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English Grammar 4 Competitive Exams & School Education

Blog, Education / Nayini Satyanarayana Reddy

english grammar 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. enhlish grammar 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

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Polytechnic Engineering Maths Feature Image for Sem 1 Concept

Engineering Mathematics SEM – 1Concept

Education / Nayini Satyanarayana Reddy

Polytechnic Sem 1 – Engineering Mathematics Polytechnic Sem 1 – Notes for Polytechnic SEM – 1 is Designed by the ” Basics in Maths” team. Here we can learn Concepts in Basic Engineering mathematics at 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 Standard formulae of logarithms: Logarithmic Function: Let a be a positive real number and a ≠ 1. The function f: (o, ∞) → R Defined by f(x) = PARTIAL FRACTIONS Fractions: If f(x) and g(x) are two polynomials, g(x) ≠ 0, then is called rational fraction. Ex: etc. are rational fractions. Proper Fraction: A rational fraction is said to be a Proper fraction if the degree of g(x) is greater than the degree of f(x). Ex: etc. are the proper fractions. Improper Fraction: A rational fraction is said to be an Improper fraction if the degree of g(x) is less than the degree of f(x). Ex: 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) = 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: 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: ∎ If R(x) = is improper fraction, then Case (i): – If degree f(x) = degree of g(x), 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) = 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: 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. Square Matrix: A matrix in which the no. of rows is equal to no. of columns is called a square matrix. 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. 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). Diagonal Matrix: If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix. 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. 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 Null Matrix or Zero Matrix: If each element of a matrix is zero, then it is called a null matrix. 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. 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. 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. 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. Product of Matrices: Let A = [aik]mxn and B = [bkj]nxp be two matrices, then the matrix C = [cij]mxp where 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,

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AS Tutorial For Math Classes / YouTube new no. 1 Channel

Blog, Education / Nayini Satyanarayana Reddy

AS Tutorial తెలుగు Hi, This is Nayini Satyanarayana Reddy, Welcome To our Channel AS Tutorial తెలుగులో. In ASTutorialతెలుగులో I will share with you the concepts of Maths in inter. AS Tutorialతెలుగులో Provides Concepts and Problems for Inter Maths Students In Telugu. హాయ్, నేను నాయిని సత్యనారాయణ రెడ్డి, వెల్ కమ్ టు మై ఛానల్ AS Tutorial తెలుగు, AS Tutorial తెలుగుల ఇంటర్ లో మ్యాథ్స్ భావనలను మీతో పంచుకుంటాను. AS Tutorial తెలుగుఇంటర్ మ్యాథ్స్ విద్యార్థులకు తెలుగులో కాన్సెప్ట్స్ అండ్ ప్రాబ్లమ్స్ ను అందిస్తోంది. Support My Channel Subscribe to my channel like my Channel Comment on My Videos Click the Bell My YouTube Links: My Youtube Channel: Simple Equations | LHS to RHS and RHS to LHS, RULE ‘Playing With Numbers||Divisibility Rules In Telugu |Even| Odd|Prime|Composite|Co-Prime &Twin- Prime Numbers TS 10TH CLASS Trigonometry Introduction, Sides of a Right Triangle and Trigonometric Ratios How To Prove Specific Angles In Trigonometry How to Remember the Specific Angles Table in Trigonometry Trigonometric Identities In Trigonometry Logarithms|| Product Rule| Quotient Rule| and Power Rule TET TS TET Child Development & Pedagogy Part 2: TS TET Child Development & Pedagogy Part 3: TS TET Child Development & Pedagogy Part 4: TS TET Child Development & Pedagogy Part 5: TS TET 2022, Mathematics Methods, paper 2, Practice Bits p – 1 TS TET 2022, Mathematics Methods, paper 2, Practice Bits, part 3 TS TET 2022, Mathematics Methods, paper 2, Practice Bits, part 4 TS TET 2022, Mathematics Methods, paper 2, Practice Bits, part 5 TS TET – 2022: VI Class Maths, Chapter 1, మనసంఖ్యలను తెలుసుకుందాం , Practice Bits TS TET – 2022: VI Class Maths, Chapter 1, మనసంఖ్యలను తెలుసుకుందాం , Practice Bits TS TET – 2022: VI Class Maths, Chapter 2,పూర్ణాంకాలు , Practice Bits TS TET 2022: VI Class Maths, Chapter 2, Whole Numbers, Practice Bits TS TET – 2022: X Class Maths, Chapter 11, Trigonometry, Practice Bits TS TET – 2022: X Class Maths, Chapter 11, త్రికోణమితి , Practice Bit TS TET 2022, X Class Maths, Chapter 1, Real Numbers, Practice Bits, Practice Bits TS TET 2022,10 th Maths, chapter 1, Real Numbers – Logarithms TS TET 2022, Paper 2, 10 th Maths, Chapter 2, SETS, Practice bits TS TET 2022, Paper 2, 10 th Maths, Chapter 14, Statistics, Practice bits MATHS 1A Trigonometry// ALL-SILVER-TEA-CUPS Rule Periodic Functions| Trigonometric Ratios Up To Transformations Matrices Basic Concepts Trace of Matrix, Addition of Matrices, Equality of Matrices Multiplication of two matrices P – 1 Multiplication of two matrices P – 2 Multiplication of two matrices P – 3 transpose of matrices Smmetric and Skew Symmetric Matrices Adjoint and Inverse of Matrices P – 1 Adjoint and Inverse of Matrices P – 2 Adjoint and Inverse of Matrices P – 3 MATHS 1B ‘Locus’ Concept and Problems For Inter First Year Maths 1B (PART – 1) ‘Locus’ Concept and Problems For Inter First Year Maths 1B(PART – 2) Transformation Of Axes’ Concept and Problems For Inter Maths 1B P 1 ‘Transformation Of Axes’ Concept and Problems For Inter Maths 1B P 2 Please Do Not Issue A “Copyright Strike” against the channel, as it affects my channel and all previous work. If I Uploaded Videos Or Music That Is Yours And You Want It Removed. Then please massage me, and I Will Remove The Whole Video In Fewer 12Hours. My YouTube Videos Playlist Simple Mistakes Made by Maths Students Inter Maths 1A Concept & Solutions Inter Maths 1B Concept & Solutions Inter Maths 2A Concept & Solutions Inter Maths 2B Concept & Solutions 10th Class Maths Concept & Solutions SSC – CGL

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Reduced Syllabus 2021 Telangana BIE Maths

Education / Nayini Satyanarayana Reddy

Reduced Syllabus 2021 Telangana BIE Maths Reduced Syllabus 2021 This content is designed by the ‘Basics in Maths‘ team. Telangana BIE Maths Reduced Syllabus(2021) very useful I.P.E exam. PDF Files || Inter Maths 1A &1B || (New) 6th maths notes|| TS 6 th class Maths Concept TS 10th class maths concept (E/M) Inter 1st year Maths Reduced Syllabus Click Here Inter 2nd year Maths Reduced Syllabus Click Here Visit my YouTube Channel: Click on Below Logo

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Errors and Appraximations vsaqs questions and solutions

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

Education, TS Inter Maths, TS Inter Maths1B / Nayini Satyanarayana Reddy

Errors and Approximations || V.S.A.Q’S|| Errors and approximations: These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help 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 = , x = 8 and ∆x =0.02 Sol: 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 the 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 = 2 A = x2 ∆A = 2x × ∆x The approximate percentage error in area A = 2 × 2 =4 https://www.basicsinmaths.com/inter-mathematics-1a-and-1b-pdf-files/ Question 4 From the following. Find the approximations (i) Sol: Let f(x) = , where x = 1000 and ∆x =– 1 f’ (x) = Approximate value is f (x + ∆x) = f(x) + f’ (x) ∆x = 10 – 0. 0033 = 9.9967 (ii) Sol: (iii) Sol: (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 Visit my Youtube Channel: Click on Below Logo

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Differentiation vsaqs questions and solutions

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

Education / Nayini Satyanarayana Reddy

Differentiation(2m Q & S) || V.S.A.Q’S|| Differentiation This content was designed by the ‘Basics in Maths‘ team. These notes to do help 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 (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) = (x2) 2x log x + x2 (2x) log x + x2 2x (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) = (x > 0) Sol: Given f(x) = f’ (x) = . log 7 (x3 + 3x) = log 7 (3×2 + 3) =3 (x2 + 1) log 7 (iv) f(x) = log (sec x + tan x) Sol: Given, f(x) = log (sec x + tan x) f’ (x) = (sec x + tan x) = (sec2 x + sec x tan x) = 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 (x2 + 1) + (x2 + 1) (ex) = ex (2x + 0) + (x2 + 1) ex = ex (x2 + 2x + 1) = ex (x + 1)2 (ii) Let y = (iii) cos (log x + ex) (iv) x = tan (e-y) e-y = tan-1 x (v) cos [log (cot x)] (vi) sin[tan-1(ex)] (vii) cos-1(4×3 – 3x) let y = cos-1(4×3 – 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 = 3 (cos-1 x) = 3 = (viii) (ix) (x) Differentiation 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 (x3 + 6 x2 + 12x – 13) = 100(x3 + 6 x2 + 12x – 13)99 (3×2 + 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 = = 50 × 101 = 5050 Question 5 From the following functions. Find their derivatives. Question 6 If y = , find Sol: Given y = Question 7 If y = log (cosh 2x), find Sol: Given y = log (cosh 2x) Question 8 If x = a cos3 t, y = a sin3 t, find Sol: Given If x = a cos3 t, y = a sin3 t Question 9 Differentiate f(x) with respect to g(x) for the following. (i) f(x) = ex, g(x) = f’ (x) = ex and g’ (x) = derivative of f(x) with respect to g(x) = (ii ) put x = tan θ ⟹ θ = tan-1 x Question 10 if y = then prove that Sol: Given y = Differentiation Basics In Maths Ts Inter Maths IA Concept Visit My YouTube Channel: Click on the logo below

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Limits (Q’s & Ans) || V.S.A.Q’S||

Education / Nayini Satyanarayana Reddy

Limits V.S.A.Q’S Limits (Q’s & Ans) || V.S.A.Q’S|| 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. Limits Question 1 Find Sol: = 9 Question 2 Compute Sol: = a + a = 2a Question 3 compute Sol: Question 4 Show that = 1and = –1 Sol: we know that if x > 0 = –x if x < 0 As x → 0+ ⟹ x > 0 ⟹ = x = 1 As x → 0+ ⟹ x < 0 ⟹ = –x = –1 Question 5 If f (x) = , then find and . Does exist? Sol: Question 6 Show that = –1 Sol: As x → 2– ⟹ x < 2 x – 2 < 0 ⟹ < 0 Question 7 Compute and Sol: As x → 2+ ⟹ x > 2 ⟹ = 2 = As x → 2– ⟹ x < 2 ⟹ = 1 Question 8 Find Sol: Question 9 Compute Sol: Question 10 Show that Sol: Question 11 Compute Sol: Question 12 Compute Sol: Question 13 Evaluate Sol: let y = x – 1 ⟹ x = y + 1 then as x → 1, y → 0 Question 14 Compute Sol: Question 15 Compute Sol: https://www.basicsinmaths.com/inter-mathematics-1a-and-1b-pdf-files/ Question 16 Compute Sol: As x → ∞, →0 = ∞ (3/4) = ∞ Question 17 Compute Sol: We know that – 1 ≤ sin x ≤ 1 x2 – 1 ≤ x2 – sin x ≤ x2 + 1 Question 18 Find Sol: Question 19 Find Sol: Question 20 Find Sol: Question 21 Find Sol: Question 22 Compute Sol: Question 23 Compute Sol: Visit my YouTube channel: Click on the Logo

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Hyperbolic Functions vsaqs questions and solutions

Hyperbolic Functions V.S.A.Q.’S || 2 Marks

Education / Nayini Satyanarayana Reddy

Hyperbolic Functions V.S.A.Q.’S || 2 Marks Hyperbolic Functions V.S.A.Q.’Sdesigned by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students. Inter Maths – 1A Hyperbolic Functionstwo 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 = , find the values of (i) cosh 2x and (ii) sinh 2x Sol: Given cosh x = Cosh 2x = 2 cosh2 x – 1 = 2. – 1 = Sinh2 2x = cosh2 2x – 1 = – 1 = – 1 = Sinh2 2x = Question 3 If cosh x = sec θ then prove that tanh2= tan2 Sol: tanh2 = = = = = tan2 Question 4 If sinh x = 5, then show that x = Sol: Given, sinh x = 5 ⟹ x = sinh-15 We know that sinh-1x = ⟹ x = x = Question 5 Show that tanh-1 = log3 Sol: Given tanh-1 We know that tanh-1 x = tanh-1 = = = 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) = = 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 that = cosh x + sinh x Sol: = 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: ∴ (cosh x – sinh x) n = cosh nx – sinh nx Visit My YouTube Channel: Click on below logo

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Trigonometric Ratios vsaqs questions and solutions

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

Education / Nayini Satyanarayana Reddy

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S designed by the ‘Basics in Maths‘ team.These notes to do help 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 =, 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 = ⇒ sec θ – tan θ = ———— (2) (1) + (2) ⇒ (sec θ + tan θ) + (sec θ – tan θ) = 2sec θ = sec θ = (1) – (2) ⇒ (sec θ + tan θ) – (sec θ – tan θ) = 2 tan θ = ⇒ tan θ = Now sin θ = tan θ ÷ sec θ = Sin θ = 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 = cos θ, then prove that cos θ – sin θ = sin θ Sol: Given, cos θ + sin θ = cos θ Sin θ = cos θ – cos θ = ( – 1) cos θ ( + 1) sin θ = ( + 1) ( – 1) cos θ sin θ + sin θ = cos θ ∴ cos θ – sin θ = 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 that Sol: Given tan 200 = λ = = = = 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 ⇒ = 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 + cos2 – tan2 Sol: 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 = (1 – sin4 α) Sol: cos4 α + 2 cos2 α = 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 =

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Product of Vectors vsaqs questions and solutions

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

Education / Nayini Satyanarayana Reddy

Product of Vectors (Qns.& Ans) V.S.A.Q.’S Product of Vectors (Qns.& Ans) V.S.A.Q.’S; These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help 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 a 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 = 7 = 11 ⟹ cos θ = θ = Question 2 If a = i +2 j –3 k, b = 3i – j+ 2k, then show that a + b and a – b are perpendicular to each other. Sol: Given vectors are a = i +2 j –3 k, b = 3i – j+ 2k a + b = (i +2 j –3 k) + (3i – j+ 2k) = 4i + j – k a – b = (i +2 j –3 k) – (3i – j+ 2k) = –2i +3 j – 5k (a + b). (a – b) = (4i + j – k). (–2i +3 j – 5k) = – 8 + 3 + 5 = 0 ∴ a + b and a – b 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 Squaring on both sides (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 If = 11, = 23 and = 30, then find the angle between the vectors a and b and also find Sol: Given = 11, = 23 and = 30 = 30 = 900 = 900 (11)2 – 2 ×11×23 cos θ + (23)2 = 900 121 – 506 cos θ + 529 = 900 650 – 506 cos θ = 900 cos θ = ⟹ θ = = (11)2 + 2 ×11×23 cos θ + (23)2 = 121 + 2 ×11×23 × + 529 = 400 = 20 Question 5 If a = i – j – k and b = 2i – 3j + k, then find the projection vector of b on a and its magnitude. Sol: Given vectors are a = i – j – k and b = 2i – 3j + k a.b = (i – j – k). (2i – 3j + k) = 2 + 3 – 1 = 4 = The projection vector of b on a = = (i – j – k) The magnitude of the projection vector = = Question 6 If the vectors λ i – 3j + 5k and 2λ i – λ j – k are perpendicular to each other, then find λ Sol: let a = λ i – 3j + 5k, b = 2λ i – λ j – k Given, that a and b are perpendicular to each other ⟹ a.b = 0 (λ i – 3j + 5k). (2λ i – λ j – k) = 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 = OP – OA = (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 = ⟹ Cos θ = ⟹ θ = Question 9 a = 2i – j + k, b = i – 3j – 5k. Find the vector c such that a, b, and c form the sides of a triangle. Sol: Given a = 2i – j + k, b = i – 3j – 5k If a, b, and c form the sides of a triangle, then a

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