Question
If ∫x0f(t)dt=x+∫1xt f(t) dt, then the value of f(1) is [IIT 1998; AMU 2005]
Answer: Option A
:
A
∫x0dt=x+∫1xtf(t)dt⇒∫1xtf(t)dt=x−∫x1tf(t)dt
Differentiating w.r.t x, we get f(x)=1+{0−xf(x)}
⇒f(x)=1−xf(x)⇒(1+x)f(x)=1⇒f(x)=11+x
∴f(1)=11+1=12
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:
A
∫x0dt=x+∫1xtf(t)dt⇒∫1xtf(t)dt=x−∫x1tf(t)dt
Differentiating w.r.t x, we get f(x)=1+{0−xf(x)}
⇒f(x)=1−xf(x)⇒(1+x)f(x)=1⇒f(x)=11+x
∴f(1)=11+1=12
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