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

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

Answers (1)

Answer:

\text { Increasing interval }(-2,1) \cup(3, \infty) \\ \text { Decreasing interval } (-\infty ,-2) \cup (1.3)

Given:

Here given that

f(x)=\frac{3}{10} x^{4}-\frac{4}{5} x^{3}-3 x^{2}+\frac{36}{5} x+11

To find:

We have to find the intervals in which f(x) is increasing or decreasing.

Hint:

Put f ‘(x) = 0 to find critical points and then use increasing and decreasing property.

Solution:

We have,

f(x)=\frac{3}{10} x^{4}-\frac{4}{5} x^{3}-3 x^{2}+\frac{36}{5} x+11

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

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{3}{10} x^{4}-\frac{4}{5} x^{3}-3 x^{2}+\frac{36}{5} x+11\right) \\ &\Rightarrow f^{\prime}(x)=\frac{3}{10}\left(4 x^{3}\right)-\frac{4}{5}\left(3 x^{2}\right)-3(2 x)+\frac{36}{5} \\ &\Rightarrow f^{\prime}(x)=\frac{12}{10} x^{3}-\frac{12}{5} x^{2}-6 x+\frac{36}{5} \\ &\Rightarrow f^{\prime}(x)=\frac{6}{5} x^{3}-\frac{12}{5} x^{2}-6 x+\frac{36}{5} \\ &\Rightarrow f^{\prime}(x)=\frac{6}{5}\left(x^{3}-2 x^{2}-5 x+6\right) \\ &\Rightarrow f^{\prime}(x)=\frac{6}{5}(x-1)\left(x^{2}-x-6\right) \\ &\Rightarrow f^{\prime}(x)=\frac{6}{5}(x-1)(x+2)(x-3) \end{aligned}

For f(x) we have to find critical points

we must have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{6}{5}(x-1)(x+2)(x-3)=0 \\ &\Rightarrow(x-1)(x+2)(x-3)=0\left\{\therefore \frac{6}{5}>0\right\} \\ &\Rightarrow x=1, \quad x=-2, \quad x=3 \end{aligned}

\begin{aligned} &\text { Now we take the intervals }(-\infty,-2) \text { i.e, }-\infty<x<-2 . \text { In this case we have } x-1<0, x+2<\\ &0 \text { and } x-3<0 . \text { Clearly, } f^{\prime}(x)<0 \text { if } x \in(-\infty,-2) \text { . } \end{aligned}

\begin{aligned} &\text { Now we take }(-2,1) . \text { In this case we have } x-1<0, x+2>0 \text { and } x-3<0 . \text { So, } f^{\prime}(x)>0 \text { if } x &\in(-2,1) \end{aligned}

\text { After that we take }(1,3) \text { i.e, } 1<x<3 \text { . } \\ \text { Clearly, we have } x-1>0, x+2>0 \text { and } x-3<0 . \text { So, } f^{\prime}(x)<0 \text { if } x \in(1,3) \text { . }

\text { Finally we take }(3, \infty) \text { i.e, } 3<x<\infty \text { . In this case we have } x-1>0, x+2>0 \text { and } x-3>0 \text { . }

\text { Clearlyf }^{\prime}(x)>0 \text { if } x>3 \text { or } x \in(3, \infty)

\text { Thus, the function is increasing on }(-2,1) \cup(3, \infty) \text { and } \\ f(x) \text { is decreasing on interval }(-\infty,-2) \cup(1,3) \text { . }

Posted by

Gurleen Kaur

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