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If `x` is so small that `x^3` and higher powers of `x` may be neglectd, then `((1+x)^(3//2)-(1+1/2x)^3)/((1-x)^(1//2))` may be approximated as `3x+3/8x^2` b. `1-3/8x^2` c. `x/2-3/xx^2` d. `-3/8x^2`

A

`3x+3/(x^(2))`

B

`1-3/8 x^(2)`

C

`x/2-3/x x^(2)`

D

`-3/8 x^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`((1-x)^(3//2)(1+(1)/(2)x)^(3))/((1-x)^(1//2))=((1+3/2x+3/8x^(2))-(1+3/2x+3(x^(2))/(4)))/((1-x)^(1//2))`
`= (-3)/(8)x^(2)(1-x)^(-1//2)`
`= - 3/8 x^(2)(1+x/2) = - 3/8 x^(2)`
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