Question

Let $f:R^{+}\to R$ be a differentiable function satisfying
$f(x)=e+(1-x)\ln \left(\frac{x}{e}\right)+\underset{1}{\overset{x}{\int }}f(t)dt\text{ for all }x\in R^{+}.$
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, $f(x)=e+(1-x)\ln \left(\frac{x}{e}\right)+\underset{1}{\overset{x}{\int }}f(t)dt,\forall x\in R^{+}$
$\Rightarrow f^{\prime }(x)=(1-x)\frac{1}{x}+(-1)\ln \left(\frac{x}{e}\right)+f(x)$
$\Rightarrow f^{\prime }(x)=\frac{1}{x}-1-\ln x+1+f(x)$
$\therefore f^{\prime }(x)-f(x)=\left(\frac{1}{x}-\ln x\right)$
Let $f(x)=y$, so that $f^{\prime }(x)=\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}-y=\left(\frac{1}{x}-\ln x\right)$....(1)
which is a linear differential equation
So, I.F. $=e^{-\int dx}=e^{-x}$
$\therefore$ Solution of $(1)$ is given by
$y\cdot e^{-x}=\int \left(\frac{1}{x}-\ln x\right)e^{-x}dx+C$...(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
$y\cdot e^{-x}=e^{-x}\ln x+C$
$\therefore y=\ln x+C\cdot e^x$
$\text{ As }y(1)=e$
$\Rightarrow =0+Ce$
$\therefore f(x)=e^x+\ln x$
As $f(1)=e$
So, $P(1,e)$
Also, ${f^{\prime }(x)]}_{P(1,e)}=\left(e^x+\frac{1}{x}\right)=(e+1)$
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