Question

Let $f\left(x\right)$ be a periodic function of period $1$ and $f\left(\frac{1}{2}\right)=\frac{1}{2}$.
Let $\varphi \left(x\right)={\int }_0^{x}f(x+n)dx$ for all $n\in N$. Then, ${\varphi }^{\prime }\left(\frac{3}{2}\right)$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $2$
• D. $1$

Answer 1

The correct option is B $\frac{1}{2}$

Since $f\left(x\right)$ has a period of $\frac{1}{2}$, therefore
$f(1+x)=f\left(x\right)$.
Now

$\varphi \left(x\right)={\int }_0^{x}f(x+n)dx$

Differentiating both sides, we get
${\varphi }^{\prime }\left(x\right)=f(x+n)$
${\varphi }^{\prime }\left(\frac{3}{2}\right)=f(\frac{3}{2}+n)$
$=f(1+(\frac{1}{2}+n))=f(\frac{1}{2}+n)$
Now $n\in N$.
Hence,
${\varphi }^{\prime }\left(\frac{3}{2}\right)=f(\frac{1}{2}+n)=f\left(\frac{1}{2}\right)=\frac{1}{2}$