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Provide solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion vi

Answers (1)


\text {Increasing interval} (-2,3) \\ \text {Decreasing interval} (-\infty,-2) \cup(3, \infty)


Here given that


To find:

We have to find out the intervals in which function is increasing and decreasing.


First, we will find critical points and then use increasing and decreasing property.


Given that


On differentiating we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(8+36 x+3 x^{2}-2 x^{3}\right) \\ &\Rightarrow f^{\prime}(x)=36+6 x-6 x^{2} \end{aligned}

For f(x), Firstly we will find critical points.

For this we have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 36+6 x-6 x^{2}=0 \\ &\Rightarrow 6\left(6+x-x^{2}\right)=0 \\ &\Rightarrow 6+x-x^{2}=0 \\ &\Rightarrow-x^{2}+x+6=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}-x-6=0 \\ &\Rightarrow x^{2}-3 x+2 x-6=0 \\ &\Rightarrow x(x-3)+2(x-3)=0 \\ &\Rightarrow(x-3)(x+2)=0 \end{aligned}

\begin{aligned} &\Rightarrow x-3=0 \text { and } x+2=0 \\ &\Rightarrow x=3 \text { and } x=-2 \\ &\text { Clearly, } f^{\prime}(x)>0, f-2<x<3 \text { or } x \in(-2,3) \end{aligned}

\text { and } f^{\prime}(x)<0 \text { if } x<-2 \text { and } x>3 \text { or } x \in(-\infty,-2) \text { and } x \in(3, \infty)

\text { Thus, } f(x) \text { is increasing on }(-2,3) \text { and decreasing on }(-\infty,-2) \cup(3, \infty)

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Gurleen Kaur

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