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