Find the value of $k$

Question

If $f:[1,2]\mapsto \Bbb R$ is a non-negative integrable function such that $\int_1^2\dfrac{f(x)}{\sqrt x}\operatorname{dx}=k\int _1^2f(x)\operatorname{dx}\neq 0$,then find $k$.

I was thinking to use integration by parts which is not working.Any hints will be helpful.

Please provide some hints only.

Answer 1

HINT:

Use the fact that $\int _a ^b f(x)g(x)=g(c)\int _a ^b f(x)$ where $c\in (a,b)$

Answer 2

by the mean value theorem for the integrals, you have $$0=\int_1^2 f(x)\left(\frac1{\sqrt x} - k\right)\, dx = \left(\frac 1{\sqrt c} -k\right)\int_1^2f(x)\, dx \text{ for some $c$ between $1$ and $2.$}$$

without any further information on $f,$ all we can conclude is that $$\frac 1{\sqrt 2 } < k < 1. $$