TS EAMCET · Maths · Binomial Theorem
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 \(\frac{1}{32}\left(1-\frac{11 x}{8}+\frac{219}{128} x^2\right)\)
- D \(\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right)\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\sqrt{4-x}(2+x)^3}=(4-x)^{-1 / 2} \cdot(2+x)^{-3}\)…
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