Question

If $f\left(x\right)={\tan }^{-1}x-\left(1/2\right)\log x$. Then

• A. the greatest value of $f\left(x\right)$ on $[1/\sqrt{3},\sqrt{3}]$ is $\pi /6+\left(1/4\right)\log 3$
• B. the least value of $f\left(x\right)$ on $[1/\sqrt{3},\sqrt{3}]$ is $\pi /3+\left(1/4\right)\log 3$
• C. $f\left(x\right)$ decreases on $(0,\infty )$
• D. $f\left(x\right)$ increases on $(-\infty ,0)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A the greatest value of $f\left(x\right)$ on $[1/\sqrt{3},\sqrt{3}]$ is $\pi /6+\left(1/4\right)\log 3$
B the least value of $f\left(x\right)$ on $[1/\sqrt{3},\sqrt{3}]$ is $\pi /3+\left(1/4\right)\log 3$
C $f\left(x\right)$ decreases on $(0,\infty )$

The domain of $f\left(x\right)$ is $(0,\infty )$.
For $x>0$,
$f^{\prime }\left(x\right)=\frac{1}{1+x^2}-\frac{1}{2x}=\frac{2x-(1+x^2)}{(1+x^2)2x}=-\frac{{(1-x)}^2}{(1+x^2)2x}$
Thus $f^{\prime }\left(x\right)<0$, i.e.$f\left(x\right)$ decreases on $(0,\infty )$.
Also $f^{\prime }\left(x\right)=0$ if $x=1$ and $f\left(1\right)=\pi /4$
$f\left(1/\sqrt{3}\right)=\pi /6+\left(1/4\right)\log 3$, $f\left(\sqrt{3}\right)=\pi /3-\left(1/4\right)\log 3$.
Thus the greatest value is $\pi /6+\log 3$ and the least value is $\pi /3-\left(1/4\right)\log 3$.
Ans: A,B,C