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

For function $f$ to be defined, $x^{2}-36\neq 0$
and $\frac{x - 1}{x + 5}>0$ and $\left(\log\right)_{0 . 4}\left(\frac{x - 1}{x + 5}\right)\geq 0$
$\Rightarrow x^{2}\neq 36$ and $\frac{x - 1}{x + 5}>0$
and $\frac{x - 1}{x + 5}\leq 1$
$\Rightarrow x\neq \pm6$ and $\left(\right.x < -5\text{ or }x>1\left.\right)$ and $\frac{x - 1 - x - 5}{x + 5}\leq 0$
$\Rightarrow x\neq \pm6$ and $\left(\right.x < -5\text{ or }x>1\left.\right)$ and $\frac{- 6}{x + 5}\leq 0$
$\Rightarrow x\neq \pm6$ and $\left(\right.x < -5\text{ or }x>1\left.\right)$ and $x+5>0$
$\Rightarrow x\neq \pm6$ and $\left(\right.x < -5\text{ or }x>1\left.\right)$ and $x>-5$
$\Rightarrow x\in \left(\right.1,\infty \left.\right)-\left\{\right.6\left.\right\}$
$\Rightarrow $ The domain is $\left(\right.1,\infty \left.\right)-\left\{\right.6\left.\right\}$
i.e.$ (1,6) \cup\left(\right.6,\infty \left.\right)$
$\Rightarrow $ The required length $=5$