Expand $\frac{\sqrt[4]{1+3{{x}^{2}}}}{2+x}$ in ascending powers of $x$, up to and including the

Question

Expand $\frac{\sqrt[4]{1+3{{x}^{2}}}}{2+x}$ in ascending powers of $x$, up to and including the term in ${{x}^{2}}$.[3] Find the set of values of $x$ for which the expansion in part (i) is valid.[2] By substituting $x=-\frac{1}{4}$ and using your result in part (i), show that $\sqrt[4]{19}\approx \frac{p}{q}$, where $p$ and $q$ are integers to be determined.[2]

Accepted Answer

(i) $\displaystyle \frac{\sqrt[4]{1+3x^2}}{2+x}=\frac12-\frac{x}{4}+\frac{x^2}{2}+O(x^3)$. (ii) The expansion is valid for $\displaystyle -\frac{1}{\sqrt3}<x<\frac{1}{\sqrt3}$. (iii) $\displaystyle \sqrt[4]{19}\approx\frac{133}{64}$, so $p=133$ and $q=64$.

2015 MI P1 Q3

2015 MI P1 Q3

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