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Question

# Let f:(0,∞)→R be a differentiable function such that f′(x)=2−f(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 limx→0+f′(1x)=1Here, f′(x)=2−f(x)xor dydx+yx=2 [ i.e. linear differential equation in y]Integrating Factor, IF=e∫1xdx=elogx=x∴ Required solution is y.(IF)=∫Q(IF)dx+C⇒y(x)=∫2(x)dx+C⇒yx=x2+C∴y=x+Cx [∵C≠0, as f(1)≠1](a) limx→0+f′(1x)=limx→0+(1−Cx2)=1∴ Option (a) is correct.(b) limx→0+xf(1x)=limx→0+(1+Cx2)=1∴ Option (b) is correct.(c) limx→0+x2f′(x)=limx→0+(x2−C)=−C≠0∴ Option (c) is correct.(d) f(x)=x+Cx,C≠0For C>0,limx→0+f(x)=∞∴ Function is not bounded in (0,2).∴ Option (d) is incorrect.

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