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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Let 0 ≤ θ ≤ 2 π and x=| θ+1|+| θ-1|+| θ-2|+| θ-3| , then product of...

Let 0≤θ≤2π0 \leq \theta \leq 2 \pi and x=∣cos⁡θ+1∣+∣cos⁡θ−1∣+∣cos⁡θ−2∣+∣cos⁡θ−3∣x=|\cos \theta+1|+|\cos \theta-1|+|\cos \theta-2|+|\cos \theta-3|, then product of the maximum and minimum values of xx is:

Correct answer

45

Step-by-step solution

x=(cos⁡θ+1)−(cos⁡θ−1)−(cos⁡θ−2)−(cos⁡θ−3)∴x=7−2cos⁡θxmax =7−2(−1)=9xmin =7−2(1)=5\begin{aligned} x & =(\cos \theta+1)-(\cos \theta-1)-(\cos \theta-2)-(\cos \theta-3) \\ \therefore \quad x & =7-2 \cos \theta \\ x_{\text {max }} & =7-2(-1)=9 \\ x_{\text {min }} & =7-2(1)=5\end{aligned}
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