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B
a minimum value
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C
a singularity
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D
a point of inflection
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Solution
The correct option is D a point of inflection Given, f(x)=x3+1 ⇒f′(x)=3x2=0 ⇒x=0 is a point of maxima or minima.
Now, f′′(x)=6x ⇒f′′(0)=0 ⇒f′′′(x)=6 ∵f′(x) and f′′(x) is zero at x=0 and f′′′(x) is positive hence it is a piont of inflection.