Question

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 $a+bx+c{x}^2$, then

• A. $ab>c$
• B. $a=b$
• C. $bc=a$
• D. $c>b=a$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $ab>c$
B $a=b$
C $bc=a$

$\frac{(1+x{)}^{3/2}-{(1+\frac{1}{2}x)}^3}{(1-x{)}^{1/2}}$
$=\frac{(1+\frac{3}{2}x+\frac{3}{8}{x}^2+\cdots )-(1+\frac{3}{2}x+\frac{3x^2}{4}+\frac{x^3}{8})}{(1-x{)}^{1/2}}$

$=-\frac{3}{8}{x}^2(1-x{)}^{-1/2}$
$=-\frac{3}{8}{x}^2(1+\frac{x}{2}+\cdots )$
$=-\frac{3}{8}{x}^2$
Now comparing it with $a+bx+c{x}^2$, we get
$a=0,b=0,c=-\frac{3}{8}$