Question

The function $f$ defined as $f\left(x\right)=\sqrt{{\left(\log \right)}_{0.4}\left(\frac{x-1}{x+5}\right)}\times \frac{1}{x^2-36}$ has domain as the union of two intervals, of unequal length. Then the length of the smaller interval will be :

Answer 1

Answer: 5

For function $f$ to be defined, $x^2-36\ne 0$
and $\frac{x-1}{x+5}>0$ and ${\left(\log \right)}_{0.4}\left(\frac{x-1}{x+5}\right)\ge 0$
$\Rightarrow x^2\ne 36$ and $\frac{x-1}{x+5}>0$
and $\frac{x-1}{x+5}\le 1$
$\Rightarrow x\ne \pm 6$ and $(x<-5\text{ or }x>1)$ and $\frac{x-1-x-5}{x+5}\le 0$
$\Rightarrow x\ne \pm 6$ and $(x<-5\text{ or }x>1)$ and $\frac{-6}{x+5}\le 0$
$\Rightarrow x\ne \pm 6$ and $(x<-5\text{ or }x>1)$ and $x+5>0$
$\Rightarrow x\ne \pm 6$ and $(x<-5\text{ or }x>1)$ and $x>-5$
$\Rightarrow x\in (1,\infty )-\{6\}$
$\Rightarrow$ The domain is $(1,\infty )-\{6\}$
i.e.$(1,6)\cup (6,\infty )$
$\Rightarrow$ The required length $=5$