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Question

Given f:[0,)R be a strictly increasing function such that the functions g(x)=f(x)3x and h(x)=f(x)x3 are both strictly increasing function. Then the function F(x)=f(x)x2x is

A
increasing in (0,1) and decreasing in (1,)
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B
decreasing in (0,1) and increasing in (1,)
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C
increasing throughout (0,)
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D
decreasing throughout (0,)
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Solution

The correct option is C increasing throughout (0,)
g(x)=f(x)3xg(x)=f(x)3...(1)
h(x)=f(x)x3h(x)=f(x)3x2...(2)
F(x)=f(x)x2xF(x)=f(x)2x1
F(x)=12(f(x)+f(x))2x1
Now using (1) and (2)
F(x)=12(g(x)+h(x)+3x24x1)
F(x)=12(g(x)+h(x)+3((x2/3)2+1/6)
Since h(x),g(x)>0
F(x)>0
Hence F(x) is increasing in (0,)

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