A) \[-\frac{3}{8}{{x}^{2}}\]
B) \[\frac{x}{2}-\frac{3}{8}{{x}^{2}}\]
C) \[1-\frac{3}{8}{{x}^{2}}\]
D) \[3x+\frac{3}{8}{{x}^{2}}\]
Correct Answer: A
Solution :
\[\frac{{{\left( 1+x \right)}^{3/2}}-{{\left( 1+\frac{1}{2}x \right)}^{3}}}{{{\left( 1-x \right)}^{1/2}}}\] = \[\frac{1+\frac{3}{2}x+\frac{\frac{3}{2}.\frac{1}{2}}{2}{{x}^{2}}-\left( 1+\frac{3x}{2}+\frac{3.2}{2}\frac{{{x}^{2}}}{4} \right)}{{{\left( 1-x \right)}^{1/2}}}\] = \[\frac{-\frac{3}{8}{{x}^{2}}}{{{\left( 1-x \right)}^{1/2}}}\] = \[-\frac{3}{8}{{x}^{2}}{{\left( 1-x \right)}^{-1/2}}\] = \[-\frac{3}{8}{{x}^{2}}\left( 1+\frac{x}{2}+.... \right)\] = \[-\frac{3}{8}{{x}^{2}}\].
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