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Question

The function f(x)=(4sin2x1)n(x2x+1),nN,has a local minimum at x=π6, then

A
n can be any even number
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B
n can be an odd number
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C
n can be odd prime number
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D
n can be any natural number
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Solution

The correct option is A n can be any even number
f(π6)=0f(x)=(4sin2x1)n((x1)2+34)(x1)2+34>0,xRf(π6+)=limxπ/6+(4sin2x1)n(x2x+1)limxπ/6+4sin2x=1+limxπ/6+f(x)=0+f(π6)=limxπ/6(4sin2x1)n(x2x+1)limxπ/64sin2x=1limxπ/6f(x)=0
But x=π6 is a minima f(π6) must be greater than zero. This can only happen for n=2k
Hence, n can be any even number.

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