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Polytechnic Engineering Mathematics – Sem -1 Solutions

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Education contributes to the development of society.
www.basicsinmaths.com website has been given material for mathematics Polytechnic Students.

These solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.


TRANSFORMATIONS

Transformations  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File  very useful for students who are prepared for polytechnic examinations.

PART – 1

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INVERSE TRIGONOMETRIC FUNCTIONS

Inverse trigonometric functions solutions  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File  very useful for students who are prepared for polytechnic examinations.

 

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SOLUTIONS OF SIMULTANEOUS EQUATIONS

Solutions of simultaneous equations  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File  very useful for students who are prepared for polytechnic examinations.

 

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SOLUTIONS OF TRIANGLES

 

Solutions of triangles  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File  very useful for students who are prepared for polytechnic examinations.

 

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Polytechnic Engineering Maths Feature Image for Sem 2 Concept

Engineering Mathematics Sem – II Concept

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www.basicsinmaths.com website has been given material for mathematics Polytechnic Students.

These solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.


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Polytechnic Engineering Maths Feature Image for Sem 2 solutions

Polytechnic Engineering Mathematics – Sem -2 Solutions

Education is the acquisition of knowledge, skill value, etc.    
Education contributes to the development of society.
www.basicsinmaths.com website has been given material for mathematics Polytechnic Students.

These solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.


Tangents and Normals

Tangents and Normals solutions  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File  very useful for students who are prepared for polytechnic examinations.

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Maxima and Minima

The Maxima and Minim solutions  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File is very useful for students who are prepared for polytechnic examinations.

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TS Inter Maths 1A &1B Practice papers (Reduced Syllabus)

TS Inter Maths 1A and 1B Practice papers as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Practice papers to do help the intermediate First-year Maths students.

TS Inter Maths 1A and 1B Practice papers as per reduced syllabus are very useful in IPE examinations.

 


MATHS 1A PRACTICE PAPER – 1

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MATHS 1A PRACTICE PAPER – 4

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TS Inter Maths 1A and 1B Practice papers as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Practice papers to do help the intermediate First-year Maths students.

TS Inter Maths 1A and 1B Practice papers as per reduced syllabus are very useful in IPE examinations.


MATHS 1B PRACTICE PAPER – 1

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Pdf Files

Inter Mathematics 1A ands 1B Pdf Files

These PDF Files were designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A & 1B   PDF Files are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.  


MATHEMATICS 1A

 

Addition Of Vectors SAQ’S

Matrices

 

MATHEMATICS 1B

DC’s and Dr’s

Tangents and Normals

Maxima and Minima


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English Grammar Feature Image

English Grammar 4 Competitive Exams & School Education

This content is designed by the ‘Basics in Maths‘ team.

English language:

The English Language is important to communicate and interact with other people around us. It keeps us in contact with other people.
An example of the importance of a language is the ‘English language’ because it is the international language and has become the most important language to people in many parts of the world.
British brought with them their language English to India.

ఇంగ్లీష్ భాష:

మన చుట్టూ ఉన్న ఇతర వ్యక్తులతో సందేశించాదానికికి  మరియు మన చుట్టూ ఉన్న ఇతర వ్యక్తులతో సంభాషించడానికి భాష ముఖ్యం. ఇది మనల్ని ఇతర వ్యక్తులతో సంప్రదించుటకు దోహదపడుతుంది.
ఒక భాష యొక్క ప్రాముఖ్యతకు ఒక ఉదాహరణ ‘ఆంగ్ల భాష’ ఎందుకంటే ఇది అంతర్జాతీయ భాష మరియు ప్రపంచంలోని అనేక ప్రాంతాల ప్రజలకు అత్యంత ముఖ్యమైన భాషగా మారింది.
బ్రిటిష్ వారు తమ భాష ఇంగ్లిష్ ను భారతదేశానికి తీసుకువచ్చారు.

English Grammar:

Grammar is the way we arrange words to make proper sentences. Grammar rules about how to speak and write in a language. English grammar is the grammar of the English language. English grammar started out based on Old English,

ఇంగ్లిష్ వ్యాకరణం:

వ్యాకరణం అనేది సరైన వాక్యాలు చేయడానికి పదాలను ఏర్పాటు చేసే విధానం. వ్యాకరణం అనగా  ఒక భాషలో ఎలా మాట్లాడాలి మరియు ఎలా రాయాలి అనే నియమాలు. ఆంగ్ల వ్యాకరణం ఆంగ్ల భాష యొక్క వ్యాకరణం. ఓల్డ్ ఇంగ్లిష్ ఆధారంగా ఇంగ్లిష్ గ్రామర్ ప్రారంభమైంది,


Introduction

There are 26 letters in the English Language. Those are called as ‘Alphabet’

There are two parts to Alphabet.

  • Vowels (a, e, i, o, u) [ 5 letters]
  • Consonants (Remaining 21 letters)

Without vowel (sound or structure) we cannot create even a single word in English.

PARTS OF SPEECH (భాషాభాగాలు)

 

NOUN (నామవాచకం):

A noun is a naming word. The noun means the name of the person, things, places, or animals

(నామవాచకం ఒక వ్యక్తి యొక్క, ఒక వస్తువు యొక్క లేదా జంతువు యొక్క పేరును తెలుపుతుంది)

Ex: Ramu goes to college by car

Seetha went to school by bus        → underlined words are nouns

Kinds of Nouns (According to their usage):

Proper Noun: A proper noun denotes one particular person, place, or thing.

(Proper Noun, ఒక ప్రత్యేక వ్యక్తి, వస్తువు లేదా జంతువు యొక్క పేరును తెలుపుతుంది)

Ex: Raju, Hyderabad, The Ganga, etc.

Common Noun: A common noun is a name given commonly to a person, place or thing.

(CommonNoun, ఒకే జాతికి చెందిన వ్యక్తి, వస్తువు లేదా జంతువు యొక్క పేరును తెలుపుతుంది)

Ex: boy, girl, animal, river, city, etc.

Collective noun:  A collective noun denotes a group or collection of persons or things taken as one.

(Collective Noun, వ్యక్తుల, వస్తువుల లేదా జంతువుల యొక్క గుంపును తెలుపుతుంది)

Ex: herd, army, committee, flock, etc.,

Material noun: A Material noun denote the name of a particular kind of metal, liquid, or substance.

(Material Noun, ఒక నిర్దిష్ట రకం లోహం, ద్రవం లేదా పదార్థం యొక్క పేరును తెలియజేస్తుంది)

Ex: salt, sand, gold, rice, paddy, etc.,

Kinds of Nouns (According to their Meaning):

Concrete Noun:  A Concrete noun denotes something that can be tasted, something that can be touched or seen, something that exists physically.

(కాంక్రీట్ నామవాచకం దేనినైనా రుచి చూడవచ్చు, ఏదైనా తాకవచ్చు లేదా చూడవచ్చు, భౌతికంగా ఉన్నదాన్ని సూచిస్తుంది.)

Ex: Pencil, boy, girl, gold, silver, rice, etc.,

Note: Proper nouns and Material nouns are Concrete nouns.

Abstract Noun: An Abstract noun denotes something maybe an idea or emotion.

Ex: born, sad, joy, bravery, freedom, etc.,

PRONOUN (సర్వనామం):

A pronoun is a word that is used instead of a noun.

సర్వనామం ను నామవాచకానికి బదులుగా వాడుతాము.

Ex:      Ramu went to the Ground, he played cricket.

పై వాక్యం లో రాము కు బదులుగా he వాడబడినది.

Types of Pronouns:

Personal Pronouns: Personal pronoun refers to a particular person or thing. (దీనిని వ్యక్తి పేరు కి బదులుగా ఉపయోగిస్తారు)

These are three types

ఇవి మూడు రకాలు

I person: Talks about himself (తన గురించి చెప్పేది. ఉదా : నేను, నాకు, మేము , మాకు, మొ ||)

Ex:  I – we – my – us etc.,

II person: what it says about others (ఎదుటి వారి గురించి చెప్పేది. ఉదా : నీవు , మీరు , మీకు  మొ ||)

Ex: you, yours

III Person: Talks about the third person between the discussion of two people (ఇద్దరి వ్యక్తుల సంభాషణ మధ్య మూడో వ్యక్తి గురించి చెప్పే ది. ఉదా : అతను , ఆమె , అతనికి  , ఆమెకి , వారికి  మొ ||)

Ex: he, she, it, they. Etc.,

Reflexive Pronoun:

Reflexive Pronouns are used when the subject and the object of a sentence are the same. They can act as either objects or indirect objects. (ఒక వ్యక్తి చేసిన పని ఫలితాన్ని తానే పొందినప్పుడు వీటిని వాడుతారు)  

Ex: myself, himself, themself, yourselves

Relative Pronouns:

A relative pronoun introduces a clause. It refers to some noun going before and also joins two sentences together.  (రెండు వాక్యములను కలుపుటకు వాడుతాము లేదా ఒక వాక్యములో అంతకుముందే చెప్పబడిన nouns ను refer చేస్తాయి)

Ex: who ……. Persons కు

       Which ……. Places కు

       That …… Things కు వాడుతారు

Demonstrative Pronoun: Demonstrative pronouns always identify nouns, whether those nouns are named specifically or not (ఇది, దేనినైనా లేక వేనినైన ఎత్తి చూపడానికి ఉపయోగపడుతుంది)

Ex: this, that, those, these, etc.,

Distributive Pronoun: Distributive pronouns refer to persons or things one at a time. (ఒకే సమయం లో ఎందరికో చెందేవి)

Ex; each, either, neither, etc.,

Indefinite Pronouns: Indefinite pronouns refer to people or things without saying exactly who or what they are. (ఫలానా వ్యక్తీ, ఫలానా వస్తువు గురించి కాకుండా ఎవరో ఒక వ్యక్తి, ఎదో ఒక వస్తువు గురించి Indefinite Pronouns తెలియజేస్తాయి)

Ex: somebody, none, all, nobody, etc.,

Interrogative Pronouns: These are used to ask questions (ప్రశ్నలు అడగడానికి వాడుతాము)

Ex: What, who, why

Subject (కర్త ): Subject means noun or pronoun or noun and pronoun.

Adjective (విశేషణం)

An adjective is used with a noun to add something to its meaning (ఒక విశేషణం నామవాచకంతో దాని అర్థానికి ఏదైనా జోడించడానికి ఉపయోగించబడుతుంది)

Ex: large, big, small, honest, wise, etc.,

Kinds Of Adjectives:

Qualitative Adjective: It indicates the characteristic of a person or an object (ఇది ఒక వ్యక్తి లేదా ఒక వస్తువు యొక్క లక్షణాన్ని తెలుపుతుంది)

Ex: honest, wise, small, big, etc.,

Quantitative adjective: It shows how much of a thing is (ఇవి ఎంత అనే అర్థంలో వాడుతాము)

Ex: some, much, little, enough, etc.,

Numeral Adjectives:  It shows how many things are meant (సంఖ్యాత్మకమైనవి. ఎన్ని అనే పదానికి సమాధానంగా వచ్చేవి)

Ex: few, many, most, five, three, etc.,

Demonstrative Adjectives:

These, that, those, this వంటి నామవాచకం తో కలిపి వస్తే వాటిని Demonstrative Adjectives అంటారు.

Ex: This boy is tall

That girt is clever

 Verb (క్రియ)

Verb (క్రియ) లేకుండా ఇంగ్లీష్ లో వాక్యము లు ఉండవు. ఇంగ్లీష్ వాక్యానికి ‘క్రియ’ ప్రాణం ‍వంటిది.

A verb is a word that tells something about an action or state.

Verb (క్రియ) లేకుండా ఇంగ్లీష్ లో వాక్యము లు ఉండవు. ఇంగ్లీష్ వాక్యానికి ‘క్రియ’ ప్రాణం ‍వంటిది.

A verb is a word that tells something about an action or state. (పనులను తెలియజేయు పదాలను verbs అంటారు)

Ex: I go to school

I am a student

He plays Cricket

We sit at the table

There are two kinds of verbs:

  1. Main verb
  2. Helping verb

Main verbs:

A verb that has an individual meaning is called the Main verb (వ్యక్తిగత అర్థాన్ని కలిగి ఉన్న క్రియను ప్రధాన క్రియ అని అంటారు)

Ex: go, come, take, sing, play, etc.,

Helping verbs:

A verb that does not have any individual meaning is called Helping verb (వ్యక్తిగత అర్థం లేని క్రియను Helping verb అని అంటారు)

Helping verb is mainly used to identify the tense. (ప్రధానంగా టెన్స్ గుర్తించడానికి Helping verb ఉపయోగించబడుతుంది)

Ex: do, does, Is, am, are, have, has, had, will, shall, etc.,

Types of Verbs:

Transitive Verb: A transitive verb is a verb that denotes an action that passes over from the subject to an object. (Object ను కలిగి యుండే   verb ను transitive verb అంటారు)

Ex: He writes a letter

Raju sings a song

In Transitive Verb:  An intransitive verb is a verb that denotes an action that does not pass over to an object. (Object లేని verb ను Intransitive verb అంటారు)

Ex: the boy plays

The Bird sings

 Adverb (క్రియా విశేషణం)

An Adverb is a word that modifies the meaning of a verb, an adjective, or another adverb. (ఒక వాక్య్తం లోని ఒక verb, adjective లేదా adverb గురించి తెలిపేది)

Ex: quickly, very, quietly, clearly etc.,

Preposition (విభక్త్యర్థ పదం)

A preposition is a word that is placed before a noun or pronoun to show the relation between a person, place, or thing. (ఒక నామవాచకముకు లేదా సర్వనామమునకు ముందున్చబడి వాక్యంలోని ఇతర పదం లేక పదాలతో అ నామవాచకం లేదా సర్వనామం యొక్క సంబందాన్ని తెలిపే పదం)

Kinds of Prepositions

Simple Prepositions:  at, by, in, for, off, of, up, to, with, etc., are Simple Prepositions

Compound Prepositions: about, across, along, among, behind, before, below, besides, inside, within, without, etc., are called Compound Prepositions.

Phrase Prepositions: according to, Infront of, in favor of, because of, with regard to, etc., are known as Phrase Prepositions.

Conjunction (సముచ్చయం):

Conjunction combines sentences or words together. (కొన్ని వాక్యాలను లేదా పదాలను కలిపే పదం)

Ex: and, or, but, so, if, as, since, when, etc.,

Interjection (భావోద్వేగ ప్రకటన):

An Interjection is a word that expresses some sudden feelings or emotions.

(మానసిక భావాలను లేక ఉద్రేకాలను తెల్పుటకు వాడే మాటలను Interjections అని అంటారు).

Note: Interjections తరువాత ఆశ్చర్యార్ధకము (!) అనే గుర్తు ఉంచి దాని తరువాత వచ్చేమాట మొదటి అక్షరము Capital letters తో ప్రారంభించవలెను.

Ex: Hello! What are you doing here?

Alas! He  injured

Hurrah! I won the game


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Polytechnic Engineering Maths Feature Image for Sem 1 Concept

Engineering Mathematics Sem – I Concept

Mathematics Notes for Polytechnic SEM – 1 is Designed by the ” Basics in Maths” team. Here we can learn Concepts in Basic Engineering mathematics Polytechnic Sem – I.

 This Material is very Useful for Basic Engineering Mathematics Polytechnic Sem – I Students.

By learning These Notes, Basic Engineering Mathematics Polytechnic Sem – I Students can Write their Exam successfully and fearlessly.


LOGARITHMS

Logarithm: For ant two positive real numbers a, b, and a ≠ 1. If the real number x such then ax = b, then x is called logarithm of b to the base a. it is denoted by Polytechnic SEM - I image 1

Polytechnic SEM - I image 2Standard formulae of logarithms:

Polytechnic SEM - I image 3

Logarithmic Function:

Let a be a positive real number and a ≠ 1. The function f: (o, ∞) → R Defined by f(x) = Polytechnic SEM - I image 4

Polytechnic SEM - I image 5

Polytechnic SEM - I image 6


PARTIAL FRACTIONS

Fractions:

If f(x) and g(x) are two polynomials, g(x) ≠ 0, then Polytechnic SEM - I Partial Fractions1  is called rational fraction.

Ex:

Polytechnic SEM - I Partial Fractions 2   etc.  are rational fractions.

Proper Fraction:

A rational fractionPolytechnic SEM - I Partial Fractions1 is said to be a Proper fraction if the degree of g(x) is greater than the degree of f(x).

Ex:

Polytechnic SEM - I Partial Fractions 3  etc. are the proper fractions.

Improper Fraction:

A rational fractionPolytechnic SEM - I Partial Fractions1 is said to be an Improper fraction if the degree of g(x) is less than the degree of f(x).

Ex:

Polytechnic SEM - I Partial Fractions 4 etc. are the Improper fractions.

Partial Fractions:

Expressing rational fractions as the sum of two or more simpler fractions is called resolving a given fraction into a partial fraction.

∎ If R(x) =Polytechnic SEM - I Partial Fractions1  is proper fraction, then

Case(i): – For every factor of g(x) of the form (ax + b) n, there will be a sum of n partial fractions of the form:

Polytechnic SEM - I Partial Fractions 5

Case(ii): – For every factor of g(x) of the form (ax2 + bx + c) n, there will be a sum of n partial fractions of the form:

Polytechnic SEM - I Partial Fractions 6

∎ If R(x) =Polytechnic SEM - I Partial Fractions1 is improper fraction, then

Case (i): – If degree f(x) = degree of g(x),Polytechnic SEM - I Partial Fractions 6   where k is the quotient of the highest degree term of f(x) and g(x).

Case (ii): – If f(x) > g(x)

R(x) = Polytechnic SEM - I Partial Fractions 8


 MATRICES AND DETERMINANTS

Matrix: A set of numbers arranged in the form of a rectangular array having rows and columns is called Matrix.

•Matrices are generally enclosed by brackets like

•Matrices are denoted by capital letters A, B, C, and so on

•Elements in a matrix are real or complex numbers; real or complex real-valued functions.

Oder of Matrix: A matrix having ‘m’ rows and ‘n’ columns is said to be of order m x n read as m by n.

Ex:

Polytechnic SEM - I Matrices 1

Types Of Matrices

Rectangular Matrix: A matrix in which the no. of rows is not equal to the no. of columns is called a rectangular matrix.                 

Polytechnic SEM - I Matrices 2

 Square Matrix: A matrix in which the no. of rows is equal to no. of columns is called a square matrix.

Polytechnic SEM - I Matrices 3

Principal diagonal (diagonal) Matrix: If A = [a ij] is a square matrix of order ‘n’ the elements a11, a22, a33, ………. An n is said to constitute its principal diagonal.

Polytechnic SEM - I Matrices 4

Trace Matrix: The sum of the elements of the principal diagonal of a square matrix A is called the trace of the matrix. It is denoted by Tr (A).

Polytechnic SEM - I Matrices 5

Diagonal Matrix: If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix.

Polytechnic SEM - I Matrices 6

Scalar Matrix: If each non-diagonal element of a square matrix is ‘zero’ and all diagonal elements are equal to each other, then it is called a scalar matrix.

Polytechnic SEM - I Matrices 7

Identity Matrix or Unit Matrix: If each of the non-diagonal elements of a square matrix is ‘zero’ and all diagonal elements are equal to ‘1’, then that matrix is called unit matrix

Polytechnic SEM - I Matrices 29

Null Matrix or Zero Matrix: If each element of a matrix is zero, then it is called a null matrix.

Polytechnic SEM - I Matrices 8

Row matrix & column Matrix: A matrix with only one row s called a row matrix and a matrix with only one column is called a column matrix.

Polytechnic SEM - I Matrices 9

Triangular matrices:

A square matrix A = [aij] is said to be upper triangular if aij = 0   ∀ i > j

A square matrix A = [aij] is said to be lower triangular matrix aij = 0  ∀ i < j

 

Equality of matrices:


matrices A and B are said to be equal if A and B are of the same order and the corresponding elements of A and B are equal.

Polytechnic SEM - I Matrices 11

Addition of matrices:

If A and B are two matrices of the same order, then the matrix obtained by adding the corresponding elements of A and B is called the sum of A and B. It is denoted by A + B.

Polytechnic SEM - I Matrices 12

Subtraction of matrices:

If A and B are two matrices of the same order, then the matrix obtained by subtracting the corresponding elements of A and B is called the difference from A to B.

Polytechnic SEM - I Matrices 13

Product of Matrices:

Let A = [aik]mxn and B = [bkj]nxp be two matrices, then the matrix C = [cij]mxp   where

Polytechnic SEM - I Matrices 14

Note: Matrix multiplication of two matrices is possible when no. of columns of the first matrix is equal to no. of rows of the second matrix.

A m x n  . Bp x q = AB mx q; n = p

Transpose of Matrix: If A = [aij] is an m x n matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of A. It is denoted by AI or AT.

Note: (i) (AI)I = A     (ii) (k AI) = k . AI    (iii)  (A + B )T = AT + BT  (iv)  (AB)T = BTAT

Symmetric Matrix: A square matrix A is said to be symmetric if AT =A

If A is a symmetric matrix, then A + AT is symmetric.

Skew-Symmetric Matrix: A square matrix A is said to be skew-symmetric if AT = -A

If A is a skew-symmetric matrix, then A – AT is skew-symmetric.

 Minor of an element: Consider a square matrix Polytechnic SEM - I Matrices 16

the minor element in this matrix is defined as the determinant of the 2×2 matrix obtained after deleting the rows and the columns in which the element is present.

Ex: – minor of a3 is Polytechnic SEM - I Matrices 17    = b1c2 – b2c1

Minor of b2 is   = a1c3 – a3c1

Cofactor of an element: The cofactor of an element in i th row and j th column of A3×3 matrix is defined as its minor multiplied by (- 1) i+j.

Properties of determinants:

If each element of a row (column) f a square matrix is zero, then the determinant of that matrix is zero.

Ex: –    A = Polytechnic SEM - I Matrices 19   ⇒ det(A) = 0

If A is a square matrix of order 3 and k is scalar then.

If two rows (columns) of a square matrix are identical (same), then Det. Of that matrix is zero.

Ex: –    A =   Polytechnic SEM - I Matrices 20  ⇒ det(A) = 0

If each element in a row (column) of a square matrix is the sum of two numbers then its determinant can be expressed as the sum of the determinants.

Ex: –  Polytechnic SEM - I Matrices 21

If each element of a square matrix are polynomials in x and its determinant is zero when x = a, then (x-a) is a factor of that matrix.

For any square matrix A Det(A) = Det (AI).

Det (AB) = Det(A). Det(B).

For any positive integer n Det (An) = (DetA)n.

Singular and non-singular matrices:

A Square matrix is said to be singular if its determinant is is zero, otherwise it is said to be non-singular matrix.

Ex: –   A =  Polytechnic SEM - I Matrices 22    det(A) = 4 – 4 = 0

∴ A is singular matrix

B = Polytechnic SEM - I Matrices 23

Det(B) = 4 + 4 = 8≠ 0

∴ B is non-singular

 Adjoint of a matrix: The transpose of the matrix formed by replacing the elements of a square matrix A with the corresponding cofactors is called the adjoint of A.

Let A = Polytechnic SEM - I Matrices 24 and     cofactor matrix of A = Polytechnic SEM - I Matrices 25

Then adj (A) =Polytechnic SEM - I Matrices 26

 Invertible matrix: Let A be a square matrix, we say that A is invertible if there exists a matrix B such that AB =BA = I, where I is a unit matrix of the same order as A and B.

  1. (A– I)– I = A
  2. (AI)– I = (A-I)I
  3. (AB)-I = B-I A-I
  4. A-I =Polytechnic SEM - I Matrices 27

Compound Angles

The algebraic sum of two or more angles is called a ‘compound angle’. Thus, angles A + B, A – B, A + B + C etc., are Compound Angles

For any two real numbers A and B

sin (A + B) = sin A cos B + cos A Cos B

sin (A − B) = sin A cos B − cos A Cos B

cos (A + B) = cos A cos B − sin A sin B

cos (A − B) = cos A cos B + sin A sin B

tan (A + B) =Polytechnic SEM - I Compound Angles 1

tan (A − B) =Polytechnic SEM - I Compound Angles 2

cot (A + B) =Polytechnic SEM - I Compound Angles 3

cot (A − B) = Polytechnic SEM - I Compound Angles 4

tan (Polytechnic SEM - I Compound Angles 5 + A) = Polytechnic SEM - I Compound Angles 6

tan ( Polytechnic SEM - I Compound Angles 5− A) =Polytechnic SEM - I Compound Angles 7

cot (Polytechnic SEM - I Compound Angles 5+ A) =Polytechnic SEM - I Compound Angles 8

cot (Polytechnic SEM - I Compound Angles 5− A) =

sin (A + B + C) = ∑sin A cos B cos C − sin A sin B sin C  

cos (A + B + C) = cos A cos B cos C− ∑cos A sin B sin C 

tan (A + B + C) =Polytechnic SEM - I Compound Angles 10

⋇ cot (A + B + C) =Polytechnic SEM - I Compound Angles 11

⋇ sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A

⋇ cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A

Polytechnic SEM - I Compound Angles 12

 

Multiple and Sub Multiple Angles

If A is an angle, then its integral multiples 2A, 3A, 4A, … are called ‘multiple angles ‘of A and the multiple of A by fraction like are called ‘submultiple angles.

Polytechnic SEM - I Multiple and Submultiple Angles 1

∎ If Polytechnic SEM - I Multiple and Submultiple Angles 2  is not an add multiple of Polytechnic SEM - I Multiple and Submultiple Angles 3

Polytechnic SEM - I Multiple and Submultiple Angles 4 Polytechnic SEM - I Multiple and Submultiple Angles 5

Polytechnic SEM - I Multiple and Submultiple Angles 8

Polytechnic SEM - I Multiple and Submultiple Angles 7

 

PROPERTIES OF TRIANGLES

In ∆ABC,

Polytechnic SEM - I Properties of triangles 1

Lengths AB = c; BC = a; AC =b

The area of the triangle is denoted by ∆.

Perimeter of the triangle = 2s = a + b + c

A = ∠CAB; B = ∠ABC; C = ∠BCA.

R is circumradius.

Sine rule:

In ∆ABC,

Polytechnic SEM - I Properties of triangles 2

⟹ a = 2R sin A; b = 2R sin B; c = 2R sin C

Where R is the circumradius and a, b, c, are lengths of the sides of ∆ABC.

Cosine rule:

In ∆ABC,

Polytechnic SEM - I Properties of triangles 3

Projection rule:

In ∆ABC,

a = b cos C + c cos B

b = a cos C + c cos A

c = a cos B + b cos A

Tangent rule (Napier’s analogy):

Polytechnic SEM - I Properties of triangles 4

 Area of the triangle:

Polytechnic SEM - I Properties of triangles 7

In ∆ABC, a, b, and c are sides

S = Polytechnic SEM - I Properties of triangles 5and area of the triangle Polytechnic SEM - I Properties of triangles 11

 

HYPERBOLIC FUNCTIONS

Polytechnic SEM - I Hyperbolic Functions 1

The function f: R→R defined by f(x) =Polytechnic SEM - I Hyperbolic Functions 2 ∀ x ∈ R is called the ‘hyperbolic sin’ function. It is denoted by Sinh x.

∴Sinh x =Polytechnic SEM - I Hyperbolic Functions 2

Similarly,

Polytechnic SEM - I Hyperbolic Functions 3

Identities:

cosh2x – sinh2 x = 1

cosh2x = 1 + sinh2 x

sinh2 x = cosh2 x – 1

sech2 x = 1 – tanh2 x

tanh2 x = 1 – sesh2 x

cosech2 x = coth2 x – 1

coth2 x = 1 + coth2 x

Addition formulas of hyperbolic functions:

Sinh (x + y) = Sinh x Cosh y + Cosh x Sinh y

Sinh (x − y) = Sinh x Cosh y − Cosh x Sinh y

Cosh (x + y) = Cosh x Cosh y + Sinh x Sinh y  

Cosh (x − y) = Cosh x Cosh y − Sinh x Sinh y 

tanh (x + y) = Polytechnic SEM - I Hyperbolic Functions 4

tanh (x − y) = Polytechnic SEM - I Hyperbolic Functions 5

coth (x + y) = Polytechnic SEM - I Hyperbolic Functions 6

sinh 2x = 2 sinh x cosh 2x =Polytechnic SEM - I Hyperbolic Functions 7

cosh 2x = cosh2x + sinh2 x = 2 cosh2x – 1 = 1 + 2 sinh2x =Polytechnic SEM - I Hyperbolic Functions 8

tanh 2x =Polytechnic SEM - I Hyperbolic Functions 9

 

Inverse hyperbolic functions:

Sinh−1x = Polytechnic SEM - I Hyperbolic Functions 10 ∀ x ∈ R

Cosh−1x = Polytechnic SEM - I Hyperbolic Functions 11  ∀ x ∈ (1, ∞)

Tanh−1x = Polytechnic SEM - I Hyperbolic Functions 15   ∀ < 1

Polytechnic SEM - I Hyperbolic Functions 13

 

COMPLEX NUMBERS

The equation x2 + 1 = 0 has no roots in real number system.

∴ scientists imagined a number ‘i’ such that i2 = − 1.

Complex number:

if x, y are any two real numbers then the general form of the complex number is

z = x + i y; where x real part and y is the imaginary part.

3 + 4i, 2 – 5i, – 3 + 2i are the examples for Complex numbers.

  • z = x +i y can be written as (x, y).
  • If z1 = x1 + i y1, z2 = x2 + i y2, then
  • z1 + z2 = (x1 + x2, y1 + y2) = (x1 + x2) + i (y1 + y2)
  • z1 − z2 = (x1 − x2, y1 − y2) = (x1 − x2) + i (y1 − y2)
  • z1∙   z2 = (x1 x2 −y1 y2, x1y2 + x2y1) = (x1x2 −y1 y2) + i (x1y2 +x2 y1)
  • z1/ z2 = (x1x2 + y1 y2/x22 +y22, x2 y1 – x1y2/ x22 +y22)

= (x1x2 + y1 y2/x22 +y22) + i (x2 y1 – x1y2/ x22 +y22)

  Multiplicative inverse of complex number:

   The multiplicative inverse of the complex number z is 1/z.

z = x + i y then 1/z = x – i y/ x2 + y2

Conjugate complex numbers:

The complex numbers x + i y, x – i y are called conjugate complex numbers.

Conjugate of z is denoted byPolytechnic SEM - I Complex Numbers 1

The sum and product of two conjugate complex numbers are real.

If z1, z2 are two complex numbers then

Polytechnic SEM - I Complex Numbers 2

 Modulus and amplitude of complex numbers:

Modulus: – If z = x + i y, then the non-negative real numberPolytechnic SEM - I Complex Numbers 3 is called modulus of z and it is denoted byPolytechnic SEM - I Complex Numbers 4 or ‘r’.

Amplitude: – The complex number z = x + i y represented by the point P (x, y) on the XOY plane. ∠XOP = θ is called amplitude of z or argument of z.

x = r cos θ, y = r sin θ

x2 + y2 = r2 cos2θ + r2 sin2θ = r2 (cos2θ + sin2θ) = r2(1)

⇒ x2 + y2 = r2

⇒ r =Polytechnic SEM - I Complex Numbers 3  and Polytechnic SEM - I Complex Numbers 4 = r.

• Arg (z) = tan−1(y/x)

• Arg (z1.z2) = Arg (z1) + Arg (z2) + nπ for some n ∈ { −1, 0, 1}

• Arg(z1/z2) = Arg (z1) − Arg (z2) + nπ for some n ∈ { −1, 0, 1}

Note:

∎ e = cos θ + i sin θ

∎ e−iθ = cos θ − i sin θ

De- Moiver’s theorem

For any integer n and real number θ, (cos θ + i sin θ) n = cos n θ + i sin n θ.

cos α + i sin α can be written as cis α

cis α.cis β= cis (α + β)

1/cisα = cis(-α)

cisα/cisβ = cis (α – β)

(cos θ + i sin θ) -n = cos n θ – i sin n θ

(cos θ + i sin θ) (cos θ – i sin θ) = cos2θ – i2 sin2θ = cos2θ + sin2θ = 1.

cos θ + i sin θ = 1/ cos θ – i sin θ and cos θ – i sin θ = 1/ cos θ + i sin θ

(cos θ – i sin θ) n = (1/ (cos θ –+i sin θ)) n = (cos θ + i sin θ)-n = cos n θ – i sin n θ

nth root of a complex number: let n be a positive integer and z0 ≠ 0 be a given complex number. Any complex number z satisfying z n = z0 is called an nth root of z0. It is denoted by z01/n orPolytechnic SEM - I Complex Numbers 5

let z = r (cos θ + i sin θ) ≠ 0 and n be a positive integer. For k∈ {0, 1, 2, 3…, (n – 1)}

letPolytechnic SEM - I Complex Numbers 6 Then a0, a1, a2, …, an-1 are all n distinct nth roots of z and any nth root of z is coincided with one of them.

nth root of unity:  Let n be a positive integer greater than 1 andPolytechnic SEM - I Complex Numbers 7 

Note:

  • The sum of the nth roots of unity is zero.
  • The product of nth roots of unity is (– 1) n – 1.
  • The nth roots of unity 1, ω, ω2, …, ωn-1 are in geometric progression with common ratio ω.

Cube root of unity:      

x3 – 1 = 0 ⇒ x3 = 1

x =11/3

cube roots of unity are: 1 Polytechnic SEM - I Complex Numbers 8    Polytechnic SEM - I Complex Numbers 9

ω = Polytechnic SEM - I Complex Numbers 8 , ω2 =Polytechnic SEM - I Complex Numbers 9

ω2 +ω + 1 = 0 and ω3 = 1

 

TRANSFORMATIONS

 

For A, B∈ R

⋇ sin (A + B) + sin (A – B) = 2sin A cos B

⋇ sin (A + B) −sin (A – B) = 2cos A sin B

⋇ cos (A + B) + cos (A – B) = 2 cos A cos B

⋇ cos (A + B) − cos (A – B) = − 2sin A sin B

For any two real numbers C and D

⋇ sin C + sin D = 2sinPolytechnic SEM - I Transformations 1 cosPolytechnic SEM - I Transformations 2

 

⋇ sin C −sin D= 2cosPolytechnic SEM - I Transformations 1  sinPolytechnic SEM - I Transformations 2

⋇ cos C + cos D = 2 cosPolytechnic SEM - I Transformations 1   cosPolytechnic SEM - I Transformations 2

⋇ cos C − cos D = − 2sinPolytechnic SEM - I Transformations 1   sin Polytechnic SEM - I Transformations 2

If A + B + C = π or 1800, then

⋇ sin (A + B) = sin C; sin (B + C) = sin A; sin (A + C) = sin B

⋇ cos (A + B) = − cos C; cos (B + C) = −cos A; cos (A + C) = − cos B

If A + B + C = 900 or Polytechnic SEM - I Transformations 3then

⋇ sinPolytechnic SEM - I Transformations 4   = cosPolytechnic SEM - I Transformations 9;   sinPolytechnic SEM - I Transformations 5   = cosPolytechnic SEM - I Transformations 8 ;  sinPolytechnic SEM - I Transformations 6   = cosPolytechnic SEM - I Transformations 7

⋇ cos Polytechnic SEM - I Transformations 4  = sinPolytechnic SEM - I Transformations 9; cosPolytechnic SEM - I Transformations 5   = sinPolytechnic SEM - I Transformations 8; cosPolytechnic SEM - I Transformations 6   = sinPolytechnic SEM - I Transformations 7

IfPolytechnic SEM - I Transformations 1    then

⋇ sin (A + B) = cos C; sin (B + C) = cos A; sin (A + C) = cos B

⋇ cos (A + B) = sin C; cos (B + C) = sin A; cos (A + C) = sin B

INVERSE TRIGONOMETRIC RATIOS

If A, B are two sets and f: A→ B is a bijection, then f-1 is existing and f-1: B → A is an inverse function.

Polytechnic SEM - I Inverse Trigonometric Ratios 1

Polytechnic SEM - I Inverse Trigonometric Ratios 2

Properties of Inverse Trigonometric functions:

Polytechnic SEM - I Inverse Trigonometric Ratios 3 Polytechnic SEM - I Inverse Trigonometric Ratios 4 Polytechnic SEM - I Inverse Trigonometric Ratios 5 Polytechnic SEM - I Inverse Trigonometric Ratios 6 Polytechnic SEM - I Inverse Trigonometric Ratios 7

Polytechnic SEM - I Inverse Trigonometric Ratios 8

Solutions of Simultaneous Equations

Matrix Inversion Method:

Let a system of simultaneous equations be

a1 x + b1 y + c1z = d1

a2 x + b2 y + c2z = d2

a3 x + b3 y + c3z = d3

The matrix form of the above equations is

Polytechnic SEM - I Equations In Matrices 1

Therefore, the matrix equation is AX = B

If Det A ≠ 0, A-1 is exists

X = A-1 B

By using above Condition, we get the values of x, y and z

This Method is called as Matrix Inversion Method

Cramer’s Method:

Let system of simultaneous equations be

a1 x + b1 y + c1z = d1

a2 x + b2 y + c2z = d2

a3 x + b3 y + c3z = d3

Polytechnic SEM - I Equations In Matrices 2

1 is obtained by replacing the coefficients of x (1st column elements of ∆) by constant values

Polytechnic SEM - I Equations In Matrices 3

2 is obtained by replacing the coefficients of y (2nd column elements of ∆) by constant values

Polytechnic SEM - I Equations In Matrices 4

3 is obtained by replacing the coefficients of z (3rd column elements of ∆) by constant values

Now Polytechnic SEM - I Equations In Matrices 5

This method is called Cramer’s Method

Gauss-Jordan Method:

Let a system of simultaneous equations be

a1 x + b1 y + c1z = d1

a2 x + b2 y + c2z = d2

a3 x + b3 y + c3z = d3

Augmented matrix: The coefficient matrix (A) augmented with the constant column matrix (B) is called the augmented matrix. It is denoted by [AD].

[AD] = Polytechnic SEM - I Equations In Matrices 6

This Matrix is reduced to the standard form ofPolytechnic SEM - I Equations In Matrices 7by using row operations

  1. Interchanging any two rows
  2. Multiplying the elements of any two elements by a constant.
  3. Adding to the elements of one row with the corresponding elements of another row multiplied by a constant.

∴ The solution of a given system of simultaneous equations is x = α, y = β, and z = γ.

Procedure to get the standard form:
  1. Take the coefficient of x as the unity as a first equation.
  2. If 1 is there in the first-row first column, then make the remaining two elements in the first column zero.
  3. After that, if one element in R2 or R3 is 1, then make the remaining two elements in that column C2 or C3 as zeroes.
  4. If any row contains two elements as zeros and only non-zero divide that row elements with the non-zero element to get unity and make the remaining two elements in that column as zeros.

 

 

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1612566621999

Errors and Approximations || V.S.A.Q’S||

These solutions designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.


Errors and Approximations

Question 1

Find dy and ∆y for the following functions for the values of x and ∆x which are shown against each of the functions

(i) y = f(x) = x2 + x at x = 10 when ∆x = 0.1.

Sol:

Given y = f(x) = x2 + x at x = 10, ∆x = 0.1

∆y = f (x + ∆x) – f (x)

     = f (10 + 0.1) – f (10)

    = f (10.1) – f (10)

    = (10. 1)2 + 10.1 – (102 + 10)

   = 102.01 + 10.1 – 100 – 10

    = 112.11 – 110

    = 2.11

dy = f’ (x) ∆x

     = (2x + 1) (0.1)

     = [2(10) + 1] (0.1)

    = 21 × 0.1

    = 2.1

(ii)  y = cos x at x = 600 with ∆x = 10 (10 = 0.0174 radians)

Sol:

Given y = cos x, x = 600 and ∆x = 10

 ∆y = f (x + ∆x) – f (x)

      = cos (600 + 10) – cos 600

      = cos (610) – 0.5

      = 0.4848 – 0.5

      = – 0.0152

dy = f’ (x) ∆x

      = – sin x (10)

      = – sin 600 × 0.0174

      =– 0.8660 × 0.0174

      = – 0.0150

(iii)  y = x2 + 3x + 6, x = 10 with ∆x = 0.01 

Sol:

   y = x2 + 3x + 6

  ∆y = f (x + ∆x) – f (x)

       = f (10 + 0.01) – f (10)

       = f (10.01) – f (10)

      = (10.01)2 + 3 (10.01) + 6 – (102 + 3 (10) + 6)

       = 100. 2001 + 30.03 + 6 – 100 – 30 – 6

       =130. 2301 – 130

       = 0.2301

dy = f’ (x) ∆x

     = (2x + 3 + 0) (0.01)

     = (2× 10 + 3) (0.01)

      = 23 × 0.01

      = 0.23

(iv)  y =Differentiation 42 , x = 8 and ∆x =0.02

Sol:

Differentiation 42

Question 2

The side of a square is increased from 3cm to 3.01cm find the approximate increase in the area of the square.

Sol:

Let x be the side of the square and area be A

Area of the square A = x2

 x = 3 and ∆x = 0.01

∆A = 2x × ∆x

      = 2(3) (0.01)

      = 6 × 0.01

      = 0.06

Question 3

If an increase in the side of a square is 2% then find the approximate percentage of increase in its area.

Sol:

Let x be the side of the square and A be its area

Differentiation 36  = 2

 A = x2

∆A = 2x × ∆x

The approximate percentage error in area A

Differentiation 37

= 2 × 2 =4

Question 4

From the following. Find the approximations 

(i) Differentiation 38

Sol:

Let f(x) =  , where x = 1000 and ∆x =– 1

f’ (x) =Differentiation 39

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

Differentiation 40 

 = 10 – 0. 0033

 = 9.9967

(ii) Differentiation 43

Sol:

Differentiation 44

(iii) Differentiation 46

Sol:

Differentiation 45

(iv) Sin 620

Sol:

Let f(x) = sin x, where x = 600 and ∆x =20

Approximate value is

 f (x + ∆x) = f(x) + f’ (x) ∆x

= sin 600 + cos x (20)

 = sin 600+ cos 600 (0.0348)

= 0.8660 + 0.5 × 0.0348

= 0.8660 + 0.0174

=0.8834

Question 5

The radius of a sphere is measured as 14cm. Later it was found that there is an error of 0.02cm in measuring the radius. Find the approximate error in the surface area of the sphere.

Sol:

Given r = 14 cm and ∆r =0.02cm

Surface area of sphere =A = 4π r2

∆A = 8π r ∆r

      = 8 ×3.14× 14 × 0.02

      = 7.0336

 


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1612566621999

Differentiation(2m Q & S) || V.S.A.Q’S||

This content designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.


Differentiation

Question 1 

Find f’ (x) for the following functions

(i) f(x) = (ax + b) (x > -b/a)

Sol:

Given f(x) = (ax + b) n

 f’ (x) = n (ax + b) n – 1 Differentiation 1(ax + b)  

            = n (ax + b) n – 1 a

            = an (ax + b) n – 1

(ii)  f(x) = x2 2x log x

Sol:

Given f(x) = x2 2x log x

f’ (x) = Differentiation 1(x2) 2x log x + x2 Differentiation 1(2x) log x + x2 2x Differentiation 1(log x).

          = 2×2x log x +x2 2x log a log x + x2 2x (1/x)

          = x 2x[log x2 + x log x log 2 + 1]

(iii)  f(x) =Differentiation 2 (x > 0)

Sol:

Given f(x) =Differentiation 2

f’ (x) = Differentiation 2  . log 7   Differentiation 1 (x3 + 3x)

          =    Differentiation 2 log 7 (3x2 + 3)

          =3 (x2 + 1) Differentiation 2 log 7

(iv) f(x) = log (sec x + tan x)

Sol:

Given, f(x) = log (sec x + tan x)

f’ (x) = Differentiation 6 Differentiation 1(sec x + tan x)

         = Differentiation 6(sec2 x + sec x tan x)

         = Differentiation 6 sec x (sec x + tan x)

        = sec x

Question 2

Find the derivative of  the following  functions

(i) f(x) = ex (x2 + 1)

Sol:

Given f(x) = ex (x2 + 1)

 f’ (x) = ex Differentiation 1(x2 + 1) + (x2 + 1) Differentiation 1 (ex)

          = ex (2x + 0) + (x2 + 1) ex

            = ex (x2 + 2x + 1)

            = ex (x + 1)2

(ii) Differentiation 11

 Let y =Differentiation 11

  Differentiation 12

(iii) cos (log x + ex) 

Differentiation 13  

(iv) x = tan (e-y)

e-y = tan-1 x

Differentiation 15 

(v) cos [log (cot x)]

Differentiation 16

(vi) sin[tan-1(ex)]

Differentiation 17

(vii) cos-1(4x3 – 3x)

let y = cos-1(4x3 – 3x)

put x = cos θ ⟹ θ = cos-1 x

y = cos-1(4 cos 3 θ – 3cos θ)

   = cos-1(cos 3θ)

= 3 θ

= 3 cos-1 x

Differentiation 8 = 3 Differentiation 1 (cos-1 x)

      = 3Differentiation 19

     =Differentiation 20

(viii) Differentiation 21

Differentiation 22

(ix)  Differentiation 23

Differentiation 24

(x) Differentiation 25

Differentiation 26

 

Question 3

Find f’ (x), If f(x) = (x3 + 6 x2 + 12x – 13)100.

Sol:

Given f(x) = (x3 + 6 x2 + 12x – 13)100

f’ (x) = 100(x3 + 6 x2 + 12x – 13)99 Differentiation 1(x3 + 6 x2 + 12x – 13)

          = 100(x3 + 6 x2 + 12x – 13)99 (3x2 + 12 x + 12 – 0)

          =100(x3 + 6 x2 + 12x – 13)99 3 (x2 + 4 x + 4)

          = 300 (x + 2)2 (x3 + 6 x2 + 12x – 13)99

Question 4

If f(x) = 1 + x + x2 + x3 + …. + x100, then find f’ (1).

Sol:

Given f(x) = 1 + x + x2 + x3 + …. + x100

           f’(x) = 0 + 1 + 2x + 3 x2 + … 100 x99

           f’(1) =  1 + 2 + 3 + … + 100

                   =Differentiation 3

                   = 50 × 101

                  = 5050

Question 5

 From the following functions. Find their derivatives.

Differentiation 5

Question 6 

If y =Differentiation 7 , findDifferentiation 8

Sol:

Given y =Differentiation 7

Differentiation 9

Question 7

If y = log (cosh 2x), find Differentiation 8

Sol:

Given y = log (cosh 2x)

Differentiation 10

Question 8

If x = a cos3 t, y = a sin3 t, find

Sol:

Given If x = a cos3 t, y = a sin3 t

Differentiation 18

Question 9

Differentiate f(x) with respect to g(x) for the following.

(i) f(x) = ex, g(x) =Differentiation 27

f’ (x) = ex and g’ (x) = Differentiation 28

derivative of f(x) with respect to g(x) =Differentiation 29

Differentiation 30 

(ii )  Differentiation 31 

  put x = tan θ ⟹ θ = tan-1 x

Differentiation 32 

Question 10

if y = Differentiation 33 then prove thatDifferentiation 34

Sol:

Given y =Differentiation 33

Differentiation 35


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1612566621999

Limits (Q’s & Ans) || V.S.A.Q’S||

These solutions designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two marks questions and solutions are very useful in IPE examinations.


Limits

Question 1

Find Limits 2

Sol:

Limits 1

= 9

Question 2

Compute Limits 3

Sol:

Limits 4

= a + a = 2a

Question 3

compute Limits 5 

Sol:

Limits 6

Question 4

Show thatLimits 7 = 1and Limits 8 = –1

Sol:

we know that if x > 0

                            Limits 9= –x if x < 0

As x → 0+ ⟹ x > 0

Limits 9 = x

Limits 10 =  1

As x → 0+ ⟹ x < 0

Limits 9 = –x

Limits 11  = –1

Question 5

If f (x) = Limits 12, then find Limits 13andLimits 14 . Does Limits 15 exist?

Sol:

Limits 16

Question 6

Show that Limits 17= –1

Sol:

As x → 2– ⟹ x < 2

   x – 2 < 0

 ⟹  Limits 18 < 0

  Limits 19

Question 7

Compute Limits 20andLimits 21

Sol:

As x → 2+ ⟹ x > 2

  ⟹ Limits 22 = 2

Limits 23=                          

As x → 2– ⟹ x < 2

  ⟹ Limits 22 = 1

Limits 24

Question 8

Find Limits 25

Sol:

Limits 26

Question 9

Compute Limits 27

Sol:

Limits 28

Question 10

Show that Limits 29

Sol:

Limits 30

Question 11

Compute Limits 31

Sol:

Limits 32

Question 12

Compute Limits 33

Sol:

Limits 34

Question 13

Evaluate Limits 35

Sol:

let y = x – 1 ⟹ x = y + 1

then as x → 1, y → 0

Limits 36

Question 14

Compute Limits 37

Sol:

Limits 38

Question 15

Compute Limits 39

Sol:

Limits 40

Question 16

ComputeLimits 41

Sol:

Limits 42 

     As x → ∞,  →0

Limits 43

                      = ∞ (3/4)

                      = ∞

Question 17

Compute Limits 44

Sol:

We know that – 1 ≤ sin x ≤ 1

x2 – 1 ≤ x2 – sin x ≤ x2 + 1

Limits 45

Question 18

Find Limits 46

Sol:

Limits 47

Question 19

Find Limits 49

Sol:

Limits 48

Question 20

Find Limits 50

Sol:

Limits 51

Question 21

Find Limits 52

Sol:  

Limits 53

Question 22

Compute Limits 54

Sol:  

Limits 55

Question 23

ComputeLimits 56

Sol:

Limits 57

 


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1612493167414

Hyperbolic Functions V.S.A.Q.’S

These solutions designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.


Hyperbolic Functions

Question 1

Prove that for any x∈ R, sinh (3x) = 3 sinh x + 4 sinh3 x

Sol:

sinh (3x) = sinh (2x + x)

                  = sinh 2x cosh x + cosh 2x sinh x

                 = (2 sinh x cosh x) cosh x + (1 + 2 sinh2 x) sinh x

                 = 2sinh x cosh2 x + sinh x + 2 sinh3 x

                 = 2 sinh x (1 + sinh2 x) + sinh x + 2 sinh3 x

                 = 2 sinh x + 2 sinh3 x+ sinh x + 2 sinh3 x

                 = 3 sinh x + 4 sinh3 x

Question 2

If cosh x =Hyperbolic Functions 2 , find the values of (i) cosh 2x and (ii) sinh 2x

Sol:

Given cosh x =Hyperbolic Functions 2

 Cosh 2x = 2 cosh2 x – 1

                 = 2. Hyperbolic Functions 2 – 1

                 =Hyperbolic Functions 3

Sinh2 2x = cosh2 2x – 1

                 = Hyperbolic Functions 4– 1

                 = Hyperbolic Functions 5 – 1

       =Hyperbolic Functions 6

  Sinh2 2x   = Hyperbolic Functions 7

Question 3

If cosh x = sec θ then prove that tanh2Hyperbolic Functions 8= tan2Hyperbolic Functions 9

Sol:

tanh2  =Hyperbolic Functions 10

              = Hyperbolic Functions 11

             =Hyperbolic Functions 12

             =Hyperbolic Functions 13

             = tan2Hyperbolic Functions 9

Question 4

If sinh x = 5, then show that x =Hyperbolic Functions 14

Sol:

Given, sinh x = 5

      ⟹ x = sinh-15    

We know that sinh-1x = Hyperbolic Functions 15 

        ⟹ x =  Hyperbolic Functions 16     

               x  =  Hyperbolic Functions 14

Question 5

Show that tanh-1 = Hyperbolic Functions 17  log3

Sol:

Given tanh-1

We know that tanh-1 x = Hyperbolic Functions 17 Hyperbolic Functions 18   

 tanh-1  = Hyperbolic Functions 17 Hyperbolic Functions 19  

                 = Hyperbolic Functions 17Hyperbolic Functions 20   

                 = Hyperbolic Functions 17  log3

Question 6

For x, y ∈ R prove that sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y)

Sol:

R.H.S = sinh (x) cosh (y) + cosh (x) sinh (y)

Hyperbolic Functions 21

= Hyperbolic Functions 22 

= sinh (x + y)

Question 7

For any x∈ R, prove that cosh4 x – sinh4 x = cosh 2x

Sol:

cosh4 x – sinh4 x = (cosh2 x)2 – (sinh2 x)2

                                 = (cosh2 x + sinh2 x) (cosh2 x – sinh2 x)

                                 = 1. cosh 2x

                                 = cosh 2x

Question 8

Prove thatHyperbolic Functions 24 = cosh x + sinh x

Sol:

Hyperbolic Functions 23 

= cosh x + sinh x

Question 9

If sin hx = ¾ find cosh 2x and sinh 2x.

Sol:

Given sin hx = ¾

We know that cosh2 x = 1 + sinh2 x

                                           = 1 + (3/4)2

                                           = 1 + 9/16

                                           = 25/16

cos hx = 5/4

cosh 2x = 2cosh2 x – 1

                = 2(25/16) – 1

                = 25/8 – 1

                = 17/8

Sinh 2x = 2 sinh x cosh x

                = 2 (3/4) (5/4)

                = 15/8

Question 10

Prove that (cosh x – sinh x) n = cosh nx – sinh nx

Sol:

  Hyperbolic Functions 25

Hyperbolic Functions 26

∴ (cosh x – sinh x) n = cosh nx – sinh nx

 


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1612493167414

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S

These solutions designed by the ‘Basics in Maths‘ team.These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in I.P.E examinations.


Trigonometric Ratios Up to Transformations

 Question 1

Find the value of sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

Sol:

 sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

= sin2(π/10) + sin2(π/2 – π/10) + sin2(π/2+ π/10) + sin2(π – π/10)

= sin2(π/10) + cos2(π/10) + cos2(π/10) + sin2(π/10)

= 1 + 1 = 2

 Question 2

If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ

Sol:

Given sin θ = 4/5 and θ not in the first quadrant

⇒ θ in the second quadrant

⇒ cos θ < 0

    cos2θ = 1 – sin2 θ

              =1 – (4/5)2

             = 1 – 16/25

∴cos θ   = – 3/5 (∵cos θ < 0)

 Question 3

If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ

Sol:

Given, 3sin θ + 4 cos θ = 5

let 4 sin θ – 3cos θ = x

  (3sin θ + 4 cos θ )2 + (4 sin θ – 3cos θ)2 = 52 + x2

 9 sin2 θ + 16 cos2 θ + 12 sin θ cos θ + 16 sin2 θ + 9 cos2 θ – 12sin θ cis θ = 25 + x2

25 sin2 θ + 25 cos2 θ = 25 + x2

25 = 25 + x2

⇒ x2 = 0

 x = 0

∴ 4 sin θ – 3cos θ = 0

 Question 4

If sec θ + tan θ =Trigonometry up to Transformations 1, find the value of sin θ and determine the quadrant in which θ lies

Sol:

Given, sec θ + tan θ =  ———— (1)

 We know that sec2 θ – tan2 θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

 sec θ – tan θ =Trigonometry up to Transformations 2

⇒ sec θ – tan θ = Trigonometry up to Transformations 3———— (2)

 (1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) = Trigonometry up to Transformations 4

2sec θ =Trigonometry up to Transformations 5

 sec θ =Trigonometry up to Transformations 6

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) = Trigonometry up to Transformations 7

2 tan θ = Trigonometry up to Transformations 8 ⇒ tan θ =Trigonometry up to Transformations 9

Now sin θ = tan θ ÷ sec θ =Trigonometry up to Transformatio

     Sin θ =Trigonometry up to Transformations 11

Since sec θ positive and tan θ is negative θ lies in the 4th quadrant.

 Question 5

Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1

Sol:

cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16)

= [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4)

= 1.1.1.1

 =1

 Question 6

If cos θ + sin θ = Trigonometry up to Transformations 12cos θ, then prove that cos θ – sin θ =  sin θ

Sol:

Given, cos θ + sin θ = Trigonometry up to Transformations 12cos θ

Sin θ = Trigonometry up to Transformations 12 cos θ – cos θ

           = ( Trigonometry up to Transformations 12 – 1) cos θ

( Trigonometry up to Transformations 12 + 1) sin θ = ( Trigonometry up to Transformations 12 + 1) ( Trigonometry up to Transformations 12 – 1) cos θ

Trigonometry up to Transformations 12 sin θ + sin θ = cos θ

∴ cos θ – sin θ = Trigonometry up to Transformations 12 sin θ

 Question 7

Find the value of 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

Sol:

2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

= 2[(sin2 θ)3 + (cos2 θ)3] – 3[(sin2 θ)2 + (cos2)2

= 2[(sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)] – 3[(sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ]

= 2[1 – 3 sin2 θ cos2 θ] – 3 [1 – 2 sin2 θ cos2 θ]

= 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ

= – 1

 Question 8

If tan 200 = λ, then show thatTrigonometry up to Transformations 13    

Sol:

Given tan 200 = λ

Trigonometry up to Transformations 19 =Trigonometry up to Transformations 14

                               =Trigonometry up to Transformations 15

                              =Trigonometry up to Transformations 16

                              =Trigonometry up to Transformations 18

 Question 9

If sin α + cosec α = 2, find the value of sinn α + cosecn α, n∈ Z

Sol:

Given sin α + cosec α = 2

 ⇒ sin α + 1/ sin α = 2

 ⇒Trigonometry up to Transformations 17  = 2

      sin2 α + 1= 2 sin α

       sin2 α – 2 sin α + 1= 0

  (sin α – 1 )2 = 0

⇒ sin α – 1 = 0

sin α = 1 ⇒ cosec α = 1

 sinn α + cosecn α = (1)n + (1)n =1 + 1 =2

∴ sinn α + cosecn α = 2

 Question 10

Evaluate sin2 Trigonometry up to Transformations 20+ cos2 Trigonometry up to Transformations 21  – tan2   Trigonometry up to Transformations 22

Sol:

 Trigonometry up to Transformations 23

Trigonometry up to Transformations 24

 Question 11

Find the value of sin 3300. cos 1200 + cos 2100. Sin 3000

Sol:

 sin 3300. cos 1200 + cos 2100. Sin 3000

=sin (3600 – 300). cos (1800 – 600) + cos (1800 + 300). sin (3600 – 600)

= (– sin 300). (– cos 600) + (– cos300). (– sin600)

= sin 300.  cos 600 + cos300.  Sin600

= sin (600 + 300) = sin 900

=1

 Question 12

Prove that cos4 α + 2 cos2 α Trigonometry up to Transformations 25= (1 – sin4 α)

Sol:

cos4 α + 2 cos2 α Trigonometry up to Transformations 25

= cos4 α + 2 cos2 α (1 – cos2 α)

= (cos2 α)2 + 2 (1 – sin2 α) (sin2 α)

= (1 – sin2 α)2 + 2 sin2 α – 2sin4 α

= 1 + sin4 α – 2 sin2 α + 2 sin2 α – 2sin4 α

= 1 – sin4 α

 Question 13

Eliminate θ from x = a cos3 θ and y = b sin3 θ

Sol:

Given x = a cos3 θ and y = b sin3 θ

 cos3 θ = x/a and sin3 θ = y/b

 cos θ = (x/a)1/3 and sin θ = (y/b)1/3

we know that sin2 θ + cos2 θ = 1

 ⇒ [(y/b)1/3]2 + [(x/a)1/3]2 = 1

   (x/a)2/3 + (y/b)2/3 = 1

 Question 14

Find the period of the following functions

Sol:

(i) f(x) = tan 5x

we know that period of tan kx =Trigonometry up to Transformations 26

⇒ period of tan 5x =Trigonometry up to Transformations 27

(ii) f(x) =Trigonometry up to Transformations 28

we know that period of Trigonometry up to Transformations 29= Trigonometry up to Transformations 31

period of Trigonometry up to Transformations 28 = Trigonometry up to Transformations 32

                                      =Trigonometry up to Transformations 33

(iii) f(x) = 2 sinTrigonometry up to Transformations 34 + 2 cosTrigonometry up to Transformations 35

period of sin Trigonometry up to Transformations 34 = Trigonometry up to Transformations 36 = 8

period of cos Trigonometry up to Transformations 35 = Trigonometry up to Transformations 37 = 6

period of given function is = LCM (8, 6) = 24

 

(iv) f(x) = tan (x + 4x + 9x +…. + n2x)

f(x) = tan (x + 4x + 9x +…. + n2x)

       = tan (1 + 4 + 9 + … + n2) x

= tanTrigonometry up to Transformations 38x

we know that period of tan kx =Trigonometry up to Transformations 26

Period of tan  =Trigonometry up to Transformations 39

                        = Trigonometry up to Transformations 40

 Question 15

Prove that sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

Sol:

We know that sin2 A – sin2B = sin (A +B) sin (A – B)

 ⇒ sin2(52 ½)0 – sin2 (22 ½)0

= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)

 = sin 750 sin 300

 = Trigonometry up to Transformations 42      

∴ sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

 Question 16

Prove that tan 700 – tan200 = 2 tan 500

Sol:

500 = 700 – 200

Tan 500 = tan (700 – 200)

We know that tan (A –B) =Trigonometry up to Transformations 43

  ⇒ Tan 500 =Trigonometry up to Transformations 44

 ⇒ tan 700 – tan 200 = tan 500 (1 + tan700 tan 200)

     tan 700 – tan 200 = tan 500 [1 + tan700 cot (900 – 200)]

     tan 700 – tan 200 = tan 500 [1 + tan700 cot 700]

     tan 700 – tan 200 = tan 500 [1 + 1]

∴ tan 700 – tan200 = 2 tan 500

 Question 17 

If sin α = Trigonometry up to Transformations 45, sin β =Trigonometry up to Transformations 46  and α, β are acute, show that α + β =Trigonometry up to Transformations 47

Sol:

Given sin α =    Trigonometry up to Transformations 45                                           sin β =Trigonometry up to Transformations 46

Trigonometry up to Transformations 48

 tan α = 1/3                                                          tan β = ½

tan (α + β) =Trigonometry up to Transformations 49

      Trigonometry up to Transformations 50 

  tan (α + β) = 1

∴ α + β =Trigonometry up to Transformations 47

 Question 18

Find tanTrigonometry up to Transformations 51 in terms of tan A

Sol:

 tan Trigonometry up to Transformations 51 =Trigonometry up to Transformations 52

                        =Trigonometry up to Transformations 53

 Question 19

Prove thatTrigonometry up to Transformations 54 = cot 360

Sol:

Trigonometry up to Transformations 54  = Trigonometry up to Transformations 55

(on dividing numerator and denominator by cos 90)

      = Trigonometry up to Transformations 56

   = tan (450 + 90)

    = tan 540

 = tan (900 – 360)

 = cot 360

 ∴Trigonometry up to Transformations 54  = cot 360      

 Question 20

Show that cos 420 + cos 780 + cos 1620 = 0

Sol:

cos 420 + cos 780 + cos 1620

= cos (600 – 180) + cos (600 + 180) + cos (1800 – 180)

=cos 600 cos180 + sin 600 sin 180 + cos 600 cos 180 – sin 600 sin 180 – cos 180

= 2 cos 600 cos 180 – cos 180

= 2 (1/2) cos 180 – cos 180

 = cos 180 – cos 180

= 0

 Question 21

Express Trigonometry up to Transformations 57sin θ + cos θ as a single of an angle

Sol:

Trigonometry up to Transformations 57sin θ + cos θ = 2(Trigonometry up to Transformations 58  sin θ + Trigonometry up to Transformations 59cos θ)

                                = 2(cos 300 sin θ + sin 300 cos θ)

                                = 2 sin (θ + 300)

 Question 22

Find the maximum and minimum value of the following functions

(i) 3 sin x –4 cos x

a= 3, b = –4 and c = 0

Trigonometry up to Transformations 60  

                                     = 5

∴ minimum value = –5 and maximum value = 5

(ii) cos (x + ) + 2  sin (x + ) – 3

a= 1, b = 2  and c = – 3

Trigonometry up to Transformations 61

 ∴ minimum value = –6 and maximum value = 0

 Question 23

Find the range of the function f(x) = 7 cos x – 24sin x + 5

Sol:

Given f(x) = 7 cos x – 24sin x + 5

a= 7, b = –24 and c = 5

Trigonometry up to Transformations 62

∴ Range = [–20, 30]    

 Question 24

Prove that sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β) is independent of α

Sol:

sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β)

= sin2α + cos (α + β) [ cos (α + β) +2 sin α sin β]

= sin2α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]

=sin2α + cos (α + β) [ cos α cos β + sin α sin β]

=sin2α + cos (α + β) cos (α –β)

= sin2 α + cos2 α – sin2 β

=1 – sin2 β

= cos2 β

 Question 25

Simplify Trigonometry up to Transformations 63

Sol:

Trigonometry up to Transformations 63   =Trigonometry up to Transformations 64

                 =Trigonometry up to Transformations 65

                 = tan θ

Question 26

For what values of x in the first quadrantTrigonometry up to Transformations 66 is positive?

Sol:

Trigonometry up to Transformations 66 > 0 ⟹ tan 2x > 0

⟹ 0 < 2x < π/2 (∵ x is in first quadrant)

⟹ 0 < x < π/4

Question 27

If cos θ = Trigonometry up to Transformations 67 and π < θ < 3π/2, find the value of tan θ/2.

Sol:

cos θ = Trigonometry up to Transformations 67

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

tan θ/2 =Trigonometry up to Transformations 68

               =– Trigonometry up to Transformations 69 (tan θ/2 < 0)

              =–Trigonometry up to Transformations 70

           = – 2

Question 28

If A is not an integral multiple of π/2, prove that cot A – tan A = 2 cot 2A.

Sol:

cot A – tan A = Trigonometry up to Transformations 71

                         =Trigonometry up to Transformations 72

                          =Trigonometry up to Transformations 73

                          =Trigonometry up to Transformations 74

                           =Trigonometry up to Transformations 75

                           = 2 cot 2A

Question 29

Evaluate 6 sin 200 – 8sin3 200

Sol:

6 sin 200 – 8sin3 200 = 2 (3 sin 200 – 4sin3 200)

                                       = 2 sin 3(200)

                              = 2 sin 600

                              = 2Trigonometry up to Transformations 76 

                              =Trigonometry up to Transformations 77

Question 30

Express cos6 A + sin6 A in terms of sin 2A.

Sol:

cos6 A + sin6 A

= (sin2 A)3 + (cos2 A)3

= (sin2 A + cos2 A)3 – 3 sin2 A cos2 A (sin2 A + cos2 A)

= 1 – 3 sin2 A cos2 A

=1 – ¾ (4 sin2 A cos2 A)

 = 1 – ¾ sin22 A

Question 31

If 0 < θ < π/8, show that Trigonometry up to Transformations 79  = 2 cos (θ/2)

Sol:

Trigonometry up to Transformations 79

Trigonometry up to Transformations 80

Trigonometry up to Transformations 81

 =2 cos (θ/2)

Question 32

Find the extreme values of cos 2x + cos2x

Sol:

cos 2x + cos2x = 2cos2 x– 1 + cos2 x

                              =3cos2 x – 1

We know that – 1 ≤ cos x ≤ 1

 ⟹ 0 ≤ cos2 x ≤ 1

      3×0 ≤ 3×cos2 x ≤ 3×1

      0– 1 ≤3 cos2 x – 1≤ 3– 1

   – 1≤3 cos2 x – 1≤ 2

     Minimum value = – 1

     Maximum value = 2

Question 33

Prove that Trigonometry up to Transformations 82 = 4

Sol:

Trigonometry up to Transformations 82

Trigonometry up to Transformations 83

= 4

Question 34

Prove that sin 780 + cos 1320 =Trigonometry up to Transformations 84

Sol:

 sin 780 + cos 1320 = sin 780 + cos (900 + 420)

                                      = sin 780 – sin 420

                                      = 2 cosTrigonometry up to Transformations 85  sinTrigonometry up to Transformations 86

                                     = 2 cos 600 sin 180

                                      = 2Trigonometry up to Transformations 87

                                      =Trigonometry up to Transformations 84

Question 35

Find the value of sin 340 + cos 640 – cos40

Sol:

sin 340 + cos 640 – cos40

= sin 340 –2 sinTrigonometry up to Transformations 88  sinTrigonometry up to Transformations 89

= sin 340 – 2sin 340 sin 300

= sin 340 – 2 sin 340 (1/2)

=sin 340 – sin 340

=0

Question 36

Prove that 4(cos 660 + sin 840) =Trigonometry up to Transformations 90

Sol:

4(cos 660 + sin 840)  

=4(cos 660 + sin (900 – 60)  

=4(cos 660 + cos (60)  

= 4[ 2 cosTrigonometry up to Transformations 91  cos Trigonometry up to Transformations 92]

=4[ 2 cosTrigonometry up to Transformations 93  cosTrigonometry up to Transformations 94 ]

=8 cos 360 cos 300

= 8Trigonometry up to Transformations 95

=Trigonometry up to Transformations 90

Question 35

Prove that (tan θ + cot θ)2 = sec2 θ + cosec2 θ = sec2 θ. cosec2 θ

Sol:

tan θ + cot θ =    Trigonometry up to Transformations 96

                         =Trigonometry up to Transformations 97

                         =Trigonometry up to Transformations 98

                         = sec θ. cosec θ

(tan θ + cot θ)2   = sec2 θ. cosec2 θ

sec2 θ + cosec2 θ =Trigonometry up to Transformations 99

                               Trigonometry up to Transformations 100

                              = sec2 θ. cosec2 θ


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1612493167414

Product of Vectors (Qns.& Ans) V.S.A.Q.’S

These solutions designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.


Product Of Vectors

Question 1

If a = 6i +2 j +3 k, b = 2i – 9 j+ 6k, then find the angle between the vectors a and b

Sol:

Given vectors are a = 6i +2 j +3 k, b = 2i – 9 j+ 6k

 If θ is the angle between the vectors Product of Vectors 1a and b, then cos θ = 

 a .b = (6i +2 j +3 k). (2i – 9 j+ 6k) = 6(2) + 2 (– 9) + 3(6)

          = 12 – 18 + 18 = 12

Product of Vectors 2      

              = 7

Product of Vectors 3      

              = 11

  ⟹ cos θ =Product of Vectors 4

 θ =Product of Vectors 5

Question 2

If a = i +2 j –3 k, b = 3ij+ 2k, then show that a + b and ab are perpendicular to each other.

Sol:

Given vectors are a = i +2 j –3 k, b = 3ij+ 2k

a + b = (i +2 j –3 k) + (3ij+ 2k) = 4i + jk

ab = (i +2 j –3 k) – (3ij+ 2k) = –2i +3 j – 5k

(a + b). (ab) = (4i + jk). (–2i +3 j – 5k)

                            = – 8 + 3 + 5

                           = 0

a + b and ab is perpendicular to each other.

Question 3

If a and b be non-zero, non-collinear vectors. If , then find the angle between a and b

Sol:

Given Product of Vectors 6

 Squaring on both sides

 Product of Vectors 7    

 (a + b) (a + b) = (a – b) (a – b)

  a2 + 2 a. b + b2 = a2 – 2 a.b + b2

 ⟹ 4 a.b = 0

  a.b = 0

∴ the angle between a and b  is 900

Question 4

IfProduct of Vectors 8 = 11, Product of Vectors 9= 23 and Product of Vectors 11 = 30, then find the angle between the vectors a and b and also find

Sol:

Given Product of Vectors 8= 11,  Product of Vectors 9 = 23 and Product of Vectors 11= 30

       Product of Vectors 11 = 30

Product of Vectors 12 = 900

Product of Vectors 13 = 900

(11)2 2 ×11×23 cos θ + (23)2 = 900

121 506 cos θ + 529   = 900

 650 506 cos θ = 900

cos θ = Product of Vectors 14 ⟹ θ =Product of Vectors 15

Product of Vectors 16   

                = (11)2 + 2 ×11×23 cos θ + (23)2

                = 121 + 2 ×11×23 ×  + 529

               = 400

 = 20

Question 5

If a = ijk and b = 2i – 3j + k, then find the projection vector of b on a and its magnitude.

Sol:

Given vectors are a = ijk and b = 2i – 3j + k

 a.b = (ij k). (2i – 3j + k) = 2 + 3 – 1 = 4

Product of Vectors 18   =Product of Vectors 19

 The projection vector of b on a =Product of Vectors 17

                  = Product of Vectors 20  (i j k)

  The magnitude of the projection vector = Product of Vectors 21 =Product of Vectors 22

Question 6

If the vectors λ i – 3j + 5k and 2λ i – λ jk are perpendicular to each other, then find λ

Sol:

let a = λ i – 3j + 5k, b = 2λ i – λ jk

Given, that a and b are perpendicular to each other

a.b = 0

i – 3j + 5k). (2λ i – λ jk) = 0

2 λ2 + 3 λ – 5 = 0

2 λ2 + 5 λ – 2 λ – 5 = 0

λ (2 λ + 5) – 1 (2 λ + 5) = 0

(2 λ + 5) ((λ – 1) = 0

λ = 1 or λ = -5/2

Question 7

Find the Cartesian equation of the plane passing through the point (– 2, 1, 3) and perpendicular to the vector 3i + j + 5k

Sol:

let P (x, y, z) be any point on the plane

 ⟹ OP = xi + yj + zk

 OA = – 2i +j +3k

AP = OPOA = (xi + yj + zk) – (– 2i +j +3k)

 AP = (x + 2) i + (y – 1) j + (z – 3) k

 AP is perpendicular to the vector 3i + j + 5k

 ⟹ 3 (x + 2) + (y – 1) + 5(z – 3) = 0

  ⟹ 3x + 6 + y – 1 + 5z – 15 = 0

 ∴   3x + y + 5z – 10 = 0 is the required Cartesian equation of the plane

Question 8

Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4

Sol:

Given plane equations are: 2x – 3y – 6z = 5,6x + 2y – 9z = 4

 Vector equations of the above planes are: r. (2i – 3j – 6k) = 5 and r. (6i + 2j – 9k) = 4

 ⟹ n1 = 2i – 3j – 6k and n2 = 6i + 2j – 9k

 If θ is the angle between the planes r. n1 = d1 and r. n2 = d2, then

 Cos θ = Product of Vectors 23

  ⟹ Cos θ =Product of Vectors 24

 ⟹ θ = Product of Vectors 25

Question 9

a = 2ij + k, b = i – 3j – 5k. Find the vector c such that a, b, and c form the sides of a triangle.

Sol:

Given a = 2ij + k, b = i – 3j – 5k

 If a, b, and c form the sides of a triangle, then a + b + c = 0

 ⟹ a + b = – c

c = – (a + b)

       = – [(2ij + k) +( i – 3j – 5k)]

      = – (3i –4 j –4k)

     ∴ c = – 3i +4 j + 4k     

Question 10

Find the equation of the plane through the point (3, –2, 1) and perpendicular to the vector (4, 7, –4).

Sol:

 Let a = 3i – 2j + k and n = 4i + 7j – 4k

 The equation of the plane passing through point A(a) and perpendicular to the vector n is (ra). n = 0

 ⟹ [r – (3i – 2j + k)]. (4i + 7j – 4k) = 0

 ⟹ r. (4i + 7j – 4k)– [(3i – 2j + k). (4i + 7j – 4k)] = 0

r. (4i + 7j – 4k)– (12 – 14 – 4) = 0

r. (4i + 7j – 4k)– 6 = 0

r. (4i + 7j – 4k) = 6

Question 11

Find the unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k

Sol:

The vector which is parallel to the XOY-plane is of the form xi + yj

The vector which is parallel to the XOY-plane and perpendicular to 4i – 3j + k

 is 3i + 4j

 its magnitude = Product of Vectors 26     = 5

∴ The unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k =Product of Vectors 27

Question 12

If a + b + c = 0, Product of Vectors 18 = 3, Product of Vectors 9 = 5 and   Product of Vectors 30 = 7, then find the angle between a and b

Sol:

Given, a + b + c = 0,  = 3,  = 5 and    = 7

 a + b = – c

 (a + b)2 =Product of Vectors 29

Product of Vectors 28 

 32 + 52 + 2  cos θ = 72

  9 + 25 + 2.3.5 cos θ = 49

 34 + 30cos θ = 49

 30cos θ = 49 – 34

 30cos θ = 15

cos θ = 15/30 = 1/2

∴ θ = π/3

Question 13

If a = 2i – 3j + 5k, b = – i + 4j + 2k, then find a × b and unit vector perpendicular to both a and b

Sol:

 Given, a = 2i – 3j + 5k, b = – i + 4j + 2k             

 a × b =Product of Vectors 40

         = i (–6 – 20) – j (4 + 5) + k (8 – 3)

         = –26i – 9j + 5k

The unit vector perpendicular to both a and b    =   Product of Vectors 31

  =Product of Vectors 32

=Product of Vectors 33

Question 14

If a = i + j + 2k and b = 3i + 5jk are two sides of a triangle, then find its area.

Sol:

Given, a = i + 2j + 3k and b = 3i + 5jk

If a, b are two sides of a triangle, then area of the triangle =

a × b =Product of Vectors 41

         = i (–2 – 15) – j (–1 – 9) + k (5 – 6)

         = –17i + 10 j k

Product of Vectors 37 =Product of Vectors 35

Area of triangle =Product of Vectors 34 =

                              =Product of Vectors 36

Question 15

Find the area of the parallelogram for which the vectors a = 2i – 3j and b = 3ik are adjacent sides.

Sol:

Given, a = 2i – 3j and b = 3ik are adjacent sides of a parallelogram

The area of the parallelogram whose vectors a , b are adjacent sides =  

  a × b=Product of Vectors 38

              = i (3 – 0) – j (–2 – 0) + k (0 + 9)

             =3 i +2 j +9 k

Product of Vectors 37= Product of Vectors 42

                = Product of Vectors 39

∴ The area of the parallelogram =Product of Vectors 39

Question 16

Let a, b be two non-collinear unit vectors. If α = a – (a . b) b and β = a × b, then show that Product of Vectors 43

Sol:

Product of Vectors 44 = Product of Vectors 45

          = Product of Vectors 46

              = 1 – cos2 θ

             = sin2 θ

Product of Vectors 47 =

        = 1 + cos2 θ – 2cos2 θ

       = 1– cos2 θ

       = sin2 θ

Product of Vectors 43

Question 17

For any vector a, show thatProduct of Vectors 48     

Sol:

Let a = xi + yj + zk

       a × i =Product of Vectors 49

                = i ( 0 – 0) – j (0 – z) + k (0 – y)

           = – yk + zj

Product of Vectors 50   = y2 + z2        

Similarly, Product of Vectors 51  = x2 + z2      and      = y2 + x2        

Product of Vectors 50 + Product of Vectors 51+ Product of Vectors 51 = y2 + z2 + x2 + z2 + y2 + x2        

                                                        = 2(x2 + y2 +z2)

                                                        =2 Product of Vectors 53

Question 18

IfProduct of Vectors 54 = 2, Product of Vectors 55 = 3 and (p, q) =Product of Vectors 56 , then find

Sol:

Given Product of Vectors 54= 2, Product of Vectors 55 = 3 and (p, q) =Product of Vectors 56

Product of Vectors 58  = Product of Vectors 54Product of Vectors 55 sin (p, q)

                = 2 × 3 sin

                = 2 × 3×1/2

               = 3

Product of Vectors 57 = 32 = 9

Question 19

If 4i + j + pk is parallel to the vector i + 2j + 3k, then find p.

Sol:

Given 4i +  j + pk is parallel to the vector i + 2j + 3k

Product of Vectors 59   = Product of Vectors 60 =Product of Vectors 61

Product of Vectors 61 = 4

⇒ p = 12

Question 20

Compute a× (b + c) + b× (c + a) + c× (a + b)

Sol:

a× (b + c) + b× (c + a) + c× (a + b)

= a× b + a× c + b× c + b× a + c × a + c × b

= a× b + a× c + b× ca × b a × c b ×c

= 0

Question 21

Compute 2j× (3i – 4k) + (i + 2j) × k

Sol:

2j× (3i – 4k) + (i + 2j) × k

= 6(j × i) – 8(j × k) + (i × k) + 2(j × k)

 = 6k – 8i – j + 2i

 = 6i –j 6 j


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1612493167414

Addition of Vectors (Qns.& Ans) V.S.A.Q.’S

These solutions designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.


Addition of Vectors

QUESTION 1

Find the unit vector in the direction of Addition of Vectors 17

Sol: Given vector isAddition of Vectors 17

 The unit vector in the direction of a vector Addition of Vectors 9 is given byAddition of Vectors 15 

Addition of Vectors 16

QUESTION 2

Find a vector in the direction of a where Addition of Vectors 19that has a magnitude of 7 units.

Sol: Given vector is Addition of Vectors 19

 The unit vector in the direction of a vector Addition of Vectors 9 is Addition of Vectors 20

 The vector having the magnitude 7 and in the direction of  is

  Addition of Vectors 21

QUESTION 3

Find the unit vector in the direction of the sum of the vectors, a = 2i + 2j – 5k and b = 2i + j + 3k

Sol Given vectors are a = 2i + 2j – 5k and b = 2i + j + 3k

a + b = (2i + 2j – 5k) + (2i + j + 3k) = 4i + 3j – 2k

Addition of Vectors 22

Addition of Vectors 23      

 QUESTION 4

Write the direction cosines of the vector Addition of Vectors 24

Sol: Given vector is Addition of Vectors 24

Addition of Vectors 25

    ∴ Direction cosines are Addition of Vectors 27

QUESTION 5

Show that the points whose position vectors are – 2a + 3b + 5c, a + 2b + 3c, 7 ac are collinear when a, b, c are non-collinear vectors

Sol: Let OA = – 2a + 3b + 5c, OB = a + 2b + 3c, OC = 7 acA

B = OB – OA = a + 2b + 3c (– 2a + 3b + 5c)

         AB = 3a b – 2c

 AC = OC – OA = 7 ac(– 2a + 3b + 5c)

 AC = 9a – 3b – 6c = 3(3a b – 2c)

          AC = 3 AB

   A, B and C are collinear

 QUESTION 6

ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) AL and AM in terms of AB and AD (ii) 𝛌, if AM = 𝛌 AD – LM

Sol: Given, ABCD is a parallelogram and L and M are middle points of BC and CD

Addition of Vectors 1

(i) Take A as the origin

 M is the midpoint of CD

  AM =Addition of Vectors 2

          = AD + ½ AB (∵ AB = DC)

 L is the midpoint of BC

  AL =Addition of Vectors 3

      = AB + ½ AD ((∵ BC = AD)

(ii) AM = 𝛌 AD – LM

AM + LM= 𝛌 AD 

AD + ½ AB + AD + ½ AB – (AB + ½ AD) = 𝛌 AD 

AD + ½ AB + AD + ½ AB –  AB – ½ AD = 𝛌 AD 

3/2 AD = 𝛌 AD 

𝛌 = 3/2

QUESTION 7

If G is the centroid of the triangle ABC, then show that OG = Addition of Vectors 4  whenAddition of Vectors 9,Addition of Vectors 10 Addition of Vectors 42 are the position vectors of the vertices of triangle ABC.

Sol: OA = a, OB = b, OC = c and OD = d

Addition of Vectors 6

 D is the midpoint of BC

OD = Addition of Vectors 7

G divides median AD in the ratio 2: 1

OG =Addition of Vectors 8

OG =Addition of Vectors 4

QUESTION 8

If Addition of Vectors 9= Addition of Vectors 11, Addition of Vectors 10 =  Addition of Vectors 12  are collinear vectors, then find m and n.

Sol: Given Addition of Vectors 9 , Addition of Vectors 10 are collinear vectors

Addition of Vectors 9 = λAddition of Vectors 10

Addition of Vectors 13

 Equating like vectors

 2 = 4 λ; 5 = m λ; 1 = n λ

 λ =Addition of Vectors 14

 5 = Addition of Vectors 14m ⟹ m =10

1 = Addition of Vectors 14n ⟹ n = 2

∴ m = 10, n = 2

QUESTION 9

Let If  Addition of Vectors 28, Addition of Vectors 29. Find the unit vector in the direction of a + b.

Sol: Given vectors are Addition of Vectors 28 and  Addition of Vectors 29

     a + b = Addition of Vectors 33

    The unit vector in the direction of a + b = Addition of Vectors 30  

       = Addition of Vectors 31

        =  Addition of Vectors 32

QUESTION 10

If the vectors – 3i + 4j + λk and μi + 8j + 6k. are collinear vectors, then find λ and μ.

Sol: let a = – 3i + 4j + λk, b = μi + 8j + 6k

     ⟹   a = tb

 – 3i + 4j + λk = t (μi + 8j + 6k)

 – 3i + 4j + λk = μt i + 8t j + 6t k

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

 t =Addition of Vectors 14

– 3 =Addition of Vectors 14 μ ⟹μ=– 6

λ = Addition of Vectors 14 6 ⟹ λ = 3

∴ μ=– 6, λ = 3

QUESTION 11

ABCD is a pentagon. If the sum of the vectors AB, AE, BC, DC, ED and AC is 𝛌 AC then find the value of 𝛌

Sol: Given, ABCD is a pentagon

        AB + AE + BC + + DC + ED + AC = 𝛌 AC

         (AB + BC) + (AE + DC + ED) + AC = 𝛌 AC

         AC + AC + AC = 𝛌 AC

         3 AC = 𝛌 AC

         𝛌 = 3

QUESTION 12

If the position vectors of the points A, B and C are – 2i + jk and –4i + 2j + 2k and 6i – 3j – 13k respectively and AB = 𝛌 AC, then find the value of 𝛌

Sol: Given, OA = – 2i + jk , OB = –4i + 2j + 2k and OC  = 6i – 3j – 13k

  AB = OB – OA = –4i + 2j + 2k – (– 2i + jk)

          = –4i + 2j + 2k +2ij + k

        = –2i + j + 3k

 AC = OC – OA = 6i – 3j – 13k – (– 2i + jk)

       = 6i – 3j – 13k +2ij + k

     = 8i –4 j –12k

     = – 4 (2i + j + 3k)

AC = – 4 AB

 Given AB = 𝛌 AC

    𝛌 = – 1/4

QUESTION 13

If OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k then find the vector OD

Sol: Given OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k

         OD = OA + AB + BC + CD

                = i + j +k + 3i – 2j + k + i + 2j – 2k + 2i + j +3k

                = 7i + 2j +4k

QUESTION 14

Let a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k, then find the unit vector in the opposite direction of a + b + c

Sol:  Given, a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k

         a + b + c = 2i +4 j –5 k + i + j+ k + j +2 k

                           = 3i +6j –2k

 The unit vector in the opposite direction of a + b + c is Addition of Vectors 34

 ⟹ Addition of Vectors 35

      =Addition of Vectors 36

QUESTION 15

Is the triangle formed by the vectors 3i +5j +2k, 2i –3j –5k, 5i – 2j +3k

Sol: Let a =3i +5j +2k, b = 2i –3j –5k, c = 5i – 2j +3k

  Addition of Vectors 37

 ∴ Given vectors form an equilateral triangle.

QUESTION 16

Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are a, b,  (3a – 2b) are collinear.

Sol: the vector equation of the straight line passing through two points a, b is

         r = (1 – t) a+ t b

         3a – 2b = (1 – t) a+ t b

          Equating like vectors

          1 – t = 3 and t = – 2

∴ Given points are collinear.

QUESTION 17

OABC is a parallelogram If OA = a and OC = c, then find the vector equation of the side BC

Sol: Given, OABC is a parallelogram and OA = a, OC = c

Addition of Vectors 38

 The vector equation of BC is a line which is passing through C(c) and parallel to OA

 ⟹ the vector equation of BC is r = c + t a

QUESTION 18

If a, b, c are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A

Sol: Given OA = a, OB = b, OC = c

Addition of Vectors 39

 D is mid of BC

  OD =Addition of Vectors 40 = Addition of Vectors 41

 The equation of AD is

  r = (1 – t) a + t ( Addition of Vectors 41 )

QUESTION 19

Find the vector equation of the line passing through the point 2i +3j +k and parallel to the vector 4i – 2j + 3k

Sol: Let a =2i +3j +k, b = 4i – 2j + 3k  

         The vector equation of the line passing through a and parallel to the vector b is r = a + tb

       r = 2i +3j +k + t (4i – 2j + 3k)

          = (2 + 4t) i + (3 – 2t) j + (1 + 3t) k

QUESTION 20

Find the vector equation of the plane passing through the points i – 2j + 5k, 2j –k and – 3i + 5j

Sol: The vector equation of the line passing through a, b and cis r = (1 – t – s) a + tb + sc

      r = (1 – t – s) (i – 2j + 5k) + t (2j –k) + s (– 3i + 5j)

              = (1 – t – 4s) i + (– 2 – 3t + 7s) j + (5 – 6t – 5s) k


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Matrices ( Qns & Solutions) || V.S.A.Q’S||

This note is designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.  


Matrices

QUESTION 1

If A = Matrices 1, then show that A2 = –I

Sol: Given A = Matrices 1

Matrices 2

  ∴  A2 = –I

QUESTION 2

If A = Matrices 3, and A2 = 0, then find the value of k.

Sol: Given A = Matrices 3 and

 A2 = 0

⟹ A. A =0 ⟹  Matrices 4   = 0

Matrices 5 = 0

     8 + 4k = 0, – 2 – k = 0 and –4 + k2 = 0

    4k = –8; k = –2; k2 = 4

       k = –2; k = –2; k = ± 2

   ∴ k =– 2

QUESTION 3

Find the Trace of A, If A = Matrices 6

Sol: Given A =Matrices 6

       Trace of A = 1 – 1 + 1 = 1

QUESTION 4

If A =Matrices 12 , B = Matrices 13and 2X + A = B, then find X.

Sol: Given A =Matrices 12 , B = Matrices 13 and 2X + A = B

        2X = B – A

        2X =Matrices 13  – Matrices 12

              = Matrices 8

              =Matrices 9

           X =   Matrices 10 Matrices 9

         ∴ X =   Matrices 11

QUESTION 5

Find the additive inverse of A, If A =Matrices 14

Sol: Given A =Matrices 14

       Additive inverse of A = – A

    = –Matrices 14

   =Matrices 15

QUESTION 6

If Matrices 16, then find the values of x, y, z and a.

Sol: Given Matrices 16

 ⟹ x- 1 = 1 – x ; y – 5 =  – y ; z = 2 ; 1 + a = 1

 ⟹ x + x = 1 + 1; y + y = 5; z = 2; a =1– 1

  ⟹ 2x = 1; 2y = 5; z = 2; a = 0

∴ x = ½ ; y = 5/2; z = 2; a = 0

QUESTION 7

Construct 3 × 2 matrix whose elements are defined by aij =Matrices 17

Sol:

Let A= Matrices 18

a11 = Matrices 19

a11 = 1

a12 = Matrices 20

a12 =Matrices 22

a21 = Matrices 23

a21 =Matrices 24

a22 = Matrices 25

a22 = 2

a31 = Matrices 26

a31 = 0

a32 = Matrices 27

a32 =Matrices 28

 ∴ A =Matrices 29

QUESTION 8

If A = Matrices 30 and B = Matrices 31, do AB and BA exist? If they exist, find them. BA and AB commutative with respect to multiplication.

Sol: Given Matrices are A = Matrices 30 B =Matrices 31

       Order of A = 2 × 3 and Order of B = 3 × 2

AB and BA exist

 

 AB =   Matrices 30Matrices 31

    Matrices 32

BA =     Matrices 31Matrices 30

Matrices 33

 AB and  BA are not Commutative under Multiplication 

QUESTION 9

Define Symmetric and Skew Symmetric Matrices

Sol:

Symmetric Matrix: Let A be any square matrix, if AT = A, then A is called Symmetric Matrix

Skew Symmetric Matrix: Let A be any square matrix if AT = –A, then A is called Skew Symmetric Matrix

QUESTION 10

If A =Matrices 34 is a symmetric matrix, then find x.

Sol: Given, A = Matrices 34 is a symmetric matrix

       ⟹ AT = A

          Matrices 35     

          ⟹ x = 6

QUESTION 11

If A =Matrices 36 is a skew-symmetric matrix, then find x

Sol: Given A = Matrices 36is a skew-symmetric matrix

       ⟹ AT = – A

     

        ⟹ x = –x

        x+ x = 0 ⟹ 2x = 0

     ⟹ x = 0

QUESTION 12

If A =Matrices 38 and B = Matrices 39, then find (A BT) T

Sol: Given A = Matrices 38   B =Matrices 39

   BT = Matrices 40   

   (A BT) =  Matrices 38   Matrices 40

                = Matrices 41

(A BT) T = Matrices 42

QUESTION 13

If A =Matrices 43 and B =Matrices 44 , then find A + BT

Sol: Given A =Matrices 43  and B =Matrices 44

 BT =Matrices 45

A + BT = Matrices 43 + Matrices 45

 Matrices 47            

QUESTION 14

If A = Matrices 48, then show that AAT = ATA = I

Sol: Given A =Matrices 48

  AT =Matrices 49

AAT =

= Matrices 51

 = Matrices 52 

ATA =Matrices 49Matrices 48

        =

       =Matrices 52

∴ AAT = ATA = I

QUESTION 15

Find the minor of – 1 and 3 in the matrixMatrices 54

Sol: Given Matrix is

       minor of – 1 = Matrices 55 = 0 + 15 = 15

     minor of 3 = Matrices 56 = – 4 + 0 = – 4

QUESTION 16

Find the cofactors 0f 2, – 5 in the matrixMatrices 57

Sol: Given matrix is

 Cofactor of 2 = (–1)2 + 2 Matrices 58= –3 + 20 = 17

  Cofactor of – 5 = (–1)3 + 2  Matrices 59= –1(2 – 5) = –1(–3) = 3

QUESTION 17

If ω is a complex cube root of unity, then show that Matrices 62= 0(where 1 + ω+ω2 = 0)

Given matrix is    Matrices 62

   R1 → R1 + R2 + R3

   Matrices 73 Matrices 74

Matrices 74 = 0 (∵ 1 + ω+ω2 = 0)

QUESTION 18

If A = Matrices 63and det A = 45, then find x.

Sol: Given A = Matrices 63

Det A = 45

Matrices 64= 45

   ⟹ 1(3x + 24) – 0 (2x – 20) + 0 (– 12 – 15) = 45

 ⟹ 3x + 24 = 45

        3x = 45 – 24

        3x = 21

         x = 7

QUESTION 19

Find the adjoint and inverse of the following matrices

(i)

A =Matrices 65

Adj A =Matrices 66

A-1 = Matrices 67

       =Matrices 68

   ∴ A-1 =Matrices 69

(ii)

A =Matrices 70

Adj A =Matrices 71

A-1 =Matrices 72

 ∴ A-1 = Matrices 71   

QUESTION 20

Find the inverse of Matrices 75 (abc ≠ 0)