The coefficient of x in the expansion of [sqr | vedclass

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(D) Given expression: $[\sqrt{1 + x^2} - x]^{-1} = \frac{1}{\sqrt{1 + x^2} - x}$.Rationalizing the denominator:$\frac{1}{\sqrt{1 + x^2} - x} 	imes \f

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$1$

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(D) Given expression: $[\sqrt{1 + x^2} - x]^{-1} = \frac{1}{\sqrt{1 + x^2} - x}$.Rationalizing the denominator:$\frac{1}{\sqrt{1 + x^2} - x} 	imes \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2} + x} = \frac{\sqrt{1 + x^2} + x}{(1 + x^2) - x^2} = \sqrt{1 + x^2} + x$.Using the binomial expansion $(1 + x^2)^{1/2} = 1 + \frac{1}{2}x^2 + \dots$ for $|x| Expression $= 1 + \frac{1}{2}x^2 + \dots + x = 1 + x + \frac{1}{2}x^2 + \dots$.The coefficient of $x$ in this expansion is $1$.

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

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