac{10}{11} \lt 1-x \lt \dfrac{9}{10} $
or, $0 \lt x \lt \dfrac{1}{(100)^{1/3}}$
or, $0 \lt x \lt 0.21544$
Thus, the required values of $ x $ are $0 \lt x \lt 0.21544$ .
a) $ Q(x)=1+kx+\dfrac{k(k-1)}{2}x^2$
b) $0 \lt x \lt 0.21544$
