Question

Let $f\left(x\right)=$ ${\int }_0^x{e}^{t-\left(t\right)}$ dt $(x>0)$, where $\left[x\right]$ denotes greatest integer less than or equal to $x$, is :

• A. Continuous and differentiable $\forall$ x $ϵ$ $(0,3]$0
• B. Continuous but not differentiable $\forall$ x $ϵ$ $(0,3]$
• C. $f\left(x\right)=e$
• D. $f\left(2\right)=2(e-1)$

Answer 1

The correct option is D $f\left(2\right)=2(e-1)$

We have, $f\left(x\right)={\int }_0^x{e}^{t-\left[t\right]}dt$

Above can be written as:

$f\left(x\right)={\int }_0^1{e}^{t}dt+{\int }_1^2{e}^{\left\{t\right\}}\cdots +{\int }_{\left[x\right]-1}^{\left[x\right]}{e}^{\left\{t\right\}}dt+{\int }_{\left[x\right]}^x{e}^{\left\{t\right\}}dt$

$=\left[x\right]{\int }_0^1{e}^{t}dt+{\int }_{\left[x\right]}^x{e}^{\left\{t\right\}}dt=\left[x\right](e-1)+(e^t{)}_0^{x-\left[x\right]}$.

For $x\in (0,3]$

$f\left(x\right)=\left[x\right](e-1)+e^{\left\{x\right\}}-1$, which is not continuos.

$f\left(2\right)=2(e-1)+0=2(e-1)$

Hence, option D is correct.