Q63 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics
If $|x|$ is so small that $x^3$ and higher powers of $x$ can be neglected, then an approximate value of $\frac{1}{\sqrt{4-x}(2+x)^3}$ is
- A$\frac{1}{16}\left(1+\frac{13 x}{8}+\frac{219}{128} x^2\right)$
- B$\frac{1}{8}\left(1+\frac{11 x}{8}-\frac{165}{128} x^2\right)$
- C$\frac{1}{32}\left(1-\frac{11 x}{8}+\frac{219}{128} x^2\right)$
- Correct answerD$\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right)$
Solution
$\frac{1}{\sqrt{4-x}(2+x)^3}=(4-x)^{-1 / 2} \cdot(2+x)^{-3}$
$\begin{aligned} & =4^{-1 / 2} \cdot 2^{-3}\left(1-\frac{x}{4}\right)^{-1 / 2}\left(1+\frac{x}{2}\right)^{-3} \\ & =\frac{i}{16}\left(1+\frac{1}{2} \cdot \frac{x}{4}+\frac{1}{2}\left(\frac{3}{2}\right) \frac{x^2}{16}+. .\right)\left(1+(-3)\left(\frac{x}{2}\right)+3(4)\left(\frac{x}{2}\right)^2+. .\right) \\ & =\frac{1}{16}\left(1-\frac{3 x}{2}+\frac{x}{8}+6 x^2-\frac{3}{16} x^2+\frac{3}{64} x^2\right) \\ & =\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right) .\end{aligned}$
Hence, the correct answer is option: 'D'.