Question

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
(A) \(g(x)\) is strictly increasing function
(B) \(g(x)\) is strictly decreasing function
(C) \(g(x)\) is constant function
(D) \(g(x)\) is not derivable function

Answer 1

Ans: (C)
Hint \(: g(x)=\int_{x}^{2 x} \frac{f(t)}{t} d t\)
$$
\begin{aligned}
&g^{\prime}(x)=\frac{f(2 x)}{2 x} \cdot 2-\frac{f(x)}{x} \cdot 1=\frac{f(2 x)-f(x)}{x}=\frac{f(x)-f(x)}{x}[\because f(2 x)=f(x)] \\
&g^{\prime}(x)=0 \\
&g(x)=\text { constant. }
\end{aligned}
$$

Answer 2

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Answer 3

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