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Question

If f1(x)=2x,f2(x)=3sinxxcosx, then for x(0,π2)

A
f1(x)<f2(x)
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B
f1(x)>f2(x)
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C
f1(|x|)<f2(|x|)
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D
none of these
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Solution

The correct option is B f1(x)>f2(x)
Let f(x)=f1(x)f2(x)
f(x)=2x3sinx+xcosx
f(x)=22cosxxsinx
f′′(x)=sinxxcosx=cosx(tanxx)
f′′(x)>0x(0,π2).
Thus f(x) is increasing in (0,π2);f(0)=0
f(x)>0x(0,π2).
f(0)=0
f(x)>0x(0,π2).
f1(x)>f2(x)
f(x) is increasing in (0,π2)

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