Question

The function $f\left(x\right)={\log }_e[x^3+\sqrt{x^6+1}]$ is an

• A. even function
• B. odd function
• C. increasing function
• D. decreasing function

Answer 1

Answer: (B), (C)

The correct options are
B odd function
C increasing function

Given that $f\left(x\right)={\log }_e[x^3+\sqrt{x^6+1}]$

we know that a function is even if $f(-x)=f\left(x\right)$ and

a function is odd if $f(-x)=-f\left(x\right)⟹f(-x)+f\left(x\right)=0$

Therefore $f(-x)={\log }_e\left[\right(-x{)}^3+\sqrt{(-x{)}^6+1}]$

$f(-x)={\log }_e[-x^3+\sqrt{x^6+1}]\ne f\left(x\right)$

Therefore, the function is not even.

Hence, $f(-x)+f\left(x\right)={\log }_e[-x^3+\sqrt{x^6+1}]+{\log }_e[x^3+\sqrt{x^6+1}]$

$={\log }_e\left[\right(-x^3+\sqrt{x^6+1}\left)\right(x^3+\sqrt{x^6+1}\left)\right]$ (since, $\log (a+b)=\log \left(ab\right)$)

$={\log }_e\left[\right(\sqrt{x^6+1}{)}^2-\left(x^3{)}^2\right]$ (since, $(a+b)(a-b)=a^2-b^2$)

$={\log }_e[x^6+1-x^6]$

$={\log }_e\left[1\right]=0$

$\text{Therefore, the given function is odd function.}$

we know that if $f^{{}^{\prime }}\left(x\right)>0$ then the function is increasing function and

if $f^{{}^{\prime }}\left(x\right)<0$ then the function is decreasing function.

Therefore, $f^{{}^{\prime }}\left(x\right)=\frac{1}{x^3+\sqrt{x^6+1}}(3x^2+\frac{1}{2\sqrt{x^6+1}}6x^5)$

if $x>0$ then $f^{{}^{\prime }}\left(x\right)>0$

$\text{Therefore, the given function is increasing function.}$