Question

If x is so small that x² and higher powers can be neglected, then $\frac{(1 + 3x/4)^{- 4}(16 - 3x)^{1/2}}{(8 + x)^{2/3}}$equals –

• A. 1 – $\frac{305}{96}$x
• B. 1 + $\frac{305}{96}$x
• C. $\frac{1}{2}$+ $\frac{7}{8}$x
• D. None

Answer 1

Answer: (A)

1 – $\frac{305}{96}$x

Answer 2

$\frac{\left( 1 + \frac{3x}{4} \right)^{- 4}(16 - 3x)^{1/2}}{(8 + x)^{2/3}}$

=$\frac{\left( 1 + ( - 4)\left( \frac{3x}{4} \right) \right)(16)^{1/2}\left( 1 - \frac{3x}{16} \right)^{1/2}}{8^{2/3}\left( 1 + \frac{x}{8} \right)^{2/3}}$

= $\frac{(1 - 3x)4\left( 1 + \frac{1}{2}\left( - \frac{3x}{16} \right) \right)\left( 1 + \frac{x}{8} \right)^{- 2/3}}{4}$

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

= $\left( 1 - \frac{x}{12} \right)$= $\left( 1 - \frac { 99 } { 32 } x \right)$ $\left( 1 - \frac{x}{12} \right)$

= 1–$\left( \frac{99}{32} \right)$x – $\frac{x}{12}$+ 0 = 1 – $\frac{305}{96}$x.