lf f(x) is a quadratic expression which is positive for all real vaues of x and g(x)=f(x)+f′(x)+f′′(x) then for any real value of x
g(x)>0
g(x)<0
g(x)=0
g(x)>––0
A
g(x)<0
B
g(x)>0
C
g(x)=0
D
g(x)>––0
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Let f(x)=ax2+bx+c According to the given condition a>0,b2−4ac<0.........(i) ∴g(x)=ax2+bx+c+2ax+b+2a=ax2+(b+2a)x+(b+c+2a) Now discriminant of g(x) is D=(b+2a)2−4a(2a+c+b)=b2+4a2+4ab−4ab−4ac−8a2=(b2−4ac)−4a2<0 using (i) Hence g(x)>0 for any x∈R
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