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

Question

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, $\frac{19}{1728}$, $\frac{-19}{1728}$, $\frac{43}{1728}$, $\frac{-43}{1728}$

MARKS App · Accepted Answer

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 $\left[2 \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(\frac{x^2}{4}\right)^2-3\left(\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(\frac{x^2}{9}\right)^2\right]\right]$ $\begin{aligned} & \quad=x^4\left[-\frac{1}{4 \times 16}+3 \times \frac{1}{8} \times \frac{1}{81}\right]=x^4\left[-\frac{1}{64}+\frac{1}{216}\right] \ & \therefore \text { Coefficient of } x^4 \text { is }\left(-\frac{1}{64}+\frac{1}{216}\right) \ & \quad=\frac{-216+64}{(64)(216)}=\frac{-152}{(64)(216)}=\frac{-19}{1728}\end{aligned}$

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