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Q.

Let f:(0,∞)→ℝ be a differentiable function such that fx=2-f(x)x for all x∈(0,∞) and f(1)≠1 . Then

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a

limx→0+f'1x=1

b

limx→0+xf 1x=2

c

limx→0+x2f'x=0

d

|f(x)|≤2 for all x∈(0,2)

answer is A.

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Detailed Solution

|f(x)|≤2 for all x∈(0,2)given equation is fx+fxx=2∴fxe∫1xdx=∫2e∫1xdxdx+c    linear differential equation⇒xfx=x2+cfx=x+cx ⇒f'x=1-cx2⇒f'1x=1−cx21option:  ∴Ltx→0+f'1x=1 2 option:  Ltx→0+xf1x  =Ltx→0+x1x+cx=1  3 option:     Ltx→0+x2f'x=Ltx→0+x21−cx2=−c
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