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) Write the Taylor series for $f(x)$ at $x=0$ . $Q(x)=f(0)+f^{,}(x-0)+\dfrac{f^{,,}(0)}{2!}(x-0)^2=1+kx+\dfrac{k(k-1)}{2}x^2$
(b) The Remainder Estimation Theorem states that
$|R_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$
and $|R_2(x)| =|\dfrac{ 3! \times x^3}{3 !}|=|x^3|$
Since, $|x| \lt \dfrac{1}{100}$ ,
This implies that $|x|^3 \lt \dfrac{1}{100}$
$\implies \dfrac{10}{11} \lt 1-x \lt \dfrac{9}{10} $
$\implies 0 \lt x \lt 0.21544$
Therefore, $x$ is $0 \lt x \lt 0.21544$.
