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 :
Solution
<p>$$\begin{aligned}
& -1 \leq\left|\frac{2-|x|}{4}\right| \leq 1 \\
& \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1 \\
& -4 \leq 2-|x| \leq 4 \\
& -6 \leq-|x| \leq 2 \\
& -2 \leq|x| \leq 6 \\
& |x| \leq 6
\end{aligned}$$</p>
<p>$\Rightarrow x \in[-6,6]$ .... (1)</p>
<p>Now, $3-x\ne 1$</p>
<p>And $x\ne2$ .... (2)</p>
<p>and $3-x>0$</p>
<p>$x<3$ .... (3)</p>
<p>$$\begin{aligned}
& \text { From (1), (2) and (3) } \\
& \Rightarrow x \in[-6,3)-\{2\} \\
& \alpha=6 \\
& \beta=3 \\
& \gamma=2 \\
& \alpha+\beta+\gamma=11
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
This question is part of PrepWiser's free JEE Main question bank. 195 more solved questions on Sets, Relations and Functions are available — start with the harder ones if your accuracy is >70%.