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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

The maximum value of the function y=e^ 5+ 3 x+ x is

\quad The maximum value of the function y=e5+3sin⁡x+cos⁡xy=e^{5+\sqrt{3} \sin x+\cos x} is

Options

  1. Ae7e^{7}
  2. Be2e^{2}
  3. Ce5e^{5}
  4. De8e^{8}

Correct answer

A. e7e^{7}

Step-by-step solution

Maximum value of 3sin⁡x+cos⁡x\sqrt{3} \sin x+\cos x is (3)2+(1)2=2\sqrt{(\sqrt{3})^{2}+(1)^{2}}=2 Hence maximum value of given function is \mathrme5+2=\mathrme7\mathrme^{5+2}=\mathrme^{7}
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