Question

Let $\int \frac{1-\ln x}{x^{2}} d x=f(x)$, for all positive $x$. If $f(e)=\frac{1}{e}$, then $f(2)+f(4)$ is equal to

• A. $\ln 4$
• B. $\ln 2$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$f(x)=\int \frac{1}{x^{2}} d x-\int \ln x \cdot \frac{1}{x^{2}} d x $
$=-\frac{1}{x}-\left\{\ln x\left(-\frac{1}{x}\right)+\int \frac{1}{x^{2}} d x\right\} $
$=-\frac{1}{x}+\frac{\ln x}{x}+\frac{1}{x}+c=\frac{\ln x}{x}+c$
Now, $f(e)=\frac{1}{e} \Rightarrow c=0$
Hence, $f(2)+f(4)=\frac{\ln 2}{2}+\frac{\ln 4}{4}=\ln 2$