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⇒[e−xf′′(x)−e−xf′(x)]−[e−xf′(x)−e−xf(x)]≥1⇒ddx[e−xf′(x)]−ddx[e−xf(x)]≥1⇒ddx[e−xf′(x)−e−xf(x)]≥1⇒ddx[ddx(e−xf(x))]≥1
Let g(x)=e−xf(x)
Then we have g′′(x)≥1>0
So g is concave upward.
Also g(0)=g(1)=0 g(x)<0∀xϵ(0,1)⇒e−xf(x)<0⇒f(x)<0,∀xϵ(0,1)