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Questions  

If 0xf(t)dt=x+x1tf(t)dt, then the value of f(1) is

a
12
b
0
c
1
d
-12

detailed solution

Correct option is A

We have,∫0x f(t)dt=x+∫x1 tf(t)dt⇒f(x)=1+0−xf(x)  [using Leibnitz's rule) ⇒f(x)=11+xf(1)=12

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