Q44 / 120 · VITEEE 2009 · Mathematics
If $x$ is numerically so small so that $x^{2}$ and higher powers of $x$ can be neglected, then
$\left(1+\frac{2 x}{3}\right)^{3 / 2} \cdot(32+5 x)^{-1 / 5}$
is approximately equal to
- Correct answerA$\frac{32+31 x}{64}$
- B$\frac{31+32 x}{64}$
- C$\frac{31-32 x}{64}$
- D$\frac{1-2 x}{64}$
Solution
$\begin{aligned} &\left(1+\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5} \\ &=\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right](32)^{-1 / 5}\left(1+\frac{5}{32} x\right)^{-1 / 5} \\ & \quad(\text { neglecting higher powers of } x) \\ &=[1+x] 2^{-1}\left[1-\frac{1}{5}\left(\frac{5}{32}\right) x\right] \\ & \quad(\text { neglecting higher powers of } x) \\ &=\frac{1}{2}(1+x)\left(1-\frac{x}{32}\right) \\ &=\frac{(1+x)(32-x)}{64}=\frac{32+31 x}{64} \quad\left(\text { neglecting } x^{2} \text { term }\right) \end{aligned}$
Hence, the correct answer is option: 'A'.