If f(x)=∫10f(xt)dt, where f(x) is a continuos function such that f(1)=2, then
A
f(x) is a periodic function
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B
f(4)=2
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C
f(x) is an even function
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D
f(4)=5
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Solution
The correct options are Af(x) is a periodic function Bf(4)=2 Cf(x) is an even function Put xt=z⇒dzdt=x f(x)=∫x0f(z)dzx⇒xf(x)=∫x0f(z)dz ⇒xf′(x)+f(x)=f(x) ⇒xf′(x)=0 ⇒f′(x)=0⇒f(x)=c⇒f(x)=2