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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

If 5 x+12 y=13 , then the maximum value of 5 x+12 y is :

If 5cos⁡x+12cos⁡y=135 \cos x+12 \cos y=13, then the maximum value of 5sin⁡x+12sin⁡y5 \sin x+12 \sin y is :

Options

  1. A12
  2. B120\sqrt{120}
  3. C20\sqrt{20}
  4. D13

Correct answer

B. 120\sqrt{120}

Step-by-step solution

∵5cos⁡x+12cos⁡y=13⇒(5cos⁡x+12cos⁡y)2=169∴(5cos⁡x+12cos⁡y)2+(5sin⁡x+12sin⁡y)2⇒(13)2+(5sin⁡x+12sin⁡y)2=25+144+120(sin⁡xsin⁡y+cos⁡xcos⁡y)⇒(5sin⁡x+12sin⁡y)2=120cos⁡(x−y)∵−1≤cos⁡(x−y)≤1∴\because 5 \cos x+12 \cos y=13\Rightarrow \quad(5 \cos x+12 \cos y)^2=169\therefore(5 \cos x+12 \cos y)^2+(5 \sin x+12 \sin y)^2\Rightarrow \quad(13)^2+(5 \sin x+12 \sin y)^2=25+144+120(\sin x \sin y+\cos x \cos y)\Rightarrow \quad(5 \sin x+12 \sin y)^2=120 \cos (x-y)\because \quad-1 \leq \cos (x-y) \leq 1\therefore Maximum value of 5sin⁡x+12sin⁡y=1205 \sin x+12 \sin y=\sqrt{120}
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