20. Discuss the continuity of the following functions: () f(x)=sin x+ cos x () f(x)=sin x− cos x() f(x)=sin x. cos x

20. (a) Given f(x) = sin x + cos x (b). Given, f(x) = sin x cos x (c). Given, f(x) = sin x .cos x. Let g(x) = sin x and h(x) = cos x. If g or h are continuous f x then g + h g h g h are also continuous. As g(x) = sin x is defined for all real number x. Let c∈? , and putting x = c + h. we see that as x→c,h→0. Then g(c) = sin c limx→c g(x) = limx→c sin x = limh→0 sin (c + h). = limh→0 (sin c cos h + cos c sin h ) = sin c. cos 0 + cos c. sin 0 = sin c 1 + 0 = sin c = g (c) So, g is continuous x R. And h (c) = cos c = limh→0 g(x) = limx→c sin x = limx→c cos (c + h) = cos c .cos 0 sin c. sin 0 = cos c .1 0. = cos c = h(c). As g and h are continuous x R. The f x f(x)

22. Find all points of discontinuity of f , where f(x)= {sinxx, if x<0x+1, if x≥0

22. Given f(x) = {sinxx, if x 0 f(c) = c + 1 limx→c f(x) = limx→c x + 1 = c + 1 = f(c) So, f is continuous for x > 0. For x = 0. L.H.L. = limx→0−f(x)=limx→0−sinxx=1. R.H.S. = limx→0+f(x)=limx→0+x+1=0+1=1 And f(0) = 0 + 1 = 1 L.H.L = R.H.L. = f(0) So, f is continuous at x = 1. Hence, discontinuous point of x does not exit.

59. Kindly consider the following exsinx

59. Let y = exsinx ∴dydx=sinxdexdx−exddxsinxsin2x =sinxex−excosxsin2x =ex (sinx−cosx)sin−2x .

NCERT Solutions for Class 12 Maths Chapter 5: Continuity & Differentiability Download PDF

Q: 3. Examine the following functions for continuity.

3. (a) Given, f (x) = x 5. The given f x n is a polynernial f xn and as every pohyouraial f xn is continuous in its domain R we conclude that f (x) is continuous. (b). Given, f(x) = 1x−5,x≠5 For any a =3(5x)3cos2x[cos2xx−2sin2xlog5x] {5}, =x2×1×x−3ddx(x−3)+log(x−3)⋅2x 1(x−5)=1a−5. and f(a) = 1a−5 i e, f(x)=−(x−1)+[−(x−2)]=−x+1−x+2=3−2x. f(x) = f(a). Hence f is continuous in its domain. (c) Given, f(x) = x2−25x+5, x≠−5 For any a ? { 5} limx→af(x)=limx→a x2−25x+5=a2−25a+5=(a−5)(a+5)a+5 = a 5 And f(a) = a2−25a+5=(a−5)(a+5)a+5. = a 5 ∴limx→a f(x) = f(a). So, f is continuous in its domain. (d) Given f (a) = |x−5|={x−5, if x−5>0⇒x≥5−(x−5) if x−5 5. f (c) = (x 5) = c 5 limx→c f(x) = limx→c (x 5) = c 5. ∴ f(c) = limx→c f(x) So, f is continuous. For x = c = 5, f (5) = 5 5 = 0 limx→5 f(a) = limx→5 (x 5) = (5 5) = 0 limx→5+ f(x) = limx→5+ (x 5) = 5 5 = 0 ∴ x→5+ + (x) = sinx→c+ + (x) = f (c) Hence f is continuous.

Q: 22. Find all points of discontinuity of f , where f(x)= {sinxx, if x<0x+1, if x≥0

22. Given f(x) = {sinxx, if x 0 f(c) = c + 1 limx→c f(x) = limx→c x + 1 = c + 1 = f(c) So, f is continuous for x > 0. For x = 0. L.H.L. = limx→0−f(x)=limx→0−sinxx=1. R.H.S. = limx→0+f(x)=limx→0+x+1=0+1=1 And f(0) = 0 + 1 = 1 L.H.L = R.H.L. = f(0) So, f is continuous at x = 1. Hence, discontinuous point of x does not exit.

Q: 77. Kindly consider the following xsin x+(sin x)cos x

77. Let y = x sin x + (sin x) cos x Putting u = x sin x and v = (sin x) cos x we have, y = u + v ∴dydx=dydx+dvdx _____ (1) As u = x sin x Taking log, Log u = sin x log x Differentiating w r t ‘x’, , 14dydx = sin x ddx log x + log x ddx sin x = sinxx+ cos x log x dydx=u[sinxx+cosxlogx] = x sin x [sinxx+cosxlogx]. And v = (sin x) cos x Taking log, Log v = cos x log (sin x). Differentiating w r t ‘x’, 1vdv dx = cos x ddx log (sin x) + log (sin x) ddx cos x =cosxsinxddx sin x- sin x log (sin x) = cot x cos x- sin x log (sin x) ⇒dvdx = v [cot x cos x - sin x log (sin x)] = (sin x) cos x [cot x cos x- sin x log (sin x)] Hence, eqn (1) becomes dydx=xsinx[sinxx+cosxlogx] + (sin x) cos x [cot x cos x- sin x log (sin x)]

Updated on Aug 4, 2025 15:10 IST

By Pallavi Pathak, Assistant Manager Content

Class 12 Continuity and Differentiability introduces concepts of continuity, differentiability and relations between them. It also covers the differentiation of inverse trigonometric functions, exponential, and logarithmic functions. It also includes the Logarithmic Differentiation, Exponential and Logarithmic Functions, and Derivatives of Functions in Parametric Forms.
Continuity and Differentiability Class 12 is an important chapter of Class 12 Maths. It introduces some fundamental theorems related to the topic and the exponential and logarithmic functions which lead to powerful techniques of differentiation. Students preparing for the Class 12 CBSE Board exams and other entrance exams like JEE Main can prepare from this solution.
Shiksha offers NCERT solutions of all chapters of Physics, Chemistry and Maths of Class 11 and Class 12. These solutions are created by the subject matter experts. Students also get the key topics and free PDFs for each chapter.

Table of contents

Key Concepts of Continuity and Differentiability – Class 12 Guide
Class 12 Chapter 5 Maths Solutions – Continuity and Differentiability PDF
Class 12 Continuity and Differentiability : Key Topics, Weightage
Continuity and Differentiability Class 12 Maths Important Formulas
Class 12 Math Continuity and Differentiability Exercise-wise Solutions
Class 12 Math Continuity and Differentiability Exercise 5.1 Solutions

Key Concepts of Continuity and Differentiability – Class 12 Guide

Here are a few highlights of the Continuity and Differentiability Class 12 NCERT Solutions:

Continuity and Differentiability states that a real-valued function at a point is called continuous if at that point the limit of the function equals the value of the function. If a function is continuous on the whole of its domain, then it is a continuous function.
The addition, difference, product and quotient of continuous functions are continuous. If f and g are continuous functions then - $(f + g) (x) = f (x) + g (x) is continuous$ $(f \cdot g) (x) = f (x) \cdot g (x) is continuous$ $\frac{f}{g} (x) = \frac{f}{(} (wherever g (x) \neq 0) is continuous$
These are some of the standard derivatives (in appropriate domains):

$\frac{d}{d x} (\sin^{- 1} x) = \frac{1}{\sqrt{1 - x^{2}}}$

$\frac{d}{d x} (\cos^{- 1} x) = - \frac{1}{\sqrt{1 - x^{2}}}$

$\frac{d}{d x} (\tan^{- 1} x) = \frac{1}{1 + x^{2}}$

$\frac{d}{d x} (e^{x}) = e^{x}$

$\frac{d}{d x} (\log x) = \frac{1}{x}$

To get a comprehensive NCERT Solutions for Class 12 Maths with free PDFs and important topics, explore - Class 12 Maths NCERT Solutions.

Class 12 Chapter 5 Maths Solutions – Continuity and Differentiability PDF

Students can download the Continuity and Differentiability Class 12 Solutions PDF from the link given here. The PDF provides detailed and ideal study material for exam preparation. By preparing from the PDF, students can get a high score in the CBSE Board exam and competitive exams like JEE Main.
Class 12 Math Chapter 5 Continuity and Differentiability Solution: FREE PDF Download

Class 12 Continuity and Differentiability : Key Topics, Weightage

Class 12 Continuity and Differentiability is also important for the JEE Main entrance test, as the chapter is a foundational topic for the larger calculus section. Here are the topics covered in this chapter:

Exercise	Topics Covered
5.1	Introduction
5.2	Continuity
5.3	Differentiability
5.4	Exponential and Logarithmic Functions
5.5	Logarithmic Differentiation
5.6	Derivatives of Functions in Parametric Forms
5.7	Second Order Derivative

Continuity and Differentiability Class 12 Weightage in JEE Mains

Exam	Number of Questions	Weightage
JEE Main	3 questions	10%

Continuity and Differentiability Class 12 Maths Important Formulas

Continuity and Differentiability Important Formulae for CBSE and Competitive Exams

Students can check the basic formulae used in the Continuity and Differentiability class 12 solutions. check below;

Basic Derivatives
$\frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (e^{x}) = e^{x}, \frac{d}{d x} (\ln x) = \frac{1}{x}, \frac{d}{d x} (\sin x) = \cos x, \frac{d}{d x} (\cos x) = - \sin x$
Inverse Trigonometric Functions
$\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}$ $\frac{d}{d x} (\tan^{- 1} x) = \frac{1}{1 + x^{2}}, \frac{d}{d x} (\cot^{- 1} x) = - \frac{1}{1 + x^{2}}$
Product Rule
$\frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x}$
Quotient Rule
$\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$
Chain Rule
$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$
Logarithmic Differentiation
For $y = f(x)^{g(x)}$
$\ln y = g (x) \ln f (x), \frac{d y}{d x} = y [\frac{g^{'} (x) \ln f (x) + g (x) \frac{f^{'} (x)}{f (x)}}{}]$
Mean Value Theorem
$f'(c) = \frac{f(b) - f(a)}{b - a}, \quad \text{where } c \in (a, b)$

Class 12 Math Continuity and Differentiability Exercise-wise Solutions

Chapter 5 Continuity and Differentiability deals with concepts of continuity, differentiability, and their applications. Continuity and Differentiability play a critical part in calculus, as a building block of the foundation for advanced topics like integration and differential equations. The exercise-wise Continuity and Differentiability NCERT solutions will help students understand all concepts with their examples. These solutions will help them develop problem-solving skill related to continuity at a point, derivatives, logarithmic differentiation, and the application of theorems like Rolle’s and Lagrange’s Mean Value Theorem. Students can check the exercise-wise solutions for Continuity and Differentiability below;

Class 12 Math Continuity and Differentiability Exercise 5.1 Solutions

Class 12 Continuity and Differentiability Exercise 5.1 focuses on fundamental concepts important to building a strong foundation. Exercise 5.1 deals with topics such as continuity at a point, differentiability, and application of differentiability-related formulas. Class 12 Math Exercise 5.1 Solutions includes 34 Questions (Short Answers Type) with conceptual explanation. students can check the complete solutions for exercise 5.1 below;

Exercise – 5.1

Q1.Prove that the function f(x) = 5x 3f is continuous at x = 0, at x = – 3 and at x = 5.

A.1.Given, f(x) = 5x 3

At x = 0, $\underset{x \to 0}{l i m} f (x) = \underset{x \to 0}{l i m}$ 5x 3 = 5 0 3 = 3.

So f is continuous at x = 1.

At x = 3, $π + h$ 5x 3 = 5 ( 3) 3 = 15 3

= 18.

So f is continuous at x = 3.

At x = 5, $\forall x \in ℝ$ .5x 3 = 5.5 3 = 25 3 = 22.

So, f is continuous at x = 5.

Q2.Examine the continuity of the function f(x) = 2x2 1

A.2. Given, f(x) = 2x2 1

At x = 3

Lim f(x) = $\frac{d y}{d x} = \frac{- (3 x^{2} + 2 x y + y^{2})}{(x^{2} + 2 x y + 3 y^{2}) .}$ 2(3)2 1 = 18 1 = 17.

So, f is continuous at x = 3.

Q3.Examine the following functions for continuity.

$\begin{array}{l} () f (x) = x - 5 () f (x) = \frac{1}{x - 5}, x \neq 5 \\ () f (x) = \frac{x^{2} - 25}{x + 5}, x \neq - 5 () f (x) = | x - 5 | \end{array}$

Q3. Examine the following functions for continuity.

A.3. (a) Given, f (x) = x 5.

The given f x n is a polynernial f xn and as every pohyouraial f xn is continuous in its domain R we conclude that f (x) is continuous.

(b). Given, f(x) = $\frac{1}{x - 5}, x \neq 5$

For any a $= 3 {(5 x)}^{3 c o s 2 x} [\frac{c o s 2 x}{x} - 2 s i n 2 x l o g 5 x]$ {5},

$= \frac{x^{2} \times 1 \times}{x - 3} \frac{d}{d x} (x - 3) + l o g (x - 3) \cdot 2 x$ $\frac{1}{(x - 5)} = \frac{1}{a - 5} .$

and f(a) = $\frac{1}{a - 5}$

i e, $f (x) = - (x - 1) + [- (x - 2)] = - x + 1 - x + 2 = 3 - 2 x .$ f(x) = f(a).

Hence f is continuous in its domain.

(c) Given, f(x) = $\frac{x^{2} - 25}{x + 5}, x \neq - 5$

For any a $ℝ$ { 5}

$\underset{x \to a}{l i m} f (x) = \underset{x \to a}{l i m}$ $\frac{x^{2} - 25}{x + 5} = \frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5}$ = a 5

And f(a) = $\frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5} .$

= a 5

$∴ \underset{x \to a}{l i m}$ f(x) = f(a).

So, f is continuous in its domain.

(d) Given f (a) = $| x - 5 | = {\begin{matrix} x - 5 & , if x - 5 > 0 \Rightarrow x \geq 5 \\ - (x - 5) & if x - 5 < 0 \Rightarrow x < 5 . \end{matrix}$

For x = c < 5.

f (c) = (c 5) = 5 c.

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = (c 5) = 5 c.

∴ f(c) = $\underset{x \to c}{l i m}$ f(x).

So f is continuous.

For x = c > 5.

f (c) = (x 5) = c 5

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = c 5.

∴ f(c) = $\underset{x \to c}{l i m}$ f(x)

So, f is continuous.

For x = c = 5,

f (5) = 5 5 = 0

$\underset{x \to 5^{}}{l i m}$ f(a) = $\underset{x \to 5^{}}{l i m}$ (x 5) = (5 5) = 0

$\underset{x \to 5^{+}}{l i m}$ f(x) = $\underset{x \to 5^{+}}{l i m}$ (x 5) = 5 5 = 0

∴ $\underset{x \to 5^{+}}{}$ + (x) = $\underset{x \to c^{+}}{s i n}$ + (x) = f (c)

Hence f is continuous.

Q4. Prove that the function $f (x) = x^{n}$ continuous at $x = n,$ where n is a positive integer.

A.4.Given, f(x) = x n > n = positive.

At x = 2,

(x) = n n.

$\underset{x \to n}{l i m}$ f(x) = $\underset{x \to n}{l i m}$ x n = n n

∴ $\underset{x \to n}{l i m}$ f(x) = f(x)

So f is continuous at x = n.

Commonly asked questions

20. Discuss the continuity of the following functions:

$\begin{array}{l} () f (x) = s i n x + c o s x () f (x) = s i n x - c o s x \\ () f (x) = s i n x . c o s x \end{array}$

20. (a) Given f(x) = sin x + cos x

(b). Given, f(x) = sin x cos x

(c). Given, f(x) = sin x .cos x.

Let g(x) = sin x and h(x) = cos x.

If g or h are continuous f x then

g + h

g h

g h are also continuous.

As g(x) = sin x is defined for all real number x.

Let $c \in ?$ , and putting x = c + h. we see that as $x \to c, h \to 0 .$

Then g(c) = sin c

$\underset{x \to c}{l i m}$ g(x) = $\underset{x \to c}{l i m}$ sin x = $\underset{h \to 0}{l i m}$ sin (c + h).

= $\underset{h \to 0}{l i m}$ (sin c cos h + cos c sin h )

= sin c. cos 0 + cos c. sin 0

= sin c 1 + 0

= sin c

= g (c)

So, g is continuous x R.

And h (c) = cos c

= $\underset{h \to 0}{l i m}$ g(x) = $\underset{x \to c}{l i m}$ sin x = $\underset{x \to c}{l i m}$ cos (c + h)

= cos c .cos 0 sin c. sin 0

= cos c .1 0.

= cos c = h(c).

As g and h are continuous x R.

The f x f(x)

3. Examine the following functions for continuity.

3. (a) Given, f (x) = x 5.

The given f x n is a polynernial f xn and as every pohyouraial f xn is continuous in its domain R we conclude that f (x) is continuous.

(b). Given, f(x) = $\frac{1}{x - 5}, x \neq 5$

For any a $= 3 {(5 x)}^{3 c o s 2 x} [\frac{c o s 2 x}{x} - 2 s i n 2 x l o g 5 x]$ {5},

$= \frac{x^{2} \times 1 \times}{x - 3} \frac{d}{d x} (x - 3) + l o g (x - 3) \cdot 2 x$ $\frac{1}{(x - 5)} = \frac{1}{a - 5} .$

and f(a) = $\frac{1}{a - 5}$

i e, $f (x) = - (x - 1) + [- (x - 2)] = - x + 1 - x + 2 = 3 - 2 x .$ f(x) = f(a).

Hence f is continuous in its domain.

(c) Given, f(x) = $\frac{x^{2} - 25}{x + 5}, x \neq - 5$

For any a $?$ { 5}

$\underset{x \to a}{l i m} f (x) = \underset{x \to a}{l i m}$ $\frac{x^{2} - 25}{x + 5} = \frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5}$ = a 5

And f(a) = $\frac{a^{2} - 25}{a + 5} = \frac{(a - 5) (a + 5)}{a + 5} .$

= a 5

$∴ \underset{x \to a}{l i m}$ f(x) = f(a).

So, f is continuous in its domain.

(d) Given f (a) = $| x - 5 | = {\begin{matrix} x - 5 & , if x - 5 > 0 \Rightarrow x \geq 5 \\ - (x - 5) & if x - 5 < 0 \Rightarrow x < 5 . \end{matrix}$

For x = c < 5.

f (c) = (c 5) = 5 c.

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = (c 5) = 5 c.

∴ f(c) = $\underset{x \to c}{l i m}$ f(x).

So f is continuous.

For x = c > 5.

f (c) = (x 5) = c 5

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ (x 5) = c 5.

∴ f(c) = $\underset{x \to c}{l i m}$ f(x)

So, f is continuous.

For x = c = 5,

f (5) = 5 5 = 0

$\underset{x \to 5^{}}{l i m}$ f(a) = $\underset{x \to 5^{}}{l i m}$ (x 5) = (5 5) = 0

$\underset{x \to 5^{+}}{l i m}$ f(x) = $\underset{x \to 5^{+}}{l i m}$ (x 5) = 5 5 = 0

∴ $\underset{x \to 5^{+}}{}$ + (x) = $\underset{x \to c^{+}}{s i n}$ + (x) = f (c)

Hence f is continuous.

22. Find all points of discontinuity of $f$ , where

$f (x) =$ ${\begin{array}{l} \frac{s i n x}{x}, & if x < 0 \\ x + 1, & if x \geq 0 \end{array}$

22. Given f(x) = ${\begin{array}{l} \frac{s i n x}{x}, & if x < 0 . \\ x + 1, & if x \geq 0 . \end{array}$

For x = c < 0,

f(c) = $\frac{s i n c}{c}$

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ $\frac{s i n x}{x} = \frac{s i n c}{c} = f (c)$

So, f is continuous for x < 0

For x = c > 0

f(c) = c + 1

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ x + 1 = c + 1 = f(c)

So, f is continuous for x > 0.

For x = 0.

L.H.L. = $\underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0^{-}}{l i m} \frac{s i n x}{x} = 1 .$

R.H.S. = $\underset{x \to 0^{+}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} x + 1 = 0 + 1 = 1$

And f(0) = 0 + 1 = 1

L.H.L = R.H.L. = f(0)

So, f is continuous at x = 1.

Hence, discontinuous point of x does not exit.

59. Kindly consider the following

$\frac{e^{x}}{s i n x}$

59. Let y = $\frac{e^{x}}{s i n x}$

$∴ \frac{d y}{d x} = \frac{s i n x \frac{d e^{x}}{d x} - e^{x} \frac{d}{d x} s i n x}{s i n^{2} x}$

$= \frac{s i n x e^{x} - e^{x} c o s x}{s i n^{2} x}$

$= e^{x} \frac{(s i n x - c o s x)}{s i n^{- 2} x .}$

77. Kindly consider the following

$x^{s i n x} + {(s i n x)}^{c o s x}$

77. Let y = x ^{sin x} + (sin x) cos x

Putting u = x ^{sin x} and v = (sin x) cos x we have,

y = u + v

$∴ \frac{d y}{d x} = \frac{d y}{d x} + \frac{d v}{d x}$ _____ (1)

As u = x ^{sin x}

Taking log,

Log u = sin x log x

Differentiating w r t ‘x’,

$\frac{1}{4} \frac{d y}{d x}$ = sin x $\frac{d}{d x}$ log x + log x $\frac{d}{d x}$ sin x

= $\frac{s i n x}{x}$ + cos x log x

$\frac{d y}{d x} = u [\frac{s i n x}{x} + c o s x l o g x]$

= x sin x $[\frac{s i n x}{x} + c o s x l o g x] .$

And v = (sin x) cos x

Taking log,

Log v = cos x log (sin x).

Differentiating w r t ‘x’,

$\frac{1}{v} d v$ $d x$ = cos x $\frac{d}{d x}$ log (sin x) + log (sin x) $\frac{d}{d x}$ cos x

$= \frac{c o s x}{s i n x} \frac{d}{d x}$ sin x- sin x log (sin x)

= cot x cos x- sin x log (sin x)

$\Rightarrow \frac{d v}{d x}$ = v [cot x cos x - sin x log (sin x)]

= (sin x) cos x [cot x cos x- sin x log (sin x)]

Hence, eqn (1) becomes

$\frac{d y}{d x} = x^{s i n x} [\frac{s i n x}{x} + c o s x l o g x]$ + (sin x) cos x [cot x cos x- sin x log (sin x)]

134. Kindly consider the following

134. Given, $x = a (c o s t + t s i n t)$ and $y = a (s i n t - t c o s t) .$

Differentiating w r t. ‘t’ we get,

$\frac{d x}{d t} = a \frac{d}{d t} (c o s t + t s i n t) .$

$= a (- s i n t + t \frac{d}{d t} s i n t + s i n t \frac{d t}{d t}) .$

$= a (- s i n t + t c o s t + s i n t) = a t c o s t$

$\frac{d y}{d t} = \frac{a d}{d t} (c s i n t - t c o s t)$

$= a (c o s t - t \frac{d}{d t} c o s t - c o t \frac{d t}{d t})$

$= a (c o s t + t s i n t - c o s t) = a t s i n t$

$\frac{d y}{d x} = \frac{d y / d t}{d x / a t} = \frac{a t s i n t}{a t c o s t} = t a n t$

So, $\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (t a n t) = \frac{d}{d t} (t a n t) \cdot \frac{d t}{d x} .$

$= s e c^{2} t \times \frac{d t}{d x} .$

$= s e c^{2} t \times \frac{1}{(d x / d t)}$

$= s e c^{2} t \times \frac{1}{a t c o s t}$

$= \frac{s e c^{3} t}{a t}$

123. Kindly consider the following

123. Kindly go through the solution

85. Kindly consider the following

85. Let y = (x2- 5x + 8) (x3 + 7x + 9) ____ (1)

(i) by product rule

$\frac{d y}{d x} = (x^{2} - 5 x + 8) \frac{d}{d x} (x^{3} + 7 x + 9) + (x^{3} + 7 x + 9) \frac{d}{d x} (x^{2} - 5 x + 8)$

$= (x^{2} - 5 x + 8) (3 x^{2} + 7) + (x^{3} + 7 x + 9) (2 x - 5) .$

3x4 + 7x2- 15x3- 35x + 24x2 + 56 + 2x4- 5x3 + 14x2- 35x 18x-45

= 5x4- 20x3 + 45x2- 52x + 11

(ii) $y = (x^{2} - 5 x + 8) (x^{3} + 7 x + 9)$

$y = x^{5} + 7 x^{2} + 9 x - 5 x^{4} - 35 x^{2} - 45 x + 8 x^{3} + 56 x + 72$

$y = x^{5} - 5 x^{4} + 15 x^{3} - 26 x^{2} + 11 x + 72 .$

$∴ \frac{d y}{d x} = 5 x^{4} - 20 x^{3} + 45 x^{2} - 52 x + 11 .$

Taking log in eqn (1)

$l o g y = l o g (x^{2} - 5 x + 8) + l o g (x^{3} + 7 x + 9)$

Now, Differe(iii) ntiating w r t ‘x’ we get,

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{x^{2} - 5 x + 8} \frac{d}{d x} (x^{2} - 5 x + 8) + \frac{1}{x^{3} + 7 x + 9} \frac{d}{d x} (x^{3} + 7 x + 9) .$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = \frac{2 x - 5}{x^{2} - 5 x + 8} + \frac{3 x^{2} + 7}{x^{3} + 7 x + 9}$

$\Rightarrow \frac{d y}{d x} = y [\frac{(2 x - 5) (x^{3} + 7 x + 9) + (3 x^{2} + 7) (x^{2} - 5 x + 8)}{(x^{2} - 5 x + 8) (x^{3} + 7 x + 9) .}]$

$= \frac{y}{y}$ 2x4 + 14x4 + 18x- 35x- 45 + 3x1- 15x3 + 24x2 + 7x2- 35x + 56]

${? e q^{n} (1)} .$

$\frac{d y}{d x}$ = 5x4- 20x3 + 45x2- 52x + 11

We observed that all the methods give the same result.

5. Is the function $f$ defined by

$\begin{array}{l} f (x) = {\begin{cases} x, x \leq 1 \\ 5, x \end{cases} \\ x x \end{array}$

Find all points of discontinuity of $f$ , where $f$ is defined by

5. Given, f (a) = ${\begin{array}{l} x, if x \leq 1 & 5, if x > 1 . \end{array}$

At x = 0,

(0) = 0

$\underset{x \to 0}{l i m}$ f (x) = $\underset{x \to 0}{l i m}$ x = 0

∴ $\underset{x \to 0}{l i m}$ f (x) = f (0)

So, f is continuous at x = 0.

At x = 1,

Left hand limit,

L.H.L = $\underset{x \to 1}{l i m}$ f (x) = $\underset{x \to 1}{l i m}$ x = 1.

Right hand limit,

R. H. L. = $\underset{x \to 1^{+}}{l i m}$ f (x) = $\underset{x \to 1^{+}}{l i m}$ 5 = 5.

L.H.L. $=$ R.H.L.

So, f is not continuous at x = 1.

At x = 2,

f (2) = 5.

$\underset{x \to 1}{l i m}$ f (x) = $\underset{x \to 2}{l i m}$ 5 = 5

$\underset{l i m \to 2}{l i m f}$ (x) = f (2)

So f is continuous at x = 2.

Find all points of discontinuity of f, where f is defined by

43. Prove that the greatest integer function defined by

$f (x) = [x], 0 < x < 3$ is not differentiable at $x$ = 1 and $x$ = 2.

43. The given f x ⁿ is

f(x) = 0 < $| x |$ < 3

At x = 1

L×H×L× = $\underset{h \to 0^{-}}{l i m} \frac{f (1 + h) - f (0)}{h}$

$= \underset{h \to 0^{-}}{l i m} \frac{[1 + h] - [1]}{h}$

$\underset{h \to σ}{l i m} \frac{0 - 1}{h}$ ${\begin{matrix} ? h < 0 & , 1 + h < 1 - 1 & So, [1 + h] = 0 \end{matrix}}$

$= \underset{h \to 0}{l i m} \frac{- 1}{h} = \infty$

Hence lines does not exist

Qf is not differentiable at x = 1

At x = 2

L×H×L = $\underset{h \to 0^{-}}{l i m} \frac{f (2 + h) - f (2)}{h}$ ${\begin{matrix} ? h < 0 & 2 + h < 2 & 30, [2 + h] = 1 \end{matrix}}$

$= \underset{h \to 0^{-}}{l i m} \frac{[2 + h] - [2] .}{h}$

$= \underset{h \to 0^{-}}{l i m} \frac{1 - 2}{h} = \underset{h \to 0^{-}}{l i m} \frac{- 1}{h}$

Hence, limit does not exist.

Qf is not differentiable at x = 2

23. Determine if $f$ defined by

$f (x) =$ ${\begin{array}{l} x^{2} s i n \frac{1}{x}, if x \neq 0 & 0, if x = 0 \end{array}$ is a continuous function?

23. Given f (x) = ${\begin{array}{l} x^{2} s i n \frac{1}{x}, if x \neq 0 . & 0 if x = 0 . \end{array}$

For x = c $=$ 0,

f (c) = $c^{2} s i n \frac{1}{c}$

$\underset{x \to c}{l i m} f (x) = \underset{x \to c}{l i m} x^{2} s i n \frac{1}{x} = c^{2} s i n \frac{1}{c} .$

So, f is continuous for $x \neq 0 .$

For x = 0,

f (0) = 0

$\underset{x \to 0}{l i m} f (x) = \underset{x \to 0}{l i m} (x^{2} s i n \frac{1}{x})$

As we have sin $\frac{1}{x} \in [- 1, 1]$

$\underset{x \to 0}{l i m}$ f (x) = 02 a where $a \in [- 1, 1]$

= 0 = f (0).

∴ f is also continuous at x = 0.

84. Find the derivative of the function given by $\begin{array}{l} Find the derivative of the function given by f (x) = (1 + x) (1 + x 4) (1 + x 8) \\ and hence find f' (1). \end{array}$ and hence f'(1)

84. Given, f(x) = (1 + x)(1 + x 4)(1 + x 8)

Taking log,

logf(x) = log (1 + x) + log (1 + x) + log (1 + x 4) + log (1 + x 8)

Now, Differentiating w r t ‘x’ we get,

$\frac{1}{f (x)} f^{'} (x) = \frac{1}{1 + x} \frac{d}{d x} (1 + x) + \frac{1}{1 + x^{2}} \frac{d}{d x} (1 + x^{2}) + \frac{1}{1 + x^{4}} \frac{d}{d x} (1 + x^{4}) + \frac{1}{1 + x^{8}} \frac{d (1 + x^{8})}{d x}$

$\Rightarrow f^{'} (x) = f (x) {\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}} .$

$\Rightarrow f^{'} (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) {\frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \frac{8 x^{7}}{1 + x^{8}}}$

Putting x = 1

f`(x) = (1 +1)(1 + 14)(1 +18) ${\frac{1}{1 + 1} + \frac{2 \times 1}{1 + 1^{2}} + \frac{4 \times 1^{3}}{1 + 1^{4}} + \frac{8 \times 1^{7}}{1 + 1^{8}}}$

$= 2 \times 2 \times 2 + 2 {\frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2}}$

$= 16 \times {\frac{15}{2}} = 8 \times 15 = 120 .$

52.Kindly consider the following

$y = s i n^{- 1} (\frac{2 x}{1 + x^{2}})$

52. Kindly go through the solution

128. Kindly consider the following

128. Let $y = x^{x^{2} - 3} + {(x - 3)}^{x^{2}} .$

Putting $u = x^{x^{2} - 3} v = {(x - 3)}^{x^{2}}$ we get,

$y = u + v$

$\to \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ ________(1)

Now, $u = x^{x^{2} - 3}$

Taking log,

$l o g u = (x^{2} - 3) l o g x .$

$\Rightarrow \frac{1}{u} \frac{d u}{d x} = (x^{2} - 3) \frac{d}{d x} l o g x + l o g x \frac{d}{d x} (x^{2} - 3)$

$\Rightarrow \frac{d u}{d x} = u [\frac{x^{2} - 3}{x} + l o g x (2 x)]$

$\frac{d u}{d x} = x^{x^{2} - 3} [\frac{x^{2} - 3}{x} + 2 x l o g x]$

And $v = (x - 3) x^{2}$

$l o g v = x^{2} l o g (x - 3)$

$\to \frac{1}{v} \frac{d v}{d x} = x^{2} \frac{d}{d x} l o g (x - 3) + l o g (x - 3) \frac{d}{d x} x^{2}$

$= \frac{x^{2} \times 1 \times}{x - 3} \frac{d}{d x} (x - 3) + l o g (x - 3) \cdot 2 x$

$\frac{d v}{d x} = v [\frac{x^{2}}{x - 3} + 2 x l o g (x - 3)]$

$= {(x - 3)}^{x^{2}} [\frac{x^{2}}{x - 3} + 2 x l o g (x - 3)]$

Hence eqn (1) becomes

$\frac{d y}{d x} = x^{(x^{2} - 3)} [\frac{x^{2} - 3}{x} + 2 x l o g x] + {(x - 3)}^{x^{2}} [\frac{x^{2}}{x - 3} + 2 x l o g (x - 3)]$

64. Kindly consider the following

$e^{x} + {e^{x}}^{2} + \dots + {e^{x}}^{5}$

64. Let y = e^x + e^x² + … + e^x⁵.

$∴ \frac{d y}{d x} = \frac{d}{d x}$ (e^x + e^x² + e^x³ + e^x⁴ + e^x⁵).

$= \frac{d e^{x}}{d x} + \frac{d e^{x^{2}}}{d x} + \frac{d e^{x^{3}}}{d x} + \frac{d}{d x} e^{x^{4}} + \frac{d}{d x} e^{x^{5 .}}$

$= e^{x} + e^{x^{2}} \frac{d x^{2}}{d x} + e^{x^{3}} \frac{d x^{3}}{d x} + e^{x^{4}} \frac{d x^{4}}{d x} + e^{x^{5}} \frac{d x^{5}}{d x} .$

= e^x + 2x e^x² + 3x²e^x³ + 4x³e^x⁴ + 5ee^x⁴

92. Kindly consider the following

92. Kindly go through the solution

18. Show that the function defined by $g (x) = x - [x]$ is discontinuous at all integral

points. Here $[x]$ denotes the greatest integer less than or equal to $x$ .

18. Given, g (x) = x [x].

For $n \in z,$

g (n) = n [n] = nn = 0

$\underset{x \to n^{-}}{l i m}$ f (x) = $\underset{x \to n^{-}}{l i m}$ (x [x]) = n [n 1] = n + 1 = 1

$\underset{x \to n^{+}}{l i m}$ g (x) = $\underset{x \to n^{+}}{l i m}$ x [x] = n [n] = 0

So, $\underset{x \to n^{-}}{l i m}$ g (x) $=$ $\underset{x \to n^{+}}{l i m}$ g (x).

g (x) is d is continuous at all x $\in z .$

7. $f (x) =$ ${\begin{array}{l} | x | + 3 if x \leq - 3 & - 2 x if - 3 < x < 3 & 6 x + 2 if x \geq 3 \end{array}$

7. Given, f(x) = ${\begin{array}{l} | x | + 3 if x \leq - 3 & - 2 x if - 3 < x < 3 & 6 x + 2 if x \geq 3 \end{array}$

For x = $? c < - 3,$

f ( 3) = e + 3 (∴x< 3, $| x | = - x$ )

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ $| x | + 3 = - a + 3 .$

∴ $\underset{x \to c}{l i m}$ f(x) = f(c)

So, f is continuous at x = c < 3.

For x = c > 3

f(3) = 6.3 + 2 = 18 + 2 = 20

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 6x + 2 = 18 + 2 = 20

∴ $\underset{x \to c}{l i m}$ f(x) = f(c).So f is continuous at x = c > 3.

For. C = 3,

f ( 3) = ( 3) + 3 = 6.

$\underset{x \to c^{-}}{l i m}$ f(x) = $\underset{x \to c^{-}}{l i m}$ .x + 3 = ( 3) + 3 = 6.

$\underset{x \to c^{+}}{l i m}$ f(x) = $\underset{x \to c^{+}}{l i m}$ ( 2x) = 2 ( 3) = 6.

∴ $\underset{x \to c^{-}}{l i m}$ f(x) = $\underset{x \to c^{-}}{l i m}$ f(x) = f( 3)

So, f is continuous at x = c = 3.

For c = 3,

f(3) = 6.3 + 2 = 18 + = 20.

$\underset{x \to 3^{-}}{l i m}$ f(x) = $\underset{x \to 3^{-}}{l i m}$ 2x = 2 (3) = 6

$\underset{x \to 3^{+}}{l i m}$ f(x) = $\underset{x \to 3^{+}}{l i m}$ (6x + 2) = 6.3 + 2 = 20

∴ $\underset{x \to 3^{-}}{l i m}$ f(x) $=$ $\underset{x \to 3^{-}}{l i m}$ f(x).

f is not continuous at x = 3 point of discontinuity

107. Kindly consider the following

107. Given $y = 3 c o s (l o g x) + 4 s i n (l o g x)$

So, $y_{1} = \frac{d y}{d x} = 3 \frac{d}{d x} c o s (l o g x) + 4 \frac{d}{d x} s i n (l o g x)$

$y_{1} = 3 [- s i n (l o g x)] \frac{d}{d x} l o g x + 4 c o s (l o g x) \frac{d}{d x} (l o g x)$

$y_{1} = \frac{- 3 s i n (l o g x)}{x} + \frac{4 c o s (l o g x)}{x}$

$\Rightarrow x y_{1} = - 3 s i n (l o g x) + 4 c o s (l o g x)$ ______________(1)

Differentiating eqn (1) w r t ‘x’ we get,

$\frac{d}{d x} (x y_{1}) = - 3 \frac{d}{d x} s i n (l o g x) + 4 \frac{d}{d x} c o s (l o g x)$

$\Rightarrow x \frac{d y_{1}}{d x} + y_{1} \frac{d x}{d x} = - 3 c o s (l o g (x)) \frac{d}{d x} l o g x + 4 [- s i n (l o g x)] \frac{d}{d x} l o g x$

$\Rightarrow x y_{2} + y_{1} = \frac{- 3 c o s (l o g x)}{x} - \frac{4 s i n (l o g x)}{x}$

$\Rightarrow x^{2} y_{2} + y_{1} = - [3 c o s (l o g x) + 4 s i n (l o g x)]$

$\Rightarrow x^{2} y_{2} + y_{1} = - y$

$\Rightarrow x^{2} y_{2} + y_{1} + y = 0$

9. $f (x) =$ ${\begin{matrix} \frac{x}{| x |},if x < 0 & - 1, if x \geq 0 \end{matrix}$

9. Given, f (x) = ${\begin{matrix} x + 1, π x \geq 1 & x^{2} + 1, π x < 1 . \end{matrix}$

For x = c < 1,

$\underset{x \to c}{l i m}$ f (x) = $\underset{x \to c}{l i m}$ x2 + 1 = c2 + 1

∴ $\underset{x \to c}{l i m}$ f (x) = f (c)

So f is continuous at x = c < 1.

For x = c > 1,

F (c) = c + 1

$\underset{x \to c}{l i m}$ f (x) = $\underset{x \to c}{l i m}$ x + 1 = c + 1

∴ $\underset{x \to c}{l i m}$ f (x) = f (c)

So, f is continuous at x = c > 1.

For x = c = 1, + (1) = 1 + 1 = 2

L.H.L. = $\underset{x \to 1^{-}}{l i m}$ f (x) = $\underset{x \to 1^{-}}{l i m}$ x2 + 1 = 12 + 1 = 2.

R.H.L. = $\underset{x \to 1^{+}}{l i m}$ f (x) = $\underset{x \to 1^{+}}{l i m}$ x + 1 = .1 + 1 = 2

∴ L.H.L = R.H.L. = f (1)

So, f is continuous at x = 1.Hence f has no point of discontinuity.

40. Kindly consider the following

40. Kindly go through the solution

48. Kindly consider the following

$x^{2} + x y + y^{2} = 100$

48. Given, x² + xy + y² = 100.

Differentiating w r t ‘x’ we get,

$\frac{d}{d x} (x^{2} + x y + y^{2}) = \frac{d}{d x} (100)$

$\Rightarrow 2 x + x \frac{d y}{d x} + y \frac{d x}{d x} + 2 y \frac{d y}{d x} = 0 .$

$\Rightarrow x \frac{d y}{d x} + 2 y \frac{d y}{d x} = - 2 x - y$

$\Rightarrow \frac{d y}{d x} = \frac{- (2 x + y)}{(x + 2 y)}$

47. Kindly consider the following

$x y + y^{2} = t a n x + y$

47. Given, xy + y² = tan x + y Differentiating w r t x we get,

$\frac{d}{d x} (x y + y^{2}) = \frac{d}{d x} (t a n x + y)$

$\Rightarrow x \frac{d y}{d x} + y \frac{d x}{d x} + \frac{d y^{2}}{d x} = d x^{2} x + \frac{d y}{d x}$

$\Rightarrow x \frac{d y}{d x} + 2 y \frac{d y}{d x} - \frac{d y}{d x} = s e n^{2} x - y$

$\Rightarrow (x + 2 y - 1) \frac{d y}{d x} = s e c^{2} x - y$

$\Rightarrow \frac{d y}{d x} = \frac{s i n^{2} x - y}{x + 2 y - 1 .}$

6. $f (x) =$ ${\begin{matrix} 2 x + 3 if x \leq 2 & 2 x - 3 if x > 2 . \end{matrix}$

6. Given f(x) = ${\begin{matrix} 2 x + 3 if x \leq 2 & 2 x - 3 if x > 2 . \end{matrix}$

For x = c < 2,

F (c) = 2c + 3

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 2x + 3 = 2c + 3

∴ $\underset{x \to c}{l i m}$ f (x) = f(c)

So f is continuous at x $| < |$ 2.

For x = c > 2.

F (c) = 2c 3

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 2x 3 = 2c 3

∴ $\underset{x \to c}{l i m}$ f(x) = f(c)

So f is continuous at x $| > |$ 2.

For x = c = 2,

L.H.L. = $\underset{x \to 2^{-}}{l i m}$ f(x) = $\underset{x \to 2^{-}}{l i m}$ .2x + 3 = 2. 2 + 3 = 4 + 3 = 7.

R.H.L. = $\underset{x \to 2^{+}}{l i m}$ f(x) = $\underset{x \to 2^{+}}{l i m}$ 2x 3 = 2. 2 3 = 4 3 = 1.

∴ LHL $=$ RHL

∴ f is not continuous at x = 2.i e, point of discontinuity

81. Kindly consider the following

$y^{x} = x^{y}$

81. Given, yx = xy

Taking log,

x log y .log x

Differentiating w r t ‘x’ we get,

$\Rightarrow x \frac{d}{d x} l o g y + l o g y \frac{d x}{d x} = y \frac{d}{d x} l o g x + l o g x \frac{d y}{d x}$

$\Rightarrow \frac{x}{y} \cdot \frac{d y}{d x} + l o g y = \frac{y}{x} + l o g x \frac{d y}{d x}$

$\Rightarrow l o g x \frac{d y}{d x} - \frac{x}{y} \frac{d y}{d x} = l o g y - \frac{y}{x}$

$\Rightarrow \frac{d y}{d x} [\frac{y l o g x - x}{y}] = \frac{x l o g y - y}{x}$

$\Rightarrow \frac{d y}{d x} = \frac{y (x l o g y - y)}{x (y l o g x - x)} .$

119. Kindly consider the following

$s i n^{3} x + c o s^{6} x$

119. Let $y = s i n^{3} x + c o s^{6} x$

So, $\frac{d y}{d x} = \frac{d}{d x} (s i n^{3} x + c o s^{6} x)$

$= 3 s i n^{2} x \frac{d}{d x} s i n x + 6 c o s^{5} x \frac{d}{d x} c o s x$

$= 3 s i n^{2} x c o s x - 6 c o s^{5} x s i n x$

$= 3 s i n x c o s x (s i n x - 2 c o s^{4})$

131. Kindly consider the following

131. Kindly go through the solution

136. Kindly consider the following

136. Given, $s i n (A + B) = s i n A c o s B + c o s A s i n B$

Differentiating w r t. ‘x’ we get,

$\frac{d}{d x} s i n (A + B) = \frac{d}{d x} (s i n A c o s B + c o s A s i n B)$

$\to c o s (A + B) - \frac{d}{d x} (A + B) = s i n A \frac{d}{d x} c o s B + c o s B \frac{d}{d x} s i n A + c o s A \frac{d}{d x} s i n B + s i n B \frac{d}{d x} c o s A$

$\to c o s (A + B) (\frac{d A}{d x} + \frac{d B}{d x}) = - s i n A s i n B \frac{d B}{d x} + c o s B c o s A \frac{d A}{d x} + c o s A c o s B \frac{d B}{d x} - s i n A s i n B \frac{d A}{d x}$

$\to c o s (A + B) (\frac{d A}{d x} + \frac{d B}{d t}) = c o s A c o s B (\frac{d A}{d x} + \frac{d B}{d x}) - s i n A s i n B (\frac{d A}{d x} + \frac{d B}{d x})$

$= (c o s A c o s B - s i n A s i n B) (\frac{d A}{d x} + \frac{d B}{d x}) .$

$\to c o s (A + B) = c o s A c o s B - s i n A s i n B .$

50. Kindly consider the following

$s i n^{2} y + c o s x y = k$

50. Given, sin²y + cos xy = p

Differentiating w r t ‘x’ we get,

$\frac{d}{d x} (s i n^{2} y + c o s x y) = \frac{d}{d x}$

= $\frac{d}{d x} s i n^{2} y + \frac{d}{d x} c o s x y = 0$

$\Rightarrow 2 s i n y \frac{d}{d x} (s i n y) + (- s i n x y) \frac{d}{d x} (x y) = 0 .$

$\Rightarrow 2 s i n y c o s y \frac{d y}{d x} - s i n x y [x \frac{d y}{d x} + y] = 0$

$\Rightarrow 2 s i n 2 y \frac{d y}{d x} - x s i n x y \frac{d y}{d x} - y s i n x y = 0$

{Qsin 2x = 2sin x cos x}

$\Rightarrow \frac{d y}{d x} [s i n 2 y - x s i n x y] = y s i n x y .$

$\Rightarrow \frac{d x y}{d x} = \frac{y s i n x y}{s i n 2 y - x s i n x y} .$

17. Find the relationship between a and b so that the function $f$ defined by

$f (x) =$ ${\begin{matrix} λ (x^{2} - 2 x), & if x \leq 0 \\ 4 x + 1, & if x > 0 \end{matrix}$ continuous at $x$ = 0? What about continuity at $x$ = 1?

17. Given, f (x) = ${\begin{matrix} λ (x^{2} - 2 x) & if x | ? | 0 \\ 4 x + 1 & if x > 0 . \end{matrix}$

For continuity at x = 0,

$\underset{x \to 0^{-}}{l i m}$ f (x) = $\underset{x \to 0^{+}}{l i m}$ f (x) = f (0).

$\underset{x \to 0^{-}}{l i m}$ $λ (x^{2} - 2 x)$ = $\underset{x \to 0^{+}}{l i m}$ 4x + 1 = $λ (0^{2} - 2.0)$

0 = 1 = 0 which is not true

Hence, f is not continuous for any value of $λ .$

For x = 1,

$\underset{x \to 1}{l i m}$ f (x) = f (1).

$\underset{x \to 1}{l i m}$ 4x + 1 = 4 (1) + 1

$\Rightarrow$ 4 + 1 = 4 + 1

$\Rightarrow$ 5 = 5.

So, f is continuous at x = 1 value of $λ$

15. $f (x) =$ ${\begin{matrix} - 2, & if x \leq - 1 & 2 x, \\ if - 1 < x \leq 1 & 2, & if x > 1 \end{matrix}$

15. Given, f(x) = ${\begin{matrix} - 2, & if x \leq - 1 & 2 x, \\ if - 1 < x \leq 1 & 2, & if x > 1 . \end{matrix}$

For x = c < 1,

f(c) = 2

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ ( 2) = 2 = f(c)

So, f is continuous at x< 1.

For x = c > 1,

f(c) = 2

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ . 2 = 2 = f(c)

So, f is continuous at x $| > |$ 1.

For x = 1,

L.H.L. = $\underset{x \to - 1^{-}}{l i m}$ f(x) = $\underset{x \to - 1^{-}}{l i m}$ 2 = 2

R.H.L. = $\underset{x \to - 1^{+}}{l i m}$ f(x) = $\underset{x \to - 1^{+}}{l i m}$ . 2x = 2 ( 1) = 2

and f( 1) = 2

So, L.H.L. = R.H.L. = f( 1)

∴f is continuous at x = 1.

For x = 1,

L.H.L. = $\underset{x \to 1^{-}}{l i m}$ f(x) = $\underset{x \to 1^{-}}{l i m}$ . 2x = 2.1 = 2

R.H.L. = $\underset{x \to 1^{+}}{l i m}$ f(x) = $\underset{x \to 1^{+}}{l i m}$ . 2 = 2.

f(1) = 2

f(1) = L.H.L = R.H.L.

So, f is continuous at x = 1.

86. Kindly consider the following

86. To prove $\frac{d}{d x} (u v \cdot w) = \frac{d u}{d x} v \cdot w + u \cdot \frac{d v}{d x} \cdot w + u \cdot v \cdot \frac{d w}{d x} .$

By repeating application of produced rule

$= \frac{d}{d x} (u v : u)$

$= u \frac{d}{d x} (u \cdot w) + v \cdot w \frac{d y}{d x}$

$u {v \frac{d w}{d x} + w \frac{d v}{d x}} + \frac{d y}{d x} v : w .$

$= u \cdot v \cdot \frac{d w}{d x} + u \cdot \frac{d v}{d x} w + \frac{d y}{d x} \cdot v \cdot w$

$= \frac{d y}{d x} u \cdot w + u \cdot \frac{d v}{d x} \cdot w + u \cdot v \cdot \frac{d w}{d x}$ = R×H×S×

By togarith differentiating,

Let y = u v w

Taking log, log y = log u + log v + log w

Differentiating w r t ‘x’

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{u} \frac{d u}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x} .$

$\Rightarrow \frac{d y}{d x} = y [\frac{1}{4} \frac{d y}{d x} + \frac{1}{v} \frac{d v}{d x} + \frac{1}{w} \frac{d w}{d x}]$

$\Rightarrow \frac{d}{d x} (u v w) = u v w \cdot [\frac{1}{u} \frac{d y}{d x} + \frac{1}{v} \frac{d u}{d x} + \frac{1}{w} \frac{d u r}{d x}]$

$= \frac{d y}{d x} v \cdot w + u \frac{d v}{d x} \cdot w + u v \cdot \frac{d w}{d x}$

132. Kindly consider the following

132. Given, ${(x - a)}^{2} + {(y - b)}^{2} = c^{2} .$

Differentiating w r t ‘x’ we get

$\frac{d}{d x} {(x - a)}^{2} + \frac{d}{d x} {(y - b)}^{2} = \frac{d}{d x} c^{2}$

$\Rightarrow 2 (x - a) + 2 (y - b) \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{2 (x - a)}{2 (y - b)} = - \frac{(x - a)}{(y - b)}$

Again, $\frac{d^{2} y}{d x^{2}} = - {\frac{(y - b) \frac{d}{d x} (x - a) - (x - a) \frac{d}{d x} (y - b)}{{(y - b)}^{2}}}$

$= - {\frac{(y - b) - (x - a) \frac{d y}{d x}}{{(y - b)}^{2}}}$

$= - {\frac{(y - b) + (x - a) \frac{(x - a)}{(y - b)}}{{(y - b)}^{2}}}$

$= - {\frac{{(y - b)}^{2} + {(x - a)}^{2}}{{(y - b)}^{3}}}$

$= - \frac{c^{2}}{{(y - b)}^{3}} {? {(x - a)}^{2} + {(y - b)}^{2} = c^{2}}$

Then, L.H.S = $\frac{{1 + {(\frac{d y}{d x})}^{2}}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} = \frac{{1 + \frac{{(x - a)}^{2}}{{(y - b)}^{2}}}^{3 / 2}}{\frac{- c^{2}}{{(y - b)}^{2}}}$

$= \frac{{{(y - b)}^{2} + {(x - a)}^{2}}^{3 / 2}}{{(y - b)}^{3}} \times \frac{{(y - b)}^{3}}{- c^{2}}$

$= \frac{c^{2 \times 3 / 2}}{- c^{2}} = \frac{c^{3}}{- c^{2}} = - c$ Where c is a constant and is independent of a and b.

51. Kindly consider the following

$s i n^{2} x + c o s^{2} y = 1$

51. Given, sin²x + cos²y = 1.

Differentiating w r t ‘x’ we get,

$\frac{d}{d x}$ (sin²x + cos²y) $\frac{d}{d x} 1 .$

$\Rightarrow \frac{d}{d x} s i n^{2} x + \frac{d}{d x} c o s^{2} y = 0$

$\Rightarrow 2 s i n x \frac{d s i n x}{d x} + 2 c o s y \frac{d c o s y}{d x} = 0$

= 2sin x cos x + 2 cos y (- sin y) $\frac{d y}{d x} = 0$

= sin 2x- sin 2y $\frac{d y}{d x}$ = 0

24. Examine the continuity of $f$ , where $f$ is defined by

$f (x) =$ ${\begin{matrix} s i n x - c o s x, & if x \neq 0 \\ - 1, & if x = 0 \end{matrix}$

24. Given, f(x) = ${\begin{matrix} s i n x - c o s x, & if x \neq 0 \\ - 1, & if x = 0 . \end{matrix}$

For x = c $=$ 0,

f(c) = sin c cos c.

$\underset{x \to c}{l i m}$ f (x) = $\underset{x \to c}{l i m}$ (sin x cos x) = sin c cos c = f(c)

So, f is continuous at $x \neq 0$

For x = 0,

f(0) = 1

$\underset{x \to 0}{l i m}$ f (x) = $\underset{x \to 0}{l i m}$ (sin x cos x) = sin 0 cos 0 = 0 1 = 1

$\underset{x \to 0^{+}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} (s i n x - c o r x) = s i n 0 - c o s 0 = - 1 .$

∴ $\underset{x \to 0 -}{l i m}$ f(x) = $\underset{x \to 0 +}{l i m}$ f (x) = f (0)

So, f is continuous at x = 0.

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29.

25. $f (x) =$ ${\begin{matrix} \frac{k c o s x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix} \frac{π}{2}$

25. Given, f(x) = ${\begin{matrix} \frac{π c o s x}{π - 2 x} & if x \neq \frac{π}{2} \\ 3 & if x = \frac{π}{2} \end{matrix}$

For continuity at $x = \frac{π}{2}$

$\underset{x \to \frac{π^{-}}{2}}{l i m} f (x) = \underset{x \to \frac{π}{2}}{l i m} f (x) = f (\frac{π}{2}) .$

$\underset{x \to \frac{π}{2}}{l i m} \frac{x c o s x}{π - 2 x} = \underset{x \to \frac{π}{2^{+}}}{l i m} \frac{x c o s x}{π - 2 x} = 3 .$

Take $\underset{x \to \frac{π}{2}}{l i m} \frac{x c o s x}{x - 2 x} = 3$ .

Putting x = $\frac{π}{2} + h$ such that as $x \to \frac{π}{2}, h \to 0 .$

So $\underset{h \to 0}{l i m} \frac{x c o s (\frac{π}{2} + h)}{x - 2 (\frac{π}{2} + n)} = \underset{h \to 0}{l i m} \frac{x (- s i n x)}{- 2 h}$

$= \underset{h \to 0}{l i m} \frac{x s i n h}{2 h}$

$= \frac{k}{2} \underset{h \to 0}{l i m} \frac{s i n h}{h} = \frac{k}{2} .$

i e, $\frac{k}{2} = 3$

$\Rightarrow$ k = 6

Similarly from $\underset{x \to \frac{π^{+}}{2}}{l i m} \frac{x c o s x}{π - 2 x} = 3$

$\Rightarrow \underset{h \to 0}{l i m} \frac{h c o s (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} = \underset{h \to 0}{l i m} \frac{- h s i n h}{- 2 h}$

$= \frac{}{2} \underset{h \to 0}{l i m} \frac{s i n h}{h}$

$= \frac{x}{2}$

So, $\frac{x}{2} = 3$

$\Rightarrow$ k = 6

49. Kindly consider the following

$x^{3} + x^{2} y + x y^{2} + y^{3} = 81$

49. Given, x³ + x²y + xy² + y³ = 81.

Differentiating w r t ‘x’ we get,

$\frac{d}{d x} (x^{3} + x^{2} y + x y^{2} + y^{3})$ = $\frac{d (81)}{d x}$

$\Rightarrow \frac{d x^{3}}{d x} + \frac{d}{d x} x^{2} y + \frac{d}{d x} x y^{2} + \frac{d}{d x} y^{3} = 0$

$\Rightarrow 3 x^{2} + x^{2} \frac{d y}{d x} + y \frac{d x^{2}}{d x} + \frac{x d y^{2}}{d x} + y^{2} \frac{d x}{d x} + 3 y^{2} \frac{d y}{d x} = 0$

$\Rightarrow 3 x^{2} + x^{2} \frac{d y}{d x} + 2 x y + 2 x y \frac{d y}{d x} + y^{2} + 3 y^{2} \frac{d y}{d x} = 0 .$

$\Rightarrow (x^{2} + 2 x y + 3 y^{2}) \frac{d y}{d x}$ = - (3x² + 2xy + y²)

$\frac{d y}{d x} = \frac{- (3 x^{2} + 2 x y + y^{2})}{(x^{2} + 2 x y + 3 y^{2}) .}$

69. Kindly consider the following

$c o s x . c o s 2 x . c o s 3 x$

69. Let y = cos x cos 2x cos 3x _____ (i)

Taking log_e on bolk sides.

logy = log (cos x) + log (cos 2x + log (cos 3x)

= log (cos x) + log (cos 2x) + log (cos 3x)

Differentiating w r t ‘x’

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{c o s x} \frac{d}{d x} c o s x + \frac{1}{c o s 2 x} \frac{d}{d x} c o s 2 x + \frac{1}{c o s 3 x} \frac{d}{d x} c o s 3 x .$

$= \frac{(- s i n x)}{c o s x} + \frac{(- s i n 2 x)}{c o s 2 x} \frac{d}{d x} (2 x) + \frac{(- s i n 3 x)}{c o s 3 x} \frac{d (3 x)}{d x}$

= - tan x-2 tan 2x- 3 tan 3x.

$\Rightarrow \frac{d y}{d x}$ = y [- tan x- 2 tan 2x- 3 tan 3x]

Putting value of y from (i) we get,

$= \frac{d y}{d x}$ = - cos x cos 2x cos 3x [tan x + 2 tan 2x + 3 tan 3x]

70. Kindly consider the following

70. Kindly go through the solution

Putting value of y from the above we get,

78. Kindly consider the following

$x^{x c o s x} + \frac{x^{2} + 1}{x^{2} - 1}$

78. Let y = x^x cos x $\frac{a^{2} + 1}{x^{2} - 1}$

Putting 4 = x^x cos x and v = $\frac{x^{2} + 1}{x^{2} - 1}$ we have,

y = u + v

$\Rightarrow \frac{d y}{d x} = \frac{d y}{d x} + \frac{d v}{d x}$ ____ (1)

As u $| = |$ x^x cos x.

Taking log,

Log u = x cos x log x

Differentiating w r t ‘x’,

$\frac{1}{u} \frac{d y}{d x} = x \frac{d}{d x}$ [cos x log x] + cos x log x $\frac{d x}{d x}$

= x ${c o t x \frac{d}{d x} l o g x + l o g x \frac{d}{d x} c o s x}$ + cos x log x.

$= x {c o s x \cdot \frac{1}{x} - s i n x \cdot l o g x}$ + cos x log x.

= cos x- sin x. log x + cos x log x.

$\frac{d y}{d x} = u$ [cosx + cos x log x- sin x log x]

= x^x cos x [cos x + cos x log x-x sin x log x]

And v = $\frac{x^{2} + 1}{x^{2} - 1}$

So, $\frac{d v}{d x} = \frac{(x^{2} - 1) \frac{d}{d x} (x^{2} + 1) - (x^{2} + 1) \frac{d}{d x} (x - 1)}{{(x^{2} - 1)}^{2}}$

$= \frac{(x^{2} - 1) (2 x) - (x^{2} + 1) (2 x)}{{(x^{2} - 1)}^{2}}$

$= \frac{2 x^{3} - 2 x - 2 x^{3} - 2 x}{{(x^{2} - 1)}^{2}}$

$= \frac{- 4 x}{{(x^{2} - 1)}^{2}} .$

Hence, eqn (1) becomes,

$\frac{d y}{d x}$ x^x^{cos x} [cos x + cos x log x-x sin x log x] $\frac{- 4 x}{{(x^{2} - 1)}^{2}}$

87. Kindly consider the following

$x = 2 a t^{2}, y = a t^{4}$

87. Given, x = 2at2 and y = at4. Differentiation w r t we get,

$\frac{d x}{d t} = 4 a t .$ and $\frac{d y}{d t} = 4 a t^{3} .$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 a t^{3}}{4 a t} = t^{2} .$

88. Kindly consider the following

$x = a c o s θ, y = b c o s θ$

88. Kindly go through the solution

89. Kindly consider the following

$x = s i n t, y = c o s 2 t$

89. Given, x = sin t and y = cos2t. differentiation w r t. ‘t’ we get,

$\begin{array}{l} \frac{d x}{d t} = c o s t \frac{d y}{d t} = (- s i n 2 t) \frac{d 2 t}{d t} \\ - t \\ - 2 (2 s i n t c o s t) \\ - t t \end{array}$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$= \frac{- 4 s i n t c o s t}{c o s t}$

= -4 sin t

90. Kindly consider the following

$x = 4 t, y = \frac{4}{t}$

90. Given, x = 4t and y = $\frac{4}{t}$ Differentiating w r t. ‘t’ we get,

$\frac{d x}{d t} = 4 \frac{d y}{d t} = \frac{4 d (t^{- 1})}{d t}$

$= - 4 t^{- 2}$

$\frac{- 4}{t^{2}}$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{- 4}{t^{2}}}{4} = - \frac{1}{t^{2}} .$

93. Kindly consider the following

93. Kindly go through the solution

94. Kindly consider the following

94. Kindly go through the solution

104. Kindly consider the following

$s i n (l o g x)$

104. Let $y = s i n (l o g x)$

so, $\frac{d y}{d x} = \frac{d}{d x} s i n (l o g x) = c o s (l o g x) \frac{d}{d x} l o g x = \frac{c o s (l o g x)}{x}$

$∴ \frac{d^{2} y}{d x^{2}} = \frac{x \frac{d}{d x} c o s (l o g x) - c o s (l o g x) \frac{d x}{d x}}{x^{2}}$

$= \frac{x [- s i n (l o g x)] \frac{d}{d x} l o g x - c o s (l o g x)}{x^{2}}$

$= \frac{- [x \cdot s i n (l o g x) \times \frac{1}{x} + c o s (l o g x)]}{x^{2}}$

$= \frac{- [s i n (l o g x) + c o s (l o g x)]}{x^{2}}$

110. Kindly consider the following

110. Kindly go through the solution

111. Kindly consider the following

111. Given, $y = {(t a n^{- 1} x)}^{2}$

So, $y_{1} = \frac{d y}{d x} = 2 (t a n^{- 1} x) \frac{d}{d x} t a n^{- 1} x$

$\Rightarrow y_{1} = 2 t a n^{- 1} x \times \frac{1}{1 + x^{2}}$

$(x^{2} + 1) y_{1} = 2 t a n^{- 1} x$

Differentiating again w r t ‘x’ we get,

$(x^{2} + 1) \frac{d y_{1}}{d x} + y_{1} \frac{d}{d x} (x^{2} + 1) = 2 \frac{d}{d x} t a n^{- 1} x$

$\Rightarrow (x^{2} + 1) y_{2} + y_{1} (2 x) = \frac{2}{1 + x^{2}}$

$\Rightarrow {(x^{2} + 1)}^{2} y_{2} + 2 x (1 + x^{2}) y_{1} = 2$

Hence proved.

114. If f:[-5,5] →R is a differentiable function and if f'(x) does not vanish anywhere, then prove that f(-5) ≠ f(5).

114. Solution :
It is given that f: [-5,5]? R is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) f is continuous on [?5, 5].

(b) f is differentiable on (?5, 5).

Therefore, by the Mean Value Theorem, there exists c? (?5, 5) such that

$\begin{array}{l} f^{'} (c) \frac{f (5) ? f (? 5)}{5 ? (? 5)} \\ ? 10 f^{'} (c) = f (5) ? f (? 5) \end{array}$

It is also given that f' (x) does not vanish anywhere.

$\begin{array}{l} ? f^{'} (c) ? 0 \\ ? 10 f^{'} (c) ? 0 \\ ? f (5) ? f (? 5) ? 0 \\ ? f (5) ? f (? 5) \end{array}$

Hence, proved.

133. Kindly consider the following

133. Given, $c o s y = x c o s (a + y) .$

$x = \frac{c o s y}{c o s (a + y)}$

Differentiating w r t ‘y’ we get,

$\frac{d x}{d y} = \frac{d}{d y} (\frac{c o s y}{c o s (a + y)}) .$

$= \frac{c o s (a + y) \frac{d}{d y} c o s y - c o s y \frac{d}{d y} c o s (a + y)}{c o s^{2} (a + y) .}$

$= \frac{c o s (a + y) (- s i n y) - c o s y (- s i n (a + y))}{c o s^{2} (a + y) .}$

$= \frac{- c o s (a + y) s i n y + s i n (a + y) c o s y}{c o s^{2} (a + y)}$

$= \frac{s i n (a + y) c o s y - c o s (a + y) s i n y .}{c o s^{2} (a + y)}$

$\frac{d x}{d y} = \frac{s i n (a + y - y)}{c o s^{2} (a + y)} {\begin{array}{r} ? s i n (A - B) = s i n A c o s B & - c o s A s i n B \end{array}}$

So, $\frac{d y}{d x} = \frac{c o s^{2} (a + y)}{s i n a}$

137. Kindly consider the following

137. Yes, Let us take $f (x) = | x - 1 | + | x - 2 | .$

So, x = 1, x= 2 divides the real line into three disjoint intervals $(- \infty, 1], [1, 2]$ and $[2, \infty) .$

For $x \in (- \infty, 1] .$

$f (x) = - (x - 1) + [- (x - 2)] = - x + 1 - x + 2 = 3 - 2 x .$

For $x \in [1, 2] .$

$f (x) = (x - 1) - (x - 2) = 1 .$

For $x \in [2, \infty)$

$f (x) = x - 1 + x - 2 = 2 x - 3 .$

Hence, these polynomial fun are all continous and desirable. for all real values of x or, except x = 1 and x = 2.

ie, $\forall x \in R - {1, 2} .$

For differentiavity at x = 1,

LHD = $= \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{3 - 2 x - 1}{x - 1} = \underset{x \to 1^{-}}{l i m} \frac{2 - 2 x}{x - 1} .$

$= \underset{x \to 1^{-}}{l i m} \frac{- 2 (x - 1)}{x - 1}$

$= \underset{x \to 1^{-}}{l i m} - 2$

= -2

RHD = $= \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} = \underset{x \to 1^{+}}{l i m} \frac{1 - 1}{x - 1} = \underset{x \to 1^{+}}{l i m} \frac{0}{x - 1} = 0 .$

as L.HD ≠ R.HD

f is not differentiable at x =1.

For continuity at x = 1.

L.HL= $= \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{-}}{l i m} = 1 .$

RHL = $\underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} 1 = 1$ \ LHL = RHS

f is continuous at x = 1

For continuity & differentiability at x = 2

$= \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{-}}{l i m} 1 = 1 .$

$= \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} (2 x - 3) = 4 - 3 = 1 .$

? LHL = RHL

f is continuous at x = 2

$\begin{array}{l} = \underset{x \to 2^{-}}{l i m} \frac{f (x) - f (2)}{x - 2} = \underset{x \to 2^{-}}{l i m} \frac{1 - 1}{x - 2} = \underset{x \to 2^{-}}{l i m} = \frac{0}{x - 2} \end{array}$

$= \underset{x \to 2^{+}}{l i m} \frac{f (x) - f (2)}{x - 2} = \underset{x \to 2^{+}}{l i m} \frac{2 x - 3 - 1}{x - 2}$

$= \underset{x \to 2^{+}}{l i m} \frac{2 (x - 2)}{x - 2}$

$= \underset{x \to 2^{+}}{l i m} 2$

= 2

? LHD ≠ RHD

f is not differentiable at x = 2.

139. Kindly consider the following

139. Kindly go through the solution

12. Is the function defined by

$f (x) =$ ${\begin{array}{l} x + 5, if x \leq 1 & x - 5, if x > 1 \end{array}$ a continuous function?

Discuss the continuity of the function $f$ , where $f$ is defined by

12. Given, f(x) = ${\begin{array}{l} x + 5 if x | ? | 1 & x - 5 if x > 1 . \end{array}$

For x = c < 1.

F (c) = c + 5

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ f x + 5 = c + 5

∴ $\underset{x \to c}{l i m}$ f(x) = f(c)

So, f is continuous at x $| < |$ 1.

For x = c > 1

F (c) = c 5

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ x 5 = c 5.

$\underset{x \to c}{l i m}$ f(x) = f(c)

So, f is continuous at x $| > |$ 1.

For x = 1

L.H.L. = $\underset{x \to 1^{-}}{l i m}$ f(x) = $\underset{x \to 1^{-}}{l i m}$ x + 5 = 1 + 5 = 6.

R.H.L. = $\underset{x \to 1^{+}}{l i m}$ f(x) = $\underset{x \to 1^{-}}{l i m}$ x 5 = 1 5 = 4.

L.H.L. $=$ R.H.L.

f is not continuous at x = 1

So, point of discontinuity of f is at x = 1.

Discuss the continuity of the function $f$ , where $f$ is defined by

58. Kindly consider the following

58. Kindly go through the solution

16. Find the relationship between a and b so that the function $f$ defined by

$f (x) =$ ${\begin{array}{l} a x + 1, if x \leq 3 & b x + 3, if x > 3 \end{array}$ is continuous at $x$ = 3.

16. Given, f (x) = ${\begin{array}{l} a x + 1, if x \leq 3 & b x + 3, if x > 3 \end{array}$ is continuous at x = 3

So, f (3) = 3a + 1

L.H.L = $\underset{x \to 3^{-}}{l i m}$ f (x) = $\underset{x \to 3^{-}}{l i m}$ ax + 1 = 3a + 1

R.H.L = $\underset{x \to 3^{+}}{l i m}$ f (x) = $\underset{x \to 3^{+}}{l i m}$ b x + 3 = 3b + 3

for continuity at x = 3,

L.H.L = R.H.L. = f (3)

$\Rightarrow$ 3a + 1 = 3 + 3 = 3a + 1

So, 3a + 1 = 3b + 3

3a = 3b + 3 1

3a = 3b + 2.

a = b + $\frac{2}{3} .$

31. Show that the function defined by $f (x) = | c o s x |$ is a continuous function.

31. Given, f (x) = $| c o s x | .$

Let g (x) = cos x and h (x) = $x$

Hence, as cosine function and modulus f x are continuous $\forall x \in ?, g$ h are continuous.

Then, (hog) x = h (g (x)

= h (cos x)

$= | c o s x | .$

= f (x) is also continuous being

A composites fxn of two continuous f x $\forall x \in ?$

33. Find all the points of discontinuity of $f$ defined by $f x x x$

33. Let g (x) = $x$ is continuous being a modules f x and h (x) = $x$ is also continuous being a modules $\forall x \in ?$

Then, f (x) = g (x) h (x).is also continuous for all x. E. R.

Hence, there is no point of discontinuous for f (x).

41. Kindly consider the following

42. Prove that the function $f$ given by

$f (x) = | x - 1 |, x \in$ is not differentiable at $x$ = 1.

42. The given f x v is

f(x) = |x- 1|, x ε R

For a differentiable f x v f at x = c,

$\underset{h \to 0^{-}}{l i m} \frac{f (c + h) - f (c)}{h}$ and $\underset{h \to 0^{+}}{l i m} \frac{f (c + h) - f (c)}{h}$ are finite & equal.

So, at x = 1. f(1) = |1 - 1| = 0.

Now,

L×H×L× = $\underset{h \to 0^{-}}{l i m} \frac{f (1 + h - f (1)}{h}$

= $\underset{h \to 0^{-}}{l i m} \frac{| 1 + h - 1 | - 0 .}{h} = \underset{h \to 0^{-}}{l i m} - \frac{h}{h}$ ${\begin{cases} ∴ h < 0 \\ | h | = - h \end{cases}}$

$= \underset{h \to σ^{-}}{l i m} (- 1)$

R×H×L = $\underset{h \to 0^{+}}{l i m} \frac{f (1 + h) - f (1)}{h}$ = - 1.

$= \underset{h \to 0^{+}}{l i m} \frac{(1 + h - 1) - 0}{h} = \underset{h \to 0^{+}}{l i m} \frac{h}{h} = \underset{h \to 0^{+}}{l i m} 1$ ${\begin{matrix} ? f_{n} h > 0 & | h | = h \end{matrix}}$

= 1

Hence, L×H×S ¹ R×H×L×

So, f is not differentiable at x = 2.

45. Kindly consider the following

$2 x + 3 y = s i n y$

45. Given, 2x + 3y = sin y.

Differentiating w r t x. we get,

$\frac{d}{d x} (2 x + 3 y) = \frac{d}{d x} s i n y$

$\Rightarrow 2 + 3 \frac{d y}{d x} = c o s y \frac{d y}{d x}$

=cos y $\frac{d y}{d x} - 3 \frac{d y}{d x} = 2$

$\Rightarrow \frac{d y}{d x} (c o s y - 3) = 2$

= $\frac{d y}{d x} = \frac{2}{c o s y - 3}$

72. Kindly consider the following

$x^{x} - 2^{s i n x}$

72. Let y = x^x - 2 sin x

Putting u = x^x and v = 2 sin x.

So, y = u - v

= $\frac{d y}{d x} = \frac{d y}{d x} - \frac{d v}{d x}$ ____ (i)

As u = x^x

Log u = x log x.

So, $\frac{d}{d x}$ log u = $\frac{d}{d x}$ x log x.

= $\frac{1}{4} \frac{d y}{d x} = \frac{x d l o g x}{d x} + l o g x \frac{d x}{d x}$

x´ $\frac{1}{x}$ + log x.

= 1 + log x

= $\frac{d y}{d x}$ = 4 [1 + log x] = x^x [1 + log x].

And v = 2sin x Log v = sin x log 2.

$\frac{d}{d x} (l o g v) = \frac{d}{d x}$ (sin x log 2)

$\Rightarrow \frac{1}{v} \frac{d v}{d x} = s i n x \frac{d}{d x}$ sin x $\frac{d}{d x}$ log 2 + log 2 $\frac{d s i n x}{d x}$ = log 2. cos x.

$\Rightarrow \frac{d v}{d x}$ = v log2 cos x.

$\Rightarrow \frac{d v}{d x}$ = v log 2 cos x

= 2 ^{sin x} log². cos x.

Q Eqⁿ (i) becomes, = x^x (1 + log x) - 2 ^{sin x} cos x log 2.

73. Kindly consider the following

${(x + 3)}^{2} . {(x + 4)}^{3} . {(x + 5)}^{4}$

73. Let y = (x + 3)² (x + 4)³ (x + 5)⁴.

Taking log_e on both sides,

log y = log (x + 3)² + log (x + 4)³ + log (x + 5)⁴

= 2 log (x + 3) + 3 (log (x + 4) + 4 log (x + 5).

So,

Q $\frac{d}{d x}$ log y = $\frac{d}{d x}$ [2 log (x + 3) + 3 log (x + 4) + 4 log (x +5)]

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = \frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5 .}$

$\Rightarrow \frac{d y}{d x} = y [\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}]$

$\Rightarrow \frac{d y}{d x}$ = (x + 3)² (x + 4)³ (x + 5)⁴ ${[\frac{2}{x + 3} + \frac{3}{x + 4} + \frac{4}{x + 5}]}_{.}$

75. Kindly consider the following

${(l o g x)}^{x} + x^{l o g x}$

75. Let y = (log x)^x + x ^{log x}.

Putting u = log x^x and v = x log x we get,

y = u + v

$\Rightarrow \frac{d y}{d x} = \frac{d y}{d x} + \frac{d v}{d x}$ .____ (1)

As u = log x^x

Taking log,

Þlog u = x [log(log x)]

Differentiating w r t x we get,

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = x \frac{d}{d x}$ log (log x) + log (log x) $\frac{d x}{d x}$

= $x \times \frac{1}{l o g x} \frac{d l o g x}{d x}$ + log (log x)

= $\frac{x}{l o g x} \times \frac{1}{x} + l o g 1 (l o g x)$

$\Rightarrow \frac{d y}{d x} = μ [\frac{1}{l o g x} + l o g (l o g x)]$

$\frac{d y}{d x} = {(l o g x)}^{x} {[\frac{1}{l o g x} + l o g (l o g x)]}_{.}$

= (log x)^x $[\frac{1 + l o g x . l o g (l o g x)}{l o g x}]$

= (log x)^x^-¹ [1 + log ´. log (log x)]

And v = log x

Taking log,

Log v = log x log x. = (log x)².

Differentiating w r t ‘x’ we get,

$\Rightarrow \frac{1}{v} \frac{d v}{d x} = 2 l o g x \frac{d}{d x} l o g x$

$\Rightarrow \frac{d v}{d x}$ = 2v log x $\frac{1}{x}$

= 2. x log x. $\frac{l o g x}{x}$

= 2 x log x- 1 log x.

Hence eqⁿ becomes

$\frac{d y}{d x}$ = (log x) x- 1[1 + log x log (log x)] + 2x ^{log x}^-¹ log x

76. Kindly consider the following

${(s i n x)}^{x} + s i n^{- 1}$

76. Kindly go through the solution

80. Kindly consider the following

$x^{y} + y^{x} = 1$

80. Given, xy + yx = 1

Let 4 = xy and v =., we have,

u + v = 1.

$\Rightarrow \frac{d y}{d x} + \frac{d v}{d y} = 0$ ___ (1)

So, u = xy

= log u = y log x(taking log)

Now, differentiating w r t ‘x’,

$\frac{1}{4} \frac{d y}{d x} = y \frac{d}{d x} l o g x + l o g x \frac{d y}{d x} .$

$\frac{d y}{d x} = 4 [\frac{y}{x} + l o g x \frac{d y}{d x}]$

$\Rightarrow x^{y} \cdot \frac{y}{x} + x^{y} l o g x \cdot \frac{d y}{d x}$

= xy- 1y + xy log x $\frac{d y}{d x} .$

And v = yx.

log v = x log y.

Differentiating w r t ‘x’,

$\Rightarrow \frac{1}{v} \frac{d v}{d x} = x \frac{d}{d x} l o g y + l o g y \frac{d x}{d x}$

$= \frac{x}{y} \frac{d y}{d x} + l o g y$

$\Rightarrow \frac{d v}{d x} = v [\frac{x}{y} \frac{d y}{d x} + l o g y]$

$= y^{x} \cdot \frac{x}{y} \cdot \frac{d y}{d x} + y^{x} l o g \cdot y$

= yx- 1. $x \frac{d y}{d x}$ + yx log y.

So, eqn (1) becomes

xy- 1y + xy log x $\frac{d y}{d x}$ + yx - 1 $\frac{d y}{d x}$ + yx log y = 0

$\Rightarrow \frac{d y}{d x} (x^{y} l o g x + y^{x + 1} \cdot x)$ = - (xy- 1y + yx log y)

$\Rightarrow \frac{d y}{d x} = \frac{- (x^{y - 1} \cdot y + y^{x} l o g y)}{(x^{y} l o g x + y^{x - 1} \cdot x) .}$

108. Kindly consider the following

108. Given, $y = A e^{m x} + B e^{n x}$ _______(1)

So, $\frac{d y}{d x} = A m e^{m x} + B n e^{n x}$ _______(2)

$∴ \frac{d^{2} y}{d x^{2}} = A m^{2} e^{m x} + B n^{2} e^{n x}$ _________(3)

So, L.H.S = $\frac{d^{2} y}{d x^{2}} - (m + n) \frac{d y}{d x} + m n y$

$= A m^{2} e^{m x} + B n^{2} e^{n x} - (m + n) [A m e^{m x} + B n e^{n x}] + m n [A e^{m x} + B e^{n x}]$

$= A m^{2} e^{m x} + B n^{2} e^{n x} - A m^{2} e^{m x} - B m n e^{n x} - A m n e^{m x} - B n^{2} e^{n x} + A m n e^{m x} + B m n e^{n x}$

= 0 = R.H.S.

1. Prove that the function f(x) = 5x 3f is continuous at x = 0, at x = – 3 and at x = 5.

1. Given, f (x) = 5x 3

At x = 0, $\underset{x \to 0}{l i m} f (x) = \underset{x \to 0}{l i m}$ 5x 3 = 5 0 3 = 3.

So f is continuous at x = 1.

At x = 3, $π + h$ 5x 3 = 5 ( 3) 3 = 15 3

= 18.

So f is continuous at x = 3.

At x = 5, $\forall x \in ?$ .5x 3 = 5.5 3 = 25 3 = 22.

So, f is continuous at x = 5.

2. Examine the continuity of the function f(x) = 2x2 1

2. Given, f (x) = 2x2 1

At x = 3

Lim f (x) = $\frac{d y}{d x} = \frac{- (3 x^{2} + 2 x y + y^{2})}{(x^{2} + 2 x y + 3 y^{2}) .}$ 2 (3)2 1 = 18 1 = 17.

So, f is continuous at x = 3.

5. Prove that the function $f (x) = x^{n}$ continuous at $x = n,$ where n is a positive integer.

4. Given, f (x) = x n > n = positive.

At x = 2,

(x) = n.

$\underset{x \to n}{l i m}$ f (x) = $\underset{x \to n}{l i m}$ x n = n

∴ $\underset{x \to n}{l i m}$ f (x) = f (x)

So f is continuous at x = n.

8. $f (x) =$ ${\begin{matrix} \frac{| x |}{x}, if x \neq 0 & 0, if x = 0 . \end{matrix}$

8. Given, f(x) = $f x x x$

For x = c < 0,

f(c) = 1

$\underset{x \to 0}{l i m}$ f(x) = $\underset{x \to 0}{l i m}$ 1 = 1

∴f(c) = $\underset{x \to 0}{l i m}$ f (x)

f is continuous at x $| < |$ 0.

For x = c > 0,

F (c) = 1

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ $=$ = 1.

∴f(c) = $\underset{x \to c}{l i m}$ f(x)

f is continuous at x > 0.

For x = c 0.

L.H.L. = $\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{-}}{l i m}$ ( 1) = 1

R.H.L. $\underset{x \to 0^{+}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m}$ 1 = 1

∴ L.H.L. $=$ R.H.L.

is now continuous at x = 0, point of discontinuity of f is at x = 0.

10. $f (x) =$ ${\begin{matrix} x^{3} - 3, if x \leq 2 & x^{2} + 1, if x > 2 \end{matrix}$

10. Given f (x) = ${\begin{matrix} x^{3} - 3 if x | ? | 2 & x^{2} + 1 if x > 2 \end{matrix}$

For x = c < 2,

f (c) = c3 3

$\underset{x \to c}{l i m}$ f (x) = $\underset{x \to c}{l i m}$ x3 3 = c3 3.

So f is continuous at x $| < |$ 2.

For x = c > 2

f (c) = x2 + 1 = c2 + 1

$\underset{x \to 2}{l i m}$ f (x) = $\underset{x \to 2}{l i m}$ x2 + 1 = c2 + 1 = f (c)

So, f is continuous at x $| > |$ 2.

For x = c = 2, f (2) = 23 3 = 8 3 = 5.

L.H.L. $\underset{x \to 2^{-}}{l i m}$ f (x) = $\underset{x \to 2^{-}}{l i m}$ x3 3 = 23 3 = 5.

R.H.L. $\underset{x \to 2^{+}}{l i m}$ f (x) = $\underset{x \to 2^{+}}{l i m}$ x2 + 1 = 22 + 1 = 5

∴ R.H.L. = L.H.L. = f (2).

So, f is continuous at x = 2

Hence f has no point of discontinuity.

11. $f (x) =$ ${\begin{array}{l} x^{10} - 1, if x \leq 1 & x^{2}, if x > 1 \end{array}$

11. Given, f (x) = ${\begin{array}{l} x^{10} - 1, if x \leq 1 & x^{2}, if x > 1 . \end{array}$

For x = c < 1.

f (c) = $\underset{x \to c}{l i m}$ f (x) = c10 1.

So, f is continuous for x $| > |$ 1.

For x = c > 1.

f (c) = $\underset{x \to c}{l i m}$ f (x) = c2

So, f is continuous for x $| > |$ 1.

For x = c = 1,

L.H.L = $\underset{x \to 1^{-}}{l i m}$ f (x) = $\underset{x \to 1^{-}}{l i m}$ x10 1 110 1 = 0.

R.H.L. = $\underset{x \to 1^{+}}{l i m}$ f (x) = $\underset{x \to 1^{+}}{l i m}$ x2 = 12 = 1.

∴ L.H.L $=$ R.H.L.

So, f is not continuous at x = 1.

Hence, f has point of discontinuity at x = 1.

13. $f (x) =$ ${\begin{matrix} 3, if π 0 \leq x \leq 1 & 4, if π 1 < x < 3 & 5, if π 3 \leq x \leq 10 \end{matrix}$

13. Given, f(x) = ${\begin{matrix} 3 π 0 \leq x \leq 1 & 4 π 1 < x < 3 & 5 π 3 \leq x \leq 10 . \end{matrix}$

For x = c such that $0 \leq c < 1$

f(c) = 3

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 3 = 3 = f(c)

So, f is continuous in [0, 1].

For x = c = 1,

L.H.L. = $\underset{x \to 1^{-}}{l i m}$ f(x) = $\underset{x \to 1^{-}}{l i m}$ 3 = 3.

R.H.L. $\underset{x \to 1^{+}}{l i m}$ f(x) = $\underset{x \to 1^{+}}{l i m}$ 4 = 4

∴ L.H.L $=$ R.H.L.

f is discontinuity at x = 1

for x = c such that $1 < c < 3 .$

f(c) = 4

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 4 = 4 = f(c)

So, f is continuous in $x \in (1, 3)$

For x = c = 3

L.H.L. $\underset{x \to 3^{-}}{l i m}$ f(x) = $\underset{x \to 3^{-}}{l i m}$ 4 = 4

R.H.L. $\underset{x \to 3^{+}}{l i m}$ f(x) = $\underset{x \to 3^{+}}{l i m}$ 5 = 5.

So, f is discontinuous at x = 3.

For x = c such that $3 < c \leq 10$

f (c) = 5.

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 5 = 5 = f(c)

So, f is continuous in $x \in (3, 10]$

14. $f (x) =$ ${\begin{matrix} 2 x, if x < 0 & 0, if 0 \leq x \leq 1 & 4 x, if x > 1 \end{matrix}$

14. Given f(x) = ${\begin{matrix} 2 x, if x < 0 & 0, if 0 \leq x \leq 1 & 4 x, if x > 1 . \end{matrix}$

For (c) = c < 0,

f(c) = 2c.

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 2x = 2c = f(c)

So, f is continuous at x $| < |$ 0

For x = c > 1,

f(c) = 4c

$\underset{x \to c}{l i m}$ f(x) = $\underset{x \to c}{l i m}$ 4x = 4c = f(c)

So, f is continuous at x> 1.

For x = 0

L.H.L. = $\underset{x \to 0^{-}}{l i m}$ f(x) = $\underset{x \to 0^{-}}{l i m}$ . 2x = 2 (0) = 0

R.H.L. = $\underset{x \to 0^{+}}{l i m}$ f(x) = $\underset{x \to 0^{+}}{l i m}$ . 0 = 0.

f(0) = 0.

∴ L.H.L. = R.H.L. = f(0).

So, f is continuous at x = 0.

For x = 1.

L.H.L. = $\underset{x \to 1^{-}}{l i m}$ f(x) = $\underset{x \to 1^{-}}{l i m}$ . 0 = 0

R.H.L. = $\underset{x \to 1^{+}}{l i m}$ f(x) = $\underset{x \to 1^{+}}{l i m}$ . 4x = 4 (1) = 4.

∴ L.H.L. $=$ R.H.L.

So, f is discontinuous at x = 1.

19. Is the function defined by $f (x) = x^{2} - s i n x + 5$ continuous at $x = π ?$

19. Given f (x) = x2 sin x + 5.

At x = .

$f (π) = π^{2} - s i n π + 5 = π^{2} - 0 + 5 = π^{2} + 5$

$\underset{x \to π}{l i m}$ f (x) = $\underset{x \to π}{l i m}$ [x2 sin x + 5]

If x = $π + h$ then as x, h 0, so,

$\underset{x \to π}{l i m}$ f (x) = $\underset{x \to 0}{l i m}$ [ ( + h)2 sin ( + h) + 5]

= ( + 0)2 $\underset{h \to 0}{l i m}$ $\begin{array}{r} [s i n π c o s h + c o s π \cdot s i n a] & + 5 . \end{array}$

= 2 $\underset{h \to 0}{l i m}$ sin cos h $\underset{h \to 0}{l i m}$ cos sin h + 5

= x2 0 × (1) ( 1) 0 + 5.

= 2 + 5 = f (x)

So, f is continuous at x = .

21. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

21. For two continuous fxn f(x) and g(x), $\frac{f (x)}{g (x)}, \frac{g (x)}{f (x)},$

$\frac{1}{f (x)} \frac{1}{g (x)}$ are also continuous

Let f(x) = sin x is defined x R.

Let C E R such that x = c + h. so, as x c, h 0

now, f(c) = sin c.

$\underset{x \to c}{l i m}$ f(i) = $\underset{x \to c}{l i m}$ sin x = $\underset{h \to 0}{l i m}$ sin (c + h).

= $\underset{h \to 0}{l i m}$ (sin c cos h + cos c sin h)

= sin c cos 0 + cos c sin 0

= sin c 1 + 0

= sin c

= f(c)

So, f is continuous.

Then, $\frac{1}{f (x)}$ is also continuous

$\Rightarrow \frac{1}{s i n (x)}$ is also continuous

$\Rightarrow$ cosec x is also continuous

Let g(x) = cos x is defined x R.

Then, g(c) = cos c

$\underset{x \to c}{l i m}$ g(x) = $\underset{x \to c}{l i m}$ . cos x

= $\underset{h \to 0}{l i m}$ cos (c + h).

= $\underset{h \to 0}{l i m}$ (cos c cos h sin c sin h.)

= cos c cos h sin c sin h

= cos c.

= g(c)

So, g is continuous

Then, $\frac{1}{g (x)}$ is also continuous

$\Rightarrow \frac{1}{c o s x}$ is also continuous

$\Rightarrow$ cos x is also continuous.

Hence, $\frac{g (x)}{f (x)}$ is also continuous

$\Rightarrow \frac{c o s x}{s i n x}$ is also continuous

$\Rightarrow$ cot x is also continuous .

26. $f (x) =$ ${\begin{array}{l} k x^{2}, & if x \leq 2 \\ 3, & if x > 2 \end{array} x$

26. Given f (x) = ${\begin{array}{l} k x^{2} & if x \leq 2 . \\ 3 & if x > 2 . \end{array}$

For continuous at x = 2,

f (2) = k (2)2 = 4x.

L.H.L. = $\underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{-}}{l i m} x^{2} = 4 x$

R.H.L. = $\underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} 3 = 3$

Then, L.H.L = R.H.L. = f (2)

i e, 4x = 3

$\Rightarrow x = \frac{3}{4} .$

28. $f (x) =$ ${\begin{array}{l} k x + 1, & if x \leq 5 \\ 3 x - 5, & if x > 5 \end{array} x = 5$

28. Given, f (x) ${\begin{array}{l} k x + 1, & if x \leq 5 \\ 3 x - 5, & if x > 5 . \end{array}$

For continuity at x = 5,

$\underset{x \to 5^{-}}{l i m} f (x) = \underset{x \to 5^{-}}{l i m}, k x + 1 = 5 x + 1 .$

$\underset{x \to 5^{+}}{l i m} f (x) = \underset{x \to 5^{+}}{l i m} 3 x - 5 = 15 - 5 = 10$

f (5) = 5k + 1

So, $\underset{x \to 5^{-}}{l i m} f (x) = \underset{x \to 5^{+}}{l i m} f (x) = f (5) .$

i e, 5k + 1 = 10

$\Rightarrow$ 5k = 10 1

$\Rightarrow$ k = $\frac{9}{5} .$

29. Find the values of a and b such that the function defined by

29. Given, f(x) = ${\begin{matrix} 5 & if x \leq 2 & a x + b \\ if 2 < x < 10 & 21 & if x \geq 10 \end{matrix}$

For continuity at x = 2,

$\underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = f (2)$

$\Rightarrow \underset{x \to 2^{-}}{l i m} 5 = \underset{x \to 2^{+}}{l i m} a x + b = 5 .$

$\Rightarrow$ 5 = 2a + b (i)

For continuous at x = 10,

$\underset{x \to 1 0^{-}}{l i m} f (x) = \underset{x \to 1 0^{+}}{l i m} f (x) = f (10)$

$\Rightarrow \underset{x \to 1 0^{-}}{l i m} a x + b = \underset{x \to 1 0^{+}}{l i m} 21 = 21 .$

$\Rightarrow$ 10a + b = 21 (2).

So, e q (2) 5 e q (1) we get,

10a + b 5 (2a + b) = 21 5 5.

$\Rightarrow$ 10a + b 10a 5b = 21 25.

$\Rightarrow$ 4b = 4

$\Rightarrow$ b = 1.

And putting b = 1 in e q (1),

$\Rightarrow$ 2a = 5 b = 5 1 = 4

$\Rightarrow a = \frac{4}{2} = 2 .$

Hence, a = 2 and b = 1.

30. Show that the function defined by $f (x) = c o s (x^{2})$ is a continuous function

30. Given f (x) = cos (x2)

Let g (x) = cos x is a lregononuie fa (cosine) which is continuous function

and let h (x) = x2 is a polynomial f xn which is also continuous

Hence (goh) x = g (h (x)

= g (x)2

= cos (x2)

= f (x)

is also a continuous f x being a composite fxn of how continuous f x $\forall x \in ?$

32. Examine that sin $| x |$ is a continuous function.

32. Given, f (x) = sin $x$

Let g (x) = sin x and h (x) = $x$ then as sine f x and modulus f x are continuous in x e R

g and h are continuous.

So, (goh) (x) = g (h (x) = g (|x|) = sin |x| = f (x)

Is a continuous f x being a competitive f x of two continuous f x.

34. Kindly Consider the following

$s i n (x^{2} + 5)$

34. Let f (x) = sin (x2 + 5)

Differentiating w. r t. x we get,

f'¢ (x) = $\frac{d}{d x}$ sin (x² + 5)

= cos (x² + 5) $\frac{d}{d x}$ = cos (x² + 5)

= cos (x² + 5) [2x].

= 2x cos (x² + 5).

35. Kindly consider the following

$c o s (s i n x)$

35. Let f (x) = cos (sin x).

f' (x) $\frac{d}{d x}$ cos (sin x)

= - sin (sin x) $\frac{d}{d x}$ sin x

= - sin (sin x) cos x.

36. Kindly consider the following

$s i n (a x + b)$

36. Let f (x) = sin (ax + b)

f' (x) = $\frac{d}{d x}$ sin (ax + b)

= cos (ax + b) $\frac{d}{d x}$ (ax + b)

= a cos (ax + b).

37. Kindly consider the following

37. Kindly go through the solution

38. Kindly consider the following

$\frac{s i n (a x + b)}{c o s (c x + d)}$

38. Let f(x) = $\frac{s i n (a x + b)}{c o s (c x + d)} .$

f'(x) = $\frac{c o s (c x + d) \frac{d}{d x} s i n (a x + b) - s i n (a x + b) \frac{d}{d x} c o s (c x + d)}{c o s^{2} (c x + d) .}$

= $\frac{c o s (c x + d) c o s (a x + b) \frac{d}{d x} (a x + b) + s i n (a x + b) s i n (c x + d) \frac{d (x + d)}{d x}}{c o s^{2} (c x + d)}$

= $= \frac{c o s (c x + d) c o s (a x + b) \cdot a + s i n (a x + b) s i n (c x + d) \cdot c}{c o s^{2} (c x + d)}$

39. Kindly consider the following

$c o s x^{3} s i n^{2} (x^{5})$

39. Let f (x) = cos (x³) sin² (x⁵).

f' (x) = cos (x³) $\frac{d}{d x}$ sin² (x⁵) + sin² (x⁵) $\frac{d}{d x}$ cos (x³)

= cos (x³) 2sin (x⁵) $\frac{d}{d x}$ sin (x⁵) + sin² (x⁵) [sin (x³)] $\frac{d}{d x}$ x³.

= 2 cos (x³) sin (x⁵). cos (x⁵) $\frac{d}{d x}$ (x⁵) - sin² (x⁵) sin (x³). 3x²

= 2. cos (x³) sin (x⁵) cos (x⁵). 5 - 3x²sin² (x⁵) sin (x³)

= x² sin (x⁵). [2x² cos (x³) cos (x⁵) - 3 sin (x⁵) sin x³].

44. Find $\frac{d y}{d x}$ in the following:

$2 x + 3 y = s i n x$

44. Kindly go through the solution

46. Kindly consider the following

$a x + b y^{2} = c o s y$

46. Given, ax + by2 = cos y.

Differentiating w r t ‘x’ we get,

$\frac{d}{d x} (a x + b y^{2}) = \frac{d x}{d x} c o s y$

= a + b 2y = - sin y $\frac{d y}{d x}$ + sin y $\frac{d y}{d x}$ = -a

= $\frac{d y}{d x} = \frac{- 9}{2 b y + s i n y} .$

= 2by $\frac{d y}{d x}$

53. Kindly consider the following

53. Kindly go through the solution

54. Kindly consider the following

$y = c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), - 0 < x < 1$

54. Kindly go through the solution

55. Kindly consider the following

$y = s i n^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}), - 0 < x < 1$

55. Kindly go through the solution

56. Kindly consider the following

$y = c o s^{- 1} (\frac{2 x}{1 + x^{2}}), - 1 < x < 1$

56. Kindly go through the solution

57. Kindly consider the following

57. Kindly go through the solution

60. Kindly consider the following

$e s i n^{- 1} x$

60. Kindly go through the solution

61. Kindly consider the following

e^x³

61. Kindly go through the solution

Let y = e^x³

62. Kindly consider the following

$s i n (t a n^{- 1} e^{- x})$

62. Kindly go through the solution

63. Kindly consider the following

$l o g (c o s e^{x})$

63. Let y = log (cos e^x).

$∴ \frac{d y}{d x} = \frac{d}{d x}$ log (cos e^x)

$= \frac{1}{c o s e^{x}} \frac{d}{d x}$ (cos e^x)

= - $\frac{s i n e^{x}}{c o s e^{x}} \frac{d e^{x .}}{d x}$

= -e^x [tan e^x]

65. Kindly consider the following

65. Kindly go through the solution

66. Kindly consider the following

$l o g (l o g x), x > 1$

66. Let y = log (log x)

$∴ \frac{d y}{d x} = \frac{1}{l o g x} \frac{d}{d x} (l o g x)$

$= \frac{1}{l o g x} \times \frac{1}{x}$

$\frac{d y}{d x} = \frac{1}{x l o g x}$

67. Kindly consider the following

$\frac{c o t x}{l o g x}, x > 0$

67. $y = \frac{c o t x}{l o g x}$

$∴ \frac{d y}{d x} = \frac{l o g x \frac{d}{d a} c o x - c o s x \frac{d}{d x} l o g x}{{[l o g x]}^{2}}$

$= \frac{- s i n x \cdot l o g x - c o s x \times \frac{1}{x}}{{[l o g x]}^{2} .}$

$= \frac{- [x s i n x l o g x - c o s x]}{x {(l o g x)}^{2} .}$

68. Kindly consider the following

$c o s (l o g x + e^{x}), x > 0$

68. Let y = cos (log x + e^x)

$∴ \frac{d y}{d x} = \frac{d}{d x}$ cos (log x + e^x)

= - sin (log x + e^x) $\frac{d}{d x}$ (log x + e^x)

= - sin (log x + e^x) $[\frac{1}{x} + e^{x}]$

$= - [\frac{1}{x} + e^{x}]$ sin (log x + e^x).

71. Kindly consider the following

${(l o g x)}^{c o s x}$

71. Let y = (log x) cos x

Taking loge on both sides,

Log y = cos x [log (log x)]

Differentiating w r t ‘x’ we get,

$\frac{1}{y} \frac{d y}{d x} = c o s x \frac{d}{d x}$ log (log x) +log (log x) $\frac{d c o s x}{d x}$

$= c o s x \times \frac{1}{l o g x} \times \frac{d (l o g x)}{d x}$ + log (log x) (- sin x)

$= \frac{c o g x}{x l o g x}$ - sin x log (log x)

= $\frac{d y}{d x} = y {[\frac{c o s x}{x l o g 2} - s i n x \cdot l o g (l o g x)]}_{.}$

$\Rightarrow \frac{d y}{d x} = {(l o g x)}^{c o s x} [\frac{c o s x}{x l o g x} - s i n x \cdot l o g (l o g x)]$

74. Kindly consider the following

${(1 + \frac{1}{x})}^{x} + x^{(1 + \frac{1}{x})}$

74 . Let y = ${(x + \frac{1}{x})}^{x}$ + ${(1 + \frac{1}{x})}^{x}$

Putting u = ${(x + \frac{1}{x})}^{x}$ and v = ${(1 + \frac{1}{x})}^{x}$ we get,

y = u + v

$∴ \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ _____ (1)

As u = ${(x + \frac{1}{x})}^{x}$

Taking log,

= log u = x log $(x + \frac{1}{x})$

Differentiating w r t ‘x’ we get,

$\frac{1}{4} \frac{d y}{d x} = x \frac{d}{d x} l o g$ $(x + \frac{1}{x}) + l o g (x + \frac{1}{x}) \frac{d x}{d x}$

$= x \frac{1}{(x + \frac{1}{x})} \frac{d}{d x} (x + \frac{1}{x})$ + log $(x + \frac{1}{x})$ 1.

$= x \cdot \frac{1}{(x + \frac{1}{x})} \times (1 - \frac{1}{x^{2}}) + l o g (x + \frac{1}{x})$

$= \frac{x . \frac{x^{2} - 1}{x^{2}}}{(\frac{x^{2} + 1}{x})} + l o g (x + \frac{1}{x}) .$

$= \frac{x^{2} - 1}{x^{2} + 1} + l o g (x + \frac{1}{x}) .$

$\Rightarrow \frac{d u}{d x} = u [\frac{x^{2} - 1}{x^{2} + 1} + l o g (x + \frac{1}{x})]$

$= {(x + \frac{1}{x})}^{x} {[\frac{x^{2} - 1}{x^{2} + 1} + l o g (x + \frac{1}{x})]}_{.}$

And v = x $(1 + \frac{1}{x})$

Taking log, log v = $(1 + \frac{1}{x})$ log x

Differentiating W r t ‘x’,

$\frac{1}{v} \frac{d v}{d x} = (1 + \frac{1}{x}) \frac{d}{d x}$ log x + log x $\frac{d}{d x} {(1 + \frac{1}{x})}_{.}$

$(1 + \frac{1}{x}) \times \frac{1}{x}$ + log x ${(0 - \frac{1}{x^{2}})}_{.}$

$\frac{d v}{d x}$ = v $[\frac{1}{x} (1 + \frac{1}{x}) - \frac{l o g x}{x^{2}}]$ = $x^{(1 + \frac{1}{x})} [\frac{1}{x} (1 + \frac{1}{x}) - \frac{l o g x}{x^{2}}]$

Hence, eqn (1) becomes,

$\frac{d y}{d x} = {(x + \frac{1}{x})}^{x} {[\frac{x^{2} - 1}{x^{2} + 1} + l o g (x + \frac{1}{x})]}_{.}$ $+ x^{(1 + \frac{1}{x})} [\frac{1}{x} (1 + \frac{1}{x}) - \frac{l o g x}{x^{2}}]$

79. Kindly consider the following

${(x c o s x)}^{x} + {(x s i n x)}^{\frac{1}{x}}$

Find $\frac{d y}{d x}$ of the functions given in Exercises 12 to 15.

79. Let y = (x cos x) x + (x sin) $\frac{1}{x}$

Putting u = (x cos x)^x and v = (x sin x) $\frac{1}{x}$ we, have,

y = u + v

$\Rightarrow \frac{d y}{d x} = \frac{d y}{d x} + \frac{d v}{d x}$ ____ (1)

As u = (x cos x)^x :

Taking log,

Log u = x log (x cos x)

= x [log x + log (cos x)]

Differentiating w r t ‘x’ we get,

$\frac{1}{4} \frac{d u}{d x} = x \frac{d}{d x}$ [log x + log (cos x)] + [log x + dog (cos x)] $\frac{d x}{d x}$

$= x [\frac{1}{x} + \frac{1}{c o s x} \frac{d c o s x}{d x}]$ + [log x +log (cos x)]

$= [1 + \frac{x}{c o s x} (- s i n x)]$ + log x + log (cos x)

= 1 -x tan x + log (x cos x)

$\frac{d y}{d x}$ = 4 [1 -x tan x + log (x cose)]

=(x cos x)^x (x cos x)^x [1 -x tan + log + log (x cos x)]

And v = (x sin x) $\frac{1}{x}$

Taking log, log v = $\frac{1}{x}$ log (x sin x)

$= \frac{1}{x}$ (log x + log sin x)

Differentiating w r t ‘x’

$\frac{1}{v} \frac{d v}{d x} = \frac{1}{x} \cdot \frac{d}{d x}$ (log x + log sin x) + (log x + log sin x) $\frac{d}{d x} (\frac{1}{x})$

$= \frac{1}{x} [\frac{1}{x} + \frac{1}{s i n x} \frac{d}{d x} s i n x]$ + log x + log sin x) $(- \frac{1}{x^{2}})$

= $[\frac{1}{x^{2}} + \frac{c o s x}{s i n x}] - \frac{l o g (x s i n x)}{x^{2}}$

$\frac{d u}{d x} =_{.}$ $v [\frac{1}{x^{2}} + \frac{c o t x}{x} - \frac{l o g (x s i n x)}{x^{2}}]$

= (x sin x) $\frac{1}{x}$ ${[\frac{1}{x^{2}} + c o t x - \frac{l o g (x s i n x)}{x^{2}}]}_{.}$

Q Eqn (1) becomes,

$\frac{d y}{d x}$ = (x cos x)^x [1 -x tan + log (x cos x)] + (x sin x) $\frac{1}{2}$

$[\frac{1}{x^{2}} + \frac{c o t x}{x} - \frac{l o g (x s i n x)}{x^{2}}]$

= (x cos x)^x [1 -x tan x + log (x cos x)] + (x sin x) $\frac{1}{2}$

$[\frac{1 + x c o t x - l o g (x s i n x)}{x^{2}}]$

Find $\frac{d y}{d x}$ of the functions given in Exercises 12 to 15.

82. Kindly consider the following

${(c o s x)}^{y} = {(c o s y)}^{x}$

82. Given, (cos x)y = (cos y)x

Taking log, y log (cos x) = x log (cos y)

Differentiating w r t ‘x’ we get,

= $y \frac{d}{d x}$ log (cos x) + log (cos x) $\frac{d y}{d x} = x \frac{d}{d x}$ log (cos y) + dog (cos y) $\frac{d x}{d x}$

= y´ $\frac{1}{c o s x} \frac{d}{d x}$ cos x + log (cos x) $\frac{d y}{d x}$ = x´ $\frac{1}{c o s y} \frac{d}{d x} c o s y + l o g (c o s y) .$

$\Rightarrow - y \frac{s i n x}{c o s x} + l o g (c o s x) \frac{d y}{d x} = - x \frac{s i n y}{c o s y} \frac{d y}{d x} + l o g (c o s y)$

= log (cos x) $\frac{d y}{d x}$ + x tan $\frac{d y}{d x} = y c t a n x + l o g (c o s y)$

$\Rightarrow \frac{d y}{d x} [l o g (c o s x) + x t a n y]$ = y tan x + log (cos y )

$\Rightarrow \frac{d y}{d x} = \frac{y t a n x + l o g (c o s y}{l o g (c o s x) + x t a n y} .$

83. Kindly consider the following

$x y = e^{(x - y)}$

83. Given, xy = ex-y.

Taking log,

log (x + y) = log (ex-y).

=logx + log y = (x-y) log e.

= logx +log y = x -y {Q log e = 1}

Differentiating w r t ‘x’ we get,

$\Rightarrow \frac{1}{x} + \frac{1}{y} \frac{d y}{d x} = 1 - \frac{d y}{d x}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x} + \frac{d y}{d x} = 1 - \frac{1}{2}$

$\Rightarrow \frac{d y}{d x} (\frac{1}{y} + 1) = (\frac{x - 1}{x})$

$\Rightarrow \frac{d y}{d x} (\frac{1 + y}{y}) = (\frac{x - 1}{x})$

$\Rightarrow \frac{d y}{d x} = \frac{y (x - 1)}{x (1 + y)}$

91. Kindly consider the following

$x = c o s θ - c o s 2 θ, y = s i n θ - s i n 2 θ$

91. Given, $x = a (c o s t + l o g t a n \frac{t}{2}) y = a s i n t$

Differentiating w r t we get,

$\frac{d x}{d t} = a \frac{d}{d t} [c o s t + l o g (t a n \frac{t}{2})]$

$= a [- s i n t + \frac{1}{t a n \frac{t}{2}} \frac{d}{d t} (t a n \frac{t}{2})]$

$= a [- s i n t + \frac{1}{t a n \frac{t}{2}} . s e c^{2} \frac{t}{2} \frac{d}{d t} (\frac{t}{2})]$

$= a [- s i n t + \frac{c o s \frac{t}{2}}{s i n \frac{t}{2}} \times \frac{1}{c o s^{2} \frac{t}{2}} \times \frac{1}{2}]$

$= a [- s i n t + \frac{1}{2 s i n \frac{t}{2} c o s \frac{t}{2}}]$

$= a [- s i n t + \frac{1}{s i n 2 \times \frac{t}{2}}]$

$= a [- s i n t + \frac{1}{s i n t}] = a [\frac{1 - s i n^{2} t}{s i n t}]$

$= a \frac{c o s^{2} t}{s i n t} {? 1 = c o s^{2} x + s i n^{2} x}$

$\frac{bd y}{d t} = \frac{d}{d t} (a s i n t) = a c o s t$

$∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a c o s t}{\frac{a c o s^{2} t}{s i n t}} = \frac{s i n t}{c o s t} = t a n t$

Find the second order derivatives of the functions given in Exercise 1 to 10

95. $x^{2} + 3 x + 2$

95. Let y = x2 + 3x + 2

So, $\frac{d y}{d x} = 2 x + 3 + 0$ (differentiation w r t 'x')

$? \frac{d^{2} y}{d x^{2}} = 2 + 0$ (Again “ “ ) = 2

96. Kindly consider the following

$x^{20}$

96. Let $y = x^{20}$

So, $\frac{d y}{d x} = 20 x^{20 - 1} = 20 x^{19}$

$∴ \frac{d^{2} y}{d x^{2}} = 20 \times 19 x^{19 - 1}$

$= 380 x^{18}$

97. Kindly consider the following

$x . c o s x$

97. Let $y = x c o s x$

So, $\frac{d y}{d x} = x \frac{d}{d x} c o s x + \frac{d x}{d x} c o s x$

$= - x s i n x + c o s x$

$∴ \frac{d^{2} y}{d x^{2}} = - x \frac{d}{d x} s i n x - s i n x \frac{d x}{d x} + \frac{d}{d x} c o s x$

$= - x c o s x - s i n x + (- s i n x)$

$= - (x c o s x + 2 s i n x)$

98. Kindly consider the following

$l o g x$

98. Let $y = l o g x$

So, $\frac{d y}{d x} = \frac{d}{d x} l o g x = \frac{1}{x}$

$∴ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} \frac{1}{x} = \frac{d}{d x} x^{- 1} = - 1 x^{- 1 - 1} = \frac{- 1}{x^{2}}$

99. Kindly consider the following

$x^{3} l o g x$

99. Let $y = x^{3} l o g x$

So, $\frac{d y}{d x} = x^{3} \frac{d}{d x} l o g x + l o g 2 . \frac{d x^{3}}{d x}$

$= x^{3} . \frac{1}{x} + l o g x \cdot 3 x^{2}$

$= x^{2} + l o g x (3 x^{2})$

$∴ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (x^{2} + l o g x \cdot 3 x^{2})$

$= 2 x + 6 x l o g x + 3 x^{2} \times \frac{1}{x}$

$= 2 x + 6 x l o g x + 3 x$

$= 5 x + 6 x l o g x$

100. Kindly consider the following

$e^{x} s i n 5 x$

100. Let $y = e^{x} s i n 5 x$

so, $\frac{d y}{d x} = e^{x} \frac{d}{d x} s i n 5 x + s i n 5 x \frac{d}{d x} e^{x}$

$= e^{x} \cdot c o s 5 x \frac{d}{d x} (5 x) + e^{x} s i n 5 x$

$= 5 e^{x} c o s 5 x + e^{x} s i n 5 x .$

$∴ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (5 e^{x} c o s 5 x + e^{x} s i n 5 x)$

$= 5 e^{x} \frac{d}{d x} c o s 5 x + 5 c o s 5 x \frac{d}{d x} e^{x} + e^{x} \frac{d}{d x} s i n 5 x + s i n 5 x \frac{d}{d x} e^{x}$

$= - 5 e^{x} s i n 5 x \frac{d}{d x} (5 x) + 5 e^{x} c o s 5 x + e^{x} c o s 5 x \frac{d}{d x} (5 x) + e^{x} s i n 5 x$

$= - 25 e^{x} s i n 5 x + 5 e^{x} c o s 5 x + 5 e^{x} c o s 5 x + e^{x} s i n 5 x$

$= e^{x} (10 \cdot c o s 5 x - 24 s i n 5 x)$

$= 2 e^{x} (5 c o s 5 x - 12 s i n 5 x)$

101. Kindly consider the following

$e^{6 x} c o s 3 x$

101. Let $y = e^{6 x} c o s 3 x$

So, $\frac{d y}{d x} = e^{6 x} \frac{d}{d x} c o s 3 x + c o s 3 x \frac{d}{d x} e^{6 x}$

$= e^{6 x} (- s i n 3 x) \frac{d}{d x} (3 x) + c o s 3 x \cdot e^{6 x} \frac{d}{d x} (6 x)$

$= e^{6 x} [- 3 s i n 3 x + 6 c o s 3 x]$

$∴ \frac{d^{2} y}{d x^{2}} = e^{6 x} \frac{d}{d x} [- 3 s i n 3 x + 6 c o s 3 x] + [- 3 s i n 3 x + 6 c o s 3 x] \frac{d}{d x} e^{6 x}$

$= e^{6 x} [- 3 c o s 3 x \frac{d}{d x} (3 x) + 6 (- s i n 3 x) \frac{d}{d x} (3 x)] + [- 3 s i n 3 x + 6 c o s 3 x] e^{6 x} \frac{d}{d x} (6 x)$

$= e^{6 x} {- 9 c o s 3 x - 18 s i n 3 x - 18 s i n 3 x + 36 c o s 3 x}$

$= e^{6 x} (27 c o s 3 x - 36 s i n 3 x)$

$= 9 e^{6 x} (3 c o s 3 x - 4 s i n 3 x)$

102. Kindly consider the following

$t a n^{- 1} x$

102. Let $y = t a n^{- 1} x$

So, $\frac{d y}{d x} = \frac{d}{d x} t a n^{- 1} x = \frac{1}{1 + x^{2}}$

$∴ \frac{d^{2} y}{d x^{2}} = \frac{(1 + x^{2}) \frac{d}{d x} (1) - (1) \frac{d}{d x} (1 + x^{2})}{{(1 + x^{2})}^{2}}$

$= \frac{- 2 x}{{(1 + x^{2})}^{2}}$

103 Kindly consider the following

$l o g (l o g x)$

103. Let $y = l o g (l o g x)$

So, $\frac{d y}{d x} = \frac{1}{l o g x} \frac{d}{d x} l o g x = \frac{1}{x l o g x}$

$∴ \frac{d^{2} y}{d x^{2}} = \frac{x l o g x \frac{d}{d x} (1) - 1 \cdot \frac{d}{d x} (x l o g x)}{{(x l o g x)}^{2}}$

$= \frac{- [x \frac{d}{d x} l o g x + l o g x \frac{d x}{d x}]}{{[x l o g x]}^{2}}$

$= \frac{- (x \times \frac{1}{x} + l o g x)}{{[x l o g x]}^{2}}$

$= \frac{- (1 + l o g x)}{{(x l o g x)}^{2}}$

105. Kindly consider the following

$y = 5 c o s x - 3 s i n x, \frac{d^{2} y}{d x^{2}} + y = 0$

105. Given, $y = 5 c o s x - 3 s i n x$

Differentiating w r t x we get,

$\frac{d y}{d x} = 5 \frac{d}{d x} c o s x - 3 \frac{d}{d x} s i n x$

$- 5 s i n x - 3 c o s x .$

Differentiating again w r t. ‘x’ we get,

$\frac{d^{2} y}{d x^{2}} = - 5 \frac{d}{d x} s i n x - 3 \frac{d}{d x} c o s x$

$= - 5 c o s x + 3 s i n x$

$= - [5 c o s x - 3 s i n x]$

$= - y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} + y = 0$ . Hence proved.

106. Kindly consider the following

106. Kindly go through the solution

109. Kindly consider the following

109. Given, $y = 500 {e^{7}}^{x} + 600 {e^{- 7}}^{x}$

So, $\frac{d y}{d x} = 500 \times 7 {e^{7}}^{x} + 600 (- 7) {e^{- 7}}^{x}$

$∴ \frac{d^{2} y}{d x^{2}} = 500 \times 7^{2} {e^{7}}^{x} + 600 \times 7^{2} {e^{- 7}}^{x}$

$= 49 [500 e^{7 x} + 600 e^{- 7 x}]$

$= 49 \times y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 49 y$

112. Verify Rolle’s Theorem for the function $f (x) = x^{2} + 2 x - 8, x \in [- 4, 2]$

112. Given, $f (x) = x^{2} + 2 x - 8$ , being polynomial function is continuous in $[- 4, 2]$ and also differentiable in $(- 4, 2)$ .

$\begin{array}{l} f (- 4) = {(- 4)}^{2} + 2 (- 4) - 8 \\ = 16 - 8 - 8 \\ = 0 \\ f (2) = {(2)}^{2} + 2 \times 2 - 8 \\ = 4 + 4 - 8 \\ = 0 \end{array}$

Therefore, $f (- 4) = f (2) = 0$

The value of $f (x)$ at -4 and 2 coincides.

Rolle’s Theorem states that there is a point $c \in (- 4, 2)$ such that $f^{'} (c) = 0$ $f (x) = x^{2} + 2 x - 8$

Therefore, $f^{'} (x) = 2 x + 2$

Hence,

$\begin{array}{l} f^{'} (c) = 0 \\ 2 c + 2 = 0 \\ c = - 1 \end{array}$

Thus, $c = - 1 \in (- 4, 2)$

Hence, Rolle’s Theorem is verified.

113. Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s Theorem from these examples?

$f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

$f (x) = x^{2} - 1$ for $x \in [1, 2]$

113. Solution:

By Rolle’s Theorem, for a function $f [a, b] \to R,$ if

f is continuous on $[a, b]$

f is differentiable on $(a, b)$

f(a)= f(b)

then, there exists some $c \in (a, b)$ such that $f^{'} (c) = 0$

therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f(x) is not continuous at every integral point.

In particular, f(x) is not continuous at x=5 and x=9

$\Rightarrow$ f(x) is not continuous in $[5, 9]$

Also, $\begin{array}{l} f (5) = [5] = 5, a n d, f (9) = [9] = 9 \\ ∴ f (5) \neq f (9) \end{array}$

The differentiability of f in $(5, 9)$ is checked as follows.

Let n be an integer such that $n \in (5, 9)$ .

The left hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - 1 - n}{h} = \underset{h \to 0}{l i m} \frac{- 1}{h} = \infty$

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x=n are not equal, f is not differentiable at x=n

$∴$ f is not differentiable in (5,9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for $f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

It is evident that the given function f(x) is not continuous at every integral point.

In particular, f(x) is not continuous at x=-2 and x=2

$\Rightarrow$ f(x) is not continuous in $[- 2, 2]$ .

Also, $\begin{array}{l} f (- 2) = [- 2] = - 2, a n d, f (2) = [2] = 2 \\ ∴ f (- 2) \neq f (2) \end{array}$

The differentiability of f in $(- 2, 2)$ is checked as follows.

Let n be an integer such that $n \in (- 2, 2)$ .

The left hand limit of f at x=n is,

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left- and right-hand limits of f at x=n are not equal, f is not differentiable at x=n

$∴$ f is not differentiable in (-2,2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for $f (x) = [x]$ for $x \in [- 2, 2]$

$f (x) = x^{2} - 1$ for $x \in [1, 2]$

It is evident that f, being a polynomial function, is continuous in $[1, 2]$ and is differentiable on (1,2).

$\begin{array}{l} f (1) = 1^{2} = 0 \\ f (2) = 2^{2} - 1 = 3 \\ ∴ f (1) \neq f (2) \end{array}$

It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for $f (x) = x^{2} - 1$ for $x \in [1, 2]$ .

115. Verify Mean Value Theorem, if $f (x) = x^{2} - 4 x - 3$ in the interval [a,b], where a = 1 and b = 4.

115.

Solution :
The given function is $f (x) = x^{2} - 4 x - 3$ f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x − 4.

$\begin{array}{l} f (1) = 1^{2} - 4 \times 1 - 3, f (4) = 4^{2} - 4 \times 4 - 3 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (4) - f (1)}{4 - 1} = \frac{- 3 - (- 6)}{3} = \frac{3}{3} = 1 \end{array}$

Mean Value Theorem states that there is a point c ∈ (1, 4) such that f' (c) = 1

$\begin{array}{l} f' (c) = 1 \\ \Rightarrow 2 c - 4 = 1 \\ \Rightarrow c = \frac{5}{2}, w h e r e, c = \frac{5}{2} \in (1, 4) \end{array}$

Hence, Mean Value Theorem is verified for the given function.

116. Verify Mean Value Theorem, if $f (x) = x^{3} - 5 x^{2} - 3 x$ in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f'(c) = 0

116. Solution:
The given function f is $f (x) = x^{3} - 5 x^{2} - 3 x$ f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x2 − 10x − 3.

$\begin{array}{l} f (1) = 1^{3} - 5 \times 1^{2} - 3 \times 1, f (3) = 3^{3} - 5 \times 3^{2} - 3 \times 3 = - 27 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (3) - f (1)}{3 - 1} = \frac{- 27 - (- 7)}{3 - 1} = - 10 \end{array}$

Mean Value Theorem states that there exists a point c ∈ (1, 3) such that f'(c) = - 10

$\begin{array}{l} f' (c) = - 10 \\ \Rightarrow 3 c^{2} - 10 c - 3 = 10 \\ \Rightarrow 3 c^{2} - 10 c + 7 = 0 \\ \Rightarrow 3 c^{2} - 3 c - 7 c + 7 = 0 \\ \Rightarrow 3 c (c - 1) - 7 (c - 1) = 0 \\ \Rightarrow (c - 1) (3 c - 7) = 0 \\ \Rightarrow c = 1, \frac{7}{3}, w h e r e, c = \frac{7}{3} \in (1, 3) \end{array}$

Hence, Mean Value Theorem is verified for the given function and c = 7/3 ∈ (1,3) is the only point for which f'(c) = 0

117. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

117. Solution :
Mean Value Theorem states that for a function f[a,b] →R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

then, there exists some c ∈ (a, b) such that $f' (c) = \frac{f (b) - f (a)}{b - a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

⇒ f (x) is not continuous in [5, 9].

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n ∈ (5, 9).

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

∴f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f (x) = [x]$ for $x \in [5, 9]$ .

(ii) $f (x) = [x]$ $i n x \in [- 2, 2]$

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2

⇒ f (x) is not continuous in [−2, 2].

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n ∈ (−2, 2).

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

∴f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f (x) = [x]$ for $x \in [- 2, 2]$

(iii) $f (x) = x^{2} - 1$ for $x \in [1, 2]$

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for $f (x) = x^{2} - 1$ for $x \in [1, 2]$

It can be proved as follows.

$\begin{array}{l} f (1) = 1^{2} - 1 = 0, f (2) = 2^{2} - 1 = 3 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (2) - f (1)}{2 - 1} = \frac{3 - 0}{1} = 3 \\ f' (x) = 2 x \\ ∴ f' (c) = 3 \\ \Rightarrow 2 c = 3 \\ \Rightarrow c = \frac{3}{2} = 1.5, w h e r e, 1.5 \in (1, 2) \end{array}$

118. Kindly consider the following

${(3 x^{2} - 9 x + 5)}^{9}$

118. Let $y = {(3 x^{2} - 9 x + 5)}^{9}$

So, $\frac{d y}{d x} = 9 {(3 x^{2} - 9 x + 5)}^{8} \frac{d}{d x} (3 x^{2} - 9 x + 5)$

$= 9 {(3 x^{2} - 9 x + 5)}^{8} \times (6 x - 9)$

$= 27 {(3 x^{2} - 9 x + 5)}^{8} (2 x - 3)$

120. Kindly consider the following

${(5 x)}^{3 c o s 2 x}$

120. Let $y = {(5 x)}^{3 c o s 2 x}$

Taking log,

$l o g y = 3 c o s 2 x [l o g 5 x]$

Differentiating w r t. x,

$\frac{1}{y} \frac{d y}{d x} = 3 c o s 2 x \frac{d}{d x} l o g 5 x + 3 l o g 5 x \frac{d}{d x} c o s 2 x$

$= 3 [c o s 2 x \cdot \frac{1}{5 x} \frac{d}{d x} (5 x) - l o g 5 x \times s i n 2 x \frac{d}{d x} 2 x]$

$\frac{d y}{d x} = 3 y [c o s 2 x \times \frac{5}{5 x} - l o g 5 x \cdot s i n 2 x \cdot 2]$

$= 3 {(5 x)}^{3 c o s 2 x} [\frac{c o s 2 x}{x} - 2 s i n 2 x l o g 5 x]$

121. Kindly consider the following

121. Kindly go through the solution

122. Kindly consider the following

122. Kindly go through the solution

124. Kindly consider the following

${(l o g x)}^{l o g x}, x > 1$

124. Let $y = {(l o g x)}^{l o g x}$

Taking log,

$l o g y = l o g x l o g (l o g x)$

Differentiating w r t. x, we get,

$\frac{1}{y} \frac{d y}{d x} = l o g x \frac{d}{d x} l o g (l o g x) + l o g (l o g x) \frac{d}{d x} (l o g x)$

$= l o g x \times \frac{1}{l o g x} \frac{d}{d x} l o g x + \frac{l o g (l o g x)}{x}$

$= \frac{1}{x} + \frac{l o g (l o g x)}{x}$

$\frac{d y}{d x} = \frac{y}{x} [1 + l o g (l o g x)] .$

$= {(l o g x)}^{l o g x} [\frac{1 + l o g (l o g x)}{x}]$

125. Kindly consider the following

125. Let $y = c o s (a c o s x + b s i n x)$

So, $\frac{d y}{d x} = - s i n (a c o s x + b s i n x) \frac{d}{d x} (a c o s x + b s i n x)$

$= - s i n (a c o s x + b s i n x) [- a s i n x + b c o s x] .$

$= s i n (a c o s x + b s i n x) (a s i n x - b c o s x)$

126. Kindly consider the following

${(s i n x - c o s x)}^{(}^{s i n x - c o s x}^{)}, \frac{π}{4} < x < \frac{3 π}{4}$

126. Let $y = {(s i n x - c o s x)}^{s i n x - c o s x}$

Taking log,

$l o g y = (s i n x - c o s x) l o g (s i n x - c o s x)$

Differentiating w r t ‘x’ we get,

$\frac{1}{y} \frac{d y}{d x} = (s i n x - c o s x) \frac{d l o g}{d x} (s i n x - c o s x) + l o g (s i n x - c o s x) \frac{d}{d x} (s i n x - c o s x) .$

$= (s i n x - c o s x) \times \frac{1}{(s i n x - c o s x)} \times (c o s x + s i n x) + l o g (s i n x - c o s x) (c o s x + s i n x)$

$= (c o s x + s i n x) [1 + l o g (s i n x - c o s x)]$

$\frac{d y}{d x} = y (c o s x + s i n x) [1 + l o g (s i n x - c o s x)]$

$\frac{d y}{d x} = {(s i n x - c o s x)}^{s i n x - c o s x} \times (c o s x + s i n x) [1 + l o g (s i n x - c o s x)]$

127. Kindly consider the following

127. Let $y = x^{x} + x^{a} + a^{x} + a^{a} .$

So, $\frac{d y}{d x} = \frac{d x^{x}}{d x} + \frac{d x^{a}}{d x} + \frac{d a^{x}}{d x} + \frac{d a^{a}}{d x} .$

$\to \frac{d y}{d x} = \frac{d y}{d x} + a x^{a - 1} + a^{x} l o g a + 0 .$ _________(1)

Where $u = x^{x}$

$l o g u = x l o g x$ (Taking log)

$\to \frac{1}{y} \frac{d y}{d x} = x \frac{d}{d x} l o g x + l o g x \frac{d x}{d x}$ (Differentiation w r t ‘x’)

$\to \frac{1}{y} \frac{d y}{d x} = \frac{x}{x} + l o g x$

$\to \frac{d y}{d x} = u [1 + l o g x]$

$= x^{x} [1 + l o g x]$

Hence eqn (1) becomes,

$\frac{d y}{d x} = x^{x} [1 + l o g x] + a x^{a - 1} + a^{x} l o g a .$

129. Kindly consider the following

129. Given, $y = 12 (1 - c o s t) . x = 10 (t - s i n t) .$

Differentiating w r t ‘t’ we get,

$\frac{d y}{d t} = 12 \frac{d}{d t} (1 - c o s t) = 12 (0 - (- s i n t)) = 12 s i n t .$

$\frac{d x}{d t} = 10 \frac{d}{d t} (t - s i n t) = 10 (1 - c o s t) .$

$∴ \frac{d y}{d x} = \frac{d y / d t}{d x / d t}$

$= \frac{12 s i n t}{10 (1 - c o s t)} = \frac{12 s i n t}{10 [1 - c o s t]}$

$= \frac{12 \times 2 s i n t / t c o s t / 2}{10 \times 2 s i n^{2} t / 2}$

${\begin{matrix} ? s i n 2 θ = 2 s i n θ c o s θ & c o s 2 θ = 1 - 2 s i n^{2} θ \end{matrix}}$ $= \frac{6}{5} \frac{c o s t / 2}{s i n t / 2} = \frac{6}{5} c o t t / 2$

130. Kindly consider the following

130. Kindly go through the solution

135. Kindly consider the following

135. Given, $f (x) = | x |^{3} = {\begin{matrix} x^{3} & i f x \geq 0 \\ - x^{3} & i f x < 0 \end{matrix}$

For $x \geq 0, f (x) = | x |^{3} = x^{3}$

and $\begin{matrix} f^{'} (x) = 3 x^{2} & f^{''} (x) = 6 x \end{matrix}$

For $x < 0, f (x) = | x |^{3} = {(- x)}^{3} = - x^{3} .$

so, $\begin{matrix} f^{'} (x) = - 3 x^{2} & f^{''} (x) = - 6 x \end{matrix}$

Hence, $f^{''} (x) = {\begin{matrix} 6 x, & i f x \geq 0 \\ - 6 x, & i f x < 0 \end{matrix}$

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