Question

Consider a function $f:(-\infty ,\frac{-A}{8})\cup (\frac{-A}{8},\frac{-B}{8})\to R$ such that $f\left(x\right)=lo{g}_{|x-1|-|x+1|-\frac{1}{2}}(x^2-x-\frac{1}{2}{)}^{x^2-3x+2}$, then the value of $A+B$ is

Answer 1

We have, $f:(-\infty ,\frac{-A}{8})\cup (\frac{-A}{8},\frac{-B}{8})\to R$ such that
$f\left(x\right)=lo{g}_{|x-1|-|x+1|-\frac{1}{2}}(x^2-x+\frac{1}{2}{)}^{x^2-3x+2}$
$LetP=|x-1|-|x+1|-\frac{1}{2}andQ=(x^2-x-\frac{1}{2}{)}^{x^2-3x+2}$
For the function $f\left(x\right)$ to be defined,
$P\ne 1,P>0andQ>0$
$P=\{\begin{array}{cc}-(x-1)-[-(x+1\left)\right]-\frac{1}{2};& x\le -1\\ x-1-[-(x+1\left)\right]-\frac{1}{2};& -1<x<1\\ (x-1)-(x+1)-\frac{1}{2};& x\ge 1\end{array}$

$P=\{\begin{array}{cc}\frac{3}{2};& x\le -1\\ 2x-\frac{1}{2};& -1<x<1\\ -\frac{5}{2};& x\ge 1\end{array}$
Now, f(x) has a valid base for $x\le -1$ and
for $-1<x<1,-2x-\frac{1}{2}>0and-2x-\frac{1}{2}\ne 0\Rightarrow x<-\frac{1}{4}andx\ne -\frac{3}{4}\therefore x\in (-\infty ,-\frac{1}{4})-\{-\frac{3}{4}\}$
$Q>0\Rightarrow (x^2-x+\frac{1}{2}{)}^{x^2-3x+2}>0\Rightarrow {(x-\frac{1}{2})}^2+\frac{1}{4}>0,\text{which is true for all}x\in R$
Therefore, the domain of function $f\left(x\right)is(-\infty ,-\frac{1}{4})-\{-\frac{3}{4}\}$
$\therefore A=6andB=2A+B=8$