Q58 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics
If $|x|$ is so small that $x^2$ and higher powers of $x$ may be neglected, then the approximate value of $\frac{\sqrt{4+x}+\sqrt[3]{8-x}}{\left(1-\frac{2 x}{3}\right)^{\frac{3}{2}}}$ when $x=\frac{6}{25}$ is
- A6
- Correct answerB5
- C$\frac{2}{3}$
- D$\frac{5}{6}$
Solution
Using binomial expansions and neglecting $x^2$ and higher powers, we get
$$
\begin{aligned}
\frac{\left[2\left(1+\frac{x}{4}\right)^{\frac{1}{2}}+2\left(1-\frac{x}{8}\right)^{\frac{1}{3}}\right]}{\left(1-\frac{2 x}{3}\right)^{\frac{3}{2}}} \\
=2\left[1+\frac{1}{2} \cdot \frac{x}{4}+1-\frac{1}{3} \cdot \frac{x}{8}\right]\left(1-\frac{2 x}{3}\right)^{-\frac{3}{2}} \\
=2\left[2+\frac{x}{8}-\frac{x}{24}\right]\left[1-\frac{3}{2} \cdot\left(\frac{-2 x}{3}\right)\right] \\
=2\left[2+\frac{2 x}{24}\right][1+x]
\end{aligned}
$$
$$
\begin{aligned}
& =2\left[2+\frac{x}{12}\right][1+x]=2\left[2+2 x+\frac{x}{12}\right] \\
& =2\left[2+\frac{25 x}{12}\right]
\end{aligned}
$$
When, $x=\frac{6}{25}$
$$
=2\left[2+\frac{25}{12} \cdot \frac{6}{25}\right]=2\left[2+\frac{1}{2}\right]=2 \cdot \frac{5}{2}=5
$$
Hence, the correct answer is option: 'B'.