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AP EAMCET · Maths · Binomial Theorem

If \(x\) is so small, that \(x^5\) and higher power of \(x\) may be neglected, then the coefficient of \(x^4\) in the expansion of \(\sqrt{x^2+4}-\sqrt{x^2+9}\), is

  1. A \(\frac{19}{1728}\)
  2. B \(\frac{-19}{1728}\)
  3. C \(\frac{43}{1728}\)
  4. D \(\frac{-43}{1728}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{-19}{1728}\)

Step-by-step Solution

Detailed explanation

Given, \(\sqrt{x^2+4}-\sqrt{x^2+9}\) \(\begin{aligned} & =\left(x^2+4\right)^{\frac{1}{2}}-\left(x^2+9\right)^{\frac{1}{2}} \\ & =2\left(1+\frac{x^2}{4}\right)^{\frac{1}{2}}-3\left(1+\frac{x^2}{9}\right)^{\frac{1}{2}} \end{aligned}\) Now, coefficient of \(x^4\) is…