Question

Let the function $f:R\to R$ satisfies $f\left(x\right)={\int }_0^x(\frac{1}{e^{-t}}+f(x-t\left)\right)dt$.
[Note : e denotes Napier's constant]
If the area enclosed by the curve $y=f\left(x\right)$ and x-axis from $x=x_1$ to $x=0$, where $x_1$ is abscissa of point of inflection on the graph of $y=f\left(x\right)$, is $1-\frac{b}{e^c}$, where $b,c\in N$, then $(b+c)$ is smaller than

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

Answer: (A)

$4$

Given : $f\left(x\right)={\int }_0^x(e^{-t}+f(x-t\left)\right)dt$

$f\left(x\right)={\int }_0^x{e}^{-t}dt+{\int }_0^{x}f\left(t\right)dt$

Using Leibnitz rule

$f^{\prime }\left(x\right)=e^x+f\left(x\right)f^{\prime }\left(x\right)-f\left(x\right)=e^x$

Multiplying with $e^{-x}$ both sides

Integrating both sides

$e^{-x}f\left(x\right)=xf\left(x\right)=x{e}^x$

Coordinate of tangent to the curve $f\left(x\right)$

$f^{\prime }\left(x\right)=x{e}^x+e^x=0x=-1y=-e^{-x}$

Inflection of the point is $x=-1$

According to question

${\int }_{-1}^{0}f\left(x\right)dx={\int }_{-1}^{0}x{e}^x={|x{e}^x-e^x|}_{-1}^0=0-1-[-e^{-1}-e^{-1}]=-1+\frac{2}{e^1}b=2c=1b+c=2+1=3<4$

Hence the correct answer is less than $4$.