Question

If $f\left(x\right)=x-\left[x\right]$ and $g\left(x\right)=\underset{0}{\overset{x}{\int }}f(t+n)dt\forall n\in N,$ then $g^{\prime }\left(\frac{5}{2}\right)$ is equal to

• A. $f\left(\frac{1}{2}\right)$
• B. $\frac{1}{2}$
• C. $f\left(\frac{3}{2}\right)$
• D. $f\left(\frac{5}{2}\right)-f\left(5\right)$

Answer 1

Answer: (A), (B), (C), (D)

$f\left(\frac{5}{2}\right)-f\left(5\right)$

$f\left(x\right)$ is periodic with period $1$
$g\left(x\right)=\underset{0}{\overset{x}{\int }}f(t+n)dt\Rightarrow g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow g^{\prime }\left(x\right)=f\left(x\right)=x-\left[x\right]\therefore g^{\prime }\left(\frac{5}{2}\right)=f\left(\frac{5}{2}\right)=\frac{1}{2}=f\left(\frac{1}{2}\right)=f\left(\frac{3}{2}\right);\text{also}f\left(5\right)=0$