Q57 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{\left(1+\frac{2 x}{3}\right)^{-4}(4+5 x)^{1 / 2}}{(9+x)^{3 / 2}}$ when $x=\frac{6}{371}$, is
- A$\frac{1}{27}$
- B$\frac{29}{378}$
- C$\frac{3}{27}$
- Correct answerD$\frac{1}{14}$
Solution
We have,
$\begin{aligned}
& \frac{\left(1+\frac{2 x}{3}\right)^{-4}(4+5 x)^{1 / 2}}{(9+x)^{3 / 2}} \\
& =\left(1+\frac{2 x}{3}\right)^{-4} \times 2\left(1+\frac{5}{4} x\right)^{1 / 2} \times(9+x)^{-3 / 2} \\
& =\left(1+\frac{2 x}{3}\right)^{-4} \times 2 \times\left(1+\frac{5}{4} x\right)^{1 / 2} \times \frac{1}{27}\left(1+\frac{x}{9}\right)^{-3 / 2} \\
& =\frac{2}{27}\left(1-\frac{8 x}{3}\right)\left(1+\frac{5}{8} x\right)\left(1-\frac{1}{6} x\right) \\
& =\frac{2}{27}\left(1-\frac{8}{3} x+\frac{5}{8} x-\frac{1}{6} x\right)=\frac{2}{27}\left(1-\frac{53}{24} x\right)
\end{aligned}$
$\begin{aligned}
\text { Put, } x=\frac{6}{371} & =\frac{2}{27}\left(1-\frac{53 \times 6}{24 \times 371}\right) \\
& =\frac{2}{27}\left(1-\frac{53}{1484}\right)=\frac{2}{27} \times \frac{1431}{1484}=\frac{1}{14}
\end{aligned}$
Hence, the correct answer is option: 'D'.