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Question

The set of real values of x, for which h(x)=1+2x2+4x4+6x6++100x100 is concave downward is

A
xϕ
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B
xW
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C
xR
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D
xR+
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Solution

The correct option is A xϕ
h(x)=1+2x2+4x4+6x6++100x100
h(x)=0+22x+42x3+62x5++1002x99
h′′(x)=22+42(3x2)+62(5x4)++1002(99x98)
So, h′′(x) is always positive for all real values of x
Hence h(x) is always concave upward

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