Question:
If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{\left(1+x\right)^{\frac{3}{2}} - \left(1+ \frac{1}{2}x\right)^{3}}{\left(1-x\right)^{\frac{1}{2}}} $ may be approximated as
If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{\left(1+x\right)^{\frac{3}{2}} - \left(1+ \frac{1}{2}x\right)^{3}}{\left(1-x\right)^{\frac{1}{2}}} $ may be approximated as
Updated On: Jul 5, 2022
- $ 1 - \frac{3}{8} x^2$
- $ 3x + \frac{3}{8} x^2$
- $ - \frac{3}{8} x^2$
- $ \frac{x}{2} - \frac{3}{8} x^2$
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The Correct Option is C
Solution and Explanation
$\because \, x^3 $ and higher powers of x may be neglected
$ \therefore \frac{\left(1+x\right) \frac{3}{2} -\left(1+\frac{x}{2}\right)^{3}}{\left(1-x^{\frac{1}{2}}\right)} $
$=\left(1-x\right)^{\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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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
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- The first and the last terms are xn and yn respectively.
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- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.