Let the domain of the function f(x)= ^-1((4x+5)/(3x-7)) be [ , ] and the domain of g(x)= _…

Question

Let the domain of the function $f(x)={\cos }^{-1}(\frac{4x+5}{3x-7})$ be $[\alpha ,\beta ]$ and the domain of $g(x)={\log }_{2}(2-6{\log }_{27}(2x+5))$ be $(\gamma ,\delta )$.

Then $|7(\alpha +\beta )+4(\gamma +\delta )|$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 96

Accepted Answer

$\begin{gathered} f(x)={\cos }^{-1}(\frac{4x+5}{3x-7}) \ \Rightarrow -1\le (\frac{4x+5}{3x-7})\le 1 \ (\frac{4x+5}{3x-7})\ge -1 \ \frac{4x+5+3x-7}{3x-7}\ge 0 \ \Rightarrow \frac{7x-2}{3x-7}\ge 0 \end{gathered}$

$\begin{gathered} x\in (-\infty,\frac{2}{7}]\cup (\frac{7}{3},\infty) \ \frac{4x+5}{3x-7}\le 1\Rightarrow \frac{x+12}{3x-7}\le 0 $

$\therefore $ Domain of $f(x)$ is

$[-12,\frac{2}{7}]\alpha =-12,\beta =\frac{2}{7}$

$g(x)={\log }_{2}(2-6{\log }_{27}(2x+5))$

Domain

$2-6{\log }_{27}(2x+5)>0$

$ $$Rightarrow \ 6{\log }_{27}(2x+5)$\2x+5>0\Rightarrow x>-\frac{5}{2}$$$

Domain is $x\in (-\frac{5}{2},-1)$

$$$ \gamma =-\frac{5}{2},\delta =-1 \ |7(\alpha +\beta )+4(\gamma +\delta )|=|\, 7(-12+\frac{2}{7}+4(-\frac{5}{2}-1)|\, \ |-82-14|=96 \end{gathered}$$

Solutions · 1