If the domain of the function f(x)= _(0.6)(|2x-5x^2-4|) is (-,a] b [c,d) (e,), then the va…

Question

If the domain of the function

$f(x)=\sqrt{{\log }_{(0.6)}(|\frac{2x-5}{{x}^{2}-4}|)}$ is $(-\infty,a]\cup {b}\cup [c,d)\cup (e,\infty)$, then the value of $a+b+c+d+e$ is \_\_\_\_\_\_\_\_.

Answer: 4

Accepted Answer

For domain ${\log }_{0.6}|\frac{2x-5}{{x}^{2}-4}|\ge 0$

$|\frac{2x-5}{{x}^{2}-4}|\le 1\, \&\, x
e \frac{5}{2}\, \ldots ..\, (1)$

$-1\le \frac{2x-5}{{x}^{2}-4}\le 1$

$\frac{2x-5}{{x}^{2}-4}+1\ge 0$

$\frac{{x}^{2}+2x-9}{{x}^{2}-4}\ge 0$

$\frac{(x+1{)}^{2}-10}{(x-2)(x+2)}\ge 0$

$x\in (-\infty,-1-\sqrt{10}]\cup (-2,2)\cup [-1+\sqrt{10},\infty)\, \ldots ..(2)$

$\frac{2x-5}{{x}^{2}-4}-1\le 0$

$\frac{2x-5-{x}^{2}+4}{{x}^{2}-4}\le 0$

$\frac{{x}^{2}-2x+1}{{x}^{2}-4}\ge 0$

$\frac{(x-1{)}^{2}}{(x-2)(x+2)}\ge 0$

$x\in (-\infty,-2)\cup (2,\infty)\cup {1}\, \ldots ..(3)$

$(1)\cap (2)\cap (3)$

$x\in (-\infty,-1-\sqrt{10}]\cup {1}\cup [-1+\sqrt{10},\frac{5}{2})\cup (\frac{5}{2},\infty)$

$a+b+c+d+e=-2+1+5=4$

Solutions · 1