A) \[1-\frac{3}{8}{{x}^{2}}\]
B) \[3x+\frac{3}{8}{{x}^{2}}\]
C) \[-\frac{3}{8}{{x}^{2}}\]
D) \[\frac{x}{2}-\frac{3}{8}{{x}^{2}}\]
Correct Answer: C
Solution :
[c] \[\because {{x}^{3}}\] and higher powers of x may be neglected \[\therefore \,\,\,\,\,\,\,\,\,\,\,\frac{{{(1+x)}^{\frac{3}{2}}}-{{\left( 1+\frac{x}{2} \right)}^{3}}}{\left( 1-{{x}^{\frac{1}{2}}} \right)}\]\[={{(1-x)}^{\frac{-1}{2}}}\left[ \left( 1+\frac{3}{2}x+\frac{\frac{3}{2}.\frac{1}{2}}{2!}{{x}^{2}} \right)-\left( 1+\frac{3x}{2}+\frac{3.2}{2!}\frac{{{x}^{2}}}{4} \right) \right]\] \[=\left[ 1+\frac{x}{2}+\frac{\frac{1}{2}.\frac{3}{2}}{2!}{{x}^{2}} \right]\left[ \frac{-3}{8}{{x}^{2}} \right]=\frac{-3}{8}{{x}^{2}}\] (as \[{{x}^{3}}\] and higher powers of x can be neglected)
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