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Question

Let f:RR be a continuous function satisfying
f(x)+x0tf(t)dt+x2=0
for all xR. Then

A
limxf(x)=2
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B
limxf(x)=2
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C
f(x) has more than one point in common with the x-axis
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D
f(x) has no point in common with the x-axis
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Solution

The correct option is D limxf(x)=2
f(x)+xf(x)+2x=0

dydx+xy=2x

I.F.=ex2/2

y.ex2/2=ex2/2+C

ex2/2.y=2ex2/2+C

y=2+Cex2/2

f(0)=0C=2

f(x)=2[ex2/21]

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