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B
f′(x)>0 if 0<x<2
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C
f′′(x)>0 if 0<x<1
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D
f′′(x)<0 if 1<x<2
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Solution
The correct option is Df′′(x)<0 if 1<x<2 From the graph, it is clear that f(x) is increasing in [0,2] ⇒f′(x)>0 if x∈[0,2]
Also, f(x) is concave up in (0,1) and concave down in (1,2) ∴f′′(x)>0 if 0<x<1
and f′′(x)<0 if 1<x<2