For |x|

Question

(C) Given that $|x| < \frac{4}{3}$.We have $\frac{1}{(4-3 x)^{\frac{1}{2}}} = (4-3 x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \left(1-\frac{3 x}{4}\right)^{

Vedclass · Accepted Answer

$\frac{1}{2}+\frac{3 x}{16}+\frac{27 x^2}{256}$

Vedclass · Answer

(C) Given that $|x| We have $\frac{1}{(4-3 x)^{\frac{1}{2}}} = (4-3 x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}}$.Using the binomial expansion $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!} u^2 + \dots$, where $u = -\frac{3x}{4}$ and $n = -\frac{1}{2}$:$\frac{1}{2} \left(1-\frac{3 x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2} \left[ 1 + \left(-\frac{1}{2}\right) \left(-\frac{3x}{4}\right) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2} \left(-\frac{3x}{4}\right)^2 + \dots \right]$$= \frac{1}{2} \left[ 1 + \frac{3x}{8} + \frac{3}{8} \cdot \frac{9x^2}{16} + \dots \right]$$= \frac{1}{2} \left[ 1 + \frac{3x}{8} + \frac{27x^2}{128} + \dots \right]$$= \frac{1}{2} + \frac{3x}{16} + \frac{27x^2}{256} + \dots$

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	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

For $|x| < \frac{4}{3}$, the approximate value of $\frac{1}{(4-3 x)^{\frac{1}{2}}}$ is

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