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  • Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxvii

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

Answers (1)

Answer:

\text { Increasing interval }(-3,0) \cup(5, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(0,5)

Given:

Here given that

f(x)=\frac{3}{2}x^{4}-4x^{3}-45x^{2}+51

To find:

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

Hint:

\text { If } f^{\prime}(x)>0 \text { for all } x \in(a ., b), \text { then } f(x) \text { is increasing on }(a, b) \text { and If } f^{\prime}(x)<0 \text { for all } x \in \text { (a., } b) \text { , then } f(x) \text { is decreasing on }(a, b) \text { . }

Solution:

We have,

f(x)=\frac{3}{2}x^{4}-4x^{3}-45x^{2}+51

Differentiating w.r.t. x, we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{3}{2} x^{4}-4 x^{3}-45 x^{2}+51\right) \\ &f^{\prime}(x)=\frac{3}{2}\left(4 x^{3}\right)-12 x^{2}-45(2 x) \\ &=6 x^{3}-12 x^{2}-90 x \end{aligned}

We have to find critical points, we must have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{3}-12 x^{2}-90 x=0 \\ &\Rightarrow 6 x\left(x^{2}-2 x-15\right)=0 \\ &\Rightarrow x\left(x^{2}-2 x-15\right)=0,\{\therefore 6=0\} \\ &\Rightarrow x\left(x^{2}-5 x+3 x-15\right)=0 \\ &\Rightarrow x(x-5)(x+3)=0 \\ &\Rightarrow x=0 ; x=5 ; x=-3 \end{aligned}

The points x=0, 5, -3 divide the real line into four disjoint intervals,

\begin{aligned} &(-\infty,-3),(-3,0),(0,5) \text { and }(5, \infty) \\ &\text { In intervals }(-3,0) \text { and }(5, \infty), f^{\prime}(x)>0 \end{aligned}

\text { Clearly, } f^{\prime}(x)>0 \text { if }-3<x<0 \text { and } 5<x<\infty \\ \text { However, }

\begin{aligned} &\text { In the intervals }(-\infty,-3) \text { and }(0,5) f^{\prime}(x)<0 \text { . }\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty<x<-3 \text { and } 0<x<5 \end{aligned}

\text { Thus, } f(x) \text { is increasing in the intervals }(-3,0) \cup(5, \infty) \text { and } \\ \text { decreasing in the intervals }(-\infty,-3) \cup(0,5) \text { . }

Posted by

Gurleen Kaur

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