RD Sharma Class 12 Exercise 16.2 Increasing and Decreasing Functions Solutions Maths

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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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RD Sharma Class 12 Solutions Chapter 16 Increasing and Decreasing Functions - Other Exercise

Increasing and Decreasing Functions Excercise16.2

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion i

Answer:
$f(x) \text { is increasing on the interval }\left(-\infty, \frac{-3}{2}\right) \text { and decreasing on the interval }\left(\frac{-3}{2}, \infty\right)$
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

Answer:
$f(x) \text { is increasing on the interval }(-1, \infty) \text { and decreasing on the interval }(-\infty,-)$
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

Answer:
$\text { Increasing interval }\left(-\infty, \frac{-9}{2}\right) \\ \text { Decreasing interval }\left(\frac{-9}{2}, \infty\right)$
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

Answer:
$\text {Increasing interval}(-\infty, 1) \cup(3, \infty) \\ \text {Decreasing interval}(1,3)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}-24 x+18=0 \\ &\Rightarrow 6\left(x^{2}-4 x+3\right)=0 \\ &\Rightarrow x^{2}-3 x-x+3=0\{\therefore 6>0\} \\ &\Rightarrow(x-3)(x-1)=0 \\ &\Rightarrow x-3=0 \text { and } x-1=0 \\ &\Rightarrow x=3 \text { and } x=1 \end{aligned}$
$\begin{aligned} &\text { Clearly, } f^{\prime}(x)>0, f(x)<1 \text { and } x>3 \text { or } x \in(-\infty, 1) \text { and } x \in(3, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &1<x<3 \text { or } x \in(1,3) \end{aligned}$
$\text { So, } f(x) \text { is increasing the interval }(-\infty, 1) \cup(3, \infty) \text { and } f(x) \text { is decreasing on interval }(1,3) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion v

Answer:
$\text {Increasing interval}\: (-2,3) \\ \text {Decreasing interval}\: (-\infty,-2) \cup(3, \infty)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 36+6 x-6 x^{2}=0 \\ &\Rightarrow 6\left(6+x-x^{2}\right)=0 \\ &\Rightarrow 6+x-x^{2}=0 \\ &\Rightarrow-x^{2}+x+6=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}-x-6=0 \\ &\Rightarrow x^{2}-3 x+2 x-6=0 \\ &\Rightarrow x(x-3)+2(x-3)=0 \\ &\Rightarrow(x-3)(x+2)=0 \end{aligned}$
$\begin{aligned} &\Rightarrow x-3=0 \text { and } x+2=0 \\ &\Rightarrow x=3 \text { and } x=-2 \\ &\text { Clearly, } f^{\prime}(x)>0, f-2<x<3 \text { or } x \in(-2,3) \end{aligned}$
$\text { and } f^{\prime}(x)<0 \text { if } x<-2 \text { and } x>3 \text { or } x \in(-\infty,-2) \text { and } x \in(3, \infty)$
$\text { Thus, } f(x) \text { is increasing on }(-2,3) \text { and decreasing on }(-\infty,-2) \cup(3, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion vi

Answer:
$\text {Increasing interval} (-2,3) \\ \text {Decreasing interval} (-\infty,-2) \cup(3, \infty)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 36+6 x-6 x^{2}=0 \\ &\Rightarrow 6\left(6+x-x^{2}\right)=0 \\ &\Rightarrow 6+x-x^{2}=0 \\ &\Rightarrow-x^{2}+x+6=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}-x-6=0 \\ &\Rightarrow x^{2}-3 x+2 x-6=0 \\ &\Rightarrow x(x-3)+2(x-3)=0 \\ &\Rightarrow(x-3)(x+2)=0 \end{aligned}$
$\begin{aligned} &\Rightarrow x-3=0 \text { and } x+2=0 \\ &\Rightarrow x=3 \text { and } x=-2 \\ &\text { Clearly, } f^{\prime}(x)>0, f-2<x<3 \text { or } x \in(-2,3) \end{aligned}$
$\text { and } f^{\prime}(x)<0 \text { if } x<-2 \text { and } x>3 \text { or } x \in(-\infty,-2) \text { and } x \in(3, \infty)$
$\text { Thus, } f(x) \text { is increasing on }(-2,3) \text { and decreasing on }(-\infty,-2) \cup(3, \infty)$

I

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion vii

Answer:
$\text {Increasing interval} (-\infty,-2) \cup(4, \infty) \\ \text {Decreasing interval} (-2,4)$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 15 x^{2}-30 x-120=0\\ &\Rightarrow 15\left(x^{2}-2 x-8\right)=0\\ &\Rightarrow x^{2}-2 x-8=0\{\therefore 15>0\}\\ &\Rightarrow x^{2}-4 x+2 x-8=0\\ &\Rightarrow x(x-4)+2(x-4)=0\\ &\Rightarrow(x-4)(x+2)=0\\ &\Rightarrow x-4=0 \text { and } x+2=0\\ &\Rightarrow x=4 \text { and } x=-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>4 \text { or } x \in(-\infty,-2) \text { and } x \in(4, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}$
$\begin{aligned} &-2<x<4 \text { or } x \in(-2,4)\\ &\text { Thus, } f(x) \text { is increasing on the interval }(-\infty,-2) \cup(4, \infty) \text { and } f(x) \text {is decreasing on interval} (-2,4). \end{aligned}$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion viii

Answer:
$\text {Increasing interval} (-\infty,-2) \cup(6, \infty) \\ \text {Decreasing interval} (-2,6)$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-12 x-36=0\\ &\Rightarrow 3\left(x^{2}-4 x-12\right)=0\\ &\Rightarrow x^{2}-4 x-12\{\therefore 3>0\}\\ &\Rightarrow x^{2}-6 x+2 x-12=0\\ &\Rightarrow(x-6)(x+2)=0\\ &\Rightarrow x-6=0 \text { and } x+2=0\\ &\Rightarrow x=6 \text { and } x=-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>6 \text { or } x \in(-\infty,-2) \text { and } x \in(6, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}$
$\begin{aligned} &-2<x<6 \text { or } x \in(-2,6)\\ &\text { Thus, } f(x) \text { is increasing on the interval }(-\infty,-2) \cup(6, \infty) \text { and } \\ &\text {f(x) is decreasing on interval (-2,6).} \end{aligned}$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion ix

Answer:
$\text { Increasing interval }(-\infty, 2) \cup(3, \infty) \\ \text { Decreasing interval (2,3) }$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}-30 x+36=0 \\ &\Rightarrow 6\left(x^{2}-5 x+6\right)=0 \\ &\Rightarrow x^{2}-5 x+6=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}-3 x-2 x+6=0 \\ &\Rightarrow(x-3)(x-2)=0 \\ &\Rightarrow x-3=0 \text { and } x-2=0 \\ &\Rightarrow x=3 \text { and } x=2 \end{aligned}$
$\begin{aligned} &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<2 \text { and } x>3 \text { or } x \in(-\infty, 2) \text { and } x \in(3, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &2<x<3 \text { or } x \in(2,3) \end{aligned}$
$\text { So, } f(x) \text { is increasing on the interval } (-\infty, 2)\: \cup\: (3, \infty) \text { and } f(x) \text { is decreasing on interval }(2,3).$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion x

Answer:
$\text {f(x) is decreasing on the interval (-2,-1) and} \\ \text { increasing on the interval} (-\infty,-2) \cup (-1, \infty)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}+18 x+12=0 \\ &\Rightarrow 6\left(x^{2}+3 x+2\right)=0 \\ &\Rightarrow x^{2}+3 x+2=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}+2 x+x+2=0 \\ &\Rightarrow(x+2)(x+1)=0 \\ &\Rightarrow x+2=0 \text { and } x+1=0 \\ &\Rightarrow x=-2 \text { and } x=-1 \\ &\text { Clearly, } f'(x)<0, \text { if }-2<x<-1 \text { or } x \in(-2,-1) \end{aligned}$
$\text { and } f^{\prime}(x)>0 \text { if } x<-1 \text { and } x>-2 \text { or } x \in(-\infty,-2) \text { and } x \in(-1, \infty)$
$\text {Thus, } f(x) \text { is decreasing on the interval (-2,-1) and} \\ \text {increasing on the interval } (-\infty,-2) \cup(-1, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xi

Answer:
$\text { Increasing interval }(-\infty, 1) \cup(2, \infty)\\ \text { Decreasing interval } (1,2)$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 6 x^{2}-18 x+12=0\\ &\Rightarrow 6\left(x^{2}-3 x+2\right)=0\\ &\Rightarrow x^{2}-3 x+2=0\{\therefore 6>0\}\\ &\Rightarrow x^{2}-2 x-x+2=0\\ &\Rightarrow(x-1)(x-2)=0\\ &\Rightarrow x-1=0 \text { and } x-2=0\\ &\Rightarrow x=1 \text { and } x=2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<1 \text { and } x>2 \text { or } x \in(-\infty, 1) \text { and } x \in(2, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}$
$\begin{aligned} &1<x<2 \text { or } x \in(1,2)\\ &\text { So, } f(x) \text { is increasing on the interval }(-\infty, 1) \cup(2, \infty) \text { and } f(x) \text { is decreasing on interval }(1,2) . \end{aligned}$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xii

Answer:
$\text { Increasing interval (-1,2)} \\ \text { Decreasing interval} (-\infty,-1) \cup(2, \infty)$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 12+6 x-6 x^{2}=0\\ &\Rightarrow 6\left(2+x-x^{2}\right)=0\\ &\Rightarrow 2+x-x^{2}=0\\ &\Rightarrow x^{2}-x-2=0\\ &\Rightarrow x^{2}-2 x+x-2=0\\ &\Rightarrow(x-2)(x+1)=0\\ &\Rightarrow x-2=0 \text { and } x+1=0\\ &\Rightarrow x=2 \text { and } x=-1\\ &\text { Clearly, } f^{\prime}(x)>0, \text { if }-1<x<2 \text { or } x \in(-1,2)\\ &\text { and } f^{\prime}(x)<0 \text { if } x<-1 \text { and } x>2 \text { or } x \in(-\infty,-1) \text { and } x \in(2, \infty) \end{aligned}$
$\text { Thus, } f(x) \text { is increasing on the interval (-1,2) and } \\ \text { decreasing on the interval } (-\infty,-1) \cup(2, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xiii

Answer:
$\text { Increasing interval }(-\infty,-2) \cup(2, \infty) \\ \text { Decreasing interval (-2,2) }$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 6 x^{2}-24=0\\ &\Rightarrow 6\left(x^{2}-4\right)=0\\ &\Rightarrow x^{2}-4=0\{\therefore 6>0\}\\ &\Rightarrow x^{2}=4\\ &\Rightarrow x=\pm 2\\ &\Rightarrow x=+2,-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>2 \text { or } x \in(-\infty,-2) \text { and } x \in(2, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &-2<x<2 \text { or } x \in(-2,2) \end{aligned}$
$\text { So, } f(x) \text { is increasing on the interval }(-\infty,-2) \cup(2, \infty) \text { and } \\ f(x) \text { is decreasing on interval (-2,2). }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xiv

Answer:
$\text { Increasing interval (-2,-1) } \\ \text { Decreasing interval }(-\infty,-2) \cup(-1, \infty)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow-6 x^{2}-18 x-12=0 \\ &\Rightarrow 6 x^{2}+18 x+12=0 \\ &\Rightarrow 6\left(x^{2}+3 x+2\right)=0 \\ &\Rightarrow x^{2}+2 x+x+2=0 \\ &\Rightarrow x(x+2)+1(x+2)=0 \\ &\Rightarrow(x+2)(x+1)=0 \\ &\Rightarrow x+2=0 \text { and } x+1=0 \\ &\Rightarrow x=-2 \text { and } x=-1 \end{aligned}$
$\begin{aligned} &\text { Clearly, } f^{\prime(x)}<0, \text { if } x<-1 \text { and } x<-2 \text { or } x \in(-\infty,-2) \text { and } x \in(-1, \infty) \\ &\text { and } f^{\prime}(x)>0 \text { if }-2<x<-1 \text { or } x \in(-2,-1) \end{aligned}$
$\text { Thus, } f(x) \text { is increasing on the interval }(-2,-1) \text { and decreasing on the interval }(-\infty,-2)\:\: \cup\:\: (-1,\infty )$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xv

Answer:
$\text { Increasing interval } \left(-\infty, \frac{4}{3}\right) \cup(2, \infty) \\ \text { Decreasing interval } \left(\frac{4}{3}, 2 \right)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow(x-2)(3 x-4)=0 \\ &\Rightarrow x-2=0 \text { and } 3 x-4=0 \end{aligned}$
$\begin{aligned} &\Rightarrow x=2 \text { and } 3 x=4\\ &\text { or } x=\frac{4}{3}\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<\frac{4}{3} \text { and } x>2 \text { and } f^{\prime}(x)<0 \text { if } \frac{4}{3}<x<2 \end{aligned}$
$\text { So, } f(x) \text { is increasing on the interval }\left(-\infty, \frac{4}{3}\right) \cup(2, \infty) \text { and decreasing on the interval }\left(\frac{4}{3}, 2\right) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xvi

Answer:
$\text { Increasing interval }(-\infty, 2) \cup(6, \infty) \\ \text { Decreasing interval(2.6) }$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-24 x+36=0\\ &\Rightarrow 3\left(x^{2}-8 x+12\right)=0\\ &\Rightarrow x^{2}-8 x+12=0\{\therefore 3>0\}\\ &\Rightarrow x^{2}-6 x-2 x+12=0\\ &\Rightarrow(x-6)(x-2)=0\\ &\Rightarrow x-6=0 \text { and } x-2=0\\ &\Rightarrow x=6 \text { and } x=2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<2 \text { and } x>6 \text { or } x \in(-\infty, 2) \text { and } x \in(6, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &2<x<6 \text { or } x \in(2,6) \end{aligned}$
$\text { So, } f(x) \text { is increasing on the interval }(-\infty, 2) \cup \: (6, \infty) \text { and } f(x) \text { is decreasing on interval }(2,6) \text {.}$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xvii

Answer:
$\text { Increasing interval }(-\infty,-2) \cup(2, \infty) \\ \text { Decreasing interval(-2,2) }$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 6 x^{2}-24=0\\ &\Rightarrow 6 x^{2}=24\\ &\Rightarrow x^{2}=\frac{24}{6}\\ &\Rightarrow x^{2}=4\\ &\Rightarrow x=\pm 2\\ &\Rightarrow x=+2,-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x>2 \text { and } x<-2 \text { or } x \in(-\infty, 2) \text { and } x \in(-2, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &-2<x<2 \text { or } x \in(-2,2) \end{aligned}$
$\text { So, } f(x) \text { is increasing on the interval }(-\infty,-2) \cup(2, \infty) \text { and }\\ f(x) \text { is decreasing on interval } (-2,2).$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xviii

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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{6}{5}(x-1)(x+2)(x-3)=0 \\ &\Rightarrow(x-1)(x+2)(x-3)=0\left\{\therefore \frac{6}{5}>0\right\} \\ &\Rightarrow x=1, \quad x=-2, \quad x=3 \end{aligned}$
$\begin{aligned} &\text { Now we take the intervals }(-\infty,-2) \text { i.e, }-\infty<x<-2 . \text { In this case we have } x-1<0, x+2<\\ &0 \text { and } x-3<0 . \text { Clearly, } f^{\prime}(x)<0 \text { if } x \in(-\infty,-2) \text { . } \end{aligned}$
$\begin{aligned} &\text { Now we take }(-2,1) . \text { In this case we have } x-1<0, x+2>0 \text { and } x-3<0 . \text { So, } f^{\prime}(x)>0 \text { if } x &\in(-2,1) \end{aligned}$
$\text { After that we take }(1,3) \text { i.e, } 1<x<3 \text { . } \\ \text { Clearly, we have } x-1>0, x+2>0 \text { and } x-3<0 . \text { So, } f^{\prime}(x)<0 \text { if } x \in(1,3) \text { . }$
$\text { Finally we take }(3, \infty) \text { i.e, } 3<x<\infty \text { . In this case we have } x-1>0, x+2>0 \text { and } x-3>0 \text { . }$
$\text { Clearlyf }^{\prime}(x)>0 \text { if } x>3 \text { or } x \in(3, \infty)$
$\text { Thus, the function is increasing on }(-2,1) \cup(3, \infty) \text { and } \\ f(x) \text { is decreasing on interval }(-\infty,-2) \cup(1,3) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xix

Answer:
$\text { Increasing interval }(1, \infty) \\ \text { Decreasing interval }(-\infty,1)$
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

Answer:
$\text { Increasing interval }(-3,-1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(-1, 2)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow(x+1)(x-2)(x+3)=0 \\ &\Rightarrow x=-1, \quad x=2, \quad x=-3 \end{aligned}$
$\text { The possible intervals are }(-3,-1),(-\infty,-3),(-1,2) \text { and }(2, \infty)$
$\text { Now we take }(-3,-1) . \text { i.e, }-3<x<-1 \text { . }$
$\text { In this case we have } x+1<0, x-2<0 \text { and } x+3>0 . \text { Clearly, } f^{\prime}(x)>0 \text { if }-3<x<-1 \text { . }$
$\begin{aligned} &\text { Now we take the intervals }(-\infty,-3) \text { i.e, }-\infty<x<-3 \text { . In this case we have } x+1<0, x-2<&0 \\ &\text { and } x+3<0 . \text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty<x<-3 \end{aligned}$
$\text { After that we take }(-1,2) \text { i.e, }-1<x<2 \text { . }$
$\text { Clearly, we have } x+1>0, x-2<0 \text { and } x+3>0 . \text { So, } f^{\prime}(x)<0 \text { if }-1<x<2 \text { . }$
$\text { Finally we take }(2, \infty) \text { i.e, } 2<x<\infty . \text { In this case we have } x+1>0, x-2>0 \text { and } x+3>0 \text { . }$
$\text { Clearly }f^{\prime}(x)>0 \text { if } 2<x<\infty$
$\text { Thus, the function is increasing on }(-3,-1) \cup(2, \infty) \text { and }$
$f(x) \text { is decreasing on interval }(-\infty,-3) \cup(-1,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxi

Answer:
$\text { Increasing interval }(0,1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,0) \cup(1,2)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 4 x^{3}-12 x^{2}+8 x=0 \\ &\Rightarrow 4 x\left(x^{2}-3 x+2\right)=0 \\ &\Rightarrow x\left(x^{2}-3 x+2\right)=0\{\therefore 4>0\} \\ &\Rightarrow x\left(x^{2}-2 x-x+2\right)=0 \\ &\Rightarrow x(x-2)(x-1)=0 \\ &\Rightarrow x=0, \quad x=1, \quad x=2 \end{aligned}$
$\begin{aligned} &\text { The possible intervals are }(-\infty, 0),(0,1),(1,2) \text { and }(2, \infty)\\ &\text { in intervals }(0,1) \text { and }(2, \infty), f^{\prime}(x) \end{aligned}$
$\text { Clearly, } f^{\prime}(x)>0 \text { if } 0<x<1 \text { and } 2<x<\infty$
$\begin{aligned} &\text { However in the intervals }(-\infty, 0) \text { and }(1,2) f^{\prime}(x) .\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty<x<0 \end{aligned}$
$\text { Thus, } f(x) \text { is increasing on }(0,1) \cup(2, \infty) \text { and decreasing on }(-\infty, 0) \cup(1,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxii

Answer:
$\text { Increasing interval }(0,1) \\ \text { Decreasing interval }(1,\infty )$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{15}{2}\left(x^{\frac{1}{2}}-x^{\frac{3}{2}}\right)=0,\left\{\therefore \frac{15}{2}=0\right\} \\ &\Rightarrow x^{\frac{1}{2}}-x^{\frac{3}{2}}=0 \\ &\Rightarrow x^{\frac{1}{2}}\left(1-x^{\frac{3}{2}}\right)=0 \end{aligned}$
$\begin{aligned} &\Rightarrow x=0 \text { and } x=1 \\ &\text { Clearly, } f^{\prime}(x)>0, \text { if } 0<x<1 \text { or } x \in(0,1) \\ &\text { and } f^{\prime}(x)<0 \text { if } x>1 \text { or } x \in(1, \infty) \end{aligned}$ $\text { Thus, } f(x) \text { is increasing on the interval }(0,1) \text { and decreasing on the interval }(1, \infty)$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxiii

Answer:
$\text { Increasing interval }(0, \infty)\\ \text { Decreasing interval }(-\infty,0)$
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

Answer:
$\text { Increasing interval }(-\infty, 1) \cup(3, \infty) \\ \text { Decreasing interval }(1,3)$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-12 x+9=0\\ &\Rightarrow 3\left(x^{2}-4 x+3\right)=0\\ &\Rightarrow x^{2}-4 x+3=0\\ &\Rightarrow x^{2}-3 x-x+3=0\\ &\Rightarrow(x-3)(x-1)=0\\ &\Rightarrow x=3,1\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<1 \text { and } x>3 \text { or } x \in(-\infty, 1) \text { and } x \in(3, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &1<x<3 \text { or } x \in(-1,3) \end{aligned}$
$\text { So, } f(x) \text { is increasing on the interval }(-\infty, 1) \cup (3, \infty) \text { and } f(x) \text { is decreasing on interval }(1,3) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxv

Answer:
$\text { Increasing interval }(0,1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,0) \cup(1,2)$
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,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 4 x(x-2)(x-1)=0 \\ &\Rightarrow x(x-2)(x-1)=0 \\ &\Rightarrow x=0,1,2 \end{aligned}$
$\text { The possible intervals are }(-\infty, 0),(0,1),(1,2) \text { and }(2, \infty)$
$\begin{aligned} &\text { In the intervals }(-\infty, 0) \text { and }(1,2) f^{\prime}(x)<0 \text { . }\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty<x<0 \text { and } 1<x<2 \end{aligned}$
$\text { However, } \\ \text { In intervals }(0,1) \text { and }(2, \infty), f^{\prime}(x)>0 \\ \text { Clearly, } f^{\prime}(x)>0 \text { if } 0<x<1 \text { and } 2<x<\infty \text { . }$
$\text { Thus, the function } f(x) \text { is increasing on }(0,1) \cup(2, \infty) \text { and decreasing on }(-\infty, 0) \cup(1,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxvi

Answer:
$\text { Increasing interval }(-1,0) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty,-1) \cup(0,2)$
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 12 x\left(x^{2}-x-2\right)=0\\ &\Rightarrow x\left(x^{2}-x-2\right)=0,\{\therefore 12=0\}\\ &\Rightarrow x\left(x^{2}-2 x+x-2\right)=0\\ &\Rightarrow x(x-2)(x+1)=0\\ &\Rightarrow x=0 ; x=-1 ; x=2\\ &\text { The points } x=0,-1,2 \text { divide the real line into four disjoint intervals, } \end{aligned}$
$\begin{aligned} &(-\infty,-1),(-1,0),(0,2) \text { and }(2, \infty)\\ &\text { In the intervals }(-\infty,-1) \text { and }(0,2) f^{\prime}(x)<0 \text { . } \end{aligned}$
$\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty<x<-1 \text { and } 0<x<2$
$\begin{aligned} &\text { However, } \\ &\text { In intervals }(-1,0) \text { and }(2, \infty), f^{\prime}(x)>0\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if }-1<x<0 \text { and } 2<x<\infty \end{aligned}$
$\text { Thus, } f(x) \text { is increasing in the intervals }(-1,0) \cup(2, \infty) \text { and } \\ \text { decreasing in the intervals }(-\infty,-1) \cup(0,2) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxvii

Answer:
$\text { Increasing interval }(-3,0) \cup(5, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(0,5)$
Given:
Here given that
$f(x)=\frac{3}{2}x^{4}-4x^{3}-45x^{2}+51$
To find:
We have to find the intervals in which function is increasing and decreasing.
Hint:
$\text { If } f^{\prime}(x)>0 \text { for all } x \in(a ., b), \text { then } f(x) \text { is increasing on }(a, b) \text { and If } f^{\prime}(x)<0 \text { for all } x \in \text { (a., } b) \text { , then } f(x) \text { is decreasing on }(a, b) \text { . }$
Solution:
We have,
$f(x)=\frac{3}{2}x^{4}-4x^{3}-45x^{2}+51$
Differentiating w.r.t. x, we get,
$\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(\frac{3}{2} x^{4}-4 x^{3}-45 x^{2}+51\right) \\ &f^{\prime}(x)=\frac{3}{2}\left(4 x^{3}\right)-12 x^{2}-45(2 x) \\ &=6 x^{3}-12 x^{2}-90 x \end{aligned}$
We have to find critical points, we must have,
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{3}-12 x^{2}-90 x=0 \\ &\Rightarrow 6 x\left(x^{2}-2 x-15\right)=0 \\ &\Rightarrow x\left(x^{2}-2 x-15\right)=0,\{\therefore 6=0\} \\ &\Rightarrow x\left(x^{2}-5 x+3 x-15\right)=0 \\ &\Rightarrow x(x-5)(x+3)=0 \\ &\Rightarrow x=0 ; x=5 ; x=-3 \end{aligned}$
The points x=0, 5, -3 divide the real line into four disjoint intervals,
$\begin{aligned} &(-\infty,-3),(-3,0),(0,5) \text { and }(5, \infty) \\ &\text { In intervals }(-3,0) \text { and }(5, \infty), f^{\prime}(x)>0 \end{aligned}$
$\text { Clearly, } f^{\prime}(x)>0 \text { if }-3<x<0 \text { and } 5<x<\infty \\ \text { However, }$
$\begin{aligned} &\text { In the intervals }(-\infty,-3) \text { and }(0,5) f^{\prime}(x)<0 \text { . }\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty<x<-3 \text { and } 0<x<5 \end{aligned}$ $\text { Thus, } f(x) \text { is increasing in the intervals }(-3,0) \cup(5, \infty) \text { and } \\ \text { decreasing in the intervals }(-\infty,-3) \cup(0,5) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxviii
Answer:

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

Answer:
$\text { Increasing interval }(-3,2) \cup(4, \infty) \\ \text { Decreasing interval }(-\infty ,-3) \cup(2, 4)$
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,
$\begin{aligned} &(-\infty,-3),(-3,2),(2,4) \text { and }(4, \infty) \\ &\text { In intervals }(-3,2) \text { and }(4, \infty), f^{\prime}(x)>0 \end{aligned}$
$\text { Clearly, } f^{\prime}(x)>0 \text { if }-3<x<2 \text { and } 4<x<\infty \\ \text { However, }$
$\begin{aligned} &\text { In the intervals }(-\infty,-3) \text { and }(2,4) f^{\prime}(x)<0\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty<x<-3 \text { and } 2<x<4 \text { . } \end{aligned}$ $\text { Thus, } f(x) \text { is increasing in the intervals }(-3,2) \cup(4, \infty) \text { and } \text { decreasing in the intervals }(-\infty,-3) \cup(2,4) \text { . }$

Increasing and Decreasing Functions exercise 16.2 question 2

Answer:
$i.\: x=3 \\ ii.\: co-ordinate (\frac{5}{2},\frac{1}{4})$
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:

First we will find critical point then find f’(x) for all the value of x.
Find m₁ and m₂ from the curve then find points

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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.
Therefore, they follow the relation
$\begin{aligned} &\Rightarrow-\frac{1}{m_{1}}=m_{2} \\ &\Rightarrow-\frac{1}{2 x-6}=1 \\ &\Rightarrow 2 x-6=-1 \\ &\Rightarrow x=\frac{5}{2} \end{aligned}$
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

Answer:
$f(x) \text { is increasing on } \left(0, \frac{3 \pi}{4}\right) \\ \text { and } f(x) \text { is decreasing on } \left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$
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.
$\text { Clearly, } f^{\prime}(x)>0 \text { if } 0<x<\frac{3 \pi}{4} \text { and } f^{\prime}(x)<0 \text { if } \frac{3 \pi}{4}<x<\frac{7 \pi}{4} \text { . }$
$\text {Thus } f(x) \text { increases on }\left(0, \frac{3 \pi}{4}\right) \cup\left(\frac{7 \pi}{4}, 2 \pi\right) \text { and } f(x) \text { is decreasing on interval } x \in\left(\frac{3 \pi}{4}, \frac{7 \pi}{4}\right)$

Increasing and Decreasing Functions exercise 16.2 question 7

Answer:
$f(x) \text { is increasing on } (0, \frac{\pi }{2}) \text { and decreasing in } (\frac{\pi }{2},\pi ) \\ \text { Hence, } f(x) \text { is neither increasing nor decreasing in } ( 0,\pi )$
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

Answer:
$f(x) \text { is increasing on } (0,\frac{\pi }{2}) \text { and decreasing on }(\frac{\pi }{2},\pi ).$
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

Answer:
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

Answer:
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

Answer:
$f(x) \text { is decreasing function on }(0,\frac{\pi }{2}).$
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

Answer:
$f(x) = sin\: x \text { is increasing function on }(-\frac{\pi }{2},\frac{\pi }{2})$
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

Answer:
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:

For increasing function f'(x)>0 then f(x) is increasing.
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

Answer:
$f(x) \text { is increasing function on } (-\frac{\pi }{2},\frac{\pi }{2})$
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
$\begin{aligned} &x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned}$
i.e, 4^th quadrant to 1^st quadrant, where
$\begin{aligned} \Rightarrow sec^{2}\: x> 0 \\ \Rightarrow f'(x)> 0 \end{aligned}$
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

Answer:
$f(x) \text { is decreasing function on } (\frac{\pi }{4},\frac{\pi }{2})$
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

Answer:
$f(x) \text { is decreasing function on }(\frac{3\pi }{8},\frac{5\pi }{8})$
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
$\begin{aligned} &x \in\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right) \\ &\Rightarrow \frac{3 \pi}{8}<x<\frac{5 \pi}{8} \\ &\Rightarrow \frac{3 \pi}{4}<2 x<\frac{5 \pi}{4} \\ &\Rightarrow \pi<2 x+\frac{\pi}{4}<\frac{3 \pi}{2} \end{aligned}$
$\begin{aligned} &\text { As here } 2 x+\frac{\pi}{4} \text { lies in } 3 \mathrm{rd} \text { quadrant. }\\ &\Rightarrow \cos \left(2 x+\frac{\pi}{4}\right)<0\\ &\Rightarrow 2 \cos \left(2 x+\frac{\pi}{4}\right)<0\\ &\Longrightarrow f^{\prime}(x)<0 \end{aligned}$
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

Answer:
$f(x) \text { is decreasing on }(0,\frac{\pi }{4}) \text { and increasing on } (\frac{\pi }{4},\frac{\pi }{2})$
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:

condition for f(x) to be increasing is f'(x)>0
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

Answer:
$f(x)=(x-1)e^{x}+1 \text { is an increasing function for all x }> 0.$
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

Answer:
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:

condition for f(x) to be increasing is f'(x)>0
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

Answer:
$f(x) \text { is an increasing function for all }x \: \in \: R$
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

Answer:
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

Answer:
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

Answer:
$f(x) \text { is increasing function on } (-\frac{\pi }{4},\frac{\pi }{4})$
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
$\begin{aligned} &x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \\ &\Rightarrow-\frac{\pi}{4}<x<\frac{\pi}{4} \\ &\Rightarrow 0<\frac{\pi}{4}+x<\frac{\pi}{2} \\ &\Rightarrow \sin 0^{\circ}<\sin \left(\frac{\pi}{4}+x\right)<\sin \frac{\pi}{2} \\ \end{aligned}$
$\begin{aligned} &\Rightarrow 0<\sin \left(\frac{\pi}{4}+x\right)<1 \\ &\Rightarrow \sqrt{2} \sin \left(\frac{\pi}{4}+x\right)>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}$ $\text { Hence } f(x) \text { is increasing on interval } x\: \in (-\frac{\pi }{4},\frac{\pi }{4})$

Increasing and Decreasing Functions exercise 16.2 question 24

Answer:
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

Answer:
$f(x) \text { is increasing on } (-\frac{\pi }{3},\frac{\pi }{3})$
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
$\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(-\frac{x}{2}+\sin x\right) \\ &\Rightarrow f^{\prime}(x)=-\frac{1}{2}+\cos x \\ &\text { Now, } x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \\ &\Rightarrow-\frac{\pi}{3}<x<\frac{\pi}{3} \end{aligned}$
$\begin{aligned} &\Rightarrow \cos \left(-\frac{\pi}{3}\right)<\cos x<\cos \frac{\pi}{3} \\ &\Rightarrow \frac{1}{2}<\cos x,\left[\therefore \cos \left(-\frac{\pi}{3}\right)=\cos \frac{\pi}{3}=\frac{1}{2}\right] \\ &\Rightarrow-\frac{1}{2}+\cos x>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}$ $\text { Hence } f(x) \text { is increasing function for x } \in \: (-\frac{\pi }{3},\frac{\pi }{3})$

Increasing and Decreasing Functions exercise 16.2 question 26

Answer:
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,
$\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow \frac{x}{(1+x)^{2}}=0\\ &\Rightarrow x=0 \text { and domain of }(1+x)^{2} \text { is }(-1, \infty)\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x>0\\ &\text { and } f^{\prime}(x)<0 \text { if }-1<x<0 \end{aligned}$
Hence,f(x) is increasing in (0, $\infty$ ), decreases in (-1,0).

Increasing and Decreasing Functions exercise 16.2 question 27

Answer:
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

Answer:
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

Answer:
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

Answer:

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

Answer:
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)

Answer:
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

Answer:
$f(x) \text { is strictly increasing on }(-\frac{\pi }{2},0) \text { and strictly decreasing on }(0,\frac{\pi }{2})$
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:

for f(x) to be increasing we must have f'(x)>0
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

Answer:
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)

Answer:
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)

Answer:
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)

Answer:
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

Answer:
$f(x) \text { is an increasing function on } (0,\frac{\pi }{2})$
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

Answer:

Answer:
$a\leq 0$
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

Increasing and Decreasing Functions exercise 16.2 question 36

Answer:
$b\geq 1$
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

Answer:
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

Answer:
$\left|f^{\prime}(x)<1\right| \text { for all } x \in[0,1]$
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)

Answer:
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)

Answer:
$f(x)=\sin x+|\sin x| \text { is } \\ \text { an increasing in interval } \left(0, \frac{\pi}{2}\right), f(x) \text { is decreasing in } \left(\frac{\pi}{2}, \pi\right) \\ \text { and neither increasing nor decreasing in }(\pi, 2 \pi).$
Given:
$f(x)=sin x+ |sin x |,0<x\leq 2\pi$
To prove:
We have to find the interval in which f(x) is an increasing or decreasing.
Hint:

for f(x) to be increasing we must have f'(x)>0
for f(x) to be decreasing we must have f'(x)<0

Solution:
Here we have
$f(x)=sin x+ |sin x |,0<x\leq 2\pi$
We know that
$f(x)=\{2 \sin x, \quad \text { if } 0<x \leq \pi \& 0, \text { if } \pi<x>2 \pi$
On differentiating f(x) w.r.t x we get
$f^{\prime}(x)=\{2 \cos x, \quad \text { if } 0<x \leq \pi \& 0, \quad \text { if } \pi<x<2 \pi$
$\text { The function } 2cos\: x \text { will be positive in }(0,\frac{\pi }{2})$
$\text { Hence the function is increasing in the interval }(0,\frac{\pi }{2})$
$\text { The function 2cos x will be negative between }(\frac{\pi }{2},\pi )$
$\text { Hence the function } f(x) \text { is decreasing in the interval }(\frac{\pi }{2},\pi ). \text { the value of } f'(x)=0 \text { when } \pi \: \leq x\:< 2\pi .$
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)

Answer:
$f(x) \text{ is increasing in }(0,\frac{\pi }{3}) \\ \text {and} f(x) \text { is decreasing in } (\frac{\pi }{3},\frac{\pi }{2})$
Given:
$f(x)=sin\: x (1+cos \: x ),0<x<\frac{\pi }{2}$
To find:
We have to find the intervals in which f(x) is increasing or decreasing.
Hint:

for f(x) to be increasing we must have f'(x) > 0
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,
$(2 \cos x-1)>0 \text { and }(\cos x+1)>0$
$\begin{aligned} &\Rightarrow 0<x<\frac{\pi}{3}\\ &\Rightarrow x \in\left(0, \frac{\pi}{3}\right)\\ &\text { So, } f(x) \text { is increasing in }\left(0, \frac{\pi}{3}\right) \end{aligned}$
For f(x) to be decreasing we must have, $\begin{aligned} &f^{\prime}(x)<0\\ &\Rightarrow f_{f}^{\prime}(x)=(2 \cos x-1)(\cos x+1)<0\\ &\text { This can only be possible when, }(2 \cos x-1)<0 \text { and }(\cos x+1)<0 \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{\pi}{3}<x<\frac{\pi}{2}\\ &\Rightarrow x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\\ &\text { Hence } f(x) \text { is decreasing in }\left(\frac{\pi}{3}, \frac{\pi}{2}\right) \end{aligned}$

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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

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RD Sharma Chapter-wise Solutions

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