If f(x)=sinx+2cos2x,π4≤x≤3π4. Then, f(x) attains its
A
minimum at x=π4
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B
maximum at x=π2
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C
minimum at π2
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D
maximum at x=sin−1(14)
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Solution
The correct option is C minimum at π2 Given f(x)=sinx+2cos2x,x∈[π4,3π4] ∴f′(x)=cosx−4cosxsinx and f′′(x)=−sinx−4cos2x For maximum or minimum of f(x) Put f′(x)=0 ⇒cosx−4cosx.sinx=0 ⇒cosx(1−4sinx)=0 ⇒cosx=0=cosπ2;sinx≠14 ∵x∈[π4,3π4]⇒x=π2 Now f′′(π2)=−sinπ2−4cosπ =−1+4=3>0(min) So, f(x) is minimum at x=π2 and its minimum value is f(π2)=sinπ2+2cos2π2=1−2×0=1