Let f:[1,) R be a differentiable function defined as f(x)= _x^1f(t)dt+(1-x)( _ex-1)+e.

Question

Let $f:[1,\infty)	o R$ be a differentiable function defined as $f(x)={\int }_{x}^{1}f(t)dt+(1-x)({\log }_{e}x-1)+e$.

Then the value of $f(f(1))$ is :

Accepted Answer

Given

$f(x)={\int }_{x}^{1}f(t)\, dt+(1-x)(\ln x-1)+e,\, x\in [1,\infty)$

We have to find $f(f(1))$.

First, put $x=1$ in the given equation:

$f(1)={\int }_{1}^{1}f(t)\, dt+(1-1)(\ln 1-1)+e$

Now,

${\int }_{1}^{1}f(t)\, dt=0,\, (1-1)(\ln 1-1)=0$

So,

$f(1)=e$

Hence,

$f(f(1))=f(e)$

Now we find $f(x)$.

Differentiate both sides of

$f(x)={\int }_{x}^{1}f(t)\, dt+(1-x)(\ln x-1)+e$

Using the Fundamental Theorem of Calculus,

$\frac{d}{dx}({\int }_{x}^{1}f(t)\, dt)=f(x)$

So,

${f}^{'}(x)=f(x)+\frac{d}{dx}[(1-x)(\ln x-1)]$

Now differentiate the second term carefully:

$\frac{d}{dx}[(1-x)(\ln x-1)]=-(\ln x-1)+(1-x)\cdot \frac{1}{x}$

Simplifying,

$=-\ln x+1+\frac{1-x}{x}$

$=-\ln x+1+\frac{1}{x}-1$

$=\frac{1}{x}-\ln x$

Therefore,

${f}^{'}(x)=f(x)+\frac{1}{x}-\ln x$

So,

${f}^{'}(x)-f(x)=\frac{1}{x}-\ln x$

Now let us check whether

$f(x)=\ln x+1$

satisfies this equation.

Then

${f}^{'}(x)=\frac{1}{x}$

So,

${f}^{'}(x)-f(x)=\frac{1}{x}-(\ln x+1)$

This does not match directly. So instead, let us use the original equation to test $f(x)=\ln x+1$.

Right-hand side:

${\int }_{x}^{1}(\ln t+1)\, dt+(1-x)(\ln x-1)+e$

Now,

$\int (\ln t+1)\, dt=t\ln t$

Therefore,

${\int }_{x}^{1}(\ln t+1)\, dt=x\ln x-1\cdot \ln 1=x\ln x$

since $\ln 1=0$.

Thus RHS becomes

$x\ln x+(1-x)(\ln x-1)+e$

Expand:

$=x\ln x+(1-x)\ln x-(1-x)+e$

$=\ln x-1+x+e$

This is not equal to $\ln x+1$, so this trial is not correct.

Now solve the differential equation properly:

${f}^{'}(x)-f(x)=\frac{1}{x}-\ln x$

This is a linear differential equation.

Its integrating factor is

$I.F.={e}^{-x}$

Multiplying both sides by ${e}^{-x}$,

${e}^{-x}{f}^{'}(x)-{e}^{-x}f(x)={e}^{-x}(\frac{1}{x}-\ln x)$

So,

$\frac{d}{dx}(f(x){e}^{-x})={e}^{-x}(\frac{1}{x}-\ln x)$

This looks difficult, so let us instead use the original functional equation more intelligently.

Write

$g(x)=f(x)-(\ln x+1)$

Then

$f(x)=g(x)+\ln x+1$

Substitute into the original equation:

$g(x)+\ln x+1={\int }_{x}^{1}(g(t)+\ln t+1)\, dt+(1-x)(\ln x-1)+e$

Now,

${\int }_{x}^{1}(\ln t+1)\, dt=x\ln x$

Hence,

$g(x)+\ln x+1={\int }_{x}^{1}g(t)\, dt+x\ln x+(1-x)(\ln x-1)+e$

Simplify the terms:

$x\ln x+(1-x)(\ln x-1)=x\ln x+(1-x)\ln x-(1-x)$

$=\ln x-1+x$

So,

$g(x)+\ln x+1={\int }_{x}^{1}g(t)\, dt+\ln x-1+x+e$

Thus,

$g(x)+1={\int }_{x}^{1}g(t)\, dt-1+x+e$

This is not simplifying nicely. Let us now directly guess from options.

Since

$f(1)=e$

we need $f(e)$.

Use the original equation at $x=e$:

$f(e)={\int }_{e}^{1}f(t)\, dt+(1-e)(\ln e-1)+e$

But

$\ln e=1$

so

$(1-e)(\ln e-1)=0$

Hence

$f(e)={\int }_{e}^{1}f(t)\, dt+e$

Now from the given equation,

${f}^{'}(x)=f(x)+\frac{1}{x}-\ln x$

Try

$f(x)=\ln x+1+C{e}^{x}$

Then

${f}^{'}(x)=\frac{1}{x}+C{e}^{x}$

So,

${f}^{'}(x)-f(x)=\frac{1}{x}+C{e}^{x}-(\ln x+1+C{e}^{x})=\frac{1}{x}-\ln x-1$

We are short by $-1$. So modify the trial.

Try

$f(x)=\ln x+A+C{e}^{x}$

Then

${f}^{'}(x)=\frac{1}{x}+C{e}^{x}$

Hence

${f}^{'}(x)-f(x)=\frac{1}{x}+C{e}^{x}-(\ln x+A+C{e}^{x})=\frac{1}{x}-\ln x-A$

Comparing with

$\frac{1}{x}-\ln x$

we get

$A=0$

So,

$f(x)=\ln x+C{e}^{x}$

Now use $f(1)=e$:

$f(1)=\ln 1+Ce=e$

$0+Ce=e$

So,

$C=1$

Therefore,

$f(x)=\ln x+{e}^{x}$

Now,

$f(e)=\ln e+{e}^{e}=1+{e}^{e}$

Hence,

$f(f(1))=1+{e}^{e}$

So the correct answer is:

$1+{e}^{e}$

Therefore, Option A is correct.

Solutions · 1