RD Sharma Class 12 Exercise 16.2 Increasing and Decreasing Function Solutions Maths - Download PDF Free Online

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# RD Sharma Class 12 Exercise 16.2 Increasing and Decreasing Function Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 05:55 PM IST

RD Sharma Class 12 Solutions Chapter 16 Exercise 16.2 Increasing and Decreasing Functions consist of formulae and concepts on tracking the homogeneous arrangement of natural conditions. The RD Sharma Class 12th Exercise 16.2 which is titled Increasing and Decreasing Functions will help students to practice these concepts promptly and solve all complex problems. RD Sharma Solutions are considered to be essential and the best reference book that students have used to score high in their board tests.

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## Increasing and Decreasing Functions Excercise16.2

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion i

$f(x) \text { is increasing on the interval }\left(-\infty, \frac{-3}{2}\right) \text { and decreasing on the interval }\left(\frac{-3}{2}, \infty\right)$
Given:
Here given that
$f(x)=10-6x-2x^{2}$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
$\text { If } f^{\prime}(x)>0 \text { for all } x \in(a ., b), \text { then } f(x) \text { is increasing on }(a, b) \text { . } \\ \text { If } f^{\prime}(x)< 0 \text { for all } x \in(a ., b), \text { then } f(x) \text { is increasing on }(a, b) \text { . }$
Solution:
Here given that
$f(x)=10-6x-2x^{2}$
On differentiating we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(10-6 x-2 x^{2}\right)\\ &\Rightarrow f^{\prime}(x)=-6-4 x\\ &\text { For } f(x) \text { to be increasing, we must have }\\ &f^{\prime}(x)>0\\ &\Rightarrow-6-4 x>0\\ &\Rightarrow-4 x>6\\ &\Rightarrow x<\frac{-6}{4} \end{aligned}
By applying (-) ve sign change comparison sign.
\begin{aligned} &\Rightarrow x<\frac{-3}{2}\\ &x \in\left(-\infty, \frac{-3}{2}\right)\\ &\text { So, } f(x) \text { is increasing on the interval }\left(-\infty, \frac{-3}{2}\right) \end{aligned}
\begin{aligned} &\text { For } \mathrm{f}(\mathrm{x}) \text { to be decreasing, we must have }\\ &f^{\prime}(x)<0\\ &\Rightarrow-6-4 x<0\\ &\Rightarrow-4 x<6\\ &\Rightarrow x>\frac{-6}{4} \end{aligned}
\begin{aligned} &\text { By applying }(-) \text { ve sign change comparison sign. }\\ &\Rightarrow x>\frac{-3}{2} \end{aligned}
\begin{aligned} &\Rightarrow x \in\left(\frac{-3}{2}, \infty\right)\\ &\text { So, } f(x) \text { is decreasing on the interval }\left(\frac{-3}{2},+\infty\right) \end{aligned}

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion ii

$f(x) \text { is increasing on the interval }(-1, \infty) \text { and decreasing on the interval }(-\infty,-)$
Given:
Here given that
$f(x)=x^{2}+2x-5$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
$\text { If } f^{\prime}(x)>0 \text { for all } x \in(a ., b), \text { then } f(x) \text { is increasing on }(a, b) \text { and If } f^{\prime}(x)<0 \text { for all } x \\ \in(a ., b), \text { then } f(x) \text { is decreasing on }(a, b) .$
Solution:
Here given that
$f(x)=x^{2}+2x-5$
On differentiating we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{2}+2 x-5\right) \\ &\Rightarrow f^{\prime}(x)=2 x+2 \end{aligned}
For f(x) to be increasing, we must have
\begin{aligned} &f^{\prime}(x)>0 \\ &\Rightarrow 2 x+2>0 \\ &\Rightarrow 2 x>-2 \\ &\Rightarrow x>-\frac{2}{2} \\ &\Rightarrow x>-1 \\ &x \in(-1, \infty) \end{aligned}
$\text { So, } f(x) \text { is increasing on the interval }(-1, \infty) \text { . }$
\text { Now, } \\ \begin{aligned} &\text { For } \mathrm{f}(\mathrm{x}) \text { to be decreasing, we must have }\\ &f^{\prime}(x)<0 \end{aligned}
\begin{aligned} &\Rightarrow 2 x+2<0 \\ &\Rightarrow 2 x<-2 \\ &\Rightarrow x<-\frac{2}{2} \\ &\Rightarrow x<-1 \\ &\Rightarrow x \in(-\infty,-1) \end{aligned}
$\text { So, } f(x) \text { is decreasing on the interval }(-\infty,-) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion iii

$\text { Increasing interval }\left(-\infty, \frac{-9}{2}\right) \\ \text { Decreasing interval }\left(\frac{-9}{2}, \infty\right)$
Given:
Here given that
$f(x)=6-9x-x^{2}$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
Use increasing and decreasing property.
Solution:
Here given that
$f(x)=6-9x-x^{2}$
On differentiating we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(6-9 x-x^{2}\right) \\ &\Rightarrow f^{\prime}(x)=-9-2 x \end{aligned}
For f(x) to be increasing, we must have
\begin{aligned} &f^{\prime}(x)>0 \\ &\Rightarrow-9-2 x> \\ &\Rightarrow-2 x>9 \\ &\Rightarrow x<\frac{-9}{2} \end{aligned}
By applying (-) ve sign change comparison sign.
\begin{aligned} &x \in\left(-\infty, \frac{-9}{2}\right)\\ &\text { So, } f(x) \text { is increasing on the interval }\left(-\infty, \frac{-9}{2}\right) \end{aligned}
For f(x) to be decreasing, we must have
\begin{aligned} &f^{\prime}(x)<0 \\ &\Rightarrow-9-2 x<0 \\ &\Rightarrow-2 x<9 \\ &\Rightarrow x>\frac{-9}{2} \end{aligned}
By applying (-) ve sign change comparison sign.
\begin{aligned} &\Rightarrow x \in\left(\frac{-9}{2}, \infty\right)\\ &\text { So, } f(x) \text { is decreasing on the interval }\left(\frac{-9}{2}, \infty\right) \end{aligned}

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion iv

$\text {Increasing interval}(-\infty, 1) \cup(3, \infty) \\ \text {Decreasing interval}(1,3)$
Given:
Here given that
$f(x)=2x^{3}-12x^{2}+18x+15$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
Firstly we will find critical points and then use increasing and decreasing property.
Solution:
Given that
$f(x)=2x^{3}-12x^{2}+18x+15$
On differentiating we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-12 x^{2}+18 x+15\right) \\ &\Rightarrow f^{\prime}(x)=6 x^{2}-24 x+18 \end{aligned}
Firstly we will find critical points for f(x).
For this we have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}-24 x+18=0 \\ &\Rightarrow 6\left(x^{2}-4 x+3\right)=0 \\ &\Rightarrow x^{2}-3 x-x+3=0\{\therefore 6>0\} \\ &\Rightarrow(x-3)(x-1)=0 \\ &\Rightarrow x-3=0 \text { and } x-1=0 \\ &\Rightarrow x=3 \text { and } x=1 \end{aligned}
\begin{aligned} &\text { Clearly, } f^{\prime}(x)>0, f(x)<1 \text { and } x>3 \text { or } x \in(-\infty, 1) \text { and } x \in(3, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &1
$\text { So, } f(x) \text { is increasing the interval }(-\infty, 1) \cup(3, \infty) \text { and } f(x) \text { is decreasing on interval }(1,3) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion v

$\text {Increasing interval}\: (-2,3) \\ \text {Decreasing interval}\: (-\infty,-2) \cup(3, \infty)$
Given:
Here given that
$f(x)=5+36x+3x^{2}-2x^{3}$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
First, we will find critical points and then use increasing and decreasing property.
Solution:
Given that
$f(x)=5+36x+3x^{2}-2x^{3}$
On differentiating we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(5+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
$\text { and } f^{\prime}(x)<0 \text { if } x<-2 \text { and } x>3 \text { or } x \in(-\infty,-2) \text { and } x \in(3, \infty)$
$\text { Thus, } f(x) \text { is increasing on }(-2,3) \text { and decreasing on }(-\infty,-2) \cup(3, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion vi

$\text {Increasing interval} (-2,3) \\ \text {Decreasing interval} (-\infty,-2) \cup(3, \infty)$
Given:
Here given that
$f(x)=8+36x+3x^{2}-2x^{3}$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
First, we will find critical points and then use increasing and decreasing property.
Solution:
Given that
$f(x)=8+36x+3x^{2}-2x^{3}$
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
$\text { and } f^{\prime}(x)<0 \text { if } x<-2 \text { and } x>3 \text { or } x \in(-\infty,-2) \text { and } x \in(3, \infty)$
$\text { Thus, } f(x) \text { is increasing on }(-2,3) \text { and decreasing on }(-\infty,-2) \cup(3, \infty)$

I

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion vii

$\text {Increasing interval} (-\infty,-2) \cup(4, \infty) \\ \text {Decreasing interval} (-2,4)$
Given:
Here given that
$f(x)=5x^{3}-15x^{2}-120x+3$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
Put f(x)=0 and solve this equation to find critical points of given function.
Solution:
We have,
$f(x)=5x^{3}-15x^{2}-120x+3$
Differentiating with respect to x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(5 x^{3}-15 x^{2}-120 x+3\right) \\ &f^{\prime}(x)=15 x^{2}-30 x-120 \end{aligned}
Now we have to find critical points for f(x).
We must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 15 x^{2}-30 x-120=0\\ &\Rightarrow 15\left(x^{2}-2 x-8\right)=0\\ &\Rightarrow x^{2}-2 x-8=0\{\therefore 15>0\}\\ &\Rightarrow x^{2}-4 x+2 x-8=0\\ &\Rightarrow x(x-4)+2(x-4)=0\\ &\Rightarrow(x-4)(x+2)=0\\ &\Rightarrow x-4=0 \text { and } x+2=0\\ &\Rightarrow x=4 \text { and } x=-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>4 \text { or } x \in(-\infty,-2) \text { and } x \in(4, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}
\begin{aligned} &-2

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion viii

$\text {Increasing interval} (-\infty,-2) \cup(6, \infty) \\ \text {Decreasing interval} (-2,6)$
Given:
Here given that.
$f(x)=x^{3}-6x^{2}-36x+2$
To find:
We have to find the intervals in which function is increasing and decreasing.
Hint:
Put f(x)=0 and solve this equation to find critical points of f(x) and use increasing and decreasing property.
Solution:
We have,
$f(x)=x^{3}-6x^{2}-36x+2$
Differentiating with respect to x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-6 x^{2}-36 x+2\right) \\ &f^{\prime}(x)=3 x^{2}-12 x-36 \end{aligned}
Now we have to find critical points for f(x).
We must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-12 x-36=0\\ &\Rightarrow 3\left(x^{2}-4 x-12\right)=0\\ &\Rightarrow x^{2}-4 x-12\{\therefore 3>0\}\\ &\Rightarrow x^{2}-6 x+2 x-12=0\\ &\Rightarrow(x-6)(x+2)=0\\ &\Rightarrow x-6=0 \text { and } x+2=0\\ &\Rightarrow x=6 \text { and } x=-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>6 \text { or } x \in(-\infty,-2) \text { and } x \in(6, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}
\begin{aligned} &-2

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion ix

$\text { Increasing interval }(-\infty, 2) \cup(3, \infty) \\ \text { Decreasing interval (2,3) }$
Given:
Here given that
$f(x)=2x^{3}-15x^{2}+36x+1$
To find:
We have to find the intervals in which function is increasing and decreasing.
Hint:
Put f ‘(x) = 0 to find critical points of f(x) and use increasing and decreasing property.
Solution:
We have,
$f(x)=2x^{3}-15x^{2}+36x+1$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-15 x^{2}+36 x+1\right) \\ &f \Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36 \end{aligned}
Now for critical points of f(x), we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}-30 x+36=0 \\ &\Rightarrow 6\left(x^{2}-5 x+6\right)=0 \\ &\Rightarrow x^{2}-5 x+6=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}-3 x-2 x+6=0 \\ &\Rightarrow(x-3)(x-2)=0 \\ &\Rightarrow x-3=0 \text { and } x-2=0 \\ &\Rightarrow x=3 \text { and } x=2 \end{aligned}
\begin{aligned} &\text { Clearly, } 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 { and } f^{\prime}(x)<0 \text { if }\\ &2
$\text { So, } f(x) \text { is increasing on the interval } (-\infty, 2)\: \cup\: (3, \infty) \text { and } f(x) \text { is decreasing on interval }(2,3).$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion x

$\text {f(x) is decreasing on the interval (-2,-1) and} \\ \text { increasing on the interval} (-\infty,-2) \cup (-1, \infty)$
Given:
Here given that
$f(x)=2x^{3}+9x^{2}+12x+20$
To find:
We have to find out the intervals in which function is increasing and decreasing.
Hint:
First, we will find critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=2x^{3}+9x^{2}+12x+20$
Differentiating w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}+9 x^{2}+12 x+20\right) \\ &\Rightarrow f^{\prime}(x)=6 x^{2}+18 x+12 \end{aligned}
For f(x), we have to find critical points.
For this we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}+18 x+12=0 \\ &\Rightarrow 6\left(x^{2}+3 x+2\right)=0 \\ &\Rightarrow x^{2}+3 x+2=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}+2 x+x+2=0 \\ &\Rightarrow(x+2)(x+1)=0 \\ &\Rightarrow x+2=0 \text { and } x+1=0 \\ &\Rightarrow x=-2 \text { and } x=-1 \\ &\text { Clearly, } f'(x)<0, \text { if }-2
$\text { and } f^{\prime}(x)>0 \text { if } x<-1 \text { and } x>-2 \text { or } x \in(-\infty,-2) \text { and } x \in(-1, \infty)$
$\text {Thus, } f(x) \text { is decreasing on the interval (-2,-1) and} \\ \text {increasing on the interval } (-\infty,-2) \cup(-1, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xi

$\text { Increasing interval }(-\infty, 1) \cup(2, \infty)\\ \text { Decreasing interval } (1,2)$
Given:
Here given that
$f(x)=2x^{3}-9x^{2}+12x-5$
To find:
We have to find the intervals in which function is increasing and decreasing interval of f(x).
Hint:
Put f ‘(x) = 0 and solve this equation to find critical points of f(x).
Solution:
We have,
$f(x)=2x^{3}-9x^{2}+12x-5$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-9 x^{2}+12 x-5\right) \\ &f \Rightarrow^{\prime}(x)=6 x^{2}-18 x+12 \end{aligned}
For f(x) we have to find critical points, for this we must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 6 x^{2}-18 x+12=0\\ &\Rightarrow 6\left(x^{2}-3 x+2\right)=0\\ &\Rightarrow x^{2}-3 x+2=0\{\therefore 6>0\}\\ &\Rightarrow x^{2}-2 x-x+2=0\\ &\Rightarrow(x-1)(x-2)=0\\ &\Rightarrow x-1=0 \text { and } x-2=0\\ &\Rightarrow x=1 \text { and } x=2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<1 \text { and } x>2 \text { or } x \in(-\infty, 1) \text { and } x \in(2, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}
\begin{aligned} &1

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xii

$\text { Increasing interval (-1,2)} \\ \text { Decreasing interval} (-\infty,-1) \cup(2, \infty)$
Given:
Here given that
$f(x)=6+12x+3x^{2}-2x^{3}$
To find:
We have to find the intervals in which function is increasing and decreasing.
Hint:
Put f ‘(x) = 0 to find the critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=6+12x+3x^{2}-2x^{3}$
Differentiating w.r.t. x we get,
$f'(x)=12+6x-6x^{2}$
For f(x), we have to find critical points.
We must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 12+6 x-6 x^{2}=0\\ &\Rightarrow 6\left(2+x-x^{2}\right)=0\\ &\Rightarrow 2+x-x^{2}=0\\ &\Rightarrow x^{2}-x-2=0\\ &\Rightarrow x^{2}-2 x+x-2=0\\ &\Rightarrow(x-2)(x+1)=0\\ &\Rightarrow x-2=0 \text { and } x+1=0\\ &\Rightarrow x=2 \text { and } x=-1\\ &\text { Clearly, } f^{\prime}(x)>0, \text { if }-12 \text { or } x \in(-\infty,-1) \text { and } x \in(2, \infty) \end{aligned}
$\text { Thus, } f(x) \text { is increasing on the interval (-1,2) and } \\ \text { decreasing on the interval } (-\infty,-1) \cup(2, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xiii

$\text { Increasing interval }(-\infty,-2) \cup(2, \infty) \\ \text { Decreasing interval (-2,2) }$
Given:
Here given that
$f(x)=2x^{3}-24x+107$
To find:
We have to find the intervals in which function is increasing and decreasing.
Hint:
Firstly, we will find critical points and then use increasing and decreasing property.
Soution:
We have,
$f(x)=2x^{3}-24x+107$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-24 x+107\right) \\ &\Rightarrow f^{\prime}(x)=6 x^{2}-24 \end{aligned}
For critical points we must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 6 x^{2}-24=0\\ &\Rightarrow 6\left(x^{2}-4\right)=0\\ &\Rightarrow x^{2}-4=0\{\therefore 6>0\}\\ &\Rightarrow x^{2}=4\\ &\Rightarrow x=\pm 2\\ &\Rightarrow x=+2,-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>2 \text { or } x \in(-\infty,-2) \text { and } x \in(2, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &-2
$\text { So, } f(x) \text { is increasing on the interval }(-\infty,-2) \cup(2, \infty) \text { and } \\ f(x) \text { is decreasing on interval (-2,2). }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xiv

$\text { Increasing interval (-2,-1) } \\ \text { Decreasing interval }(-\infty,-2) \cup(-1, \infty)$
Given:
Here given that
$f(x)=-2x^{3}-9x^{2}-12x+1$
To find:
We have to find the increasing and decreasing intervals of f(x).
Hint:
Put f ‘(x) = 0 to find the critical points.
Solution:
We have,
$f(x)=-2x^{3}-9x^{2}-12x+1$
On differentiating we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(-2 x^{3}-9 x^{2}-12 x+1\right) \\ &\Rightarrow f^{\prime}(x)=-6 x^{2}-18 x-12 \end{aligned}
For f(x), we have to find critical points.
We must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow-6 x^{2}-18 x-12=0 \\ &\Rightarrow 6 x^{2}+18 x+12=0 \\ &\Rightarrow 6\left(x^{2}+3 x+2\right)=0 \\ &\Rightarrow x^{2}+2 x+x+2=0 \\ &\Rightarrow x(x+2)+1(x+2)=0 \\ &\Rightarrow(x+2)(x+1)=0 \\ &\Rightarrow x+2=0 \text { and } x+1=0 \\ &\Rightarrow x=-2 \text { and } x=-1 \end{aligned}
\begin{aligned} &\text { Clearly, } f^{\prime(x)}<0, \text { if } x<-1 \text { and } x<-2 \text { or } x \in(-\infty,-2) \text { and } x \in(-1, \infty) \\ &\text { and } f^{\prime}(x)>0 \text { if }-2
$\text { Thus, } f(x) \text { is increasing on the interval }(-2,-1) \text { and decreasing on the interval }(-\infty,-2)\:\: \cup\:\: (-1,\infty )$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xv

$\text { Increasing interval } \left(-\infty, \frac{4}{3}\right) \cup(2, \infty) \\ \text { Decreasing interval } \left(\frac{4}{3}, 2 \right)$
Given:
Here given that
$f(x)=(x-1)(x-2)^{2}$
To find:
We have to find the intervals in which f(x) is increasing or decreasing.
Hint:
Use product rule of differentiation
$\text { i.e } \frac{d}{d x}(u \cdot v)=u \cdot \frac{d v}{d x}+v \frac{d u}{d x}$
Solution:
We have,
$f(x)=(x-1)(x-2)^{2}$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left[(x-1)(x-2)^{2}\right] \\ &\Rightarrow f^{\prime}(x)=(x-1) \frac{d}{d x}\left[(x-2)^{2}\right]+(x-2)^{2} \frac{d}{d x}(x-1) \\ &=2(x-1)(x-2)+(x-2)^{2} \\ &=(x-2)[2(x-1)+(x-2)] \\ &=(x-2)[2 x-2+x-2] \\ &=(x-2)(3 x-4) \end{aligned}
For f(x) we have to find critical points, for this we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow(x-2)(3 x-4)=0 \\ &\Rightarrow x-2=0 \text { and } 3 x-4=0 \end{aligned}
\begin{aligned} &\Rightarrow x=2 \text { and } 3 x=4\\ &\text { or } x=\frac{4}{3}\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<\frac{4}{3} \text { and } x>2 \text { and } f^{\prime}(x)<0 \text { if } \frac{4}{3}
$\text { So, } f(x) \text { is increasing on the interval }\left(-\infty, \frac{4}{3}\right) \cup(2, \infty) \text { and decreasing on the interval }\left(\frac{4}{3}, 2\right) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xvi

$\text { Increasing interval }(-\infty, 2) \cup(6, \infty) \\ \text { Decreasing interval(2.6) }$
Given:
Here given that
$f(x)=x^{3}-12x^{2}+36x+17$
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,
$f(x)=x^{3}-12x^{2}+36x+17$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-12 x^{2}+36 x+17\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-24 x+36 \end{aligned}
For f(x) we have to find critical points, for this we must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-24 x+36=0\\ &\Rightarrow 3\left(x^{2}-8 x+12\right)=0\\ &\Rightarrow x^{2}-8 x+12=0\{\therefore 3>0\}\\ &\Rightarrow x^{2}-6 x-2 x+12=0\\ &\Rightarrow(x-6)(x-2)=0\\ &\Rightarrow x-6=0 \text { and } x-2=0\\ &\Rightarrow x=6 \text { and } x=2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<2 \text { and } x>6 \text { or } x \in(-\infty, 2) \text { and } x \in(6, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &2
$\text { So, } f(x) \text { is increasing on the interval }(-\infty, 2) \cup \: (6, \infty) \text { and } f(x) \text { is decreasing on interval }(2,6) \text {.}$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xvii

$\text { Increasing interval }(-\infty,-2) \cup(2, \infty) \\ \text { Decreasing interval(-2,2) }$
Given:
Here given that
$f(x)=2x^{3}-24x+7$
To find:
We have to find the intervals in which f(x) is increasing or decreasing.
Hint:
First we will find critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=2x^{3}-24x+7$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-24 x+7\right) \\ &\Rightarrow f^{\prime}(x)=6 x^{2}-24 \end{aligned}
For f(x) we have to find critical points,
We must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 6 x^{2}-24=0\\ &\Rightarrow 6 x^{2}=24\\ &\Rightarrow x^{2}=\frac{24}{6}\\ &\Rightarrow x^{2}=4\\ &\Rightarrow x=\pm 2\\ &\Rightarrow x=+2,-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x>2 \text { and } x<-2 \text { or } x \in(-\infty, 2) \text { and } x \in(-2, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &-2
$\text { So, } f(x) \text { is increasing on the interval }(-\infty,-2) \cup(2, \infty) \text { and }\\ f(x) \text { is decreasing on interval } (-2,2).$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xviii

$\text { Increasing interval }(-2,1) \cup(3, \infty) \\ \text { Decreasing interval } (-\infty ,-2) \cup (1.3)$
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
\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}
$\text { After that we take }(1,3) \text { i.e, } 10, x+2>0 \text { and } x-3<0 . \text { So, } f^{\prime}(x)<0 \text { if } x \in(1,3) \text { . }$
$\text { Finally we take }(3, \infty) \text { i.e, } 30, x+2>0 \text { and } x-3>0 \text { . }$
$\text { Clearlyf }^{\prime}(x)>0 \text { if } x>3 \text { or } x \in(3, \infty)$
$\text { Thus, the function is increasing on }(-2,1) \cup(3, \infty) \text { and } \\ f(x) \text { is decreasing on interval }(-\infty,-2) \cup(1,3) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xix

$\text { Increasing interval }(1, \infty) \\ \text { Decreasing interval }(-\infty,1)$
Given:
Here given that
$f(x)=x^{4}-4x$
To find:
We have to find the intervals in which f(x) is increasing and decreasing.
Hint:
We will find critical points.
Solution:
We have,
$f(x)=x^{4}-4x$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{4}-4 x\right) \\ &\Rightarrow f^{\prime}(x)=4 x^{3}-4 \end{aligned}
For critical points we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 4\left(x^{3}-1\right)=0 \\ &\Rightarrow x^{3}-1=0\{\therefore 4>0\} \\ &\Rightarrow x^{3}=1 \end{aligned}
Taking cube root on both sides.
\begin{aligned} &\Rightarrow \sqrt[3]{x}=\sqrt[3]{1} \\ &\Rightarrow x=1 \\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x>1 \text { or } x \in(1, \infty) \text { and } f^{\prime}(x)<0 \text { if } x<1 \text { or } x \in(-\infty, 1) . \end{aligned}
$\text { Thus, } f(x) \text { is increasing on }(1, \infty) \text { and } f(x) \text { is decreasing on interval }(-\infty, 1) .$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xx

$\text { Increasing interval }(-3,-1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(-1, 2)$
Given:
Here given that
$f(x)=\frac{x^{4}}{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7$
To find:
We have to find the increasing or decreasing interval for f(x).
Hint:
First we will find critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=\frac{x^{4}}{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{x^{4}}{4}+\frac{2}{3} x^{3}-\frac{5}{2} x^{2}-6 x+7\right) \\ &\Rightarrow f^{\prime}(x)=\frac{4 x^{3}}{4}+\frac{2}{3}\left(3 x^{2}\right)-\frac{5}{2}(2 x)-6 \\ &\Rightarrow f^{\prime}(x)=x^{3}+2 x^{2}-5 x-6 \\ &\Rightarrow f^{\prime}(x)=(x+1)\left(x^{2}+x-6\right) \\ &\Rightarrow f^{\prime}(x)=(x+1)(x-2)(x+3) \end{aligned}
For critical points. we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow(x+1)(x-2)(x+3)=0 \\ &\Rightarrow x=-1, \quad x=2, \quad x=-3 \end{aligned}
$\text { The possible intervals are }(-3,-1),(-\infty,-3),(-1,2) \text { and }(2, \infty)$
$\text { Now we take }(-3,-1) . \text { i.e, }-3
$\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 }-3
\begin{aligned} &\text { Now we take the intervals }(-\infty,-3) \text { i.e, }-\infty
$\text { After that we take }(-1,2) \text { i.e, }-1
$\text { Clearly, we have } x+1>0, x-2<0 \text { and } x+3>0 . \text { So, } f^{\prime}(x)<0 \text { if }-1
$\text { Finally we take }(2, \infty) \text { i.e, } 20, x-2>0 \text { and } x+3>0 \text { . }$
$\text { Clearly }f^{\prime}(x)>0 \text { if } 2
$\text { Thus, the function is increasing on }(-3,-1) \cup(2, \infty) \text { and }$
$f(x) \text { is decreasing on interval }(-\infty,-3) \cup(-1,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxi

$\text { Increasing interval }(0,1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,0) \cup(1,2)$
Given:
Here given that
$f(x)=x^{4}-4x^{3}+4x^{2}+15$
To find:
We have to find the increasing or decreasing interval for f(x).
Hint:
Put f ‘(x) = 0 to find critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=x^{4}-4x^{3}+4x^{2}+15$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{4}-4 x^{3}+4 x^{2}+15\right) \\ &\Rightarrow f^{\prime}(x)=4 x^{3}-12 x^{2}+8 x \end{aligned}
For critical points. we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 4 x^{3}-12 x^{2}+8 x=0 \\ &\Rightarrow 4 x\left(x^{2}-3 x+2\right)=0 \\ &\Rightarrow x\left(x^{2}-3 x+2\right)=0\{\therefore 4>0\} \\ &\Rightarrow x\left(x^{2}-2 x-x+2\right)=0 \\ &\Rightarrow x(x-2)(x-1)=0 \\ &\Rightarrow x=0, \quad x=1, \quad x=2 \end{aligned}
\begin{aligned} &\text { The possible intervals are }(-\infty, 0),(0,1),(1,2) \text { and }(2, \infty)\\ &\text { in intervals }(0,1) \text { and }(2, \infty), f^{\prime}(x) \end{aligned}
$\text { Clearly, } f^{\prime}(x)>0 \text { if } 0
\begin{aligned} &\text { However in the intervals }(-\infty, 0) \text { and }(1,2) f^{\prime}(x) .\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty
$\text { Thus, } f(x) \text { is increasing on }(0,1) \cup(2, \infty) \text { and decreasing on }(-\infty, 0) \cup(1,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxii

$\text { Increasing interval }(0,1) \\ \text { Decreasing interval }(1,\infty )$
Given:
Here given that
$f(x)=5 x^{\frac{3}{2}}-3 x^{\frac{5}{2}}, x>0$
To find:
We have to find the increasing and decreasing intervals.
Hint:
We will find the critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=5 x^{\frac{3}{2}}-3 x^{\frac{5}{2}}$
Differentiating w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(5 x^{\frac{3}{2}}-3 x^{\frac{5}{2}}\right) \\ &\Rightarrow f^{\prime}(x)=5\left(\frac{3}{2} x^{\frac{1}{2}}\right)-3\left(\frac{5}{2} x^{\frac{3}{2}}\right) \\ &\Rightarrow f^{\prime}(x)=\frac{15}{2} x^{\frac{1}{2}}-\frac{15}{2} x^{\frac{3}{2}} \\ &\Rightarrow f^{\prime}(x)=\frac{15}{2}\left(x^{\frac{1}{2}}-x^{\frac{3}{2}}\right) \end{aligned}
For f(x), we have to find critical points.
We must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{15}{2}\left(x^{\frac{1}{2}}-x^{\frac{3}{2}}\right)=0,\left\{\therefore \frac{15}{2}=0\right\} \\ &\Rightarrow x^{\frac{1}{2}}-x^{\frac{3}{2}}=0 \\ &\Rightarrow x^{\frac{1}{2}}\left(1-x^{\frac{3}{2}}\right)=0 \end{aligned}
\begin{aligned} &\Rightarrow x=0 \text { and } x=1 \\ &\text { Clearly, } f^{\prime}(x)>0, \text { if } 01 \text { or } x \in(1, \infty) \end{aligned}$\text { Thus, } f(x) \text { is increasing on the interval }(0,1) \text { and decreasing on the interval }(1, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxiii

$\text { Increasing interval }(0, \infty)\\ \text { Decreasing interval }(-\infty,0)$
Given:
Here given that
$f(x)=x^{8}+6x^{z}$
To find:
We have to find the increasing and decreasing intervals.
Hint:
First find the critical points by using f ‘(x) and then apply increasing and decreasing property.
Solution:
We have,
$f(x)=x^{8}+6x^{z}$
Differentiating w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{8}+6 x^{2}\right) \\ &=8 x^{7}+12 x \end{aligned}
For critical points. We must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 8 x^{7}+12 x=0 \\ &\Rightarrow 4 x\left(2 x^{6}+3\right)=0,\{\therefore 4=0\} \\ &\Rightarrow x\left(2 x^{6}+3\right)=0 \\ &\Rightarrow x=0,\left\{\therefore 2 x^{6}+3>0\right\} \\ &\text { Clearly, } f^{\prime}(x)>0, \text { if } x>0 \text { or } x \in(0, \infty) \\ &\text { and } f^{\prime}(x)<0 \text { if } x<0 \text { or } x \in(-\infty, 0) \end{aligned}
$\text { Thus, } f(x) \text { is increasing on the interval }(0, \infty) \text { and decreasing on the interval }(-\infty, 0)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxiv

$\text { Increasing interval }(-\infty, 1) \cup(3, \infty) \\ \text { Decreasing interval }(1,3)$
Given:
Here given that
$f(x)=x^{3}-6x^{2}+9x+15$
To find:
We have to find the intervals in which function is increasing or decreasing.
Hint:
Put f ‘(x) = 0 to find critical points of f(x) and then use increasing and decreasing property.
Solution:
We have,
$f(x)=x^{3}-6x^{2}+9x+15$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-6 x^{2}+9 x+15\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+9 \end{aligned}
For f(x) we have to find critical points,
We must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-12 x+9=0\\ &\Rightarrow 3\left(x^{2}-4 x+3\right)=0\\ &\Rightarrow x^{2}-4 x+3=0\\ &\Rightarrow x^{2}-3 x-x+3=0\\ &\Rightarrow(x-3)(x-1)=0\\ &\Rightarrow x=3,1\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<1 \text { and } x>3 \text { or } x \in(-\infty, 1) \text { and } x \in(3, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &1
$\text { So, } f(x) \text { is increasing on the interval }(-\infty, 1) \cup (3, \infty) \text { and } f(x) \text { is decreasing on interval }(1,3) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxv

$\text { Increasing interval }(0,1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,0) \cup(1,2)$
Given:
Here given that
$f(x)=[x(x-2)]^{2}$
To find:
We have to find the increasing or decreasing interval for f(x).
Hint:
Put f ‘(x) = 0 to find critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=[x(x-2)]^{2}$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left[x^{2}-2 x\right]^{2} \\ &=2\left(x^{2}-2 x\right)(2 x-2) \\ &=4 x(x-2)(x-1) \end{aligned}
For f(x), we have to find critical points.
we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 4 x(x-2)(x-1)=0 \\ &\Rightarrow x(x-2)(x-1)=0 \\ &\Rightarrow x=0,1,2 \end{aligned}
$\text { The possible intervals are }(-\infty, 0),(0,1),(1,2) \text { and }(2, \infty)$
\begin{aligned} &\text { In the intervals }(-\infty, 0) \text { and }(1,2) f^{\prime}(x)<0 \text { . }\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty
$\text { However, } \\ \text { In intervals }(0,1) \text { and }(2, \infty), f^{\prime}(x)>0 \\ \text { Clearly, } f^{\prime}(x)>0 \text { if } 0
$\text { Thus, the function } f(x) \text { is increasing on }(0,1) \cup(2, \infty) \text { and decreasing on }(-\infty, 0) \cup(1,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxvi

$\text { Increasing interval }(-1,0) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty,-1) \cup(0,2)$
Given:
Here given that
$f(x)=3x^{4}-4x^{3}-12x^{2}+5$
To find:
We have to find the increasing and decreasing interval.
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,
$f(x)=3x^{4}-4x^{3}-12x^{2}+5$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=12 x^{3}-12 x^{2}-24 x \\ &=12 x\left(x^{2}-x-2\right) \end{aligned}
Now,
For critical points, we must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 12 x\left(x^{2}-x-2\right)=0\\ &\Rightarrow x\left(x^{2}-x-2\right)=0,\{\therefore 12=0\}\\ &\Rightarrow x\left(x^{2}-2 x+x-2\right)=0\\ &\Rightarrow x(x-2)(x+1)=0\\ &\Rightarrow x=0 ; x=-1 ; x=2\\ &\text { The points } x=0,-1,2 \text { divide the real line into four disjoint intervals, } \end{aligned}
\begin{aligned} &(-\infty,-1),(-1,0),(0,2) \text { and }(2, \infty)\\ &\text { In the intervals }(-\infty,-1) \text { and }(0,2) f^{\prime}(x)<0 \text { . } \end{aligned}
$\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty
\begin{aligned} &\text { However, } \\ &\text { In intervals }(-1,0) \text { and }(2, \infty), f^{\prime}(x)>0\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if }-1
$\text { Thus, } f(x) \text { is increasing in the intervals }(-1,0) \cup(2, \infty) \text { and } \\ \text { decreasing in the intervals }(-\infty,-1) \cup(0,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxvii

$\text { Increasing interval }(-3,0) \cup(5, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(0,5)$
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}
$\text { Clearly, } f^{\prime}(x)>0 \text { if }-3
\begin{aligned} &\text { In the intervals }(-\infty,-3) \text { and }(0,5) f^{\prime}(x)<0 \text { . }\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty$\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 { . }$

$\text { Increasing interval }(2, \infty) \\ \text { Decreasing interval }(-\infty , 2)$
Given:
Here given that
$f(x)=log(2+x)-\frac{2x}{2+x}$
To find:
We have to find the increasing and decreasing intervals of f(x).
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,
$f(x)=log(2+x)-\frac{2x}{2+x}$
Differentiating w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left\{\log (2+x)-\frac{2 x}{2+x}\right\} \\ &=\frac{1}{2+x}-\frac{(2+x) 2-2 x \times 1}{(2+x)^{2}} \\ &=\frac{1}{2+x}-\frac{4+2 x-2 x}{(2+x)^{2}} \\ &=\frac{1}{2+x}-\frac{4}{(2+x)^{2}} \\ &=\frac{2+x-4}{(2+x)^{2}} \\ &=\frac{x-2}{(2+x)^{2}} \end{aligned}
For critical points. We must have,
\begin{aligned} &f'(x)> 0\\ &\Rightarrow \frac{x-2}{(2+x)^{2}}>0 \\ &\Rightarrow x-2>0 \\ &\Rightarrow x>2 \\ &x \in(2, \infty) \end{aligned}
Thus, f(x) is increasing on the interval (2,$\infty$)
\begin{aligned} &f^{\prime}(x)<0 \\ &\Rightarrow \frac{x-2}{(2+x)^{2}}<0 \\ &\Rightarrow x-2<0 \\ &\Rightarrow x<2 \\ &x \in(-\infty, 2) \end{aligned}
So,f(x) is decreasing on the interval ($-\infty$,2).

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxvix

$\text { Increasing interval }(-3,2) \cup(4, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(2, 4)$
Given:
Here given that
$f(x)=\frac{x^{4}}{4}-x^{3}-5x^{2}+24x+12$
To find:
We have to find the intervals in which function is increasing and decreasing.
Hint:
Put f ‘(x) = 0 to find critical points and then use increasing and decreasing property.
Solution:
We have,
$f(x)=\frac{x^{4}}{4}-x^{3}-5x^{2}+24x+12$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{x^{4}}{4}-x^{3}-5 x^{2}+24 x+12\right) \\ &=\frac{4 x^{3}}{4}-3 x^{2}-10 x+24 \\ &=x^{3}-3 x^{2}-10 x+24 \end{aligned}
At critical points,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow x^{3}-3 x^{2}-10 x+24=0 \\ &\Rightarrow x^{3}-2 x^{2}-x^{2}+2 x-12 x+24=0 \\ &\Rightarrow x^{2}(x-2)-x(x-2)-12(x-2)=0 \\ &\Rightarrow(x-2)\left(x^{2}-x-12\right)=0 \\ &\Rightarrow(x-2)\left(x^{2}-4 x+3 x-12\right)=0 \\ &\Rightarrow(x-2)(x-4)(x+3)=0\\ &\Rightarrow x=2;\: x=4;\: x=-3 \\ &\Rightarrow x=-3,2,4 \end{aligned}
The points x=-3, 2, 4 divide the real line into four disjoint intervals,
\begin{aligned} &(-\infty,-3),(-3,2),(2,4) \text { and }(4, \infty) \\ &\text { In intervals }(-3,2) \text { and }(4, \infty), f^{\prime}(x)>0 \end{aligned}
$\text { Clearly, } f^{\prime}(x)>0 \text { if }-3
\begin{aligned} &\text { In the intervals }(-\infty,-3) \text { and }(2,4) f^{\prime}(x)<0\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty$\text { Thus, } f(x) \text { is increasing in the intervals }(-3,2) \cup(4, \infty) \text { and } \text { decreasing in the intervals }(-\infty,-3) \cup(2,4) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 2

$i.\: x=3 \\ ii.\: co-ordinate (\frac{5}{2},\frac{1}{4})$
Given:
$f(x)=(x^{2}-6x+9)$
To find:
We have to find the value of x, also we have to find co-ordinate of the point on the given curve where the normal is parallel to the line y = x+5.
Hint:
1. First we will find critical point then find f’(x) for all the value of x.
2. Find m1 and m2 from the curve then find points
Solution:
We have
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{2}-6 x+9\right) \\ &\Rightarrow f^{\prime}(x)=2 x-6 \\ &\Rightarrow f^{\prime}(x)=2(x-3) \end{aligned}
For f(x) let us find critical point, we must have
\begin{aligned} &\Rightarrow f^{\prime}(x)=0 \\ &\Rightarrow 2(x-3)=0 \\ &\Rightarrow x-3=0 \\ &\Rightarrow x=3 \\ &\text { Clearly, } f^{\prime}(x)>0 i f x>3 \text { and } f^{\prime}(x)<0 \text { if } x<3 \end{aligned}
Thus f(x) increases on 3, $\infty$ and f(x) is decreasing on interval x ∈ (-$\infty$, 3)
Now,let us find co-ordinates of point on the equation of curve is
$f(x)=(x^{2}-6x+9)$
Slope of this curve is given by
\begin{aligned} &\Rightarrow m_{1}=\frac{d y}{d x} \\ &\Rightarrow m_{1}=\frac{d}{d x}\left(x^{2}-6 x+9\right) \\ &\Rightarrow m_{1}=2x-6 \end{aligned}
Equation of line is y=x+5
Slope of this curve is
\begin{aligned} &\Rightarrow m_{2}=\frac{d y}{d x} \\ &\Rightarrow m_{2}=\frac{d}{d x}(x+5) \\ &\Rightarrow m_{2}=1 \end{aligned}
Since slope of curve is parallel to line.
\begin{aligned} &\Rightarrow-\frac{1}{m_{1}}=m_{2} \\ &\Rightarrow-\frac{1}{2 x-6}=1 \\ &\Rightarrow 2 x-6=-1 \\ &\Rightarrow x=\frac{5}{2} \end{aligned}
Thus putting the value of x in equation of curve
We get
\begin{aligned} &\Rightarrow y=x^{2}-6 x+9 \\ &\Rightarrow y=\left(\frac{5}{2}\right)^{2}-6\left(\frac{5}{2}\right)+9 \\ &\Rightarrow y=\frac{25}{4}-15+9 \\ &\Rightarrow y=\frac{25}{4}-6 \\ &\Rightarrow y=\frac{1}{4} \end{aligned}
Hence the required co-ordinate is$(\frac{5}{2},\frac{1}{4})$

Increasing and Decreasing Functions exercise 16.2 question 3

$f(x) \text { is increasing on } \left(0, \frac{3 \pi}{4}\right) \\ \text { and } f(x) \text { is decreasing on } \left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$
Given:
$\text { given } f(x)=sin\: x-cos\: x$
To find:
We have to interval in which f(x) is increasing or decreasing.
Hint:
f'(x)=0 to find critical point then use increasing or decreasing property.
Solution:
We have
$f(x)=sin\: x-cos\: x$
On differentiating both sides we get
$f'(x)=cos\: x+sin\: x$
For f(x) let us find critical point, we must have
\begin{aligned} &\Rightarrow f^{\prime}(x)=0 \\ &\Rightarrow \cos x+\sin x=0 \end{aligned}
On dividing by cos x we get
\begin{aligned} &\Rightarrow \tan x=-1 \\ &\Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4} \end{aligned}
Here the points divide the angle range from 0 to 2π.
Since we have x as angle.
$\text { Clearly, } f^{\prime}(x)>0 \text { if } 0
$\text {Thus } f(x) \text { increases on }\left(0, \frac{3 \pi}{4}\right) \cup\left(\frac{7 \pi}{4}, 2 \pi\right) \text { and } f(x) \text { is decreasing on interval } x \in\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$

Increasing and Decreasing Functions exercise 16.2 question 7

$f(x) \text { is increasing on } (0, \frac{\pi }{2}) \text { and decreasing in } (\frac{\pi }{2},\pi ) \\ \text { Hence, } f(x) \text { is neither increasing nor decreasing in } ( 0,\pi )$
Given:
$f(x)=sin\: x$
To show:
$\text { We have to show that } f(x) \text { is increasing on } (0,\frac{\pi }{2}) \text { and decreasing on } (\frac{\pi }{2},\pi ) \text { and neither increasing nor decreasing on } (0,\pi ).$
Hint:
$1) \text { We know that if } f^{\prime}(x)>0 \text { for all } x \in(a, b) \text { then } f(x) \text { is increasing. } \\ 2) \text { We know that if } f^{\prime(x)}<0 \text { for all } x \in(a, b) \text { then } f (x) \text { is decreasing on } (a, b).$
Solution:
Given
$f(x)=sin\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\sin x) \\ &\Rightarrow f^{\prime}(x)=\cos x \end{aligned}
Taking different region from 0 to 2π.
Let
\begin{aligned} x\in (0,\frac{\pi }{2}) \end{aligned}
\begin{aligned} &\Longrightarrow \cos x<0\\ &\Longrightarrow f^{\prime}(x)<0\\ &\text { Thus } \mathrm{f}(\mathrm{x}) \text { is decreasing }\left(\frac{\pi}{2}, \pi\right) \text { . } \end{aligned}
Therefor the above condition we find that
$\Rightarrow f(x) \text { is increasing on }\left(0, \frac{\pi}{2}\right) \text { and decreasing in }\left(\frac{\pi}{2}, \pi\right)$
Hence,f(x) is neither increasing nor decreasing in (0, π)

Increasing and Decreasing Functions exercise 16.2 question 8

$f(x) \text { is increasing on } (0,\frac{\pi }{2}) \text { and decreasing on }(\frac{\pi }{2},\pi ).$
Given:
$f(x)=log\: sin\: x$
To show:
$\text { We have to show that }f(x) \text { is increasing on } (0,\frac{\pi }{2}) \text { and decreasing on }(\frac{\pi }{2},\pi ).$
Hint:
Use increasing and decreasing property to find increasing and decreasing.
Solution:
Given
$f(x)=log\: sin\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\log \sin x) \\ &\Rightarrow f^{\prime}(x)=\frac{1}{\sin x} \times \cos x \\ &f^{\prime}(x)=\cot x,\left[\therefore \frac{\cos x}{\sin x}=\cot x\right] \end{aligned}
Taking different region from 0 to $\pi$
\begin{aligned} &\text { Let } x \in\left(0, \frac{\pi}{2}\right) \\ &\Rightarrow \cot x>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
$\text { Thus }f(x) \text { increasing }(0,\frac{\pi }{2} ).$
\begin{aligned} &\text { Let } x \in\left( \frac{\pi}{2},\pi \right) \\ &\Rightarrow \cot x< 0 \\ &\Rightarrow f^{\prime}(x)< 0 \end{aligned}
$\text { Thus }f(x) \text { decreasing }(\frac{\pi }{2},\pi ).$
Hence proved.

Increasing and Decreasing Functions exercise 16.2 question 9

f(x) is increasing on interval x$\in$R
Given:
$f(x)=x-sin\: x$
To prove:
We have to show that f(x) is increasing for all x$\in$R
Hint:
Show f’(x)>0 for f(x) to be increasing.
Solution:
Given
$f(x)=x-sin\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(x-\sin x) \\ &\Rightarrow f^{\prime}(x)=1-\cos x \end{aligned}
Now, as given x$\in$R
\begin{aligned} &\Rightarrow-1<\cos x<1 \\ &\Rightarrow-1>1-\cos x>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Hence condition of f(x) to be increasing.
Thus f(x) is increasing on interval x$\in$R

Increasing and Decreasing Functions exercise 16.2 question 10

f(x) is increasing for all x $\in$ R
Given:
$f(x)=x^{3}-15x^{2}+75x-50$
To prove:
We have to show that f(x) is increasing for all x $\in$ R
Hint:
Show f’(x)>0 for f(x) to be increasing.
Solution:
Given
$f(x)=x^{3}-15x^{2}+75x-50$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-15 x^{2}+75 x-50\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-30 x+75 \\ &\Rightarrow f^{\prime}(x)=3\left(x^{2}-10 x+25\right) \\ &\Rightarrow f^{\prime}(x)=3(x-5)^{2} \end{aligned}
Now, as given x $\in$ R
\begin{aligned} &\Rightarrow(x-5)^{2}>0 \\ &\Rightarrow 3(x-5)^{2}>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Hence condition for f(x) to be increasing.
Thus f(x) is increasing on interval x $\in$ R.

Increasing and Decreasing Functions exercise 16.2 question 11

$f(x) \text { is decreasing function on }(0,\frac{\pi }{2}).$
Given:
$f(x) =cos^{2}\: x$
To prove:
$\text { We have to show that }f(x) \text { is decreasing function on }(0,\frac{\pi }{2}).$
Hint:
Condition for f(x) to be decreasing that f’(x)<0.
Solution:
Given
$f(x) =cos^{2}\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(\cos ^{2} x\right) \\ &=2 \cos x(-\sin x) \\ &\Rightarrow f^{\prime}(x)=-2 \sin x \cos x \\ &\Rightarrow f^{\prime}(x)=-\sin 2 x \end{aligned}
Now, as given
$x\: \in \: (0,\frac{\pi }{2})$
\begin{aligned} &\Rightarrow 2 x \in(0, \pi) \\ &\Rightarrow \sin (2 x)>0 \\ &\Rightarrow-\sin (2 x)<0 \end{aligned}
By applying negative sign, change in comparison sign.
\begin{aligned} &\Rightarrow f'(x)< 0 \end{aligned}
Hence condition of f(x) to be decreasing.
Thus f(x) is decreasing on interval
$x\: \in \: (0,\frac{\pi }{2})$.

Increasing and Decreasing Functions exercise 16.2 question 12

$f(x) = sin\: x \text { is increasing function on }(-\frac{\pi }{2},\frac{\pi }{2})$
Given:
$f(x) = sin\: x$
To prove:
$\text { We have to show that } f(x) = sin\: x \text { is increasing function on }(-\frac{\pi }{2},\frac{\pi }{2})$
Hint:
Condition for f(x) to be increasing that f’(x)>0.
Solution:
Given
$f(x) = sin\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\sin x)\\ &\Longrightarrow f^{\prime}(x)=\cos x\\ &\text { Now, as given } x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text { . }\\ &\Longrightarrow \cos x>0\\ &\Longrightarrow f^{\prime}(x)>0 \end{aligned}
Hence condition of f(x) to be increasing.$\text { Hence }f(x) \text { is increasing on interval }x\: \in (-\frac{\pi }{2},\frac{\pi }{2})$

Increasing and Decreasing Functions exercise 16.2 question 13

f(x) is decreasing on (0,$\pi$) and increasing on (-$\pi$,0) and neither increasing nor decreasing on (-$\pi$,$\pi$).
Given:
$f(x)=cos\: x$
To show:
We have to show that f(x) is decreasing on (0,$\pi$) and increasing on (-$\pi$,0) and neither increasing nor decreasing on (-$\pi$,$\pi$).
Hint:
1. For increasing function f'(x)>0 then f(x) is increasing.
2. For f(x) to be decreasing f'(x)<0
Solution:
Given
$f(x)=cos\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\cos x) \\ &\Rightarrow f^{\prime}(x)=-\sin x \end{aligned}
Taking different region from -$\pi$ to $\pi$.
\begin{aligned} &\text { Let } x \in(0, \pi) \\ &\Rightarrow \sin x>0 \\ &\Rightarrow-\sin x<0 \end{aligned}
By applying negative sign change comparison sign.
\begin{aligned} &\Rightarrow f^{\prime}(x)<0\\ \end{aligned}
Thus f(x) is decreasing (0,$\pi$)
\begin{aligned} &\text { Let } x \in(-\pi, 0)\\ &\Longrightarrow \sin x<0\\ &\Rightarrow-\sin x>0 \end{aligned}
By applying negative sign change comparison sign.
$\Rightarrow f'(x)> 0$
Thus f(x) is increasing (-$\pi$,0).
Therefore from above condition we find that
?f(x) is decreasing in (0, $\pi$) and increasing in (-$\pi$,0)
Hence, f(x) is neither increasing nor decreasing in (-$\pi$, $\pi$)

Increasing and Decreasing Functions exercise 16.2 question 14

$f(x) \text { is increasing function on } (-\frac{\pi }{2},\frac{\pi }{2})$
Given:
$f(x) =tan \: x$
To prove:
$\text { We have to show that } f(x) \text { is increasing function on } (-\frac{\pi }{2},\frac{\pi }{2})$
Hint:
Condition for increasing function f’(x)>0.
Solution:
Given
$f(x) =tan \: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\tan x)\\ &\Rightarrow f^{\prime}(x)=\sec ^{2} x\\ \end{aligned}
Now, as given
\begin{aligned} &x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned}
\begin{aligned} \Rightarrow sec^{2}\: x> 0 \\ \Rightarrow f'(x)> 0 \end{aligned}
Hence condition of f(x) to be increasing.
$\text { Hence } f(x) \text { is increasing on interval } x \in (-\frac{\pi }{2},\frac{\pi }{2}).$
Hence proved.

Increasing and Decreasing Functions exercise 16.2 question 15

$f(x) \text { is decreasing function on } (\frac{\pi }{4},\frac{\pi }{2})$
Given:
$f(x) =tan^{-1}(sin\: x+cos\: x)$
To prove:
$\text { We have to show that } f(x) \text { is decreasing function on } (\frac{\pi }{4},\frac{\pi }{2})$
Hint:
Condition for decreasing function f’(x)<0.
Solution:
Given
$f(x) =tan^{-1}(sin\: x+cos\: x)$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(\tan ^{-1}(\sin x+\cos x)\right) \\ &\Rightarrow f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^{2}} \times(\cos x-\sin x),\left[\therefore \frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}\right] \\ &\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} \\ &\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{2(1+\sin x \cos x)},\left[\therefore \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}
Now, as given
\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}
\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ \end{aligned}
as here cosine values are smaller than sine values for same angle.
\begin{aligned} &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}
Hence condition for f(x) to be decreasing.
$\text { Thus }f(x) \text { is decreasing on interval } x\: \in \: (\frac{\pi }{4},\frac{\pi }{2})$
\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}
\begin{aligned} &x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right) \\ &\Rightarrow \cos x-\sin x<0 \\ \end{aligned}
as here cosine values are smaller than sine values for same angle.
\begin{aligned} &\Rightarrow \frac{\cos x-\sin x}{2(1+\sin x \cos x)}<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}
Hence condition for f(x) to be decreasing.
$\text { Thus }f(x) \text { is decreasing on interval } x\: \in \: (\frac{\pi }{4},\frac{\pi }{2})$

Increasing and Decreasing Functions exercise 16.2 question 16

$f(x) \text { is decreasing function on }(\frac{3\pi }{8},\frac{5\pi }{8})$
Given:
$f(x)=sin(2x+\frac{\pi }{4})$
To prove:
$\text { We have to show that } f(x) \text { is decreasing function on }(\frac{3\pi }{8},\frac{5\pi }{8})$
Hint:
Condition for f(x) to be decreasing is f’(x)<0.
Solution:
Given
$f(x)=sin(2x+\frac{\pi }{4})$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x} \sin \left(2 x+\frac{\pi}{4}\right) \\ &\Rightarrow f^{\prime}(x)=\cos \left(2 x+\frac{\pi}{4}\right) \cdot 2 \\ &\Rightarrow f^{\prime}(x)=2 \cos \left(2 x+\frac{\pi}{4}\right) \end{aligned}
Now, as given
\begin{aligned} &x \in\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right) \\ &\Rightarrow \frac{3 \pi}{8}
\begin{aligned} &\text { As here } 2 x+\frac{\pi}{4} \text { lies in } 3 \mathrm{rd} \text { quadrant. }\\ &\Rightarrow \cos \left(2 x+\frac{\pi}{4}\right)<0\\ &\Rightarrow 2 \cos \left(2 x+\frac{\pi}{4}\right)<0\\ &\Longrightarrow f^{\prime}(x)<0 \end{aligned}
Hence condition for f(x) to be decreasing.
$\text { Thus } f(x) \text { is decreasing on interval }x\: \in \: (\frac{3\pi }{8},\frac{5\pi }{8})$

Increasing and Decreasing Functions exercise 16.2 question 17

$f(x) \text { is decreasing on }(0,\frac{\pi }{4}) \text { and increasing on } (\frac{\pi }{4},\frac{\pi }{2})$
Given:
$f(x)=cot^{-1}(sin\: x+cos\: x)$
To prove:
$\text { We have to show that } f(x) \text { is decreasing on }(0,\frac{\pi }{4}) \text { and increasing on } (\frac{\pi }{4},\frac{\pi }{2})$
Hint:
1. condition for f(x) to be increasing is f'(x)>0
2. condition for f(x) to be decreasing f'(x)<0
Solution:
Given
$f(x)=cot^{-1}(sin\: x+cos\: x)$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left[\cot ^{-1}(\sin x+\cos x)\right] \\ &\Rightarrow f^{\prime}(x)=\frac{-1}{1+(\sin x+\cos x)^{2}} \times(\cos x-\sin x),\left[\therefore \frac{d}{d x} \cot ^{-1} x=\frac{-1}{1+x^{2}}\right] \\ &\Rightarrow f^{\prime}(x)=\frac{\sin x-\cos x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} \end{aligned}
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{\sin x-\cos x}{2+2 \sin x \cos x},\left[\therefore \sin ^{2} x+\cos ^{2} x=1\right] \\ &\Rightarrow f^{\prime}(x)=\frac{\sin x-\cos x}{2(1+\sin x \cos x)} \end{aligned}
Now, as given
\begin{aligned} &x \in\left(0, \frac{\pi}{4}\right) \\ &\Rightarrow \frac{\sin x-\cos x}{2(1+\sin x \cos x)}<0 \: \: \: \text{(for some values of x in first quadrant )}\\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}
$\text { Thus }f(x) \text { is decreasing on interval } x \: \in \: (0,\frac{\pi }{4})$
\begin{aligned} &\text { Now, for } x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\\ &\Rightarrow \sin x-\cos x>0 \: \: \: \:\; (\text { for some values of x in first quadrant })\\ &\Rightarrow \frac{\sin x-\cos x}{2(1+\sin x \cos x)}>0\\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}$\text { Thus }f(x) \text { is decreasing on interval } x \: \in \: (\frac{\pi }{4},\frac{\pi }{2})$

Increasing and Decreasing Functions exercise 16.2 question 18

$f(x)=(x-1)e^{x}+1 \text { is an increasing function for all x }> 0.$
Given:
$f(x)=(x-1)e^{x}+1$
To prove:
$\text { We have to show that } f(x)=(x-1)e^{x}+1 \text {is an increasing function for all x}>0.$
Hint:
f’(x) > 0 is condition for increasing function of f(x).
Solution:
Given
$f(x)=(x-1)e^{x}+1$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left[(x-1) e^{x}+1\right] \\ &\Rightarrow f^{\prime}(x)=e^{x}+(x-1) e^{x} \\ &\Rightarrow f^{\prime}(x)=e^{x}(1+x-1) \\ &\Rightarrow f^{\prime}(x)=x e^{x} \\ &\text { As given } x>0 \\ &\Rightarrow e^{x}>0 \\ &\Rightarrow x e^{x}>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Hence condition of f(x) to be increasing is satisfied.
Thus f(x) is increasing on interval x>0

Increasing and Decreasing Functions exercise 16.2 question 19

f(x) is neither decreasing nor increasing on (0,1)
Given:
$f(x)=x^{2}-x+1$
To prove:
We have to show that f(x) is neither decreasing nor increasing on (0,1).
Hint:
1. condition for f(x) to be increasing is f'(x)>0
2. condition for f(x) to be decreasing is f'(x)<0
Solution:
Given
$f(x)=x^{2}-x+1$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{2}-x+1\right) \\ &\Rightarrow f^{\prime}(x)=2 x-1 \end{aligned}
Taking different region from (0,1)
\begin{aligned} &\text { Let } x \in\left(0, \frac{1}{2}\right) \\ &\Rightarrow 2 x-1<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}
\begin{aligned} &\text { Thus } f(x) \text { is decreasing }\left(0, \frac{1}{2}\right) \\ \end{aligned}
\begin{aligned} &\text { Let } x \in\left(\frac{1}{2}, 1\right) \\ &\Rightarrow 2 x-1>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
\begin{aligned} &\text { Thus } f(x) \text { is increasing }\left(\frac{1}{2},1\right) \\ \end{aligned}
Therefor from above condition we find that
$\Rightarrow f(x) \text { is decreasing in } (0,\frac{1}{2}) \text { and increasing in } (\frac{1}{2},1)$
Hence, f(x) is neither increasing nor decreasing in (0,1)

Increasing and Decreasing Functions exercise 16.2 question 20

$f(x) \text { is an increasing function for all }x \: \in \: R$
Given:
$f(x)= x^{9}-4x^{7}+11$
To prove:
$\text { We have to show that }f(x) \text { is an increasing function for all }x \: \in \: R$
Hint:
f’(x) > 0 is condition for increasing function of f(x).
Solution:
Given
$f(x) x^{9}-4x^{7}+11$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{9}-4 x^{7}+11\right) \\ &\Rightarrow f^{\prime}(x)=9 x^{8}-28 x^{6} \\ &\Rightarrow f^{\prime}(x)=x^{6}\left(9 x^{2}+28\right) \\ &\text { As given, } x \in R \\ &\Rightarrow x^{6}>0 \text { and } 9 x^{2}+28>0 \\ &\Rightarrow x^{6}\left(9 x^{2}+28\right)>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Hence, condition for fx to be increasing$\text { Thus }f(x) \text { is an increasing on interval }x \: \in \: R$

Increasing and Decreasing Functions exercise 16.2 question 21

f(x) is an increasing on R.
Given:
$f(x)=x^{3}-6x^{2}+12x-18$
To prove:
We have to prove that f(x) is an increasing on R.
Hint:
Show f’(x) > 0 for increasing function.
Solution:
Given
$f(x)=x^{3}-6x^{2}+12x-18$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-6 x^{2}+12 x-18\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+12 \\ &\Rightarrow f^{\prime}(x)=3\left(x^{2}-4 x+4\right) \\ &\Rightarrow f^{\prime}(x)=3(x-2)^{2} \\ \end{aligned}
\begin{aligned} &\text { As given, } x \in R \\ &\Rightarrow(x-2)^{2}>0 \\ &\Rightarrow 3(x-2)^{2}>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Thus f(x) is increasing function x $\in$ R

Increasing and Decreasing Functions exercise 16.2 question 22

f(x) is an increasing on the interval [4, 6].
Given:
$f(x)=x^{2}-6x+3$
To prove:
We have to prove that f(x) is an increasing on the interval [4, 6].
Hint:
A function f(x) is said to be increasing on [a, b] if f’(x) > 0.
Solution:
Given
$f(x)=x^{2}-6x+3$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{2}-6 x+38\right) \\ &\Rightarrow f^{\prime}(x)=2 x-6 \\ &\Rightarrow f^{\prime}(x)=2(x-3) \\ \end{aligned}
\begin{aligned} &\text { Again, } x \in[4,6] \\ &\Rightarrow 4 \leq x \leq 6 \\ &\Rightarrow 1 \leq x-3 \leq 3 \\ &\Rightarrow(x-3)>0 \\ &\Rightarrow 2(x-3)>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Thus f(x) is an increasing function x $\in$ [4, 6]

Increasing and Decreasing Functions exercise 16.2 question 23

$f(x) \text { is increasing function on } (-\frac{\pi }{4},\frac{\pi }{4})$
Given:
$f(x)=sin\: x-cos\: x$
To prove:
$\text { We have to prove that } f(x) \text { is increasing function on } (-\frac{\pi }{4},\frac{\pi }{4})$
Hint:
If f ’(x) > 0 ∀ x $\in$ (a,b) then f(x) is increasing on (a,b).
Solution:
Given
$f(x)=sin\: x-cos\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\sin x-\cos x) \\ &\Rightarrow f^{\prime}(x)=\cos x+\sin x \\ &=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right) . .\{\text { multiply and divide by } \sqrt{2}\} \\ \end{aligned}
\begin{aligned} &=\sqrt{2}\left(\sin \frac{\pi}{4} \cos x+\cos \frac{\pi}{4} \sin x\right) \\ &\Rightarrow f^{\prime}(x)=\sqrt{2} \sin \left(\frac{\pi}{4}+x\right) \end{aligned}
Now, as given
\begin{aligned} &x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \\ &\Rightarrow-\frac{\pi}{4}
\begin{aligned} &\Rightarrow 0<\sin \left(\frac{\pi}{4}+x\right)<1 \\ &\Rightarrow \sqrt{2} \sin \left(\frac{\pi}{4}+x\right)>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}$\text { Hence } f(x) \text { is increasing on interval } x\: \in (-\frac{\pi }{4},\frac{\pi }{4})$

Increasing and Decreasing Functions exercise 16.2 question 24

f(x) is decreasing function on R
Given:
$f(x)=tan^{-1}x-x$
To prove:
We have to show that f(x) is decreasing function on R.
Hint:
A function f(x) to be decreasing if f ’(x) > 0
Solution:
We have
$f(x)=tan^{-1}x-x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(\tan ^{-1} x-x\right) \\ &=\frac{1}{1+x^{2}}-1,\left[\therefore \frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}\right] \\ &\Rightarrow f^{\prime}(x)=\frac{-x^{2}}{1+x^{2}} \end{aligned}
\begin{aligned} &\text { Now, } x \in R \\ &\Rightarrow x^{2}>0 \text { and } 1+x^{2}>0 \\ &\Rightarrow \frac{x^{2}}{1+x^{2}}>0 \\ &\Rightarrow \frac{-x^{2}}{1+x^{2}}<0 \end{aligned}
By applying negative sign change comparison sign
\begin{aligned} &\Rightarrow f'(x)< 0 \end{aligned}
Hence f(x) is decreasing function for x $\in$R

Increasing and Decreasing Functions exercise 16.2 question 25

$f(x) \text { is increasing on } (-\frac{\pi }{3},\frac{\pi }{3})$
Given:
$f(x)=-\frac{x}{2}+sin\: x$
To prove:
$\text { We have to determine whether } f(x) \text { is increasing or decreasing on } (-\frac{\pi }{3},\frac{\pi }{3})$
Hint:
For f(x) to be increasing we must have f ’(x) > 0 and f ‘(x) < 0 for decreasing.
Solution:
We have
$f(x)=-\frac{x}{2}+sin\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(-\frac{x}{2}+\sin x\right) \\ &\Rightarrow f^{\prime}(x)=-\frac{1}{2}+\cos x \\ &\text { Now, } x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \\ &\Rightarrow-\frac{\pi}{3}
\begin{aligned} &\Rightarrow \cos \left(-\frac{\pi}{3}\right)<\cos x<\cos \frac{\pi}{3} \\ &\Rightarrow \frac{1}{2}<\cos x,\left[\therefore \cos \left(-\frac{\pi}{3}\right)=\cos \frac{\pi}{3}=\frac{1}{2}\right] \\ &\Rightarrow-\frac{1}{2}+\cos x>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}$\text { Hence } f(x) \text { is increasing function for x } \in \: (-\frac{\pi }{3},\frac{\pi }{3})$

Increasing and Decreasing Functions exercise 16.2 question 26

f(x) is increasing in (0,$\infty$)
and f(x) is decreasing (-1,0)
Given:
$f(x)=log(1+x)-\frac{x}{1+x}$
To find:
We have to find the intervals in which f(x) is increasing and decreasing.
Hint:
First, we find critical point then use property of increasing and decreasing.
Solution:
We have,
$f(x)=log(1+x)-\frac{x}{1+x}$
Differentiating w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left\{\log (1+x)-\frac{x}{1+x}\right\} \\ &f^{\prime}(x)=\frac{1}{1+x}-\left[\frac{(1+x)-x}{(1+x)^{2}}\right] \end{aligned}
\begin{aligned} &=\frac{1}{1+x}-\frac{1}{(1+x)^{2}},\left[\therefore \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x} \cdot u \frac{d v}{d x}}{v^{2}}\right] \\ &=\frac{x}{(1+x)^{2}} \end{aligned}
For critical points. We must have,
\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow \frac{x}{(1+x)^{2}}=0\\ &\Rightarrow x=0 \text { and domain of }(1+x)^{2} \text { is }(-1, \infty)\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x>0\\ &\text { and } f^{\prime}(x)<0 \text { if }-1
Hence,f(x) is increasing in (0,$\infty$), decreases in (-1,0).

Increasing and Decreasing Functions exercise 16.2 question 27

f(x) is increasing in (-$\infty$,-1)
and f(x) is decreasing in (-1,$\infty$)
Given:
$f(x)=(x+2)e^{-x}$
To find:
We have to find the intervals in which f(x) is increasing and decreasing.
Hint:
First, we find critical point then find property of increasing and decreasing intervals.
Solution:
we have
$f(x)=(x+2)e^{-x}$
Differentiating w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=e^{-x}-e^{-x}(x+2) \\ &=e^{-x}(1-x-2) \\ &=-e^{-x}(x+1) \end{aligned}
For critical points.
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow-e^{-x}(x+1)=0 \\ &\Rightarrow x=-1 \\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-1 \\ &\text { and } f^{\prime}(x)<0 \text { if } x<-1 \end{aligned}
Hence, f(x) is increasing in (-$\infty$,-1), decreasing in (-1,$\infty$)

Increasing and Decreasing Functions exercise 16.2 question 28

f(x) is increasing for all x.
Given:
$f(x)=10^{x}$
To find:
We have to show that f(x) is increasing for all x.
Hint:
Condition to be function is increasing i.e, f ‘(x) > 0
Solution:
We have,
$f(x)=10^{x}$
On differentiating both sides w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=10^{x} \times \log 10,\left[\therefore \frac{d}{d x} a^{x}=a^{x} \log a\right] \\ &\text { Now, } x \in R \\ &\Rightarrow 10^{x}>0 \\ &\Rightarrow 10^{x} \log 10>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Hence, f(x) is an increasing function for all x.

Increasing and Decreasing Functions exercise 16.2 question 29

f(x)=x-[x] is increasing in (0,1).
Given:
$f(x)=x-[x]$
To find:
We have to prove that f(x)=x-[x] is increasing in (0,1).
Hint:
For increasing function f ‘(x) > 0
Solution:
We have,
$f(x)=x-[x]$
We know that
\begin{aligned} &\text { For } x \in(0,1) \\ &\Rightarrow[x]=0 \\ &\therefore f(x)=x \end{aligned}
On differentiating both sides w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=1>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Hence,f(x) is an increasing function for all x $\in$ (0,1).

Increasing and Decreasing Functions exercise 16.2 question 29 textbook solutio
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### Question:29

Increasing and Decreasing Functions exercise 16.2 question 29

f(x)=x-[x] is increasing in (0,1).
Given:
$f(x)=x-[x]$
To find:
We have to prove that f(x)=x-[x] is increasing in (0,1).
Hint:
For increasing function f ‘(x) > 0
Solution:
We have,
$f(x)=x-[x]$
We know that
\begin{aligned} &\text { For } x \in(0,1) \\ &\Rightarrow[x]=0 \\ &\therefore f(x)=x \end{aligned}
On differentiating both sides w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=1>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Hence,f(x) is an increasing function for all x $\in$ (0,1).

Increasing and Decreasing Functions exercise 16.2 question 30 subquestion (i)

f(x) is an increasing on R
Given:
$f(x)=3x^{5}+40x^{3}+240x$
To prove:
We have to prove that f(x) is an increasing on R.
Hint:
If f’(x) > 0 then f(x) is increasing function.
Solution:
Given
$f(x)=3x^{5}+40x^{3}+240x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(3 x^{5}+40 x^{3}+240 x\right) \\ &\Rightarrow f^{\prime}(x)=15 x^{4}+120 x^{2}+240 \\ &\Rightarrow f^{\prime}(x)=15\left(x^{4}+8 x^{2}+16\right) \\ &\Rightarrow f^{\prime}(x)=15\left(x^{2}+4\right)^{2} \end{aligned}
\begin{aligned} &\text { Now, } x \in R \\ &\Rightarrow\left(x^{2}+4\right)^{2}>0 \\ &\Rightarrow 15\left(x^{2}+4\right)^{2}>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Thus f(x) is increasing function for all x $\in$ R.

Increasing and Decreasing Functions exercise 16.2 question 30 subquestion (ii)

f(x) is an increasing on R.
Given:
$f(x)=4x^{3}-18x^{2}+27x-27$
To prove:
We have to prove that f(x) is an increasing on R.
Hint:
If f’(x) > 0 then f(x) is increasing function.
Solution:
Here we have
$f(x)=4x^{3}-18x^{2}+27x-27$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(4 x^{3}-18 x^{2}+27 x-27\right) \\ &\Rightarrow f^{\prime}(x)=12 x^{2}-36 x+27 \\ &\Rightarrow f^{\prime}(x)=3\left(4 x^{2}-12 x+9\right) \\ &\Rightarrow f^{\prime}(x)=3(2 x-3)^{2} \end{aligned}
\begin{aligned} &\text { Now, } x \in R \\ &\Rightarrow(2 x-3)^{2}>0 \\ &\Rightarrow 3(2 x-3)^{2}>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
Thus f(x) is increasing function on R.

Increasing and Decreasing Functions exercise 16.2 question 31

$f(x) \text { is strictly increasing on }(-\frac{\pi }{2},0) \text { and strictly decreasing on }(0,\frac{\pi }{2})$
Given:
$f(x)=log\: cos\: x$
To prove:
$\text { We have to prove that } f(x) \text { is strictly increasing on }(-\frac{\pi }{2},0) \text { and strictly decreasing on }(0,\frac{\pi }{2})$
Hint:
1. for f(x) to be increasing we must have f'(x)>0
2. for f(x) to be decreasing we must have f'(x)<0
Solution:
Given
$f(x)=log\: cos\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}[\log \cos x] \\ &\Rightarrow f^{\prime}(x)=\frac{1}{\cos x} \times(-\sin x) \\ &\Rightarrow f^{\prime}(x)=-\tan x \\ &\text { In interval }\left(0, \frac{\pi}{2}\right), \tan x>0 \\ &\Rightarrow-\tan x<0 \end{aligned}
By applying negative sign change comparison sign
$\therefore f^{\prime}(x)<0 \text { on }\left(0, \frac{\pi}{2}\right)$
$\text { Hence } f(x) \text { is strictly decreasing on }(0,\frac{\pi }{2})$
\begin{aligned} &\text { In interval }\left(\frac{\pi}{2}, \pi\right), \tan x<0 \\ &\Rightarrow-\tan x>0 \end{aligned}
By applying negative sign change comparison sign
$\therefore f^{\prime}(x)>0 \text { on }\left(\frac{\pi}{2}, \pi\right)$
$\text { Hence } f(x) \text { is strictly increasing on }(\frac{\pi }{2},\pi )$

I

Increasing and Decreasing Functions exercise 16.2 question 32

f(x) is strictly increasing on R.
Given:
$f(x)=x^{3}-3x^{2}+4$
To prove:
We have to prove that f(x) is strictly increasing on R.
Hint:
For f(x) to be increasing we must have f’(x) > 0.
Solution:
Here we have
$f(x)=x^{3}-3x^{2}+4$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-3 x^{2}+4 x\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-6 x+4 \\ &\Rightarrow f^{\prime}(x)=3\left(x^{2}-2 x+1\right)+1 \\ &f^{\prime}(x)=3(x-1)^{2}+1 \\ &\text { Here } 3(x-1)^{2}+1>0 \text { for all } x \in R \end{aligned}
Hence f(x) is strictly increasing function on R.

Increasing and Decreasing Functions exercise 16.2 question 33 subquestion (i)

f(x) is strictly decreasing on R
Given:
$f(x)=cos\: x$
To prove:
We have to prove that f(x) is strictly decreasing on R.
Hint:
For f(x) to be decreasing we must have f’(x) < 0.
Solution:
Given
$f(x)=cos\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\cos x)\\ &\Rightarrow f^{\prime}(x)=-\sin x\\ &\text { Since for each } x \in(0, \pi), \sin x>0\\ &\Rightarrow-\sin x<0 \end{aligned}
By applying negative sign change comparison sign.
\begin{aligned} &\Rightarrow f^{\prime}(x)< 0 \end{aligned}
Hence f(x) is strictly decreasing (0, $\pi$).

Increasing and Decreasing Functions exercise 16.2 question 33 subquestion (ii)

f(x) is strictly increasing in ($\pi$,2$\pi$)
Given:
$f(x)=cos\: x$
To prove:
We have to prove that f(x) is strictly increasing in ($\pi$,2$\pi$)
Hint:
For f(x) to be increasing we must have f’(x) > 0.
Solution:
Given
$f(x)=cos\: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\cos x)\\ &\Rightarrow f^{\prime}(x)=-\sin x\\ &\text { Since for each } x \in(\pi, 2 \pi), \sin x<0\\ &\Rightarrow-\sin x>0 \end{aligned}
By applying negative sign change comparison sign.
\begin{aligned} &\Rightarrow f^{\prime}(x)> 0 \end{aligned}
Hence f(x) is strictly increasing on ($\pi$,2$\pi$)

Increasing and Decreasing Functions exercise 16.2 question 33 subquestion (iii)

f(x) is neither increasing nor decreasing in (0,2$\pi$)
Given:
$f(x)=cos \: x$
To prove:
We have to prove that f(x) is neither increasing nor decreasing in (0,2$\pi$)
Hint:
For f(x) to be increasing we must have f’(x) > 0
and for f(x) to be decreasing we must have f’(x) < 0
Solution:
Given
$f(x)=cos \: x$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\cos x) \\ &\Rightarrow f^{\prime}(x)=-\sin x \\ &\text { Since for each } x \in(0, \pi), \\ &\sin x>0 \\ &\Rightarrow-\sin x<0 \end{aligned}
By applying negative sign change comparison sign.

$\Rightarrow f'(x)<0$
Hence f(x) is decreasing function in (0, $\pi$)
Again,
\begin{aligned} &f^{\prime}(x)=-\sin x\\ &\text { Since for each } x \in(\pi, 2 \pi), \sin x<0\\ &\Longrightarrow-\sin x>0 \end{aligned}
By applying negative sign change comparison sign.
$\Rightarrow f'(x)>0$
Clearly, from above we get f(x) is neither increasing nor decreasing in (0,2$\pi$)

Increasing and Decreasing Functions exercise 16.2 question 34

$f(x) \text { is an increasing function on } (0,\frac{\pi }{2})$
Given:
$f(x)=x^{2}-xsin\: x$
To prove:
$\text { We have to show that } f(x) \text { is an increasing function on } (0,\frac{\pi }{2})$
Hint:
For f(x) to be increasing we must have f’(x) > 0.
Solution:
Given
$f(x)=x^{2}-xsin\: x$
On differentiating both sides w.r.t x we get
$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{2}-x \sin x\right)$
\begin{aligned} &\Rightarrow f^{\prime}(x)=2 x-\sin x-x \cos x,\left[\therefore \frac{d}{d x} u v=v \frac{d u}{d x} \cdot u \frac{d v}{d x}\right] \\ &\text { Now, } x \in\left(0, \frac{\pi}{2}\right) \\ &\Rightarrow 0 \leq \sin x \leq 1 \\ &\Rightarrow 0 \leq \cos x \leq 1 \\ &\Rightarrow 2 x-\sin x-x \cos x>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}
$\text { Hence } f(x) \text { is an increasing function on } (0,\frac{\pi }{2})$

### Question:35

Increasing and Decreasing Functions exercise 16.2 question 35

$a\leq 0$
Given:
$f(x)=x^{3}-ax$
To prove:
We have to find the value of a for which f(x) is an increasing function on R.
Hint:
Given f(x) is increasing function that means f’(x) > 0.
Solution:
Here we have
$f(x)=x^{3}-ax$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-a x\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-a \end{aligned}
Given f(x) is increasing function on R
\begin{aligned} &\Rightarrow f^{\prime}(x)>0 \text { for all } x \in R\\ &\Rightarrow 3 x^{2}-a>0 \text { for all } x \in R\\ &\Rightarrow a<3 x^{2} \text { for all } x \in R\\ &\text { But the least value of } 3 x^{2}=0 \text { for } x=0 \end{aligned}
Hence a ≤ 0 is required value of a.

Increasing and Decreasing Functions exercise 16.2 question 35
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### Question:35

Increasing and Decreasing Functions exercise 16.2 question 35

$a\leq 0$
Given:
$f(x)=x^{3}-ax$
To prove:
We have to find the value of a for which f(x) is an increasing function on R.
Hint:
Given f(x) is increasing function that means f’(x) > 0.
Solution:
Here we have
$f(x)=x^{3}-ax$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-a x\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-a \end{aligned}
Given f(x) is increasing function on R
\begin{aligned} &\Rightarrow f^{\prime}(x)>0 \text { for all } x \in R\\ &\Rightarrow 3 x^{2}-a>0 \text { for all } x \in R\\ &\Rightarrow a<3 x^{2} \text { for all } x \in R\\ &\text { But the least value of } 3 x^{2}=0 \text { for } x=0 \end{aligned}
Hence a ≤ 0 is required value of a.

Increasing and Decreasing Functions exercise 16.2 question 36

$b\geq 1$
Given:
$f(x)=sin\: x -bx+c$
To prove:
We have to find the value of b for which f(x) is decreasing function on R.
Hint:
We will apply f’(x) < 0 for decreasing then evaluate the value of b.
Solution:
We have
$f(x)=sin\: x -bx+c$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\sin x-b x+c) \\ &\Rightarrow f^{\prime}(x)=\cos x-b \end{aligned}
Given f(x) is decreasing function on R

\begin{aligned} &\Rightarrow f^{\prime}(x)<0 \text { for all } x \in R \\ &\Rightarrow \cos x-b<0 \text { for all } x \in R \\ &\Rightarrow b>\cos x \text { for all } x \in R \end{aligned}
But the least value of cos x is 1
Hence b≥1 is required value of b.

Increasing and Decreasing Functions exercise 16.2 question 37

f(x) is an increasing function on R or all the value of a
Given:
$f(x)=x+cos\: x -a$
To prove:
We have to show that f(x) is an increasing function on R or all the value of a.
Hint:
For f(x) to be increasing function we must have f’(x) > 0.
Solution:
Here we have
$f(x)=x+cos\: x -a$
On differentiating both sides w.r.t x we get
\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(x+\cos x-a) \\ &\Rightarrow f^{\prime}(x)=1-\sin x,\left[\frac{d}{d x}(\text { constant })=0\right] \\ &\text { Since }-1 \leq \sin x \leq 1 \end{aligned}
Then we have,
\begin{aligned} &1-\sin x \geq 0 \text { for all value of } x\\ &\operatorname{So}f^{\prime}(x) \geq 0 \text { forall } x \in R \end{aligned}
Hence the function is increasing on Rf or all value of a.

Increasing and Decreasing Functions exercise 16.2 question 38

$\left|f^{\prime}(x)<1\right| \text { for all } x \in[0,1]$
Given:
$\text { Let } f \text { defined on } [0,1] \text { be twice differentiable such that } \left|f^{\prime \prime}(x) \leq 1\right| \text { for all } x \in[0,1] \text { and } f(0)=f(1)$
To prove:
$\text { We have to show that }\left|f^{\prime}(x)<1\right| \text { for all } x \in[0,1]$
Hint:
Using mean value theorem
$\frac{\left|f^{\prime}(x)-f^{\prime}(c)\right|}{x-c}=f^{\prime \prime}(d)$
Solution:
As f(0) = f(1) and f is differentiable
Hence by rolls theorem
$f^{\prime}(c)=0 \text { for some } c \in[0,1]$
Let us now apply mean value theorem for point 0 and x $\in$ [0,1]
Hence
\begin{aligned} &\frac{\left|f^{\prime}(x)-f^{\prime}(c)\right|}{x-c}=f^{\prime \prime}(d) \\ &\Rightarrow \frac{\left|f^{\prime}(x)-0\right|}{x-c}=f^{\prime \prime}(d), \quad[\therefore f(c)=0] \\ &\Rightarrow \frac{\left|f^{\prime}(c)\right|}{x-c}=f^{\prime \prime}(d) \end{aligned}
\begin{aligned} &\text { As given that } f^{\prime \prime}(d) \leq 1 \text { for } x \in[0,1]\\ &\Rightarrow \frac{\left|f^{\prime}(c)\right|}{x-c} \leq 1\\ &\Longrightarrow\left|f^{\prime}(c)\right| \leq x-c \end{aligned}
Now, as both x and c lies in [0,1]\begin{aligned} &\text { Hence } x-c \in[0,1] \\ &\Rightarrow\left|f^{\prime}(x)\right|<1 \text { for all } x \in[0,1] \end{aligned}

Increasing and Decreasing Functions exercise 16.2 question 39 subquestion (i)

f(x)=x |x| is an increasing function for all real values.
Given:
f(x)=x |x|
To prove:
We have to find the interval in which f(x) is an increasing or decreasing.
Hint:
For f(x) to be increasing function we must have f’(x) > 0.
Solution:
Here we have
$f(x)=f(x)=x|x|, x \in R$
We know that
$f(x)=\left\{-x^{2}, \quad \text { if } x<0 \& x^{2}, \quad \text { if } x>0\right.$
On differentiating f(x) w.r.t x we get
\begin{aligned} &f^{\prime}(x)=\{-2 x, \quad \text { if } x<0 \& 2 x, \quad \text { if } x>0 \\ &\Rightarrow f^{\prime}(x)>0 \text { for all value of } x . \end{aligned}
Hence f(x) is increasing function for all value of real values.

Increasing and Decreasing Functions exercise 16.2 question 39 subquestion (ii)

$f(x)=\sin x+|\sin x| \text { is } \\ \text { an increasing in interval } \left(0, \frac{\pi}{2}\right), f(x) \text { is decreasing in } \left(\frac{\pi}{2}, \pi\right) \\ \text { and neither increasing nor decreasing in }(\pi, 2 \pi).$
Given:
$f(x)=sin x+ |sin x |,0
To prove:
We have to find the interval in which f(x) is an increasing or decreasing.
Hint:
1. for f(x) to be increasing we must have f'(x)>0
2. for f(x) to be decreasing we must have f'(x)<0
Solution:
Here we have
$f(x)=sin x+ |sin x |,0
We know that
$f(x)=\{2 \sin x, \quad \text { if } 02 \pi$
On differentiating f(x) w.r.t x we get
$f^{\prime}(x)=\{2 \cos x, \quad \text { if } 0
$\text { The function } 2cos\: x \text { will be positive in }(0,\frac{\pi }{2})$
$\text { Hence the function is increasing in the interval }(0,\frac{\pi }{2})$
$\text { The function 2cos x will be negative between }(\frac{\pi }{2},\pi )$
$\text { Hence the function } f(x) \text { is decreasing in the interval }(\frac{\pi }{2},\pi ). \text { the value of } f'(x)=0 \text { when } \pi \: \leq x\:< 2\pi .$
Therefore the function fx is neither increasing nor decreasing in the interval ($\pi$,2$\pi$)

Increasing and Decreasing Functions exercise 16.2 question 39 subquestion (iii)

$f(x) \text{ is increasing in }(0,\frac{\pi }{3}) \\ \text {and} f(x) \text { is decreasing in } (\frac{\pi }{3},\frac{\pi }{2})$
Given:
$f(x)=sin\: x (1+cos \: x ),0
To find:
We have to find the intervals in which f(x) is increasing or decreasing.
Hint:
1. for f(x) to be increasing we must have f'(x) > 0
2. for f(x) to be decreasing we must have f'(x) < 0
Solution:
We have,
$f(x)=sin\: x (1+cos\, x )$
Differentiating w.r.t. x we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}[\sin x(1+\cos x)] \\ &{\left[\therefore \frac{d}{d x} u v=v \frac{d u}{d x} \cdot u \frac{d v}{d x}\right]} \\ &\Rightarrow f^{\prime}(x)=\cos x-\sin x \cdot \sin x+\cos x \cdot \cos x \end{aligned}
\begin{aligned} &\Rightarrow f^{\prime}(x)=\cos x+\cos ^{2} x-\sin ^{2} x,\left[\therefore \sin ^{2} x=1-\cos ^{2} x\right] \\ &\Rightarrow f^{\prime}(x)=\cos x+\cos ^{2} x-1+\cos ^{2} x \\ &\Rightarrow f^{\prime}(x)=2 \cos ^{2} x+\cos x-1 \end{aligned}
\begin{aligned} &\Rightarrow f^{\prime}(x)=2 \cos ^{2} x+2 \cos x-\cos x-1 \\ &\Rightarrow f^{\prime}(x)=2 \cos x(\cos x+1)-1(\cos x+1) \\ &\Rightarrow f^{\prime}(x)=(2 \cos x-1)(\cos x+1) \end{aligned}
For f(x) to be increasing, we must have,
\begin{aligned} &f^{\prime}(x)>0 \\ &\Rightarrow f^{\prime}(x)=(2 \cos x-1)(\cos x+1)>0 \end{aligned}
This can only be possible when,
$(2 \cos x-1)>0 \text { and }(\cos x+1)>0$
\begin{aligned} &\Rightarrow 0
For f(x) to be decreasing we must have,\begin{aligned} &f^{\prime}(x)<0\\ &\Rightarrow f_{f}^{\prime}(x)=(2 \cos x-1)(\cos x+1)<0\\ &\text { This can only be possible when, }(2 \cos x-1)<0 \text { and }(\cos x+1)<0 \end{aligned}
\begin{aligned} &\Rightarrow \frac{\pi}{3}

RD Sharma Class 12 Solutions Increasing and Decreasing Functions exercise 16.2 should be thoroughly followed by all students if they want to have impressive scores. Students should start their preparations beforehand and practice on all days to score high grades and surpass their peers.

The book contains information on various mathematical formulae, issues, and solutions which will give students a clear understanding of all chapters. Consequently, the students use the data provided in the book to test their own scores and track their performance.

Rd Sharma class 12 chapter 16 exercise 16.2 will always have the latest syllabus present in NCERT Books and prescribed by CBSE. The book will further help students to learn new methods of solving questions and expand their knowledge on the subject. RD Sharma Class 12th Exercise 16.2 solution will have answers on the following concepts: -

• The solution of judicious logarithmic disparities

• Stringently increasing functions

• Stringently decreasing functions

• Monotonic functions

• Monotonically increasing function

• Monotonically decreasing functions

• Fundamental and adequate conditions for monotonicity

• Discovering the stretches in which a job is increasing or decreasing

• Demonstrating the monotonicity of a situation on a given stretch

• Discovering the stretch where a cycle is increasing or decreasing

Benefits of using Rd Sharma class 12 chapter 16 exercise 16.2

There are certain benefits that are associated with using RD Sharma solutions for exam preparations. Some of the advantages include:-

• One book covers all chapters and includes answers to all questions provided in NCERT maths books.

• Simple solutions given to help understand the ideas better.

• Zero Cost and no investment required to avail the pdf of the book.

• A detailed solving of problems and formulae to teach students the basics of maths.

• Great for last minute revisions and to test self-performance.

## RD Sharma Chapter-wise Solutions

It is pretty easy to download the RD Sharma Class 12th Exercise 16.2 pdf online. Students need to navigate to the webpage of Career360 to find the free pdf.

2. Is RD Sharma useful to score high in the board exams?

There are numerous exam resources that students can use for preparations. However, preparing with RD Sharma Book Solutions will help them the most to understand everything and score better in class 12 exams.

3. Which is the best site to get class 12 rd Sharma chapter 16 exercise 16.2 solution for Class 12?

Students should use the Career360 website to get the Rd Sharma class 12 chapter 16 exercise 16.2 pdf as it's the one-stop destination for all RD Sharma solutions.

4. Are the RD Sharma Class 12 Maths Solutions Chapter 16.2 sufficient for CBSE students?

It's significantly recommended that class 12 RD Sharma chapter 16.2 exercise 16.2 solution students pick the RD Sharma Class 12 Solutions from Career360 as reference material for solutions.

5. List the number of questions in class 12 RD Sharma chapter 16.2?

There are 70 questions and answers in the RD Sharma Solutions Class 12 RD Sharma chapter 16.2 exercise 16.2.

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2 Jobs Available
##### Test Manager

A Test Manager is a professional responsible for planning, coordinating and controlling test activities. He or she develops test processes and strategies to analyse and determine test methods and tools for test activities. The test manager jobs involve documenting tests that have been carried out, analysing and evaluating software quality to determine further recommended procedures.

2 Jobs Available
##### Azure Developer

A career as Azure Developer comes with the responsibility of designing and developing cloud-based applications and maintaining software components. He or she possesses an in-depth knowledge of cloud computing and Azure app service.

2 Jobs Available
##### Deep Learning Engineer

A Deep Learning Engineer is an IT professional who is responsible for developing and managing data pipelines. He or she is knowledgeable about analyzing and storing data collected from various sources.  A Career as a Deep Learning Engineer needs to help the  data scientists and analysts to create effective data sets.

2 Jobs Available