Medium MCQ +4 / -1 PYQ · JEE Mains 2020

The minimum value of 2sinx + 2cosx is :

  1. A ${2^{-1 + \sqrt 2 }}$
  2. B ${2^{1 - {1 \over {\sqrt 2 }}}}$ Correct answer
  3. C ${2^{1 - \sqrt 2 }}$
  4. D ${2^{-1 + {1 \over {\sqrt 2 }}}}$

Solution

Using AM $\ge$ GM<br><br>$$ \Rightarrow {{{2^{\sin \,x}} + {2^{\cos \,x}}} \over 2} \ge \sqrt {{2^{\sin x}}{{.2}^{\cos x}}} $$<br><br>$$ \Rightarrow {2^{\sin x}} + {2^{\cos x}} \ge {2^{1 + \left( {{{\sin x + \cos x} \over 2}} \right)}}$$<br><br>$\Rightarrow \min ({2^{\sin x}} + {2^{\cos x}}) = {2^{1 - {1 \over {\sqrt 2 }}}}$ <br><br>As we know range of sin x + cos x is : <br><br>$- \sqrt 2$ $\le$ sin x + cos x $\le$ $\sqrt 2$. <br><br>So Minimum value of sin x + cos x = $- \sqrt 2$

About this question

Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression

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