If x is so small that x3 and higher powers of x may be neglected- then 1-x3-2-1-1-2x31-x1-2 may be approximated as

The correct option is C 38x^2 (1+x)^3/2 (1+x/2)^3(1 x)^1/2=(1+32x+38x^2+...) (1+32x+34x)+...(1 x2 x^28+...) =( 3/8x^2)(1 x2+x^28+....) = 38x^ ...

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Byju's Answer

Standard XII

Mathematics

Greatest Binomial Coefficients

If x is so sm...

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

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

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

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

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

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Solution

The correct option is C $- \frac{3}{8} x^{2}$
$\frac{(1 + x)^{3 / 2} - (1 + x / 2)^{3}}{(1 - x)^{1 / 2}} = \frac{(1 + \frac{3}{2} x + \frac{3}{8} x^{2} + . . .) - (1 + \frac{3}{2} x + \frac{3}{4} x) + . . .}{(1 - \frac{x}{2} - \frac{x^{2}}{8} + . . .)} = (- 3 / 8 x^{2}) (1 - \frac{x}{2} + \frac{x^{2}}{8} + . . . .) = - \frac{3}{8} x^{2} (1 + \frac{x}{2} - \frac{x^{2}}{8} + . . .) = \frac{- 3 x^{2}}{8}$

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Similar questions

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

Q. If

x

is so small that

x^{3}

and higher powers of

x

may be neglected, then

\frac{(1 + x)^{3 / 2} - {(1 + \frac{1}{2} x)}^{3}}{(1 - x)^{1 / 2}}

may be approximated as ___________________.

Q. If x is so small that

x^{3}

and higher powers of x may be neglected, then

\frac{(1 + x)^{3 / 2} - {(1 + \frac{1}{2} x)}^{3}}{(1 - x)^{1 / 2}}

may be approximately as:

Q. If

x

is so small that

x^{3}

and higher powers of

x

may be neglected and

\frac{(1 + x)^{3 / 2} - {(1 + \frac{1}{2} x)}^{3}}{(1 - x)^{1 / 2}}

may be approximated as

a + b x + c x^{2}

, then

Q. 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{1}{2} x)}^{3}}{(1 - x)^{\frac{1}{2}}}

may be approximated as

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If x is so small that x3 and higher powers of x may be neglected, then(1+x)32−(1+12x)3(1−x)12 may be approximated as

The correct option is C −38x2(1+x)3/2−(1+x/2)3(1−x)1/2=(1+32x+38x2+...)−(1+32x+34x)+...(1−x2−x28+...)=(−3/8x2)(1−x2+x28+....)=−38x2(1+x2−x28+...)=−3x28

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