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Question

Let f(x) be a polynomial of degree two which is positive for all xR
If g(x)=f(x)+f(x)+f′′(x)+xf′′′(x)+x2fIV(x) then for any real x

A
g(x)<0
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B
g(x)>0
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C
g(x)=0
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D
g(x)0
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Solution

The correct option is B g(x)>0
Let f(x)=ax2+bx+c
According to the given condition a>0,b24ac<0.........(i)
g(x)=ax2+bx+c+2ax+b+2a+0+0=ax2+(b+2a)x+(b+c+2a)
Now discriminant of g(x) is D=(b+2a)24a(2a+c+b)=b2+4a2+4ab4ab4ac8a2=(b24ac)4a2<0 Using (i)
Hence g(x)>0 for any xR

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