Question

The function $f\left(x\right)={\log }_e\left(x^3+\sqrt{x^6+1}\right)$ is of the following types:
(a) even and increasing
(b) odd and increasing
(c) even and decreasing
(d) odd and decreasing

Answer 1

(b) odd and increasing

$$\begin{gathered} f\left(x\right)={\log }_e\left(x^3+\sqrt{x^6+1}\right) \\ \Rightarrow f(-x)={\log }_e\left(-x^3+\sqrt{x^6+1}\right) \\ ={\log }_e\left\{\frac{\left(-x^3+\sqrt{x^6+1}\right)\left(x^3+\sqrt{x^6+1}\right)}{x^3+\sqrt{x^6+1}}\right\} \\ ={\log }_e\left(\frac{x^6+1-x^6}{x^3+\sqrt{x^6+1}}\right) \\ ={\log }_e\left(\frac{1}{x^3+\sqrt{x^6+1}}\right) \\ =-{\log }_e\left(x^3+\sqrt{x^6+1}\right) \\ =-f\left(x\right) \\ \mathrm{Hence},f(-x)=-f\left(x\right) \\ \mathrm{Therefore},\mathrm{it}\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}. \end{gathered}$$

$$\begin{gathered} f\left(x\right)={\log }_e\left(x^3+\sqrt{x^6+1}\right) \\ \frac{d}{dx}\left\{f\left(x\right)\right\}=\left(\frac{1}{x^3+\sqrt{x^6+1}}\right)\times \left(3x^2+\frac{1}{2\sqrt{x^6+1}}\times 6x^5\right) \\ =\left(\frac{1}{x^3+\sqrt{x^6+1}}\right)\times \left(\frac{6x^2\sqrt{x^6+1}+6x^5}{2\sqrt{x^6+1}}\right) \\ =\left(\frac{1}{x^3+\sqrt{x^6+1}}\right)\times \left\{\frac{6x^2\left(\sqrt{x^6+1}+x^3\right)}{2\sqrt{x^6+1}}\right\} \\ =\left(\frac{6x^2}{2\sqrt{x^6+1}}\right)>0 \\ \mathrm{Therefore},\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{an}\mathrm{increasing}\mathrm{function}. \end{gathered}$$