Question

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

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

Answer 1

Answer: (D)

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