Question

An odd function is symmetric about the vertical line $x=a\left(a>0\right)$ and if $\sum _{r=0}^{\infty }{\left[f\left(1+4ar\right)\right]}^r=8$ , then find the numerical value of $8f\left(1\right)$ $($ where $0<f\left(1\right)<1)$ .

Answer 1

Answer: 7

Given that,
$f\left(-x\right)=-f\left(x\right)$
and $f\left(2a-x\right)=f\left(x\right)$
By adding both equations,
$\Rightarrow f\left(-x\right)+f\left(2a-x\right)=0$
Replace $x$ with $\left(-x\right)$
$f\left(x\right)+f\left(x+2a\right)=\mathrm{0....}\left(1\right)$
Replace $x$ with $x+2a$
$\Rightarrow f\left(x+2a\right)+f\left(x+4a\right)=\mathrm{0....}\left(2\right)$
Subtract equation $\left(1\right)$ and equation $\left(2\right)$
$\Rightarrow f\left(x\right)-f\left(x+4a\right)=0$
$\Rightarrow f\left(x\right)=f\left(x+4a\right)$
$\Rightarrow f\left(x\right)$ is a periodic function and its period is $4a$
$\Rightarrow f\left(1\right)=f\left(1+4a\right)=f\left(1+8a\right)$
Now,
$\sum _{r=0}^{\infty }f{\left(1+4ar\right)}^r=8$
$\Rightarrow {\left[f\left(1\right)\right]}^0+\left[f\left(1+4a\right)\right]+{\left[f\left(1+8a\right)\right]}^2+...........=8$
$\Rightarrow 1+f\left(1\right)+{\left[f\left(1\right)\right]}^2+{\left[f\left(1\right)\right]}^3+..........=8$
$\Rightarrow \frac{1}{1-f\left(1\right)}=8$
$\Rightarrow f\left(1\right)=\frac{7}{8}$
$\Rightarrow 8f\left(1\right)=7$