This document provides an overview of functions, limits, and continuity. It defines key concepts such as domain and range of functions, and examples of standard real functions. It also covers even and odd functions, and how to calculate limits, including left and right hand limits. Methods for evaluating algebraic limits using substitution, factorization, and rationalization are presented. The objectives are to understand functions, domains, ranges, and how to evaluate limits of functions.
This document discusses functions, limits, and continuity. It begins by defining functions, domains, ranges, and some standard real functions like constant, identity, modulus, and greatest integer functions. It then covers limits of functions including one-sided limits and properties of limits. Examples are provided to illustrate evaluating limits using substitution and factorization methods. The overall objectives are to understand functions, domains, ranges, limits of functions and methods to evaluate limits.
This document provides examples and exercises on inverse functions. It first shows an example of finding the inverse of the function f(x) = 2x-3/(x+2). It gives the steps to solve for x in terms of y and obtain the inverse function f^-1(x) = -2x-3/(x-2). It then asks the reader to verify that f(f^-1(x)) = x. The exercises provide 20 functions and ask the reader to find their inverses and verify the inverses are correct. It also gives graphs of 8 functions and asks the reader to determine properties of the inverse graphs, including their domains and ranges and any fixed or end points.
This module discusses methods for finding the zeros of polynomial functions of degree greater than 2, including: factor theorem, factoring, synthetic division, and depressed equations. It introduces the number of roots theorem, which states that a polynomial of degree n has n roots. It also discusses determining the rational zeros of a polynomial using the rational roots theorem and factor theorem. Examples are provided to illustrate these concepts and methods.
The document discusses interpolation, which involves using a function to approximate values between known data points. It provides examples of Lagrange interpolation, which finds a polynomial passing through all data points, and Newton's interpolation, which uses divided differences to determine coefficients for approximating between points. The examples demonstrate constructing Lagrange and Newton interpolation polynomials using given data sets.
The document is an advertisement for Vedantu, an online education platform, promoting their free online admission test to win scholarships for classes 6-12, JEE, and NEET. It highlights success stories of students who scored well in board exams and engineering/medical entrance exams after taking online classes on Vedantu. It encourages students to register now for the admission test to secure limited seats and chance at 100% scholarship.
1) The document explains how to evaluate functions by plugging values into the function.
2) It provides examples of evaluating different functions such as f(x)=2x-3 at x=2, f(x)=3x+7 at x=-1, and f(x)=x^2+x-2 at x=5.
3) The final example shows completing a function table for f(x)=3x^2-5x+10 by evaluating it at x values from -1 to 3.
This document provides an overview of functions, limits, and continuity. It defines key concepts such as domain and range of functions, and examples of standard real functions. It also covers even and odd functions, and how to calculate limits, including left and right hand limits. Methods for evaluating algebraic limits using substitution, factorization, and rationalization are presented. The objectives are to understand functions, domains, ranges, and how to evaluate limits of functions.
This document discusses functions, limits, and continuity. It begins by defining functions, domains, ranges, and some standard real functions like constant, identity, modulus, and greatest integer functions. It then covers limits of functions including one-sided limits and properties of limits. Examples are provided to illustrate evaluating limits using substitution and factorization methods. The overall objectives are to understand functions, domains, ranges, limits of functions and methods to evaluate limits.
This document provides examples and exercises on inverse functions. It first shows an example of finding the inverse of the function f(x) = 2x-3/(x+2). It gives the steps to solve for x in terms of y and obtain the inverse function f^-1(x) = -2x-3/(x-2). It then asks the reader to verify that f(f^-1(x)) = x. The exercises provide 20 functions and ask the reader to find their inverses and verify the inverses are correct. It also gives graphs of 8 functions and asks the reader to determine properties of the inverse graphs, including their domains and ranges and any fixed or end points.
This module discusses methods for finding the zeros of polynomial functions of degree greater than 2, including: factor theorem, factoring, synthetic division, and depressed equations. It introduces the number of roots theorem, which states that a polynomial of degree n has n roots. It also discusses determining the rational zeros of a polynomial using the rational roots theorem and factor theorem. Examples are provided to illustrate these concepts and methods.
The document discusses interpolation, which involves using a function to approximate values between known data points. It provides examples of Lagrange interpolation, which finds a polynomial passing through all data points, and Newton's interpolation, which uses divided differences to determine coefficients for approximating between points. The examples demonstrate constructing Lagrange and Newton interpolation polynomials using given data sets.
The document is an advertisement for Vedantu, an online education platform, promoting their free online admission test to win scholarships for classes 6-12, JEE, and NEET. It highlights success stories of students who scored well in board exams and engineering/medical entrance exams after taking online classes on Vedantu. It encourages students to register now for the admission test to secure limited seats and chance at 100% scholarship.
1) The document explains how to evaluate functions by plugging values into the function.
2) It provides examples of evaluating different functions such as f(x)=2x-3 at x=2, f(x)=3x+7 at x=-1, and f(x)=x^2+x-2 at x=5.
3) The final example shows completing a function table for f(x)=3x^2-5x+10 by evaluating it at x values from -1 to 3.
The document discusses operations that can be performed on functions, including addition, subtraction, multiplication, division, and composition. It provides examples of evaluating each type of operation on functions by first evaluating the individual functions at a given value or variable and then performing the indicated operation on the results. Composition involves evaluating the inner function first and substituting its result into the outer function.
The remainder theorem states that when a polynomial f(x) is divided by (x - c), the remainder is equal to f(c). This is based on the concept of synthetic division, where a polynomial is divided into a quotient and remainder. Some examples are worked out to demonstrate evaluating a polynomial f(x) at a value c to find the remainder when divided by (x - c), in accordance with the remainder theorem.
The document is a math worksheet containing calculus problems involving functions. It includes 21 problems involving operations on functions such as composition, inversion and transformations of function graphs. The problems involve determining expressions for composed functions, inverses, graphs of related functions obtained through transformations of an original function graph. The document also provides answers to the problems.
The document discusses functions and their properties. It defines a function as a rule that assigns exactly one output value to each input value in its domain. Functions can be represented graphically, numerically in a table, or with an algebraic rule. The domain of a function is the set of input values, while the range is the set of output values. Basic operations like addition, subtraction, multiplication and division can be performed on functions in the same way as real numbers. Composition of functions is defined as evaluating one function using the output of another as the input.
The document discusses using the remainder theorem to find the remainder of a polynomial function divided by a binomial function without fully performing the division. It provides two examples:
1) When f(x) = x^3 - 3x^2 + 7 is divided by g(x) = x + 2, the remainder is -13.
2) When f(x) = x^3 - 3x^2 + 5x - 1 is divided by g(x) = x - 1, the remainder is 2.
The key steps are to set the binomial factor equal to 0 to find the value to substitute into the polynomial, then evaluate the polynomial at that value to determine the remainder.
This document provides information about polynomial operations including:
1) Defining polynomials as algebraic expressions involving integer powers of a variable and real number coefficients.
2) Examples of adding, subtracting, and multiplying polynomials by using vertical or FOIL methods.
3) Important formulas for polynomial operations such as (a + b)(a - b) = a2 - b2 and (a + b)2 = a2 + 2ab + b2.
4) Worked examples of applying these formulas and methods to polynomials involving single and multiple variables.
The document discusses operations that can be performed on functions, including addition, subtraction, multiplication, and division. Definitions of each operation are provided, along with examples of applying the operations to specific functions. Addition of functions involves adding the outputs of each function, subtraction involves subtracting the outputs, multiplication involves multiplying the outputs, and division involves dividing the outputs given the denominator function is not equal to 0. Several examples are worked through applying the different operations to functions like f(x)=2x and g(x)=-x+5. The examples also demonstrate evaluating composite functions and restricting domains as needed.
1) The document contains solutions to 10 calculus problems involving finding derivatives and evaluating functions. The problems include finding derivatives of trigonometric, logarithmic, and exponential functions, as well as evaluating functions at given values.
2) The solutions show the steps taken to solve each problem through substitution and application of derivative rules.
3) The techniques demonstrated include finding derivatives using basic rules, evaluating functions by substitution, and simplifying expressions involving trigonometric identities.
The document discusses limits and continuity of functions. It provides examples of computing one-sided limits, limits at points of discontinuity, and limits involving algebraic, trigonometric, exponential and logarithmic functions. The key rules for limits include the properties of limits, the sandwich theorem, and limits of compositions of functions. Continuity of functions is defined as a function having a limit equal to its value at a point. Polynomials, trigonometric functions and exponentials are shown to be continuous everywhere they are defined.
This document provides information and examples on multiplying polynomials, including:
1) Multiplying a monomial and polynomial using the distributive property.
2) Multiplying two polynomials using both the horizontal and vertical methods.
3) Factoring trinomials and identifying similar and conjugate binomials. Methods like FOIL and grouping are discussed.
The document provides information about functions. It defines a function as a relation where each input has exactly one output. A function has a domain, which is the set of all legal inputs, and a range, which is the set of all possible outputs. Notation for a function is y=f(x), where y depends on x. Examples are provided of evaluating functions at different inputs and finding the domains of functions. Piecewise functions are also introduced.
This document provides information about derivatives and their applications:
1. It defines the derivative as the limit of the difference quotient, and explains how to calculate derivatives using first principles. It also covers rules for finding derivatives of sums, products, quotients, exponentials, and logarithmic functions.
2. Higher order derivatives are introduced, with examples of how to take second and third derivatives.
3. Applications of derivatives like finding velocity and acceleration from a position-time function are demonstrated. Maximum/minimum values and how to find local and absolute extrema are also discussed with an example.
This document discusses inverse functions. It provides examples of functions and their inverses, including:
- A function f(x) = {(1, 4), (2, 6), (5, 3)} and its inverse f-1(x) = {(4, 1), (6, 2), (3, 5)}.
- Showing that composing a function with its inverse results in the identity function f-1(f(x)) = x and f(f-1(x)) = x.
- Finding the inverse of the function f(x) = 2/x and graphing a function and its inverse.
This document discusses inverse functions. It provides examples of functions and their inverses, including:
- A function f(x) = {(1, 4), (2, 6), (5, 3)} and its inverse f-1(x) = {(4, 1), (6, 2), (3, 5)}.
- Showing that composing a function with its inverse results in the identity function: f-1(f(x)) = x and f(f-1(x)) = x.
- Finding the inverse of the function f(x) = 2x^2 and graphing a function and its inverse.
1. The document is a mathematics assignment on differentiation from pages 33-40. It was prepared by 4 students for their 1st semester class at the Polytechnic Manufacturing State University of Bangka Belitung.
2. The assignment contains the solutions to 10 differentiation problems finding the derivatives of various functions.
Level 3 NCEA - NZ: A Nation In the Making 1872 - 1900 SML.pptHenry Hollis
The History of NZ 1870-1900.
Making of a Nation.
From the NZ Wars to Liberals,
Richard Seddon, George Grey,
Social Laboratory, New Zealand,
Confiscations, Kotahitanga, Kingitanga, Parliament, Suffrage, Repudiation, Economic Change, Agriculture, Gold Mining, Timber, Flax, Sheep, Dairying,
This presentation was provided by Racquel Jemison, Ph.D., Christina MacLaughlin, Ph.D., and Paulomi Majumder. Ph.D., all of the American Chemical Society, for the second session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session Two: 'Expanding Pathways to Publishing Careers,' was held June 13, 2024.
🔥🔥🔥🔥🔥🔥🔥🔥🔥
إضغ بين إيديكم من أقوى الملازم التي صممتها
ملزمة تشريح الجهاز الهيكلي (نظري 3)
💀💀💀💀💀💀💀💀💀💀
تتميز هذهِ الملزمة بعِدة مُميزات :
1- مُترجمة ترجمة تُناسب جميع المستويات
2- تحتوي على 78 رسم توضيحي لكل كلمة موجودة بالملزمة (لكل كلمة !!!!)
#فهم_ماكو_درخ
3- دقة الكتابة والصور عالية جداً جداً جداً
4- هُنالك بعض المعلومات تم توضيحها بشكل تفصيلي جداً (تُعتبر لدى الطالب أو الطالبة بإنها معلومات مُبهمة ومع ذلك تم توضيح هذهِ المعلومات المُبهمة بشكل تفصيلي جداً
5- الملزمة تشرح نفسها ب نفسها بس تكلك تعال اقراني
6- تحتوي الملزمة في اول سلايد على خارطة تتضمن جميع تفرُعات معلومات الجهاز الهيكلي المذكورة في هذهِ الملزمة
واخيراً هذهِ الملزمة حلالٌ عليكم وإتمنى منكم إن تدعولي بالخير والصحة والعافية فقط
كل التوفيق زملائي وزميلاتي ، زميلكم محمد الذهبي 💊💊
🔥🔥🔥🔥🔥🔥🔥🔥🔥
The document discusses operations that can be performed on functions, including addition, subtraction, multiplication, division, and composition. It provides examples of evaluating each type of operation on functions by first evaluating the individual functions at a given value or variable and then performing the indicated operation on the results. Composition involves evaluating the inner function first and substituting its result into the outer function.
The remainder theorem states that when a polynomial f(x) is divided by (x - c), the remainder is equal to f(c). This is based on the concept of synthetic division, where a polynomial is divided into a quotient and remainder. Some examples are worked out to demonstrate evaluating a polynomial f(x) at a value c to find the remainder when divided by (x - c), in accordance with the remainder theorem.
The document is a math worksheet containing calculus problems involving functions. It includes 21 problems involving operations on functions such as composition, inversion and transformations of function graphs. The problems involve determining expressions for composed functions, inverses, graphs of related functions obtained through transformations of an original function graph. The document also provides answers to the problems.
The document discusses functions and their properties. It defines a function as a rule that assigns exactly one output value to each input value in its domain. Functions can be represented graphically, numerically in a table, or with an algebraic rule. The domain of a function is the set of input values, while the range is the set of output values. Basic operations like addition, subtraction, multiplication and division can be performed on functions in the same way as real numbers. Composition of functions is defined as evaluating one function using the output of another as the input.
The document discusses using the remainder theorem to find the remainder of a polynomial function divided by a binomial function without fully performing the division. It provides two examples:
1) When f(x) = x^3 - 3x^2 + 7 is divided by g(x) = x + 2, the remainder is -13.
2) When f(x) = x^3 - 3x^2 + 5x - 1 is divided by g(x) = x - 1, the remainder is 2.
The key steps are to set the binomial factor equal to 0 to find the value to substitute into the polynomial, then evaluate the polynomial at that value to determine the remainder.
This document provides information about polynomial operations including:
1) Defining polynomials as algebraic expressions involving integer powers of a variable and real number coefficients.
2) Examples of adding, subtracting, and multiplying polynomials by using vertical or FOIL methods.
3) Important formulas for polynomial operations such as (a + b)(a - b) = a2 - b2 and (a + b)2 = a2 + 2ab + b2.
4) Worked examples of applying these formulas and methods to polynomials involving single and multiple variables.
The document discusses operations that can be performed on functions, including addition, subtraction, multiplication, and division. Definitions of each operation are provided, along with examples of applying the operations to specific functions. Addition of functions involves adding the outputs of each function, subtraction involves subtracting the outputs, multiplication involves multiplying the outputs, and division involves dividing the outputs given the denominator function is not equal to 0. Several examples are worked through applying the different operations to functions like f(x)=2x and g(x)=-x+5. The examples also demonstrate evaluating composite functions and restricting domains as needed.
1) The document contains solutions to 10 calculus problems involving finding derivatives and evaluating functions. The problems include finding derivatives of trigonometric, logarithmic, and exponential functions, as well as evaluating functions at given values.
2) The solutions show the steps taken to solve each problem through substitution and application of derivative rules.
3) The techniques demonstrated include finding derivatives using basic rules, evaluating functions by substitution, and simplifying expressions involving trigonometric identities.
The document discusses limits and continuity of functions. It provides examples of computing one-sided limits, limits at points of discontinuity, and limits involving algebraic, trigonometric, exponential and logarithmic functions. The key rules for limits include the properties of limits, the sandwich theorem, and limits of compositions of functions. Continuity of functions is defined as a function having a limit equal to its value at a point. Polynomials, trigonometric functions and exponentials are shown to be continuous everywhere they are defined.
This document provides information and examples on multiplying polynomials, including:
1) Multiplying a monomial and polynomial using the distributive property.
2) Multiplying two polynomials using both the horizontal and vertical methods.
3) Factoring trinomials and identifying similar and conjugate binomials. Methods like FOIL and grouping are discussed.
The document provides information about functions. It defines a function as a relation where each input has exactly one output. A function has a domain, which is the set of all legal inputs, and a range, which is the set of all possible outputs. Notation for a function is y=f(x), where y depends on x. Examples are provided of evaluating functions at different inputs and finding the domains of functions. Piecewise functions are also introduced.
This document provides information about derivatives and their applications:
1. It defines the derivative as the limit of the difference quotient, and explains how to calculate derivatives using first principles. It also covers rules for finding derivatives of sums, products, quotients, exponentials, and logarithmic functions.
2. Higher order derivatives are introduced, with examples of how to take second and third derivatives.
3. Applications of derivatives like finding velocity and acceleration from a position-time function are demonstrated. Maximum/minimum values and how to find local and absolute extrema are also discussed with an example.
This document discusses inverse functions. It provides examples of functions and their inverses, including:
- A function f(x) = {(1, 4), (2, 6), (5, 3)} and its inverse f-1(x) = {(4, 1), (6, 2), (3, 5)}.
- Showing that composing a function with its inverse results in the identity function f-1(f(x)) = x and f(f-1(x)) = x.
- Finding the inverse of the function f(x) = 2/x and graphing a function and its inverse.
This document discusses inverse functions. It provides examples of functions and their inverses, including:
- A function f(x) = {(1, 4), (2, 6), (5, 3)} and its inverse f-1(x) = {(4, 1), (6, 2), (3, 5)}.
- Showing that composing a function with its inverse results in the identity function: f-1(f(x)) = x and f(f-1(x)) = x.
- Finding the inverse of the function f(x) = 2x^2 and graphing a function and its inverse.
1. The document is a mathematics assignment on differentiation from pages 33-40. It was prepared by 4 students for their 1st semester class at the Polytechnic Manufacturing State University of Bangka Belitung.
2. The assignment contains the solutions to 10 differentiation problems finding the derivatives of various functions.
Similar to mathongo.com-NCERT-Solutions-Class-12-Maths-Chapter-5-Continuity-and-Differentiability.pdf (15)
Level 3 NCEA - NZ: A Nation In the Making 1872 - 1900 SML.pptHenry Hollis
The History of NZ 1870-1900.
Making of a Nation.
From the NZ Wars to Liberals,
Richard Seddon, George Grey,
Social Laboratory, New Zealand,
Confiscations, Kotahitanga, Kingitanga, Parliament, Suffrage, Repudiation, Economic Change, Agriculture, Gold Mining, Timber, Flax, Sheep, Dairying,
This presentation was provided by Racquel Jemison, Ph.D., Christina MacLaughlin, Ph.D., and Paulomi Majumder. Ph.D., all of the American Chemical Society, for the second session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session Two: 'Expanding Pathways to Publishing Careers,' was held June 13, 2024.
🔥🔥🔥🔥🔥🔥🔥🔥🔥
إضغ بين إيديكم من أقوى الملازم التي صممتها
ملزمة تشريح الجهاز الهيكلي (نظري 3)
💀💀💀💀💀💀💀💀💀💀
تتميز هذهِ الملزمة بعِدة مُميزات :
1- مُترجمة ترجمة تُناسب جميع المستويات
2- تحتوي على 78 رسم توضيحي لكل كلمة موجودة بالملزمة (لكل كلمة !!!!)
#فهم_ماكو_درخ
3- دقة الكتابة والصور عالية جداً جداً جداً
4- هُنالك بعض المعلومات تم توضيحها بشكل تفصيلي جداً (تُعتبر لدى الطالب أو الطالبة بإنها معلومات مُبهمة ومع ذلك تم توضيح هذهِ المعلومات المُبهمة بشكل تفصيلي جداً
5- الملزمة تشرح نفسها ب نفسها بس تكلك تعال اقراني
6- تحتوي الملزمة في اول سلايد على خارطة تتضمن جميع تفرُعات معلومات الجهاز الهيكلي المذكورة في هذهِ الملزمة
واخيراً هذهِ الملزمة حلالٌ عليكم وإتمنى منكم إن تدعولي بالخير والصحة والعافية فقط
كل التوفيق زملائي وزميلاتي ، زميلكم محمد الذهبي 💊💊
🔥🔥🔥🔥🔥🔥🔥🔥🔥
A Visual Guide to 1 Samuel | A Tale of Two HeartsSteve Thomason
These slides walk through the story of 1 Samuel. Samuel is the last judge of Israel. The people reject God and want a king. Saul is anointed as the first king, but he is not a good king. David, the shepherd boy is anointed and Saul is envious of him. David shows honor while Saul continues to self destruct.
Leveraging Generative AI to Drive Nonprofit InnovationTechSoup
In this webinar, participants learned how to utilize Generative AI to streamline operations and elevate member engagement. Amazon Web Service experts provided a customer specific use cases and dived into low/no-code tools that are quick and easy to deploy through Amazon Web Service (AWS.)
1. Exercise 5.1
1. Prove that the function f (x) = 5x – 3 is continuous at x = 0,
at x = – 3 and at x = 5.
Sol. Given: f (x) = 5x – 3 ...(i)
Continuity at x = 0
0
lim ( )
x
f x
→
=
0
lim (5 3)
x
x
→
− (By (i))
Putting x = 0, = 5(0) – 3 = 0 – 3 = – 3
Putting x = 0 in (i), f (0) = 5(0) – 3 = – 3
∴
0
lim ( )
x
f x
→
= f (0) (= – 3) ∴ f (x) is continuous at x = 0.
Continuity at x = – 3
3
lim ( )
x
f x
→ −
=
3
lim (5 3)
x
x
→ −
− (By (i))
Putting x = – 3, = 5(– 3) – 3 = – 15 – 3 = – 18
Putting x = – 3 in (i), f (– 3) = 5(– 3) – 3 = – 15 – 3 = – 18
∴
3
lim ( )
x
f x
→ −
= f (– 3)(= – 18)
∴ f (x) is continuous at x = – 3.
Continuity at x = 5
5
lim ( )
x
f x
→
=
5
lim (5 3)
x
x
→
− (By (i))
Putting x = 5, 5(5) – 3 = 25 – 3 = 22
Putting x = 5 in (i), f (5) = 5(5) – 3 = 25 – 3 = 22
∴
5
lim (5 3)
x
x
→
− = f (5) (= 22) ∴ f (x) is continuous at x = 5.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 1
2. 2. Examine the continuity of the function
f (x) = 2x2
– 1 at x = 3.
Sol. Given: f (x) = 2x2
– 1 ...(i)
Continuity at x = 3
3
lim ( )
x
f x
→
= 2
3
lim (2 1)
x
x
→
− [By (i)]
Putting x = 3, = 2.32
– 1 = 2(9) – 1 = 18 – 1 = 17
Putting x = 3 in (i), f (3) = 2.32
– 1 = 18 – 1 = 17
∴
3
lim ( )
x
f x
→
= f (3) (= 17) ∴ f (x) is continuous at x = 3.
3. Examine the following functions for continuity:
(a) f (x) = x – 5 (b) f (x) =
1
– 5
x
, x ≠
≠
≠
≠
≠ 5
(c) f (x) =
2
– 25
+ 5
x
x
, x ≠
≠
≠
≠
≠ – 5 (d) f (x) = |
||
|| x – 5 |
||
||.
Sol. (a) Given: f (x) = x – 5 ...(i)
The domain of f is R
(... f (x) is real and finite for all x ∈ R)
Let c be any real number (i.e., c ∈ domain of f ).
lim ( )
x c
f x
→
= lim ( 5)
x c
x
→
− [By (i)]
Putting x = c, = c – 5
Putting x = c in (i), f (c) = c – 5
∴ lim ( )
x c
f x
→
= f (c) (= c – 5)
∴ f is continuous at every point c in its domain (here R).
Hence f is continuous.
Or
Here f (x) = x – 5 is a polynomial function. We know that
every polynomial function is continuous (see note below).
Hence f (x) is continuous (in its domain R)
Very important Note. The following functions are
continuous (for all x in their domain).
1. Constant function
2. Polynomial function.
3. Rational function
( )
( )
f x
g x
where f (x) and g(x) are
polynomial functions of x and g (x) ≠ 0.
4. Sine function (⇒ sin x).
5. cos x. 6. ex
.
7. e– x
. 8. log x (x > 0).
9. Modulus function.
(b) Given: f (x) =
1
5
x −
, x ≠ 5 ...(i)
Given: The domain f is R – (x ≠ 5) i.e., R – {5}
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 2
3. (... For x = 5, f (x) =
1
5
x −
=
1
5 5
−
=
1
0
→ ∞
∴ 5 ∉ domain of f )
Let c be any real number such that c ≠ 5
lim ( )
x c
f x
→
=
1
lim
5
x c x
→ −
[By (i)]
Putting x = c, =
1
5
c −
Putting x = c in (i), f (c) =
1
5
c −
∴ lim ( )
x c
f x
→
= f (c)
1
5
c
=
−
∴ f (x) is continuous at every point c in the domain of f.
Hence f is continuous.
Or
Here f (x) =
1
5
x −
, x ≠ 5 is a rational function
Polynomial 1 of degree 0
=
Polynomial ( 5) of degree 1
x
−
and its denominator
i.e., (x – 5) ≠ 0 (... x ≠ 5). We know that every rational
function is continuous (By Note below Solution of Q. No.
3(a)). Therefore f is continuous (in its domain R – {5}).
(c) f (x) =
2
25
5
x
x
−
+
, x ≠ – 5
Here f (x) =
2
25
5
x
x
−
+
, x ≠ – 5 is a rational function and
denominator x + 5 ≠ 0 (... x ≠ – 5).
(In fact f (x) =
2
25
5
x
x
−
+
, (x ≠ – 5) =
( 5)( 5)
5
x x
x
+ −
+
= x – 5, (x ≠ – 5) is a polynomial function). We know that
every rational function is continuous. Therefore f is
continuous (in its domain R – {– 5}).
Or
Proceed as in Method I of Q. No. 3(b).
(d) Given: f (x) = | x – 5 |
Domain of f (x) is R (... f (x) is real and finite for all real
x in (– ∞, ∞))
Here f (x) = | x – 5 | is a modulus function.
We know that every modulus function is continuous.
(By Note below Solution of Q. No. 3(a)). Therefore f is
continuous in its domain R.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 3
4. 4. Prove that the function f (x) = xn
is continuous at
x = n where n is a positive integer.
Sol. Given: f (x) = xn
where n is a positive integer. ...(i)
Domain of f (x) is R (... f (x) is real and finite for all real x)
Here f (x) = xn
, where n is a positive integer.
We know that every polynomial function of x is a continuous
function. Therefore, f is continuous (in its whole domain R) and
hence continuous at x = n also.
Or
lim ( )
x n
f x
→
= lim n
x n
x
→
[By (i)]
Putting x = n, = nn
Again putting x = n in (i), f (n) = nn
∴ lim ( )
x n
f x
→
= f (n) (= nn
) ∴ f (x) is continuous at x = n.
5. Is the function f defined by
f (x) =
, if 1
5, if >1
x x
x
≤
continuous at x = 0?, At x = 1?, At x = 2 ?
Sol. Given: f (x) =
, if 1 ...( )
5, if 1 ...( )
x x i
x ii
≤
>
(Read Note (on continuity) before the solution of Q. No. 1 of this
exercise)
Continuity at x = 0
Left Hand Limit =
0
lim
x −
→
f (x) =
0
lim
x −
→
x [By (i)]
(x → 0–
⇒ x < slightly less than 0 ⇒ x < 1)
Putting x = 0, = 0
Right hand limit =
0
lim
x +
→
f (x) =
0
lim
x +
→
x [By (i)]
(x → 0+
⇒ x is slightly greater than 0 say x = 0.001 ⇒ x < 1)
Putting x = 0,
0
lim
x +
→
f (x) = 0 ∴
0
lim
x −
→
f (x) =
0
lim
x +
→
f (x) = 0
∴
0
lim
x →
f (x) exists and = 0 = f (0)
(... Putting x = 0 in (i), f (0) = 0)
∴ f (x) is continuous at x = 0.
Continuity at x = 1
Left Hand Limit = –
1
lim
x →
f (x) = –
1
lim
x →
x [By (i)]
Putting x = 1, = 1
Right Hand Limit =
1
lim
x +
→
f (x) =
1
lim
x +
→
5
Putting x = 1,
1
lim
x +
→
f (x) = 5
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 4
5. ∴ –
1
lim
x →
f (x) ≠
1
lim
x +
→
f (x) ∴
1
lim
x →
f (x) does not exist.
∴ f (x) is discontinuous at x = 1.
Continuity at x = 2
Left Hand Limit =
2
lim
x −
→
f (x) =
2
lim
x −
→
5 [By (ii)]
(x → 2 – ⇒ x is slightly < 2 ⇒ x = 1.98 (say) ⇒ x > 1)
Putting x = 2, = 5
Right Hand Limit =
2
lim
x +
→
f (x) =
2
lim
x +
→
5 [By (ii)]
(x → 2 + ⇒ x is slightly > 2 and hence x > 1 also)
Putting x = 2, = 5
∴
2
lim
x −
→
f (x) =
2
lim
x +
→
f (x) (= 5)
∴
2
lim
x →
f (x) exists and = 5 = f (2)
(Putting x = 2 > 1 in (ii), f (2) = 5)
∴ f (x) is continuous at x = 2
Answer. f is continuous at x = 0 and x = 2 but not continuous
at x = 1.
Find all points of discontinuity of f, where f is defined by
(Exercises 6 to 12)
6. f (x) =
2 + 3, 2
2 – 3, > 2
x x
x x
≤
.
Sol. Given: f (x) = 2x + 3, x ≤ 2 ...(i)
= 2x – 3 x > 2 ...(ii)
To find points of discontinuity of f (in its domain)
Here f (x) is defined for x ≤ 2 i.e., on (– ∞, 2]
and also for x > 2 i.e., on (2, ∞)
∴ Domain of f is (– ∞, 2] ∪ (2, ∞) = (– ∞, ∞) = R
By (i), for all x < 2 (x = 2 being partitioning point can’t be
mentioned here) f (x) = 2x + 3 is a polynomial and hence
continuous.
By (ii), for all x > 2, f (x) = 2x – 3 is a polynomial and hence
continuous. Therefore f (x) is continuous on R – {2}.
Let us examine continuity of f at partitioning point
x = 2
Left Hand Limit =
2
lim
x −
→
f (x) =
2
lim
x −
→
(2x + 3) [By (i)]
Putting x = 2, = 2(2) + 3 = 4 + 3 = 7
Right Hand Limit =
2
lim
x +
→
f (x) =
2
lim
x +
→
(2x – 3) [By (ii)]
Putting x = 2, = 2(2) – 3 = 4 – 3 = 1
∴
2
lim
x −
→
f (x) ≠
2
lim
x +
→
f (x)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 5
6. ∴
2
lim
x →
f (x) does not exist and hence f (x) is discontinuous at
x = 2 (only).
7. f (x) =
| |+ 3, if – 3
– 2 , if – 3 < < 3
6 + 2, if 3
x x
x x
x x
≤
≥
.
Sol. Given: f (x) =
| | 3, if – 3 ...( )
– 2 , if – 3 3 ...( )
6 2, if 3 ...( )
x x i
x x ii
x x iii
+ ≤
< <
+ ≥
Here f (x) is defined for x ≤ – 3 i.e., (– ∞, – 3] and also for
– 3 < x < 3 and also for x ≥ 3 i.e., on [3, ∞).
∴ Domain of f is (– ∞, – 3] ∪ (– 3, 3) ∪ [3, ∞) = (– ∞, ∞) = R.
By (i), for all x < – 3, f (x) = | x | + 3 = – x + 3
(... x < – 3 means x is negative and hence | x | = – x)
is a polynomial and hence continuous.
By (ii), for all x (– 3 < x < 3) f (x) = – 2x is a polynomial and
hence continuous.
By (iii), for all x > 3, f (x) = 6x + 2 is a polynomial and hence
continuous. Therefore, f (x) is continuous on R – {– 3, 3}.
From (i), (ii) and (iii) we can observe that x = – 3 and
x = 3 are partitioning points of the domain R.
Let us examine continuity of f at partitioning point
x = – 3
Left Hand Limit =
3
lim
x −
→ −
f (x) =
3
lim
x −
→ −
(| x | + 3) [By (i)]
(... x → – 3–
⇒ x < – 3)
=
3
lim
x −
→ −
(– x + 3)
(... x → – 3–
⇒ x < – 3 means x is negative and hence
| x | = – x)
Put x = – 3, = 3 + 3 = 6
Right Hand Limit =
3
lim
x +
→ −
f (x) =
3
lim
x +
→ −
(– 2x) [By (ii)]
(... x → – 3+
⇒ x > – 3)
Putting x = – 3, = – 2(– 3) = 6
∴
3
lim
x +
→ −
f (x) =
3
lim
x +
→ −
f (x) (= 6)
∴
3
lim
x → −
f (x) exists and = 6
Putting x = – 3 in (i), f (– 3) = | – 3 | + 3 = 3 + 3 = 6
∴
3
lim
x → −
f (x) = f (– 3) (= 6)
∴ f (x) is continuous at x = – 3.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 6
7. Now let us examine continuity of f at partitioning point
x = 3
Left Hand Limit =
3
lim
x −
→
f (x) =
3
lim
x −
→
(– 2x) [By (ii)]
(... x → 3–
⇒ x < 3)
Putting x = 3, = – 2(3) = – 6
Right Hand Limit =
3
lim
x +
→
f (x) =
3
lim
x +
→
(6x + 2) [By (iii)]
(... x → 3+
⇒ x > 3)
Putting x = 3, = 6(3) + 2 = 18 + 2 = 20
∴
3
lim
x −
→
f (x) ≠
3
lim
x +
→
f (x)
∴
3
lim
x →
f (x) does not exist and hence f (x) is discontinuous at
x = 3 (only).
8. f (x) =
| |
, if 0
0, if = 0
x
x
x
x
≠
.
Sol. Given: f (x) =
| |
x
x
if x ≠ 0
[i.e., =
x
x
= 1 if x > 0 (... For x > 0, | x | = x)
and = –
x
x
= – 1 if x < 0 (... For x < 0, | x | = – x)
i.e., f (x) = 1 if x > 0 ...(i)
= – 1 if x < 0 ...(ii)
= 0 if x = 0 ...(iii)
Clearly domain of f (x) is R (... f (x) is defined for x > 0, for x < 0
and also for x = 0)
By (i), for all x > 0, f (x) = 1 is a constant function and hence
continuous.
By (ii), for all x < 0, f (x) = – 1 is a constant function and hence
continuous.
Therefore f (x) is continuous on R – {0}.
Let us examine continuity of f at the partitioning point x = 0
Left Hand Limit =
0
lim
x −
→
f (x) =
0
lim
x −
→
– 1 [By (ii)]
(... x → 0–
⇒ x < 0)
Put x = 0, = – 1
Right Hand Limit =
0
lim
x +
→
f (x) =
0
lim
x +
→
1 [By (i)]
(... x → 0+
⇒ x > 0)
Put x = 0, = 1
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 7
8. ∴
0
lim
x −
→
f (x) ≠
0
lim
x +
→
f (x)
∴
0
lim
x →
f (x) does not exist and hence f (x) is discontinuous at
x = 0 (only).
Note. It may be noted that the function given in Q. No. 8 is
called a signum function.
9. f (x) =
, if < 0
| |
– 1, if 0
x
x
x
x ≥
.
Sol. Given:
f (x) =
| |
x
x
, if x < 0 =
x
x
−
= – 1 if x < 0 ...(i)
(... For x < 0, | x | = – x)
– 1 if x ≥ 0 ...(ii)
Here f (x) is defined for x < 0 i.e., on (– ∞, 0) and also for x ≥ 0
i.e., on [0, ∞).
∴ Domain of f is (– ∞, 0) ∪ [0, ∞) = (– ∞, ∞) = R.
From (i) and (ii), we find that
f (x) = – 1 for all real x (< 0 as well as ≥ 0)
Here f (x) = – 1 is a constant function.
We know that every constant function is continuous.
∴ f is continuous (for all real x in its domain R)
Hence no point of discontinuity.
10. f (x) =
2
+ 1, if 1
+ 1, if < 1
x x
x x
≥
.
Sol. Given: 2
1, if 1 ...( )
...( )
1, if 1
x x i
ii
x x
+ ≥
+ <
Here f (x) is defined for x ≥ 1 i.e., on [1, ∞) and also for
x < 1 i.e., on (– ∞, 1).
Domain of f is (– ∞, 1) ∪ [1, ∞) = (– ∞, ∞) = R
By (i), for all x > 1, f (x) = x + 1 is a polynomial and hence
continuous.
By (ii), for all x < 1, f (x) = x2
+ 1 is a polynomial and hence
continuous. Therefore f is continuous on R – {1}.
Let us examine continuity of f at the partitioning point
x = 1.
Left Hand Limit = –
1
lim
x →
f (x) = –
1
lim
x →
(x2
+ 1) [By (ii)]
(... x → 1–
⇒ x < 1)
Putting x = 1, = 12
+ 1 = 1 + 1 = 2
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 8
9. Right Hand Limit =
1
lim
x +
→
f (x) =
1
lim
x +
→
(x + 1) [By (i)]
(... x → 1+
⇒ x > 1)
Putting x = 1, = 1 + 1 = 2
∴ –
1
lim
x →
f (x) =
1
lim
x +
→
f (x) (= 2)
∴
1
lim
x →
f (x) exists and = 2
Putting x = 1 in (i), f (1) = 1 + 1 = 2
∴
1
lim
x →
f (x) = f (1) (= 2)
∴ f (x) is continuous at x = 1 also.
∴ f is be continuous on its whole domain (R here).
Hence no point of discontinuity.
11. f (x) =
3
2
– 3, if 2
+ 1, if > 2
x x
x x
≤
.
Sol. Given: f (x) =
3
2
3, if 2 ...( )
...( )
1, if 2
x x i
ii
x x
− ≤
+ >
Here f (x) is defined for x ≤ 2 i.e., on
(– ∞, 2] and also for x > 2 i.e., on (2, ∞).
∴ Domain of f is (– ∞, 2] ∪ (2, ∞) = (– ∞, ∞) = R
By (i), for all x < 2, f (x) = x3
– 3 is a polynomial and hence
continuous.
By (ii), for all x > 2, f (x) = x2
+ 1 is a polynomial and hence
continuous.
∴ f is continuous on R – {2}.
Let us examine continuity of f at the partitioning point x = 2.
Left Hand Limit =
2
lim
x −
→
f (x) =
2
lim
x −
→
(x3
– 3) [By (i)]
(... x → 2–
⇒ x < 2)
Putting x = 2, = 23
– 3 = 8 – 3 = 5
Right Hand Limit =
2
lim
x +
→
f (x) =
2
lim
x +
→
(x2
+ 1) [By (ii)]
(... x → 2+
⇒ x > 2)
Putting x = 2, = 22
+ 1 = 4 + 1 = 5
∴
2
lim
x −
→
f (x) =
2
lim
x +
→
f (x) (= 5)
∴
2
lim
x →
f (x) exists and = 5
Putting x = 2 in (i), f (2) = 23
– 3 = 8 – 3 = 5
∴
2
lim
x →
f (x) = f (2) (= 5)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 9
10. ∴ f (x) is continuous at x = 2 (also).
Hence no point of discontinuity.
12. f (x) =
10
2
– 1, if 1
, if > 1
x x
x x
≤
.
Sol. Given: f (x) =
10
2
1, if 1 ...( )
...( )
, if 1
x x i
ii
x x
− ≤
>
Here f (x) is defined for x ≤ 1 i.e., on (– ∞, 1] and also for
x > 1 i.e., on (1, ∞).
∴ Domain of f is (– ∞, 1] ∪ (1, ∞) = (– ∞, ∞) = R
By (i), for all x < 1, f (x) = x10
– 1 is a polynomial and hence
continuous.
By (ii), for all x > 1, f (x) = x2
is a polynomial and hence
continuous.
∴ f (x) is continuous on R – {1}.
Let us examine continuity of f at the partitioning point
x = 1.
Left Hand Limit = –
1
lim
x →
f (x) = –
1
lim
x →
(x10
– 1) [By (i)]
(... x → 1–
⇒ x < 1)
Putting x = 1, = (1)10
– 1 = 1 – 1 = 0
Right Hand Limit =
1
lim
x +
→
f (x) =
1
lim
x +
→
x2
[By (ii)]
Putting x = 1, = 12
= 1
∴ –
1
lim
x →
f (x) ≠
1
lim
x +
→
f (x)
∴
1
lim
x →
f (x) does not exist.
Hence the point of discontinuity is x = 1 (only).
13. Is the function defined by
f (x) =
+ 5 if 1
– 5 if >1
x x
x x
≤
a continuous function?
Sol. Given: f (x) =
5, if 1 ...( )
5, if 1 ...( )
x x i
x x ii
+ ≤
− >
Here f (x) is defined for x ≤ 1 i.e., on (– ∞, 1] and also for
x > 1 i.e., on (1, ∞)
∴ Domain of f is (– ∞, 1] ∪ (1, ∞] = (– ∞, ∞) = R.
By (i), for all x < 1, f (x) = x + 5 is a polynomial and hence
continuous.
By (ii), for all x > 1, f (x) = x – 5 is a polynomial and hence
continuous.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 10
11. ∴ f is continuous on R – {1}.
Let us examine continuity at the partitioning point x = 1.
Left Hand Limit = –
1
lim
x →
f (x) = –
1
lim
x →
(x + 5) [By (i)]
Putting x = 1, = 1 + 5 = 6
Right Hand Limit =
1
lim
x +
→
f (x) =
1
lim
x +
→
(x – 5) [By (ii)]
Putting x = 1, = 1 – 5 = – 4
∴ –
1
lim
x →
f (x) ≠
1
lim
x +
→
f (x)
∴
1
lim
x →
f (x) does not exist.
Hence f (x) is discontinuous at x = 1.
∴ x = 1 is the only point of discontinuity.
Discuss the continuity of the function, f, where f is defined by
14. f (x) =
3, if 0 1
4, if 1 < < 3
5, if 3 10
x
x
x
≤ ≤
≤ ≤
.
Sol. Given: f (x) =
3, if 0 1 ...( )
4, if 1 3 ...( )
5, if 3 10 ...( )
x i
x ii
x iii
≤ ≤
< <
≤ ≤
From (i), (ii) and (iii), we can see that f (x) is defined in [0, 1]
∪ (1, 3) ∪ [3, 10] i.e., f (x) is defined in [0, 10].
∴ Domain of f (x) is [0, 10].
From (i), for 0 ≤ x < 1, f (x) = 3 is a constant function and hence
is continuous for 0 ≤ x < 1.
From (ii), for 1 < x < 3, f (x) = 4 is a constant function and hence
is continuous for 1 < x < 3.
From (iii), for 3 < x ≤ 10, f (x) = 5 is a constant function and
hence is continuous for 3 < x ≤ 10.
Therefore, f (x) is continuous in the domain [0, 10] – {1, 3}.
Let us examine continuity of f at the partitioning point
x = 1.
Left Hand Limit = –
1
lim
x →
f (x) = –
1
lim
x →
3 [By (i)]
(... x → 1–
⇒ x < 1)
Putting x = 1; = 3
Right Hand Limit =
1
lim
x +
→
f (x) =
1
lim
x +
→
4 [By (ii)]
(... x → 1+
⇒ x > 1)
Putting x = 1, = 4
∴ –
1
lim
x →
f (x) ≠
1
lim
x +
→
f (x)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 11
12. ∴
1
lim
x →
f (x) does not exist and hence f (x) is discontinuous at
x = 1.
Let us examine continuity of f at the partitioning point x = 3.
Left Hand Limit =
3
lim
x −
→
f (x) =
3
lim
x −
→
4 [By (ii)]
(... x → 3–
⇒ x < 3)
Putting x = 3, = 4
Right Hand Limit =
3
lim
x +
→
f (x) =
3
lim
x +
→
5 [By (iii)]
(... x → 3+
⇒ x > 3)
Putting x = 3; = 5
∴
3
lim
x −
→
f (x) ≠
3
lim
x +
→
f (x)
∴
3
lim
x →
f (x) does not exist and hence f (x) is discontinuous at
x = 3 also.
∴ x = 1 and x = 3 are the two points of discontinuity of the
function f in its domain [0, 10].
15. f (x) =
2 , if < 0
0, if 0 1
4 , if > 1
x x
x
x x
≤ ≤ .
Sol. The domain of f is {x ∈ R : x < 0} ∪ {x ∈ R : 0 ≤ x ≤ 1}
∪ {x ∈ R : x > 1} = R
x = 0 and x = 1 are partitioning points for the domain of this
function.
For all x < 0, f (x) = 2x is a polynomial and hence continuous.
For 0 < x < 1, f (x) = 0 is a constant function and hence
continuous.
For all x > 1, f (x) = 4x is a polynomial and hence continuous.
Let us discuss continuity at partitioning point x = 0.
At x = 0, f (0) = 0 [... f (x) = 0 if 0 ≤ x ≤ 1]
0
lim
x −
→
f (x) =
0
lim
x −
→
2x[... x → 0–
⇒ x < 0 and f (x) = 2x for x < 0]
= 2 × 0 = 0
0
lim
x +
→
f (x) =
0
lim
x +
→
0[... x → 0+
⇒ x > 0 and f (x) = 0 if 0 ≤ x ≤ 1]
= 0
∴
0
lim
x −
→
f (x) =
0
lim
x +
→
f (x) = 0
Thus
0
lim
x →
f (x) = 0 = f (0) and hence f is continuous at 0.
Let us discuss continuity at partitioning point x = 1.
At x = 1, f (1) = 0 [... f (x) = 0 if 0 ≤ x ≤ 1]
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 12
13. –
1
lim
x →
f (x) = –
1
lim
x →
0 [x → 1– ⇒ x < 1 and f (x) = 0 if 0 ≤ x ≤ 1]
= 0
1
lim
x +
→
f (x) =
1
lim
x +
→
4x [x → 1+ ⇒ x > 1 and f (x) = 4x for x > 1]
= 4 × 1 = 4
The left and right hand limits of f at x = 1 do not coincide i.e.,
are not equal.
∴
1
lim
x →
f (x) does not exist and hence f (x) is
discontinuous at x = 1.
Thus f is continuous at every point in the domain except x = 1.
Hence, f is not a continuous function and x = 1 is the only point
of discontinuity.
16. f (x) =
– 2, if – 1
2 , if – 1 < 1
2, if > 1
x
x x
x
≤
≤ .
Sol. Given: f (x) =
– 2, if – 1 ...( )
2 , if – 1 1 ...( )
2, if 1 ...( )
x i
x x ii
x iii
≤
< ≤
>
From (i), (ii) and (iii) we can see that f (x) is defined for
{x : x ≤ – 1} ∪ {x : – 1 < x ≤ 1} ∪ {x : x > 1}
i.e., for (– ∞, – 1] ∪ (– 1, 1] ∪ (1, ∞) = (– ∞, ∞) = R
∴ Domain of f (x) is R.
From (i), for x < – 1, f (x) = – 2 is a constant function and hence
is continuous for x < – 1.
From (ii), for – 1 < x < 1, f (x) = 2x is a polynomial function and
hence is continuous for – 1 < x < 1.
From (iii), for x > 1, f (x) = 2 is a constant function and hence is
continuous for x > 1.
Therefore f (x) is continuous in R – {– 1, 1}.
Let us examine continuity of f at the partitioning
point x = – 1.
Left Hand Limit = –
1
lim
x → −
f (x) = –
1
lim
x → −
(– 2) [By (i)]
(... x → – 1–
⇒ x < – 1)
Putting x = – 1, = – 2
Right Hand Limit =
1
lim
x +
→ −
f (x) =
1
lim
x +
→ −
2x (By (ii)]
(... x → – 1+
⇒ x > – 1)
Putting x = – 1, = 2(– 1) = – 2
∴ –
1
lim
x → −
f (x) =
1
lim
x +
→ −
f (x) (= – 2) ∴
1
lim
x → −
f (x) exists and = – 2.
Putting x = – 1 in (i), f (– 1) = – 2
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 13
14. ∴
1
lim
x → −
f (x) = f (– 1) (= – 2) ∴ f (x) is continuous at x = – 1.
Let us examine continuity of f at the partitioning point x = 1
Left Hand Limit = –
1
lim
x →
f (x) = –
1
lim
x →
(2x) [By (ii)]
(... x → 1–
⇒ x < 1)
Putting x = 1, = 2(1) = 2
Right Hand Limit =
1
lim
x +
→
f (x) =
1
lim
x +
→
2 [By (iii)]
(... x → 1+
⇒ x > 1)
Putting x = 1, = 2
∴ –
1
lim
x →
f (x) =
1
lim
x +
→
f (x) (= 2) ∴
1
lim
x →
f (x) exists and = 2.
Putting x = 1 in (ii), f (1) = 2(1) = 2
∴
1
lim
x →
f (x) = f (1) (= 2) ∴ f (x) is continuous at x = 1 also.
Therefore f is continuous for all x in its domain R.
17. Find the relationship between a and b so that the function
f defined by
f (x) =
+1, if 3
+ 3, if > 3
ax x
bx x
≤
is continuous at x = 3.
Sol. Given: f (x) =
1 if 3 ...( )
3 if 3 ...( )
ax x i
bx x ii
+ ≤
+ >
and f (x) is continuous at x = 3.
Left Hand Limit =
3
lim
x −
→
f (x) =
3
lim
x −
→
(ax + 1) [By (i)]
(x → 3–
⇒ x < 3)
Putting x = 3, = 3a + 1 ...(iii)
Right Hand Limit =
3
lim
x +
→
f (x) =
3
lim
x +
→
(bx + 3) [By (ii)]
(... x → 3+
⇒ x > 3)
Putting x = 3, = 3b + 3 ...(iv)
Putting x = 3 in (i), f (3) = 3a + 1 ...(v)
Because f (x) is continuous at x = 3 (given)
∴
3
lim
x −
→
f (x) =
3
lim
x +
→
f (x) = f (3)
Putting values from (iii), (iv) and (v) we have
3a + 1 = 3b + 3 (= 3a + 1)
∴ 3a + 1 = 3b + 3 [... First and third members are equal]
⇒ 3a = 3b + 2
Dividing by 3, a = b +
2
3
.
18. For what value of λ
λ
λ
λ
λ is the function defined by
f (x) =
2 if 0
( – 2 ),
if > 0
4 + 1,
x
x x
x
x
≤
λ
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 14
15. continuous at x = 0? What about continuity at x = 1?
Sol. Given: f (x) =
2 if 0 ...( )
( – 2 ),
if 0 ...( )
4 1,
x i
x x
x ii
x
≤
λ
>
+
Given: f (x) is continuous at x = 0. To find λ.
Left Hand Limit =
0
lim
x −
→
f (x) =
0
lim
x −
→
λ(x2
– 2x) [By (i)]
(... x → 0–
⇒ x < 0)
Putting x = 0, = λ(0 – 0) = 0
Right Hand Limit =
0
lim
x +
→
f (x) =
0
lim
x +
→
(4x + 1) [By (ii)]
(... x → 0+
⇒ x > 0)
Putting x = 0, = 4(0) + 1 = 1
∴
0
lim
x −
→
f (x) (= 0) ≠
0
lim
x +
→
f (x) (= 1)
∴
0
lim
x →
f (x) does not exist whatever λ may be
(... Neither left limit nor right limit involves λ)
∴ For no value of λ, f is continuous at x = 0.
To examine continuity of f at x = 1
Left Hand Limit = –
1
lim
x →
f (x) = –
1
lim
x →
(4x + 1) [By (ii)]
(x → 1–
⇒ x is slightly < 1 say x = 0.99 > 0)
Put x = 1, = 4 + 1 = 5
Right Hand Limit =
1
lim
x +
→
f (x) =
1
lim
x +
→
(4x + 1) [By (ii)]
(x → 1+
⇒ x is slightly > 1 say x = 1.1 > 0)
Put x = 1, = 4 + 1 = 5
∴
–
1
lim
x →
f (x) =
1
lim
x +
→
f (x) (= 5)
∴
1
lim
x →
f (x) exists and = 5
Putting x = 1 in (ii) (... 1 > 0), f (1) = 4 + 1 = 5)
∴
1
lim
x →
f (x) = f (1) (= 5)
∴ f (x) is continuous at x = 1 (for all real values of λ).
19. 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.
Sol. Given: g(x) = x – [x]
Let x = c be any integer (i.e., c ∈ Z (= I))
Left Hand Limit = lim
x c−
→
g(x) = lim
x c−
→
(x – [x])
Put x = c – h, h → 0+
=
0
lim
h +
→
(c – h – [c – h]) c – 1 c – h c
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 15
16. =
0
lim
h +
→
(c – h – (c – 1))
[... If c ∈ Z and h → 0+
, then [c – h] = c – 1]
=
0
lim
h +
→
(c – h – c + 1) =
0
lim
h +
→
(1 – h)
Put h = 0, = 1 – 0 = 1
Right Hand Limit = lim
x c+
→
g(x) = lim
x c+
→
(x – [x])
Put x = c + h, h → 0+
=
0
lim
h +
→
(c + h – [c + h]) =
0
lim
h +
→
(c + h – c)
(... If c ∈ Z and h → 0+
, then [c + h] = c)
=
0
lim
h +
→
h
Put h = 0; = 0
∴ lim
x c−
→
g(x) ≠ lim
x c+
→
g(x)
∴ lim
x c
→
g(x) does not exist and hence g(x) is discontinuous at
x = c (any integer).
∴ g(x) = x – [x] is discontinuous at all integral points.
Very Important Note. If two functions f and g are continuous in
a common domain D,
then (i) f + g (ii) f – g (iii) fg are continuous in the same domain D.
(iv)
f
g
is also continuous at all points of D except those where
g(x) = 0.
20. Is the function f (x) = x2
– sin x + 5 continuous at x = π
π
π
π
π?
Sol. Given: f (x) = x2
– sin x + 5 = (x2
+ 5) – sin x
= g(x) – h(x) ...(i)
where g(x) = x2
+ 5 and h(x) = sin x
We know that g(x) = x2
+ 5 is a polynomial function and hence is
continuous (for all real x)
Again h(x) = sin x being a sine function is continuous (for all real x)
∴ By (i) f (x) = x2
– sin x + 5 = g(x) – h(x)
being the difference of two continuous functions is also continuous
for all real x (see Note above) and hence continuous at x = π(∈ R)
also.
Or
Given: f (x) = x2
– sin x + 5 ...(i)
To examine continuity at x = π
π
π
π
π
lim
x → π
f (x) = lim
x → π
(x2
– sin x + 5) [By (i)]
Putting x = π, = π2
– sin π + 5
c + 1
c + h
c
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 16
17. = π2
+ 5
[... sin π = sin 180° = sin (180° – 0°) = sin 0° = 0]
Again putting x = π in (i), f (π) = π2
– sin π + 5
= π2
– 0 + 5 = π2
+ 5
∴ lim
x → π
f (x) = f (π)
∴ f (x) is continuous at x = π.
21. Discuss the continuity of the following functions:
(a) f (x) = sin x + cos x (b) f (x) = sin x – cos x
(c) f (x) = sin x . cos x.
Sol. We know that sin x is a continuous function for all real x
Also we know that cos x is a continuous function for all real x
(see solution of Q. No. 22(i) below)
∴ By Note at the end of solution of Q. No. 19,
(i) their sum function f (x) = sin x + cos x is also continuous
for all real x.
(ii) their difference function f (x) = sin x – cos x is also
continuous for all real x.
(iii) their product function f (x) = sin x . cos x is also continuous
for all real x.
Note. To find lim
x c
→
f (x), we can also start with putting x = c + h
where h → 0 (and not only h → 0+
)
∴ lim
x c
→
f (x) =
0
lim
h →
f (c + h).
(Please note that this method of finding the limits makes us find
both lim
x c−
→
f (x) and lim
x c+
→
f (x) simultaneously).
22. Discuss the continuity of the cosine, cosecant, secant and
cotangent functions.
Sol. (i) Let f (x) be the cosine function
i.e., f (x) = cos x ...(i)
Clearly, f (x) is real and finite for all real values of
x i.e., f (x) is defined for all real x. Therefore domain of
f (x) is R.
Let x = c ∈ R.
→
lim
x c
f (x) = lim
x c
→
cos x
Put x = c + h where h → 0
=
0
lim
h →
cos (c + h) =
0
lim
h →
(cos c cos h – sin c sin h)
Putting h = 0, = cos c cos 0 – sin c sin 0
= cos c (1) – sin c (0)
= cos c
∴ lim
x c
→
f (x) = cos c
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 17
18. Putting x = c in (i), f (c) = cos c
∴ lim
x c
→
f (x) = f (c) (= cos c)
∴ f (x) is continuous at (every) x = c ∈ R
∴ f (x) = cos x is continuous on R.
(ii) Let f (x) be cosecant function
i.e., f (x) = cosec x =
1
sin x
f (x) is not finite i.e., → ∞
when sin x = 0 i.e., when x = nπ, n ∈ Z.
∴ Domain of f (x) = cosec x is D = R – {x = nπ; n ∈ Z}.
(... f (x) is real and finite V x ∈ D).
Now f (x) = cosec x =
1
sin x
=
( )
( )
g x
h x
...(i)
Now g(x) = 1 being constant function is continuous on
domain D and h(x) = sin x is non-zero and continuous on
Domain D.
Therefore by (i), f (x) = cosec x
1 ( )
sin ( )
g x
x h x
= =
is continuous
on domain D = R – {x = nπ, n ∈ Z}
(Also read Note at the end of solution of Q. No. 19).
(iii) Let f (x) be the secant function
i.e.,f (x) = sec x =
1
cos x
f (x) is not finite i.e., → ∞
When cos x = 0 i.e., when x = (2n + 1)
2
π
, n ∈ Z.
∴ Domain of f (x) = sec x is
D = R – {x = (2n + 1)
2
π
; n ∈ Z}
Now f (x) = sec x =
1
cos x
=
( )
( )
g x
h x
...(i)
Now g(x) = 1 being constant function is continuous on
domain D and h(x) = cos x is non-zero and continuous on
domain D.
Therefore by (i), f (x) = sec x
1 ( )
cos ( )
g x
x h x
= =
is continuous
on domain D = R – {x : x = (2n + 1)
2
π
; n ∈ Z}.
(iv) Let f (x) be the cotangent function i.e., f (x) = cot x =
cos
sin
x
x
.
f (x) is not finite i.e., → ∞
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 18
19. When sin x = 0 i.e., when x = nπ, n ∈ Z.
∴ Domain of f (x) = cot x is
D = R – {x = nπ; n ∈ Z}
Now f (x) = cot x =
cos
sin
x
x
=
( )
( )
g x
h x
...(i)
Now g(x) = cos x being cosine function is continuous on D
and is non-zero on D.
Therefore by (i), f (x) = cot x
cos ( )
sin ( )
x g x
x h x
= =
is continuous
on domain D = R – {x : x = nπ, n ∈ Z}.
23. Find all points of discontinuity of f, where
f (x) =
sin
, if < 0
+ 1, if 0
x
x
x
x x ≥
.
Sol. The domain of f = {x ∈ R : x < 0} ∪ {x ∈ R : x ≥ 0} = R
x = 0 is the partitioning point of the domain of the given function.
For all x < 0, f (x) =
sin x
x
(given)
Since sin x and x are continuous for x < 0 (in fact, they are
continuous for all x) and x ≠ 0
∴ f is continuous when x < 0
For all x > 0, f (x) = x + 1 is a polynomial and hence continuous.
∴ f is continuous when x > 0.
Let us discuss the continuity of f (x) at the partitioning
point x = 0.
At x = 0, f (0) = 0 + 1 = 1 [... f (x) = x + 1 for x ≥ 0]
0
lim
x −
→
f (x) =
0
lim
x −
→
sin x
x
sin
0 0 and ( ) for 0
x
x x f x x
x
−
→ ⇒ < = <
∵
= 1
0
lim
x +
→
f (x) =
0
lim
x +
→
(x + 1)
0 0 and ( ) 1 for 0
x x f x x x
+
→ ⇒ > = + >
∵
= 0 + 1 = 1
Since
0
lim
x −
→
f (x)=
0
lim
x +
→
f (x) = 1 ∴
0
lim
x →
f (x) = 1
Thus
→ 0
lim
x
f (x) = f (0) and hence f is continuous at x = 1.
Now f is continuous at every point in its domain and hence f is
a continuous function.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 19
20. 24. Determine if f defined by
f (x) =
2 1
sin , if 0
0, if = 0
x x
x
x
≠
is a continuous function?
Sol. For all x ≠ 0, f (x) = x2
sin
1
x
being the product function of two
continuous functions x2
(polynomial function) and sin
1
x
(a sine
function) is continuous for all real x ≠ 0.
Now let us examine continuity at x = 0.
→ 0
lim
x
f (x) =
→ 0
lim
x
x2
sin
1
x
Putting x = 0 = 0 × A finite quantity between – 1 and 1 = 0
1
sin ( sin ) always lies between 1 and 1
x
= θ −
∵
Also f (x) = 0 at x = 0 i.e., f (0) = 0
∴
→ 0
lim
x
f (x) = f (0), therefore function f is continuous at
x = 0 (also).
Hence f (x) continuous on domain R of f.
25. Examine the continuity of f, where f is defined by
f (x) =
sin – cos , if 0
– 1, if = 0
x x x
x
≠
.
Sol. Given: f (x) =
sin – cos if 0 ...( )
– 1 if 0 ...( )
x x x i
x ii
≠
=
From (i), f (x) is defined for x ≠ 0 and from (ii) f (x) is defined
for x = 0.
∴ Domain of f (x) is {x : x ≠ 0} ∪ {0} = R.
From (i), for x ≠ 0, f (x) = sin x – cos x being the difference of two
continuous functions sin x and cos x is continuous for all x ≠ 0.
Hence f (x) is continuous on R – {0}.
Now let us examine continuity at x = 0.
0
lim
x →
f (x) =
0
lim
x →
(sin x – cos x)
[By (i) as x → 0 means x ≠ 0]
Putting x = 0, = sin 0 – cos 0 = 0 – 1 = – 1
From (ii) f (x) = – 1 when x = 0
i.e., f (0) = – 1
∴
0
lim
x →
f (x) = f (0) (= – 1)
∴ f (x) is continuous at x = 0 (also).
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 20
21. Hence f (x) is continuous on domain R of f.
Find the values of k so that the function f is continuous at the
indicated point in Exercises 26 to 29.
26. f (x) =
cos
, if
– 2 2
at =
2
3, if =
2
k x
x
x
x
x
π
≠
π
π
π
.
Sol. Left Hand Limit =
2
lim
x
π
→
f (x) =
2
lim
x
π
→
cos
2
k x
x
π −
Put x =
2
π
– h where h → 0+
=
0
lim
h +
→
cos –
2
2 –
2
k h
h
π
π
π −
=
0
lim
h +
→
sin
2
π − π +
k h
h
=
0
lim
h +
→
sin
2
k h
h
=
2
k
×
0
lim
h +
→
sin h
h
=
2
k
× 1 =
2
k
...(i)
Right Hand Limit = +
π
→
2
lim
x
f (x) = +
π
→
2
lim
x
cos
2
k x
x
π −
Put x =
2
π
+ h where h → 0+
=
0
lim
h +
→
cos
2
2
2
k h
h
π
+
π
π − +
=
0
lim
h +
→
sin
2
k h
h
−
π − π −
=
0
lim
h +
→
sin
2
k h
h
−
−
=
2
k
×
0
lim
h +
→
sin h
h
=
2
k
× 1 =
2
k
...(ii)
Also f
2
π
= 3 ...(iii) ... f (x) = 3 when x =
2
π
(given)
Because f (x) is continuous at x =
2
π
(given)
∴
–
2
lim
x
π
→
f (x) =
2
lim
x
+
π
→
f (x) = f
2
π
Putting values from (i), (ii), and (iii),
2
k
= 3 or k = 6.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 21
22. 27. f (x) =
2
, if 2
at = 2
3, if > 2
kx x
x
x
≤
.
Sol. Given: f (x) =
2 ...( )
, if 2
...( )
3, if 2
i
kx x
ii
x
≤
>
Given: f (x) is continuous at x = 2.
Left Hand Limit =
2
lim
x −
→
f (x) =
2
lim
x −
→
kx2
[By (i)]
(... x → 2–
⇒ x is < 2)
Put x = 2, = k(2)2
= 4k
Right Hand Limit =
2
lim
x +
→
f (x) =
2
lim
x +
→
3 [By (ii)]
(... x → 2+
⇒ x > 2)
Putting x = 2, = 3
Putting x = 2 in (i) f (2) = k(2)2
= 4k.
Because f (x) is continuous at x = 2 (given),
therefore
2
lim
x −
→
f (x) =
2
lim
x +
→
f (x) = f (2)
Putting values, 4k = 3 = 3 ⇒ k =
3
4
.
28. f (x) =
1, if
at =
cos , if >
kx + x
x
x x
≤ π
π
π
.
Sol. Given: f (x) =
1, if ...( )
cos , if ...( )
kx x i
x x ii
+ ≤ π
> π
Given: f (x) is continuous at x = π.
Left Hand Limit = lim
x −
→ π
f (x) = lim
x −
→ π
(kx + 1) [By (i)]
(... x → π–
⇒ x < π)
Putting x = π, = kπ + 1
Right Hand Limit = lim
x +
→ π
f (x) = lim
x +
→ π
cos x [By (ii)]
(... x → π+
⇒ x > π)
Putting x = π, = cos π = cos 180° = cos (180° – 0)
= – cos 0 = – 1
Putting x = π in (i), f (π) = kπ + 1
But f (x) is continuous at x = π (given), therefore
lim
x −
→ π
f (x) = lim
x +
→ π
f (x) = f (π)
Putting values kπ + 1 = – 1 = kπ + 1
⇒ kπ + 1 = – 1 [... First and third members are same]
⇒ kπ = – 2 ⇒ k = –
2
π
.
29. f (x) =
1, if 5
at = 5
3 – 5, if > 5
kx + x
x
x x
≤
.
Sol. Given: f (x) =
1 if 5 ...( )
3 – 5 if 5 ...( )
kx x i
x x ii
+ ≤
>
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 22
23. 246 MATHEMATICS–XII
Given: f (x) is continuous at x = 5.
Left Hand Limit =
5
lim
x −
→
f (x) =
5
lim
x −
→
(kx + 1) [By (i)]
Putting x = 5, = k(5) + 1 = 5k + 1
Right Hand Limit =
5
lim
x +
→
f (x) =
5
lim
x +
→
(3x – 5) [By (ii)]
Putting x = 5, = 3(5) – 5 = 15 – 5 = 10
Putting x = 5 in (i), f (5) = 5k + 1
But f (x) is continuous at x = 5 (given)
∴
5
lim
x −
→
f (x) =
5
lim
x +
→
f (x) = f (5)
Putting values 5k + 1 = 10 = 5k + 1
⇒ 5k + 1 = 10 ⇒ 5k = 9 ⇒ k =
9
5
.
30. Find the values of a and b such that the function defined
by
f (x) =
5, if 2
+ , if 2 < < 10
21, if 10
x
ax b x
x
≤
≥
.
is a continuous function.
Sol. Given: f (x) =
5 if 2 ...( )
if 2 10 ...( )
21 if 10 ...( )
x i
ax b x ii
x iii
≤
+ < <
≥
From (i), (ii) and (iii), f (x) is defined for {x ≤ 2} ∪ {2 < x < 10}
∪ {x ≥ 10} i.e., for (– ∞, 2] ∪ (2, 10) ∪ [10, ∞) i.e., for (– ∞, ∞) i.e.,
on R. ∴ Domain of f (x) is R.
Given: f (x) is a continuous function (of course on its domain here
R), therefore f (x) is also continuous at partitioning points x = 2
and x = 10 of the domain.
Because f (x) is continuous at partitioning point x = 2, therefore
2
lim
x −
→
f (x) =
2
lim
x +
→
f (x) = f (2) ...(iv)
Now
2
lim
x −
→
f (x) =
2
lim
x −
→
5 [By (i)]
(... x → 2–
⇒ x < 2)
Putting x = 2, = 5
Again
2
lim
x +
→
f (x) =
2
lim
x +
→
(ax + b) [By (ii)]
(... x → 2+
⇒ x > 2)
Putting x = 2, = 2a + b
Putting x = 2 in (i), f (2) = 5.
Putting these values in eqn. (iv), we have
5 = 2a + b = 5 ⇒ 2a + b = 5 ...(v)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 23
24. Again because f (x) is continuous at partitioning point x = 10,
therefore
10
lim
x −
→
f (x) =
10
lim
x +
→
f (x) = f (10) ...(vi)
Now
10
lim
x −
→
f (x) =
10
lim
x −
→
(ax + b) [By (ii)]
(x → 10–
⇒ x < 10)
Putting x = 10, = 10a + b
Again
10
lim
x +
→
f (x) =
10
lim
x +
→
21 [By (iii)]
(... x → 10+
⇒ x > 10)
Putting x = 10; = 21
Putting x = 10 in Eqn. (iii), f (10) = 21
Putting these values in eqn. (vi), we have
10a + b = 21 = 21
⇒ 10a + b = 21 ...(vii)
Let us solve eqns. (v) and (vii) for a and b.
Eqn. (vii) – eqn. (v) gives 8a = 16 ⇒ a =
16
8
= 2
Putting a = 2 in (v), 4 + b = 5 ∴ b = 1.
∴ a = 2, b = 1.
Very Important Result: Composite function of two continuous
functions is continuous.
We know by definition that ( fog)x = f ( g(x))
and ( gof)x = g( f (x))
31. Show that the function defined by f (x) = cos (x2
) is a
continuous function.
Sol. Given: f (x) = cos (x2
) ...(i)
f (x) has a real and finite value for all x ∈ R.
∴ Domain of f (x) is R.
Let us take g (x) = cos x and h(x) = x2
.
Now g(x) = cos x is a cosine function and hence is continuous.
Again h(x) = x2
is a polynomial function and hence is continuous.
∴ ( goh)x = g(h(x)) = g(x2
) [... h(x) = x2
]
= cos (x2
) (Changing x to x2
in g(x) = cos x)
= f (x) (By (i)) being the composite function of two
continuous functions is continuous for all x in
domain R.
Or
Take g(x) = x2
and h(x) = cos x.
Then (hog)x = h( g(x)) = h(x2
)
= cos (x2
) = f (x).
32. Show that the function defined by f (x) = |
||
|| cos x |
||
|| is a
continuous function.
Sol. f (x) = | cos x | ...(i)
f (x) has a real and finite value for all x ∈ R.
∴ Domain of f (x) is R.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 24
25. Let us take g(x) = cos x and h(x) = | x |
We know that g(x) and h(x) being cosine function and modulus
function are continuous for all real x.
Now ( goh)x = g(h(x)) = g(| x |) = cos | x | being the composite
function of two continuous functions is continuous (but ≠ f (x))
Again (hog)x = h(g(x)) = h(cos x)
= | cos x | = f (x) [By (i)]
[Changing x to cos x in h(x) = | x |, we have h(cos x) = | cos x |]
Therefore f (x) = | cos x | (= (hog)x) being the composite function
of two continuous functions is continuous.
33. Examine that sin |
||
|| x |
||
|| is a continuous function.
Sol. Let f (x) = sin x and g(x) = | x |
We know that sin x and |
||
|| x |
||
|| are continuous functions.
∴ f and g are continuous.
Now ( fog ) (x) = f { g (x)} = sin { g(x)} = sin | x |
We know that composite function of two continuous functions is
continuous.
∴ fog is continuous. Hence, sin | x | is continuous.
34. Find all points of discontinuity of f defined by
f (x) = |
||
|| x |
||
|| – |
||
|| x + 1 |
||
|| .
Sol. Given: f (x) = | x | – | x + 1 | ...(i)
This f (x) is real and finite for every x ∈ R.
∴ f is defined for all x ∈ R i.e., domain of f is R.
Putting each expression within modulus equal to 0
i.e., x = 0 and x + 1 = 0 i.e., x = 0 and x = – 1.
– ∞ – 1 0 ∞
Marking these values of x namely – 1 and 0 (in proper ascending
order) on the number line, domain R of f is divided into three
sub-intervals (– ∞, – 1], [– 1, 0] and [0, ∞).
On the sub-interval (–∞, –1] i.e., for x ≤ –1, (say for x = – 2 etc.)
x < 0 and (x + 1) is also < 0 and therefore
| x | = – x and | x + 1 | = – (x + 1)
Hence (i) becomes f (x) = | x | – | x + 1 |
= – x – (– (x + 1)) = – x + x + 1
i.e., f (x) = 1 for x ≤ – 1 ...(ii)
On the sub-interval [– 1, 0] i.e.,for – 1 ≤ x ≤ 0
1
say for
2
x
−
=
x < 0 and (x + 1) > 0 and therefore | x | – x and | x + 1 |
= x + 1.
Hence (i) becomes f (x) = | x | – | x + 1 |
= – x – (x + 1) = – x – x – 1
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 25
26. = – 2x – 1
i.e., f (x) = – 2x – 1 for – 1 ≤ x ≤ 0 ...(iii)
On the sub-interval [0, ∞) i.e., for x ≥ 0,
x ≥ 0 and also x + 1 > 0 and therefore
| x | = x and | x + 1 | = x + 1
Hence (i) becomes f (x) = | x | – | x + 1 | = x – (x + 1)
= x – x – 1 = – 1
i.e., f (x) =– 1 for x ≥ 0 ...(iv)
From (ii), for x < – 1, f (x) = 1 is a constant function and hence is
continuous for x < – 1.
From (iii), for – 1 < x < 0, f (x) = – 2x – 1 is a polynomial
function and hence is continuous for – 1 < x < 0.
From (iv), for x > 0, f (x) = – 1 is a constant function and hence is
continuous for x > 0.
∴ f is continuous in R – {– 1, 0}.
Let us examine continuity of f at partitioning point
x = – 1.
–
1
lim
x → −
f (x) = –
1
lim
x → −
1 [By (ii)]
(... x → – 1–
⇒ x < – 1)
Putting x = – 1, = 1
1
lim
x +
→ −
f (x) =
1
lim
x +
→ −
(– 2x – 1) (By (iii))
(... x → – 1+
⇒ x > – 1)
Putting x = – 1, = – 2(– 1) – 1 = 2 – 1 = 1
∴ –
1
lim
x → −
f (x) =
1
lim
x +
→ −
f (x) (= 1)
∴
1
lim
x → −
f (x) exists and = 1.
Putting x = – 1 in (ii) or (iii), f (– 1) = 1
∴
1
lim
x → −
f (x) = f (– 1) (= 1)
∴ f is continuous at x = – 1 also.
Let us examine continuity of f at partitioning point
x = 0.
0
lim
x −
→
f (x) =
0
lim
x −
→
(– 2x – 1) (By (iii))
(... x → 0
–
⇒ x < 0)
Putting x = 0, = – 2(0) – 1 = – 1
0
lim
x +
→
f (x) =
0
lim
x +
→
(– 1) [By (iv)]
(... → 0+
⇒ x > 0)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 26
27. Putting x = 0, = – 1 ∴ 0
lim
x −
→
f (x) =
0
( )
x
Lt f x
+
→
(= – 1)
∴
→ 0
lim
x
f (x) exists and = – 1
Putting x = 0 in (iii) or (iv), f (0) = – 1
∴
→ 0
lim
x
f (x) = f (0) (= – 1)
∴ f is continuous at x =0 also.
∴ f is continuous on the domain R.
∴ There is no point of discontinuity.
Second Solution
We know that every modulus function is continuous for all real x.
Therefore |x| and |x + 1| are continuous for all real x.
Also, we know that difference of two continuous functions is
continuous.
∴ f (x) = |x| – |x + 1| is also continuous for all real x.
∴ There is no point of discontinuity.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 27
28. Exercise 5.2
Differentiate the functions w.r.t. x in Exercises 1 to 8.
1. sin (x2
+ 5).
Sol. Let y = sin (x2
+ 5)
∴
dy
dx
=
d
dx
sin (x2
+ 5) = cos (x2
+ 5)
d
dx
(x2
+ 5)
=
∵ sin ( ) cos ( ) ( )
d d
f x f x f x
dx dx
= cos (x2
+ 5) . (2x + 0)
...
d
dx
xn
= n xn – 1
and
d
dx
(c) = 0
= 2x cos (x2
+ 5).
Caution. sin (x2
+ 5) is not the product of two functions. It is
composite function: sine of (x2
+ 5).
2. cos (sin x).
Sol. Let y = cos (sin x)
∴
dy
dx
=
d
dx
cos (sin x) = – sin (sin x)
d
dx
sin x
= −
∵ cos ( ) sin ( ) ( )
d d
f x f x f x
dx dx
= – sin (sin x) . cos x = – cos x sin (sin x).
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 28
29. 3. sin (ax + b).
Sol. Let y = sin (ax + b)
∴
dy
dx
=
d
dx
sin (ax + b) = cos (ax + b)
d
dx
(ax + b)
= cos (ax + b)
+
( ) ( )
d d
a x b
dx dx
= cos (ax + b) [a(1) + 0]
= a cos (ax + b).
Note. It may be noted that letters a to q of English Alphabet
are treated as constants (similar to 3, 5 etc.) as per convention.
4. sec (tan x ).
Sol. Let y = sec (tan x )
∴
dy
dx
=
d
dx
sec (tan x )
= sec (tan x ) tan (tan x )
d
dx
(tan x )
=
∵ sec ( ) sec ( ) tan ( ) ( )
d d
f x f x f x f x
dx dx
= sec (tan x ) tan (tan x ) sec2
( x )
d
dx
x
=
∵ 2
( ) sec ( ) ( )
d d
f x f x f x
dx dx
= sec (tan x ) tan (tan x ) sec2 x
1
2 x
− −
= = = =
∵ 1/2 1/2 1 1/2
1 1 1
2 2 2
d d
x x x x
dx dx x
5.
sin ( )
cos ( )
ax + b
cx + d
.
Sol. Let y =
+
+
sin ( )
cos ( )
ax b
cx d
∴
dy
dx
=
+ + − + +
+
2
cos ( ) sin ( ) sin ( ) cos ( )
cos ( )
d d
cx d ax b ax b cx d
dx dx
cx d
2
(DEN.) (NUM) NUM (DEN)
By Quotient Rule
(DEN)
d d
d u dx dx
dx v
−
=
∵
=
+ + + − + − +
+
+
2
cos ( ) cos ( ) ( ) sin ( ) ( sin ( ))
( )
cos ( )
d
cx d ax b ax b ax b cx d
dx
d
cx d
dx
cx d
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 29
30. =
+ + + + +
+
2
cos ( ) cos ( ) sin ( ) sin ( )
cos ( )
a cx d ax b c ax b cx d
cx d
+ = + = + = =
∵ ( ) ( ) ( ) ( ) 0 . 1
d d d d
ax b ax b a x a a
dx dx dx dx
+ =
Similarly ( )
d
cx d c
dx
6. cos x3
sin2
(x5
).
Sol. Let y = cos x3
sin2
(x5
) = cos x3
(sin x5
)2
∴
dy
dx
= cos x3
d
dx
(sin x5
)2
+ (sin x5
)2
d
dx
cos x3
=
∵ By Product Rule ( ) I (II) + II (I)
d d d
uv
dx dx dx
= cos x3
. 2 (sin x5
)
d
dx
sin x5
+ (sin x5
)2
(– sin x3
)
d
dx
x3
= cos x3
. 2 (sin x5
) cos x5
(5x4
) + sin2
x5
(– sin x3
) 3x2
= =
∵ 5 5 5 5 4
sin cos cos (5 )
d d
x x x x x
dx dx
= 10x4
cos x3
sin x5
cos x5
– 3x2
sin2
x5
sin x3
= x2
sin x5
[10x2
cos x3
cos x5
– 3 sin x5
sin x3
].
7. 2 2
cot ( )
x .
Sol. Let y = 2
2
cot ( )
x = 2 (cot (x2
))1/2
∴
dy
dx
= 2 .
1
2
(cot x2
)1/2 – 1
d
dx
(cot (x2
))
−
=
∵ 1
( ( )) ( ( )) ( )
n n
d d
f x n f x f x
dx dx
= (cot x2
)–1/2
−
2 2 2
cosec ( )
d
x x
dx
= −
∵ 2
cot ( ) cosec ( ( )) ( )
d d
f x f x f x
dx dx
=
− 2 2
2
cosec ( )
cot
x
x
(2x) =
− 2 2
2
2 cosec ( )
cot ( )
x x
x
.
8. cos ( x ).
Sol. Let y = cos ( x )
∴
dy
dx
=
d
dx
cos ( x ) = – sin x
d
dx
x
= −
∵ cos ( ) sin ( ) ( )
d d
f x f x f x
dx dx
= – sin x
1
2 x
− −
= = = =
∵ 1/2 1/2 1 1/2
1 1 1
2 2 2
d d
x x x x
dx dx x
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 30
31. 9. Prove that the function f given by f (x) = |
||
|| x – 1 |
||
||,
x ∈
∈
∈
∈
∈ R is not differentiable at x = 1.
Sol. Definition. A function f (x) is said to be differentiable
at a point x = c if lim
x c
→
( ) – ( )
–
f x f c
x c
exists
(and then this limit is called f ′(c) i.e., value of f ′(x) or
dy
dx
at x = c)
Here f (x) = | x – 1 |, x ∈ R ...(i)
To prove: f (x) is not differentiable at x = 1.
Putting x = 1 on (i), f (1) = | 1 – 1 | = | 0 | = 0
Left Hand Derivative = Lf ′(1) =
→ –
1
lim
x
−
−
( ) (1)
1
f x f
x
=
→ –
1
lim
x
− −
−
| 1| 0
1
x
x
=
→ –
1
lim
x
− −
−
( 1)
1
x
x
[... x → 1–
⇒ x < 1 ⇒ x – 1 < 0 ⇒ | x – 1 | = –(x – 1)]
=
→ –
1
lim
x
(– 1) = – 1 ...(ii)
Right Hand derivative = Rf ′(1) = +
→ 1
lim
x
−
−
( ) (1)
1
f x f
x
= +
→ 1
lim
x
− −
−
| 1| 0
1
x
x
= +
→ 1
lim
x
−
−
( 1)
1
x
x
(... x → 1+
⇒ x > 1 ⇒ x – 1 > 0 ⇒ | x – 1 | = x – 1)
=
+
→ 1
lim
x
1 = 1 ...(iii)
From (ii) and (iii), Lf ′(1) ≠ Rf ′(1)
∴ f (x) is not differentiable at x = 1.
Note. In problems on limits of Modulus function, and bracket
function (i.e., greatest Integer Function), we have to find both left
hand limit and right hand limit (we have used this concept quite
few times in Exercise 5.1).
10. Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2.
Sol. Given: f (x) = [x], 0 < x < 3 ...(i)
Differentiability at x = 1
Putting x = 1 in (i), f (1) = [1] = 1
Left Hand derivative = Lf ′(1) =
→ –
1
lim
x
−
−
( ) (1)
1
f x f
x
=
→ –
1
lim
x
−
−
[ ] 1
1
x
x
Put x = 1 – h, h → 0+
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 31
32. = +
→ 0
lim
h
− −
− −
[1 ] 1
1 1
h
h
=
+
→ 0
lim
h
−
−
0 1
h
= +
→ 0
lim
h
1
h
[We know that as h → 0+
, [c – h] = c – 1 if c is an integer.
Therefore [1 – h] = 1 – 1 = 0]
Put h = 0, =
1
0
= ∞ does not exist.
∴ f (x) is not differentiable at x = 1.
(We need not find Rf ′(1) as Lf ′(1) does not exist).
Differentiability at x = 2
Putting x = 2 in (i), f (2) = [2] = 2
Left Hand derivative = Lf ′(2) =
2
lim
x −
→
−
−
( ) (2)
2
f x f
x
=
−
→ 2
lim
x
−
−
[ ] 2
2
x
x
Put x = 2 – h as h → 0+
=
+
→ 0
lim
h
− −
− −
[2 ] 2
2 2
h
h
= +
→ 0
lim
h
−
−
1 2
h
= +
→ 0
lim
h
−
−
1
h
(For h → 0+
, [2 – h] = 2 – 1 = 1)
= +
→ 0
lim
h
1
h
=
1
0
= ∞ does not exist.
∴ f (x) is not differentiable at x = 2.
Note. For h → 0+
, [c + h] = c if c is an integer.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 32
33. Exercise 5.3
Find
dy
dx
in the following Exercises 1 to 15.
1. 2x + 3y = sin x.
Sol. Given: 2x + 3y = sin x
Differentiating both sides w.r.t. x, we have
d
dx
(2x) +
d
dx
(3y) =
d
dx
sin x
∴ 2 + 3
dy
dx
= cos x ⇒ 3
dy
dx
= cos x – 2 ∴
dy
dx
=
−
cos 2
3
x
.
2. 2x + 3y = sin y.
Sol. Given: 2x + 3y = sin y
Differentiating both sides w.r.t. x, we have
d
dx
(2x) +
d
dx
(3y) =
d
dx
sin y ∴ 2 + 3
dy
dx
= cos y
dy
dx
⇒ – cos y
dy
dx
+ 3
dy
dx
= – 2 ⇒ –
dy
dx
(cos y – 3) = – 2
⇒
dy
dx
=
2
cos 3
y −
.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 33
34. 3. ax + by2
= cos y.
Sol. Given: ax + by2
= cos y
Differentiating both sides w.r.t. x, we have
d
dx
(ax) +
d
dx
(by2
) =
d
dx
(cos y) ∴ a + b . 2y
dy
dx
= – sin y
dy
dx
⇒ 2by
dy
dx
+ sin y
dy
dx
= – a
⇒
dy
dx
(2by + sin y) = – a ⇒
dy
dx
=
−
+
2 sin
a
by y
4. xy + y2
= tan x + y.
Sol. Given: xy + y2
= tan x + y
Differentiating both sides w.r.t. x, we have
d
dx
(xy) +
d
dx
y2
=
d
dx
tan x +
d
dx
y
Applying product rule,
x
d
dx
y + y
d
dx
x + 2y
dy
dx
= sec2
x +
dy
dx
⇒ x
dy
dx
+ y + 2y
dy
dx
= sec2
x +
dy
dx
⇒ x
dy
dx
+ 2y
dy
dx
–
dy
dx
= sec2
x – y
⇒ (x + 2y – 1)
dy
dx
= sec2
x – y ∴
dy
dx
=
−
+ −
2
sec
2 1
x y
x y
.
5. x2
+ xy + y2
= 100.
Sol. Given: x2
+ xy + y2
= 100
Differentiating both sides w.r.t. x,
d
dx
x2
+
d
dx
xy +
d
dx
y2
=
d
dx
(100)
∴ 2x +
+
d d
x y y x
dx dx
+ 2y
dy
dx
= 0
⇒ 2x + x
dy
dx
+ y + 2y
dy
dx
= 0
⇒ (x + 2y)
dy
dx
= – 2x – y ⇒
dy
dx
= –
+
+
(2 )
2
x y
x y
.
6. x3
+ x2
y + xy2
+ y3
= 81.
Sol. Given: x3
+ x2
y + xy2
+ y3
= 81
Differentiating both sides w.r.t. x,
d
dx
x3
+
d
dx
x2
y +
d
dx
xy2
+
d
dx
y3
=
d
dx
81
∴ 3x2
+
+
2 2
.
dy d
x y x
dx dx
+ x
d
dx
y2
+ y2 d
dx
(x) + 3y2 dy
dx
= 0
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 34
35. ⇒ 3x2
+ x2 dy
dx
+ y . 2x + x . 2y
dy
dx
+ y2
.1 + 3y2 dy
dx
= 0
⇒
dy
dx
(x2
+ 2xy + 3y2
) = – 3x2
– 2xy – y2
⇒
dy
dx
= –
2 2
2 2
(3 2 )
2 3
x xy y
x xy y
+ +
+ +
.
7. sin2
y + cos xy = π
π
π
π
π.
Sol. Given: sin2
y + cos xy = π
Differentiating both sides w.r.t. x,
d
dx
(sin y)2
+
d
dx
cos xy =
d
dx
(π)
∴ 2 (sin y)1 d
dx
sin y – sin xy
d
dx
(xy) = 0
⇒ 2 sin y cos y
dy
dx
– sin xy
+
. 1
dy
x y
dx
= 0
⇒ sin 2y
dy
dx
– x sin xy
dy
dx
– y sin xy = 0
⇒ (sin 2y – x sin xy)
dy
dx
= y sin xy
∴
dy
dx
=
−
sin
sin 2 sin
y xy
y x xy
.
8. sin2
x + cos2
y = 1.
Sol. Given: sin2
x + cos2
y = 1
Differentiating both sides w.r.t. x,
d
dx
(sin x)2
+
d
dx
(cos y)2
=
d
dx
(1)
∴ 2 (sin x)1
d
dx
sin x + 2 (cos y)1
d
dx
cos y = 0
⇒ 2 sin x cos x + 2 cos y
−
sin
dy
y
dx
= 0
⇒ 2 sin x cos x – 2 sin y cos y
dy
dx
= 0
⇒ sin 2x – sin 2y
dy
dx
= 0
⇒ – sin 2y
dy
dx
= – sin 2x ⇒
dy
dx
=
sin 2
sin 2
x
y
.
9. y = sin–1
2
2
1 +
x
x
.
Sol. Given: y = sin–1
2
2
1
x
x
+
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 35
36. To simplify the given Inverse T-function, put x = tan θ
θ
θ
θ
θ.
∴ y = sin–1
2
2 tan
1 tan
θ
+ θ
= sin–1
(sin 2θ) = 2θ
⇒ y = 2 tan–1
x (... x = tan θ ⇒ θ = tan–1
x)
∴
dy
dx
= 2 . 2
1
1 x
+
= 2
2
1 x
+
.
10. y = tan–1
3
2
3 –
1 – 3
x x
x
,
– 1
3
< x <
1
3
.
Sol. Given: y = tan–1
3
2
3
1 3
x x
x
−
−
,
1
3
−
< x <
1
3
To simplify the given Inverse T-function, put x = tan θ
θ
θ
θ
θ.
∴ y = tan–1
3
2
3 tan tan
1 3 tan
θ − θ
− θ
= tan–1
(tan 3θ) = 3θ
⇒ y = 3 tan–1
x (... x = tan θ ⇒ θ = tan–1
x)
∴
dy
dx
= 3 . 2
1
1 x
+
= 2
3
1 x
+
.
11. y = cos–1
2
2
1 –
1 +
x
x
, 0 < x < 1.
Sol. Given: y = cos–1
2
2
1
1
x
x
−
+
, 0 < x < 1
To simplify the given Inverse T-function, put x = tan θ
θ
θ
θ
θ.
∴ y = cos–1
2
2
1 tan
1 tan
− θ
+ θ
= cos–1
(cos 2θ)
= 2θ = 2 tan–1
x (... x = tan θ ⇒ θ = tan–1
x)
∴
dy
dx
= 2 . 2
1
1 x
+
= 2
2
1 x
+
.
12. y = sin–1
2
2
1 –
1 +
x
x
, 0 < x < 1.
Sol. Given: y = sin–1
2
2
1
1
x
x
−
+
To simplify the given Inverse T-function, put x = tan θ
θ
θ
θ
θ.
∴ y = sin–1
2
2
1 tan
1 tan
− θ
+ θ
= sin–1
(cos 2θ)
= sin–1
sin 2
2
π
− θ
=
2
π
– 2θ
⇒ y =
2
π
– 2 tan–1
x (... x = tan θ ⇒ θ = tan–1
x)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 36
37. ∴
dy
dx
= 0 – 2 . 2
1
1 x
+
= 2
2
1 x
−
+
.
13. y = cos–1
2
2
1 +
x
x
, – 1 < x < 1.
Sol. Given: y = cos–1
2
2
1
x
x
+
To simplify the given Inverse T-function put x = tan θ
θ
θ
θ
θ.
∴ y = cos–1
2
2 tan
1 tan
θ
+ θ
= cos–1
(sin 2θ)
= cos–1
cos 2
2
π
− θ
=
2
π
– 2θ
⇒ y =
2
π
– 2 tan–1
x (... x = tan θ ⇒ θ = tan–1
x)
∴
dy
dx
= 0 – 2 . 2
1
1 x
+
= 2
2
1 x
−
+
.
14. y = sin–1
(2x 2
1 – x ), –
1
2
< x <
1
2
.
Sol. Given: y = sin–1
(2x
2
1 x
− )
Put x = sin θ
θ
θ
θ
θ
To simplify the given Inverse T-function,
put x = sin θ
θ
θ
θ
θ (For 2 2
a x
− , put x = a sin θ)
∴ y = sin–1
(2 sin θ 2
1 sin
− θ )
= sin–1
(2 sin θ 2
cos θ ) = sin–1
(2 sin θ cos θ)
y = sin–1
(sin 2θ) = 2θ = 2 sin–1
x
[... x = sin θ ⇒ θ = sin–1
x]
∴
dy
dx
= 2 . 2
1
1 x
−
.
15. y = sec–1
2
1
2 – 1
x
0 < x <
1
2
.
Sol. Given: y = sec–1
2
1
2 1
x
−
To simplify the given inverse T-function, put x = cos θ
θ
θ
θ
θ.
∴ y = sec–1
2
1
2 cos 1
θ −
= sec–1
1
cos 2
θ
= sec–1
(sec 2θ) = 2θ = 2 cos–1
x (... x = cos θ ⇒ θ = cos–1
x)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 37
38. ∴
dy
dx
= 2
2
1
1 x
−
−
=
2
2
1 x
−
−
.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 38
39. Exercise 5.4
Differentiate the following functions 1 to 10 w.r.t. x
1.
sin
x
e
x
.
Sol. Let y =
sin
x
e
x
∴
dy
dx
= 2
(DEN) (NUM) (NUM) (DEN)
(DEN)
d d
dx dx
−
= 2
sin sin
sin
x x
d d
x e e x
dx dx
x
−
= 2
sin . cos
sin
x x
x e e x
x
−
= ex
2
(sin cos )
sin
x x
x
−
.
2.
–1
sin x
e .
Sol. Let y =
1
sin x
e
−
∴
dy
dx
=
1
sin x
e
− d
dx
sin–1
x
( ) ( )
( )
f x f x
d d
e e f x
dx dx
=
∵
=
1
sin x
e
−
.
2
1
1 x
−
.
3.
3
x
e .
Sol. Let y =
3
x
e =
3
( )
x
e
∴
dy
dx
=
3
( )
x
e
d
dx
x3 ( ) ( )
( )
f x f x
d d
e e f x
dx dx
=
∵
=
3
( )
x
e 3x2
= 3x2 3
( )
x
e .
4. sin (tan–1
e– x
).
Sol. Let y = sin (tan–1
e– x
)
∴
dy
dx
= cos (tan–1
e– x
)
d
dx
(tan–1
e– x
)
sin ( ) cos ( ) ( )
d d
f x f x f x
dx dx
=
∵
= cos (tan–1
e– x
) 2
1
1 ( )
x
e−
+
d
dx
e– x
1
2
1
tan ( ) ( )
1 ( ( ))
d d
f x f x
dx dx
f x
−
=
+
∵
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 39
40. = cos (tan–1
e– x
) 2
1
1 x
e−
+
e– x
d
dx
(– x)
= –
1
2
cos (tan )
1
x x
x
e e
e
− − −
−
+
( ) 1
d
x
dx
− = −
∵
5. log (cos ex
).
Sol. Let y = log (cos ex
)
∴
dy
dx
=
1
cos x
e
d
dx
(cos ex
)
1
log ( ) ( )
( )
d d
f x f x
dx f x dx
=
∵
=
1
cos x
e
(– sin ex
)
d
dx
ex cos ( ) sin ( ) ( )
d d
f x f x f x
dx dx
= −
∵
= – (tan ex
) ex
= – ex
(tan ex
)
6. ex
+ ex
2
+ ... + ex
5
.
Sol. Let y = ex
+ ex
2
+ ... + ex
5
= ex
+ ex
2
+ ex
3
+ ex
4
+ ex
5
∴
dy
dx
=
d
dx
ex
+
d
dx
ex
2
+
d
dx
ex
3
+
d
dx
ex
4
+
d
dx
ex
5
= ex
+ ex
2 d
dx
x2
+ ex
3 d
dx
x3
+ ex
4 d
dx
x4
+ ex
5 d
dx
x5 ( ) ( )
( )
f x f x
d d
e e f x
dx dx
=
∵
= ex
+ ex
2
. 2x + ex
3
. 3x2
+ ex
4
. 4x3
+ ex
5
5x4
= ex
+ 2x ex
2
+ 3x2
ex
3
+ 4x3
ex
4
+ 5x4
ex
5
.
7. x
e , x > 0.
Sol. Let y = x
e = ( x
e )1/2
∴
dy
dx
=
1
2
(
x
e )–1/2
d
dx
x
e
1
( ( )) ( ( )) ( )
n n
d d
f x n f x f x
dx dx
−
=
∵
=
1
2 x
e
x
e
d
dx
x
( ) ( )
( )
f x f x
d d
e e f x
dx dx
=
∵
=
1
2 x
e
x
e
1
2 x
1/2 1/2
1 1
2 2
d d
x x x
dx dx x
−
= = =
∵
=
4
x
x
e
x e
.
8. log (log x), x > 1.
Sol. Let y = log (log x)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 40
41. ∴
dy
dx
=
1
log x
d
dx
(log x)
1
log ( ) ( )
( )
d d
f x f x
dx f x dx
=
∵
=
1
log x
1
x
=
1
log
x x
.
9.
cos
log
x
x
, x > 0.
Sol. Let y =
cos
log
x
x
∴
dy
dx
=
2
(DEN) (NUM) (NUM) (DEN)
(DEN)
d d
dx dx
−
= 2
log (cos ) cos log
(log )
d d
x x x x
dx dx
x
−
= 2
1
log ( sin ) cos .
(log )
x x x
x
x
− −
= 2
cos
sin log
(log )
x
x x
x
x
− +
= – 2
( sin log cos )
(log )
x x x x
x x
+
.
10. cos (log x + ex
), x > 0.
Sol. Let y = cos (log x + ex
)
∴
dy
dx
= – sin (log x + ex
)
d
dx
(log x + ex
)
cos ( ) sin ( ) ( )
d d
f x f x f x
dx dx
= −
∵
= – sin (log x + ex
) .
1 x
e
x
+
= –
1 x
e
x
+
sin (log x + ex
).
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 41
42. Exercise 5.5
Note. Logarithmic Differentiation.
The process of differentiating a function after taking its logarithm
is called logarithmic differentiation.
This process of differentiation is very useful in the following
situations:
(i) The given function is of the form ( f (x)) g(x)
(ii) The given function involves complicated (as per our thinking)
products (or and) quotients (or and) powers.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 42
43. Remark 1. log
( )
m n p
q k
a b c
d l
= m log a + n log b + p log c – q log d – k log l
Remark 2. log (u + v) ≠ log u + log v
and log (u – v) ≠ log u – log v.
Differentiate the following functions given in Exercises
1 to 5 w.r.t. x.
1. cos x cos 2x cos 3x.
Sol. Let y = cos x cos 2x cos 3x ...(i)
Taking logs on both sides, we have (see Note, (ii) page 261)
log y = log (cos x cos 2x cos 3x)
= log cos x + log cos 2x + log cos 3x
Differentiating both sides w.r.t. x, we have
d
dx
log y =
d
dx
log cos x +
d
dx
log cos 2x +
d
dx
log cos 3x
∴
1
y
dy
dx
=
1
cos x
d
dx
cos x +
1
cos 2x
d
dx
cos 2x
+
1
cos 3x
d
dx
cos 3x
1
log ( ) ( )
( )
d d
f x f x
dx f x dx
=
∵
=
1
cos x
(– sin x) +
1
cos 2x
(– sin 2x)
d
dx
(2x)
+
1
cos 3x
(– sin 3x)
d
dx
3x
= – tan x – (tan 2x) 2 – tan 3x (3)
∴
dy
dx
= – y (tan x + 2 tan 2x + 3 tan 3x).
Putting the value of y from (i),
dy
dx
= – cos x cos 2x cos 3x (tan x + 2 tan 2x + 3 tan 3x).
2.
( – 1)( – 2)
( – 3)( – 4)( – 5)
x x
x x x
.
Sol. Let y =
( 1)( 2)
( 3)( 4)( 5)
x x
x x x
− −
− − −
=
1/2
( 1)( 2)
( 3)( 4)( 5)
x x
x x x
− −
− − −
...(i)
Taking logs on both sides, we have
log y =
1
2
[log (x – 1) + log (x – 2) – log (x – 3)
– log (x – 4) – log (x – 5)] (By Remark I above)
Differentiating both sides w.r.t. x, we have
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 43
44. 1
y
dy
dx
=
1
2
1 1 1
( 1) ( 2) ( 3)
1 2 3
d d d
x x x
x dx x dx x dx
− + − − −
− − −
1 1
( 4) ( 5)
4 5
d d
x x
x dx x dx
− − − −
− −
∴
dy
dx
=
1
2
y
1 1 1 1 1
1 2 3 4 5
x x x x x
+ − − −
− − − − −
Putting the value of y from (i),
dy
dx
=
1
2
( 1)( 2)
( 3)( 4)( 5)
x x
x x x
− −
− − −
1 1 1 1 1
1 2 3 4 5
x x x x x
+ − − −
− − − − −
.
3. (log x)cos x
.
Sol. Let y = (log x)cos x
...(i) [Form ( f (x)) g(x)
]
Taking logs on both sides of (i), we have (see Note (i) page 261)
log y = log (log x)cos x
= cos x log (log x)
[... log mn
= n log m]
∴
d
dx
log y =
d
dx
[cos x . log (log x)]
⇒
1
y
dy
dx
= cos x
d
dx
log (log x) + log (log x)
d
dx
cos x
[By Product Rule]
= cos x
1
log x
d
dx
log x + log (log x)(– sin x)
=
cos
log
x
x
.
1
x
– sin x log (log x)
∴
dy
dx
= y
cos
sin log (log )
log
x
x x
x x
−
.
Putting the value of y from (i),
dy
dx
= (log x)cos x
cos
sin log (log )
log
x
x x
x x
−
.
Very Important Note.
To differentiate y = ( f (x)) g(x)
± (l(x))m(x)
or y = ( f (x)) g(x)
± h(x)
or y = ( f (x)) g(x)
± k where k is a constant;
Never start with taking logs of both sides, put one term
= u and the other = v
∴ y = u ± v
∴
dy
dx
=
du
dx
±
dv
dx
Now find
du
dx
and
dv
dx
by the methods already learnt.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 44
45. 4. x x
– 2sin x
.
Sol. Let y = xx
– 2sin x
Put u = xx
and v = 2sin x
(See Note)
∴ y = u – v
∴
dy
dx
=
du
dx
–
dv
dx
...(i)
Now u = xx
[Form (f (x))g(x)
]
∴ log u = log xx
= x log x [... log mn
= n log m]
∴
d
dx
log u =
d
dx
(x log x)
⇒
1
u
du
dx
= x
d
dx
log x + log x
d
dx
x
= x .
1
x
+ log x . 1 = 1 + log x
∴
du
dx
= u (1 + log x) = xx
(1 + log x) ...(ii)
Again v = 2sin x
∴
dv
dx
=
d
dx
2sin x
= 2sin x
log 2
d
dx
sin x
( ) ( )
log ( )
f x f x
d d
a a a f x
dx dx
=
∵
⇒
dv
dx
= 2sin x
(log 2) cos x = cos x . 2sin x
log 2 ...(iii)
Putting values from (ii) and (iii) in (i),
dy
dx
= xx
(1 + log x) – cos x . 2sin x
log 2.
5. (x + 3)2
(x + 4)3
(x + 5)4
.
Sol. Let y = (x + 3)2
(x + 4)3
(x + 5)4
...(i)
Taking logs on both sides of eqn. (i) (see Note (ii) page 261)
we have log y =2 log (x + 3) + 3 log (x + 4)
+ 4 log (x + 5) (By Remark I page 262)
∴
d
dx
log y = 2
d
dx
log (x + 3) + 3
d
dx
log (x + 4) + 4
d
dx
log (x + 5)
⇒
1
y
dy
dx
= 2 .
1
3
x +
d
dx
(x + 3) + 3
1
4
x +
d
dx
(x + 4)
+ 4 .
1
5
x +
d
dx
(x + 5)
=
2
3
x +
+
3
4
x +
+
4
5
x +
∴
dy
dx
= y
2 3 4
3 4 5
x x x
+ +
+ + +
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 45
46. Putting the value of y from (i),
dy
dx
= (x + 3)2
(x + 4)3
(x + 5)4
2 3 4
3 4 5
x x x
+ +
+ + +
.
Differentiate the following functions given in Exercises 6 to 11 w.r.t. x.
6.
1
+
x
x
x
+
1
1+
x
x .
Sol. Let y =
1
x
x
x
+
+
1
1
x
x
+
Putting
1
x
x
x
+
= u and
1
1
x
x
+
= v,
We have y = u + v ∴
dy
dx
=
du
dx
+
dv
dx
...(i)
Now u =
1
x
x
x
+
Taking logarithms, log u = log
1
x
x
x
+
= x log
1
x
x
+
[Form uv]
Differentiating w.r.t. x, we have
1
u
du
dx
= x .
1
1
x
x
+
d
dx
1
x
x
+
+ log
1
x
x
+
. 1
1
u
du
dx
= x .
1
1
x
x
+
. 2
1
1
x
−
+ log
1
x
x
+
. 1
–1 –2
2
1 – 1
(– 1)
d d
x x
dx x dx x
= = =
∵
⇒
du
dx
= u
2
2
1 1
log
1
x
x
x
x
−
+ +
+
=
1
x
x
x
+
2
2
1 1
log
1
x
x
x
x
−
+ +
+
...(ii)
Also v =
1
1
x
x
+
Taking logarithms, log v = log
1
1
x
x
+
=
1
1
x
+
log x
Differentiating w.r.t. x, we have
1
v
.
dv
dx
=
1
1
x
+
.
1
x
+ log x .
2
1
x
−
–1 –2
2
1 – 1
(– 1)
d d
x x
dx x dx x
= = =
∵
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 46
47. ⇒
dv
dx
= v 2
1 1 1
1 log x
x x x
+ −
=
1
1
x
x
+
2
1 1 1
1 log x
x x x
+ −
...(iii)
Putting the values of
du
dx
and
dv
dx
from (ii) and (iii) in (i), we
have
dy
dx
=
1
x
x
x
+
2
2
1 1
log
1
x
x
x
x
−
+ +
+
+
1
1
x
x
+
2
1 1 1
1 log x
x x x
+ −
7. (log x)x
+ xlog x
.
Sol. Let y = (log x)x
+ xlog x
= u + v where u = (log x)x
and v = xlog x
∴
dy
dx
=
du
dx
+
dv
dx
...(i)
Now u = (log x)x
[(f (x)) g(x)
]
∴ log u = log (log x)x
= x log (log x) [... log mn
= n log m]
∴
d
dx
log u =
d
dx
[x log (log x)]
∴
1
u
du
dx
= x
d
dx
log (log x) + log (log x)
d
dx
x (By product rule)
= x .
1
log x
d
dx
log x + log (log x) . 1
= x .
1
log x
.
1
x
+ log (log x)
∴
du
dx
= u
1
log (log )
log
x
x
+
= (log x)x
1
log (log )
log
x
x
+
= (log x)x (1 log log (log ))
log
x x
x
+
= (log x)x – 1
(1 + log x log (log x)) ...(ii)
Again v = xlog x
[Form ( f (x)) g(x)
]
∴ log v = log xlog x
= log x . log x [... log mn
= n log m]
= (log x)2
∴
d
dx
log v =
d
dx
(log x)2
∴ 1
v
dv
dx
= 2 (log x)1 d
dx
log x
1
( ( )) ( ( )) ( )
n n
d d
f x n f x f x
dx dx
−
=
∵
= 2 log x .
1
x
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 47
48. ∴
dv
dx
= v
2
log x
x
= xlog x
.
2
x
log x
= 2xlog x – 1
log x ...(iii)
Putting values of
du
dx
and
dv
dx
from (ii) and (iii) in (i), we have
dy
dx
= (log x)x – 1
(1 + log x log (log x)) + 2xlog x – 1
log x.
8. (sin x)x
+ sin–1
x .
Sol. Let y = (sin x)x
+ sin–1
x
= u + v where u = (sin x)x
and v = sin–1
x
∴
dy
dx
=
du
dx
+
dv
dx
...(i)
Now u = (sin x)x
[Form ( f (x)) g(x)
]
∴ log u = log (sin x)x
= x log sin x
∴
d
dx
(log u) =
d
dx
(x log sin x)
⇒
1
u
du
dx
= x
d
dx
log sin x + log sin x
d
dx
x
= x .
1
sin x
d
dx
sin x + (log sin x) . 1
= x
1
sin x
cos x + log sin x = x cot x + log sin x
∴
du
dx
= u (x cot x + log sin x) = (sin x)x
(x cot x + log sin x)...(ii)
Again v = sin–1 x
∴
dv
dx
=
2
1
1 ( )
x
−
d
dx x
1
2
1
sin ( ) ( )
1 ( ( ))
d d
f x f x
dx dx
f x
−
=
−
∵
=
1
1 x
−
1
2 x
1/2 1/2
1 1
2 2
d d
x x x
dx dx x
−
= = =
∵
or
dv
dx
=
1
2 1
x x
−
=
1
2 (1 )
x x
−
=
2
1
2 x x
−
...(iii)
Putting values of
du
dx
and
dv
dx
from (ii) and (iii) in (i),
dy
dx
= (sin x)x
(x cot x + log sin x) +
2
1
2 x x
−
.
9. xsin x
+ (sin x)cos x
.
Sol. Let y = xsin x
+ (sin x)cos x
= u + v where u = xsin x
and v = (sin x)cos x
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 48
49. ∴
dy
dx
=
du
dx
+
dv
dx
...(i)
Now u = xsin x
[Form ( f (x)) g(x)
]
∴ log u = log xsin x
= sin x log x
∴
d
dx
log u =
d
dx
(sin x log x)
⇒
1
u
du
dx
= sin x
d
dx
log x + log x
d
dx
sin x
= sin x .
1
x
+ (log x) cos x =
sin x
x
+ cos x log x
∴
du
dx
= u
sin
cos log
x
x x
x
+
= xsin x
sin
cos log
x
x x
x
+
...(ii)
Again v = (sin x)cos x
[Form f (x)g(x)
]
∴ log v = log (sin x)cos x
= cos x log sin x
∴
d
dx
(log v) =
d
dx
[cos x log sin x]
⇒
1
v
dv
dx
= cos x
d
dx
log sin x + log sin x
d
dx
cos x
= cos x
1
sin x
d
dx
(sin x) + log sin x (– sin x)
= cot x . cos x – sin x log sin x
∴
dv
dx
= v (cos x cot x – sin x log sin x)
= (sin x)cos x
(cos x cot x – sin x log sin x) ...(iii)
Putting values of
du
dx
and
dv
dx
from (ii) and (iii) in (i),
we have
dy
dx
= xsin x sin
cos log
x
x x
x
+
+ (sin x)cos x
(cos x cot x – sin x log sin x)
10. xx cos x
+
2
2
+1
– 1
x
x
.
Sol. Let y = xx cos x
+
2
2
1
1
x
x
+
−
Putting xx cos x
= u and
2
2
1
1
x
x
+
−
= v
We have y = u + v ∴
dy
dx
=
du
dx
+
dv
dx
...(i)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 49
50. Now u = xx cos x
Taking logarithms, log u = log xx cos x
= x cos x log x
Differentiating w.r.t. x, we have
1
u
.
du
dx
=
d
dx
(x cos x log x)
=
d
dx
(x) . cos x log x + x
d
dx
(cos x) . log x
+ x cos x
d
dx
(log x)
( ) .
d du dv dw
uvw vw u w uv
dx dx dx dx
= + +
∵
= 1 cos x log x + x (– sin x) log x + x cos x .
1
x
⇒
du
dx
= u [cos x log x – x sin x log x + cos x]
= xx cos x
[cos x log x – x sin x log x + cos x] ...(ii)
Also v =
2
2
1
1
x
x
+
−
. Using quotient rule, we have
dv
dx
=
2 2 2 2
2 2
( 1) ( 1) ( 1) . ( 1)
( 1)
d d
x x x x
dx dx
x
− + − + −
−
=
2 2
2 2
( 1) . 2 ( 1) . 2
( 1)
x x x x
x
− − +
−
=
3 3
2 2
2 2 2 2
( 1)
x x x x
x
− − −
−
= – 2 2
4
( 1)
x
x −
...(iii)
Putting the values of
du
dx
and
dv
dx
from (ii) and (iii) in (i), we have
dy
dx
= xx cos x
[cos x log x – x sin x log x + cos x] – 2 2
4
( 1)
x
x −
.
11. (x cos x)x
+ (x sin x)1/x
.
Sol. Let y = (x cos x)x
+ (x sin x)1/x
Putting (x cos x)x
= u and (x sin x)1/x
= v,
We have y = u + v ∴
dy
dx
=
du
dx
+
dv
dx
...(i)
Now u = (x cos x)x
Taking logarithms, log u = log (x cos x)x
= x log (x cos x)
= x (log x + log cos x)
Differentiating w.r.t. x, we have
1
u
.
du
dx
= x
1 1
. ( sin )
cos
x
x x
+ −
+ (log x + log cos x) . 1
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 50
51. ⇒
du
dx
= u [1 – x tan x + log (x cos x)]
[... log x + log cos x = log (x cos x)]
= (x cos x)x
[1 – x tan x + log (x cos x)] ...(ii)
Also v = (x sin x)1/x
Taking logarithms, log v = log (x sin x)1/x
=
1
x
log (x sin x)
=
1
x
(log x + log sin x)
Differentiating w.r.t. x, we have
1
v
.
dv
dx
=
1
x
1 1
. cos
sin
x
x x
+
+ (log x + log sin x) 2
1
x
−
⇒
dv
dx
= v 2 2
1 cot log ( sin )
x x x
x
x x
+ −
= (x sin x)1/x
. 2
1 cot log ( sin )
x x x x
x
+ −
...(iii)
Putting the values of
du
dx
and
dv
dx
from (ii) and (iii) in (i), we
have
dy
dx
= (x cos x)x
[1 – x tan x + log (x cos x)]
+ (x sin x)1/x
2
1 cot log ( sin )
x x x x
x
+ −
.
Find
dy
dx
of the functions given in Exercises 12 to 15:
12. xy
+ yx
= 1.
Sol. Given : xy
+ yx
= 1
⇒ u + v = 1 where u = xy
and v = yx
∴
d
dx
(u) +
d
dx
(v) =
d
dx
(1)
i.e.,
du
dx
+
dv
dx
= 0 ...(i)
Now u = xy
[(Variable)variable
= ( f (x)) g(x)
]
∴ log u = log xy
= y log x
∴
d
dx
log u =
d
dx
( y log x)
⇒
1
u
du
dx
= y
d
dx
log x + log x
dy
dx
= y .
1
x
+ log x .
dy
dx
∴
du
dx
= u log .
y dy
x
x dx
+
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 51
52. or
du
dx
= xy log
y dy
x
x dx
+
= xy y
x
+ xy
log x
dy
dx
or
du
dx
= xy – 1
y + xy
log x
dy
dx
...(ii)
1
1
y y
y
x x
x
x x
−
= =
∵
Again v = yx
∴ log v = log yx
= x log y ∴
d
dx
log v =
d
dx
(x log y)
⇒
1
v
dv
dx
= x
d
dx
(log y) + log y
d
dx
x = x .
1
y
dy
dx
+ log y . 1
⇒
dv
dx
= v log
x dy
y
y dx
+
= yx
log
x dy
y
y dx
+
= yx
x
y
dy
dx
+ yx
log y
⇒
dv
dx
= yx – 1
x
dy
dx
+ yx
log y ...(iii)
Putting values of
du
dx
and
dv
dx
from (ii) and (iii) in (i), we have
xy – 1
y + xy
log x
dy
dx
+ yx – 1
x
dy
dx
+ yx
log y = 0
or
dy
dx
(xy
log x + yx – 1
x) = – xy – 1
y – yx
log y
∴
dy
dx
= –
1
1
( log )
log
y x
y x
x y y y
x x y x
−
−
+
+
.
13. yx
= xy
.
Sol. Given: yx
= xy
⇒ xy
= yx
.
| Form on both sides is (f (x))g(x)
Taking logarithms, log xy
= log yx
⇒ y log x = x log y
Differentiating w.r.t. x, we have
y .
1
x
+ log x .
dy
dx
= x .
1
y
.
dy
dx
+ log y . 1
⇒ log
x
x
y
−
dy
dx
= log y –
y
x
⇒
log
y x x
y
−
.
dy
dx
=
log
x y y
x
−
∴
dy
dx
=
( log )
( log )
y x y y
x y x x
−
−
.
14. (cos x)y
= (cos y)x
.
Sol. Given: (cos x)y
= (cos y)x
[Form on both sides is ( f (x)) g(x)
]
∴ Taking logs on both sides, we have
log (cos x)y
= log (cos y)x
⇒ y log cos x = x log cos y [... log mn
= n log m]
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 52
53. Differentiating both sides w.r.t. x, we have
d
dx
( y log cos x) =
d
dx
(x log cos y)
Applying Product Rule on both sides,
⇒ y
d
dx
log cos x + log cos x
dy
dx
= x
d
dx
log cos y + log cos y
d
dx
x
⇒ y .
1
cos x
d
dx
cos x + log cos x
dy
dx
= x .
1
cos y
d
dx
cos y + log cos y
⇒ y
1
cos x
(– sin x) + log cos x
dy
dx
= x
1
cos y
sin
dy
y
dx
−
+ log cos y
⇒ – y tan x + log cos x .
dy
dx
= – x tan y
dy
dx
+ log cos y
⇒ x tan y
dy
dx
+ log cos x .
dy
dx
= y tan x + log cos y
⇒
dy
dx
(x tan y + log cos x) = y tan x + log cos y
⇒
dy
dx
=
tan log cos
tan log cos
y x y
x y x
+
+
.
15. xy = ex – y
.
Sol. Given: xy = ex – y
Taking logs on both sides, we have
log (xy) = log ex – y
⇒ log x + log y = (x – y) log e
⇒ log x + log y = x – y (... log e = 1)
Differentiating both sides w.r.t. x, we have
d
dx
log x +
d
dx
log y =
d
dx
x –
d
dx
y
⇒
1
x
+
1
y
dy
dx
= 1 –
dy
dx
⇒
1
y
dy
dx
+
dy
dx
= 1 –
1
x
⇒
dy
dx
1
1
y
+
=
1
x
x
−
⇒
1 y
y
+
dy
dx
=
1
x
x
−
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 53
54. Cross-multiplying x(1 + y)
dy
dx
= y(x – 1)
⇒
dy
dx
=
( 1)
(1 )
y x
x y
−
+
.
16. Find the derivative of the function given by
f (x) = (1 + x)(1 + x2
)(1 + x4
)(1 + x8
) and hence find f ′
′
′
′
′(1).
Sol. Given: f (x) = (1 + x)(1 + x2
)(1 + x4
)(1 + x8
) ...(i)
Taking logs on both sides, we have
log f (x) = log (1 + x) + log (1 + x2
) + log (1 + x4
) + log (1 + x8
)
Differentiating both sides w.r.t. x, we have
1
( )
f x
d
dx
f (x) =
1
1 x
+
d
dx
(1 + x) + 2
1
1 x
+
d
dx
(1 + x2
)
+ 4
1
1 x
+
d
dx
(1 + x4
) + 8
1
1 x
+
d
dx
(1 + x8
)
⇒
1
( )
f x
f ′(x) =
1
1 x
+
. 1 + 2
1
1 x
+
. 2x + 4
1
1 x
+
. 4x3
+ 8
1
1 x
+
8x7
∴ f ′(x) = f (x)
3 7
2 4 8
1 2 4 8
1 1 1 1
x x x
x x x x
+ + +
+ + + +
Putting the value of f (x) from (i),
f ′(x) = (1 + x)(1 + x2
)(1 + x4
)(1 + x8
)
3 7
2 4 8
1 2 4 8
1 1 1 1
x x x
x x x x
+ + +
+ + + +
Putting x = 1,
f ′(1) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)
1 2 4 8
1 1 1 1 1 1 1 1
+ + +
+ + + +
= 2.2.2.2
1 2 4 8
2 2 2 2
+ + +
= 16
15
2
= 8 × 15 = 120.
17. Differentiate (x2
– 5x + 8)(x3
+ 7x + 9) in three ways
mentioned below:
(i) by using product rule.
(ii) by expanding the product to obtain a single
polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
Sol. Given: Let y = (x2
– 5x + 8)(x3
+ 7x + 9) ...(1)
(i) To find
dy
dx
by using Product Rule
dy
dx
= (x2
– 5x + 8)
d
dx
(x3
+ 7x + 9)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 54
55. + (x3
+ 7x + 9)
d
dx
(x2
– 5x + 8)
= (x2
– 5x + 8)(3x2
+ 7) + (x3
+ 7x + 9)(2x – 5)
= 3x4
+ 7x2
– 15x3
– 35x + 24x2
+ 56
+ 2x4
– 5x3
+ 14x2
– 35x + 18x – 45
= 5x4
– 20x3
+ 45x2
– 52x + 11 ...(2)
(ii) To find
dy
dx
by expanding the product to obtain a
single polynomial.
From (i), y = (x2
– 5x + 8) (x3
+ 7x + 9)
= x5
+ 7x3
+ 9x2
– 5x4
– 35x2
– 45x
+ 8x3
+ 56x + 72
or y = x5
– 5x4
+ 15x3
– 26x2
+ 11x + 72
∴
dy
dx
= 5x4
– 20x3
+ 45x2
– 52x + 11 ...(3)
(iii) To find
dy
dx
by logarithmic differentiation
Taking logs on both sides of (i), we have
log y = log (x2
– 5x + 8) + log (x3
+ 7x + 9)
∴
d
dx
log y =
d
dx
log (x2
– 5x + 8) +
d
dx
log (x3
+ 7x + 9)
⇒
1
y
dy
dx
= 2
1
5 8
x x
− +
d
dx
(x2
– 5x + 8)
+ 3
1
7 9
x x
+ +
.
d
dx
(x3
+ 7x + 9)
= 2
1
5 8
x x
− +
(2x – 5) + 3
1
7 9
x x
+ +
(3x2
+ 7)
∴
dy
dx
= y
2
2 3
(2 5) 3 7
5 8 7 9
x x
x x x x
− +
+
− + + +
= y
3 2 2
2 3
(2 5)( 7 9) (3 7)( 5 8)
( 5 8)( 7 9)
x x x x x x
x x x x
− + + + + − +
− + + +
= y
4 2 3 4 3
2 2
2 3
[2 14 18 5 35 45 3 15
24 7 35 56]
( 5 8)( 7 9)
x x x x x x x
x x x
x x x x
+ + − − − + −
+ + − +
− + + +
or
dy
dx
= y
4 3 2
2 3
(5 20 45 52 11)
( 5 8)( 7 9)
x x x x
x x x x
− + − +
− + + +
Putting the value of y from (i),
dy
dx
= (x2
– 5x + 8)(x3
+ 7x + 9)
4 3 2
2 3
(5 20 45 52 11)
( 5 8)( 7 9)
x x x x
x x x x
− + − +
− + + +
= 5x4
– 20x3
+ 45x2
– 52x + 11 ...(4)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 55
56. From (2), (3) and (4), we can say that value of
dy
dx
is same
obtained by three different methods used in (i), (ii) and (iii).
18. If u, v and w are functions of x, then show that
d
dx
(u . v . w) =
du
dx
v . w + u .
dv
dx
. w + u . v
dw
dx
in two ways-first by repeated application of product rule,
second by logarithmic differentiation.
Sol. Given: u, v and w are functions of x.
To prove:
d
dx
(u . v . w) =
du
dx
. v . w + u .
dv
dx
. w + u . v .
dw
dx
...(i)
(i) To prove eqn. (i): By repeated application of product
rule
L.H.S. =
d
dx
(u . v . w)
Let us treat the product uv as a single function
=
d
dx
[(uv)w] = uv
d
dx
(w) + w
d
dx
(uv)
Again Applying Product Rule on
d
dx
(uv)
L.H.S. =
d
dx
(uvw) = uv
dw
dx
+ w
d d
u v v u
dx dx
+
= uv
dw
dx
+ uw
dv
dx
+ vw
du
dx
Rearranging terms
or
d
dx
(uvw) =
du
dx
. v . w + u .
dv
dx
. w + u . v .
dw
dx
which proves eqn. (i)
(ii) To prove eqn. (i): By Logarithmic differentiation
Let y = uvw
Taking logs on both sides
log y = log (u . v . w) = log u + log v + log w
∴
d
dx
log y =
d
dx
log u +
d
dx
log v +
d
dx
log w
⇒
1
y
dy
dx
=
1
u
du
dx
+
1
v
dv
dx
+
1
w
dw
dx
⇒
dy
dx
= y
1 1 1
du dv dw
u dx v dx w dx
+ +
Putting y = uvw,
d
dx
(uvw) = uvw
1 1 1
du dv dw
u dx v dx w dx
+ +
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 56
57. =
du
dx
. v . w + u .
dv
dx
. w + u .v.
dw
dx
which proves eqn. (i).
Remark. The result of eqn. (i) can be used as a formula
for derivative of product of three functions.
It can be used as a formula for doing Q. No. 1 and Q. No. 5
of this Exercise 5.5.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 57
58. Exercise 5.6
If x and y are connected parametrically by the equations given in
dy
Exercises 1 to 5, without eliminating the parameter, find
dx
.
1. x = 2at2
, y = at4
.
Sol. Given: x = 2at2
and y = at4
Differentiating both eqns. w.r.t. t, we have
dx
dt
=
d
dt
(2at2
) and
dy
dt
=
d
dt
(at4
)
= 2a
d
dt
t2
= a
d
dt
t4
= a.4t3
= 2a.2t = 4at = 4at3
We know that
dy
dx
=
/
/
dy dt
dx dt
=
3
4
4
at
at
= t2
.
2. x = a cos θ
θ
θ
θ
θ, y = b cos θ
θ
θ
θ
θ.
Sol. Given: x = a cos θ and y = b cos θ
Differentiating both eqns. w.r.t. θ, we have
dx
dθ
=
d
dθ
(a cos θ) and
dy
dθ
=
d
dθ
(b cos θ)
= a
d
dθ
cos θ = b
d
dθ
cos θ
= – a sin θ = – b sin θ
We know that
dy
dx
=
/
/
dy d
dx d
θ
θ
=
sin
sin
b
a
− θ
− θ
=
b
a
.
3. x = sin t, y = cos 2t.
Sol. Given: x = sin t and y = cos 2t
Differentiating both eqns. w.r.t. t, we have
dx
dt
= cos t and
dy
dt
= – sin 2t
d
dt
(2t)
= – (sin 2t) 2 = – 2 sin 2t
We know that
dy
dx
=
/
/
dy dt
dx dt
= –
2 sin 2
cos
t
t
−
= – 2 .
2 sin cos
cos
t t
t
= – 4 sin t.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 58
59. 4. x = 4t, y =
4
t
.
Sol. Given: x = 4t and y =
4
t
∴
dx
dt
=
d
dt
(4t) and
dy
dt
=
d
dt
4
t
= 4
d
dt
t = 2
(4) 4
d d
t t
dt dt
t
−
= 4(1) = 4 = 2
(0) 4(1)
t
t
−
= – 2
4
t
We know that
dy
dx
=
/
/
dy dt
dx dt
=
2
4
4
t
−
= 2
1
( )
t
−
.
5. x = cos θ
θ
θ
θ
θ – cos 2θ
θ
θ
θ
θ, y = sin θ
θ
θ
θ
θ – sin 2θ
θ
θ
θ
θ.
Sol. Given: x = cos θ – cos 2θ and y = sin θ – sin 2θ
∴
dx
dθ
=
d
dθ
(cos θ) –
d
dθ
cos 2θ and
dy
dθ
= cos θ –
d
dθ
sin 2θ
= – sin θ – (– sin 2θ)
d
dθ
2θ = cos θ – cos 2θ
d
dθ
2θ
= – sin θ + (sin 2θ) 2 = cos θ – cos 2θ(2)
= 2 sin 2θ – sin θ = cos θ – 2 cos 2θ.
We know that
dy
dx
=
/
/
dy d
dx d
θ
θ
=
cos 2 cos 2
2 sin 2 sin
θ − θ
θ − θ
.
If x and y are connected parametrically by the equations given in
Exercises 6 to 10, without eliminating the parameter, find
dy
dx
.
6. x = a(θ
θ
θ
θ
θ – sin θ
θ
θ
θ
θ), y = a(1 + cos θ
θ
θ
θ
θ).
Sol. x = a(θ – sin θ) and y = a (1 + cos θ)
Differentiating both eqns. w.r.t. θ, we have
dx
dθ
= a
d
dθ
(θ – sin θ) and
dy
dθ
= a
d
dθ
(1 + cos θ)
= a sin
d d
d d
θ − θ
θ θ
and
dy
dθ
= a (1) cos
d d
d d
+ θ
θ θ
⇒
dx
dθ
= a(1 – cos θ) and
dy
dθ
= a(0 – sin θ) = – a sin θ
We know that
dy
dx
=
/
/
dy d
dx d
θ
θ
=
sin
(1 cos )
a
a
− θ
− θ
= –
sin
1 cos
θ
− θ
= –
2
2 sin cos
2 2
2 sin
2
θ θ
θ
= –
cos
2
sin
2
θ
θ
= – cot
2
θ
.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 59
60. 7. x =
3
sin
cos 2
t
t
, y =
3
cos
cos 2
t
t
.
Sol. We have x =
3
sin
cos 2
t
t
and y =
3
cos
cos 2
t
t
∴
dx
dt
=
3 3
2
cos 2 . (sin ) sin . ( cos 2 )
( cos 2 )
d d
t t t t
dt dt
t
−
[By Quotient Rule]
=
2 3 1/2
1
cos 2 . 3 sin . (sin ) sin . (cos 2 ) . (cos 2 )
2
cos 2
d d
t t t t t t
dt dt
t
−
−
=
3
2 sin
cos 2 . 3 sin cos . ( 2 sin 2 )
2 cos 2
cos 2
t
t t t t
t
t
− −
=
2 3
3/2
3 sin cos cos 2 sin sin 2
(cos 2 )
t t t t t
t
+
=
2 3
3/2
3 sin cos cos 2 sin . 2 sin cos
(cos 2 )
t t t t t t
t
+
=
2 2
3/2
sin cos (3 cos 2 2 sin )
(cos 2 )
t t t t
t
+
and
dy
dt
=
3 3
2
cos 2 . (cos ) cos . ( cos 2 )
( cos 2 )
d d
t t t t
dt dt
t
−
[By Quotient Rule]
=
2 3 1/2
1
cos 2 . 3 cos . (cos ) cos . (cos 2 ) . (cos 2 )
2
cos 2
d d
t t t t t t
dt dt
t
−
−
=
3
2 cos
cos 2 . 3 cos ( sin ) ( 2 sin 2 )
2 cos 2
cos 2
t
t t t t
t
t
− − −
=
2 3
3/2
3 cos sin cos 2 cos . sin 2
(cos 2 )
t t t t t
t
− +
=
2 3
3/2
3 cos sin cos 2 cos . 2 sin cos
(cos 2 )
t t t t t t
t
− +
=
2 2
3/2
sin cos (2 cos 3 cos 2 )
(cos 2 )
t t t t
t
−
∴
dy
dx
=
/
/
dy dt
dx dt
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 60
61. =
2 2
3/2
sin cos (2 cos 3 cos 2 )
(cos 2 )
t t t t
t
−
.
3/2
2 2
(cos 2 )
sin cos (3 cos 2 2 sin )
t
t t t t
+
=
2 2
2 2
cos [2 cos 3(2 cos 1)]
sin [3(1 2 sin ) 2 sin ]
t t t
t t t
− −
− +
=
2
2
cos (3 4 cos )
sin (3 4 sin )
t t
t t
−
−
=
3
3
(4 cos 3 cos )
3 sin 4 sin
t t
t t
− −
−
=
cos 3
sin 3
t
t
−
= – cot 3t
Hence
dy
dx
= – cot 3t.
8. x = a
cos + log tan
2
t
t , y = a sin t.
Sol. x = a cos log tan
2
t
t
+
⇒
dx
dt
= a
1
sin . tan
2
tan
2
d t
t
t dt
− +
= a
2
1 1
sin . sec .
2 2
tan
2
t
t
t
− +
= a
2
cos
1 1
2
sin . .
2
sin cos
2 2
t
t
t t
− +
= a
1
sin
2 sin cos
2 2
t
t t
− +
= a
1
sin
sin
t
t
− +
= a
1
sin
sin
t
t
−
= a
2
1 sin
sin
t
t
−
=
2
cos
sin
a t
t
y = a sin t ⇒
dy
dt
= a cos t
∴
dy
dx
=
/
/
dy dt
dx dt
= 2
cos
cos
sin
a t
a t
t
=
sin
cos
t
t
= tan t.
9. x = a sec θ
θ
θ
θ
θ, y = b tan θ
θ
θ
θ
θ.
Sol. x = a sec θ and y = b tan θ
Differentiating both eqns. w.r.t. θ, we have
dx
dθ
= a sec θ tan θ and
dy
dθ
= b sec2
θ
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 61
62. We know that
dy
dx
=
/
/
dy d
dx d
θ
θ
=
2
sec
sec tan
b
a
θ
θ θ
=
sec
tan
b
a
θ
θ
=
1
.
cos
sin
.
cos
b
a
θ
θ
θ
=
cos
b
θ
.
cos
sin
a
θ
θ
=
sin
b
a θ
=
b
a
cosec θ.
10. x = a(cos θ
θ
θ
θ
θ + θ
θ
θ
θ
θ sin θ
θ
θ
θ
θ), y = a(sin θ
θ
θ
θ
θ – θ
θ
θ
θ
θ cos θ
θ
θ
θ
θ).
Sol. We have x = a(cos θ + θ sin θ) and y = a(sin θ – θ cos θ)
∴
dx
dθ
= a(– sin θ + θ cos θ + sin θ . 1) = aθ cos θ
and
dy
dθ
= a[cos θ – (θ(– sin θ) + cos θ . 1)]
= a [cos θ + θ sin θ – cos θ] = aθ sin θ
∴
dy
dx
=
dy
d
dx
d
θ
θ
=
sin
cos
a
a
θ θ
θ θ
= tan θ.
11. If x =
–1
sin t
a , y =
–1
cos t
a , show that
dy
dx
= –
y
x
.
Sol. Given: x =
1
sin t
a
−
=
1
sin 1/2
( )
t
a
−
=
1
1/2 sin t
a
−
...(i)
∴
dx
dt
=
1
1/2 sin t
a
−
log a
d
dt
1
1
sin
2
t
−
( ) ( )
log and log ( )
x x f x f x
d d d
a a a a a a f x
dx dx dx
= =
∵
⇒
dx
dt
=
1
1/2 sin t
a
−
log a .
1
2 2
1
1 t
−
...(ii)
Again given: y =
1
cos t
a
−
=
1
cos 1/2
( )
t
a
−
=
1
1/2 cos t
a
−
...(iii)
∴
dy
dt
=
1
1/2 cos t
a
−
log a
d
dt
1
1
cos
2
t
−
=
1
1/2 cos t
a
−
log a .
1
2 2
1
1 t
−
−
...(iv)
We know that
dy
dx
=
/
/
dy dt
dx dt
Putting values from (iv) and (ii),
dy
dx
=
1
1
1/2 cos
2
1/2 sin
2
1 1
log
2 1
1 1
log .
2 1
t
t
a a
t
a a
t
−
−
−
−
−
=
1
1
1/2 cos
1/2 sin
t
t
a
a
−
−
−
= –
y
x
(By (iii) and (i))
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 62
63. Exercise 5.7
Find the second order derivatives of the functions given in
Exercises 1 to 5.
1. x2
+ 3x + 2.
Sol. Let y = x2
+ 3x + 2
∴
dy
dx
= 2x + 3.1 + 0 = 2x + 3
Again differentiating w.r.t. x,
2
2
d y
dx
=
d
dx
dy
dx
= 2(1) + 0 = 2.
2. x20
.
Sol. Let y = x20
∴
dy
dx
= 20x19
Again differentiating w.r.t. x,
2
2
d y
dx
= 20.19x18
= 380x18
.
3. x cos x.
Sol. Let y = x cos x
∴
dy
dx
= x
d
dx
cos x + cos x
d
dx
x [By Product Rule]
= – x sin x + cos x
Again differentiating w.r.t. x,
2
2
d y
dx
= –
d
dx
(x sin x) +
d
dx
cos x
= – sin sin ( )
d d
x x x x
dx dx
+
– sin x
= – (x cos x + sin x) – sin x = – x cos x – sin x – sin x
= – x cos x – 2 sin x = – (x cos x + 2 sin x).
4. log x.
Sol. Let y = log x ∴
dy
dx
=
1
x
Again differentiating w.r.t. x,
2
2
d y
dx
=
d
dx
1
x
=
d
dx
x–1
= (– 1) x–2
= 2
1
x
−
.
5. x3
log x.
Sol. Let y = x3
log x
∴
dy
dx
= x3 d
dx
log x + log x
d
dx
x3
[By Product Rule]
= x3
.
1
x
+ (log x) 3x2
= x2
+ 3x2
log x
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 63
64. Again differentiating w.r.t. x,
2
2
d y
dx
=
d
dx
x2
+ 3
d
dx
(x2
log x)
= 2x + 3
2 2
log log
d d
x x x x
dx dx
+
= 2x + 3 2 1
. (log ) 2
x x x
x
+
= 2x + 3(x + 2x log x)
= 2x + 3x + 6x log x = 5x + 6x log x
= x(5 + 6 log x).
Find the second order derivatives of the functions given in
exercises 6 to 10.
6. ex
sin 5x.
Sol. Let y = ex
sin 5x
∴
dy
dx
= ex d
dx
sin 5x + sin 5x
d
dx
ex
[By Product Rule]
= ex
cos 5x
d
dx
5x + sin 5x . ex
= ex
cos 5x . 5 + ex
sin 5x
or
dy
dx
= ex
(5 cos 5x + sin 5x)
Again applying Product Rule of derivatives
2
2
d y
dx
= ex d
dx
(5 cos 5x + sin 5x) + (5 cos 5x + sin 5x)
d
dx
ex
= ex
(5(– sin 5x) . 5 + (cos 5x) . 5) + (5 cos 5x + sin 5x) ex
= ex
(– 25 sin 5x + 5 cos 5x + 5 cos 5x + sin 5x)
= ex
(10 cos 5x – 24 sin 5x)
= 2ex
(5 cos 5x – 12 sin 5x).
7. e6x
cos 3x.
Sol. Let y = e6x
cos 3x
∴
dy
dx
= e6x d
dx
cos 3x + cos 3x
d
dx
e6x
= e6x
(– sin 3x)
d
dx
(3x) + cos 3x . e6x
d
dx
6x
= – e6x
sin 3x . 3 + cos 3x . e6x
. 6
⇒
dy
dx
= e6x
(– 3 sin 3x + 6 cos 3x)
Again applying Product Rule of derivatives,
2
2
d y
dx
= e6x
d
dx
(– 3 sin 3x + 6 cos 3x)
+ (– 3 sin 3x + 6 cos 3x)
d
dx
e6x
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 64
65. = e6x
[– 3 . cos 3x . 3 – 6 sin 3x . 3]
+ (– 3 sin 3x + 6 cos 3x) e6x
. 6
= e6x
(– 9 cos 3x – 18 sin 3x – 18 sin 3x + 36 cos 3x)
= e6x
(27 cos 3x – 36 sin 3x)
= 9e6x
(3 cos 3x – 4 sin 3x).
8. tan–1
x.
Sol. Let y = tan–1
x
∴
dy
dx
= 2
1
1 x
+
Again differentiating w.r.t. x,
2
2
d y
dx
=
d
dx 2
1
1 x
+
=
2 2
2 2
(1 ) (1) 1 (1 )
(1 )
d d
x x
dx dx
x
+ − +
+
=
2
2 2
(1 )0 (2 )
(1 )
x x
x
+ −
+
= 2 2
2
(1 )
x
x
−
+
.
9. log (log x).
Sol. Let y = log (log x)
∴
dy
dx
=
1
log x
d
dx
log x
1
log ( ) ( )
( )
d d
f x f x
dx f x dx
=
∵
=
1
log x
1
x
=
1
log
x x
Again differentiating w.r.t. x,
2
2
d y
dx
= 2
( log ) (1) 1 ( log )
( log )
d d
x x x x
dx dx
x x
−
= 2
( log ) 0 log log ( )
( log )
d d
x x x x x x
dx dx
x x
− +
= – 2
1
. log . 1
( log )
x x
x
x x
+
= – 2
(1 log )
( log )
x
x x
+
.
10. sin (log x).
Sol. Let y = sin (log x)
∴
dy
dx
= cos (log x)
d
dx
(log x) = cos (log x) .
1
x
=
cos (log )
x
x
Again differentiating w.r.t. x,
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 65
66. 2
2
d y
dx
= 2
cos (log ) cos (log ) ( )
d d
x x x x
dx dx
x
−
= 2
[ sin (log )] log cos (log )
d
x x x x
dx
x
− −
= 2
1
sin (log ) . cos (log )
x x x
x
x
− −
= 2
[sin (log ) cos (log )]
x x
x
− +
.
11. If y = 5 cos x – 3 sin x, prove that
2
2
d y
dx
+ y = 0.
Sol. Given: y = 5 cos x – 3 sin x ...(i)
∴
dy
dx
= – 5 sin x – 3 cos x
Again differentiating w.r.t. x,
2
2
d y
dx
= – 5 cos x + 3 sin x
= – (5 cos x – 3 sin x) – y (By (i))
or
2
2
d y
dx
= – y ∴
2
2
d y
dx
+ y = 0.
12. If y = cos–1
x. Find
2
2
d y
dx
in terms of y alone.
Sol. Given: y = cos–1
x ⇒ x = cos y ...(i)
∴
dy
dx
=
2
1
1 x
−
−
=
2
1
1 cos y
−
−
(By (i))
=
2
1
sin y
−
=
1
sin y
−
= – cosec y
or
dy
dx
= – cosec y ...(ii)
Again differentiating both sides w.r.t. x,
2
2
d y
dx
= –
d
dx
(cosec y) = – cosec cot
dy
y y
dx
−
= cosec y cot y (– cosec y) (By (ii))
= – cosec2
y cot y.
13. If y = 3 cos (log x) + 4 sin (log x), show that x2
y2 + xy1 + y = 0.
Sol. Given: y = 3 cos (log x) + 4 sin (log x) ...(i)
∴
dy
dx
= (y1) = – 3 sin (log x)
d
dx
log x + 4 cos (log x)
d
dx
log x
or y1 = – 3 sin (log x) .
1
x
+ 4 cos (log x) .
1
x
Multiplying both sides by L.C.M. = x,
xy1 = – 3 sin (log x) + 4 cos (log x)
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 66
67. Again differentiating both sides w.r.t. x,
d
dx
(xy1) = – 3 cos (log x)
d
dx
log x – 4 sin (log x)
d
dx
log x
⇒ x
d
dx
y1 + y1
d
dx
x = – 3 cos (log x) .
1
x
– 4 sin (log x) .
1
x
(By Product Rule)
⇒ xy2 + y1 = –
[3 cos (log ) 4 sin (log )]
x x
x
+
Cross-multiplying
x(xy2 + y1) = – [3 cos (log x) + 4 sin (log x)]
⇒ x2
y2 + xy1 = – y (By (i))
⇒ x2
y2 + xy1 + y = 0.
14. If y = Aemx
+ Benx
, show that
2
2
d y
dx
– (m + n)
dy
dx
+ mny = 0.
Sol. Given: y = Aemx
+ Benx
...(i)
∴
dy
dx
= Aemx
d
dx
(mx) + Benx
d
dx
(nx)
( ) ( )
( )
f x f x
d d
e e f x
dx dx
=
∵
or
dy
dx
= Am emx
+ Bn enx
...(ii)
∴
2
2
d y
dx
= Am.emx
.m + Bnenx
.n
= Am2
emx
+ Bn2
enx
...(iii)
Putting values of y,
dy
dx
and
2
2
d y
dx
from (i), (ii) and (iii) in
L.H.S. =
2
2
d y
dx
– (m + n)
dy
dx
+ mny
= Am2
emx
+ Bn2
enx
– (m + n) (Am emx
+ Bn enx
) + mn(Aemx
+ Benx
)
= Am2
emx
+ Bn2
enx
– Am2
emx
– Bmn enx
– Amn emx
– Bn2
enx
+ Amn emx
+ Bmn enx
= 0 = R.H.S.
15. If y = 500 e7x
+ 600 e–7x
, show that
2
2
d y
dx
= 49y.
Sol. Given: y = 500 e7x
+ 600 e–7x
...(i)
∴
dy
dx
= 500 e7x
(7) + 600 e–7x
(– 7) = 500(7) e7x
– 600(7) e–7x
∴
2
2
d y
dx
= 500(7) e7x
(7) – 600(7)e–7x
(– 7)
= 500(49) e7x
+ 600(49) e–7x
or
2
2
d y
dx
= 49[500 e7x
+ 600 e–7x
] = 49y (By (i))
or
2
2
d y
dx
= 49y.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 67
68. 16. If ey
(x + 1) = 1, show that
2
2
d y
dx
=
2
dy
dx
.
Sol. Given: ey
(x + 1) = 1
⇒ ey
=
1
1
x +
Taking logs of both sides, log ey
= log
1
1
x +
or y log e = log 1 – log (x + 1)
or y = – log (x + 1) [... log e = 1 and log 1 = 0]
∴
dy
dx
= –
1
1
x +
d
dx
(x + 1) =
1
1
x
−
+
= – (x + 1)–1
∴
2
2
d y
dx
= – (– 1)(x + 1)–2
d
dx
(x + 1)
1
( ( )) ( ( )) ( )
n n
d d
f x n f x f x
dx dx
−
=
∵
= 2
1
( 1)
x +
( 1) 1 0 1
d
x
dx
+ = + =
∵
L.H.S. =
2
2
d y
dx
= 2
1
( 1)
x +
R.H.S. =
2
dy
dx
=
2
1
1
x
−
+
= 2
1
( 1)
x +
∴ L.H.S. = R.H.S. i.e.,
2
2
d y
dx
=
2
dy
dx
.
17. If y = (tan–1
x)2
, show that (x2
+ 1)2
y2 + 2x(x2
+ 1)y1 = 2.
Sol. Given: y = (tan–1
x)2
...(i)
∴ y1 = 2(tan–1
x)
d
dx
tan–1
x
1
( ( )) ( ( )) ( )
n n
d d
f x n f x f x
dx dx
−
=
∵
⇒ y1 = 2 (tan–1
x) 2
1
1 x
+
⇒ y1 =
1
2
2 tan
1
x
x
−
+
Cross-multiplying, (1 + x2
) y1 = 2 tan–1
x
Again differentiating both sides w.r.t. x,
(1 + x2
)
d
dx
y1 + y1
d
dx
(1 + x2
) = 2 . 2
1
1 x
+
⇒ (1 + x2
) y2 + y1 . 2x = 2
2
1 x
+
Multiplying both sides by (1 + x2
),
(x2
+ 1)2
y2 + 2xy1 (1 + x2
) = 2.
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 68
69. Exercise 5.8
( )
( )
f x
g x (g(x) ≠
≠
≠
≠
≠ 0), sin x, cos x, ex
, e–x
, log x (x > 0) are conti-
nuous and derivable for all real x.
Note 2: Sum, difference, product of two continuous (derivable)
functions is continuous (derivable).
1. Verify Rolle’s theorem for f (x) = x2
+ 2x – 8, x ∈
∈
∈
∈
∈ [– 4, 2].
Sol. Given: f (x) = x2
+ 2x – 8; x ∈ [– 4, 2] ...(i)
Here f (x) is a polynomial function of x (of degree 2).
∴ f (x) is continuous and derivable everywhere i.e., on (– ∞, ∞).
Hence f (x) is continuous in the closed interval [– 4, 2] and
derivable in open interval (– 4, 2).
Putting x = – 4 in (i), f (– 4) = 16 – 8 – 8 = 0
Putting x = 2 in (i), f (2) = 4 + 4 – 8 = 0
∴ f (– 4) = f (2) (= 0)
∴ All three conditions of Rolle’s Theorem are satisfied.
From (i), f ′(x) = 2x + 2.
Putting x = c, f ′(c) = 2c + 2 = 0 ⇒ 2c = – 2
⇒ c = –
2
2
= – 1 ∈ open interval (– 4, 2).
∴ Conclusion of Rolle’s theorem is true.
∴ Rolle’s theorem is verified.
2. Examine if Rolle’s theorem is applicable to any of the following
functions. Can you say some thing about the converse of Rolle’s
theorem from these examples?
(i) f (x) = [x] for x ∈
∈
∈
∈
∈ [5, 9] (ii) f (x) = [x] for x ∈
∈
∈
∈
∈ [– 2, 2]
(iii) f (x) = x2
– 1 for x ∈
∈
∈
∈
∈ [1, 2].
Sol. (i) Given: f (x) = [x] for x ∈ [5, 9] ...(i)
(of course [x] denotes the greatest integer ≤ x)
We know that bracket function [x] is discontinuous at all the
integers (See Ex. 15, page 155, NCERT, Part I). Hence
f (x) = [x] is discontinuous at all integers between 5 and 9 i.e.,
discontinuous at x = 6, x = 7 and x = 8 and hence discontinuous
in the closed interval [5, 9] and hence not derivable in the open
interval (5, 9). ...(ii) (... discontinuity ⇒ Non-derivability)
Again from (i), f (5) = [5] = 5 and f (9) = [9] = 9
∴ f (5) ≠ f (9)
∴ Conditions of Rolle’s Theorem are not satisfied.
∴ Rolle’s Theorem is not applicable to f (x) = [x] in the
closed interval [5, 9].
But converse (conclusion) of Rolle’s theorem is true for this
function f (x) = [x].
i.e., f ′(c) = 0 for every real c belonging to open interval
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 69
70. (5, 9) other than integers. (i.e., for every real c ≠ 6, 7, 8)
(even though conditions are not satisfied).
Let us prove it.
Left Hand derivative = Lf ′(c) = lim
x c−
→
( ) ( )
f x f c
x c
−
−
= lim
x c−
→
[ ] [ ]
x c
x c
−
−
(By (i))
Put x = c – h, h → 0+
, =
0
lim
h +
→
[ ] [ ]
c h c
c h c
− −
− −
=
0
lim
h +
→
[ ] [ ]
c c
h
−
−
[... We know that for c ∈ R – Z, as h → 0+
, [c – h] = [c]]
=
0
lim
h +
→
0
h
−
=
0
lim
h +
→
0
(... h → 0+
⇒ h > 0 and hence h ≠ 0)
= 0 ...(iii)
Right Hand derivative = Rf ′(c) = lim
x c+
→
( ) ( )
f x f c
x c
−
−
= lim
x c+
→
[ ] [ ]
x c
x c
−
−
(By (i))
Put x = c + h, h → 0+
, =
0
lim
h +
→
[ ] [ ]
c h c
c h c
+ −
+ −
=
0
lim
h +
→
[ ] [ ]
c c
h
−
[... We know that for c ∈ R – Z, as h → 0+
, [c + h] = [c]]
=
0
lim
h +
→
0
h
=
0
lim
h +
→
0
(... h → 0+
⇒ h > 0 and hence h ≠ 0)
= 0 ...(iv)
From (iii) and (iv) Lf ′(c)=R f ′(c) = 0
∴ f ′(c) = 0 V real c ∈ open interval (5, 9) other than integers
c = 6, 7, 8.
(ii) Given: f (x) = [x] for x ∈ [– 2, 2].
Reproduce the solution of (i) part replacing closed interval [5, 9]
by [– 2, 2] and integers 6, 7, 8 by – 1, 0 and 1 lying between – 2
and 2.
(iii) Given: f (x) = x2
– 1 for x ∈ [1, 2] ...(i)
Here f (x) is a polynomial function of x (of degree 2).
∴ f (x) is continuous and derivable everywhere i.e., on
(– ∞, ∞).
Hence f (x) is continuous in the closed interval [1, 2] and
derivable in the open interval (1, 2).
Again from (i), f (1) = 1 – 1 = 0
Class 12 Chapter 5 - Continuity and Differentiability
MathonGo 70