Question

When x is so small that its square and higher powers may be neglected, find the value of
$\frac{{(1+\frac{2}{3}x)}^{-5}+\sqrt{4+2x}}{\sqrt{(4+x{)}^3}}$.

• A. $=\frac{1}{4}(3-\frac{95}{24}x)$
• B. $=\frac{1}{8}(3-\frac{95}{24}x)$
• C. $=\frac{1}{8}(3-\frac{24}{95}x)$
• D. $=\frac{1}{4}(3-\frac{9}{24}x)$

Answer 1

The correct option is B

$=\frac{1}{8}(3-\frac{95}{24}x)$

Since $x^2$ and the higher powers may be neglected, it will be sufficient to retain the first two terms in the expansion of each binomial.
Therefore the expression $=\frac{{(1+\frac{2}{3}x)}^{-5}+2{(1+\frac{x}{2})}^{\frac{1}{2}}}{8{(1+\frac{x}{4})}^{\frac{3}{2}}}$
$=\frac{(1-\frac{10}{3}x)+2(1+\frac{1}{4}x)}{8(1+\frac{3}{8}x)}$
$=\frac{1}{8}(3-\frac{17}{6}x){(1+\frac{3}{8}x)}^{-1}=\frac{1}{8}(3-\frac{17}{6}x)(1-\frac{3}{8}x)=\frac{1}{8}(3-\frac{95}{24}x)$
The term involving $x^2$ being neglected.