If the domain of the function f(x)= ^-1((2-|x|)/(4))+ _e(3-x)^-1 is [- , )- , then + + is …

Question

If the domain of the function $f(x)={\cos }^{-1}(\frac{2-|x|}{4})+{{{\log }_{e}(3-x)}}^{-1}$ is $[-\alpha ,\beta )-{\gamma }$, then $\alpha +\beta +\gamma $ is equal to :

Accepted Answer

$\begin{gathered} -1\le |\frac{2-|x|}{4}|\le 1 \ \Rightarrow |\frac{2-|x|}{4}|\le 1 \ -4\le 2-|x|\le 4 \ -6\le -|x|\le 2 \ -2\le |x|\le 6 \ |x|\le 6 \end{gathered}$

$\Rightarrow x\in [-6,6]$ .... (1)

Now, $3-x\ne 1$

And $x\ne 2$ .... (2)

and $3-x>0$

$x$$$\begin{gathered} \Rightarrow x\in [-6,3)-{2} \ \alpha =6 \ \beta =3 \ \gamma =2 \ \alpha +\beta +\gamma =11 \end{gathered}$$

Solutions · 1