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Let $g(x)=\int_{x}^{2 x} \frac{f(t)}{t} d t$ where $x>0$ and $f$ be continuous function and $f(2 x)=f(x)$, then
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The correct answer is:
$\mathrm{g}(\mathrm{x})$ is constant function
$g(x)=\int_{x}^{2 x} \frac{f(t)}{t} d t$
$g^{\prime}(x)=\frac{f(2 x)}{2 x} \cdot 2^{\prime}-\frac{f(x)}{x} \cdot 1=\frac{f(2 x)-f(x)}{x}=\frac{f(x)-f(x)}{x}[\because f(2 x)=f(x)]$
$g^{\prime}(x)=0$
$g(x)=$ constant.
$g^{\prime}(x)=\frac{f(2 x)}{2 x} \cdot 2^{\prime}-\frac{f(x)}{x} \cdot 1=\frac{f(2 x)-f(x)}{x}=\frac{f(x)-f(x)}{x}[\because f(2 x)=f(x)]$
$g^{\prime}(x)=0$
$g(x)=$ constant.
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