Verified Solution • 2025–26 Session moderate QJEEM24Mab7df: JEE Main 2024 Mathematics - Class 11 JEEM24Mab7df Solution: If the domain of the function is , then is equal to : If the domain of the function f(x)=cos−1(2−∣x∣4)+{loge(3−x)}−1f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left\{\log _e(3-x)\right\}^{-1}f(x)=cos−1(42−∣x∣)+{loge(3−x)}−1 is [−α,β)−{γ}[-\alpha, \beta)-\{\gamma\}[−α,β)−{γ} , then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to : Multiple Choice Options A 11 B 12 C 9 D 8 Step-by-Step Verified Solution amp;−1≤∣2−∣x∣4∣≤1amp;⇒∣2−∣x∣4∣≤1amp;−4≤2−∣x∣≤4amp;−6≤−∣x∣≤2amp;−2≤∣x∣≤6amp;∣x∣≤6\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}amp;−1≤42−∣x∣≤1amp;⇒42−∣x∣≤1amp;−4≤2−∣x∣≤4amp;−6≤−∣x∣≤2amp;−2≤∣x∣≤6amp;∣x∣≤6 ⇒x∈[−6,6]\Rightarrow x \in[-6,6]⇒x∈[−6,6] .... (1) Now, 3−x≠13-x\ne 13−x=1 And x≠2x\ne2x=2 .... (2) and 3-x>0 x<3 .... (3) amp; From (1), (2) and (3) amp;⇒x∈[−6,3)−{2}amp;α=6amp;β=3amp;γ=2amp;α+β+γ=11\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}amp; From (1), (2) and (3) amp;⇒x∈[−6,3)−{2}amp;α=6amp;β=3amp;γ=2amp;α+β+γ=11 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring