The set of real values of x, for which h(x)=1+2x2+4x4+6x6+⋯+100x100 is concave downward is
A
x∈ϕ
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B
x∈W
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C
x∈R−
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D
x∈R+
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Solution
The correct option is Ax∈ϕ 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