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Question

f(x)=1+2sinx+2cos2x, 0xπ/2 is maximum at-

A
x=π/2
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B
x=π/6
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C
x=π/3
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D
No where
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Solution

The correct option is B x=π/6
f=1+2sinx+2cos2x
f=2cosx4cosxsinx
=2cosx2sin2x
f=2cosx2sin2x
f(x)=0cosxsin2x=0
(cosx)(12sinx)=0
xcosx=0 or sinx=12
x=π2 x=π6
f(π2)=2+4=2
min at
$x=\dfrac{\pi}{2}
f(π6)=2(12)4(12)
=2 max at
x=π6
local min at x=π2
local max at x=π6
look at critical point and end points
f(0)=3,f(π2)=3
f(π6)=3.5
Global more at x=π6

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