Question

Find the domain of the function

$\sqrt{ log_{0.4}( \frac{x - 1}{x + 5} ) } \times \frac{1}{ {x}^{2} - 36}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: \sqrt{ log_{0.4}\bigg( \dfrac{x - 1}{x + 5} \bigg) } \times \dfrac{1}{ {x}^{2} - 36}$

Let assume that

$\rm \:f(x) = \sqrt{ log_{0.4}\bigg( \dfrac{x - 1}{x + 5} \bigg) } \times \dfrac{1}{ {x}^{2} - 36}$

Let further assume that

$\rm \: f(x) = g(x) \times h(x)$

where,

$\rm \: g(x) = \sqrt{ log_{0.4}\bigg( \dfrac{x - 1}{x + 5} \bigg) }$

and

$\rm \: h(x) = \dfrac{1}{ {x}^{2} - 36}$

So, Domain of f(x) is given by

$\rm \: D_{f} \: = \: D_{g} \: \cap \: D_{h}$

Now, Consider

$\rm \: h(x) = \dfrac{1}{ {x}^{2} - 36}$

can be rewritten as

$\rm \: h(x) = \dfrac{1}{ {x}^{2} - {6}^{2} }$

$\rm \: h(x) = \dfrac{1}{(x - 6)(x + 6)}$

$\rm\implies \:D_h \: : \: x \: \ne \: \{ - 6, \: 6 \}$

Now, Consider

$\rm \: g(x) = \sqrt{ log_{0.4}\bigg( \dfrac{x - 1}{x + 5} \bigg) }$

We know,

$\boxed{\sf{  \: log_{a}(x) \: if \: 0 < a < 1 \: \: then \: 0 < x < 1 \: }}$

So,

$\rm \: \sqrt{ log_{0.4}\bigg( \dfrac{x - 1}{x + 5} \bigg) } \: exist \: \: when$

$\rm \: 0 < \dfrac{x - 1}{x + 5} < 1$

So, it means

$\rm \: \dfrac{x - 1}{x + 5} > 0 \: and \: \dfrac{x - 1}{x + 5} < 1$

$\rm \: \dfrac{x - 1}{x + 5} > 0 \: and \: \dfrac{x - 1}{x + 5} - 1 < 0$

$\rm \: \dfrac{x - 1}{x + 5} > 0 \: and \: \dfrac{x - 1 - x - 5}{x + 5} < 0$

$\rm \: \dfrac{x - 1}{x + 5} > 0 \: and \: \dfrac{ - 6}{x + 5} < 0$

$\rm \: \dfrac{x - 1}{x + 5} > 0 \: and \: \dfrac{6}{x + 5} > 0$

$\rm \: x < - 5 \: or \: x > 1 \: \: and \: \: x > - 5$

$\rm\implies \:x > 1$

$\rm\implies \:D_g \: : x \: \in \: (1, \infty )$

Now,

$\rm \: D_f \: = \: D_g \: \cap \: D_h$

$\rm \: D_f \:: \: x \: \in \: (1, \: \infty ) \: \cap \: \{ - 6, \: 6 \}$

$\rm \: D_f \:: \: x \: \in \: (1, \: \infty ) \: - \{6 \}$

Hence,

$\rm \: Domain \: of \: \sqrt{ log_{0.4}\bigg( \dfrac{x - 1}{x + 5} \bigg) } \times \dfrac{1}{ {x}^{2} - 36} \: is \: \: x \: \in \: (1, \: \infty ) \: - \{6 \}$