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Question

Let f:(0,)R be a differential function such that f(x)=2f(x)x for all x(0,) and f(1)1. Then:

A
limx0+f(1x)=1
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B
limx0+xf(1x)=2
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C
limx0+x2f(x)=0
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D
|f(x)|2 for all x(0,2)
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Solution

The correct option is A limx0+f(1x)=1
dydx=2yx
dydx+yx=2
Integrating factor =e1xdx=x

y(x)=2x dx+c
f(x)=x+cx
Given that f(1)1c0

f(x)=x+cx, cR{0}

Option 1:
f(x)=1cx2
limx0+f(1x)=limx0+(1cx2)=1

Option 2:
limx0+xf(1x)=limx0+x(1x+cx)=limx0+(1+cx2)=1

Option 3:
limx0+x2f(x)=limx0+x2(1cx2)=limx0+(x2c)0

Option 4:
|f(x)|2c
For c=4,|f(x)|4, which contradicts the given option.

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