If x is so small that x^5 and higher powers o | vedclass

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(B) Given expression: $\sqrt{x^2+4}-\sqrt{x^2+9}$ $= 2(1+\frac{x^2}{4})^{1/2} - 3(1+\frac{x^2}{9})^{1/2}$ Using the binomial expansion $(1+u)^n = 1 +

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$\frac{-19}{1728}$

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(B) Given expression: $\sqrt{x^2+4}-\sqrt{x^2+9}$ $= 2(1+\frac{x^2}{4})^{1/2} - 3(1+\frac{x^2}{9})^{1/2}$ Using the binomial expansion $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots$ $= 2[1 + \frac{1}{2}(\frac{x^2}{4}) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(\frac{x^2}{4})^2 + \dots] - 3[1 + \frac{1}{2}(\frac{x^2}{9}) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(\frac{x^2}{9})^2 + \dots]$ The $x^4$ term is: $2[\frac{-1/4}{2}(\frac{x^4}{16})] - 3[\frac{-1/4}{2}(\frac{x^4}{81})]$ $= 2[-\frac{1}{8} \cdot \frac{x^4}{16}] - 3[-\frac{1}{8} \cdot \frac{x^4}{81}]$ $= -\frac{x^4}{64} + \frac{x^4}{216} = x^4(\frac{-216+64}{13824}) = x^4(\frac{-152}{13824}) = -\frac{19}{1728}x^4$ Thus,the coefficient of $x^4$ is $-\frac{19}{1728}$.

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

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