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Question

Mathematics Question on Differential equations

The minimum value of f(x)=sin4⁡x+cos4⁡x,0≤x≤π2

A

(A) 122

B

(B) 1414

C

(C) −12

D

(D) 12

Answer

(D) 12

Explanation

Solution

Explanation:
∵f(x)=sin4⁡x+cos4⁡x=(sin2⁡x+cos2⁡x)2−2sin2⁡x⋅cos2⁡x⇒f(x)=1−12sin2⁡2x Also 0≤sin2⁡2x≤1∴ Minimum value of f(x)1−12=12