Question

The coefficient of $x$ in the expansion of ${[\sqrt{1+x^2}-x]}^{-1}$ in ascending powers of $x$, when $\left|x\right|<1$, is

• A. $0$
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $1$

Answer 1

The correct option is D $1$

We have ${(\sqrt{1+x^2}-x)}^{-1}=\frac{1}{(\sqrt{1+x^2}-x)}\times \frac{(\sqrt{1+x^2}+x)}{(\sqrt{1+x^2}+x)}$

$=\frac{\sqrt{1+x^2}+x}{1+x^2-x^2}={(1+x^2)}^{\frac{1}{2}}+x$

$\therefore$ coefficients of $x$ in the expansion of ${(\sqrt{1+x^2}-x)}^{-1}$

$=$ coefficients of $x$ in the expansion of $[{(1+x^2)}^{\frac{1}{2}}+x]$

$=1$