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Question

Question: For the function $f(x) = (x^2 + bx + c)e^x$ and $g(x) = (x^2 + bx + c)e^x + e^{x(2x + b)}$. Which of...

For the function f(x)=(x2+bx+c)exf(x) = (x^2 + bx + c)e^x and g(x)=(x2+bx+c)ex+ex(2x+b)g(x) = (x^2 + bx + c)e^x + e^{x(2x + b)}. Which of the following holds good?

A

if f(x)>0f(x) > 0 for all real xg(x)>0x \nRightarrow g(x) > 0

B

if f(x)>0f(x) > 0 for all real xg(x)>0x \Rightarrow g(x) > 0

C

if g(x)>0g(x) > 0 for all real xf(x)>0x \Rightarrow f(x) > 0

D

if g(x)>0g(x) > 0 for all real xf(x)>0x \nRightarrow f(x) > 0

Answer

if f(x)>0f(x) > 0 for all real xg(x)>0x \Rightarrow g(x) > 0

Explanation

Solution

Given f(x)=(x2+bx+c)exf(x) = (x^2 + bx + c)e^x and g(x)=(x2+bx+c)ex+ex(2x+b)g(x) = (x^2 + bx + c)e^x + e^{x(2x + b)}, we can rewrite g(x)g(x) as g(x)=f(x)+ex(2x+b)g(x) = f(x) + e^{x(2x + b)}.

Since ex(2x+b)>0e^{x(2x + b)} > 0 for all real xx, if f(x)>0f(x) > 0 for all real xx, then g(x)=f(x)+ex(2x+b)>0g(x) = f(x) + e^{x(2x + b)} > 0 for all real xx.

Therefore, if f(x)>0f(x) > 0 for all real xx, then g(x)>0g(x) > 0 for all real xx.