The function $x-\frac{\log(1+x)}{x}(x>0)$ is increasing in:
Answer & explanation
Correct answer: option 2
$f'(x)=\frac{x-\frac{\log(1+x)}{x}+\log(1+x)}{x^2}>0$ for increasing function $f'(x)>0$
$⇒x^2-\frac{x}{1+x}+\log(1+x)>0⇒x>0\,x∈(0, ∞)$
The function $x-\frac{\log(1+x)}{x}(x>0)$ is increasing in:
Correct answer: option 2
$f'(x)=\frac{x-\frac{\log(1+x)}{x}+\log(1+x)}{x^2}>0$ for increasing function $f'(x)>0$
$⇒x^2-\frac{x}{1+x}+\log(1+x)>0⇒x>0\,x∈(0, ∞)$