View
275
Download
2
Category
Preview:
Citation preview
No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanicalincluding photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.
MATHEMATICS First Year Diploma Semester - I
Printed at: Repro India Ltd., Mumbai
TEID : 920
Written as per the revised ‘G’ Scheme syllabus prescribed by the Maharashtra State Board of Technical Education (MSBTE) w.e.f. academic year 2012-2013
First Edition: June 2015
Basic
Salient Features
• Concise content with complete coverage of revised G-scheme syllabus. • Simple & lucid language.
• Illustrative examples showing detailed solution of problems.
• MSBTE Questions from Winter-2006 to Summer-2015.
• MSBTE Question Papers of Summer-2014, Winter-2014, Summer-2015. • Three Model Question Papers for practice.
• List of formulae for quick reference.
PREFACE Target’s “Basic Mathematics” is compiled with an aim of shaping engineering minds of students while catering to their needs. It is a complete & thorough book designed as per the new revised G-scheme of MSBTE curriculum effective from June 2012. Each unit from the syllabus is divided into chapters bearing ‘specific objectives’ in mind. The sub-topic wise classification of this book helps the students in easy comprehension.
Each chapter includes the following features:
Theory of each mathematical concept is explained with appropriate references. Diagrams and illustrations are provided wherever necessary. Italicized definitions are mentioned for important topics.
Illustrative Examples are provided in order to explain the method of solving the problems.
The detailed step-by-step solution in these problems helps students to understand and remember each minute step with proper justification of the same.
Solved Problems covering every type of MSBTE question gives students the confidence to
attempt all the questions in the examination. Exercise (With final answers) covers a variety of questions from simple to complex to help the
students gain thorough revision in solving various types of problems. MSBTE Problems covering questions from year 2006 to 2015 are solved exactly as they are
expected to be solved by the students in the examination. Formulae section is provided for quick recap and last minute revision of all the formulae at one glance. MSBTE Question Papers of year 2014 and Summer-2015 are added at the end to make students familiar with the MSBTE examination pattern. A set of three Model Question Papers are designed as per MSBTE Pattern for thorough revision and to prepare the students for the final examination.
Best of luck to all the aspirants! From, Publisher
SYLLABUS
Topic and Contents Hours Marks
Topic 1 - Algebra
1.1 Determinant [04] Specific Objectives:
04
32
Solve simultaneous equations in three variables using Cramer’s rule .
Definition and expansion of determinant of order 3.
Cramer’s rule to solve simultaneous equations in three variables.
1.2 Matrices [16] Specific Objectives:
10
Perform all algebraic operations on matrices. Solve simultaneous equations in three variables.
Definition of a matrix of order m x n and types of matrices.
Algebra of matrices with properties and examples.
Transpose of a matrix with properties.
Cofactor of an element of a matrix.
Adjoint of matrix and inverse of matrix by adjoint method.
Solution of simultaneous equations containing two and three unknowns by matrix inversion method.
1.3 Partial Fraction [12] Specific Objectives:
08
Find partial fraction of proper and improper fraction.
Definition of fraction, proper, improper fraction and partial fraction.
Resolve proper fractions into partial fraction with denominator containing
i. non repeated linear factors, ii. repeated linear factors, iii. non repeated quadratic irreducible factors.
To resolve improper fraction into partial fraction.
Topic 2- Trigonometry
40
2.1 Trigonometric Ratios of Allied, Compound, Multiple and Sub-Multiple Angles [16]
Specific objectives: 10
Solve examples of allied angle, compound angle, multiple and sub-multiple angles.
Trigonometric ratios of any angle.
Definition of allied angle, compound, multiple and sub-multiple angles.
Trigonometric ratios of above angles with proofs. Simple examples
2.2 Factorization and De-factorization Formulae [12] Specific objectives:
08 Derive factorization and de-factorization formulae to solve examples.
Formulae for factorization and de-factorization with proof and examples.
2.3 Inverse Trigonometric Ratios [12] Specific objectives:
08 Solve examples of inverse trigonometric ratios.
Definition of inverse trigonometric ratios.
Principal value of inverse trigonometric ratios.
Relation between inverse trigonometric ratios with proof and examples.
Topic 3 - Co-ordinate Geometry
3.1 Straight Line [16] Specific objectives:
10 16
Solve problems with given condition.
Angle between two lines with proof. Examples.
Condition of parallel and perpendicular lines.
Point of intersection of two lines, equation of line passing through point of intersection with given condition.
Perpendicular distance between point and line with proof and examples.
Distance between two parallel line with proof and examples.
Topic 4 - Statistics
4.1 Measures of Dispersion [12] Specific objectives:
06 12
Find the range, mean deviation, standard deviation and consistency of any data.
Measures of dispersion - range, mean deviation from mean and median and standard deviation.
Variance and its coefficient.
Comparisons of two sets of observations.
TOTAL 64 100
Contents
Chapter No. Topic Page No.
Unit - I: Algebra 1
1 Determinant 2
2 Matrices 45
3 Partial Fraction 146
Unit - II: Trigonometry 212
4 Trigonometric Ratios of Allied, Compound, Multiple and Sub-Multiple Angles
213
5 Factorization and De-factorization Formulae 277
6 Inverse Trigonometric Ratios 297
Unit - III: Co-ordinate Geometry 327
7 Straight Line 328
Unit - IV: Statistics 373
8 Measures of Dispersion 374
Formulae and Model Question Papers
Formulae 407
Model Question Paper I 417
Model Question Paper II 420
Model Question Paper III 423
MSBTE Question Papers
Question Paper – Summer 2014 426
Question Paper – Winter 2014 429
Question Paper – Summer 2015 432
1
Basic Physics (F.Y.Dip.Sem.-1) MSBTEUnit I: Algebra
Publications Pvt. Ltd. Target
Algebra UNIT
I
1.1 Introduction 1.2 Determinant of Second Order 1.3 Determinant of Third Order 1.3 (a) Definition and elements of
determinant of third order 1.3 (b) Minors and Cofactors 1.3 (c) Value of a determinant of third
order
1.4 Properties of Determinants 1.5 Cramer’s Rule 1.5 (a) Cramer’s Rule for equations
involving two unknown variables 1.5 (b) Cramer’s Rule for equations
involving three unknown variables
2.1 Introduction 2.2 Definition of a Matrix 2.3 Types of Matrices 2.3.(a) Rectangular Matrices 2.3.(b) Square Matrices 2.3.(c) Zero Matrices / Null Matrices 2.4 Algebra of Matrices/Operation on
Matrices 2.4.(a) Scalar Multiplication of Matrices 2.4.(b) Equality of Matrices 2.4.(c) Addition of Matrices
2.4.(d) Subtraction of Matrices 2.4.(e) Multiplication of Matrices 2.5 Transpose of Matrices 2.6 Determinant of a Square Matrix 2.7 Minors and Cofactors 2.7.(a) Minor of an element of a Matrix 2.7.(b) Cofactor of an element of a Matrix 2.8 Adjoint of a Matrix 2.9 Inverse of a Matrix 2.10 Matrix Form of System of
Equations
3.1 Introduction 3.2 Polynomials 3.3 Proper and Improper Fractions 3.4 Factorization of the Denominator of
a proper fraction and determining its type
3.5 Resolve a given proper fraction into partial fractions
3.5.(a) Resolve a proper fraction with Distinct linear factors as denominator, into partial fractions
3.5.(b) Resolve a proper fraction with Repeated linear factors as denominator, into partial fractions
3.5.(c) Resolve a proper fraction with Distinct Irreducible Quadratic factor as denominator, into Partial fractions
3.6 Resolve a given Improper fraction into partial fractions
Chapter-1 Determinant
Chapter-2 Matrices
Chapter-3 Partial Fraction
Basic Physics (F.Y.Dip.Sem.-1) MSBTEBasic Mathematics (F.Y.Dip.Sem.-1) MSBTE
2
Publications Pvt. Ltd. Target
1.1 Introduction
1.2 Determinant of Second Order
Determinants are arrangement of numbers (real, imaginary, complex) in equal number of rows (horizontal arrangement) and columns (vertical arrangement) enclosed between a pair of vertical segments.They are denoted by or D or |A|. Determinants are scalar quantities. A determinant of order 2 is an arrangement of four numbers in two rows and two columns, enclosed between two vertical segments. Following is an example of a determinant of order 2.
D = 1 2
3 4
D = 1 2
3 4
Thus; 1 and 2 are elements of Row 1, 3 and 4 are elements of Row 2, 1 and 3 are elements of Column 1, 2 and 4 are elements of Column 2. The set of elements from top left corner to the bottom right corner in a determinant is called the Main Diagonal or Leading Diagonal or Principal Diagonal of that determinant.
D = 1 2
3 4 Here, 1 and 4 are elements of Principal Diagonal.
The set of elements from bottom left corner to the top right corner in a determinant is called the Reverse or Secondary Diagonal of that determinant.
D = 1 2
3 4 Here, 3 and 2 are elements of Secondary Diagonal.
To calculate the value of the determinant, consider the determinant of numbers a, b, c, d as given below,
D =a b
c d
= a d c b = ad cb.
Determinant 01
Row 1
Row 2
Column 1 Column 2
Principal Diagonal
Secondary Diagonal
Basic Physics (F.Y.Dip.Sem.-1) MSBTEChapter 01: Determinant
3
Publications Pvt. Ltd. Target i.e. Value of determinant D = Product of elements in the Principal Diagonal Product of elements in the Secondary Diagonal.
D = 1 2
3 4
= (1)(4) (3)(2) = 4 6 = 2
1. Evaluate:
2 5
3 4
Solution:
Let D =2 5
3 4
Now, D = Product of elements in the Principal Diagonal Product of elements in the Secondary Diagonal. D = [2(4)] [3(5)] = 8 15 D = 23
1. Evaluate:2 3
4 8
Solution:
Let D = 2 3
4 8
= (2)(8) (4)(3) = 16 + 12 = 28
2. Find x, if
3
1 2
x
x= 0.
Solution:
Given,3
1 2
x
x= 0
x(x + 2) (1)(3) = 0 x2 + 2x 3 = 0 x2 + 3x x 3 = 0 x(x + 3) 1(x + 3) = 0 (x + 3)(x 1) = 0 x = 3 or x = 1
3. Solve : 2 7
3 12= 1
9 2
x
Solution:
Given, 2 7
3 12= 1
9 2
x
(2)(12) (3)(7) = (x)(2) (9)(1) 24 21 = 2x 9 3 + 9 = 2x 2x = 12 x = 6
Illustrative Example
Solved Problems
Recommended