Question

Let $f\left(x\right)={\int }_1^x\left(t\ln \right(t)-\frac{\ln \left(t\right)}{t})dt$, where $x>1$, then

• A. $f\left(x\right)$ has one point of maxima and no point of minima.
• B. $f^{\prime }\left(x\right)$ has two distinct roots
• C. $f\left(x\right)$ has one point of minima and no point of maxima
• D. $f\left(x\right)$ is monotonic

Answer 1

The correct option is D $f\left(x\right)$ is monotonic

$f\left(x\right)={\int }_1^x(t\ln \left(t\right)-\frac{\ln \left(t\right)}{t})dt$ and $(x>1)$

$f^{\prime }\left(x\right)=[x\times \ln x-\frac{\ln \left(x\right)}{x}]\times \frac{d}{dx}\left(x\right)-0$

$f^{\prime }\left(x\right)=[x\ln x-\frac{\ln x}{x}]$

$f^{\prime \prime }\left(x\right)\Rightarrow x\times \frac{1}{x}+\ln x-[\ln x\times \left(\frac{-1}{x^2}\right)+\frac{1}{x}\times \frac{1}{x}]$

$f^{\prime \prime }\left(x\right)=1+\ln x-[-\frac{\ln x}{x^2}+\frac{1}{x^2}]$

$=1+\ln x+\frac{\ln x}{x^2}-\frac{1}{x^2}>0$

Hence $f^{\prime \prime }\left(x\right)$ will be positive.

Therefore, $f\left(x\right)$ has a minimum and it is an increasing function.

Thus, $f\left(x\right)$ is monotonic.