Binomial-Theorem-And-Its-Simple-Applications Question 214
Question: If x is so small that $ x^{3} $ and higher powers of x may be neglected, then $ \frac{{{(1+x)}^{\frac{3}{2}}}-{{( 1+\frac{1}{2}x )}^{3}}}{{{(1-x)}^{\frac{1}{2}}}} $ may be approximated as
Options:
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
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Answer:
Solution:
$ \therefore ,\frac{{{(1+x)}^{\frac{3}{2}}}-{{( 1+\frac{x}{2} )}^{3}}}{( 1-{x^{\frac{1}{2}}} )} $ $ ={{(1-x)}^{\frac{-1}{2}}}[ ( 1+\frac{3}{2}x+\frac{\frac{3}{2}.\frac{1}{2}}{2!}x^{2} )-( 1+\frac{3x}{2}+\frac{3.2}{2!}\frac{x^{2}}{4} ) ] $ $ =[ 1+\frac{x}{2}+\frac{\frac{1}{2}.\frac{3}{2}}{2!}x^{2} ][ \frac{-3}{8}x^{2} ]=\frac{-3}{8}x^{2} $ (as $ x^{3} $ and higher powers of x can be neglected)
BETA
