Let f(x) be a quadratic expression which is positive for all real x. If g(x)=f(x)−f′(x)+f′′(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 Bg(x)≥0 Let f(x)=ax2+bx+c>0∀x∈R ⇒b2−4ac<0 and a>0 ...(1) Now g(x)=f(x)−f′(x)+f′′(x)=(ax2+bx+c)−(2ax+b)+2a =ax2+(b−2a)x−(2a−b+c) ⇒Δ=(b−2a)2−4a(2a−b+c)=(b2−4ac)−4a2<0 (Using (1)) ⇒g(x)>0∀x∈R⇒g(x)≥0∀x∈R