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

• A. $\frac{19}{1728}$
• B. $\frac{-19}{1728}$
• C. $\frac{43}{1728}$
• D. $\frac{-43}{1728}$

Answer 1

Answer: (B)

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 $\left(-\frac{1}{64}+\frac{1}{216}\right)$
$=\frac{-216+64}{(64)(216)}=\frac{-152}{(64)(216)}=\frac{-19}{1728}$