A) \[\frac{32+31x}{64}\]
B) \[\frac{31+32x}{64}\]
C) \[\frac{31-32x}{64}\]
D) \[\frac{1-2x}{64}\]
Correct Answer: A
Solution :
\[{{\left( 1+\frac{2x}{3} \right)}^{3/2}}{{(32+5x)}^{-1/5}}\] \[=\left[ 1+\frac{3}{2}\left( \frac{2x}{3} \right) \right]{{(32)}^{-1/5}}{{\left( 1+\frac{5}{32}x \right)}^{-1/5}}\] (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) \[=\frac{1}{2}(1+x)\left( 1-\frac{x}{32} \right)\] \[=\frac{(1+x)\,(32-x)}{64}=\frac{32+31x}{64}\] (neglect\[{{x}^{2}}\]term)
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