AP EAMCET · Maths · Binomial Theorem
If \(x\) is so small that \(x^2\) and higher powers of \(x\) can be neglected, then the approximate value of \(\left(1+\frac{3}{4} x\right)^{\frac{1}{2}}\left(1-\frac{2 x}{3}\right)^{-2}\) is
- A \(\frac{41+24 x}{41}\)
- B \(\frac{41-24 x}{41}\)
- C \(\frac{24+41 x}{24}\)
- D \(\frac{24-41 x}{24}\)
Answer & Solution
Correct Answer
(C) \(\frac{24+41 x}{24}\)
Step-by-step Solution
Detailed explanation
If \(x\) is so small that \(x^2\) and higher powers of \(x\) can be neglected, then \[ (1+x)^n=1+n x \] So, \(\left(1+\frac{3}{4} x\right)^{1 / 2}\left(1-\frac{2}{3} x\right)^{-2}=\left(1+\frac{3}{8} x\right)\left(1+\frac{4 x}{3}\right)\) \(=1+\frac{3}{8} x+\frac{4 x}{3} \quad\)…
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