Question

Let $f:R^{+}\to R$ be a differentiable function such that $f\left(x\right)=e-(x-1)\ln \left(\frac{x}{e}\right)+\underset{1}{\overset{x}{\int }}f\left(x\right)dx.$ Then

• A. the equation of normal at $x=1$ is
  $x+(1+e)y+(e^2+e+1)=0.$
• B. the equation of tangent at $x=1$ is $x(1-e)-y-1=0.$
• C. the equation of tangent at $x=1$ is $x(1+e)-y-1=0.$
• D. the equation of normal at $x=1$ is
  $x+(1+e)y-(e^2+e+1)=0.$

Answer 1

Answer: (C), (D)

the equation of normal at $x=1$ is

$x+(1+e)y-(e^2+e+1)=0.$
Given, $f\left(x\right)=y=e-(x-1)\ln \left(\frac{x}{e}\right)+\underset{1}{\overset{x}{\int }}f\left(x\right)dx$
Differentiating w.r.t. $x,$ we get
$\frac{dy}{dx}=-\frac{(x-1)}{x}-\ln \left(\frac{x}{e}\right)+y$
$\Rightarrow \frac{dy}{dx}-y=\frac{1}{x}-\ln x$
$I.F.=e^{-\int 1dx}=e^{-x}$
Now, the general solution is
$y{e}^{-x}=\int (\frac{1}{x}-\ln x)e^{-x}dx+C$
$\Rightarrow y{e}^{-x}=\int \frac{e^{-x}}{x}dx-\int \left(\ln x\right)e^{-x}dx+C$
$\Rightarrow y{e}^{-x}=e^{-x}\ln x+\int \left(e^{-x}\right)\ln x-\int \ln x{e}^{-x}dx+C$
$\Rightarrow y{e}^{-x}=e^{-x}\ln x+C$
$\Rightarrow y=\ln x+C{e}^x=f\left(x\right)$

since $f\left(1\right)=e,$ therefore $C=1$
$\therefore f\left(x\right)=\ln x+e^x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}+e^x$
Slope of tangent at point $(1,e)$ is $(1+e).$
and slope of normal at point $(1,e)$ is $-\frac{1}{1+e}.$

Equation of tangent:
$y-e=(1+e)(x-1)$
$\Rightarrow (1+e)x-y-1=0$

Equation of normal:
$y-e=-\frac{1}{1+e}(x-1)$
$\Rightarrow (1+e)(y-e)=-x+1$
$\therefore x+(1+e)y-(e^2+e+1)=0$