If x sqrt{3} and frac{x^2+1}{(x^2+2)(x^2+3) | vedclass

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(D) Let $u = x^{-2}$. Since $x > \sqrt{3}$,we have $x^2 > 3$,so $u = \frac{1}{x^2} < \frac{1}{3}$.We rewrite the expression as $\frac{x^2+1}{(x^2+2)(x

Vedclass · Accepted Answer

$-46$

Vedclass · Answer

(D) Let $u = x^{-2}$. Since $x > \sqrt{3}$,we have $x^2 > 3$,so $u = \frac{1}{x^2} We rewrite the expression as $\frac{x^2+1}{(x^2+2)(x^2+3)}$.Dividing numerator and denominator by $x^4$,we get $\frac{x^{-2} + x^{-4}}{(1 + 2x^{-2})(1 + 3x^{-2})} = (u + u^2)(1 + 2u)^{-1}(1 + 3u)^{-1}$.Using the binomial expansion $(1+z)^{-1} = 1 - z + z^2 - z^3 + \dots$,we have:$(1+2u)^{-1} = 1 - 2u + 4u^2 - 8u^3 + \dots$$(1+3u)^{-1} = 1 - 3u + 9u^2 - 27u^3 + \dots$Multiplying these: $(1+2u)^{-1}(1+3u)^{-1} = (1 - 2u + 4u^2 - 8u^3 + \dots)(1 - 3u + 9u^2 - 27u^3 + \dots) = 1 - 5u + 19u^2 - 65u^3 + \dots$Now,$(u + u^2)(1 - 5u + 19u^2 - 65u^3 + \dots) = u - 5u^2 + 19u^3 - 65u^4 + u^2 - 5u^3 + 19u^4 - 65u^5 + \dots$$= u - 4u^2 + 14u^3 - 46u^4 + \dots$The term $x^{-8}$ corresponds to $u^4$. The coefficient is $-46$.

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $\|x\| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $\|x\| < 1$
$(D)$ If $\|x\|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

If $x > \sqrt{3}$ and $\frac{x^2+1}{(x^2+2)(x^2+3)}$ is expanded in terms of powers of $x^{-2}$,then the coefficient of $x^{-8}$ is

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