Let f(x) be a quadratic expression which is positive for all x. If g(x)=f(x)+f′(x)+f′′(x) then for allreal x,
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a
g(x)<0
b
g(x)>0
c
g(x)=0
d
g(x)≥0
answer is B.
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Detailed Solution
Let f(x)=ax2+bx+cAs f(x)>0∀x∈R, we must have a>0 and b2−4ac<0.Also, f′(x)=2ax+b and f′′(x)=2aThus, g(x)=ax2+(2a+b)x+2a+b+cSince a>0 and discriminant (2a+b)2−4a(2a+b+c)=4a2+4ab+b2−8a2−4ab−4ac=−4a2+b2−4ac<0 ∵b2−4ac<0We get g(x)>0∀x∈R