Question

If x is so small that its square and higher powers may be neglected, then the value of $\frac{(8 + 3x)^{2/3}}{(2 + 3x)(4 - 5x)^{1/2}}$ is –

• A. 1 – $\frac{3}{2}$x
• B. 1 + $\frac{5}{8}$x
• C. 1 – $\frac{5}{8}$x
• D. None

Answer 1

Answer: (C)

1 – $\frac{5}{8}$x

Answer 2

We have, $\frac{(8 + 3x)^{2/3}}{(2 + 3x)\sqrt{(4 - 5x)}}$

= $\frac{8^{2/3}\left( 1 + \frac{3x}{8} \right)^{2/3}}{2\left( 1 + \frac{3}{2}x \right)2\sqrt{\left( 1 - \frac{5x}{4} \right)}}$

= $\left( 1 + \frac{3x}{8} \right)^{2/3}$ $\left( 1 + \frac { 3 } { 2 } x \right) ^ { - 1 }$ $\left( 1 - \frac{5x}{4} \right)^{- 1/2}$

= $\left( 1 + \frac{2}{3} \cdot \frac{3x}{8} + ... \right)$ $\left( 1 - \frac { 3 } { 2 } x + \ldots \right)$ $\left( 1 + \frac{5}{8}x + ... \right)$

= 1 + $\left( \frac{1}{4} - \frac{3}{2} + \frac{5}{8} \right)$x = 1 – $\frac{5}{8}$x.