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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Given that x ∈ R , find the minimum value of (3 5-4 x + 13-12 x )^2...

Given that x∈Rx \in R, find the minimum value of (35−4cos⁡x+13−12sin⁡x)2(3 \sqrt{5-4 \cos x}+\sqrt{13-12 \sin x})^2.

Correct answer

40

Step-by-step solution

E=(35−4cos⁡x+13−12sin⁡x)=(45−36cos⁡x+13−12sin⁡x)=((6−3cos⁡x)2+(3sin⁡x)2+(2−3sin⁡x)2+(3cos⁡x)2)=PA+PB∣minimum =AB=40\begin{aligned} E & =(3 \sqrt{5-4 \cos x}+\sqrt{13-12 \sin x}) \\ & =(\sqrt{45-36 \cos x}+\sqrt{13-12 \sin x}) \\ & =\left(\sqrt{(6-3 \cos x)^2+(3 \sin x)^2}+\sqrt{(2-3 \sin x)^2+(3 \cos x)^2}\right) \\ & =P A+\left.P B\right|_{\text {minimum }}=A B=\sqrt{40}\end{aligned} Where \begin{aligned} & P=3 \cos x, 3 \sin x \\ & A=(6,0) \text { and } B=(0,2) \end{aligned} Hence, minimum value of E2=40E^2=40
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