Let $f:\lbrack 1,\infty) \rightarrow R$ be a differentiable function defined as $f(x) = \int_{1}^{x}

Question

Let $f:\lbrack 1,\infty) \rightarrow R$ be a differentiable function defined as $f(x) = \int_{1}^{x}\mspace{2mu} f(t)dt + (1 - x)\left( \log_{e}x - 1 \right) + e$. Then the value fo $f(f(1))$ is :

Accepted Answer

Sol. [Key Concept: Put $x = 1$ to get $f(1) = e$; solve the ODE (I.F. $= e^{- x}$) for $f(x)$, then evaluate $f(e)$.]$${f(1) = \int_{1}^{1}\mspace{2mu} f(t)dt + (1 - 1)(ln1 - 1) + e }{f(1) = e }{f(x) = \int_{1}^{x}\mspace{2mu} f(t)dt + (1 - x)(lnx - 1) + e }$$diff. w.r.t. x$$f^{'}(x) = f(x) + \frac{1}{x} - lnx $$I.F. $= e^{\int - 1dx} = e^{- x}$$${f(x)e^{- x} = \int e^{- x}\left( \frac{1}{x} - \mathcal{l}nx \right)dx }{f(x)e^{- x} = e^{- x}lnx + C }$$put $x = 1,y = e$$${C = 1 }{f(x)e^{- x} = e^{- x} \mid nx + 1 }{f(x) = e^{x} + lnx }{f(e) = e^{e} + 1 }$$

Question

Options

Solution

More Continuity and Differentiability Questions