Practicing Success
Practicing Success
CUET Preparation Today
The function $f(x)=\frac{x}{\log x}$ increases on the interval |
$(0, \infty)$ $(0, e)$ $(e, \infty)$ none of these |
$(e, \infty)$ |
Clearly, f(x) is defined for x > 0 Now, $f(x)=\frac{x}{\log x} \Rightarrow f'(x)=\frac{\log x-1}{(\log x)^2}$ ∴ $f'(x)>0 \Rightarrow \log x-1>0 \Rightarrow \log x>1 \Rightarrow x>e \Rightarrow x \in(e, \infty)$ |
