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Question

The absolute minimum and maximum values of f(x)=sinx+12cos2x,xϵ[0,π2] are respectively

A
34,12
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B
0,12
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C
12,34
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D
12,34
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Solution

The correct option is C 34,12

Differentiate the given function f(x)=sinx+12cos2x.

f(x)=cosxsin2x

Put f(x)=0,

cosxsin2x=0

cosx=2sinxcosx

sinx=12

x=sin1(12)

x=π6

Put the value of x and the end points of the given interval in the given function.

f(π6)=sin(π6)+12cos2(π6)

=34

f(0)=sin(0)+12cos2(0)

=12

f(π2)=sin(π2)+12cos2(π2)

=12

The maximum value of the given function is 34 and the minimum value is 12.


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