Question

The domain of function $f\left(x\right)=\frac{1}{[|x-1|]+[|7-x|]-6}$ where [ ] denotes the greatest integral function is $x\in R-\left\{\right(0,1]\cup \{1,a,3,b,c,6,7\}\cup [d,e\left)\right\}$. Find $\frac{a+b+c+d+e}{13}$

Answer 1

$f\left(x\right)$ is defined when
$[|x-1|]+[|7-x|]-6\ne 0$
$\Rightarrow \{\begin{array}{ccc}[1-x]+[7-x]\ne 6,& when& x\le 1.....\left(i\right))\\ [x-1]+[7-x]\ne 6,& when& 1\le x\le 7....\left(ii\right)\\ [x-1]+[x-7]\ne 6,& when& x\ge 7.....\left(iii\right)\end{array}$
Taking Eq.(1), we have
$[1-x]+[7-x]\ne 6$
$1+[-x]+7+[-x]\ne 6$
$\Rightarrow 2[-x]\ne -2\Rightarrow [-x]\ne -1$
$\Rightarrow x\notin (0,1]$ .....(a)
From Eq. (ii), we have
$[x-1]+[7-x]\ne 6$
$\Rightarrow \left[x\right]-1+7+[-x]\ne 6\Rightarrow \left[x\right]+[-x]\ne 0$
$\Rightarrow x\notin I$
$\Rightarrow x\notin \{1,2,3,4,5,6,7\}$ ....(b)
From Eq. (iii), we have $[x-1]+[x-7]\ne 6$
$\Rightarrow \left[x\right]-1+\left[x\right]-7\ne 6\Rightarrow 2\left[x\right]\ne 14$
$\Rightarrow \left[x\right]\ne 7\Rightarrow x\notin [7,8)$ ....(c)
Hence, from statements (a), (b) and (c), we have
Domain $f\left(x\right)\in R-\left\{\right(0,1]\cup \{1,2,3,4,5,6,7\}\cup [7,8\left)\right\}$
$\Rightarrow a+b+c+d+e=26$

Hence, $\frac{a+b+c+d+e}{13}=2$