If the domain of the function f(x) = cos^{-1} | vedclass

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(C) For the function $f(x) = \cos^{-1}\left(\frac{2-|x|}{4}\right) + (\log_e(3-x))^{-1}$,we need to satisfy the domain conditions for both parts.$1$.

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) For the function $f(x) = \cos^{-1}\left(\frac{2-|x|}{4}ight) + (\log_e(3-x))^{-1}$,we need to satisfy the domain conditions for both parts.$1$. For $\cos^{-1}\left(\frac{2-|x|}{4}ight)$,the argument must be in $[-1, 1]$:$-1 \leq \frac{2-|x|}{4} \leq 1$$-4 \leq 2-|x| \leq 4$$-6 \leq -|x| \leq 2$Since $|x| \geq 0$,we have $-|x| \leq 2$ always true. Thus,$|x| \leq 6$,which implies $x \in [-6, 6]$.$2$. For $(\log_e(3-x))^{-1}$,we need $\log_e(3-x) 
eq 0$ and $3-x > 0$:$3-x > 0 \Rightarrow x $\log_e(3-x) 
eq 0 \Rightarrow 3-x 
eq 1 \Rightarrow x 
eq 2$.Combining these conditions:$x \in [-6, 6] \cap (-\infty, 3) \setminus \{2\} = [-6, 3) \setminus \{2\}$.Comparing this with $[-\alpha, \beta) \setminus \{\gamma\}$,we get $\alpha = 6$,$\beta = 3$,and $\gamma = 2$.Therefore,$\alpha + \beta + \gamma = 6 + 3 + 2 = 11$.

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

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