1)
Let $x\neq0, |x|<\frac{1}{2}$ and
$ f(x)= 1+2x+4x^{2}+8x^{3}+....$ Then $f^{-1}(x)= $
Answer:
Option A
Explanation:
We have
$ f(x)= 1+2x+4x^{2}+8x^{3}+....$ and |x| < $\frac{1}{2}$
Let f(x)=y
$y= 1+2x+4x^{2}+8x^{3}+.....$
= $1+(2x)+(2x)^{2}+(2x)^{3}+....$
$y= \frac{1}{1-2x}$ [ $\because$ sum of infinite G.P, $S_{\infty}= \frac{a}{1-r}]$
$\Rightarrow$ $ y-2xy=1 $ $\Rightarrow$ $2xy=y-1$
$\Rightarrow$ $x=\frac{y-1}{2y} \Rightarrow f^{-1}(y)=\frac{y-1}{2y}$
hence, $f^{-1}= \frac{x-1}{2x}$
