The interval of f(x)=x3+2x2+5x,x<0 for which it is concave upwards is
A
(0,∞)
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B
(−23,0)
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C
(−23,∞)
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D
(−∞,0)
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Solution
The correct option is B(−23,0) f(x)=x3+2x2+5x,∀x<0⋯(i) f′(x)=3x2+4x+5 ⇒f′′(x)=6x+4
For f to be concave upward, f′′(x)>0 ⇒x>−23⋯(ii)
From (i)and (ii), we get
Interval of concave upwards is (−23,0)