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Question

Let f(x)=x3+6x2+12x+15xϵR, at x=2

A
f(x) has a maximum
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B
f(x) has minimum
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C
f(x) has a maximum
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D
f(x) has minimum
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Solution

The correct option is D f(x) has minimum
f(x)=x3+6x2+12x+15
f(x)=3x2+12x+12=3(x+2)2
f′′(x)=6(x+2)
f′′′(x)=6
clearly at x=2,f(x)=0=f′′(x) and f′′′(x)>0
Hence only conclusion drawn from here is that f(x) has minima at x=2

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