If x is so small that x3 and higher power of x may be neglected , then (1+x)32−(1+x2)3(1−x)12may be.

1+32x+32.122x2−1+342+3.22.x24(1−x)12=−3x28(1−x)−12[ neglecting higher powers of x]=−3x281+12x+12.322.x2=−3x28

If x is so small that x3 and higher power of x may be neglected , then (1+x)32−(1+x2)3(1−x)12may be.

1+32x+32.122x2−1+342+3.22.x24(1−x)12=−3x28(1−x)−12[ neglecting higher powers of x]=−3x281+12x+12.322.x2=−3x28

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If $x$ is so small that $x^{3}$ and higher power of $x$ may be neglected , then $\frac{{(1 + x)}^{\frac{3}{2}} - {(1 + \frac{x}{2})}^{3}}{{(1 - x)}^{\frac{1}{2}}}$ may be.

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$\frac{x}{2} - \frac{3}{8} x^{2}$

$- \frac{3}{8} x^{2}$

$3 x + \frac{3}{8} x^{2}$

$1 - \frac{3}{8} x^{2}$

answer is B.

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Detailed Solution

$\frac{(1 + \frac{3}{2} x + \frac{\frac{3}{2} . \frac{1}{2}}{2} x^{2}) - (1 + \frac{34}{2} + \frac{3.2}{2} . \frac{x^{2}}{4})}{{(1 - x)}^{\frac{1}{2}}}$

$= - \frac{3 x^{2}}{8} {(1 - x)}^{\frac{- 1}{2}}$ [ neglecting higher powers of $x$ ]

$= - \frac{3 x^{2}}{8} (1 + \frac{1}{2} x + \frac{\frac{1}{2} . \frac{3}{2}}{2} . x^{2})$

$= - \frac{3 x^{2}}{8}$

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If x is so small that x3 and higher power of x may be neglected , then (1+x)32−(1+x2)3(1−x)12may be.

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If $x$ is so small that $x^{3}$ and higher power of $x$ may be neglected , then $\frac{{(1 + x)}^{\frac{3}{2}} - {(1 + \frac{x}{2})}^{3}}{{(1 - x)}^{\frac{1}{2}}}$ may be.