Question

Let the function $f:(-\infty ,\infty )\to \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ be given by $f(x)={\sin }^{-1}\left({\log }_3\left(\frac{x^2-x+1}{x^2+x+1}\right)\right)$, then

• A. $f\left(\frac{1}{x}\right)=-f(-x)$
• B. $f(x)$ is a strictly increasing function in $(-\infty ,\infty )$
• C. f(x) is a surjective function
• D. f (x) is a injective function.

Answer 1

Answer: (A), (C)

(A) $f(x)={\sin }^{-1}\left({\log }_3\left(\frac{x^2-x+1}{x^2+x+1}\right)\right)$
$f(-x)={\sin }^{-1}\left({\log }_3\left(\frac{x^2+x+1}{x^2-x+1}\right)\right)=-f(x)$
$f\left(\frac{1}{x}\right)=f(x)=-f(-x)$
(B) $f^{\prime }(x)=\frac{1}{\sqrt{1-{\left({\log }_3\left(\frac{x^2-x+1}{x^2+x+1}\right)\right)}^2}}\cdot \frac{{\log }_{3}e}{\left(\frac{x^2-x+1}{x^2+x+1}\right)}\cdot \frac{2\left(x^2-1\right)}{{\left(x^2+x+1\right)}^2}$

Since $\frac{x^2-x+1}{x^2+x+1}\in \left[\frac{1}{3},3\right]$
$\therefore f(x)=\left[\frac{-\pi }{2},\frac{\pi }{2}\right]$