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

(D) $\frac{x}{2}-\frac{3}{8} x^{2}$

JEE Main · 2005 · 08 May 2005 · Offline · Binomial Theorem

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

Answer: (C)

For small

x

, expand each factor to second order:

(1 + x)^{3/2} \approx 1 + \frac{3}{2} x + \frac{3}{8} x^{2}

(1 + \frac{x}{2})^{3} \approx 1 + \frac{3}{2} x + \frac{3}{4} x^{2}

, and

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

, which is

1

to leading order. The numerator is

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

, and dividing by the denominator (effectively

1

at this order) leaves

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

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option