Question

Let the function $f :(-\infty, \infty) \rightarrow\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ 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 $\frac{x^2-x+1}{x^2+x+1} \in\left[\frac{1}{3}, 3\right]$
$\therefore f ( x )=\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$