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Question

Let f:[0,1]R (the set of all real numbers) be a function. Suppose the function f is twice differentiable,f(0)=f(1)=0 and satisfies f"(x)2f(x)+f(x)ex,x[0,1]
Which of the following is true for 0<x<1?

A
0<f(x)<
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B
12<f(x)<12
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C
14<f(x)<1
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D
<f(x)<0
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Solution

The correct option is D <f(x)<0
We have f′′(x)2f(x)+f(x)ex
[f′′(x)f(x)][f(x)f(x)]ex[exf′′(x)exf(x)][exf(x)exf(x)]1ddx[exf(x)]ddx[exf(x)]1ddx[exf(x)exf(x)]1ddx[ddx(exf(x))]1
Let g(x)=exf(x)
Then we have g′′(x)1>0
So g is concave upward.
Also g(0)=g(1)=0
g(x)<0xϵ(0,1)exf(x)<0f(x)<0,xϵ(0,1)

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