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    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.

    This Story also Contains
    1. RD Sharma Class 12 Solutions Chapter 16 Increasing and Decreasing Functions - Other Exercise
    2. Increasing and Decreasing Functions Excercise16.2
    3. RD Sharma Chapter-wise Solutions

    RD Sharma Class 12th Exercise 16.2 is an excellent study material and a super useful course book for class 12 . It is ideal for students preparing for challenging tests and exams like JEE mains. To get an excellent score in their exams, practice, students need to have a clear idea of all concepts. The RD Sharma Class 12 solutions will help students in this endeavor to learn the essential concepts of . The staff at CAREER360 have created some amazing techniques to help students accelerate their speed and complete answers faster in their exam. These techniques also help in getting accurate or correct answers and further develops their chance of being on the leaderboards.

    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:
    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.
    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\inR
    Given:
    f(x)=x-sin\: x
    To prove:
    We have to show that f(x) is increasing for all x\inR
    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\inR
    \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\inR

    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:
    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

    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, 4th quadrant to 1st 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:
    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

    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:
    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

    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 \inR

    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
    Edit Q



    Question:29

    Increasing and Decreasing Functions exercise 16.2 question 29

    Answer:

    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 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:
    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

    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
    Edit Q



    Question:35

    Increasing and Decreasing Functions exercise 16.2 question 35

    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 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:
    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<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:
    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<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}

    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

    Frequently Asked Question (FAQs)

    1. How can I download Sharma class 12th exercise 16.2 pdf?

    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.

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    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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    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Social Media Manager

    A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

    2 Jobs Available
    Choreographer

    The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

    2 Jobs Available
    Talent Director

    Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots. 

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Content Writer

    Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

    2 Jobs Available
    Reporter

    Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

    2 Jobs Available
    Linguist

    Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

    Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian. 

    2 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Engineer

    A career as Production Engineer is crucial in the manufacturing industry. He or she ensures the functionality of production equipment and machinery to improve productivity and minimize production costs in order to drive revenues and increase profitability. 

    2 Jobs Available
    Automation Test Engineer

    An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

    2 Jobs Available
    Product Designer

    Individuals who opt for a career as product designers are responsible for designing the components and overall product concerning its shape, size, and material used in manufacturing. They are responsible for the aesthetic appearance of the product. A product designer uses his or her creative skills to give a product its final outlook and ensures the functionality of the design. 

    Students can opt for various product design degrees such as B.Des and M.Des to become product designers. Industrial product designer prepares 3D models of designs for approval and discusses them with clients and other colleagues. Individuals who opt for a career as a product designer estimate the total cost involved in designing.

    2 Jobs Available
    R&D Personnel

    A career as R&D Personnel requires researching, planning, and implementing new programs and protocols into their organization and overseeing new products’ development. He or she uses his or her creative abilities to improve the existing products as per the requirements of the target market.

    2 Jobs Available
    Commercial Manager

    A Commercial Manager negotiates, advises and secures information about pricing for commercial contracts. He or she is responsible for developing financial plans in order to maximise the business's profitability.

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Computer System Analyst

    Individuals in the computer systems analyst career path study the hardware and applications that are part of an organization's computer systems, as well as how they are used. They collaborate closely with managers and end-users to identify system specifications and business priorities, as well as to assess the efficiency of computer systems and create techniques to boost IT efficiency. Individuals who opt for a career as a computer system analyst support the implementation, modification, and debugging of new systems after they have been installed.

    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

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