JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 18)
If the domain of the function $$f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left\{\log _e(3-x)\right\}^{-1}$$ is $$[-\alpha, \beta)-\{\gamma\}$$, then $$\alpha+\beta+\gamma$$ is equal to :
11
12
9
8
Explanation
$$\begin{aligned} & -1 \leq\left|\frac{2-|x|}{4}\right| \leq 1 \\ & \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1 \\ & -4 \leq 2-|x| \leq 4 \\ & -6 \leq-|x| \leq 2 \\ & -2 \leq|x| \leq 6 \\ & |x| \leq 6 \end{aligned}$$
$$\Rightarrow x \in[-6,6]$$ .... (1)
Now, $$3-x\ne 1$$
And $$x\ne2$$ .... (2)
and $$3-x>0$$
$$x<3$$ .... (3)
$$\begin{aligned} & \text { From (1), (2) and (3) } \\ & \Rightarrow x \in[-6,3)-\{2\} \\ & \alpha=6 \\ & \beta=3 \\ & \gamma=2 \\ & \alpha+\beta+\gamma=11 \end{aligned}$$
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