# NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.2) Exercise 6.2

The subject specialists at the Extramarks’ website have created the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2  and the solutions for all the other chapters as well to help equip students with a clear and accurate understanding of all the topics taught in Class 12. These NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 include thorough, step-by-step explanations for all the problems found in the textbooks. These can be of great help for board examination preparations.

First Derivative Test, Maximum and Lowest Values of a Function in a Closed Interval, Approximations, Maxima and Minima, Growing and Decreasing Functions, Tangents and Normals, and many such topics are all covered in Chapter 6 of the NCERT textbook. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 contain solutions to all the questions related to these topics that are there in the second exercise of this chapter.

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Founded in 1961 as a literary, scientific, and philanthropic society under the Societies Registration Act, the National Council of Educational Research and Training (NCERT) is a stand-alone entity of the Government of India. It is responsible for establishing the curriculum for all nationwide schools that adhere to the Central Board of Secondary Education (CBSE). One of the ideal resources to aid in a student’s preparation for the CBSE exam, as well as engineering entrance exams like the JEE etc., is the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2.

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## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.2) Exercise 6.2

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 are the solutions to the second exercise of Chapter 6 of Class 12 Mathematics. In this chapter, students get to examine the derivative’s uses in a variety of academic domains, including engineering, science, social science, and many more.

For instance, students will discover how the derivative can be used to 1) calculate the rate of change of quantities, 2, find the equations of tangent and normal to a curve at a point, and 3, find turning points on the graph of a function. All of this will assist the students in identifying the points at which the largest or smallest value (locally) of a function occurs.

Additionally, students will get to use a function’s derivative to determine the intervals at which it is increasing or decreasing.

The derivative is then used to determine the approximate value of various quantities.

The chapter is divided into different exercises that contain questions that the students can practice understanding the concepts that are taught in the chapter thoroughly. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 are solutions to these questions. These are specifically for the Class 12 maths Chapter 6 Ex 6.2, but there are solutions available on the Extramarks’ website to all the exercises in the NCERT books.

### What Is The Distance Between Two Points

In each chapter there will be some topics that the students depending on their calibre might find more difficult than other topics. Finding the distance between two points can be one of those difficult topics but with the help of the right tools, the students can achieve good marks in it as well.

### Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives

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### NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2

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Q.1 Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Ans

$\begin{array}{l}\mathrm{Function}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=3\mathrm{x}+17\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=3>0\mathrm{ }\mathrm{in}\mathrm{every}\mathrm{interval}\mathrm{of}\mathrm{R}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{function}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathbf{R}\mathrm{.}\end{array}$

Q.2 Show that the function given by f(x) = e2x is strictly increasing on R

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{x}}_{\mathrm{1}}\mathrm{and}{\mathrm{x}}_{\mathrm{2}}\mathrm{be}\mathrm{any}\mathrm{two}\mathrm{numbers}\mathrm{in}\mathbf{R}.\\ \mathrm{Then},\mathrm{ }\mathrm{we}\mathrm{have}:\\ {\mathrm{x}}_{\mathrm{1}}<{\mathrm{x}}_{\mathrm{2}}⇒2{\mathrm{x}}_{\mathrm{1}}< 2{\mathrm{x}}_{\mathrm{2}}\\ ⇒ \mathrm{ }{\mathrm{e}}^{2{\mathrm{x}}_{\mathrm{1}}}<{\mathrm{e}}^{2{\mathrm{x}}_{\mathrm{2}}}\\ ⇒\mathrm{f}\left({\mathrm{x}}_{1}\right)<\mathrm{f}\left({\mathrm{x}}_{2}\right)\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathbf{R}\mathrm{.}\end{array}$

Q.3 Show that the function given by f(x) = sin x is (a) strictly increasing in ( 0 , π 2 ) , (b) strictly decreasing in ( π 2 , π ) , (c) neither increasing nor decreasing in ( 0 , π )

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{f}\left(\mathrm{x}\right) =\mathrm{sin}\mathrm{x}\mathrm{.}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cosx}\\ \left(\mathrm{a}\right)\mathrm{Since}\mathrm{for}\mathrm{each} \mathrm{x}\in \left(0,\frac{\mathrm{\pi }}{2}\right),\mathrm{ }\mathrm{cosx}>0\\ \mathrm{ }\mathrm{we}\mathrm{ }\mathrm{have}\mathrm{f}‘\left(\mathrm{x}\right)>0.\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{ }\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \left(\mathrm{b}\right)\mathrm{Since}\mathrm{for}\mathrm{each} \mathrm{x}\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right),\mathrm{ }\mathrm{cosx}<0\\ \mathrm{ }\mathrm{we}\mathrm{ }\mathrm{have}\mathrm{f}‘\left(\mathrm{x}\right)<0.\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{ }\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\\ \left(\mathrm{c}\right)\mathrm{From}\mathrm{the}\mathrm{results}\mathrm{obtained}\mathrm{in}\left(\mathrm{a}\right)\mathrm{and}\left(\mathrm{b}\right),\mathrm{it}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{f}\mathrm{is}\\ \mathrm{ }\mathrm{neither}\mathrm{increasing}\mathrm{nor}\mathrm{ }\mathrm{decreasing}\mathrm{in}\left(0,\mathrm{\pi }\right).\end{array}$

Q.4 Find the intervals in which the function f given by f(x) = 2x2 − 3x is

(a) strictly increasing (b) strictly decreasing

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^{\mathrm{2}}-3\mathrm{x}\mathrm{.}\\ \mathrm{f}‘\left(\mathrm{x}\right)=4\mathrm{x}-3\\ \mathrm{Therefore},\mathrm{f}‘\left(\mathrm{x}\right)=0⇒4\mathrm{x}-3=0\\ ⇒ \mathrm{ }\mathrm{x}=\frac{3}{4}\\ \mathrm{Now}\mathrm{the}\mathrm{point}\mathrm{x}=\frac{3}{4}\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{ }\mathrm{namely},\left(-\mathrm{\infty },\mathrm{ }\frac{3}{4}\right)\mathrm{and}\left(\frac{3}{4},\mathrm{\infty }\right).\end{array}$

$\begin{array}{l}\text{In the interval}\left(-\infty ,\text{\hspace{0.17em}}\frac{3}{4}\right),\\ \text{}\text{}\text{f’}\left(x\right)=4x-3<0\\ Therefore,\text{f is strictly deacreasing in this interval}\text{. Also, in}\\ \text{the interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\frac{3}{4},\infty \right)\text{, f’}\left(x\right)>0\text{​}\text{\hspace{0.17em}}\text{and so the function f is strictly}\\ \text{increasing in this interval}\text{.}\end{array}$

Q.5 Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing

Ans

$\begin{array}{l}\therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒6\mathrm{ }\left(\mathrm{x}-3\right)\mathrm{ }\left(\mathrm{x}+2\right)=0\\ ⇒\mathrm{x}=-2,3\\ \mathrm{The}\mathrm{points}\mathrm{x}=-2\mathrm{and}\mathrm{x}= 3\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\infty ,-2\right),\mathrm{ }\left(-2,3\right)\mathrm{and}\left(3,\infty \right).\end{array}$

$\begin{array}{l}\text{In intervals}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\infty ,-2\right)\text{and}\left(3,\infty \right),\text{\hspace{0.17em}}f‘\left(x\right)\text{\hspace{0.17em}}\text{is positive while in interval}\\ \left(-2,3\right),\text{\hspace{0.17em}}f‘\left(x\right)\text{is negative}\text{.}\\ \text{Hence, the given function (f) is strictly increasing in intervals}\\ \left(-\infty ,-2\right)\text{and}\left(3,\infty \right),\text{while function (f) is strictly decreasing in}\\ \text{interval}\text{\hspace{0.17em}}\left(-2,3\right).\end{array}$

Q.6 Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x2 + 2x − 5 (b)10 − 6x − 2x2

(c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9x − x2

(e) (x + 1)3 (x − 3)3

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{2}}+2\mathrm{x}-\mathrm{5}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒2\mathrm{x}+2=0\\ ⇒ \mathrm{x}=-1\\ \mathrm{Point}\mathrm{x}=-1\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-1\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(-1,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{interval}\left(-\mathrm{\infty },-1\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2<0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-\mathrm{\infty },-1\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{for}\mathrm{x}<-1.\\ \mathrm{In}\mathrm{interval} \left(-1,\mathrm{\infty }\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2>0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-1,\mathrm{\infty }\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}\mathrm{x}>-1.\\ \left(\mathrm{b}\right)\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{10}-6\mathrm{x}-2{\mathrm{x}}^{\mathrm{2}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{ }6-4\mathrm{x}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒-\mathrm{ }6-4\mathrm{x}=0\\ ⇒ \mathrm{x}=-\frac{3}{2}\\ \mathrm{Point}\mathrm{x}=-\frac{3}{2}\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\mathrm{intervals}\\ \mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-\frac{3}{2}\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(-\frac{3}{2},\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{interval}\left(-\mathrm{\infty },-\frac{3}{2}\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{ }6-4\mathrm{x}>0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-\mathrm{\infty },-\frac{3}{2}\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}\mathrm{x}<-\frac{3}{2}\mathrm{.}\\ \mathrm{In}\mathrm{interval} \left(-\frac{3}{2},\mathrm{\infty }\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=-6-4\mathrm{x}<0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(-\frac{3}{2},\mathrm{\infty }\right).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{for}\mathrm{x} >-\frac{3}{2}\mathrm{.}\\ \left(\mathrm{c}\right)\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=-2{\mathrm{x}}^{\mathrm{3}}-9{\mathrm{x}}^{\mathrm{2}}-12\mathrm{x}+\mathrm{1}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{ }6{\mathrm{x}}^{2}-18\mathrm{x}-12=-6\left({\mathrm{x}}^{2}+3\mathrm{x}+2\right)\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0⇒-6\left({\mathrm{x}}^{2}+3\mathrm{x}+2\right)=0\\ ⇒\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)=0\\ ⇒ \mathrm{x}=-1,-2\\ \mathrm{The}\mathrm{points}\mathrm{x}=-1\mathrm{and}\mathrm{x}=-2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-2\right),\left(-2,-1\right)\mathrm{and}\left(-1,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{intervals} \left(-\mathrm{\infty },-2\right)\mathrm{and}\left(-1,\mathrm{\infty }\right),\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)\mathrm{ }\mathrm{is}\mathrm{negative}.\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\\ \left(-\mathrm{\infty },-2\right)\mathrm{and}\left(-1,\mathrm{\infty }\right),\\ \mathrm{i}.\mathrm{e}.,\mathrm{ }\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{when}\mathrm{x}<-2\mathrm{and}\mathrm{x}>-1.\end{array}$

$\begin{array}{l}\text{In intervals}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-2,-1\right),\text{\hspace{0.17em}}f‘\left(x\right)\text{\hspace{0.17em}}\text{is positive}\text{.}\\ \text{So, function (f) is strictly increasing in interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-2,-1\right),\\ i.e.,\text{\hspace{0.17em}}\text{function (f) is strictly increasing}\text{\hspace{0.17em}}when\text{\hspace{0.17em}}-20\\ \therefore \text{f is strictly increasing in interval}\text{\hspace{0.17em}}\left(-\infty ,-\frac{9}{2}\right).\\ \text{Thus, f is strictly increasing for x <}-\frac{9}{2}\text{.}\\ \text{In interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\frac{9}{2},\infty \right),\text{\hspace{0.17em}}f‘\left(x\right)=-\text{9}-2\text{x}<0\\ \therefore \text{f is strictly decreasing in interval}\text{\hspace{0.17em}}\left(-\frac{9}{2},\infty \right).\\ \text{Thus, f is strictly decreasing for x >}-\frac{9}{2}\text{.}\end{array}$

$\begin{array}{l}\left(e\right)\text{We have,}\\ \text{f(x)}={\text{(x + 1)}}^{\text{3}}\text{(x}-{\text{3)}}^{\text{3}}\\ Differentiating\text{both sides with respect to x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f’}\left(x\right)={\left(\text{x + 1}\right)}^{\text{3}}\frac{d}{dx}{\left(\text{x}-\text{3}\right)}^{\text{3}}+{\left(\text{x}-\text{3}\right)}^{\text{3}}\frac{d}{dx}{\left(\text{x + 1}\right)}^{\text{3}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\left[By\text{product rule}\right]\\ ⇒\text{f’}\left(x\right)={\left(\text{x + 1}\right)}^{\text{3}}3{\left(\text{x}-\text{3}\right)}^{\text{2}}\frac{d}{dx}\left(\text{x}-\text{3}\right)+{\left(\text{x}-\text{3}\right)}^{\text{3}}3{\left(\text{x + 1}\right)}^{\text{2}}\frac{d}{dx}\left(\text{x + 1}\right)\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\left[By\text{chain rule}\right]\\ ⇒\text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{3}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(1-0\right)+3{\left(\text{x}-\text{3}\right)}^{\text{3}}{\left(\text{x + 1}\right)}^{\text{2}}\left(1+0\right)\\ ⇒\text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x+1+x-3\right)\\ ⇒\text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(2x-2\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)\\ Now,\text{f’}\left(x\right)=0⇒x=-1,\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\\ \text{The points x}=-1,\text{}x=1,\text{}and\text{}x=3\text{divide the real line into}\\ \text{four disjoint intervals i}\text{.e}\text{.,}\left(-\infty ,-1\right),\left(-1,\text{\hspace{0.17em}}1\right),\left(1,3\right)\text{and}\left(3,\infty \right).\\ \text{In intervals}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(-\infty ,-1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and (}-\text{1, 1),}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{f’}\left(x\right)=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)<0\\ \therefore \text{f is strictly decreasing in intervals}\text{\hspace{0.17em}}\left(-\infty ,-1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\left(-\text{1, 1}\right).\\ \text{In intervals (1, 3) and}\text{\hspace{0.17em}}\left(3,\infty \right),\text{\hspace{0.17em}}\text{f’}\left(x\right)=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)>0\\ \therefore \text{f is strictly increasing in intervals}\left(\text{1, 3}\right)\text{and}\text{\hspace{0.17em}}\left(3,\infty \right).\end{array}$

Q.7

$\begin{array}{l}\mathrm{Show}that\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{y}=\mathrm{log}\left(1+\mathrm{x}\right)-\frac{2\mathrm{x}}{2+\mathrm{x}},\phantom{\rule{0ex}{0ex}}\mathrm{x}>-1,\mathrm{is}\mathrm{an}\mathrm{increasing}\\ \mathrm{function}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{of}\mathrm{x}\mathrm{throughout}\mathrm{its}domain.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}, \mathrm{y}=\mathrm{log}\left(1+\mathrm{x}\right)-\frac{2\mathrm{x}}{2+\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{log}\left(1+\mathrm{x}\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{2+\mathrm{x}}\right)\\ =\frac{1}{1+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{x}\right)-\frac{\left(2+\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}2\mathrm{x}-2\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}\left(2+\mathrm{x}\right)}{{\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{1}{1+\mathrm{x}}×\left(0+1\right)-\frac{2\left(2+\mathrm{x}\right)-2\mathrm{x}\left(0+1\right)}{{\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{1}{1+\mathrm{x}}-\frac{4}{{\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{4+4\mathrm{x}+{\mathrm{x}}^{2}-4-4\mathrm{x}}{\left(1+\mathrm{x}\right){\left(2+\mathrm{x}\right)}^{2}}\\ =\frac{{\mathrm{x}}^{2}}{\left(1+\mathrm{x}\right){\left(2+\mathrm{x}\right)}^{2}}\\ \mathrm{Now}, \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ ⇒\mathrm{ }\frac{{\mathrm{x}}^{2}}{\left(1+\mathrm{x}\right){\left(2+\mathrm{x}\right)}^{2}}=0\\ ⇒ \mathrm{ }{\mathrm{x}}^{2}=0\left[\left(1+\mathrm{x}\right)\ne 0\mathrm{and}{\left(2+\mathrm{x}\right)}^{2}\ne 0\right]\\ ⇒ \mathrm{ }\mathrm{x}=0\\ \mathrm{Since}\mathrm{x}>-1,\mathrm{point}\mathrm{x}= 0\mathrm{divides}\mathrm{the}\mathrm{domain}\left(-1,\mathrm{\infty }\right)\mathrm{in}\mathrm{two}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,-1<\mathrm{ }\mathrm{x}< 0\mathrm{and}\mathrm{x}> 0\mathrm{.}\\ \mathrm{When}-1<\mathrm{x}< 0,\mathrm{we}\mathrm{have}:\\ \mathrm{x}< 0⇒{\mathrm{x}}^{2}>0\\ \mathrm{x}>-1⇒\mathrm{x}+2>0⇒{\left(\mathrm{x}+2\right)}^{2}>0\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^{2}}{{\left(\mathrm{x}+2\right)}^{2}}>0\\ \mathrm{Also},\mathrm{when}\mathrm{x}> 0:\\ \mathrm{x}>0⇒{\mathrm{x}}^{2}>0, \left(2+{\mathrm{x}}^{2}\right)>0\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^{2}}{{\left(\mathrm{x}+2\right)}^{2}}>0\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{increasing}\mathrm{throughout}\mathrm{this}\mathrm{domain}\mathrm{.}\end{array}$

Q.8

$\begin{array}{l}\mathrm{Find}\phantom{\rule{0ex}{0ex}}\mathrm{the}\phantom{\rule{0ex}{0ex}}\mathrm{values}\phantom{\rule{0ex}{0ex}}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{x}\phantom{\rule{0ex}{0ex}}\mathrm{for}\phantom{\rule{0ex}{0ex}}\mathrm{which}\phantom{\rule{0ex}{0ex}}\mathrm{y}={\left[\mathrm{x}\left(\mathrm{x}-2\right)\right]}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{an}\phantom{\rule{0ex}{0ex}}\mathrm{increasing}\\ \mathrm{function}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{y}={\left[\mathrm{x}\left(\mathrm{x}-\mathrm{2}\right)\right]}^{2}\\ ={\mathrm{x}}^{2}{\left(\mathrm{x}-2\right)}^{2}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}{\left(\mathrm{x}-2\right)}^{2}\\ ={\mathrm{x}}^{2}\frac{\mathrm{d}}{\mathrm{dx}}{\left(\mathrm{x}-2\right)}^{2}+{\left(\mathrm{x}-2\right)}^{2}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}\\ ={\mathrm{x}}^{2}.2\left(\mathrm{x}-2\right)\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}-2\right)+{\left(\mathrm{x}-2\right)}^{2}.2\mathrm{x}\\ =2{\mathrm{x}}^{2}\left(\mathrm{x}-2\right)\left(1-0\right)+2\mathrm{x}{\left(\mathrm{x}-2\right)}^{2}\\ =2\mathrm{x}\left(\mathrm{x}-2\right)\left(\mathrm{x}+\mathrm{x}-2\right)\\ =2\mathrm{x}\left(\mathrm{x}-2\right)\left(2\mathrm{x}-2\right)\\ =4\mathrm{x}\left(\mathrm{x}-2\right)\left(\mathrm{x}-1\right)\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0⇒\mathrm{x}=0,1,2\\ \mathrm{The}\mathrm{points}\mathrm{x}= 0,\mathrm{x}= 1,\mathrm{and}\mathrm{x}= 2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{four}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\left(-\mathrm{\infty },0\right),\left(0,1\right),\left(1,2\right),\left(2,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{intervals}\left(-\mathrm{\infty },0\right)\mathrm{and} \left(1,2\right),\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}<0\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\mathrm{ }\left(-\mathrm{\infty },0\right)\mathrm{and} \left(1,2\right).\\ \mathrm{However},\mathrm{in}\mathrm{intervals}\left(0, 1\right)\mathrm{and}\left(2,\mathrm{\infty }\right),\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}>0\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{intervals}\left(0, 1\right)\mathrm{and}\left(2,\infty \right)\mathrm{.}\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}0 <\mathrm{x}< 1\mathrm{and}\mathrm{x}> 2\mathrm{.}\end{array}$

Q.9

$\begin{array}{l}\mathrm{Prove}\phantom{\rule{0ex}{0ex}}\mathrm{that}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{y}\phantom{\rule{0ex}{0ex}}=\phantom{\rule{0ex}{0ex}}\frac{4\mathrm{sin\theta }}{\left(2+\mathrm{cos\theta }\right)}-\mathrm{\theta }\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{an}\phantom{\rule{0ex}{0ex}}\mathrm{increasing}\phantom{\rule{0ex}{0ex}}\mathrm{function}\phantom{\rule{0ex}{0ex}}\mathrm{of}\\ \mathrm{\theta }\phantom{\rule{0ex}{0ex}}\mathrm{in}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[0,\frac{\mathrm{\pi }}{2}\right].\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have}, \mathrm{y}=\frac{4\mathrm{sin\theta }}{\left(\mathrm{2}+\mathrm{cos\theta }\right)}-\mathrm{\theta }\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{\theta },\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{\mathrm{d}}{\mathrm{d\theta }}\left\{\frac{4\mathrm{sin\theta }}{\left(\mathrm{2}+\mathrm{cos\theta }\right)}-\mathrm{\theta }\right\}\\ =\frac{\left(2+\mathrm{cos\theta }\right)\frac{\mathrm{d}}{\mathrm{d\theta }}\left(4\mathrm{sin\theta }\right)-4\mathrm{sin\theta }\frac{\mathrm{d}}{\mathrm{d\theta }}\left(2+\mathrm{cos\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{\theta }\\ =\frac{4\left(2+\mathrm{cos\theta }\right)\mathrm{cos\theta }-4\mathrm{sin\theta }\left(0-\mathrm{sin\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{4\left(2+\mathrm{cos\theta }\right)\mathrm{cos\theta }+4{\mathrm{sin}}^{2}\mathrm{\theta }}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{8\mathrm{cos\theta }+4\left({\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ \mathrm{Now},\mathrm{}\frac{\mathrm{dy}}{\mathrm{d\theta }}=0\\ ⇒\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1=0⇒\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}=1\\ ⇒8\mathrm{cos\theta }+4={\left(2+\mathrm{cos\theta }\right)}^{2}\\ ⇒8\mathrm{cos\theta }+\overline{)4}=\overline{)4}+4\mathrm{cos\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }\\ ⇒{\mathrm{cos}}^{2}\mathrm{\theta }-4\mathrm{cos\theta }=0\\ ⇒\mathrm{cos\theta }\left(\mathrm{cos\theta }-4\right)=0\\ ⇒\mathrm{cos\theta }=0 \mathrm{or}\mathrm{ }\mathrm{cos\theta }=4\mathrm{ }\\ \mathrm{Since}\mathrm{cos}\mathrm{\theta }\ne \mathrm{4}\\ \therefore \mathrm{cos\theta }=0⇒\mathrm{\theta }=\frac{\mathrm{\pi }}{2}\\ \mathrm{Now}, \mathrm{ }\frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{8\mathrm{cos\theta }+4}{{\left(2+\mathrm{cos\theta }\right)}^{2}}-1\\ =\frac{8\mathrm{cos\theta }+4-{\left(2+\mathrm{cos\theta }\right)}^{2}}{{\left(2+\mathrm{cos\theta }\right)}^{2}}\\ =\frac{8\mathrm{cos\theta }+4-4-4\mathrm{cos\theta }-{\mathrm{cos}}^{2}\mathrm{\theta }}{{\left(2+\mathrm{cos\theta }\right)}^{2}}\\ =\frac{\mathrm{cos\theta }\left(4-\mathrm{cos\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}\\ \mathrm{In}\mathrm{interval} \left(0,\frac{\mathrm{\pi }}{2}\right), \mathrm{we}\mathrm{have}\mathrm{cos\theta }> 0.\mathrm{Also}, 4 >\mathrm{cos\theta }\\ ⇒\mathrm{4}-\mathrm{cos\theta }\mathrm{ }>0\mathrm{.}\\ \therefore \mathrm{cos\theta }\left(4-\mathrm{cos\theta }\right)>0\mathrm{and}\mathrm{also}{\left(2+\mathrm{cos\theta }\right)}^{2}>0\\ ⇒\frac{\mathrm{cos\theta }\left(4-\mathrm{cos\theta }\right)}{{\left(2+\mathrm{cos\theta }\right)}^{2}}>0⇒\frac{\mathrm{dy}}{\mathrm{d\theta }}>0\\ \mathrm{Therefore},\mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval} \left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{Also},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{at} \mathrm{x}=0\mathrm{and}\mathrm{x}=\frac{\mathrm{\pi }}{2}.\\ \mathrm{Hence},\mathrm{y}\mathrm{is}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{ }\left(0,\frac{\mathrm{\pi }}{2}\right).\end{array}$

Q.10

$\begin{array}{l}\mathrm{Provethatthelogarithmicfunctionisstrictly}\phantom{\rule{0ex}{0ex}}\mathrm{increasing}\\ \mathrm{on}\left(0,\mathrm{\infty }\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\\ \mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{for}\mathrm{x}> 0,\mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}>0\\ \mathrm{Hence},\mathrm{f}\left(\mathrm{x}\right) =\mathrm{log}\mathrm{x}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0,\infty \right)\mathrm{.}\\ \end{array}$

Q.11 Prove that the function f given by f(x)= x2 − x + 1 is neither strictly increasing
nor strictly decreasing on (−1, 1).

Ans

$\begin{array}{l}{\text{The given function is f(x) = x}}^{\text{2}}-\text{x + 1}\text{.}\\ Differentiating\text{with respect to x, we get}\\ \text{f’}\left(x\right)=2x-1\\ Now,\text{\hspace{0.17em}}f‘\left(x\right)=0⇒2x-1=0⇒x=\frac{1}{2}\\ \text{The point}\frac{1}{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{divides the interval}\left(-\text{1, 1}\right)\text{into two disjoint}\\ \text{intervals i}\text{.e}\text{.,}\left(-1,\text{}\frac{1}{2}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\left(\frac{1}{2}\text{\hspace{0.17em}}\text{, 1}\right)\text{.}\\ \text{Now, for interval}\left(-1,\text{}\frac{1}{2}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}f‘\left(x\right)=2x-1<0\\ \text{Therefore, f is strictly decreasing in interval}\text{\hspace{0.17em}}\left(-1,\text{}\frac{1}{2}\right).\\ \text{Now, for interval}\left(\frac{1}{2}\text{\hspace{0.17em}}\text{, 1}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}}f‘\left(x\right)=2x-1>0\\ \text{So, f is strictly increasing in interval}\left(\frac{1}{2}\text{\hspace{0.17em}}\text{, 1}\right)\text{.}\\ \text{Hence, f is neither strictly increasing nor decreasing in}\\ \text{interval}\left(-\text{1, 1}\right).\end{array}$

Q.12

$\begin{array}{l}\mathrm{Which of the following functions are}strictly\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{decreasing}\\ \mathrm{on}\left(0,\frac{\mathrm{\pi }}{2}\right)?\end{array}$

$\begin{array}{l}\left(\mathrm{A}\right)\mathrm{cosx}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{B}\right)\mathrm{cos}2\mathrm{x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}\mathrm{cos}3\mathrm{x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{D}\right)\mathrm{tan}x\end{array}$

Ans

$\begin{array}{l}\left(A\right)\text{\hspace{0.17em}}Let\text{\hspace{0.17em}}{f}_{1}\left(x\right)=\mathrm{cos}x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f’}}_{\text{1}}\left(x\right)=\frac{d}{dx}\mathrm{cos}x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\mathrm{sin}x\\ In\text{interval}\left(0,\frac{\pi }{2}\right),\text{sinx is positive in first quadrant}\text{.}\\ \text{So,}\text{\hspace{0.17em}}\text{}\text{}{\text{f’}}_{\text{1}}\left(x\right)<0\\ Therefore,\text{\hspace{0.17em}}{f}_{1}\left(x\right)=\mathrm{cos}x\text{\hspace{0.17em}}\text{is strictly decreasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \left(B\right)Let\text{\hspace{0.17em}}{f}_{2}\left(x\right)=\mathrm{cos}2x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f’}}_{\text{2}}\left(x\right)=\frac{d}{dx}\mathrm{cos}2x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-2\mathrm{sin}2x\\ In\text{interval}\left(0,\frac{\pi }{2}\right),\text{sin2x is positive in first quadrant}\text{.}\\ \text{So,}\text{\hspace{0.17em}}\text{}\text{}{\text{f’}}_{\text{2}}\left(x\right)<0\\ Therefore,\text{\hspace{0.17em}}{f}_{2}\left(x\right)=\mathrm{cos}2x\text{\hspace{0.17em}}\text{is strictly decreasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \left(C\right)Let\text{\hspace{0.17em}}{f}_{3}\left(x\right)=\mathrm{cos}3x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f’}}_{\text{3}}\left(x\right)=\frac{d}{dx}\mathrm{cos}3x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-3\mathrm{sin}3x\\ Now,\text{\hspace{0.17em}}{\text{f’}}_{\text{3}}\left(x\right)=0⇒-3\mathrm{sin}3x=0\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x=\pi ⇒x=\frac{\pi }{3}\\ \text{The point}x=\frac{\pi }{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{divides the interval}\left(0,\frac{\pi }{2}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{into two disjoint}\\ \text{intervals i}\text{.e}\text{.,}\text{\hspace{0.17em}}\left(0,\frac{\pi }{3}\right)\text{and}\left(\frac{\pi }{3},\frac{\pi }{2}\right).\\ Now,\text{​}\text{in interval}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(0,\frac{\pi }{3}\right),\\ {\text{f’}}_{\text{3}}\left(x\right)=-3\mathrm{sin}3x<0\text{}\text{}\left[\because 00\\ Therefore,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{f}}_{\text{3}}\text{is strictly increasing in interval}\text{\hspace{0.17em}}\left(\frac{\pi }{3},\frac{\pi }{2}\right).\\ {\text{Hence, f}}_{\text{3}}\text{is neither increasing nor decreasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \left(D\right)Let\text{\hspace{0.17em}}{f}_{4}\left(x\right)=\mathrm{tan}x\\ Differentiating\text{w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}{\text{f’}}_{\text{4}}\left(x\right)=\frac{d}{dx}\mathrm{tan}x\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\mathrm{sec}}^{2}x\\ In\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{int}erval\text{}\left(0,\frac{\pi }{2}\right),\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sec}x>0⇒{\mathrm{sec}}^{2}x>0\\ \therefore {f}_{4}\left(x\right)>0\\ \therefore \text{f4 is strictly increasing in interval}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \text{Therefore, functions cos x and cos 2x are strictly decreasing}\\ \text{in}\text{\hspace{0.17em}}\left(0,\frac{\pi }{2}\right).\\ \text{Hence, the correct answers are A and B}\text{.}\end{array}$

Q.13

$\begin{array}{l}\mathrm{On}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{intervals}\mathrm{is}\mathrm{the}\mathrm{function}\mathrm{f}\mathrm{given}\\ \mathrm{by}f\left(\mathrm{x}\right)={\mathrm{x}}^{100}+\mathrm{sinx}– 1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{strictly}\mathrm{decreasing}?\\ \left(\mathrm{A}\right)\left(0,1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{B}\right)\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{C}\right)\left(0,\frac{\mathrm{\pi }}{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{D}\right)\phantom{\rule{0ex}{0ex}}\mathrm{None}\mathrm{of}these\end{array}$

Ans

$\begin{array}{l}\mathrm{We} \mathrm{have},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{100}+\mathrm{sinx}-1\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=100{\mathrm{x}}^{99}+\mathrm{cosx}\\ \mathrm{f}‘\left(\mathrm{x}\right)>0\left[\because 100{\mathrm{x}}^{99}>0\mathrm{and}\mathrm{cosx}>0\mathrm{in}\mathrm{interval}\left(0,1\right)\right]\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0, 1\right)\mathrm{.}\\ \mathrm{In}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right),\\ \mathrm{cosx}<0\mathrm{and}100{\mathrm{x}}^{\mathrm{99}}>0\\ \mathrm{So}, \mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\\ \mathrm{In}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right),\\ \mathrm{cosx}>0\mathrm{and}100{\mathrm{x}}^{\mathrm{99}}>0\\ ⇒100{\mathrm{x}}^{99}+\mathrm{cosx}>0\\ \mathrm{So}, \mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{intervals}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.14 Find the least value of a such that the function f given f(x) =x2+ax+1 is strictly increasing on (1, 2).

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+\mathrm{ax}+1\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+\mathrm{a}\\ \mathrm{Now},\mathrm{ }\mathrm{function}\mathrm{f}\mathrm{will}\mathrm{be}\mathrm{increasing}\mathrm{in}\mathrm{ }\left(1,2\right),\mathrm{if}\\ \mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0⇒2\mathrm{x}+\mathrm{a}>0\\ \mathrm{ }\mathrm{x}>-\frac{\mathrm{a}}{2}\\ \mathrm{Therefore},\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{such}\mathrm{that}\\ \mathrm{ }\mathrm{x}>-\frac{\mathrm{a}}{2}, \mathrm{ }\mathrm{when}\mathrm{ }\mathrm{x}\in \left(1,2\right)\\ ⇒\mathrm{ }\mathrm{x}>-\frac{\mathrm{a}}{2}, \mathrm{ }\left(\mathrm{when}\mathrm{ }1<\mathrm{x}<2\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{for}\mathrm{f}\mathrm{to}\mathrm{be}\mathrm{increasing}\mathrm{on}\left(1, 2\right)\mathrm{is}\\ \mathrm{given}\mathrm{by},\\ \mathrm{–}\frac{\mathrm{a}}{2}=1⇒\mathrm{a}=-2\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{is}-2.\end{array}$

Q.15

$\begin{array}{l}\mathrm{Let}\mathrm{I}\mathrm{be}\mathrm{any}\mathrm{interval}\mathrm{disjoint}\mathrm{from}\left(-1,1\right).\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{function}\\ \mathrm{fgiven}by\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\frac{1}{\mathrm{x}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{is}\mathrm{strictly}\mathrm{increasing}on\phantom{\rule{0ex}{0ex}}\mathrm{I}.\end{array}$

Ans

$\begin{array}{l}\mathrm{we}\mathrm{}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\frac{1}{\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^{2}}\\ \mathrm{Now},\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)=0⇒1-\frac{1}{{\mathrm{x}}^{2}}=0\\ ⇒1=\frac{1}{{\mathrm{x}}^{2}}⇒\mathrm{x}=±1\\ \mathrm{The}\mathrm{points}\mathrm{x}=1\mathrm{and}\mathrm{x}=-1\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{in}\mathrm{three}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{ }\left(-\mathrm{\infty },-1\right),\left(-1,1\right) \mathrm{and} \left(1,\mathrm{\infty }\right).\\ \mathrm{In}\mathrm{interval}\left(-1, 1\right),\mathrm{it}\mathrm{is}\mathrm{noticed}\mathrm{that}:\\ -1<\mathrm{x}<1⇒{\mathrm{x}}^{2}<1\\ ⇒1<\frac{1}{{\mathrm{x}}^{2}}, \mathrm{x}\ne 0\\ ⇒1-\frac{1}{{\mathrm{x}}^{2}}<0, \mathrm{x}\ne 0\\ \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^{2}}<0 \mathrm{on}\mathrm{}\left(-1,1\right)~\left\{0\right\}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{on} \left(-1,1\right)~\left\{0\right\}\\ \mathrm{In}\mathrm{intervals}\mathrm{ }\left(-\mathrm{\infty },-1\right) \mathrm{and} \left(1,\mathrm{\infty }\right),\mathrm{it}\mathrm{is}\mathrm{seen}\mathrm{that}\\ \mathrm{x}<-1 \mathrm{or}\mathrm{x}>1\\ ⇒{\mathrm{x}}^{\mathrm{2}}>\mathrm{1}⇒1>\frac{1}{{\mathrm{x}}^{2}}\\ ⇒1-\frac{1}{{\mathrm{x}}^{2}}>0\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^{2}}>0\mathrm{on}\mathrm{ }\left(-\mathrm{\infty },-1\right) \mathrm{and} \left(1,\mathrm{\infty }\right).\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathrm{ }\left(-\mathrm{\infty },-1\right) \mathrm{and} \left(1,\mathrm{\infty }\right).\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{I}\mathrm{disjoint}\mathrm{from}\left(-1, 1\right)\mathrm{.}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Provethatthefunctionfgivenbyf}\left(\mathrm{x}\right) =\mathrm{logsinxisstrictly}\\ \mathrm{increasingon}\left(0,\frac{\mathrm{\pi }}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{andstrictlydecreasing}on\phantom{\rule{0ex}{0ex}}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{logsinx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{sinx}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}\left[\mathrm{By}\mathrm{ }\mathrm{chain}\mathrm{rule}\right]\\ \mathrm{ }=\frac{1}{\mathrm{sinx}}\mathrm{cosx}=\mathrm{cotx}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cotx}>0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cotx}<0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\end{array}$

Q.17 Prove that the function f given by f(x) = logcos x is strictly decreasing on ( 0 , π 2 ) and strictly increasing on ( π 2 , π ) .

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{logcosx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{cosx}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}\left[\mathrm{By}\mathrm{ }\mathrm{chain}\mathrm{rule}\right]\\ \mathrm{ }=\frac{1}{\mathrm{cosx}}×-\mathrm{sinx}=-\mathrm{tanx}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{tanx}<0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(0,\frac{\mathrm{\pi }}{2}\right).\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{tanx}>0 \mathrm{ }\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{function}\mathrm{in}\mathrm{interval}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\end{array}$

Q.18

$\begin{array}{l}\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{f}\mathrm{given}\mathrm{byf}\left(\mathrm{x}\right) ={\mathrm{x}}^{2}-3{\mathrm{x}}^{2}+3\mathrm{x}-100\\ \mathrm{is}\mathrm{increasingin}R.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+3\mathrm{x}-100\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=3{\mathrm{x}}^{2}-6\mathrm{x}+3\\ =3\left({\mathrm{x}}^{2}-2\mathrm{x}+1\right)\\ =3{\left(\mathrm{x}-1\right)}^{2}\\ \mathrm{For}\mathrm{any}\mathrm{x}\in \mathbf{R}, {\left(\mathrm{x}-1\right)}^{2}>0.\\ \mathrm{Thus},\mathrm{f}‘\left(\mathrm{x}\right) \mathrm{is}\mathrm{always}\mathrm{positive}\mathrm{in}\mathbf{R}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{increasing}\mathrm{in}\mathbf{R}\mathrm{.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{The}\phantom{\rule{0ex}{0ex}}\mathrm{interval}\phantom{\rule{0ex}{0ex}}\mathrm{in}\phantom{\rule{0ex}{0ex}}\mathrm{which}\phantom{\rule{0ex}{0ex}}\mathrm{y}={\mathrm{x}}^{2}{\mathrm{e}}^{-\mathrm{x}}\phantom{\rule{0ex}{0ex}}\mathrm{is}\phantom{\rule{0ex}{0ex}}\mathrm{increasing}\phantom{\rule{0ex}{0ex}}\mathrm{is}\\ \left(\mathrm{A}\right)\left(-\mathrm{\infty },\mathrm{\infty }\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{B}\right)\left(-2,0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{C}\right)\left(2,\mathrm{\infty }\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{D}\right)\left(0,2\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{y}={\mathrm{x}}^{2}{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^{2}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^{2}\\ \mathrm{ }=-{\mathrm{x}}^{2}{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}\left(2\mathrm{x}\right)\\ \mathrm{ }={\mathrm{e}}^{-\mathrm{x}}\left(-{\mathrm{x}}^{2}+2\mathrm{x}\right)\\ \mathrm{Now},\mathrm{ }\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ ⇒{\mathrm{e}}^{-\mathrm{x}}\left(-{\mathrm{x}}^{2}+2\mathrm{x}\right)=0\\ ⇒\mathrm{x}=0,2\left[\because {\mathrm{e}}^{-\mathrm{x}}\ne 0\right]\\ \mathrm{The}\mathrm{points}\mathrm{x}=0\mathrm{and}\mathrm{x}=2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\left(-\mathrm{\infty },0\right),\left(0,2\right)\mathrm{ }\mathrm{and} \left(2,\mathrm{\infty }\right).\\ \mathrm{f}‘\left(\mathrm{x}\right)<0\mathrm{in}\mathrm{intervals}\left(-\mathrm{\infty },0\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(2,\mathrm{\infty }\right).\\ \mathrm{So},\mathrm{ }\mathrm{f}\left(\mathrm{x}\right)\mathrm{}\mathrm{is}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\left(-\mathrm{\infty },0\right)\mathrm{ }\mathrm{and}\mathrm{ }\left(2,\mathrm{\infty }\right).\\ \mathrm{Now},\mathrm{ }\mathrm{f}‘\left(\mathrm{x}\right)>0\mathrm{in}\mathrm{interval}\left(0,2\right)\\ \mathrm{So},\mathrm{ }\mathrm{f}\left(\mathrm{x}\right)\mathrm{}\mathrm{is}\mathrm{increasing}\mathrm{in}\mathrm{intervals}\left(0,2\right)\mathrm{ }\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\left(0, 2\right)\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$