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

Answers (1)

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.

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