AP EAMCET · Maths · Binomial Theorem
When \(\mathrm{x}\) is so small that its square and its higher powers may be neglected, then the value of \(\frac{\left(1+\frac{3}{4} x\right)^{-4} \sqrt{(3+x)}}{\sqrt{(3-x)^3}}\) is approximately equal to
- A \(\frac{1}{3}-\frac{7 x}{9}\)
- B \(\frac{1}{3}+\frac{7 x}{9}\)
- C \(\frac{1}{3}+\frac{11 x}{18}\)
- D \(\frac{1}{3}-\frac{11 x}{18}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}-\frac{7 x}{9}\)
Step-by-step Solution
Detailed explanation
Given that \(x^n \approx 0, n=2,3,4 \ldots\) Let \(y=\frac{\left(1+\frac{3}{4} x\right)^{-4} \sqrt{(3+x)}}{\sqrt{(3-x)^3}}\)…
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