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 +

Vedclass · Accepted Answer

$\frac{-19}{1728}$

Vedclass · Answer

(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 $-\frac{19}{1728}$.

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $\|x\| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $\|x\| < 1$
$(D)$ If $\|x\|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

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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