JEE Mains · Maths · STD 11 - 8. sequence and series
The minimum value of \(2^{sin x}+2^{cos x}\) is
- A \(2^{1-\frac{1}{\sqrt{2}}}\)
- B \(2^{-1+\sqrt{2}}\)
- C \(2^{1-\sqrt{2}}\)
- D \(2^{-1+\frac{1}{\sqrt{2}}}\)
Answer & Solution
Correct Answer
(A) \(2^{1-\frac{1}{\sqrt{2}}}\)
Step-by-step Solution
Detailed explanation
Usnign \(AM \geq GM\) \(\Rightarrow \frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}}\) \(\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1+\left(\frac{\sin x+\cos x}{2}\right)}\) \(\Rightarrow \min \left(2^{\sin x}+2^{\cos x}\right)=2^{1-\frac{1}{\sqrt{2}}}\)
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