#### please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise Multiple choice question , question 7 maths textbook solution

Correct option (b)

Hint: Use that, a function $f(x)$ is odd if $f(-x)=-f(x)$ & even if $f(-x)=f(x)$

& If $f(x)$  is increasing function ${f}'(x)>0$ , decreasing if ${f}'(x)<0$ .

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

Explanation: It is given that

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

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

\begin{aligned} \therefore 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)} \times\left(x^{3}+\sqrt{x^{6}+1}\right)\right] \end{aligned}

\begin{aligned} &=\log _{e}\left[\frac{1}{\left(x^{3}+\sqrt{x^{6}+1}\right)}\right] \\ &=-\log _{e}\left(x^{3}+\sqrt{x^{6}+1}\right) \\ &=-f(x) \\ \because f(-x) &=-f(x) \end{aligned}

So, $f(x)$ is odd function

Now,Differentiate (i) w.r.t  x

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

\begin{aligned} &=\frac{1}{\left(x^{3}+\sqrt{x^{6}+1}\right)} \times 3 x^{2}\left(\frac{\sqrt{x^{6}+1}}{\sqrt{x^{6}+1}}+x^{3}\right) \\ &f^{\prime}(x)=\frac{3 x^{2}}{\sqrt{x^{6}+1}} \end{aligned}

Here x2 & x6 are even power of x

So, $f^{\prime}(x)=\frac{3 x^{2}}{\sqrt{x^{6}+1}}>0$

Thus,$f(x)$  is odd and increasing function.