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.

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Polytechnic Engineering Mathematics – Sem -2 Solutions Read More »

The post Polytechnic Engineering Mathematics – Sem -2 Solutions first appeared on Basics In Maths.]]>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 Polytechnic Maths SEM – 2 Solutions

Ratre Measure Polytechnic Maths SEM – 2 Solutions

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

The post TS Inter Maths 1A &1B Practice papers (Reduced Syllabus) first appeared on Basics In Maths.]]>These Practice papers to do help the intermediate First-year Maths students.

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

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The post TS Inter Maths 1A &1B Practice papers (Reduced Syllabus) first appeared on Basics In Maths.]]>

The post Inter Mathematics 1A ands 1B Pdf Files first appeared on Basics In Maths.]]>

English Grammar 4 Competitive Exams & School Education Read More »

The post English Grammar 4 Competitive Exams & School Education first appeared on Basics In Maths.]]>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.

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,

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.

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**

**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.,

**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.,

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.

**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 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

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.

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.,

**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) 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:

- Main verb
- Helping verb

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.,

**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

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.,

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)

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

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

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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**Logarithm:** For ant two positive real numbers a, b, and a ≠ 1. If the real number x such then a^{x} = b, then x is called logarithm of b to the base a. it is denoted by

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

If f(x) and g(x) are two polynomials, g(x) ≠ 0, then is called rational fraction.

Ex:

_{ }etc. are rational fractions.

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.

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.

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 (ax^{2} + 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)

**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:

**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 a_{11}, a_{22}, a_{33}, ………. A_{n 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.

A square matrix A = [a_{ij}] is said to be upper triangular if a_{ij} = 0 ∀ i > j

A square matrix A = [a_{ij}] is said to be lower triangular matrix a_{ij} = 0 ∀ i < j

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.

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.

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.

Let A = [a_{ik}]_{mxn} and B = [b_{kj}]_{nxp} be two matrices, then the matrix C = [c_{ij}]_{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 } . B_{p x q} = AB _{mx q}; n = p

**Transpose of Matrix:** If A = [a_{ij}] 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 A^{I} or A^{T}.

**Note:** (i) (A^{I})^{I} = A (ii) (k A^{I}) = k . A^{I} (iii) (A + B )^{T} = A^{T} + B^{T} (iv) (AB)^{T} = B^{T}A^{T}

**Symmetric Matrix:** A square matrix A is said to be symmetric if A^{T }=A

If A is a symmetric matrix, then A + A^{T} is symmetric.

**Skew-Symmetric Matrix:** A square matrix A is said to be skew-symmetric if A^{T }= -A

If A is a skew-symmetric matrix, then A – A^{T} is skew-symmetric.

** Minor of an element:** Consider a square matrix

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 a_{3} is = b_{1}c_{2} – b_{2}c_{1}

Minor of b_{2 }is = a_{1}c_{3} – a_{3}c_{1}

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

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

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.

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.

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 (A^{I}).

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

For any positive integer n Det (A^{n}) = (DetA)^{n}.

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

∴ A is singular matrix

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 = and cofactor matrix of A =

** ****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.

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

**⋇ **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** **

⋇ sin (A + B) sin (A – B) = sin^{2 }A – sin^{2} B = cos^{2 }B – cos^{2} A

⋇ cos (A + B) cos (A – B) = cos^{2 }A – sin^{2} B = cos^{2 }B – sin^{2} A

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.

∎ If is not an add multiple of

In ∆ABC,

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.

In ∆ABC,

⟹ 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.

In ∆ABC,

In ∆ABC,

a = b cos C + c cos B

b = a cos C + c cos A

c = a cos B + b cos A

In ∆ABC, a, b, and c are sides

**⨂ **The function f: R→R defined by f(x) = ∀ x ∈ R is called the ‘hyperbolic sin’ function. It is denoted by Sinh x.

Similarly,

**⨂ **cosh^{2}x – sinh^{2} x = 1

cosh^{2}x = 1 + sinh^{2} x

sinh^{2} x = cosh^{2} x – 1

**⨂ **sech^{2} x = 1 – tanh^{2} x

tanh^{2} x = 1 – sesh^{2} x

**⨂ **cosech^{2} x = coth^{2} x – 1

coth^{2} x = 1 + coth^{2} x

**⨂ **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** **

**⨂ **sinh 2x = 2 sinh x cosh 2x** = **

**⨂ **cosh 2x = cosh^{2}x + sinh^{2} x = 2 cosh^{2}x – 1 = 1 + 2 sinh^{2}x =

**Inverse hyperbolic functions:**

The equation x^{2} + 1 = 0 has no roots in real number system.

∴ scientists imagined a number ‘i’ such that i^{2} = − 1.

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 z
_{1}= x_{1}+ i y_{1}, z_{2}= x_{2}+ i y_{2}, then - z
_{1 + }z_{2}= (x_{1}+ x_{2}, y_{1}+ y_{2}) = (x_{1}+ x_{2}) + i (y_{1}+ y_{2}) - z
_{1 − }z_{2}= (x_{1}− x_{2}, y_{1}− y_{2}) = (x_{1}− x_{2}) + i (y_{1}− y_{2}) - z
_{1∙ }z_{2}= (x_{1}x_{2}−y_{1}y_{2}, x_{1}y_{2 }+ x_{2}y_{1}) = (x_{1}x_{2}−y_{1}y_{2}) + i (x_{1}y_{2}+x_{2}y_{1}) - z
_{1/ }z_{2 = (}x_{1}x_{2}+ y_{1}y_{2}/x_{2}^{2}+y_{2}^{2}, x_{2}y_{1 }– x_{1}y_{2}/ x_{2}^{2}+y_{2}^{2})

= (x_{1}x_{2} + y_{1} y_{2}/x_{2}^{2} +y_{2}^{2}) + i (x_{2} y_{1 }– x_{1}y_{2}/ x_{2}^{2} +y_{2}^{2})

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

z = x + i y then 1/z = x – i y/ x^{2} + y^{2}

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

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

If z_{1}, z_{2} are two complex numbers then

**Modulus: –** If z = x + i y, then the non-negative real number is called modulus of z and it is denoted by 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

x^{2} + y^{2} = r^{2} cos^{2}θ + r^{2} sin^{2}θ = r^{2} (cos^{2}θ + sin^{2}θ) = r^{2}(1)

⇒ x^{2} + y^{2} = r^{2}

• Arg (z) = tan^{−1}(y/x)

• Arg (z_{1}.z_{2}) = Arg (z_{1}) + Arg (z_{2}) + nπ for some n ∈ { −1, 0, 1}

• Arg(z_{1}/z_{2}) = Arg (z_{1}) − Arg (z_{2}) + nπ for some n ∈ { −1, 0, 1}

Note:

∎ e^{iθ} = cos θ + i sin θ

∎ e^{−iθ} = cos θ − i sin θ

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) = cos^{2}θ – i^{2} sin^{2}θ = cos^{2}θ + sin^{2}θ = 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

**n ^{th} root of a complex number: **let n be a positive integer and z

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

let Then a_{0}, a_{1}, a_{2, …, }a_{n-1} are all n distinct n^{th} roots of z and any n^{th} root of z is coincided with one of them.

**n ^{th} root of unity: **Let n be a positive integer greater than 1 and

**Note:**

- The sum of the n
^{th}roots of unity is zero. - The product of n
^{th}roots of unity is (– 1)^{ n – 1}. - The n
^{th}roots of unity 1, ω, ω^{2}, …, ω^{n-1}are in geometric progression with common ratio ω.

x^{3} – 1 = 0 ⇒ x^{3} = 1

x =1^{1/3}

ω^{2} +ω + 1 = 0 and ω^{3} = 1

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

If A + B + C = π or 180^{0}, 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

⋇ sin = cos; sin = cos ; sin = cos

⋇ cos = sin; cos = sin; cos = sin

⋇ 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

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.

Let a system of simultaneous equations be

a_{1 }x + b_{1} y + c_{1}z = d_{1}

a_{2 }x + b_{2} y + c_{2}z = d_{2}

a_{3 }x + b_{3} y + c_{3}z = d_{3}

The matrix form of the above equations is

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

Let system of simultaneous equations be

a_{1 }x + b_{1} y + c_{1}z = d_{1}

a_{2 }x + b_{2} y + c_{2}z = d_{2}

a_{3 }x + b_{3} y + c_{3}z = d_{3}

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

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

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

This method is called Cramer’s Method

Let a system of simultaneous equations be

a_{1 }x + b_{1} y + c_{1}z = d_{1}

a_{2 }x + b_{2} y + c_{2}z = d_{2}

a_{3 }x + b_{3} y + c_{3}z = d_{3}

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

This Matrix is reduced to the standard form ofby using row operations

- Interchanging any two rows
- Multiplying the elements of any two elements by a constant.
- 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.

- Take the coefficient of x as the unity as a first equation.
- If 1 is there in the first-row first column, then make the remaining two elements in the first column zero.
- After that, if one element in R
_{2}or R_{3}is 1, then make the remaining two elements in that column C_{2}or C_{3}as zeroes. - 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.

PDF FILE: Mathematics Notes 4 Polytechnic SEM – I

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**Errors and Approximations**

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

**Sol: **

Given y = f(x) = x^{2} + 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 – (10^{2} + 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

**Sol: **

Given y = cos x, x = 60^{0} and ∆x = 1^{0}

∆y = f (x + ∆x) – f (x)

= cos (60^{0} + 1^{0}) – cos 60^{0}

= cos (61^{0}) – 0.5

= 0.4848 – 0.5

= – 0.0152

dy = f’ (x) ∆x

= – sin x (1^{0})

= – sin 60^{0} × 0.0174

=– 0.8660 × 0.0174

= – 0.0150

**Sol: **

y = x^{2} + 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 – (10^{2} + 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

**Sol: **

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 = x^{2}

** **x = 3 and ∆x = 0.01

∆A = 2x × ∆x

= 2(3) (0.01)

= 6 × 0.01

= 0.06

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

A = x^{2}

∆A = 2x × ∆x

The approximate percentage error in area A

= 2 × 2 =4

From the following. Find the approximations

**Sol: **

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

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= 10 – 0. 0033

= 9.9967

**Sol:**

**Sol:**

**(iv) **Sin 62^{0}

**Sol: **

Let f(x) = sin x, where x = 60^{0} and ∆x =2^{0}

Approximate value is

f (x + ∆x) = f(x) + f’ (x) ∆x

= sin 60^{0} + cos x (2^{0})

= sin 60^{0}+ cos 60^{0} (0.0348)

= 0.8660 + 0.5 × 0.0348

= 0.8660 + 0.0174

=0.8834

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 r^{2}

∆A = 8π r ∆r

= 8 ×3.14× 14 × 0.02

= 7.0336

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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) = x^{2} 2^{x} log x

**Sol:**

Given f(x) = x^{2} 2^{x} log x

f’ (x) = ^{ }(x^{2}) 2^{x} log x + x^{2} (2^{x}) log x + x^{2} 2^{x} (log x).

= 2×2^{x} log x +x^{2} 2^{x} log a log x + x^{2} 2^{x} (1/x)

= x 2^{x}[log x^{2} + x log x log 2 + 1]

**Sol:**

f’ (x) = . log 7 ^{ }(x^{3} + 3x)

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

**Sol:**

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

= sec x

Find the derivative of the following functions

**(i) **f(x) = e^{x} (x^{2} + 1)

**Sol:**

Given f(x) = e^{x} (x^{2} + 1)

f’ (x) = e^{x} (x^{2} + 1) + (x^{2} + 1) ^{ }(e^{x})

= e^{x} (2x + 0) + (x^{2} + 1) e^{x}

= e^{x} (x^{2} + 2x + 1)

= e^{x} (x + 1)^{2}

**(iii) **cos (log x + e^{x})

**(iv) **x = tan (e^{-y})

e^{-y} = tan^{-1} x

**(v) **cos [log (cot x)]

**(vi) **sin[tan^{-1}(e^{x})]

**(vii) **cos^{-1}(4x^{3} – 3x)

let y = cos^{-1}(4x^{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

Find f’ (x), If f(x) = (x^{3} + 6 x^{2} + 12x – 13)^{100}.

**Sol:**

Given f(x) = (x^{3} + 6 x^{2} + 12x – 13)^{100}

f’ (x) = 100(x^{3} + 6 x^{2} + 12x – 13)^{99} (x^{3} + 6 x^{2} + 12x – 13)

= 100(x^{3} + 6 x^{2} + 12x – 13)^{99} (3x^{2} + 12 x + 12 – 0)

=100(x^{3} + 6 x^{2} + 12x – 13)^{99} 3 (x^{2} + 4 x + 4)

= 300 (x + 2)^{2} (x^{3} + 6 x^{2} + 12x – 13)^{99}

If f(x) = 1 + x + x^{2} + x^{3} + …. + x^{100}, then find f’ (1).

**Sol:**

Given f(x) = 1 + x + x^{2} + x^{3} + …. + x^{100}

f’(x) = 0 + 1 + 2x + 3 x^{2} + … 100 x^{99}

f’(1) = 1 + 2 + 3 + … + 100

= 50 × 101

= 5050

From the following functions. Find their derivatives.

**Sol:**

**Sol:**

Given y = log (cosh 2x)

If x = a cos^{3} t, y = a sin^{3} t, find

**Sol:**

Given If x = a cos^{3} t, y = a sin^{3} t

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

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

put x = tan θ ⟹ θ = tan^{-1} x

**Sol: **

The post Differentiation(2m Q & S) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>

**Limits**

**Question 1**

**Sol:**

= 9

**Question 2**

**Sol:**

= a + a = 2a

**Question 3**

**Sol:**

**Question 4**

**Sol:**

we know that if x > 0

As x → 0+ ⟹ x > 0

As x → 0+ ⟹ x < 0

**Question 5**

If f (x) = , then find and . Does exist?

**Sol:**

**Question 6**

**Sol:**

As x → 2– ⟹ x < 2

x – 2 < 0

**Question 7**

**Sol: **

As x → 2+ ⟹ x > 2

As x → 2– ⟹ x < 2

**Question 8**

**Sol:**

**Question 9**

**Sol:**

**Question 10**

**Sol: **

**Question 11**

**Sol: **

**Question 12**

**Sol:**

**Question 13**

**Sol:**

let y = x – 1 ⟹ x = y + 1

then as x → 1, y → 0

**Question 14**

**Sol:**

**Question 15**

**Sol:**

**Question 16**

**Sol: **

As x → ∞, →0

= ∞ (3/4)

= ∞

**Question 17**

**Sol:**

We know that – 1 ≤ sin x ≤ 1

x^{2} – 1 ≤ x^{2} – sin x ≤ x^{2} + 1

**Question 18**

**Sol:**

**Question 19**

**Sol:**

**Question 20**

**Sol:**

**Question 21**

**Sol: **

**Question 22**

**Sol: **

**Question 23**

**Sol:**

The post Limits (Q’s & Ans) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>

**Hyperbolic Functions**

Prove that for any x∈ R, sinh (3x) = 3 sinh x + 4 sinh^{3} x

**Sol:**

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

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

= (2 sinh x cosh x) cosh x + (1 + 2 sinh^{2} x) sinh x

= 2sinh x cosh^{2} x + sinh x + 2 sinh^{3} x

= 2 sinh x (1 + sinh^{2} x) + sinh x + 2 sinh^{3} x

= 2 sinh x + 2 sinh^{3} x+ sinh x + 2 sinh^{3} x

= 3 sinh x + 4 sinh^{3} x

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

**Sol:**

Cosh 2x = 2 cosh^{2} x – 1

Sinh^{2} 2x = cosh^{2} 2x – 1

If cosh x = sec θ then prove that tanh^{2}= tan^{2}

**Sol: **

If sinh x = 5, then show that x =

**Sol: **

Given, sinh x = 5

⟹ x = sinh^{-1}5

**Sol: **

Given tanh^{-1}

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)

For any x∈ R, prove that cosh^{4} x – sinh^{4} x = cosh 2x

**Sol:**

cosh^{4} x – sinh^{4} x = (cosh^{2} x)^{2} – (sinh^{2} x)^{2}

= (cosh^{2} x + sinh^{2} x) (cosh^{2} x – sinh^{2} x)

= 1. cosh 2x

= cosh 2x

**Sol: **

**=** cosh x + sinh x

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

**Sol: **

Given sin hx = ¾

We know that cosh^{2} x = 1 + sinh^{2} x

= 1 + (3/4)^{2}

= 1 + 9/16

= 25/16

cos hx = 5/4

cosh 2x = 2cosh^{2} 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

Prove that (cosh x – sinh x)^{ n} = cosh nx – sinh nx

**Sol: **

∴ (cosh x – sinh x)^{ n} = cosh nx – sinh nx

The post Hyperbolic Functions V.S.A.Q.’S first appeared on Basics In Maths.]]>

**Trigonometric Ratios Up to Transformations**

Find the value of sin^{2}(π/10) + sin^{2}(4π/10) + sin^{2}(6π/10) + sin^{2}(9π/10)

**Sol:**

** **sin^{2}(π/10) + sin^{2}(4π/10) + sin^{2}(6π/10) + sin^{2}(9π/10)

= sin^{2}(π/10) + sin^{2}(π/2 – π/10) + sin^{2}(π/2+ π/10) + sin^{2}(π – π/10)

= sin^{2}(π/10) + cos^{2}(π/10) + cos^{2}(π/10) + sin^{2}(π/10)

= 1 + 1 = 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

cos^{2}θ = 1 – sin^{2} θ

=1 – (4/5)^{2}

= 1 – 16/25

∴cos θ = – 3/5 (∵cos θ < 0)

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} = 5^{2} + x^{2}

9 sin^{2} θ + 16 cos^{2} θ + 12 sin θ cos θ + 16 sin^{2} θ + 9 cos^{2} θ – 12sin θ cis θ = 25 + x^{2}

25 sin^{2} θ + 25 cos^{2} θ = 25 + x^{2}

25 = 25 + x^{2}

⇒ x^{2} = 0

x = 0

∴ 4 sin θ – 3cos θ = 0

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 sec^{2} θ – tan^{2} θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

(1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) =

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) =

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

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

If cos θ + sin θ = cos θ, then prove that cos θ – sin θ = sin θ

**Sol:**

( + 1) sin θ = ( + 1) ( – 1) cos θ

Find the value of 2(sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ)

**Sol:**

2(sin^{6} θ + cos^{6} θ) – 3 (sin^{4} θ + cos^{4} θ)

= 2[(sin^{2} θ)^{3} + (cos^{2} θ)^{3}] – 3[(sin^{2} θ)^{2} + (cos^{2})^{2}

= 2[(sin^{2} θ + cos^{2} θ)^{3} – 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ)] – 3[(sin^{2} θ + cos^{2} θ)^{2} – 2 sin^{2} θ cos^{2} θ]

= 2[1 – 3 sin^{2} θ cos^{2} θ] – 3 [1 – 2 sin^{2} θ cos^{2} θ]

= 2 – 6 sin^{2} θ cos^{2} θ – 3 + 6 sin^{2} θ cos^{2} θ

= – 1

If tan 20^{0} = λ, then show that

**Sol:**

Given tan 20^{0} = λ

If sin α + cosec α = 2, find the value of sin^{n} α + cosec^{n} α, n∈ Z

**Sol:**

Given sin α + cosec α = 2

⇒ sin α + 1/ sin α = 2

sin^{2} α + 1= 2 sin α

sin^{2} α – 2 sin α + 1= 0

(sin α – 1 )^{2} = 0

⇒ sin α – 1 = 0

sin α = 1 ⇒ cosec α = 1

sin^{n} α + cosec^{n} α = (1)^{n} + (1)^{n} =1 + 1 =2

∴ sin^{n} α + cosec^{n} α = 2

Evaluate sin^{2} **+ cos**^{2} ** – tan**^{2}

**Sol:**

Find the value of sin 330^{0}. cos 120^{0} + cos 210^{0}. Sin 300^{0}

**Sol:**

** **sin 330^{0}. cos 120^{0} + cos 210^{0}. Sin 300^{0}

**=**sin (360^{0} – 30^{0}). cos (180^{0} – 60^{0}) + cos (180^{0} + 30^{0}). sin (360^{0} – 60^{0})

**= **(– sin 30^{0}). (– cos 60^{0}) + (– cos30^{0}). (– sin60^{0})

**= **sin 30^{0}. cos 60^{0} + cos30^{0}. Sin60^{0}

**= **sin (60^{0} + 30^{0}) = sin 90^{0}

**=**1

Prove that cos^{4} α + 2 cos^{2} α = (1 – sin^{4} α)

**Sol: **

= cos^{4} α + 2 cos^{2} α (1 – cos^{2} α)

= (cos^{2} α)^{2} + 2 (1 – sin^{2} α) (sin^{2} α)

= (1 – sin^{2} α)^{2} + 2 sin^{2} α – 2sin^{4} α

= 1 + sin^{4} α – 2 sin^{2} α + 2 sin^{2} α – 2sin^{4} α

= 1 – sin^{4} α

Eliminate θ from x = a cos^{3} θ and y = b sin^{3} θ

**Sol: **

Given x = a cos^{3} θ and y = b sin^{3} θ

cos^{3} θ = x/a and sin^{3} θ = y/b

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

we know that sin^{2} θ + cos^{2} θ = 1

⇒ [(y/b)^{1/3}]^{2} + [(x/a)^{1/3}]^{2} = 1

(x/a)^{2/3} + (y/b)^{2/3} = 1

Find the period of the following functions

**Sol:**

**(i)** f(x) = tan 5x

we know that period of tan kx =

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

**(iv)** f(x) = tan (x + 4x + 9x +…. + n^{2}x)

f(x) = tan (x + 4x + 9x +…. + n^{2}x)

= tan (1 + 4 + 9 + … + n^{2}) x

we know that period of tan kx =

Prove that sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0} =

**Sol:**

We know that sin^{2} A – sin^{2}B = sin (A +B) sin (A – B)

⇒ sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0}

= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)

= sin 75^{0} sin 30^{0}

∴ sin^{2}(52 ½)^{0} – sin^{2} (22 ½)^{0} =

Prove that tan 70^{0} – tan20^{0} = 2 tan 50^{0}

**Sol:**

50^{0} = 70^{0} – 20^{0}

Tan 50^{0 }= tan (70^{0} – 20^{0})

⇒ tan 70^{0} – tan 20^{0} = tan 50^{0} (1 + tan70^{0} tan 20^{0})

tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + tan70^{0} cot (90^{0} – 20^{0})]

tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + tan70^{0} cot 70^{0}]

tan 70^{0} – tan 20^{0} = tan 50^{0} [1 + 1]

∴ tan 70^{0} – tan20^{0} = 2 tan 50^{0}

If sin α = , sin β = and α, β are acute, show that α + β =

**Sol: **

tan α = 1/3 tan β = ½

tan (α + β) = 1

**Sol:**

**Sol:**

(on dividing numerator and denominator by cos 9^{0})

= tan (45^{0} + 9^{0})

= tan 54^{0}

= tan (90^{0} – 36^{0})

= cot 36^{0}

Show that cos 42^{0} + cos 78^{0} + cos 162^{0} = 0

**Sol:**

cos 42^{0} + cos 78^{0} + cos 162^{0}

= cos (60^{0} – 18^{0}) + cos (60^{0} + 18^{0}) + cos (180^{0} – 18^{0})

=cos 60^{0} cos18^{0} + sin 60^{0} sin 18^{0} + cos 60^{0} cos 18^{0} – sin 60^{0} sin 18^{0} – cos 18^{0}

= 2 cos 60^{0} cos 18^{0} – cos 18^{0}

= 2 (1/2) cos 18^{0} – cos 18^{0}

= cos 18^{0} – cos 18^{0}

= 0

Express sin θ + cos θ as a single of an angle

**Sol:**

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

** **= 2(cos 30^{0} sin θ + sin 30^{0} cos θ)

= 2 sin (θ + 30^{0})

Find the maximum and minimum value of the following functions

**(i) **3 sin x –4 cos x

a= 3, b = –4 and c = 0

= 5

∴ minimum value = –5 and maximum value = 5

(**ii) **cos (x + ) + 2 sin (x + ) – 3

a= 1, b = 2 and c = – 3

∴ minimum value = –6 and maximum value = 0

** Question 23**

**Sol: **

Given f(x) = 7 cos x – 24sin x + 5

a= 7, b = –24 and c = 5

∴ Range = [–20, 30]

Prove that sin^{2}α + cos^{2} (α + β) + 2 sin α sin β cos (α + β) is independent of α

**Sol:**

sin^{2}α + cos^{2} (α + β) + 2 sin α sin β cos (α + β)

= sin^{2}α + cos (α + β) [ cos (α + β) +2 sin α sin β]

= sin^{2}α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]

=sin^{2}α + cos (α + β) [ cos α cos β + sin α sin β]

=sin^{2}α + cos (α + β) cos (α –β)

= sin^{2} α + cos^{2} α – sin^{2} β

=1 – sin^{2} β

= cos^{2} β

**Sol:**

= tan θ

For what values of x in the first quadrant is positive?

**Sol:**

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

⟹ 0 < x < π/4

If cos θ = and π < θ < 3π/2, find the value of tan θ/2.

**Sol:**

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

= – 2

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

**Sol:**

= 2 cot 2A

Evaluate 6 sin 20^{0} – 8sin^{3} 20^{0}

**Sol:**

6 sin 20^{0} – 8sin^{3} 20^{0} = 2 (3 sin 20^{0} – 4sin^{3} 20^{0})

= 2 sin 3(20^{0})

= 2 sin 60^{0}

Express cos^{6} A + sin^{6} A in terms of sin 2A.

**Sol:**

cos^{6} A + sin^{6} A

**= **(sin^{2} A)^{3} + (cos^{2} A)^{3}

= (sin^{2} A + cos^{2} A)^{3} – 3 sin^{2} A cos^{2} A (sin^{2} A + cos^{2} A)

= 1 – 3 sin^{2} A cos^{2} A

=1 – ¾ (4 sin^{2} A cos^{2} A)

= 1 – ¾ sin^{2}2 A

If 0 < θ < π/8, show that = 2 cos (θ/2)

**Sol:**

=2 cos (θ/2)

Find the extreme values of cos 2x + cos^{2}x

**Sol:**

cos 2x + cos^{2}x = 2cos^{2} x– 1 + cos^{2} x

=3cos^{2} x – 1

We know that – 1 ≤ cos x ≤ 1

⟹ 0 ≤ cos^{2} x ≤ 1

3×0 ≤ 3×cos^{2} x ≤ 3×1

0– 1 ≤3 cos^{2} x – 1≤ 3– 1

– 1≤3 cos^{2} x – 1≤ 2

Minimum value = – 1

Maximum value = 2

**Sol:**

**= **4

Prove that sin 78^{0} + cos 132^{0} =

**Sol:**

** **sin 78^{0} + cos 132^{0} = sin 78^{0} + cos (90^{0} + 42^{0})

= sin 78^{0} – sin 42^{0}

= 2 cos 60^{0} sin 18^{0}

Find the value of sin 34^{0} + cos 64^{0} – cos4^{0}

**Sol:**

sin 34^{0} + cos 64^{0} – cos4^{0}

= sin 34^{0} – 2sin 34^{0} sin 30^{0}

= sin 34^{0} – 2 sin 34^{0} (1/2)

=sin 34^{0} – sin 34^{0}

=0

Prove that 4(cos 66^{0} + sin 84^{0}) =

**Sol:**

4(cos 66^{0} + sin 84^{0})

=4(cos 66^{0} + sin (90^{0 }– 6^{0})

=4(cos 66^{0} + cos (6^{0})

=8 cos 36^{0} cos 30^{0}

Prove that (tan θ + cot θ)^{2} = sec^{2} θ + cosec^{2} θ = sec^{2} θ. cosec^{2} θ

**Sol:**

= sec θ. cosec θ

(tan θ + cot θ)^{2} = sec^{2} θ. cosec^{2} θ

= sec^{2} θ. cosec^{2} θ

The post Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>

**Question 1**

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

**Sol: **

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

If θ is the angle between the vectors **a** and **b**, then cos θ =

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

** = **12 – 18 + 18 = 12

= 7

= 11

**Question 2**

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

**Sol: **

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

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

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

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

= – 8 + 3 + 5

= 0

∴ **a** + **b **and **a** – **b** is perpendicular to each other.

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

**Sol: **

Squaring on both sides

(**a** + **b**) (**a** + **b**) = (**a – b**) (**a – b**)

**a**^{2} + 2 **a**. **b** + **b**^{2} = **a**^{2} – 2 **a**.**b** + **b**^{2}

** **⟹ 4 **a**.**b** = 0

**a**.**b** = 0

∴ the angle between **a** and **b ** is 90^{0}

If = 11, = 23 and = 30, then find the angle between the vectors** a** and **b** and also find

**Sol: **

(11)^{2} **– **2 ×11×23 cos θ + (23)^{2} = 900

121 **– **506 cos θ + 529 = 900

650 **– **506 cos θ = 900

= (11)^{2} **+ **2 ×11×23 cos θ + (23)^{2}

= 121 + 2 ×11×23 × + 529

= 400

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

**Sol: **

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

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

** **The projection vector of **b** on **a = **

** The magnitude** of the projection vector = =

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

**Sol:**

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

Given, that **a **and **b **are perpendicular to each other

⟹ **a**.**b** = 0

(λ **i** – 3**j** + 5**k**). (2λ **i** – λ **j** – **k**) = 0

2 λ^{2} + 3 λ – 5 = 0

2 λ^{2} + 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

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

**Sol:**

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

⟹ **OP** = x**i** + y**j** + z**k**

**OA** = – **2i** +**j** +3**k**

**AP** = **OP** – **OA** = (x**i** + y**j** + z**k**) – (– **2i** +**j** +3**k**)

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

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

⟹ 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

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**. (2**i** – 3**j** – 6**k**) = 5 and **r**. (6**i** + 2**j** – 9**k**) = 4

⟹ **n _{1}** = 2

If θ is the angle between the planes **r**. **n _{1}** = d

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

**Sol:**

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

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

⟹ **a** + **b **= – c

⟹ **c** = – (**a** + **b**)

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

= – (3**i** –4 **j** –4**k**)

∴ **c** = – 3**i** +4 **j** + 4**k**

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

**Sol:**

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

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

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

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

**r**. (4**i **+ 7**j **– 4**k**)– (12 – 14 – 4) = 0

**r**. (4**i **+ 7**j **– 4**k**)– 6 = 0

**r**. (4**i **+ 7**j **– 4**k**) = 6

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

**Sol: **

The vector which is parallel to the XOY-plane is of the form x**i** + y**j **

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

is 3**i** + 4**j **

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

If **a** + **b** + **c** = 0, = 3, = 5 and = 7, then find the angle between **a** and **b**

**Sol:**

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

**a** + **b** = – **c**

3^{2} + 5^{2} + 2 cos θ = 7^{2}

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

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

**Sol:**

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

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

** = **–26**i – **9**j **+ 5**k**

The unit vector perpendicular to both **a** and **b** =

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

**Sol:**

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

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

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

** = **–17**i + **10** j **– **k**

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

**Sol: **

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

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

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

** ** **=**3** i +**2 **j** +9 **k**

∴ The area of the parallelogram =

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

**Sol:**

= 1 – cos^{2} θ

= sin^{2} θ

= 1 + cos^{2} θ – 2cos^{2} θ

= 1– cos^{2} θ

= sin^{2} θ

**Sol: **

Let **a** = x**i** + y**j** + z**k**

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

** = **– y**k **+ z**j**

Similarly, ** ** = x^{2} + z^{2}** **and** ** = y^{2} + x^{2}** **

** + ****+** ** = y**^{2} + z^{2} + x^{2} + z^{2} + y^{2} + x^{2}** **

= 2(x^{2} + y^{2} +z^{2})

If = 2, = 3 and (**p**, **q**) = , then find

**Sol:**

= 2 × 3 sin

= 2 × 3×1/2

= 3

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

**Sol:**

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

⇒ p = 12

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**× **c** –**a **× **b **–** a** × **c **–** b** ×**c**

**= 0**

Compute 2**j**× (3**i – **4**k**) + (**i** + 2**j**) × **k**

**Sol:**

2**j**× (3**i – **4**k**) + (**i** + 2**j**) × **k **

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

= **– **6**k – **8**i** **– j **+ 2**i**

= **– **6**i –j** **–**6** j**

The post Product of Vectors (Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>

**Addition of Vectors**

Find the unit vector in the direction of

The unit vector in the direction of a vector is given by

Find a vector in the direction of **a where** that has a magnitude of 7 units.

The unit vector in the direction of a vector is

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

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

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

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

Write the direction cosines of the vector

**QUESTION 5**

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

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

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

** AB = **3**a **–** b **– 2**c**

** AC **= **OC – OA **= 7** a** – **c** – **(**– 2**a **+ 3**b **+ 5**c)**

** AC = **9**a **– 3**b **– 6**c =** 3**(**3**a **–** b **– 2**c)**

** AC = **3 **AB**

** **A, B and C are collinear

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

(i) Take A as the origin

M is the midpoint of CD

** = AD **+ ½** AB **(∵ AB = DC)

L is the midpoint of BC

** = 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**

If G is the centroid of the triangle ABC, then show that **OG = ** when, are the position vectors of the vertices of triangle ABC.

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

** **D is the midpoint of BC

G divides median AD in the ratio 2: 1

If = , = are collinear vectors, then find m and n.

**Sol:** Given , are collinear vectors

Equating like vectors

2 = 4 λ; 5 = m λ; 1 = n λ

∴ m = 10, n = 2

Let If , . Find the unit vector in the direction of **a + b.**

** **The unit vector in the direction of **a + b **=

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

**Sol:** let **a **= – 3**i** + 4**j** + λ**k**, **b** = μ**i** + 8**j** + 6**k**

** ⟹ **** a **= t**b**

– 3**i** + 4**j** + λ**k** = t (μ**i** + 8**j** + 6**k)**

– 3**i** + 4**j** + λ**k** = μt **i + **8t **j + **6t **k**

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

∴ μ=– 6, λ = 3

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

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

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

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

= –4**i** + 2**j** + 2**k **+2**i** – **j** + **k**

= –2**i** + **j** + 3**k **

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

= 6**i** – 3**j** – 13**k **+2**i** – **j** + **k**

= 8**i** –4 **j** –12**k**

** = **– 4 (2**i** + **j** + 3**k)**

**AC = **– 4** AB**

Given** AB = **𝛌** AC**

** **𝛌 = – 1/4

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

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

** OD = OA **+** AB **+** BC **+** CD **

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

** = **7**i** + 2**j** +4**k**

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

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

** = **3**i** +6**j** –2**k**

** **The unit vector in the opposite direction of **a **+ **b** + **c **is

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

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

∴ Given vectors form an equilateral triangle.

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

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

**r** = (1 – t) **a**+ t **b **

** **3**a – **2**b = (**1 – t) **a**+ t **b **

** **Equating like vectors

1 – t = 3 and t = – 2

∴ Given points are collinear.

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**

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**

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

D is mid of BC

The equation of AD is

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

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

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

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

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

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

Sol: The vector equation of the line passing through **a,** **b and c**is **r** = (1 – t – s) a + t**b **+ s**c**

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

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

The post Addition of Vectors (Qns.& Ans) V.S.A.Q.’S first appeared on Basics In Maths.]]>

**Matrices**

If A = , then show that A^{2} = –I

∴ A^{2} = –I

If A = , and A^{2} = 0, then find the value of k.

A^{2} = 0

8 + 4k = 0, – 2 – k = 0 and –4 + k^{2} = 0

4k = –8; k = –2; k^{2} = 4

k = –2; k = –2; k = ± 2

∴ k =– 2

Trace of A = 1 – 1 + 1 = 1

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

**Sol:** Given A = , B = and 2X + A = B

2X = B – A

Find the additive inverse of A, If A =

Additive inverse of A = – A

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

⟹ 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

Construct 3 × 2 matrix whose elements are defined by a_{ij} =

**Sol:**

Let A= _{ }

a_{11} = 1

a_{22} = 2

a_{31} = 0

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

**Sol:** Given Matrices are A = B =

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

AB and BA exist

AB and BA are not Commutative under Multiplication

Define Symmetric and Skew Symmetric Matrices

**Sol:**

Symmetric Matrix: Let A be any square matrix, if A^{T} = A, then A is called Symmetric Matrix

Skew Symmetric Matrix: Let A be any square matrix if A^{T} = –A, then A is called Skew Symmetric Matrix

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

**Sol:** Given, A = is a symmetric matrix

⟹ A^{T} = A

⟹ x = 6

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

**Sol:** Given A = is a skew-symmetric matrix

⟹ A^{T} = – A

⟹ x = –x

x+ x = 0 ⟹ 2x = 0

⟹ x = 0

If A = and B = , then find (A B^{T})^{ T}

If A = and B = , then find A + B^{T}

If A = , then show that AA^{T} = A^{T}A = I

∴ AA^{T} = A^{T}A = I

Find the minor of – 1 and 3 in the matrix

**Sol:** Given Matrix is

Find the cofactors 0f 2, – 5 in the matrix

**Sol:** Given matrix is

Cofactor of 2 = (–1)^{2 + 2} = –3 + 20 = 17

Cofactor of – 5 = (–1)^{3 + 2} = –1(2 – 5) = –1(–3) = 3

If ω is a complex cube root of unity, then show that = 0(where 1 + ω+ω^{2} = 0)

R_{1} → R_{1} + R_{2} + R_{3}

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

Det A = 45

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

⟹ 3x + 24 = 45

3x = 45 – 24

3x = 21

x = 7

Find the adjoint and inverse of the following matrices

(i)

(ii)

Det A = a (bc – 0) – 0(0 – 0) + 0(0 – 0)

Det A = abc ≠ 0

Adj A = (Cofactor matrix of A)^{ T}

Find the rank of the following matrices.

Det A = 1 (0 – 2) – 2(1 – 0) + 1(– 1 – 0)

= – 2– 2– 1

= – 5 ≠ 0

∴ Rank of A = 3

Det A = – 1 (24 – 25) + 2(18 – 20) + – 3(15 – 16)

= – 1– 4 + 3

= – 0

Det B = – 4 + 6 = 2 ≠ 0

∴ Rank of A = 2

Det of Sub matrix of A = – 1 – 0 = – 1 ≠ 0

∴ Rank of A = 2

Det of Sub matrix of A =1 (1 – 0) – 0(0 – 0) + 0(0 – 0)

= 1≠ 0

∴ Rank of A = 3

The post Matrices ( Qns & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>

The Plane (2m Questions & Solutions) || V.S.A.Q’S|| Read More »

The post The Plane (2m Questions & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>Inter Maths – 1B two mark 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.

Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (2, 3, – 5).

Sol:

The plane passes through A and perpendicular to OA, then the line segment OA is normal to the plane.

Dr’s of OA = (2, 3, – 5)

The equation of the plane passing through point (x_{1}, y_{1}, z_{1}) and dr’s (a, b, c) is

a(x – x_{1}) + b (y – y_{1}) + c (z – z_{1}) = 0

⟹ 2(x – 2) + 3 (y – 3) – 5 (z + 5) = 0

2x – 4 + 3y – 9 – 5z – 25 = 0

2x + 3y – 5z – 38 = 0

Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4)

Sol:

The equation of the plane passing through the points (x_{1}, y_{1}, z_{1}) (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}) is

The plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4) is

x (30 – 20) – (y + 1) (20 – 6) + (z + 1) (40 – 18) = 0

x (10) – (y + 1) (14) + (z + 1) (22) = 0

10x – 14y – 14 + 22z + 22 = 0

10x – 14y + 22z + 8 = 0

2(5x – 7y + 11z + 4) = 0

∴ the equation of the plane is 5x – 7y + 11z + 4 = 0

Find the equation to the plane parallel to the ZX-plane and passing through (0, 4, 4).

Sol:

Equation of ZX-plane is y = 0

The equation of the plane parallel to the ZX-plane is y = k

But it is passing through (0, 4, 4)

⟹ y = 4

Find the equation to the plane passing through the point (α, β, γ) and parallel to the plane axe + by + cz + d = 0.

Sol:

The equation of the plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0

But it is passing through the point (α, β, γ)

a α + b β + c γ + k = 0

⟹ k = – a α – b β – c γ

The equation of the plane is ax + by + cz – a α – b β – c γ = 0

⟹ a(x – α) + b (y – β)+ c (z – γ) = 0

Find the angle between the planes 2x – y + z = 6 and x + y + 2z = 7.

Sol: If θ is the angle between the planes a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 and a_{2}x + b_{2} y + c_{2} z + d_{2} = 0, then cos θ =

Cos θ = cos 60^{0}

θ = 60^{0} =

Reduce the equation x + 2y – 2z – 9 = 0 to the normal form and hence find the dc’s of the normal to the plane.

Sol: Given plane is x + 2y – 2z – 9 = 0

x + 2y – 2z = 9

dc’s of the normal to the plane are

Suppose a plane makes intercepts 2, 3, 4 on X, Y, Z axes respectively. Find the equation of the plane in the intercept form.

Sol: Given a = 2, b = 3, c = 4

The equation of the line in the intercept form is

Express x – 3y + 2z = 9 in the intercept form

Sol: Given plane is x – 3y + 2z = 9

a = 9, b = – 3, c = 9/2

Find the direction cosine of the normal to the plane x + 2y + 2z – 4 = 0.

Sol: Given plane is x + 2y + 2z – 4 = 0

We know that Dr’s of the normal to the plane ax + by + cz + d = 0 are (a, b, c)

⟹ dc’s of the normal to the plane =

⟹ dr’s of the normal to the plane x + 2y + 2z – 4 = 0 are (1, 2, 2)

⟹ dc’s of the normal to the plane are

Find the midpoint of the line joining the points (1, 2, 3) and (–2, 4, 2)

Sol: Given points are A (1, 2, 3), B (–2, 4, 2)

Show that the points A ( – 4, 9, 6), B (– 1, 6, 6), and C(0, 7, 10) right-angled isosceles triangle.

Sol:

Distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is PQ =

AB^{2} + BC^{2 } = 18 + 18 = 36 = AC^{2}

∴ ∠ B = 90^{0}

points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle

Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 = 0

Sol:

Let P (x, y, z) be the locus of the point

A (0, y, 0) be any point on Y – axis

B = (1, 2, – 1)

Condition is PA = 3PB

PA^{2} = (3PB)^{2}

z^{2} = 9[x^{2 }– 2x + 1 + y^{2} – 4y + 4 + z^{2} + 2z + 1]

x^{2} + z^{2} = 9x^{2 }– 18x + 9 + 9y^{2} – 36y + 36 +9 z^{2} + 18z + 9

∴ 8x ^{2} + 9y^{2} + 8 z^{2} – 18x – 36y + 18z + 54 =0

A, B, C are three points on OX, OY and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C and O

Sol:

Let P (x, y, z) be the required point

O = (0, 0, 0) A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)

Given, AP = BP = CP = OP

AP = OP

⟹ AP^{2} =OP^{2}

(x – a )^{2 }+ y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

x^{2} + a^{2} – 2ax + y^{2} + z^{2} = x^{2} + y^{2} + z^{2}

a^{2} – 2ax = 0

a (a – 2x) = 0

a – 2x = 0 (∵ a≠0)

a = 2x ⟹ a/2

Similarly, y = b/2 and z = c/2

∴ P = (a/2, b/2, c/2)

Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear

Sol:

Given points are A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2)

AB + BC = AC

∴ A B and C are collinear

Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.

Sol:

Let A = (5, – 1, 7), B = (x, 5, 1)

Given AB = 9

⟹ AB^{2} = 81

(5 – x)^{2} + (– 1 – 5)^{2} + (7 – 1)^{2} = 81

(5 – x)^{2 }+ 36 + 36 = 81

(5 – x)^{2} + 72 = 81

(5 – x)^{2 }= 81 – 72 = 9

(5 – x) = ± 3

5 – x = 3 or 5 – x = – 3

5 – 3 = x or 5 + 3 = x

x = 2 or x = 8

If the point (1, 2, 3) is changed to the point (2, 3, 1) through the translation of axes. Find a new origin.

Sol:

Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)

x = X + h, y = Y + k, z = Z + l

h = x – X, k = y – Y, l = z – Z

h = 1 – 2, k = 2 – 3, l = 3 – 1

h = – 1, k = – 1, l = 2

New origin is (– 1, – 1, 2)

By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

Sol:

If a point P divides the line segment joining the points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) in the ratio, then P =

Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).

Sol:

Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)

Suppose P divides AB in the ratio k : 1

7 – 2k = k + 1

7 – 1 = k + 2k

6 = 3k

k = 2

∴ P divides AB in the Ratio 1 : 2

Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not.

Sol:

Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)

Let C divides AB in the ratio k : 1

k = 4 (k + 1)

2 – 4k = 4k + 4

– 4k– 4k = 4 – 2

– 8k = 2

K = -1/4

C divides AB in the Ratio 1 : 4 externally

∴ A, B, C are collinear

Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)

Sol:

The centroid of the triangle whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) is

∴ The centroid of the triangle =

Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)

Sol:

The centroid of the triangle whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is

Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)

Sol:

Let P be any point on the YZ-plane

P = (0, y, z)

Let P divides AB in the ratio k:1

3k + 2 = 0

3k =– 2

YZ-plane divides AB in the ratio – 2:3

Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).

Sol:

let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)

ABCD is a parallelogram

The midpoint of AC = Midpoint of BD

⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0

x = 3, y = 3, z = 1

∴ The fourth vertex D = (3, 3, 1)

A (5, 4, 6), B (1, –1, 3), C (3, 3, 1) are three points. Find the coordinates of the point in which the bisector of ∠BAC meets the side BC.

Sol:

We know that the bisector of ∠BAC divides BC in the ratio AB:AC

= 5:3

If D is a point where the bisector of ∠BAC meets BC

⟹ D divides BC in the ratio 5:3

If M (α, β, γ) is the midpoint of the line joining the points (x_{1}, y_{1}, z_{1}) and B, then find B

Sol:

Let B (x, y, z) be the required point

M is the midpoint of AB

⟹ 2 α = x + x_{1}; 2 β = y + y_{1}; 2 γ = z + z_{1}

x =2 α – x_{1}; x =2 β – y_{1;} x =2 γ – z_{1}

_{ }∴ B = (2 α – x_{1}, 2 β – y_{1}, 2 γ –)

If H, G, S and I respectively denote orthocenter, centroid, circumcenter and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.

Sol:

Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)

AB = BC = AC

⟹ ∆ ABC is an equilateral triangle

We know that, in an equilateral triangle orthocenter, centroid, circumcenter and incentre are same

= (2, 2, 2)

∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)

Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).

Sol:

Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)

a = 5

b = 4

c = 3

∴ I = (1, 1,0)

Find the ratio in which the point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also, find the harmonic conjugate of P

Sol:

Harmonic Conjugate: If P divides AB in the ratio m: n, then the Harmonic Conjugate of P (i.e., Q) divides AB in the ratio –m: n.

Given points are A (3, 2, – 4), B (9, 8, –10) and P (5, 4, – 6)

P divides AB in the ratio is x_{2} – x : x – x_{2}

= 3 – 5 : 5 – 9

= 1 : 2 (internally)

Let Q be the harmonic conjugate of P

⟹ Q divides AB in the ratio –1: 2

= (–3, –4, –2)

Q (–3, –4, –2) is the harmonic conjugate of P (5, 4, – 6)

If (3, 2, – 1), (4, 1,1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of the tetrahedron, find the fourth Vertex.

Sol:

Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)

Let the fourth vertex is D = (x, y, z)

The centroid of the tetrahedron whose vertices are (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) is

13 + x = 16, 5 + y = 8, 5 + z = 2

x = 3, y = 3, z = 3

∴ the fourth vertex D = (3, 3, 3)

Show that the points A(3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear and find the ratio in which B divides AC.

Sol:

Given points are A(3, 2, –4), (5, 4, –6) and C(9, 8, –10)

∴ the points A (3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear

B divides AB in the ratio is AB:BC = = 1:2

The post 3D Coordinates (Q’s & Ans) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>

Straight Lines (2m Questions & Solutions) || V.S.A.Q’S|| Read More »

The post Straight Lines (2m Questions & Solutions) || V.S.A.Q’S|| first appeared on Basics In Maths.]]>Prove that the points (1, 11), (2, 15), and (– 3, – 5) are collinear, and find the equation of the line containing them.

Sol:

Let A (1, 11), B (2, 15), and C (– 3, – 5)

** The slope of the line segment joining the points (x _{1}, y_{1}) and (x_{2}, y_{2}) is **

Slope of AB = = 4

**QUESTION2.**

Find the condition for the points (a, 0), (h, k), and (0, b) to be collinear.

Sol:

Let A (a, 0), B (h, k) and C (0, b)

**The slope of the line segment joining the points (x _{1}, y_{1}) and (x_{2}, y_{2}) is **Given that A, B, and C are collinear points

The slope of AB = The slope of BC

⟹ – hk = (h – a) ( b – k)

– hk = hb – hk – ab + ak

⟹ 0 = hb + ak – ab

Find the equations of the straight lines which makes intercepts whose sum is sum is 5 and product is 6.

Sol:

The equation of the line in the intercept form is

Given that, a + b = 5 and ab = 6

⟹ b = 5 – a

a(5 – a) = 6

5a – a^{2} = 6

a^{2} – 5a + 6 = 0

a – 3a – 2a + 6 = 0

a (a – 3) – 2(a – 3) = 0

(a – 3) (a – 2) = 0

a = 3 or a = 2

case (i) if a = 3 ⟹ b = 2

case (ii) if a = 2 ⟹ b = 3

Find the equation of the straight line which makes an angle 135^{0} with the positive X – axis measured countered clockwise and passing through the point (– 2, 3).

Sol:

Slope of the line m = tan 135^{0} = – 1

The point is (– 2, 3)

**The equation of the straight line in slope point form is (y – y _{1}) = m (x – x_{1})**

The equation of the line passing through the point (– 2, 3) with slope – 1 is

y – 3 = – 1 (x + 2)

y – 3 =– x – 2

x + y – 1 = 0

Find the equation of the straight line passing through the points (1, – 2) and (– 2, 3).

Sol:

Given points are (1, – 2), (– 2, 3)

**The equation of the straight line in two points form is (y – y _{1}) = **

The equation of required straight line is

– 3 (y + 2) = 5 (x – 1)

– 3y – 6 = 5x – 5

5x + 3y + 1 = 0

**QUESTION6.**

Find the slopes of the line x + y = 0 and x – y = 0

Sol:

**The slope of the line ax + by + c = 0 is ** ** **

The slope of the line x + y = 0 is = – 1

The slope of the line x – y = 0 is = 1

Find the angle which the straight-line y = x – 4 makes with the Y-axis.

Sol:

Compare with y = mx + c

Angle made by the line with X-axis is

Angle made by the line with Y-axis is

Find the equation of the reflection of the line x = 1 in the Y-axis.

Sol:

Given equation is x = 1

Reflection about the Y-axis is x =– 1

Required equation of the line is x + 1 = 0

**QUESTION9.**

Write the equations of the straight lines parallel to X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis.

Sol:

(i) The equation of the straight line parallel to X-axis which is at a distance of 3 units above the X-axis is y = 3

⟹ y – 3 = 0

(ii) The equation of the straight line parallel to X-axis which is at a distance of 4 units below the X-axis is y = – 4

⟹ y + 4 = 0

**QUESTION10.**

Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it.

Sol:

(i) The equation of the straight line parallel to the Y-axis which is at a distance of 2 units from the Y-axis to the right of it is x = 2

⟹ x – 2 = 0

(ii) The equation of the straight line parallel to the Y-axis which is at a distance of 5 units from the Y-axis to the left of it is x =– 5

⟹ x + 5 = 0

Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2.

Sol:

Given the slope of the line passing through (2, 5) and (x, 3) is 2.

**The slope of the line segment joining the points (x _{1}, y_{1}) and (x_{2}, y_{2}) is **

** **⟹ 2(x – 2) = – 2

x – 2 = – 1

x = 1

Find the equation of the straight line passing through (– 4, 5) and making non – zero intercepts on the coordinate axes whose sum is zero.

Sol:

The equation of a line in the intercept form is

Given a = b

x + y = a

but it is passing through (– 4, 5)

– 4+ 5 = a

a = 1

∴ The equation of the required straight line is x + y = 1

x + y – 1 = 0

**QUESTION13.**

Find the equation of the straight line passing through (– 2, 5) and cutting off equal and non – zero intercepts on the coordinate axes.

Sol:

The equation of a line in the intercept form is

Given a + b = 0

b = – a

x – y = a

but it is passing through (– 2, 4)

– 2– 4 = a

a = –6

∴ The equation of the required straight line is x – y = –6

x + y + 6 = 0

Sol:

The equation of the straight line in the normal form is x cos α + y sin α = p

p = 4 and α = 135^{0}

x cos 135^{0} + y sin 135^{0} = 4

**QUESTION15.**

Sol:

Given θ = 135^{0} and (h, k) = (3, 2)

The parametric equations are: x = h + r Cos θ, y = k +r Sin θ

x = 3 + r Cos 135^{0}, y = 2 + r Sin 135^{0}

(i) x + y +1 = 0

Sol:

Given equation is x + y +1 = 0

x + y =– 1

– x – y = 1

(ii) x + y = 2

Sol:

Given equation is x + y = 2

Sol:

Given equation is 3x + 4y = a

The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes is

a^{2 }= 24 ⟹ a = 12(∵ a > 0)

**QUESTION18.**

Find the sum of the squares of the intercepts of the line 4x – 3y = 12

Sol:

Given equation is 4x – 3y = 12

a = 3 and b =– 4

the sum of the squares of intercepts = a^{2} + b ^{2}

= 3^{2} + (– 4)^{2}

= 9 + 16 = 25

Find the angle made by the straight – line y = – x + 3 with the positive direction of the X-axis measured in the counter-clockwise direction.

Sol:

Give equation of straight line is y = x + 3

It is in the form of y = mx + c

**QUESTION 20.**

Find the equation of the straight line in the symmetric form, given the slope and point on the line (2, 3).

Sol:

Equation of the line in the symmetric form is

Given Point (x_{1}, y_{1}) = (2, 3) and slope m =

⟹ θ = 60^{0}

If the product of the intercepts made by the straight-line x tan α + y sec α = 1 (0≤ α < π/2) on the coordinate axes is equal to sin α, find α.

Sol:

Given equation is x tan α + y sec α = 1

a = cot α, b = cos α

product of intercepts = sin α

⟹ ab = sin α

cot α.cos α = sin α

(cos α)/ (sin α). cos α = sin α

⟹ cos^{2} α = sin^{2} α

⟹ tan^{2} α = 1

∴ α = 45^{0}

**QUESTION22.**

If the sum of the reciprocals of the intercepts made by a variable straight–line on the axes of coordinates is a constant, then prove that the line always passes through a fixed point.

Sol:

The equation of the line in the intercept form is

Let the sum of the reciprocals of intercepts is k

⟹

∴ the line always passes through a fixed point (1/k, 1/k).

**QUESTION23.**

Find the ratio in which the straight-line 2x + 3y – 20 = 0 divides the join of the points (2, 3) and (2, 10).

Sol:

Given equation is L ≡ 2x + 3y – 20 = 0

L_{11} = 2(2) + 3(3) = 4 + 9– 20 = – 7

L_{22} = 2(2) + 3(10) = 4 + 30 – 20 = 14

We know that the line L = 0 divides the line joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) in the ratio – L_{11} : L_{22}

= – (– 7) : 14

= 1 : 2

**QUESTION24.**

State whether (3, 2) (– 4, – 3) are on the same or opposite sides of the straight-line 2x – 3y + 4 = 0.

Sol:

Given equation is L ≡ 2x – 3y + 4 = 0

L_{11} = 2(3) – 3(2) + 4 = 6 – 6 + 4 = 4 > 0

L_{22} = 2(– 4) – 3(–3) + 4 = – 8 + 9 + 4 = 5 > 0

L_{11}, L_{22} has the Same sign

∴ Given points lies the same side of the line 2x – 3y + 4 = 0.

**QUESTION25.**

Find the ratio in which (i) X-axis and (ii) Y-axis divide the line segment joining A (2, –3) and B (3, – 6)

Sol:

(i) X – axis divide in the ratio – y_{1} : y_{2 } = – ( –3) : – 6 = – 1 : 2

(ii) Y – axis in the ratio – x_{1} : x_{2 = } – 2 : 3 = – 2 : 3

Find the equation of the straight line passing through the point of intersection of the lines x + y + 1 = 0 and 2x – y + 5 = 0 and containing the point (5, – 2).

Sol:

Given equations are

x + y + 1 = 0 ……… (1)

2x – y + 5 = 0……. (2)

The equation of the line passing through the point of intersection of the lines L_{1} = 0 and L_{2} = 0 is L_{1} + λ L_{2} = 0

The equation of the line passing through the point of intersection of the lines (1) and (2) is

x + y + 1 + λ (2x – y + 5) = 0

but the line passes through (5, – 2)

5 – 2 + 1 + λ (2(5) – (– 2) + 5) = 0

4 + λ (10 + 2 + 5) = 0

4 + λ (17) = 0

λ = –4 /17

∴ the equation of required line is

x + y + 1 + (–4 /17) (2x – y + 5) = 0

17(x + y + 1) –4 (2x – y + 5) = 0

17x + 17y + 17 – 8x + 4y – 20 = 0

9x + 21y – 3 = 0

3x + 7y – 1 = 0

**QUESTION27.**

If a, b, and c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.

Sol:

Given, a, b, and c are in A.P. and the equation is ax + by + c = 0

a, b, and c are in A.P.

⟹ b = (a + c)/2

Substitute b value in the equation ax + by + c = 0

⟹ ax + [ (a + c)/2]y + c = 0

ax + ay/2 + cy/2 + c = 0

a(x + y/2) + c(y/2 + 1) = 0

dividing on both sides by a

⟹ (x + y/2) + (c/a) (y/2 + 1) = 0

It is in the form of L_{1} + λ L_{2} = 0

∴ given equation represent a family of concurrent lines

x + y/2 = 0 and y/2 + 1 = 0 are the lines

y/2 + 1 = 0 ⟹ y/2 = – 1

y =– 2 ⟹ x = 1

The point of concurrency is (– 2, 1)

**QUESTION28.**

Find the point of intersection of the line 7x + y + 3 = 0 and x + y = 0

Sol:

Given equations are 7x + y + 3 = 0 and x + y = 0

∴ The point of intersection is

Transform the following equations into the form L_{1} + λ L_{2} = 0 and find the point of concurrency of the family of straight lines represented by the equation.

(i) (2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0

Given equation is

(2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0

2x + 5kx – 3y – 6ky + 2 – k = 0

2x – 3y + 2 + k (5x – 6y – 1) = 0

It is in the form of L_{1} + λ L_{2} = 0 L_{1 }≡ 2x – 3y + 2 = 0 and L_{2}≡ 5x – 6y – 1 = 0

By solving above equations, we get point of concurrency

∴ The point of concurrency is (5, 4)

(ii) ( k + 1) x + (k + 2) y + 5 = 0

Given equation is (k + 1) x + (k + 2) y + 5 = 0

kx + x + ky + 2y + 5 = 0

x + 2y + 5 + k (x + y) = 0

It is in the form of L_{1} + λ L_{2} = 0

L_{1 }≡ x + 2y + 5 = 0 and L_{2}≡ x + y = 0

By solving the above equations, we get point of concurrency

∴ The point of concurrency is (5, – 5)

**QUESTION30.**

Find the value of ‘p’, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.

Sol:

Given that x + p = 0 —— (1), y + 2 = 0 —— (2) and 3x + 2y + 5 = 0 —— (3) are concurrent.

From (1) x + p = 0 ⟹ x = – p

From (2) y + 2 = 0 ⟹ y = – 2

Point of intersection of (1) and (2) is (– p, – 2)

From (3)

3y + 2x + 5 = 0

3(– p) + 2 (– 2) + 5 = 0

– 3p – 4 + 5 = 0

– 3p + 1 = 0

3p = 1

p = 1/3

Find the area of the triangle formed by the following straight lines and the coordinate axes.

(i) x – 4y + 2 = 0

The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes is

The area of the triangle formed by the straight-line x – 4y + 2 = 0 with the coordinate axes is =

(ii) 3x – 4y + 12 = 0

The area of the triangle formed by the straight-line 3x – 4y + 12 = 0 with the coordinate axes is =

**QUESTION32.**

Find the angle between the lines 2x + y + 4 = 0 and y – 3x =7

Sol:

Given Equations are 2x + y + 4 = 0 and y – 3x =7

If θ is the angle between the lines a_{1}x + b_{1}y + c = 0 and a_{2}x + b_{2}y + c = 0, then cos θ =

∴ θ = 45^{0}

**QUESTION33.**

Find the perpendicular distance from the point (– 3, 4) to the straight line 5x – 12y = 2

Sol:

Given equation is 5x – 12y = 2 ⟹ 5x – 12y – 2 = 0

The perpendicular distance from the point (x_{1}, y_{1}) to the straight-line ax + by + c = 0 is

The perpendicular distance from the point (– 3, 4) to the straight-line 5x – 12y – 2 = 0 is

= 5

**QUESTION34.**

Find the distance between the parallel lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0

Sol:

Given equations are 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0

3x + 4y – 3 = 0 ⟹ 2(3x + 4y – 3) = 2(0) (multiplying on both sides by 2)

⟹ 6x + 8y – 6 = 0

The distance between two parallel lines ax + by + c_{1} =0 and ax + by + c_{2} =0 is

The distance between two parallel lines 6x + 8y – 6 = 0 and 6x + 8y – 1 = 0 is

Find the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0

Sol:

Let (h, k) be the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0

If (h, k) is the foot of the perpendicular from (x_{1}, y_{1}) on the straight-line ax + by + c = 0, then

h+ 1 = 5 and k – 3 = – 1

h = 4 and k = 2

∴ (h, k) = (4, 2)

**QUESTION36.**

Find the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0

Sol:

Let (h, k) be the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0

If (h, k) is the image of the point (x_{1}, y_{1}) with respect to the straight line ax + by + c = 0, then

If (h, k) is the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0,

h – 1 = – 4 and k + 2 = 6

h =– 3 and k = 4

∴ (h, k) = (– 3, 4)

**QUESTION37.**

If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β), find α + β.

Sol:

Given 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β)

⟹ (α, β) is the image of (3, –4) concerning the straight line 2x – 3y – 5 = 0

α – 3 = – 4 and β + 4 = 6

α = – 1 and β = 2

α + β = – 1 + 2 = 1

**QUESTION38.**

Find the incenter of the triangle whose vertices are (1, ), (2, 0) and (0, 0)

Sol:

Let O = (0, 0), A = (1, ) and B = (2, 0)

in centre of a triangle with sides a, b and c, whose vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is

= =

**QUESTION39.**

Find the ortho centre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0 and 2x + y – 7 = 0.

Sol:

Given equations are

x + y + 10 = 0 …… (1)

x – y – 2 = 0 …… (2)

2x + y – 7 = 0. …… (3)

Slope of the line (1) is – 1

Slope of the line (2) is 1

The straight lines (1) and (2) are perpendicular lines

The point of intersection of (1) and (2) is the orthocenter

x + y = – 10 ⟹ y = – 10 – x

from (2) x – y – 2 = 0

x – (– 10 – x) – 2 = 0

x + 10 + x – 2 = 0

2x + 8 = 0 ⟹ x = – 4

⟹ y = – 10 + 4 = – 6

∴ Orthocenter is (– 4, – 6)

**QUESTION40.**

Find the circum centre of the triangle whose sides are x = 1, y = 1 and x + y = 1

Sol:

Given equations are x = 1, y = 1 and x + y = 1

A = (1, 0), B = (1, 1) and C = (0, 1)

AB and BC are perpendicular lines

Circum centre is the midpoint of hypotenuse

i.e., midpoint of AC

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Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| Read More »

The post Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| first appeared on Basics In Maths.]]>**QUESTION 1**

Find the Domain of the following real-valued functions.

It is defined when 6x – x^{2} – 5 ≠ 0

⇒ x^{2} – 6x + 5 ≠ 0

x^{2} – 5x – x + 5 ≠ 0

x (x – 5) –1(x – 5) ≠ 0

(x – 5) (x – 1) ≠ 0

x ≠ 5 or x ≠ 1

∴ domain = R – {1, 5}

It is defined when 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0

⇒ x ≥ –3, x ≤ 3 and x ≠ 0

⇒ –3≤ x, x ≤ 3 and x ≠ 0

⇒ –3≤ x ≤ 3 and x ≠ 0

x ∈ [–3, 3] – {0}

∴ domain = [–3, 3] – {0}

It is defined when x + 2 ≥ 0, 1 – x > 0 and 1 – x ≠ 0

⇒ x ≥ –2, x < 1 and x ≠ 0

x ∈ [–2, ∞) ∩ (– ∞, 1) – {0}

⇒ x ∈ [–2, 1) – {0}

∴ domain = [–2, 1) – {0}

It is defined when 4x – x^{2} ≥ 0

⇒ x^{2} – 4x ≤ 0

x (x – 4) ≤ 0

(x – 0) (x – 4) ≤ 0

x ∈ [0, 4]

∴ domain = [0, 4]

(v) f(x) = log (x^{2} – 4x + 3)

Given f(x) = log (x^{2} – 4x + 3)

It is defined when x^{2} – 4x + 3 > 0

⇒ x^{2 }– 3x – x + 3 > 0

x (x – 3) –1(x – 3) > 0

(x – 3) (x – 1) > 0

x ∈ (–∞, 1) ∪ (3, ∞)

x ∈ R – [1, 3]

∴ domain = R – [1, 3]

Given f(x) = It is defined when x^{2} – 1 ≥ 0 and x^{2} – 3x + 2 > 0

(x + 1)(x – 1) ≥ 0 and x^{2} – 2x – x + 2 > 0

(x + 1) (x – 1) ≥ 0 and x (x – 2) (x – 1) > 0

x∈ (–∞, –1) ∪ (1, ∞) and x ∈ (–∞, –1) ∪ (2, ∞)

∴ domain = R – (–1, 2]

⇒ x < 0

∴ domain = (–∞, 0)

⇒ x > 0

∴ domain = (0, ∞)

**QUESTION 2**

If f : R→ R , g : R → R defined by f (x ) = 4x – 1, g(x) = x^{2} + 2 then find (i) (gof) (x) (ii) (gof) () (iii) (fof) (x) (iv) go(fof) (0).

**Sol:** Given f(x) = 4x – 1, g(x) = x^{2} + 2

(i) (gof) (x) = g (f (x))

= g (4x – 1)

= (4x – 1)^{2} + 2

= 16x^{2} – 8x + 1 + 2

= 16x^{2} – 8x + 3

= g (a + 1 – 1)

= g(a)

= a^{2} + 2

(iii) (fof) (x) = f (f (x))

= f (4x – 1)

= 4 (4x – 1) – 1

= 16x – 4 – 1

= 16x – 5

(iv) go(fof) (0) = g(fof) (0)

= g (f (f (0)))

= g (f (– 1))

= g (– 4 – 1)

= g (– 5)

= (– 5)^{2} + 2

= 25 + 2 = 27

**QUESTION 3**

If f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x^{2}, then find (i) (3f – 2g) (x) (ii) (fg)(x) (iii) (x) (iv) (f + g+ 2) (x)

**Sol:** Given f(x) = 2x – 1 and g(x) = x^{2}

(i) (3f – 2g) (x) = 3f(x) – 2 g(x)

= 3(2x – 1) – 2(x^{2})

= 6x – 3 – 2x^{2}

(ii) (fg)(x) = f(x) g(x)

= (2x – 1) (x^{2})

= 2x^{3} – 3x^{2}

(iv) (f + g+ 2) (x) = f(x) + g(x) + 2

= 2x – 1 + x^{2} + 2

= x^{2} + 2x + 1

**QUESTION 4**

If f = {(4, 5), (5, 6), (6, – 4)} g = {(4, – 4), (6, 5), (8,5)}, then find (i) f + g (ii) f – g (iii) 2f +4g (iv) f + 4 (v) fg (vi) f/g (vii) (viii) (ix) f^{2} (x) f^{3}

**Sol:** Given f = {(4, 5), (5, 6), (6, – 4)}

g = {(4, – 4), (6, 5), (8,5)}

The domain of f ∩ The Domain of g = {4, 6}

(i) (f + g) (4) = f (4) + g (4)

= 5 – 4 = 1

(f + g) (6) = f (6) + g (6)

=– 4 + 5 = 1

∴ f + g = {(4, 1), (6, 1)}

(ii) (f – g) (4) = f (4) – g (4)

= 5 – (– 4) = 5 + 4 = 9

(f – g) (6) = f (6) – g (6)

= – 4– 5 = – 9

∴ f – g = {(4, 9), (6, – 9}

(iii) (2f +4g) (4) = 2 f (4) + 2 g (4)

= 2(5) + 4 (– 4)

= 10 – 16

=– 6

(2f +4g) (6) = 2 f (6) + 2 g (6)

= 2(– 4) + 4 (5)

= – 8 + 20

=12

∴ (2f +4g) = {(4, – 6), (6, 12)}

(iv) (f + 4) (4) = f (4) + 4 = 5 + 4 = 9

(f + 4) (5) = f (5) + 4 = 6 + 4 = 10

(f + 4) (6) = f (6) + 4 = – 4 + 4 = 0

∴ (f + 4) = {(4, 9), (5, 10), (6, 0)}

(v) fg (4) = f (4) g (4) = (5) (– 4) =– 20

fg (6) = f (6) g (6) = (– 4) (5) =– 20

∴ fg = {(4, – 20), (6, – 20)}

(vi) f/g (4) = f(4)/g(4) = 5/ – 4 = – 5/4

f/g (6) = f(6)/g(6) = – 4/ 5

∴ f/g = {(4, – 5/4), (6, – 4/ 5)}

(ix) f^{2}(4) = (f (4))^{2} = (5)^{2} = 25

f^{2}(5) = (f (5))^{2} = (6)^{2} = 36

f^{2}(6) = (f (6))^{2} = (– 4)^{2} = 16

∴ f^{2} = {(4, 25), (5, 36), (6, 16)}

(x) f^{3}(4) = (f (4))^{3} = (5)^{3} = 125

f^{3}(5) = (f (5))^{3} = (6)^{3} = 216

f^{3}(6) = (f (6))^{3} = (– 4)^{3} = –64

∴ f^{3} = {(4, 125), (5, 216), (6, –64)}

**QUESTION 5**

If A = {0, π/6, π/4, π/3, π/2} and f: A→ B is a surjection defined by f(x) = cos x, then find B.

** Sol:** Given A = {0, π/6, π/4, π/3, π/2}

f(x) = cos x

f (0) = cos (0) = 1

f(π/3) = cos (π/3) = 1/2

f(π/2) = cos (π/2) = 0

**QUESTION 6**

If A = {–2, –1, 0, 1, 2} and f: A→ B is a surjection defined by f(x) =x^{2} + x + 1, then find B.

**Sol:** Given A = {–2, –1, 0, 1, 2} and f(x) = x^{2} + x + 1

f (–2) = (–2)^{2} + (–2) + 1= 4 – 2 + 1 = 3

f (–1) = (–1)^{2} + (–1) + 1= 1 – 1 + 1 = 1

f (0) = (0)^{2} + (0) + 1= 0 + 0 + 1 = 1

f (1) = (1)^{2} + (1) + 1= 1 + 1 + 1 = 3

f (2) = (2)^{2} + (2) + 1= 4 + 2 + 1 = 7

∴ B = {1, 3, 7}

**QUESTION 7**

If A = {1, 2, 3, 4} and f: A→ B is a surjection defined by f(x) = then find B.

** Sol:** Given A = {1, 2, 3, 4} and f(x) =

**QUESTION 8**

If f(x) = 2, g(x) = x^{2}, h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x)

**Sol:** Given f(x) = 2, g(x) = x^{2}, h(x) = 2x

(fo(goh)) (x) = fo (g (h (x))

= f (g (h (x))

= f(g(2x)

= f((2x)^{2})

= f(4x^{2}) = 2

**QUESTION 9**

If f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x^{2} +1 then find (i) (gof^{-1}) (2) (ii) (gof) (x – 1)

** Sol:** Given f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x^{2} +1

(i) Let y = f(x) ⟹ x = f^{-1}(y)

y = 3x – 2

y + 2 = 3x

x = (y + 2)/3

f^{-1}(y) = (y + 2)/3

∴ f^{-1}(x) = (x + 2)/3

Now

(gof^{-1}) (2) = g(f^{-1}(2))

= g ((2 + 2)/3)

= g (4/3)

= (4/3)^{2} + 1 = 16/9 + 1 = 25/9

(ii) (gof) (x – 1) = g (f (x – 1))

= g [ 3(x – 1) – 2)]

= g (3x – 3 – 2)

=g (3x – 5)

= (3x – 5)^{2} + 1

= 9x^{2} – 30x + 25 + 1

= 9x^{2} – 30x + 26

**QUESTION 10**

If f: N→ N defined by f (x) = 2x + 5, Is onto? Explain with reason.

**Sol:** Given f (x) = 2x + 5

Let y = f(x) ⟹ x = f^{-1}(y)

y = 2x + 5

2x = y – 5

x = (y – 5)/2 ∉ N

∴ f(x) is not onto

**QUESTION 11**

Find the inverse of the following functions

(i) If a, b ∈ R, f: R→ R defined by f(x) = ax + b

Given function is f(x) = ax + b

Let y = f(x) ⟹ x = f^{-1}(y)

y = ax + b

y – b = ax

x = (y – b)/a

f^{-1}(y) = (y – b)/a

∴ f^{-1}(x) = (x – b)/a

(ii) f: R→ (0, ∞) defined by f(x) = 5^{x}

Given function is f(x) = 5^{x}

Let y = f(x) ⟹ x = f^{-1}(y)

y = 5^{x}

(iii) f: (0, ∞) → R defined by f(x) =

Let y = f(x) ⟹ x = f^{-1}(y)

x = 2^{y}

f^{-1}(y) = 2^{y}

∴ f^{-1}(x) = 2^{x}

(iv) f: R→ R defined by f(x) = e^{4x + 7}

Given function is f(x) = e^{4x + 7}

Let y = f(x) ⟹ x = f^{-1}(y)

y = e^{4x + 7}

(v) f: R→ R defined by f(x) = (2x + 1)/3

Given function is f(x) = (2x + 1)/3

Let y = f(x) ⟹ x = f^{-1}(y)

y= (2x + 1)/3

3y = 2x + 1

2x = 3y – 1

x = (3y – 1)/2

f^{-1}(y) = (3y – 1)/2

∴ f^{-1}(x) = (3x – 1)/2

**QUESTION 12**

If f: R→ R defined by f(x) = , then show that f (tan θ) = cos 2θ

∴ f (tan θ) = cos 2θ

**QUESTION 13**

If f: R – {±1} → R defined by f(x) = , then show that = 2f (x)

**QUESTION 14**

If the function f: R→ R defined by f(x) = , then show that f (x + y) + f (x – y) = 2 f(x) f(y).

∴ f (x + y) + f (x – y) = 2 f(x) f(y)

**QUESTION 15**

If f(x) = cos (log x), then show that = 0

**Sol:** Given function is f(x) = cos (log x)

= cos (log x) cos (log y) – cos (log x) cos (log y)

= 0

Hence proved

**QUESTION 16**

Find the range of the following real-valued functions

**Sol:**

e^{y} > 0

∴ Range of f(x) is R

It is defined when x – 2 ≠ 0

⟹ x ≠ 2

Domain = R – {2}

y = x + 2

if x = 2 ⟹ y = 4

∴ Range of f(x) is R – {4}

**QUESTION 17**

Find the domain and range of the following real valued functions

**Sol:**

It is defined when 1 + x^{2} ≠ 0

⟹ x^{2} ≠ – 1

x ∈ R

∴ domain of f(x) is R

y (1 + x^{2}) = x

y + x^{2}y = x

x^{2} y – x + y = 0

It is defined when 1 – 4y^{2} ≥ 0 and 2y ≠ 0

⟹ 4y^{2} – 1 ≤ 0 and y ≠ 0

(2y – 1) (2y + 1) ≤ 0 and y ≠ 0

(y – 1/2) (y + 1/2) ≤ 0 and y ≠ 0

– ½≤ y ≤ ½ and y ≠ 0

∴ Range of f(x) is [– ½, ½] – {0}

It is defined when 9 – x^{2 }≥ 0

⟹ x^{2} – 9 ≤ 0

(x + 3) (x – 3) ≤ 0

– 3 ≤ y ≤ 3

∴ the domain of f (x) is [– 3, 3]

y^{2} = 9 – x^{2}

x^{2} = 9 – y^{2 }

It is defined when 9 – y^{2 }≥ 0

⟹ y^{2} – 9 ≤ 0

(y + 3) (y – 3) ≤ 0

– 3 ≤ y ≤ 3

y ∈ [– 3, 3]

∴ Range of f (x) is [0, 3] (∵ y ≥ 0)

clearly, x ∈ R

∴ domain of f(x) is R

If x = 0, then y = 1

If x = – 1, then y = 1

If x = 1, then y = 3

If x = – 2, then y = 3

If x = 2, then y = 5

∴ Range of f (x) is [1, ∞)

(iv) f(x) = [x]

clearly Domain = R and Range = Z

**QUESTION 18.**

Define (i) One – One function (ii) Onto function (iii) Bijection (iv) Even and Odd functions

(i) One – One Function: one – one, if every element of A has a unique image in B.

(ii) Onto function: A function f: A→ B is said to be onto if ∀ y ∈ B there exists x ∈ A such that f(x) = y.

(iii) Bijection: A function f: A→ B is said to be Bijection if it is both one-one and onto.

(iv) Even and Odd functions:

If f(–x) = f(x), then f(x) is even function

If f(– x)

= – f(x), then f(x) is odd function

The post Functions (2M Questions &Solutions)|| V.S.A.Q.’S|| first appeared on Basics In Maths.]]>

** Natural numbers:** All the counting numbers starting from 1 are called Natural numbers.

1, 2, 3… Etc.

** Whole numbers:** Whole numbers are the collection of natural numbers including zero.

0, 1, 2, 3 …

** Integers: **integers are the collection of whole numbers and negative numbers.

….,-3, -2, -1, 0, 1, 2, 3,…..

** ****Integers on a number lin**e:

** Operations on integers:**

3 + 4 = 7

-2 + 4 = 2

** Subtraction of integers on a number line:-**

6 – 3 = 3

** Multiplication of integers on a number line:-**

2 × 3 ( 2 times of 3) = 6

3 × (- 4 ) ( 3 times of -4) = -12

** Multiplication of two negative integers:**

- To multiply two negative integers, first, we multiply them as whole numbers and put plus sign before the result.
- The multiplication of two negative integers is always negative.

Ex:- -3 × -2 = 6, -10 × -2 = 20 and so on.

** Multiplication of more than two negative integers:**

• If we multiply three negative integers, then the result will be a negative integer.

Ex:- -3 × -4 × -5 = -60, -1× -7 × -4 = -28 and so on.

• If we multiply four negative integers, then the result will be a positive integer.

Ex:- -3 × -4 × -5 × -2 = 120, -1× -7 × -4 × -2 = 56 and so on.

** Note:-**

1. If the no. of negative integers is even, then the result will be positive.

2. If the no. of negative integers is odd, then the result will be negative.

** Division of integers:**

- The division is the inverse of multiplication.
- When we divide a negative integer by a positive integer or a positive integer by a negative integer, we divide them as whole numbers then put negative signs for the quotient.

Ex:- -3 ÷ 1 = 3, 4 ÷ -2 = -2 and so on.

- When we divide a negative integer by a negative integer, we get a positive number as the quotient.

Ex:- -3 ÷ -1 = 3, -4 ÷ -2 = 2 and so on.

** Properties of integers:**

** 1.Closure property:-**

** 2.commutative property:-**

**3.associative property:-**

**Additive identity:-**

1 + 0 = 0 + 1 = 1, 10 + 0 = 0 + 10 = 10

•For any integer ‘a’, a + 0 = 0 + a

•0 is the additive identity.

**Additive inverse:- **

2 + (-2) = (-2) + 2 = 0, 5 + (-5) = (-5) + 5 = 0

•For any integer ‘a’, a+ (-a) = (-a) + a = 0

•Additive inverse of a = -a and additive inverse of (-a) = a

**Multiplicative identity:-**

2 × 1 = 1 × 2 = 2, 5 × 1 = 1 × 5 = 5

•For any integer ‘a’, a × 1 = 1 × a = a

•1 is the multiplicative identity.

**multiplicative inverse:-**

For any integer ‘a’, 1/a × a = a × 1/a = 1

- multiplicative inverse of a = 1/a
- Multiplicative inverse of 1/a = a.

**distributive property:- **

For any three integers a, b and c, a × (b + c) = (a × b) + (a × c).

3 × (2 + 4) = 18

(3 × 2) + (3 × 4) = 6 + 12 = 18

∴ 3 × (2 + 4) = (3 × 2) + (3 × 4).

**Fraction: A fraction** is a number that represents a part of the whole. A group of objects is divided into equal parts, then each part is called a fraction.

** The proper and improper fractions:**

In a **proper fraction,** the numerator is less than the denominator.

Ex: – 1/5, 2/3, and so on.

In an **improper fraction,** the numerator is greater than the denominator.

Ex: – 5/2,11/5 and so on.

**Comparing fractions:**

**Like fractions: – **We have to compare the like fractions with the numerator only because the like fractions have the same denominator. The fraction with the greater numerator is greater and the fraction with the smaller numerator is smaller.

**Unlike fractions: – **

**With the same numerator: **For comparing unlike fractions, we have to compare denominators when the numerator is the same. The fraction with a greater denominator is smaller and the fraction with a smaller denominator is smaller.

**Note: – **To find the equivalent fractions of both the fractions with the same denominator, we have to take the LCM of their denominators.

**Addition of fractions:**

∗ **Like Fractions:**

∗ Unlike** fractions:**

**Subtraction of fractions:**

**∗ Like fractions:**

**Unlike fractions: –** Firs**t,** we have to find the equivalent fraction of given fractions and then subtract them as like fractions

**Multiplication of fractions:**

**Multiplication of fraction by a whole number: –**

Multiplication of numbers means adding repeatedly.

• To multiply a whole number with a proper or improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator the same.

2.**Multiplication of fraction with a fraction: –**

multiplication of two fractions =

**Division of fractions:**

⇒ 6 one-thirds in two wholes

**Reciprocal of fraction: **reciprocal of a fraction is .

**Note:**

- dividing by a fraction is equal to multiplying the number by its reciprocal.
- For dividing a number by mixed fraction, first, convert the mixed fraction into an improper fraction and then solve it.

**1.Division of a whole number by a fraction: –**

**2.Division of a fraction by another fraction: –**

** ** **Decimal number or fractional decimal: **

In a decimal number, a dot(.) or a decimal point separates the whole part of the number from the fractional part.

The part right side of the decimal point is called the **decimal part** of the number as it represents a part of 1. The part left to the decimal point is called the **integral part** of the number.

Note: –

- while adding or subtracting decimal numbers, the digits in the same places must be added or subtracted.
- While writing the numbers one below the other, the decimal points must become one below the other. Decimal places are made equal by placing zeroes on the right side of the decimal numbers.

**Comparison of decimal numbers:**

while comparing decimal numbers, first we compare the integral parts. If the integral parts are the same, then compare the decimal part.

Ex: – which is bigger: 13.5 or 14.5

Ans: 14.5

Which is bigger: 13.53 or 13. 25

Ans: 13.53

**Multiplication of decimal numbers:**

For example, we multiply 0.1 × 0.1

**Multiplication of decimal numbers by 10, 100, and 1000: –**

Here, we notice that the decimal point in the product shifts to the right side by as many zeroes as in 10, 100, and 1000.

**Division of decimal number**:

**Division of decimal number by 10,100 and 1000: –**

Here, we notice that the decimal point in the product shifts to the left side by as many zeroes as in 10, 100, and 1000.

**Rational numbers:**

** **The numbers which are written in the form of p/q, where p, q are integers, and q ≠ 0, are called rational numbers.

Rational numbers are a bigger collection of integers, negative fractional numbers, positive fractional numbers.

Ex: – 1, 2, -1/2, 0 etc.

**Equation: **Equation is the condition of a variable. It says that two expressions are equal.

- An equation has two sides LHS and RHS, on both sides of the equality of sign.
- One of the expressions of the equation is must have a variable.
- If we interchange the expressions from LHS to RHS, the equation remains the same

Ex: – x + 2 = 5; 2 = x + 3

**Balanced equation:**

In an equation, if LHS =RHS, then that equation is balanced.

If the same number is added or subtracted on both sides of the balanced equation, the equation remains will the same.

Ex: 8 + 3 = 11

If add 2 on both sides ⇒ LHS = 8 + 3 + 2 = 13

RHS = 11 + 2 = 13

∴ LHS = RHS

8 +3 = 11 if subtract 2 on both sides

LHS = 8 + 3 – 2 = 9

RHS = 11 – 2 = 9

∴ LHS = RHS

**Using algebraic equations in solving day to day problems:**

- Read the problem carefully.
- Denote the unknown or quantity to be found with some letters such as x, y, z …etc.
- Write the problem in the form of an algebraic equation by making a relation among the quantities.
- Solve the equation.
- Check the solution

**Complimentary angles: **When the sum of the angles is 90^{0}, the angles are called complementary angles.

**Ex:** 30^{0}, 60^{0}; 20^{0}, 70^{0} and soon.**Supplementary angles: **When the sum of the angles is 180^{0}, the angles are called Supplementary angles.

**Ex:** 120^{0}, 60^{0}; 110^{0}, 70^{0} and soon.

**Adjacent angles: **The angle having a common Arm and a common vertex are called Adjacent angles.

⇒ ∠AOC and ∠BOC adjacent angles.

**Vertically opposite angle: **If two lines are intersecting at a point, then the angles that are formed opposite to each other at that point are called vertically opposite angles.

**Transversal: **A line that intersects two or more lines at distinct points is called a transversal.

**Corresponding angles: **

Two angles that lie on the same side of the transversal and one interior and another one exterior are called corresponding angles.

∠1, ∠5; ∠2, ∠6; ∠3, ∠7 and ∠4, ∠8

**Alternate angles: **

Two angles which are the lies opposite side of the transversal and both interior or exterior are called Alternate angles.

∠1, ∠7; ∠2, ∠8 are exterior alternate angles

∠3, ∠5; ∠4, ∠6 are interior alternate angles.

∠3, ∠6; ∠4, ∠5 interior angles same side of the transversal.

**Transversal on parallel lines:**

• If pair of parallel lines are intersected by a transversal then the angles of each pair of **corresponding angles** are equal

⇒ ∠1, =∠5; ∠2= ∠6; ∠3= ∠7 and ∠4= ∠8

•If pair of parallel lines are intersected by a transversal then the angles of each pair of **interior alternate angles** are equal.

∠3= ∠5; ∠4= ∠6

•If pair of parallel lines are intersected by a transversal then the angles of each pair of **exterior alternate angles** are equal.

∠1= ∠7; ∠2= ∠8

•If pair of parallel lines are intersected by a transversal then the angles of each pair of **interior angles on the same side of the transversal **are supplementary.

∠3+∠6= 180^{0}; ∠4+ ∠5 = 180^{0}

**Note:**

- If a transversal intersects two lines and the pair of corresponding angles are equal, then the lines are parallel.
- If a transversal intersects two lines and the pair of alternate angles are equal, then the lines are parallel.
- If a transversal intersects two lines and the interior angles on the same side of the transversal are supplementary, then the lines are parallel.

**
Triangle: ** A closed figure formed by three-line segments is called a triangle.

** **In ∆ABC,

**Classification of triangles: **

Triangles can be classified according to the properties of their sides and angles.

**According to sides:**

Based on sides triangles are three types:

- Scanlan triangle (ii) Isosceles Triangle (iii) equilateral triangle

**According to angles:**

- Acute-angled triangle (ii) Right-angled triangle (iii) Obtuse-angled triangle

**Relationship between the sides of a triangle:**

- The sum of the lengths of any two sides of a triangle is greater than the third side.

- The difference between the lengths of any two sides of a triangle is less than the third side.

**The altitude of a triangle:**

We can draw three altitudes in a triangle. A perpendicular line drawn from a vertex to its opposite side of a triangle is called the Altitude of the triangle.

**Median of a triangle:**

**
**In a triangle, a line drawn from the vertex to the mid-point of its opposite side is called the median of the triangle.

Medians of a triangle are concurrent. We can draw three medians in a triangle.

The point of concurrence of medians is called the centroid of the triangle. It is denoted by G

**Angle-sum property of a triangle:**

Some of the angles in a triangle is 180^{0}

∠A + ∠B + ∠C = 180^{0}

**An exterior angle of a triangle:**

When one side of the triangle is produced, the angle thus formed is called an exterior triangle.

**Exterior angle property:-** The exterior angle of a triangle is equal to the sum of two interior opposite angles. ** **

x^{0}+ y^{0} = z^{0}

**Ratio: **Comparison of two quantities of the same kind is called ‘Ratio.

The ratio is represented by the symbol ‘:’

If the ratio of two quantities ‘a’ and ‘b’ is a : b, then we read this as ‘a is to b’

The quantities ‘a’ and ‘b’ are called terms of the ratio.

**Proportion: **if two ratios are equal, then they are said to be proportional.

‘a’ is called as first term or antecedent and ‘b’ is called a second term or consequent.

If a: b = c : d, then a, b, c, d are in proportion and ⇒ ad = bc.

The product of means = the product of extremes

**Unitary method:** The method in which we first find the value of one unit and then the value of the required no. of units is known as the unitary method.

**Direct proportion: **In two quantities, when one quantity increase(decreases) the other quantity also increases(decreases) then two quantities are in direct proportion.

**Percentages: **

‘per cent’ means for a hundred or per every hundred. The symbol % is used to denote the percentage.

1% means 1 out of 100, 17% means 17 out of 100.

**Profit and Loss: **

Selling price = SP; Cost price = CP

If SP > CP, then we get profit

Profit = SP – CP

SP = CP + profit

If SP < CP, then we get a loss

Loss = CP – SP

SP = CP – Loss** **

**Simple interest: **

**Principle: – **The money borrowed or lent out for a certain period is called the Principle.

**Interest:** – The extra money, for keeping the principle paid by the borrower is called interest.

**Amount:** – The amount that is paid back is equal to the sum of the borrowed principal and the interest.

Amount = principle + interest

Interest (I) = where R is the rate of interest.

**Data:** The information which is in the form of numbers or words and helps in taking decisions or drawing conclusions is called data.

**Observations: **The numerical entries in the data are called observations.

**Arithmetic Mean: **The average data is also called an Arithmetic mean.

The arithmetic mean always lies between the highest and lowest observations of the data.

When all the values of the data set are increased or decreased by a certain number, the mean also increases or decreases by the same number.

**Mode: **The most frequently occurring observation in data is called Mode.

If data has two modes, then it is called bimodal data.

**Note:** If each observation in a data is repeated an equal no. of times, then the data has no mode.

**Median: **The middlemost observation in data is called the Median.

Arrange given data in ascending or descending order.

If a data has an odd no. of observations, then the middle observation is the median.

If a data has even no. of observations, then the median is the average of middle observations.

**Bar graph:**

Bar graphs are made up of uniform width which can be drawn horizontally or vertically with equal spacing between them.

The length of each bar tells us the frequency of the particular item.

**Ex:**

It represents two observations side by side.

**Ex:**

** ****Pie chart: **A circle can be divided into sectors to represent the given data

**Ex:**

Budget | Amount in rupees |

Food | 1200 |

Education | 800 |

Others | 2000 |

Savings | 5000 |

Total income | 9000 |

**Congruent figures: **Two figures are said to be congruent if they have the same shape and size.

**Congruency of line segments: ** If two-line segments have the same length, then they are congruent. Conversely, if two-line segments are congruent, then they have the same length.

** ****Congruency of Triangles:**

Two triangles are said to be congruent if (i) their corresponding angles are equal (ii) their corresponding sides are equal.

Ex: In ∆ ABC, ∆ DEF

∠A ≅ ∠D; ∠B≅ ∠E; ∠C ≅ ∠F

AB ≅ DE; BC≅ EF; AC ≅ DF

∴∆ABC ≅ ∆DEF

**The criterion for congruency of Triangles:**

**1.Side -Side -Side congruency (SSS): –**

If three side of a triangle is equal to the corresponding three sides of another triangle, then the triangles are congruent.

∴∆ABC ≅ ∆DEF

**2.Side -Angle -Side congruency (SAS): –**

If two sides and the angle included between the two sides of a triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent.

∴ ∆ABC ≅ ∆DEF

**3.Angle – Side -Angle congruency (ASA): –**

If two angles and included side of a triangle are equal to the corresponding two angles and included side of another triangle, then the triangles are congruent.

∴ ∆ABC ≅ ∆DEF

**4.Right angle – Hypotenuse – Side congruence (RHS): –**

** **If the hypotenuse and one side of a right-angled triangle are equal to the corresponding hypotenuse and side of the other right-angled triangle, then the triangles are Equal.

∴∆ABC ≅ ∆DEF

The no. of measurements required to construct a triangle = 3

A triangle can be drawn in any of the situations given below:

- Three sides of a triangle
- Two sides and the angle included between them.
- Two angles and the side included between them.
- The hypotenuse and one adjacent side of the right-angled triangle.

**Construction of a triangle when measurements of the three sides are given:**

**Ex:** construct a triangle ABC with sides AB = 4cm, BC = 7cm and AC = 5cm

**Step of constructions:**

Step -1: Draw a rough sketch of the triangle and label it with the given measurements.

Step -2: Draw a line segment of BC of length 7cm.

Step -3: with centre B, draw an arc of radius 4cm, draw another arc from C with radius 5cm such that it intersects first at A.

Step -4: join A, B and A, C. The required triangle ABC is constructed.

**Construction of a triangle when two sides and the included angle given:**

**EX: **construct a triangle ABC with sides AB = 4cm, BC = 6cm and ∠B=60^{0}

**Step of constructions:**

Step -1: Draw a rough sketch of the triangle and label it with the given measurements.

Step -2: Draw a line segment of AB of length 4cm.

Step -3: draw a ray BX making an angle 60^{0} with AB.

Step -4: draw an arc of radius 5cm from B, which cuts ray BX at C.

Step -5: join C and A, we get the required ∆ABC.

**Construction of a triangle when two angles and the side between the angles given:**

**Ex: **construct a triangle PQR with sides QR = 4cm, ∠Q= 120^{0} and ∠R= 40^{0}

**Step of constructions:**

Step -1: Draw a rough sketch of a triangle and label it with the given measurements.

Step -2: Draw a line segment QR of length 4 cm.

Step -3: Draw a ray RX, making an angle 40^{0} with QR.

Step -4: Draw a ray QY, making an angle 100^{0} with QR, which intersects ray RX.

Step -5: Mark the intersecting point of the two rays as P. Required triangle PQR is constructed.

**Construction of a triangle when two sides and the non-included angles are given**:

**Ex: **construct a triangle MAN with sides MN = 4cm, AM = 3cm and ∠A= 40^{0}

**Step of constructions:**

Step -1: Draw a rough sketch of a triangle and label it with the given measurements.

Step -2: Draw a line segment MA of length 0f 5cm.

Step -3: Draw a ray AX making an angle 40^{0} with the line segment MA.

Step -4: With M as the centre and radius 3 cm draw an arc to cut ray AX. Mark the intersecting point as N.

Step -5: join M, N, then we get the required triangle MAN.

**Construction of a right-angled triangle when hypotenuse and sides are given:**

**Ex: **construct a triangle ABC, right angle at B and AB = 4cm, Ac = 5cm

**Step of constructions:**

Step -1: Draw a rough sketch of a triangle and label it with the given measurements.

Step -2: Draw a line segment BC of length 0f 4cm.

Step -3: Draw a ray BX perpendicular to BC at B

Step -4: Draw an arc from C with a radius of 5cm to intersect ray BX at A.

Step -5: Join A, C, then we get the required triangle ABC.

**Variable: **It is a dependent term. It takes different value.

Ex: m, x, a, etc.

**Constant: ** It is an independent term. It has a fixed value.

**Like terms and Unlike terms: **If the terms contain the same variable with the same exponents, then they are like terms otherwise, unlike terms.

Ex: 3x, –4x, x are like terms

3x, 4y, 4 are unlike terms

**Coefficient: **Coefficient is a term which the multiple of another term (s)

EX: In 5x. 5 is the coefficient of x and x is the coefficient of 5

5 is a numerical coefficient

x is the literal coefficient

**Expression: **An expression is a single term or a combination of terms connected by the symbols ‘+’ (plus) or ‘−’ (minus).

Ex: 2x – 3. 3x, 2 +3 – 4 etc.

**Numerical Expressions: ** If every term of an expression is constant, then the expression is called numerical expression.

Ex: 2 + 3 + 5, 2 – 4 – 7, 1 + 5 – 4 etc.

**Algebraic expression: ** If an expression at least one algebraic term, then the expression is called an algebraic expression.

Ex: x + y, xy, x – 3, 4x + 2 etc.

**Note: **Plus (+) and Minus (−) separate the terms

Multiplication (×) and Division (÷) do not separate the terms.

**Types of Algebraic expressions:**

**Monomial: – **If an expression has only one term, then it is called a monomial.

Ex: 2x^{2}, 3y, x, y, xyz etc.

**Binomial: **If an expression has two unlike terms, then it is called binomial.

Ex: 2x+ 3y, x^{2}+ y, x +yz^{2} etc.

**Trinomial: **If an expression has three unlike terms, then it is called trinomial. Ex: 2x+ 3y + 4xy, x^{2}+ y + z, x^{2} y +yz^{2} + xy^{2} etc.

**Multinomial: **If an expression has more than three unlike terms, then it is called multinomial.

Ex: 2x+ 3y + 4xy +5, x^{2}+ y + z – 4y + 6 ,

x^{2} y +yz^{2} + xy^{2} – 4xy + 8yz etc.

**Degree of a monomial: **The sum of all exponents of the variables present in a monomial is called the degree of the monomial.

Ex: Degree of 5xy^{3 }

An exponent of x is 1 and an exponent of y is 3** **

Sum of exponents = 1 + 3 = 4

∴ degree of 5xy^{3} is 4

**Degree of an Algebraic Expression: **The highest exponent of all the terms of an expression is called the degree of an Algebraic expression.

** Ex: **degree of x^{2} + 3x + 4x^{3 }is 3

degree of 3xy + 6x^{2}y + 5x^{2}y^{2} is 4

**Addition of like terms: **

The sum of two or more like terms is a like term with a numerical coefficient that is equal to the sum of the numerical coefficients of all the like terms in addition.

Ex: 3x + 2x = (3 + 2) x = 5x

4x^{2}y + x^{2}y = (4 + 1) x^{2}y = 5x^{2}y

**Subtraction of like terms: **

The difference of two like terms is a like term with a numerical coefficient is equal to the difference between the numerical coefficients of the two like terms.

Ex: 3x − 2x = (3 − 2) x = x

4x^{2}y −2 x^{2}y = (4 −2) x^{2}y = 2x^{2}y

**Note: (**i) addition and subtraction are not done for unlike terms.** (ii) **If no terms of an expression are alike then it is said to be in the simplified form.

**The standard form of an Expression: **

In an expression, if the terms are in such a way that the degree of the terms is in descending order, then the expression is said to be in standard form.

Ex: 5 – 2x^{2}