Question

If $x$ is numerically so small so that $x^{2}$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}$ is approximately equal to

• A. $\frac{32+31 x}{64}$
• B. $\frac{31+32 x}{64}$
• C. $\frac{31-32 x}{64}$
• D. $\frac{1-2 x}{64}$

Answer 1

Answer: (A)

$\frac{32+31 x}{64}$

Answer 2

$\left(1+\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5}$
$=\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right](32)^{-1 / 5}\left(1+\frac{5}{32} x\right)^{-1 / 5}$
(Neglect higher powers of x)
$=[1+x] 2^{-1}\left[1-\frac{1}{5}\left(\frac{5}{32}\right) x\right]$
(Neglect higher powers of x)
$=\frac{1}{2}(1+x)\left(1-\frac{x}{32}\right)$
$=\frac{(1+x)(32-x)}{64}=\frac{32+31 x}{64}$
(Neglect $x^{2}$ term)