Q87 / 155 · WBJEE 2021 · Mathematics

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
    $\mathrm{g}(\mathrm{x})$ is strictly decreasing function
  • Correct answer
    C
    $\mathrm{g}(\mathrm{x})$ is constant function
  • D
    $\mathrm{g}(\mathrm{x})$ is not derivable function

Solution

$g(x)=\int_{x}^{2 x} \frac{f(t)}{t} d t$ $g^{\prime}(x)=\frac{f(2 x)}{2 x} \cdot 2^{\prime}-\frac{f(x)}{x} \cdot 1=\frac{f(2 x)-f(x)}{x}=\frac{f(x)-f(x)}{x}[\because f(2 x)=f(x)]$ $g^{\prime}(x)=0$ $g(x)=$ constant.

Hence, the correct answer is option: 'C'.

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