# Mark the correct alternative in the following:

Question:

Mark the correct alternative in the following:

The function $f(x)=\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right)$ is of the following types:

A. even and increasing

B. odd and increasing

C. even and decreasing

D. odd and decreasing

Solution:

Formula:- (i) if $f(-x)=f(x)$ then function is even

(ii) if $f(-x)=-f(x)$ then function is odd

(iii) The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)>0$ for all $x \in(a, b)$

Given:-

$f(x)=\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right)$

$\mathrm{d}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}^{3}\left(\mathrm{x}^{6}+1\right)^{\frac{1}{2}}}\left(3 \mathrm{x}^{2}+\frac{6 \mathrm{x}^{5}}{2\left(\mathrm{x}^{6}+1\right)^{\frac{1}{2}}}\right)$

$f^{\prime}(x)>0$

hence function is increasing function

$f(-x)=-\log \left(\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right)\right.$

$\Rightarrow f(-x)=-f(x)$ is odd function