Question

Two functions $f\left(x\right)$ and $g\left(x\right)$ are defined as $f\left(x\right)=\{\begin{array}{cc}[x{]}^{\left\{x\right\}},& x\ne 0\\ 0,& x=0\end{array}$ and $g\left(x\right)=\{\begin{array}{cc}\{x{\}}^{\left[x\right]},& x\ne 0\\ 0,& x=0\end{array}$ (where $[.]$ and $\{.\}$ refer to the greatest integer function and the fractional part function respectively and $n\in N).$ Then which of the following statements is/are TRUE?

• A. $\underset{0}{\overset{n+1}{\int }}f\left(x\right)dx-\underset{0}{\overset{n}{\int }}f\left(x\right)dx=\frac{n-1}{\ln n}$
• B. $\underset{0}{\overset{n+1}{\int }}g\left(x\right)dx-\underset{0}{\overset{n}{\int }}g\left(x\right)dx=\frac{1}{n+1}$
• C. $\underset{0}{\overset{t}{\int }}f\left(x\right)dx=\underset{0}{\overset{t}{\int }}g\left(x\right)dx$ has at least one solution in $t\in (0,\frac{\pi }{2})$
• D. $e<\underset{n\to \infty }{lim}\underset{0}{\overset{n}{\int }}g\left(x\right)dx<e^2$

Answer 1

Answer: (A), (B)

$\underset{0}{\overset{n+1}{\int }}g\left(x\right)dx-\underset{0}{\overset{n}{\int }}g\left(x\right)dx=\frac{1}{n+1}$

$\underset{0}{\overset{n+1}{\int }}f\left(x\right)dx-\underset{0}{\overset{n}{\int }}f\left(x\right)dx=\underset{n}{\overset{n+1}{\int }}f\left(x\right)dx$
$=\underset{n}{\overset{n+1}{\int }}{n}^{x-n}dx=\frac{1}{n^n}{\left[\frac{n^x}{\ln n}\right]}_n^{n+1}$
$=\frac{n-1}{\ln n}$

$\underset{0}{\overset{n+1}{\int }}g\left(x\right)dx-\underset{0}{\overset{n}{\int }}g\left(x\right)dx=\underset{n}{\overset{n+1}{\int }}f\left(x\right)dx$
$=\underset{n}{\overset{n+1}{\int }}(x-n{)}^{n}dx={\left[\frac{(x-n{)}^{n+1}}{n+1}\right]}_n^{n+1}$
$=\frac{1}{n+1}$

$t\in (0,1)\Rightarrow \underset{0}{\overset{t}{\int }}f\left(x\right)dx=0$
$t\in [1,\frac{\pi }{2})\Rightarrow \underset{0}{\overset{t}{\int }}f\left(x\right)dx<\frac{\pi }{2}-1$
$t\in (0,1)\Rightarrow \underset{0}{\overset{t}{\int }}g\left(x\right)dx=1$
$t\in [1,\frac{\pi }{2})\Rightarrow \underset{0}{\overset{t}{\int }}g\left(x\right)dx>0$,
Hence, $\underset{0}{\overset{t}{\int }}f\left(x\right)dx=\underset{0}{\overset{t}{\int }}g\left(x\right)dx$ $t\in (0,\frac{\pi }{2}),$ has no solution

$\underset{n\to \infty }{lim}\underset{0}{\overset{n}{\int }}g\left(x\right)dx=\underset{n\to \infty }{lim}\sum _{i=0}^n\frac{1}{i+1}$
It is divergent series.