Binomial Theorem And Its Simple Applications Question 7
Question: If x is very small in magnitude compared with a, then $ {{( \frac{a}{a+x} )}^{\frac{1}{2}}}+{{( \frac{a}{a-x} )}^{\frac{1}{2}}} $ can be approximately equal to
Options:
A) $ 1+\frac{1}{2}\frac{x}{a} $
B) $ \frac{x}{a} $
C) $ 1+\frac{3}{4}\frac{x^{2}}{a^{2}} $
D) $ 2+\frac{3}{4}\frac{x^{2}}{a^{2}} $
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Answer:
Correct Answer: D
Solution:
- [d] $ {{( \frac{a}{a+x} )}^{\frac{1}{2}}}+{{( \frac{a}{a-x} )}^{\frac{1}{2}}} $
$ ={{( \frac{a+x}{a} )}^{-\frac{1}{2}}}+{{( \frac{a-x}{a} )}^{-\frac{1}{2}}} $
$ ={{( 1+\frac{x}{a} )}^{-\frac{1}{2}}}+{{( 1-\frac{x}{a} )}^{-\frac{1}{2}}} $
$ =[ 1-\frac{1}{2}\frac{x}{a}+\frac{3}{8}\frac{x^{2}}{a^{2}} ]+[ 1+\frac{1}{2}\frac{x}{a}+\frac{3}{8}\frac{x^{2}}{a^{2}} ] $
$ [ \because ,x«a,\therefore \frac{x}{a}«1 ]=2+\frac{3}{4}.\frac{x^{2}}{a^{2}} $
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