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  4. If x is so small, that x5 and higher power of x may be neglected, then the coefficient of x4 in the expansion of √x2+4-√x2+9, is

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Q. 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

AP EAMCETAP EAMCET 2019

Solution:

Given, $\sqrt{x^{2}+4}-\sqrt{x^{2}+9}$
$=\left(x^{2}+4\right)^{\frac{1}{2}}-\left(x^{2}+9\right)^{\frac{1}{2}} $
$=2\left(1+\frac{x^{2}}{4}\right)^{\frac{1}{2}}-3\left(1+\frac{x^{2}}{9}\right)^{\frac{1}{2}}$
Now, coefficient of $x^{4}$ is
$\left[2 \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(\frac{x^{2}}{4}\right)^{2}-3\left(\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}\left(\frac{x^{2}}{9}\right)^{2}\right)\right]$
$=x^{4}\left[-\frac{1}{4 \times 16}+3 \times \frac{1}{8} \times \frac{1}{81}\right]=x^{4}\left[-\frac{1}{64}+\frac{1}{216}\right]$
$\therefore $ Coefficient of $x^{4}$ is $\left(-\frac{1}{64}+\frac{1}{216}\right)$
$=\frac{-216+64}{(64)(216)}=\frac{-152}{(64)(216)}=\frac{-19}{1728}$