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 :
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).
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The correct answer is( option 3) , $ k\in (\frac {2}{3},1]$.
We are given $$\int_1^2 \frac{f(x)}{\sqrt{x}}=k\int_1^2 f(x)\neq 0 .$$ we need to find the range of $k$.
Note that $$ 1\le x \le 2 $$ which implies $$1\le \sqrt x\le \sqrt 2 $$ and $$ \frac {\sqrt 2}{2} \le \frac {1}{ \sqrt x}\le 1 $$Note that $ f(x)\ge 0 $.
Thus the range of $ k$ is $$ \frac {\sqrt 2}{2} \le k \le 1 $$ That implies that $k\in (\frac {2}{3},1]$.
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 $\dfrac{1}{\sqrt{2}}>\dfrac{2}{3}$.