Q:

$\int_1^2 \frac{f(x)}{\sqrt{x}}=k\int_1^2 f(x)$ then $k$ lies in the interval

Let $f:[1,2]\to \mathbb{R}$ be Riemann integrable non-negative function such that $$ \int_1^2 \frac{f(x)}{\sqrt{x}}=k\int_1^2 f(x)\neq 0 .$$  Then $k$ lies in the interval :

$\left[0,\frac{1}{3}\right]$
$\left(\frac{1}{3},\frac{2}{3}\right]$
$\left(\frac{2}{3},1\right]$
$\left(1,\frac{4}{3}\right]$

By checking, $f(x)=1 $, then $k\in \left(\frac{2}{3},1\right]$ which tells that option (1) and (2) are incorrect. But I don't have any idea about the option (4). 

A:

Note that $\displaystyle\int_{1}^{2}\dfrac{f(x)}{\sqrt{x}}dx\leq\int_{1}^{2}\dfrac{f(x)}{\sqrt{1}}dx$, so $k\leq 1$.
Also, $\displaystyle\int_{1}^{2}\dfrac{f(x)}{\sqrt{x}}dx\geq\int_{1}^{2}\dfrac{f(x)}{\sqrt{2}}dx$, so $k\geq\dfrac{1}{\sqrt{2}}$, one can test that $\dfrac{1}{\sqrt{2}}>\dfrac{2}{3}$.

