Question

Let $f: R^{+} \rightarrow R$ be a differentiable function satisfying
$f(x)=e+(1-x) \ln \left(\frac{x}{e}\right)+\int\limits_1^x f(t) d t \text { for all } x \in R^{+} .$
The $x$-intercept of normal drawn to the curve $y=f(x)$ at point $P$ where $y=f(x)$ crosses the line $x =1$, is equal to

• A. $e ^2+ e -1$
• B. $\frac{ e ^2+ e +1}{ e +1}$
• C. $\frac{ e ^2- e +1}{ e +1}$
• D. $ e ^2+ e +1$

Answer 1

Answer: (D)

Given, $f ( x )= e +(1- x ) \ln \left(\frac{ x }{ e }\right)+\int\limits_1^{ x } f ( t ) dt , \forall x \in R ^{+}$
$\Rightarrow f ^{\prime}( x )=(1- x ) \frac{1}{ x }+(-1) \ln \left(\frac{ x }{ e }\right)+ f ( x ) $
$\Rightarrow f ^{\prime}( x )=\frac{1}{ x }-1-\ln x +1+ f ( x ) $
$\therefore f ^{\prime}( x )- f ( x )=\left(\frac{1}{ x }-\ln x \right)$
Let $f(x)=y$, so that $f^{\prime}(x)=\frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}-y=\left(\frac{1}{x}-\ln x\right)$....(1)
which is a linear differential equation
So, I.F. $= e ^{-\int d x}= e ^{- x }$
$\therefore$ Solution of $(1)$ is given by
$y \cdot e^{-x}=\int\left(\frac{1}{x}-\ln x\right) e^{-x} d x+C$...(2)
Put $- x = t \Rightarrow dx =- dt$
Let $I _1=\int\left(\frac{1}{ x }-\ln x \right) e ^{- x } dx =\int\left(\frac{1}{- t }-\ln (- t )\right) e ^{ t }(-1) dt =\int\left(\frac{1}{ t }+\ln (- t )\right) e ^{ t } dt$
$\therefore I _1= e ^{ t } \ln (- t )= e ^{- x } \ln x$
So, equation (2) becomes
$ y \cdot e^{-x}=e^{-x} \ln x+C $
$\therefore y=\ln x+C \cdot e^x $
$\text { As } y(1)=e$
$\Rightarrow =0+C e$
$\therefore f(x)=e^x+\ln x $
As $f(1)=e$
So, $P (1, e )$
Also, $\left.f ^{\prime}( x )\right]_{ P (1, e )}=\left( e ^{ x }+\frac{1}{ x }\right)=( e +1)$
So, equation of normal at $P(1, e)$ is
$(y-e)=\frac{-1}{e+1}(x-1)$
For $x$-intercept, put $y =0$
$\Rightarrow - e ( e +1)=-( x -1) \Rightarrow e ^2+ e = x -1$
\therefore x=e^2+e+1