Question

If $x$ is so small that $x^{2}$ and higher powers of $x$
may be neglected, then the approximate value of
$\frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}}{(2-3 x)^{4}}$ is

• A. $\frac{1}{8}(1+7 x)$
• B. $\frac{1}{16}(1-7 x)$
• C. $1-7 x$
• D. $\frac{1}{16}(1+7 x)$

Answer 1

Answer: (D)

We have,
$\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5} $
$= \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}}{2^{4}\left(1-\frac{3}{2} x\right)^{4}}$
$= \frac{\left(1+\frac{2}{3} x\right)^{-3}(1-15 x)^{-1 / 5}\left(1-\frac{3}{2} x\right)^{-4}}{16}$
$= \frac{1}{16}(1-2 x)(1+3 x)(1+6 x)$
$[\because$ neglecting higher powers $]$
$=\frac{1}{16}(1+x)(1+6 x)$
$=\frac{1}{16}(1+7 x)$