TS EAMCET · Maths · Binomial Theorem
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
- A \(\frac{32+31 x}{64}\)
- B \(\frac{31+32 x}{64}\)
- C \(\frac{31-32 x}{64}\)
- D \(\frac{1-2 x}{64}\)
Answer & Solution
Correct Answer
(A) \(\frac{32+31 x}{64}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} (1+ & \left.\frac{2 x}{3}\right)^{3 / 2}(32+5 x)^{-1 / 5} \\ & =\left[1+\frac{3}{2}\left(\frac{2 x}{3}\right)\right](32)^{-1 / 5}\left(1+\frac{5}{32} x\right)^{-1 / 5} \end{aligned}\) (Neglect higher powers of \(x\) )…
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