rational and irrational numberkvbhanu.org/userfiles/file/sample papers class xi and xii... · web...
TRANSCRIPT
RESOURCE GENERATION FOR BRIDGE COURSE IN CLASS XISUBJECT – MATHEMATICS
COURSE COORDINATOR
Mr ARUN KUMARTRAINING ASSOCIATES PHYSICS
ZIET CHANDIGARH RESOURCE PERSON
1.Mr RAJEEV RANJAN PGT MATHS
KV SEC 31 CHANDIGARH2. Mr SANJAY KUMAR PGT MATHS
KV CRPF PINJORE
LIST OF PARTICIPANTS
Workshop on Resource Generation for Bridge Course of Class XI for PGT(Maths) 01.08.2017 to 03.08.2017
S.No Name of the Participant Name of KV
1. Mrs.PoonamPassi 3 BRD2. Mr.Surinder Kumar Duggal No.4, Batinda3. Mr.Ravinder Singh No.3, Amristar4. Mr.Sawinder No.1,Hussainpu
r5. Ms.ArunaKhera No.1Ferozpur6 Mr.Pradeep Kumar No.1, Pathonkot7 Mr.Dharmender Kumar No.2, Delhi
Cantt.8 Mr.Kulwinder Singh SainikVihar9 Mrs.Charu Sharma Tagore Garden10 Mrs. RuchiSalaria PaschimVihar11 Mr.Ajeet Kumar Singh GoleMarket
Shift I12 Mr.Ajay Kumar Giri AFS Bawana13 Mr.Manish Kumar Sharma Sector12
Dwarka14 Mr.Praveen Kumar 1HBK,Dehradu
n15 Mr.Kochin Kumar Haidwani16 Mr.Pradeep Chandra Saxena Gauchar17 Mr.Birampal Singh Kausani18 Mr.NaveenWadhera SSBSrinagar19 Mr.B.S.Manral Ranikhet20 Mr.AnilRawat Uttarkashi21 Mr.S.N.Sharma Faridabad22 Mr. R.D.Sharma Nahara23 Mr.Vikas Kumar No.2, Ambala24 Mr. Raman Kumar ITBP,Bhanu25 Mr.ChanchalKalra Kasauli26 Mr . Rajiv Ranjan
(Resource Person)Sector 31,
Chandigarh27 Mr Sanjay Kumar
(Resource Person)Pinjore
STANDARDIZATION OF CONCEPTS CLASS X
1. REAL NUMBERSA. Introduction B. Euclid’s Division Lemma C. The Fundamental Theorem of Arithmetic D. Revisiting Irrational Numbers E. Revisiting Rational Numbers and Their Decimal Expansions 2. POLYNOMIALS A. Introduction B. Geometrical Meaning of the Zeroes of a Polynomial C. Relationship between Zeroes and Coefficients of a Polynomial D. Division Algorithm for Polynomials3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLESA. Introduction B. Pair of Linear Equations in Two Variables C. Consistency of System of linear equations &Graphical Method of Solution of a Pair of
Linear Equations D. Algebraic Methods of Solving a Pair of Linear Equations
a) Substitution Method b) Elimination Method c) Cross-Multiplication Method d) Equations Reducible to a Pair of Linear Equations in Two Variables
4. QUADRATIC EQUATIONS A. Introduction B. Quadratic Equations C. Solution of a Quadratic Equation by FactorisationD. Solution of a Quadratic Equation by using quadratic formulaE. Solution of a Quadratic Equation by Completing the Square F. Nature of Roots G. Applications of quadratic equations5. ARITHMETIC PROGRESSIONS A. IntroductionB. Arithmetic Progressions C. nth Term of an AP D. Sum of First n Terms of an AP 6. TRIANGLES A. Introduction B. Similar Figures C. Similarity of Triangles D. Criteria for Similarity of Triangles E. Areas of Similar Triangles F. Pythagoras Theorem 7. COORDINATE GEOMETRYA. Introduction B. Distance Formula C. Section Formula D. Area of a Triangle 8. INTRODUCTION TO TRIGONOMETRY
A. Introduction B. Trigonometric Ratios C. Trigonometric Ratios of Some Specific Angles D. Trigonometric Ratios of Complementary Angles E. Trigonometric Identities 9. SOME APPLICATIONS OF TRIGONOMETRYA. Introduction B. Heights and Distances 10. CIRCLES A. Introduction B. Tangent to a Circle C. Number of Tangents from a Point on a Circle 11. CONSTRUCTIONS A. Introduction B. Division of a Line Segment C. Construction of Tangents to a Circle 12. AREAS RELATED TO CIRCLESA. Introduction B. Perimeter and Area of a Circle — A Review C. Areas of Sector and Segment of a CircleD. Areas of Combinations of Plane Figures 13. SURFACE AREAS AND VOLUMESA. Introduction B. Surface Area of a Combination of SolidsC. Volume of a Combination of Solids D. Conversion of Solid from One Shape to Another E. Frustum of a Cone 14. STATISTICS A. Introduction B. Mean of Grouped Data C. Mode of Grouped Data D. Median of Grouped Data E. Graphical Representation of Cumulative Frequency Distribution 15. PROBABILITYA. Introduction B. Probability — A Theoretical Approach
STANDARDIZATION OF CONCEPTS CLASS IX1. NUMBER SYSTEMS A. IntroductionB. Irrational Numbers C. Real Numbers and their Decimal Expansions D. Representing Real Numbers on the Number Line E. Operations on Real NumbersF. Laws of Exponents for Real Numbers 2. POLYNOMIALS A. Introduction B. Polynomials in One Variable
C. Zeroes of a Polynomial D. Remainder Theorem E. Factorisation of Polynomials F. Algebraic Identities3. COORDINATE GEOMETRYA. Introduction B. Cartesian System C. Plotting a Point in the Plane if its Coordinates are given 4. LINEAR EQUATIONS IN TWO VARIABLESA. Introduction B. Linear EquationsC. Solution of a Linear Equation D. Graph of a Linear Equation in Two Variables E. Equations of Lines Parallel to x-axis and y-axis 5. INTRODUCTION TO EUCLID’S GEOMETRYA. IntroductionB. Euclid’s Definitions, Axioms and Postulates C. Equivalent Versions of Euclid’s Fifth Postulate 6. LINES AND ANGLES A. Introduction B. Basic Terms and DefinitionsC. Intersecting Lines and Non-intersecting Lines D. Pairs of Angles E. Parallel Lines and a TransversalF. Lines Parallel to the same Line G. Angle Sum Property of a Triangle7. TRIANGLESA. IntroductionB. Congruence of Triangles C. Criteria for Congruence of Triangles D. Some Properties of a Triangle E. Some More Criteria for Congruence of Triangles F. Inequalities in a Triangle 8. QUADRILATERALS A. Introduction B. Angle Sum Property of a QuadrilateralC. Types of Quadrilaterals D. Properties of a Parallelogram E. Another Condition for a Quadrilateral to be a Parallelogram F. The Mid-point Theorem 9. AREAS OF PARALLELOGRAMS AND TRIANGLESA. Introduction B. Figures on the same Base and Between the same Parallels C. Parallelograms on the same Base and between the same Parallels D. Triangles on the same Base and between the same Parallels 10. CIRCLES A. Introduction
B. Circles and its Related Terms : A Review C. Angle Subtended by a Chord at a Point D. Perpendicular from the Centre to a Chord E. Circle through Three Points F. Equal Chords and their Distances from the Centre G. Angle Subtended by an Arc of a Circle H. Cyclic Quadrilaterals 11. CONSTRUCTIONS A. Introduction B. Basic ConstructionsC. Some Constructions of Triangles12. HERON’S FORMULA A. IntroductionB. Area of a Triangle – by Heron’s FormulaC. Application of Heron’s Formula in finding Areas of Quadrilaterals 13. SURFACE AREAS AND VOLUMES A. Introduction B. Surface Area of a Cuboid and a CubeC. Surface Area of a Right Circular Cylinder D. Surface Area of a Right Circular Cone E. Surface Area of a Sphere F. Volume of a Cuboid G. Volume of a Cylinder H. Volume of a Right Circular ConeI. Volume of a Sphere14. STATISTICS A. Introduction B. Collection of Data C. Presentation of Data D. Geographical Representation of Data E. Measures of Central Tendency15. PROBABILITY A. Introduction B. Probability – an Experimental Approach
STANDARDIZATION OF CONCEPTS CLASS VIII
1. RATIONAL NUMBERS
A. Rational Numbers between Two Rational Numbers
B. Properties of Rational Number
2. LINEAR EQUATIONS IN ONE VARIABLE
A. Introduction to Linear Equations
B. Application of Linear Equations
3. UNDERSTANDING QUADRILATERALS
A. Polygons
B. Kinds of Quadrilaterals
C. Some Special Parallelograms
4. PRACTICAL GEOMETRY
A. Construction of Quadrilaterals
5. DATA HANDLING
B. Data Handling
C. Probability
6. SQUARES AND SQUARE ROOTS
A. Properties of Square Numbers
B. Finding Square and Square roots
7. CUBES AND CUBE ROOTS
A. Cubes and Cube Roots
8.COMPARING QUANTITIES
B. Percentage and Discount
C. Simple and Compound Interest
9.ALGEBRAIC EXPRESSIONS AND IDENTITIES
A. Algebraic Expressions
B. Algebraic Identities
C. Multiplication of Algebraic Expressions
10. VISUALISING SOLID SHAPES
A. Visualising Solid Shapes
11. MENSURATION
B. Area of Quadrilaterals and Polygons
C. Surface Area of Solids
D. Volume of Solids
12. EXPONENTS AND POWERS
A. Exponents
13. DIRECT AND INVERSE PROPORTIONS
B. Direct and Inverse Proportions
14.FACTORISATION
C. Factors of Algebraic Expressions
D. Division of Algebraic Expressions
15. INTRODUCTION TO GRAPHS
A. Graphs
B. Linear Graphs
16. PLAYING WITH NUMBERS
A. General Form of Numbers
B. Tests of Divisibility
STANDARDIZATION OF CONCEPTS CLASS VII1.INTEGERS
A. Operations on Integers
B. Properties of Integers
2.FRACTIONS AND DECIMALS
A. Multiplication and Division on Fraction
B. Multiplication and Division on Decimals
3.DATA HANDLING
A. Data Representation
B. Data Value
4.SIMPLE EQUATIONS
A. Introduction to Simple Equations
B. Application of Simple Equations
5.LINES AND ANGLES
A. Angles
B. Pairs of Lines
6.THE TRIANGLE AND ITS PROPERTIES
A. Triangles
B. Properties of Triangles
7.CONGRUENCE OF TRIANGLES
A. Congruence of Plane Figures
B. Criteria for Congruence of Triangles
8.COMPARING QUANTITIES
A. Ratios and Proportions
B. Percentages
C. Conversions
D. Application of Percentages
9.RATIONAL NUMBERS
A. Introduction to Rational Numbers
B. Comparison of Rational Numbers
C. Operations on Rational Numbers
D. 10.Practical Geometry
E. Construction of Triangles
F. Construction of Parallel Lines
11.PERIMETER AND AREA
A. Plane Figures
B. Circles
12.ALGEBRAIC EXPRESSIONS
A. Understanding Algebraic Expressions
B. Operations on Algebraic Expressions
C. Application of Algebraic Expressions
13.EXPONENTS AND POWERS
A. Exponents and Powers
B. Laws of Exponents
14.SYMMETRY
A. Line Symmetry
B. Rotational Symmetry
15.VISUALISING SOLID SHAPES
A. Introduction to Solid Shapes
B. Viewing the Different Section of Solids
STANDARDIZATION OF CONCEPTS CLASS VI
1.KNOWING OUR NUMBERS
A. Comparing Numbers
B. Estimation of the Numbers
C. Roman Numerals
D. Importance of Brackets
2.WHOLE NUMBERS
A. Whole Numbers
B. Properties of Whole Numbers
3.PLAYING WITH NUMBERS
A. Prime and Composite Numbers
B. Divisibility of Numbers
C. Prime Factorisation, HCF and LCM
4.BASIC GEOMETRICAL IDEAS
A. Points, Lines and Curves
B. Angles, Polygons and Circles
5.UNDERSTANDING ELEMENTARY SHAPES
A. Lines and Angles
B. Two Dimensional Figures
C. Three Dimensional Shapes
6.INTEGERS
A. Integers
7.FRACTIONS
B. Types of Fractions
C. Comparing Fractions
D. Addition and Subtraction of Fractions
8.DECIMALS
A. Comparing Decimals
B. Addition and Subtraction of Decimals
9.DATA HANDLING
A. Data Handling
B. Pictograph
C. Bar Graph
10.MENSURATION
A. Perimeter
B. Area
11.ALGEBRA
A. Variables
B. Use of Variables
C. Equations
12.RATIOS AND PROPORTIONS
A. Ratios
B. Proportions
13.SYMMETRY
A. Line Symmetry
B. Mirror Symmetry
14.PRACTICAL GEOMETRY
A. Basic Constructions
B. Construction of Lines
C. Construction of Angles
sym Meaning Example
bol
+ add 3+7 = 10
− subtract 5−2 = 3
× multiply 4×3 = 12
/ divide 20/5 = 4
( ) grouping symbols 2(a−3)
[ ] grouping symbols 2[ a−3(b+c) ]
{ } set symbols {1,2,3}
π pi A = πr2
∞ infinity ∞ is endless
= equals 1+1 = 2
approximately equal to π 3.14
≠ not equal to π ≠ 2
< ≤ less than, less than or equal to 2 < 3
> ≥ greater than, greater than or equal to 5 > 1
square root ("radical") √4 = 2
° Degrees 20°
Therefore a=b b=a
implies (if ... then) a and b are odd a+b is even
"if and only if" or iff or "is equivalent to" x=y+1 y=x−1
Therefore a=b b=a
! Factorial 4! = 4×3×2×1 = 24
GREEK ALPHABET CHART
NUMBER SYSTEM
1. Types of NumbersI. Natural NumbersCounting numbers 1,2,3,4,5,... are called natural numbersII. Whole NumbersAll counting numbers together with zero form the set of whole numbers.Thus,(i) 0 is the only whole number which is not a natural number.(ii) Every natural number is a whole number.III. IntegersAll natural numbers, 0 and negatives of counting numbers i.e., ...,
−3,−2,−1,0,1,2,3,..... ...,−3,−2,−1,0,1,2,3,..... together form the set of integers.
(i) Positive Integers: 1,2,3,4,... is the set of all positive integers.(ii) Negative Integers: −1,−2,−3,... is the set of all negative integers.(iii) Non-Positive and Non-Negative Integers: 0 is neither positive
nor negative.So, 0,1,2,3,.. represents the set of non-negative integers,while 0,−1,−2,−3,... represents the set of non-positive integers.IV. Even NumbersA number divisible by 2 is called an even number, e.g.,2,4,6,8 etc.
V. Odd NumbersA number not divisible by 2 is called an odd number. e.g.,1,3,5,7,9,11 etc.VI. Prime NumbersA number greater than 1 is called a prime number, if it has exactly two
factors, namely 1 and the number itself. Prime numbers up to 100
are :2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.
Prime numbers less than 14 are 2,3,5,7,11,13191 is not divisible by any of them. So, 191 is a prime number.VII. Composite NumbersNumbers greater than 1 which are not prime, are known as composite
numbers, e.g., 4,6,8,9,10Note:(i) 1 is neither prime nor composite.(ii) 2 is the only even number which is prime.(iii) There are 25 prime numbers between 1 and 100.
VII. Co- Prime Two numbers are said to be co- prime if they have no common
factors.Example:2 &3 ; 6 &11;11&15 etcNote: (i) Two prime numbers are always co prime .(ii) Two consecutive numbers arec always co-prime
Rational and Irrational NumberTopic OverviewA rational number is part of a whole expressed as a fraction, decimal or a percentage.A number is rational if we can write it as a fraction where the top number of the fraction and bottom number are both whole numbers.The term rational is derived from the word 'ratio' because the rational numbers are figures which can be written in the ratio form.Every whole number, including negative numbers and zero, is a rational number. This is because every whole number ‘n’ can be written in the form n/1For example, 3 = 3/1 and therefore 3 is a rational number.
Numbers such as 3/8 and -4/9 are also rational because their numerators and denominators are both whole numbers.Recurring decimals such as 0.26262626…, all integers and all finite decimals, such as 0.241, are also rational numbers.Alternatively, an irrational number is any number that is not rational. It is a number that cannot be written as a ratio of two integers (or cannot be expressed as a fraction).For example, the square root of 2 is an irrational number because it cannot be written as a ratio of two integers.The square root of 2 is not a number of arithmetic: no whole number, fraction, or decimal has a square of 2. Irrational numbers are square roots of non-perfect squares. Only the square roots of square numbers are rational.Similarly Pi (π) is an irrational number because it cannot be expressed as a fraction of two whole numbers and it has no accurate decimal equivalent.Pi is an unending, never repeating decimal, or an irrational number. The value of Pi is actually 3.14159265358979323… There is no pattern to the decimals and you cannot write down a simple fraction that equals Pi.Euler's Number (e) is another famous irrational number. Like Pi, Euler's Number has been calculated to many decimal places without any pattern showing. The value of e is 2.7182818284590452353… and keeps going much like the value of Pi.The golden ratio (whose symbol is the Greek letter "phi") is also an irrational number. It is a special number approximately equal to 1.618 but again its value is never ending: 1.61803398874989484820...
2.Complex NumbersA complex number is a number that can be expressed in the form a +
bi, where a and b are real and i is imaginary unit, satisfying the equation i2 = −1 In this expression, a is called the real part and b is the imaginary part of the complex number.
3. Remainder and Quotient"The remainder is r when p is divided by k" means p=kq+r the integer q
is called the quotient.For instance, "The remainder is 1 when 7 is divided by 3" means
7=3∗2+1 Dividing both sides of p=kq+r p=kq+r by k gives the following alternative form pk =q+rk
Example:The remainder is 57 when a number is divided by 10,000. What is the
remainder when the same number is divided by 1,000?(A) 5 (B) 7 (C) 43 (D) 57 (E) 570Solution:Since the remainder is 57 when the number is divided by 10,000, the
number can be expressed as 10,000n+57 , where n is an integer.Rewriting 10,000 as 1,000∗10 1,000∗10 yields 10,000n+57=1,000(10n)
+57 10,000n+57=1,000(10n)+57Now, since n is an integer, 10n is an integer. Letting 10n=q , we get10,000n+57=1,000∗q+57 10,000n+57=1,000∗q+57Hence, the remainder is still 57 (by the p=kq+r form) when the number
is divided by 1,000. The answer is (D).
Method II (Alternative form):Since the remainder is 57 when the number is divided by 10,000, the
number can be expressed as 10,000n+57 10,000n+57. Dividing this number by 1,000 yields
10,000n+571000 10,000n+571000 =10,000n1000 +571000 =10,000n1000+571000 =10n+571000 =10n+571000
Hence, the remainder is 57 (by the alternative form pk =q+rk ), and the answer is (D).
4. Even, Odd NumbersA number n is even if the remainder is zero when n n is divided by
2:n=2z+0, or n=2z A number n n is odd if the remainder is one when n n is divided by
2:n=2z+1 The following properties for odd and even numbers are very useful - you
should memorize them:even * even=even,odd * odd=odd,even * odd=even,even + even=even,odd + odd=even,even + odd=odd
Example:If n n is a positive integer and (n+1)(n+3) (n+1)(n+3) is odd, then (n+2)
(n+4) (n+2)(n+4) must be a multiple of which one of the following?(A) 3 (B) 5 (C) 6 (D) 8 (E) 16Solution:(n+1)(n+3) is odd only when both (n+1) and (n+3) are odd. This is
possible only when n is even.Hence, n=2m , where m is a positive integer. Then,(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)'(n+2)(n+4)
=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2) =4 * (product of two consecutive positive integers, one which must be even) =4 * (product of two consecutive positive integers, one which must be even) =4 * (an even number), and this equals a number that is at least a multiple of 8 =4 * (an even number), and this equals a number that is at least a multiple of 8
Hence, the answer is (D).
5. Tests of Divisibility5.1. Divisibility By 2A number is divisible by 2 if its unit's digit is any of 0,2,4,6,8 0,2,4,6,8.Example:84932 is divisible by 2, while 65935 is not.5.2. Divisibility By 3A number is divisible by 3 if the sum of its digits is divisible by 3.
Example:592482 is divisible by 3, since sum of its digits =(5+9+2+4+8+2)=30 ,
which is divisible by 3.But, 864329 is not divisible by 3, since sum of its digits
=(8+6+4+3+2+9)=32 , which is not divisible by 3.
5.3. Divisibility By 4A number is divisible by 4 if the number formed by the last two digits is
divisible by 4.Example:892648 is divisible by 4 since the number formed by the last two digits is
48, which is divisible by 4. But, 749282 is not divisible by 4, since the number formed by the last two digits is 82, which is not divisible by 4
5.4. Divisibility By 5A number is divisible by 5 if its unit's digit is either 0 or 5. Thus, 20820
and 50345 are divisible by 5, while 30934 and 40946 are not.5.5. Divisibility By 6A number is divisible by 6 if it is divisible by both 2 and 3.Example:The number 35256 is clearly divisible by 2.Sum of its digits
=(3+5+2+5+6)=21 , which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.
5.6. Divisibility By 8A number is divisible by 8 if the number formed by the last Three digits of
the given number is divisible by 8.Example:953360 is divisible by 8 since the number formed by last three digits is
360, which is divisible by 8. But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.
5.7. Divisibility By 9A number is divisible by 9 if the sum of its digits is divisible by 9.Example:60732 is divisible by 9, since sum of digits =(6+0+7+3+2)=18 , which is
divisible by 9.But, 68956 is not divisible by 9, since sum of digits =(6+8+9+5+6)=34,
which is not divisible by 9.
5.8. Divisibility By 10A number is divisible by 10 if it ends with 0.Example:96410, 10480 are divisible by 10, while 96375 is not.
5.9. Divisibility By 11A number is divisible by 11, if the difference of the sum of its digits at odd
places and the sum of its digits at even places, is either 0 or a number divisible by 11.
Example:The number 4832718 is divisible by 11, since :(sum of digits at odd
places) - (sum of digits at even places) = :(sum of digits at odd places) - (sum of digits at even places) =
=(8+7+3+4)−(1+2+8)=11 , which is divisible by 11.5.10. Divisibility By 12A number is divisible by 12 if it is divisible by both 4 and3.Example:Consider the number 34632.(i) The number formed by last two digits is 32, which is divisible by 4,(ii) Sum of digits =(3+4+6+3+2)=18 , which is divisible by 3. Thus,
34632 is divisible by 4 as well as 3. Hence, 34632 is divisible by 12.
5.11. Divisibility By 14A number is divisible by 14 if it is divisible by 2 as well as 7.5.12. Divisibility By 15A number is divisible by 15 if it is divisible by both 3 and 5.5.13. Divisibility By 16A number is divisible by 16 if the number formed by the last4 digits is
divisible by 16.Example:7957536 is divisible by 16 since the number formed by the last four digits
is 7536, which is divisib5.14. Divisibility By 24A given number is divisible by 24 if it is divisible by both 3 and 8.5.15. Divisibility By 40A given number is divisible by 40 if it is divisible by both 5 and 8.5.16. Divisibility By 80A given number is divisible by 80 if it is divisible by both 5 and 16.Note:If a number is divisible by p p as well as q q, where p p and q q are co-
primes, then the given number is divisible by pq pq. If p p and q q are not co-primes, then the given number need not be divisible by pq pq, even when it is divisible by both p p and q q.
Example:36 is divisible by both 4 and 6, but it is not divisible by (4∗6)=24
(4∗6)=24 since 4 and 6 are not co - primes.
I
6 .Highest common factor and lowest common multiple
We can use prime factors to find the highest common factor (HCF) and lowest common multiple (LCM).
Highest common factor
Here are the list of prime factors of 24 and 36:
24 = 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
If we write down the numbers that are the same in both lists, they will give us the highest common factor of 24 and 36:
HCF of 24 and 36 is 2 x 2 x 3 = 12
Lowest common multiple
To find the lowest common multiple, we need to think about which list has the most of each factor.
24 = (2 x 2 x 2) x (3)
36 = (2 x 2) x (3 x 3)
24 has the most 2s. (it has three 2s). And 36 has the most 3s (it has two 3s).
So the LCM of 24 and 36 is (2 x 2 x 2) x (3 x 3) = 72.
Indices IntroductionThe index of a number says how many times to use the number in a
multiplication.
It is written as a small number to the right and above the base number.
In this example: 82 = 8 × 8 = 64
The plural of index is indices
Six rules of the Law of Indices
Rule 1: Any number, except 0, whose index is 0 is always equal to 1, regardless of the
value of the base.
Example: Simplify 20 = 1
Rule 2 : a -m = 1/am
Example : 3-2 = 132
Rule 3 : x a ×x b =x a+b
Example : Simplify 5 6 ×5 7 .When multiplying, add together the powers: 5 6 ×5 7 =5 6+7 =5 13 .
Rule 4 : x a ÷x b =x a−b
Example : Simplify x8 ÷x2 . When dividing, subtract the powers: x 8 ÷ x 2 =x 8−2 =x 6 .
Rule 5 : (x a ) b =x a×b
Example : Simplify (x 2 ) 6 . When raising a power to another power, multiply the powers: (x 2 ) 6 =x 2×6 =x 12 .
Common error : ( 23 )2 and 232
are different.
Rule 6 : Anything to the power of 1 is itself, i.e. x 1 =x
Practice Questions
1. In each case choose an appropriate law to simplify the expression:
a) 53 × 513 b) 813 ÷ 85 c) x6 × x5 d) (a3)4 e) y7
y3 f) x8
x10 .
2. Use one of the laws to simplify, if possible, a6 × b5.
Surds
IntroductionSurds are numbers left in root form (√) to express its exact value. It has an
infinite number of non-recurring decimals. Therefore, surds are irrational numbers.
where n is an integer and a can any number real , imaginary or complex.
There are certain rules that we follow to simplify an expression involving surds. Rationalising the denominator is one way to simplify these expressions. It is done by eliminating the surd in the denominator.
Six Rules of Surds
Rule 1:
Where at least one, either a or b is +ve integer.
Example: Simplify :
Since , as 9 is the largest perfect square factor of 18.
Rule 2:
Where at least one, either a or b is +ve integer.
Example: Simplify :
Rule 3:
By multiplying both the numberator and denominator by the denominator you can rationalise the denominator.
Example: Rationalise
:
Rule 4:
Example: Simplify :
Rule 5:
Example: Rationalise :
Rule 6:
Example: Rationalise :
Common error during rationalization is committing mistake in using identities :1) a2 – b2 = a - b 2) (a+b)2= a2+b2−¿2ab
Practice Questions
Write each of the following as a radical and simplify where possible:a) 18
12
b) (4)32
c) (−125)−13
d) (−81)12
e) (.008)13
f) Rationalise the denominator √3−√2√3+√2 .
POLYNOMIAL-The word "polynomial" is originated from two words - "poly" and "nomial". Poly has a
meaning of "many", while nomial does refer to terms. In this way, the meaning of polynomial is an expression that has many term. A polynomial is an expression or function of the following form – General representation of polynomial:
F(x) = an xn + an – 1xn - 1 + an - 2 xn - 2 ……………. + a1x + a0
Where, n = non negative integer and ai = real numbers
The constants a0, a1, a2… are called as coefficients.
Degree of a Polynomial
The highest value of exponents is called degree of polynomial.Polynomials are classified according to the degree. Polynomials are of following types:
Constant PolynomialPolynomial with degree 0.
Linear Polynomial
Linear polynomial is a polynomial whose degree is 1.
The general form of a linear polynomial is ax + b.
Quadratic PolynomialQuadratic polynomial is a polynomial of degree 2.
The general form of a quadratic polynomial is ax2 + bx + c, a ≠ 0.
Example: x2 - 1 is a quadratic polynomial.
Cubic Polynomial
Cubic polynomial is a polymial of degree 3.The general form of the cubic polynomial is ax3 + bx2 + cx + d, a ≠0
Example: 3x3 - 2x + 4 is a cubic polynomial.
Roots or zeros of the polynomial
It is a solution to the polynomial equation, P(x) = 0. It is that value of x that makes the polynomial equal to 0.
In other words, the number r is a root of a polynomial P(x)if and only if P(r) = 0.
The roots of this quadratic
x2 −x − 6 = (x + 2)(x − 3)
are −2 and 3. These are the values of x that will make the polynomial equal to 0.
GEOMETRIC REPRESENTATION OF ZEROS
x-intercept and y-intercept of a graph?
The x-intercept is that value of x where the graph crosses or touches the x-axis. At the x-intercept -- on the x-axis -- y = 0.
The y-intercept is that value of y where the graph crosses the y-axis. At the y-intercept, x = 0.
What is the relationship between the root of a polynomial and the x-intercepts of its graph?
The roots are the x-intercepts!
The roots of x2 −x − 6 are −2 and 3. Therefore, the graph of y = x2 −x − 6
will have the value 0 -- it will cross the x-axis -- at −2 and 3.
How do we find the x-intercepts of the graph of any function y = f(x)?
Solve the equation, f(x) = 0.
Problem 1.
a) Find the root of the polynomial 2x + 10.
We must solve the equation, 2x + 10 = 0. The solution is x = −5.
b) Where is the x-intercept of the graph of y = 2x + 10?
At x = −5. The x-intercept is the root.
Basic Identities
1. (a+b) 2 =a2 +b2 +2ab
2. (a−b)2 =a2 +b2 −2ab
3. (a+b) 2 −(a−b)2= 4ab
4. (a+b) 2 +(a−b)2=2(a2 +b2 )
5. (a2 −b2 )=(a+b)(a−b) 6. (a+b+c)2 =a2 +b2 +c2 +2(ab+bc+ca)
7. (a3 +b 3 )=(a+b)(a 2 −ab+b2 ) 8. (a3 - b 3 )=(a- b)(a 2 +ab+b2 )
9. (a3 +b3 +c3 −3abc)=(a+b+c)(a 2 +b 2 +c2 −ab−bc−ca) 10. If a+b+c=0 ,then a3 +b3 +c3=3abc
MR AJAY KUMAR GIRI AFS BawanaMR MANISH KUMAR SHARMA Sector12 DwarkaMR SURENDER Kumar Duggal No.4, BatindaMr.Kochin Kumar HaidwaniMr.S.N.Sharma NO. 1 Faridabad
Linear equation in one variable and two variable:
EquationA statement of equality which contains one or more unknown quantity or variable is called an equation.
Linear EquationAn equation involving only linear polynomials is called linear equation
Linear equation in one variable:
Algebraic expressions with one variable with degree one are called linear expressions.
All expressions that have degree greater than one are not linear. An equation can be defined as a mathematical statement that uses symbols to express equality between mathematical expressions.
The expression on the left of the equality sign is the ‘LHS’ of the equation, while that on the right is its ‘RHS.
Solution of the equation:The value of variable, for which the LHS and RHS of an equation give the same result, is called the Solution of the equation.
An algebraic equation can be solved by performing mathematical operations such as +, - , x and ÷ to both sides of the equation or by transposing a number to the other side and performing the opposite operation on the number so transposed.
A linear equation may have a rational number as its solution.
Rules for solving Linear equation in one variable
If you add on the left side of the equal sign, you must add on the right side.If you add a term on the left side, you must add the same term on the right side.
If you multiply a term on the left side, you must multiply the same term on the right side.
Example 6: Solve for the variable. 10 - 3x = 7.
Example 7: Solve for the variable. 2(x + 5) - 7 = 3(x - 2).
Example 8: Solve for the variable: .
Example Solve the equation 2x +3=6 − (2x − 3).
Solution From 2x +3=6 − (2x − 3) we first remove the brackets on the right to give 2x +3=6 − 2x + 3 so that 2x +3=9 − 2x We need to get the x’s together by adding 2x to each side. 4x +3=9 Now take 3 away from each side: 4x = 6 so that x = 6/ 4 ( by dividing both sides by 4) = 3 /2 = 1 1
2
When solving simple equations we should always check the solution by taking our answer and substituting it in the original equation to check that the left- and right- hand sides are the same. Substituting x = 1 1
2 in the left-
hand side gives: 2 ×1 1
2 +3=3+3=6
Substituting x = 1 12 in the right-hand side gives:
6 − 2 × 1 12 − 3 = 6 − 0=6
So again, the left- and right- hand sides are equal - we’ve got that balance, so we know that we’ve got the right answer. Exercises 1. Solve the following equations. a) x + 5 = 9 b) 12 − x = 7 c) 5x = 3 d) 4x + 10 = 2 e) 5 − 3x = −4 f) 2 + 14x = 30 g) 9 + 5x = 3x + 13 h) 4 − 3x =8+ x i) 5 + 3(x − 1) = 5x − 6 3.
Pair of Linear Equations in Two Variables
Introduction-Linear Equation in Two variables
If two linear equations have the two variables, they are called a pair of linear equations in two variables. Following is the most general form of linear equations:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Here, a1, a2, b1, b2, c1 and c2 are real numbers such that;
A pair of linear equations can be represented and solved by the following methods:
a. Graphical methodb. Algebraic method
Graphical Method:
For a given pair of linear equations in two variables, the graph is represented by two lines.
a. If the lines intersect at a point, that point gives the unique solution for the two equations. If there is a unique solution of the given pair of equations, the equations are called consistent.
b. If the lines coincide, there are indefinitely many solutions for the pair of linear equations. In this case, each point on the line is a solution. If there are infinitely many solutions of the given pair of linear equations, the equations are called dependent (consistent).
c. If the lines are parallel, there is no solution for the pair of linear equations. If there is no solution of the given pair of linear equations, the equations are called inconsistent.
Algebraic Method:
There are following methods for finding the solutions of the pair of linear equations:
a. Substitution methodb. Elimination methodc. Cross-multiplication method
If a pair of linear equations is given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 then following situations can arise.
Situation 1:
In this case, the pair of linear equations is consistent. This means there is unique solution for the given pair of linear equations. The graph of the linear equations would be two intersecting lines.
Situation 2:
In this case, the pair of linear equations is inconsistent. This means there is no solution for the given pair of linear equations. The graph of linear equations will be two parallel lines.
Situation 3:
In this case, the pair of linear equations is dependent and consistent. This means there are infinitely many solutions for the given pair of linear equations. The graph of linear equations will be coincident lines.
Observe the steps how to solve the system of linear equations by using the substitution method.
(i) Find the value of one variable in terms of the other from one of the given equations.
(ii) Substitute the value of this variable in the other equation.
(iii) Solve the equation and get the value of one of the variables.
(iv) Substitute the value of this variable in any of the equation to get the value of other variable.
Follow the instructions along with the method of solution of the two simultaneous equations given below to find the value of x and y.
7x – 3y = 31 --------- (i)
9x – 5y = 41 --------- (ii)
Step I: From equation (i) 7x – 3y = 31, express y in terms of x
From equation (i) 7x – 3y = 31, we get;
– 3y = 31 – 7x
or, 3y = 7x – 31
or, 3y/3 = (7x – 31)/3
Therefore, y = (7x – 31)/3 --------- (iii)
Step II: Substitute the value of y obtained from equation (iii) (7x – 31)/3 in equation (ii) 9x – 5y = 41
Putting the value of y obtained from equation (iii) in equation (ii) we get;
9x – 5 × (7x – 31)/3 = 41 --------- (iv)
Step III: Now, solve equation (iv) 9x – 5 × (7x – 31)/3 = 41
Simplifying equation (iv) 9x – 5 × (7x – 31)/3 = 41 we get;
(27x – 35x + 155)/3 = 41
or, 27x – 35x + 155 = 41 × 3
or, 27x – 35x + 155 = 123
or, –8x + 155 = 123
or, –8x + 155 – 155 = 123 – 155
or, –8x = –32
or, 8x/8 = 32/8
Therefore, x = 4
Step IV: Putting the value of x in equation (iii)
y = (7x – 31)/3, find the value of y
Putting x = 4 in equation (iii), we get;
y = (7 × 4 – 31)/3
or, y = (28 – 31)/3
or, y = –3/3
Therefore, y = –1
Step V: Write down the required solution of the two simultaneous linear equations by using the substitution method
Therefore, x= 4 and y = –1
In this case, the general method obtained for solving simultaneous equations as follows:
1. To express y in terms of x from any one of the equations.
2. To substitute this value of y in the other equation.
3. One value of x will be obtained, by solving the equation in x thus obtained.
4. Substituting this value of x in any of the equations, we will get the corresponding value of y.
5. Solution of the two given simultaneous equations will be given by this pair of values of x and y.6. Similarly expressing x in terms of y from an equation and substituting in the other, we can find the value of y. Putting this value of yin any one of the equations, we can find the value of x and thus we can solve the two linear simultaneous equations. As in this method of solution, we express one unknown quantity in terms of the other and substitute in an equation; o we call this method as ‘Method of Substitution’.
Keep these instructions in your mind and notice how the following simultaneous equations can be solved.
Worked-out examples on two variables linear equations by using the substitution method:
2/x + 3/y = 2 --------- (i)
5/x + 10/y = 5⁵/₆ --------- (ii)
From equation (i), we get:
3/y = 2 – 2/x
or, 3/y = (2x – 2)/x
or, y/3 = x/(2x – 2)
or, y = 3x/(2x – 2) --------- (iii)
Substituting 3x/(2x – 2) in place of y in equation (ii),
or, 5/x + 10 ÷ 3x/(2x – 2) = 35/6
or, 5/x + 10(2x – 2)/3x = 35/6
or, 1/x + 2(2x – 2)/3x = 7/6
or, (3 + 4x – 4)/3x = 7/6
or, (4x – 1)/3x = 7/6
or, (4x – 1)/x = 7/2
or, 8x – 2 = 7x
or, 8x – 2 + 2 = 7x + 2 or, 8x – 7x = 7x – 7x + 2
or, x = 2
Putting the value of x = 2 in equation (iii),
or, y = 3 ∙ 2/2 ∙ 2 – 2
or, y = 6/4 – 2
or, y = 6/2
or, y = 3
Therefore, the required solution is x = 2 and y = 3.
Follow the steps to solve the system of linear equations by using the elimination method:
(i) Multiply the given equation by suitable constant so as to make the coefficients of the variable to be eliminated equal.
(ii) Add the new equations obtained if the terms having the same coefficient are opposite signs and subtract if they are of the same sign.
(iii) Solve the equation thus obtained.
(iv) Substitute the value found in any one the given equations.
(v) Solve it to get the value of the other variable.
Worked-out examples on elimination method:
1. Solve the system of equation 2x + y = -4 and 5x – 3y = 1 by the method of elimination.
Solution:
The given equations are:
2x + y = -4 …………… (i)
5x – 3y = 1 …………… (ii)
Multiply equation (i) by 3, we get;
{2x + y = -4} …………… {× 3}
6x + 3y = -12 …………… (iii)
Adding (ii) and (iii), we get;
or, x = -11/11
or, x = -1
Substituting the value of x = -1 in equation (i), we get;
2 × (-1) + y = -4
-2 + y = -4
-2 + 2 + y = -4 + 2
y = -4 + 2
y = -2
Therefore, x = -1 and y = -2 is the solution of the system of equations 2x + y = -4 and 5x – 3y = 1
2. Solve the system of equation 2x + 3y = 11, x + 2y = 7 by the method of elimination.
Solution:
The given equations are:
2x + 3y = 11 …………… (i)
x + 2y = 7 …………… (ii)
Multiply the equation (ii) by 2, we get
{x + 2y = 7} …………… (× 2)
2x + 4y = 14 …………… (iii)
Subtract equation (i) and (ii), we get
Substituting the value of y = 3 in equation (i), we get
2x + 3y = 11
or, 2x + 3 × 3 = 11
or, 2x + 9 = 11
or, 2x + 9 – 9 = 11 – 9
or, 2x = 11 – 9
or, 2x = 2
or, x = 2/2
or, x = 1
Therefore, x = 1 and y = 3 is the solution of the system of the given equations.
3. Solve 2a – 3/b = 12 and 5a – 7/b = 1
Solution:
The given equations are:
2a – 3/b = 12 …………… (i)
5a – 7/b = 1 …………… (ii)
Put 1/b = c, we have
2a – 3c = 12 …………… (iii)
5a – 7c = 1 …………… (iv)
Multiply equation (iii) by 5 and (iv) by 2, we get
10a – 15c = 60 …………… (v)
10a + 14c = 2 …………… (vi)
Subtracting (v) and (vi), we get
or, c = 58 /-29
or, c = -2
But 1/b = c
Therefore, 1/b = -2 or b = -1/2
Subtracting the value of c in equation (v), we get
10a – 15 × (-2) = 60
or, 10a + 30 = 60
or, 10a + 30 - 30= 60 - 30
or, 10a = 60 – 30
or, a = 30/10
or, a = 3
Therefore, a = 3 and b = 1/2 is the solution of the given system of equations.
4. x/2 + 2/3 y = -1 and x – 1/3 y = 3
Solution:
The given equations are:
x/2 + 2/3 y = -1 …………… (i)
x – 1/3 y = 3 …………… (ii)
Multiply equation (i) by 6 and (ii) by 3, we get;
3x + 4y = -6 …………… (iii)
3x – y = 9 …………… (iv)
Solving (iii) and (iv), we get;
or, y = -15/5
or, y = -3
Subtracting the value of y in (ii), we get;
x - 1/3 × -3 = 3
or, x + 1 = 3
or, x = 3 – 1
or, x = 2
Therefore, x = 2 and y = -3 is the solution of the equation.
x/2 + 2/3 y = -1 and x - y/3 = 3
Here we will discuss about simultaneous linear equations by using cross-multiplication method.
General form of a linear equation in two unknown quantities:
ax + by + c = 0, (a, b ≠ 0)
Two such equations can be written as:
a₁x + b₁y + c₁ = 0 ----------- (i)
a₂x + b₂y + c₂ = 0 ----------- (ii)
The formula for cross-multiplication and its use in solving two simultaneous equations can be presented as:
If (a₁b₂ - a₂b₁) ≠ 0 from the two simultaneous linear equations
a₁x + b₁y + c₁ = 0 ----------- (i)
a₂x + b₂y + c₂ = 0 ----------- (ii)
we get, by the cross-multiplication method:
x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁) ---------- (A)
That means, x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)
y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)
Worked-out examples on simultaneous linear equations by using cross-multiplication method:
1. Solve the two variables linear equation:
8x + 5y = 11
3x – 4y = 10
Solution:
On transposition, we get
8x + 5y – 11 = 0
3x – 4y – 10 = 0
Writing the co-efficient in the following way, we get:
Note: The above presentation is not compulsory for solving.
By cross-multiplication method:
x/[(5) (-10) – (-4) (-11)] = y/[(-11) (3) – (-10) (8)] = 1/[(8) (-4) – (3) (5)]
or, x/[-50 – 44] = y/[-33 + 80] = 1/[-32 – 15]
or, x/-94 = y/47 = 1/-47
or, x/-2 = y/1 = 1/-1 [multiplying by 47]
or, x = -2/-1 = 2 and y = 1/-1 = -1
Therefore, required solution is x = 2, y = -1
1. Use the method of substitution to solve each other of the pair of simultaneous equations:
(a) x + y = 15 x - y = 3
(b) x + y = 0 x - y = 2
(c) 2x - y = 3 4x + y = 3
(d) 2x - 9y = 9 5x + 2y = 27
(e) x + 4y = -4 3y - 5x = -1
(f) 2x - 3y = 2 x + 2y = 8
(g) x + y = 7 2x - 3y = 9
(h) 11y + 15x = -23 7y - 2x = 20
(i) 5x - 6y = 2 6x - 5y = 9
2. Solve each other pair of equation given below using elimination method:
(a) x + 2y = -4 3x - 5y = -1
(b) 4x + 9y = 5 -5x + 3y = 8
(c) 9x - 6y = 12 4x + 6y = 14
(d) 2y - (3/x) = 12 5y + (7/x) = 1
(e) (3/x) + (2/y) = (9/xy) (9/x) + (4/y) = (21/xy)
(f) (4/y) + (3/x) = 8 (6/y) + (5/x) = 13
(g) 5x + (4/y) = 7 4x + (3/y) = 5
(h) x + y = 3 -3x + 2y = 1
(i) -3x + 2y = 5 4x + 5y = 2
3. Solve the following simultaneous equations:
(a) 3a + 4b = 43 -2a + 3b = 11
(b) 4x - 3y = 23 3x + 4y = 11
(c) 5x + (4/y) = 7 4x + (3/y) = 5
(d) 4/(p - 3) + 6/(q - 4) = 5 5/(p - 3) - 3/(q - 4) = 1
(e) (l/6) - (m/15) = 4 (l/3) - (m/12) = 19/4
(f) 3x + 2y = 8 4x + y = 9
(g) x - y = -1 2y + 3x = 12
(h) (3y/2) - (5x/3) = -2 (y/3) + (x/3) = 13/6
(i) x - y = 3 (x/3) + (y/2) = 6
(j) (2x/3) + (y/2) = -1 (-x/3) + y = 3
(k) 5x + 8y = 9 2x + 3y = 4
(l) 3 - 2(3a - 4b) = -59 (a - 3)/4 - (b - 4)/5 = 2¹/₁₀
Answers:
1. (a) x = 9, y = 6 (b) x = 1, y = -1 (c) x = 1, y = -1
(d) x = 261/49, y = 9/49 (e) x = -8/23, y = -21/23 (f) x = 4, y = 2
(g) x = 6, y = 1 (h) x = -3, y = 2 (i) x = 4, y = 3
2. (a) x = -2, y = -1 (b) x = -1, y = 1 (c) x = 2, y = 1
(d) x = -1/2, y = 3 (e) x = 3, y = 1 (f) x = 1/2, y = 2
(g) x = -1, y = 1/3 (h) x = 1, y = 2 (i) x = -21/23, y = 26/23
3. (a) a = 5, b = 7 (b) x = 5, y = -1 (c) x = -1, y = 1/3
(d) p = 5, q = 6 (e) l = -2, m = -65 (f) x = 2, y = 1
(g) x = 2, y = 3 (h) x = 141/38, y = 53/19 (i) x = 9, y = 6
(j) x = -3, y = 2 (k) x = 5, y = -2 (l) a = 5, b = -4
Introduction of Quadratic Equation
Definition of a quadratic equation. A quadratic equation in x is an equation that can be written in the form a x2+bx+c=0, where a b and c are real numbers with a≠ 0. A quadratic equation in x also called a second degree polynomial equation in x.
We will discuss about the introduction of quadratic equation.
A polynomial of second degree is generally called a quadratic polynomial.
If f(x) is a quadratic polynomial, then f(x) = 0 is called a quadratic equation.
An equation in one unknown quantity in the form ax2 + bx + c = 0 is called quadratic equation.
A quadratic equation is an equation of the second degree.
The general form of a quadratic equation is ax2 + bx + c = 0 where a, b, c are real numbers (constants) and a ≠ 0, while b and c may be zero.Here, x is the variable, a is called the coefficient of x2, b the coefficient of x and c the constant (or absolute) term.
The values of x which satisfy the equation are called the roots of the quadratic equation.
Examples of quadratic equation:
(i) 5x2 + 3x + 2 = 0 is an quadratic equation.Here, a = the coefficient of x2 = 5,
b = coefficient of x = 3 and
c = constant = 2
(ii) 2m2 - 5 = 0 is an quadratic equation.Here, a = the coefficient of m2 = 2,
b = coefficient of m = 0 and
c = constant = -5
(iii) (x - 2)(x - 1) = 0 is an quadratic equation.
(x - 2)(x - 1) = 0
⇒ x2 - 3x + 2 = 0Here, a = the coefficient of x2 = 1,
b = coefficient of x = -3 and
c = constant = 2
(iv) x2 = 1 is an quadratic equation.x2 = 1⇒ x2 - 1 = 0Here, a = the coefficient of x2 = 1,
b = coefficient of x = 0 and
c = constant = -1
(v) p2 - 4p + 4 = 0 is an quadratic equation.Here, a = the coefficient of p2 = 1,
b = coefficient of p = -4 and
c = constant = 4
There are several methods you can use to solve a quadratic equation:
1. Factoring
2. Completing the Square
3. Quadratic Formula
4. Graphing
All methods start with setting the equation equal to zero.
Solve for x in the following equation.
Example 1: 2 x2−x−1=0
The equation is already set to zero.
Method 1: Factoring
Method 2: Completing the square
Divide both sides of the equation 2 x2−x−1=0 by 2.
Add 12 to both sides of the equation.
Add to both sides of the equation:
Factor the left side and simplify the right side :
Take the square root of both sides of the equation :
x−14=± 3
4 Add 14 to both sides of the equation :
Method 3: Quadratic Formula
The quadratic formula is
In the equation 2 x2−x−1=0 ,a is the coefficient of the term, b is the coefficient of the x term, and c is the constant. Substitute 2 for a, -1 for b, and -1 for c in the quadratic formula and simplify.
Solved examples of Quadratic equationsProblem 1: Solve for x: x2-3x-10 = 0Solution:Let us express -3x as a sum of -5x and +2x.→ x2-5x+2x-10 = 0→ x(x-5)+2(x-5) = 0→ (x-5)(x+2) = 0→ x-5 = 0 or x+2 = 0→ x = 5 or x = -2
Problem 2: Solve for x: x2-18x+45 = 0Solution:The numbers which add up to -18 and give +45 when multiplied are -15
and -3.Rewriting the equation,→ x2-15x-3x+45 = 0→ x(x-15)-3(x-15) = 0→ (x-15) (x-3) = 0→ x-15 = 0 or x-3 = 0→ x = 15 or x = 3 Till now, the coefficient of x2 was 1. Let us see how to solve the equations
where the coefficient of x2 is greater than 1.
Problem 3: Solve for x: 3x2+2x =1Solution:Rewriting our equation, we get 3x2+2x-1= 0Here, the coefficient of x2 is 3. In these cases, we multiply the constant c
with the coefficient of x2. Therefore, the product of the numbers we choose should be equal to -3 (-1*3).
Expressing 2x as a sum of +3x and –x→ 3x2+3x-x-1 = 0→ 3x(x+1)-1(x+1) = 0→ (3x-1)(x+1) = 0→ 3x-1 = 0 or x+1 = 0→ x = 1/3 or x = -1
Problem 4: Solve for x: 11x2+18x+7 = 0Solution:In this case, the sum of the numbers we choose should equal to 18 and
the product of the numbers should equal 11*7 = 77.This can be done by expressing 18x as the sum of 11x and 7x.→ 11x2+11x+7x+7 = 0→ 11x(x+1) +7(x+1) = 0→ (x+1)(11x+7) = 0→ x+1 = 0 or 11x+7 = 0→ x = -1 or x = -7/11. The factoring method is an easy way of finding the roots. But this method
can be applied only to equations that can be factored.For example, consider the equation x2+2x-6=0.If we take +3 and -2, multiplying them gives -6 but adding them doesn’t
give +2. Hence this quadratic equation cannot be factored. For this kind of equations, we apply the quadratic formula to find the
roots.The quadratic formula to find the roots,x = [-b ± √(b2-4ac)] / 2aNow, let us find the roots of the equation above.x2+2x-6 = 0Here, a = 1, b=2 and c= -6.Substituting these values in the formula,x = [-2 ± √(4 – (4*1*-6))] / 2*1→ x = [-2 ± √(4+24)] / 2→ x = [-2 ± √28] / 2
When we get a non-perfect square in a square root, we usually try to express it as a product of two numbers in which one is a perfect square. This is for simplification purpose. Here 28 can be expressed as a product of 4 and 7.
→ x = [-2 ± √(4*7)] / 2→ x = [-2 ± 2√7] / 2→ x = 2[ -1 ± √7] / 2→ x = -1 ± √7Hence, √7-1 and -√7-1 are the roots of this equation. Let us consider another example.Solve for x: x2 = 24 – 10xSolution:Rewriting the equation into the standard quadratic form,x2 +10x-24 = 0What are the two numbers which when added give +10 and when
multiplied give -24? 12 and -2.So this can be solved by the factoring method. But let’s solve it using the
new method, applying the quadratic formula.Here, a = 1, b = 10 and c = -24.x = [-10 ± √(100 – 4*1*-24)] / 2*1x = [-10 ± √(100-(-96))] / 2x = [-10 ± √196] / 2x = [-10 ± 14] / 2x = 2 or x= -12 are the roots.
DiscriminantFor an equation ax2+bx+c = 0, b2-4ac is called the discriminant and helps
in determining the nature of the roots of a quadratic equation.If b2-4ac > 0, the roots are real and distinct.If b2-4ac = 0, the roots are real and equal.If b2-4ac < 0, the roots are not real (they are complex). Consider the following example:Problem: Find the nature of roots for the equation x2+x+12 = 0.Solution:b2-4ac = -47 for this equation. So it has complex roots. Let us verify this.
→ [ -1±√(1-48)] / 2(1)→ [-1±√-47] / 2√-47 is usually written as i √47 indicating it’s an imaginary number.Hence verified.
Arithmetic ProgressionWhat is a sequence?
Here are a few lists of numbers: 3, 5, 7 ...
21, 16, 11, 6 ...
1, 2, 4, 8 ...
Ordered lists of numbers like these are called sequences. Each number in a sequence is called a term.
Sequences usually have patterns that allow us to predict what the next term might be.
For example, in the sequence 3, 5, 7 ..., you always add two to get the next term:
The three dots that come at the end indicate that the sequence can be extended, even though we only see a few terms.
We can do so by using the pattern.For example, the fourth term of the sequence should be nine, the fifth
term should be 11, etc.Sequence: A sequence is an arrangement of numbers in a definite
order according to some rule.This rule need not be a specific formula.For example. (i) 2, 4, 6, 8, 10,…… ( i.e. a sequence of even natural
numbers) Where first terma1=2 , second terma2=4 , third terma3=6……. , nthterman=2n(ii) 2, 3, 5, 7, 11, 13,……… (i.e. a sequence of prime numbers)Where first terma1=2 , second terma2=3 , third terma3=5 ……. ,
nth terman=can' t beexpressedSeries: A sequence in which terms are attached by + sign is series. For example 3 +8 +13 + 18 ……Progression: A progression is a sequence as per a definite rule. That
rule is expressed as a specific formula to calculate nth term.For example.(i) 5,8,11,……. Whose nth term = 3n+2
(ii) 2,-4, 8,-16, 32, whose nth term = (−1 )n−1 .2n
ExerciseExamine whether the following sequences are progressions. If yes, write
their nth term in the form of a specific formula in terms of n.
(i) 3, 9, 27, 81,………(ii) 2, 5, 10, 17,……..(iii) 1, 2, 2, 3, 3, 3,………..What is an arithmetic sequence?
For many of the examples above, the pattern involves adding or subtracting a number to each term to get the next term. Sequences with such patterns are called arithmetic sequences.
In an arithmetic sequence, the difference between consecutive terms is always the same.
For example, the sequence 3, 5, 7, 9 ... is arithmetic because the difference between consecutive terms is always two.
The sequence 21, 16, 11, 6 ... is arithmetic as well because the difference between consecutive terms is always five.
. The sequence 1, 2, 4, 8 ... is not arithmetic because the difference between consecutive terms is not the same.
Arithmetic progression: A sequence of numbers in which each number differs from the preceding one by a constant quantity.
ORA sequence of numbers such that the difference of any two successive
numbers is a constant. OR A sequence a1, a2 , a3 , a4 , ………an , is an A . P . , if thereexist aconstant termd suchthata2=a1+da3=a2+da4=a3+d
an=an−1+dThe constant d is the common difference of A.P.Thus, if the constant difference d and first term a, then a, a+d, a+ 2d, a+
3d,……… is an A.P.
The common difference
The common difference of an arithmetic sequence is the constant difference between consecutive terms.
For example, the common difference of 10, 21, 32, 43 ... is 11:
The common difference of –2, –5, –8, –11 ... is -3.Exercise
1. Find out which of the following sequences is A.P. For those are in A.P., find the common difference and write next two terms.(i) 3,6,12,24…..(ii) 4,4,4,4,4,……
(iii) 12
, 14
, 16
, 18
, …. .
2. Show that the sequence defined by an=3 n2−4 is not an A.P
General Term of an A.Pan=a+ (n−1 )d, where a = first term, n = no. of terms, d = common difference,
an=nthtermn th term from the end: Suppose there are m terms in the A.P. then
nth term from the end = a + (m-n)d, if last term l is mentioned in the A.P. then nth from the end = l−(n−1 )d
Exercise:1. Find 12th, 35th, and nth term of A.P. 8, 12,16,20,…….2. Find the 10th term from the end of the A.P.4,9,14,……254.3. Which term of the sequence -1,3,7,11,……. is 95?4. How many terms are there in the sequence 3,6,9,12,…….111?5. For what value of n, the nth term of A.Ps.63,65,67,……. and
3,10,17….. are equal?6. Which term of the A.P.5,15,25,…..will be 130 more than its 31st
term?7. If 2nd term of an AP 13and 5th term is 25. Find its 7th term.8. If 7 times the 7th term of an AP is equal to 11 times its 11th term,
find its 18th term.9. Which term of the AP 53, 48, 43,… is first negative term?10. How many numbers lie between 10 and 3000, which when
divided by 4 leave a remainder 3?
Sum of first n terms in an AP:Sn=n
2 {2a+ (n−1 )d }= n2(a+l ) , where Sn=∑ of first n terms ,
❑
a = first term, d = common difference, n = number of terms
l=t n=nth term=a+(n−1 )d
Here we will learn how to solve different types of problems on sum of n terms of Arithmetic Progression.1. Find the sum of the first 35 terms of an Arithmetic Progression whose third term is 7 and seventh term is two more than thrice of its third term.
Solution:
Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the given Arithmetic Progression.
According to the problem,
3rd term of an Arithmetic Progression is 7
i.e., 3th term = 7
⇒ a + (3 - 1)d = 7
⇒ a + 2d = 7 ................... (i)
and seventh term is two more than thrice of its third term.
i.e., 7th term = 3 × 3rd term + 2
⇒ a + (7 - 1)d = 3 × [a + (3 - 1)d] + 2
⇒ a + 6d = 3 × [a + 2d] + 2
Substitute the value of a + 2d = 7 we get,
⇒ a + 6d = 3 × 7 + 2
⇒ a + 6d = 21 + 2
⇒ a + 6d = 23 ................... (ii)
Now, subtract the equation (i) from (ii) we get,
4d = 16
⇒ d = 164164
⇒ d = 4
Substitute the value of d = 4 in the equation (i) we get,
⇒ a + 2 × 4 = 7
⇒ a + 8 = 7
⇒ a = 7 - 8
⇒ a = -1
Therefore, the first term of the Arithmetic Progression is -1 and common difference of the Arithmetic Progression is 4.
Now, sum of the first 35 terms of an Arithmetic Progression S35 = 35235/2[2 × (-1) + (35 - 1) × 4], [Using the Sum of the First n Terms of an Arithmetic Progression Sn = n2n
2[2a + (n - 1)d]= 35235/2[-2 + 34 × 4]= 35235/2[-2 + 136]= 35235/2[134]
= 35 × 67
= 2345.
2. If the 5th term and 12th term of an Arithmetic Progression are 30 and 65 respectively, find the sum of its 26 terms.
Solution:
Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the given Arithmetic Progression.
According to the problem,
5th term of an Arithmetic Progression is 30
i.e., 5th term = 30
⇒ a + (5 - 1)d = 30
⇒ a + 4d = 30 ................... (i)
and 12th term of an Arithmetic Progression is 65
i.e., 12th term = 65
⇒ a + (12 - 1)d = 65
⇒ a + 11d = 65 .................... (ii)
Now, subtract the equation (i) from (ii) we get,
7d = 35
⇒ d = 357357
⇒ d = 5
Substitute the value of d = 5 in the equation (i) we get,
a + 4 × 5 = 30
⇒ a + 20 = 30
⇒ a = 30 - 20
⇒ a = 10
Therefore, the first term of the Arithmetic Progression is 10 and common difference of the Arithmetic Progression is 5.
Now, sum of the first 26 terms of an Arithmetic Progression S2626 = 26226/2[2 × 10 + (26 - 1) × 5], [Using the Sum of the First n Terms of an Arithmetic Progression Snn = n2n/2[2a + (n - 1)d]
= 13[20 + 25 × 5]
= 13[20 + 125]
= 13[145]
= 1885
Exercise:1. Find sum of 20 terms of AP 4, 7, 10, 13, 16……2. Find the sum of the series 7+ 13+ 19+ 25 …up to 30 terms.3. Find the sum of series 1 +(-2) +(-5) + (-8)+………+(-236)4. If an=3−4 n , show that a1 ,a2 , a3 ,−−−form an AP . Alsofind S20
5. In an AP, if Sn=n (4 n+1 ) , find the AP .
6. In an AP, Sn=3n2+5 n∧ak=164 , find the value of k7. Find S17 ,if a4=−15∧a9=−308. Find the sum of last ten terms of the AP: 8, 10, 12, …, 126.9. How many terms of the AP: 15, -13, -11,… are needed to make the
sum -55? Explain the reason for double answer.10. Solve the equation: −4+(−1 )+2+…+ x=43711. Find the sum of integers between 100 and 300 that are (i)
divisible by 9, (ii) not divisible by 9.
Mr.Dharmender Kumar No.2, Delhi Cantt.Mrs.PoonamPassi 3 BRD
Mr.Kulwinder Singh SainikViharMr.Praveen Kumar 1HBK,Dehradun
Mr. R.D.Sharma Nahara
BASIC CONCEPTS OF
Triangles,Coordinate Geometry,Circles
Similar Figures:Two figures having the same shape but not of the same size are
called similar figures.Criterian of similarity of triangles:(i) SSS Rule of similarity : If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar.(ii)ASA Rule of similarity : If one angle of a triangle is equal to one angle of another triangle and the sides includingthese angles are in the same ratio (proportional), then the triangles are similar(iii)AAA Rule of similarity : If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.
Coordinate Geometry
Distance formula :
The distance between P(x1 ,y1) and Q(x2 ,y2) is √( x2−x1 )2+( y2− y1)
2
Section Formula : (i)The coordinates of the point P(x, y) which divides the line segment joining the points A (x1 ,y1) and
B (x2 ,y2) internally in the ratio m1 : m2 are (m1 x2+m2 x1
m1+m2,
m1 y2+m2 y1
m1+m2)
(ii)The coordinates of the point P(x, y) which divides the line segment joining the points A (x1 ,y1) and B
(x2 ,y2) externally in the ratio m1 : m2 are (m1 x2−m2 x1
m1−m2,m1 y2−m2 y1
m1−m2)
Area of triangles :
TOPIC AREAS TO BE COVERED
Triangles (i)Similar Figures(ii)Similarity of triangles
(iii)Criteria of similarity of triangles(iv)Pythagoras Theorem
Coordinate Geometry
(i)Introduction(ii)Distance formula
(iii)Section formula(iv)Area of triangles
Circles (i)Perimeter and Area of circle (ii) Tangent and normal to a circle
The area of the triangle formed by the points P(x1 ,y1) and Q(x2 ,y2) and R(x3 ,y3) is the numerical value
of the expression
12 [ x1(y2 - y1 ) + x2(y3 - y1) + x3 (y1 – y2)]
Circles
Perimeter and Area of circle :Circumference (Perimeter) of a circle of radius r is 2π r.Area of circle: Area of circle of radius r is π r2
Tangent to a circle :The line touching the circle only at one point is called the tangent to the circle.Normal to the circle: The line perpendicular to the tangent at point of contact is called the normal.
Questions for Practice
Q1. if PQ || RS, prove that Δ POQ ~ Δ SOR.
Q2. In the adjoining figure OA . OB = OC . OD. Show that ∠∠A = ∠∠C and ∠B = ∠D.
Q3. PQR is a triangle right angled at P and M is a point on QR such that PM QR. Show that PM2 = QM . MR.
Q 4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2
Q 5. In an equilateral triangle ABC, D is a point on side BC such that BD =1/3 BC. Prove that 9 AD2 = 7 AB2.Q 6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Q7. Fill in the blanks :(i) A tangent to a circle intersects it in …………point (s).(ii) A line intersecting a circle in two points is called a……. .(iii) A circle can have parallel tangents at the most…………..(iv) The common point of a tangent to a circle and the circle is called ……………….Q8. Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).Q9. Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.Q.10. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.Q.11. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.Q.12. Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.Q.13. In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?Q.14. Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).Q.15. Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.
Q.16. Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, – 4).Q.17. Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.
Mr.Vikas Kumar No.2, AmbalaMr.Pradeep Chandra Saxena GaucharMrs.Charu Sharma Tagore GardenMr.Ravinder Singh No.3, AmristarMr.B.S.Manral Ranikhet
Sr.No. Topics Areas to be covered
BASIC CONCEPTS OF TRIGONOMETRY
Introduction to Trigonometry
Right triangle trigonometry and trigonometric ratios
Trigonometric ratios of common angles
Trigonometric ratios of complementary angles
Trigonometric identities
BASIC CONCEPTS OF TRIGONOMETRY
Trigonometry
Trigomometry from Greek trigõnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships involving lengths and angles of triangles.
Right triangle trigonometry and Trigonometric ratios
In a right angle triangle ABC where B=90° ,
We can define following term for angle A
We can define the trigonometric ratios for angle A as
Notice that each ratio in the right-hand column is the inverse, or the reciprocal, of the ratio in the left-hand column.The reciprocal of sin A is cosec A ; and vice-versa.The reciprocal of cos A is sec AAnd the reciprocal of tan A is cot AThese are valid for acute angles.We can define tan A = sin A/cos AAnd Cot A =cos A/ Sin A
Important NoteSince the hypotenuse is the longest side in a right triangle, the value ofsin A or cos A is always less than 1 (or, in particular, equal to 1).
Similarly we can have define these for angle C
We can define the trigonometric ratios for angle C assin C= Perpendicular/Hypotenuse =AB/ACcosec C= Hypotenuse/Perpendicular =AC/ABcos C= Base/Hypotenuse =BC/ACsec C= Hypotenuse/Base=AC/BCtan C= Perpendicular/Base =AB/BCcot C= Base/Perpendicular=BC/AB
Trigonometric Ratios of Common angles
We can find the values of trigonometric ratio’s various angle
Question 1 State whether the following are true or false. Justify your answer.(i) The value of tan A is always less than 1.(ii) sec A =12/5 for some value of angle A.(iii) cos A is the abbreviation used for the cosecant of angle A.(iv) tan A is the product of tan and A.(v) sin x =5/3 for some angleSolutioni) False . The value of tan A increase from 0 to infinityii) True . The value of sec A increase from 1 to infinityiii) False .Cos A is the abbreviation used for the cosine of angle Aiv) False .cot A is one symbol. We cannot separate itv) False. The value of sin ? always lies between 0 and 1 and 5/3 > 1Question 1The value of (sin30 + cos30) � (sin60 + cos60) is (A) � 1 (B) 0 (C) 1 (D) 2 SolutionAnswer (B)
Trigonometric ratio’s of complimentary angles
We know that for Angle A, the complementary angle is 90 – AIn a right angle triangle ABCA+B+C=180Now B=90So A +C =90Or C=90-AWe have see in Previous section the value for trigonometric ratios for angle Csin C= Perpendicular/Hypotenuse =AB/ACcosec C= Hypotenuse/Perpendicular =AC/ABBcos C= Base/Hypotenuse =BC/ACsec C= Hypotenuse/Base=AC/BCtan C= Perpendicular/Base =AB/BCcot C= Base/Perpendicular=BC/AB
This can be rewritten assin (90-A) =AB/ACcosec (90-A) =AC/ABcos (90-A) =BC/ACsec (90-A)=AC/BCtan (90-A)=AB/BCcot( 90-A) =BC/ABAlso we know thatsin A=BC/ACcosec A= AC/BCcos A= AB/ACsec A =AC/ABtan A =BC/ABcot A =AB/BCFrom both of these, we can easily make it outSin (90-A) =cos(A)Cos(90-A) = sin ATan(90-A) =cot ASec(90-A)= cosec ACosec (90-A) =sec A
Cot(90- A) =tan ATrigonometric identities
We have studied pythagorus theorem in earliar classes which states that for a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sideIf a is the hypthonues and b and c are other two sides,thena2= b2 + c2
This same theorem can be used to proved the below trigonometric identitiessin2 A + cos2 A =11 + tan2 A =sec2 A1 + cot2 A =cosec2 A
How to solve Trigonometric identities Problems
1) Learn well the formulas for Trigonometric identities, trigonometric ratios,reciprocals The better you know the basic identities, the easier it will be to recognise what is going on in the
problems. 2) Generally RHS( Right hand side) would be more complex. So start from there and
simplify it to the same form as LHS(Left hand side) 3) It becomes many times easy, if Convert all sec, csc, cot, and tan to sin and cos. Most of
this can be done using the quotient and reciprocal identities. 4) If you see power 2 or more, it will involve using the below identities mostly
sin2 A + cos2 A =11 + tan2 A =sec2 A
1 + cot2 A =cosec2 A5) Once you get the hang of it, you will begin to see patterns and it will be easy to solve these
Trigonometric identities Problems 6) Practice and Practice. You will soon start figuring out the equation and there symmetry to
resolve them fast
EXERCISE
1 .) If sin A=2425
then find thevalue of cos A and tan A .
2) Deter min e the value of 2tan 30 ° ¿1+ tan230° ¿¿
¿3) 9 sec2 A − 9 tan2 A = .. .. . .. .. . .. .. . .. .. .. . .. .. . .. . ¿ 4 )( secθ+ tan θ)(1−sinθ )=. . .. .. . .. .. . .. .. . . ¿5) If 15cot A= 8 thenfindthevalueof sin A . ¿6 )If sin ( A−B) = 12
and cos (A+B ) = 12
then find A andB . ¿7 ) If sin A = 12
then find the value of 3cos A−4 cos3 A . ¿8 ) Express sin67°+ cos 75° in terms of trigonometric ratios of angles between 0 ° and 45° . ¿9 ) Pr ove that cosθ1−tan θ
+sin θ1− cot θ
= sin θ+cosθ . ¿10 ) Pr ove that tan A+sec A−1tan A−sec A+1
= sec A+ tan A = 1+sin Acos A
¿ ¿¿
Sr.No.
Topics Areas to be covered
BASIC CONCEPTS OF SRFACE AREA AND
VOLUMES
Introduction
SURFACE AREA :
A) CUBIODB) CUBEC) RIGHT CIRCULAR
CYLINDERD) RIGHT CIRCULAR
CONEE) SPHERE
VOLUME :
A) CUBIODB) CUBEC) RIGHT CIRCULAR
CYLINDERD) RIGHT CIRCULAR
CONEE) SPHERE
BASIC CONCEPTS OF SURFACE AREA AND VOLUMES
For Cube :-If each edge of cube ‘l’ Then1. Total surface area of cube = 6l2
2. Lateral surface area of cube = 4l2
3. Volume of cube = l3
4. Length of diagonal = √ 3 l
For Cuboid :- If length, breath , and Height of a Cuboid ‘l’ ,’b’,’h’Then 5. Total surface area of cuboid = 2 (lb + bh +hl )6. Lateral surface area of cuboid = 2(bh+hl)7. Volume of cuboid = l.b.h8. Length of diagonal =√ l2+b2+h2
For Cylinder :- If Radius of circle ‘r’and height ‘h’Then9. Curve surface area of Cylinder = 2 π rh10. Total surface area of Cylinder = 2 π r (r+h)11. Volume of Cylinder = πr2h12. Base Area of Cylinder = πr2
For Cone:- If Radius of cone ‘r’ and height ‘h’ and slant height ‘l’ Then 1. Curve surface area of Cone = π r l2. Total surface area of Cone = π r (r+l)
3. Volume of Cone =13πr2h
4. Base Area of Cone = πr2
For Sphere :-
If Radius of Sphere ‘r’ Then1. Total surface area of Sphere = 4 π r2
2. Volume of Sphere=43 πr3
For Hemisphere :- If Radius of hemisphere ‘r’ Then1. Curve surface area of hemisphere = 2 π r2
2. Total surface area of hemisphere = 3 πr2
3. Volume of hemisphere =23πr3
For Frustum of cone
: If Radius of frustum of cone ‘r1’ , ‘r2’ , height ‘h’ and slant height ‘l’ then1. l2= (r1- r2)2+ h2
2. Curve surface area of frustum of cone = π(r1+r2)l3. Total surface area of frustum of cone = π(r1+r2)l+ πr1
2+πr22.
4. Volume of frustum of cone = 13π(r1
2+r1r2+r22) h
EXERCISE1) The surface area of cubiod is………..2) The surface area of cube of edge “a” is …….3) The curved surface (lateral ) area of cylinder of height h and radius r will be ……4) Total surface area of hemisphere of radius r is……..5) The volume of sphere is…………6) The curved surface area of sphere is…..7) The area of three adjacent face of cube is x, y and z then find its volume.8) The slant height of a cone is 26cm and its base diameter is 20 cm find its height.9) The circumference of base of a cylindrical vessel is 132cm and its height is 25cm .How
many litres of water it can hold?
10) If the radii of the circular bases of a frustum of a cone are 6cm , 8cm and its slant height is 5 cm,then the curved surface area of the frustum is…… 11) If a sphere and a cube have equal areas , then the ratio of the diameter of the sphere to the edge of the cube is….. 12) If the surface area of a sphere is 324π cm2 , then its volume is …..
13) If the base area of a cone is 51 cm2 and its volume is 85 cm3 , then its vertical height is.. 14) If the vertical height of a cone is 8 cm and the area of its base is 156cm2, then its volume is.15) If a cone of height 8 cm has base diameter 12 cm, then its curved surface area is …
Mr.Sawinder No.1,HussainpurMrs. RuchiSalaria PaschimVihar
Mr.Birampal Singh KausaniMr. Raman Kumar ITBP,Bhanu
Mr.Pradeep Kumar No.1, Pathonkot
TOPIC: STATISTICS AND PROBABILITY
Topic Area to be covered
Probability
Random experiment Outcome
TrialEvent
Equally likely eventsEmpirical probability
Sample Space Sample pointSure Events
Impossible eventElementary events
Complementary eventCompound events
Some basics concepts of Probability 1. Random experiment :- An experiment is called random experiment if it satisfies the
following two conditions:
(i) It has more than one possible outcome.
(ii) It is not possible to predict the outcome in advance.
2. Outcome.:- A possible result of a random experiment is called its outcome.
3. Trial : - A trial is an action which results in one or several outcomes.
4. Event : - An event for an experiment is the collection of some outcomes of the experiment.
5. Equally likely events - If one event cannot be expected in preference to other event then they
are said to be equally likely.
6. Empirical probability : - The empirical probability P(E) of an event E happening, is given
by
P(E) = Number of trials in which the event happenedThe total number of trials
Where we assume that the outcomes of the experiment are equally likely.
7. Sample Space : - the set of all possible outcomes of a random experiment is called the
sample
space associated with the experiment. Sample space is denoted by the symbol S.
8. Sample point : - Each element of the sample space is called a sample point.
9. Sure Events : The whole sample space is called the sure event.
Remark:- The probability of a sure event (or certain event) is 1.
10. Impossible event : - The event which is not possible in an experiment is called
impossible event .
Remarks:-
i. The probability of an impossible event is 0.
ii. The probability of an Event E is number P (E) such that 0≤P (E) ≤1.
11. Elementary events: - An event having only one outcome is called an elementary event.
Remark:- 1. The sum of the probabilities of all the elementary events of an experiment is 1.
12. Complementary event :- For any event E, P (E) +P (E−
) =1, where E−
stands for not E, E and
are called complementary event.
13.Compound Event:
If an event has more than one sample point, it is called a Compound event.
For example, in the experiment of “tossing a coin thrice” the events
E: ‘Exactly one head appeared’
F: ‘Atleast one head appeared’
G: ‘Atmost one head appeared’ etc.
are all compound events. The subsets of S associated with these events are
E={HTT,THT,TTH} , F={HTT,THT, TTH, HHT, HTH, THH, HHH}
G= {TTT, THT, HTT, TTH}
Each of the above subsets contain more than one sample point, hence they are all
compound events.
Exercise : Question 1:
Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = _______.
(ii) The probability of an event that cannot happen is _________. Such as event is called _________.
(iii) The probability of an event that is certain to happen is _________. Such as event is called ________.
(iv) The sum of the probabilities of all the elementary events of an experiment is _________.
(v) The probability of an event is greater than or equal to _______ and less than or equal to _______.
Question 2:
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Question 3:
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Question 4:
Which of the following cannot be the probability of an event?
Question 5:
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Question 6:
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Question 7:
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white? (iii) not green?
Question 8:
A die is thrown once. Find the probability of getting
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Question 9:
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Question 10:
Five cards−−the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Question 11:
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Question 12:
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Question 13:
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
14. Describe the sample space for the indicated experiment.a. A coin is tossed three times.b. A die is thrown two times.c. A coin is tossed four times.
d. A coin is tossed and a die is thrown.e. A coin is tossed and then a die is rolled only in case a head is shown on the coin
15. Describe the events Number of tails is exactly 2 A = {TT}Number of tails is atleast one B = {HT, TH, TT}Number of heads is atmost one C = {HT, TH, TT}Second toss is not head D = { HT, TT}Number of tails is atmost two S = {HH, HT, TH, TT}Number of tails is more than two P = { }
STATISTICSTopic Area to be covered
Statistics
DataTypes of Data
RangeFrequency
Class-Intervals Class-size or Class width
lower class limit and upper class limit Class-Marks
Measures of central tendencyEmpirical relationship between the three measures of central tendency. Discrete frequency distribution Continuous frequency distribution
1.Data : - The facts or figures, which are numerical or otherwise, collected with a definite purpose are
called data.
2.Types of Data :-
Primary Data : - When the information was collected by the investigator herself or himself with a
definite objective in her or his mind, the data obtained is called primary data.
Secondary data : - When the information was gathered from a source which already had the
information stored, the data obtained is called secondary data.
3.Range :- The difference of the highest and the lowest values in the data is called the range of the
data.
4. Frequency : - The number of times that an observation repeats is called the frequency.
5. Class-Intervals : - The groupings of data in the form of intervals are called classes or class-
intervals. 6. Class-size or Class width : - The size of class interval is called the class-size or class width.
7. lower class limit and upper class limit : - In each of the classes, the least number is called the
lower class limit and the greatest number is called the upper class limit,
e.g., in 20-29, 20 is the ‘lower class limit’ and 29 is the ‘upper class limit’.
8. Class-Marks: - The mid-points of the class-intervals are called class-marks.
To find the class-mark of a class interval, we find the sum of the upper limit and lower limit of a class
and divide it by 2. Thus,
Class-mark =
Upper Limit + Lower limit2
9. Discrete frequency distribution : - Let the given data consist of n distinct values x1, x2, ..., xn
occurring with frequencies f1, f2 , ..., fn respectively. This data can be represented in the tabular form as given below, and is called discrete frequency distribution :
x x1
x2
x3
……. xn
f f1
f2
f3
……. fn
10.Continuous frequency distribution : -A continuous frequency distribution is a seriesin which the data are classified into different class-intervals without gaps alongwith
their respective frequencies.11.Measures of central tendency : - The three measures of central tendency are :
i. Mean : - It is found by adding all the values of the observations and dividing it by
the total number of observations. It is denoted by x−
.
for n observations x−=∑i=1
n
x i
nMean of grouped frequency distribution can be calculated by the following methods. a. Direct Method
Mean = X = ∑i=1
n
fixi
∑i=1
n
fi
Where Xi is the class mark of the ith class interval and fi frequency of that class
b. Assumed Mean method or Shortcut method
Mean = X = a +∑i=1
n
fidi
∑i=1
n
fi
Where a = assumed meanAnd di= Xi - a
c. Step deviation method .
Mean = X = a +∑i=1
n
fiui
∑i=1
n
fix h
Where a = assumed meanh = class sizeAnd ui= (Xi – a)/h
ii. Median :- It is the value of the middle-most observation (s).Median is that value of the given number of observations, which divides it into
exactly two parts. So, when the data is arranged in ascending (or descending) order the median of
ungrouped data is calculated as follows:
(i) When the number of observations (n) is odd, the median is the value of the ( n+12 )
th
observation .
(ii) When the number of observations (n) is even, the median is the mean of the ( n2 )th
and ( n2 )th
+1 observations.
Median of a grouped frequency distribution can be calculated by
Median = l + ( n2−cf
f ) x h
Where l = lower limit of median classn = number of observationscf = cumulative frequency of class preceding the median classf = frequency of median classh = class size of the median class.
iii. Mode :- The mode is the most frequently occurring observation. Mode of grouped data can be calculated by the following formula.
Mode = l + ( f 1−fo2 f 1−fo−f 2 )h
Where l = lower limit of modal classh = size of class intervalf1 = Frequency of the modal classfo = frequency of class preceding the modal classf2= frequency of class succeeding the modal class
12. Empirical relationship between the three measures of central tendency. 3 Median = Mode + 2 MeanOr, Mode = 3 Median – 2 Mean
Slno
Question Ans
1 What is the mean of 1st ten prime numbers ? 12.92 For data 5, 3 , 8 ,7 ,9 , 10 , 15 , find range. 123 If the mode of a data is 45 and mean is 27, then median is
___________. 33
4 Find the mode of the followingXi 35 38 40 42 44fi 5 9 10 7 2
Mode =40
5 Write the median class of the following distribution. Class 0-10 10-
2020-30
30-40
40-50
50-60
60-70
Frequency
4 4 8 10 12 8 4
30-40
6 For data 15, 23 , 18 ,17 ,19 , 20 , 25 , find range. 10
7 Classify the data given below as primary or secondary data.(i) Heights of 20 students of your class.(ii) Number of absentees in each day in your class for a month.(iii) Number of members in the families of your classmates.
(iv) Heights of 15 plants in or around your school.(v) Data of Humidity of a town(vi) Data of temperature of various cities in a month
primary
primary
primary
primarysecondarysecondary
Slno
Question Ans
1 Calculate the mean of the following distributionClass
interval50-60 60-70 70-80 80-90 90-100
Frequency 8 6 12 11 13
78
2 Find the mode of the following frequency distributionMarks 10-20 20-30 30-40 40-50 50-60No. of
students12 35 45 25 13
33.33
3 Find the median of the following distributionClass
interval0-10 10-20 20-30 30-40 40-50 50-60
Frequency 5 8 20 15 7 5
28.5
4 A class teacher has the following absentee record of 40 students of a class for the whole term.
No. of days
0-6 6-10 10-14
14-20
20-28 28-38 38-40
No. of students
11 10 7 4 4 3 1
Write the above distribution as less than type cumulative frequency distribution.Answer :
No. of days
Less Than
6
Less Than 10
Less Than 14
Less Than 20
Less Than 28
Less Than 38
Less Than 40
No. of students
11 21 28 32 36 39 40
Slno
Question Ans
1 If the mean distribution is 25Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 18 15 P 6 Then find p.
P=16
2 Find the mean of the following frequency distribution using step deviation method
Class 0-10 10-20 20-30 30-40 40-50Frequency 7 12 13 10 8
25
3 Find the value of p if the median of the following frequency distribution is 50
Class 20-30 30-40 40-50 50-60 60-70
70-80
80-90
Frequency
25 15 P 6 24 12 8
P=10
4 Find the median of the following dataMarks Less
Than 10
Less Than 30
Less Than 50
Less Than 70
Less Than 90
Less Than
110
Less Than
130
Less than 150
Frequency
0 10 25 43 65 87 96 100
76.36
5 Give five examples of data that you can collect from your day-to-day life.
Ans Five examples of data that we can gather from our day-to-day life are :(i) Number of students in our class.(ii) Number of fans in our school.(iii) Electricity bills of our house for last two years.(iv) Election results obtained from television or newspapers.
(v) Literacy rate figures obtained from Educational Survey.6 Classify the data in Q.5 above as primary or secondary data.
Ans Primary data; (i), (ii) and (iii) and Secondary data; (iv) and (v)
Slno
Question Ans
1 The mean of the following frequency distribution is 57.6 and the sum of the observations is 50. Find the missing frequencies f1 and f2.
Class 0-20 20-40 40-60 60-80 80-100
100-120
Total
Frequency
7 f1 12 f2 8 5 50
f1 =8 andf2 =10
2 The following distribution give the daily income of 65 workers of a factory
Daily income (in Rs)
100-120
120-140
140-160
160-180
180-200
No. of workers
14 16 10 16 9
Convert the above to a more than type cumulative frequency distribution and draw its ogive.
3 Draw a less than type and more than type ogives for the following distribution on the same graph. Also find the median from the graph.
Marks 30-39
40-49 50-59 60-69 70-79 80-89
90-99
No. of students
14 6 10 20 30 8 12
4 Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
0 , 1 , 2 , 2 , 1 , 2 , 3 , 1 , 3 , 0 ,1 , 3 ,1 , 1 , 2 , 2 , 0 , 1 , 2 , 1 , 3 , 0 , 0 , 1 , 1 , 2 , 3, 2 , 2 ,0Prepare a frequency distribution table for the data given above.
Ans.
No.of heads
Frequency
0 61 102 93 5
SELF – EVALUATION1. If mean =60 and median =50, then find mode using empirical
relationship. 2. Find the value of p, if the mean of the following distribution is 18. Variate (xi) 13 15 17 19 20+p 23Frequency (fi)
8 2 3 4 5p 6
3. Find the mean, mode and median for the following data. Classes 0-10 10-20 20-30 30-40 40-50 50-60 60-70
frequency 5 8 15 20 14 8 5
4. The median of the following data is 52.5. find the value of x and y, if the total frequency is 100. Class
Interval0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
frequency 2 5 X 12 17 20 Y 9 7 4
5. Find the mean marks for the following data. Marks Belo
w 10Below 20
Below 30
Below 40
Below 50
Below 60
Below 70
Below 80
Below 90
Below
100No. of studen
ts
5 9 17 29 45 60 70 78 83 85
6. The following table shows age distribution of persons in a particular region. Calculate the median age.
Age in years
Below 10
Below 20
Below 30
Below 40
Below 50
Below 60
Below 70
Below 80
No. of persons
200 500 900 1200 1400 1500 1550 1560
7. If the median of the following data is 32.5. Find the value of x and y. Class
Interval0-10 10-20 20-
3030-40
40-50 50-60 60-70
Total
frequency x 5 9 12 y 3 2 40
Mr.NaveenWadhera SSBSrinagarMr.Chanchal Kalra Kasauli
Mr.Ajeet Kumar Singh GoleMarket Shift IMs.Aruna Khera No.1Ferozpur
Mr.Anil Rawat Uttarkashi
Methods of Proof
With
Focus on Mathematical Olympiads.
By
Prof. V.K. GroverDepartment of Mathematics,
Panjab University,
Chandigarh.
We cannot just list all the methods of proof, as new methods are developed from time to
time and this process is never ending. However we list here a few of these methods which are
commonly used in problems related to mathematical Olympiads irrespective of the branch of
mathematics to which the problem is related to.
Argument by Contradiction.
Mathematical induction.
No square is negative or completing squares.
Telescopic sums and products.
Arranging numbers in order.
Parity considerations.
Looking at extreme situation or solution.
Doing a problem in two different ways.
Relating two different situations
Pigeon hole principle.
Symmetric considerations.
Argument by contradiction
In this method first the statement is assumed to be false and then a sequence of
logical deductions yield a conclusion that contradicts either the hypothesis or a fact which
is known to be true.
Examples1. Prove that there are infinitely many prime numbers
Solution:- Assume to the contrary that there are only finitely many prime
numbers. List them all as p1=2 , p2=3 , p3=5 , ………. , pn. Consider the number
N=p1 . p2 . p3 …… pn+1Clearly this number N is greater than all the prime
numbers listed above hence cannot be one of these, so it must be a composite
number and hence divisible by one of the prime numbers, say pi as
p1 . p2 . p3 …… pn is divisible by pi this implies that 1 is also divisible by pi , a
contradiction and hence the fact is proved.
2. Prove that there is no Polynomial with integer coefficients and of degree at least 1
with the property that P (0 ) , P (1 ) ,P (2 ) , ……… ..are all prime numbers.
Solution :-Assume the contrary and let there be a polynomial
P ( x )=an xn+an−1 xn−1+…+a0 such that P(n) is a prime for all natural numbers n,
P(0) = a0 = p, so that p is a prime number and hence P (kp ) is divisible by p for all
k ≥ 1. But as P (kp ) is prime we must have P (kp )=p for allk ≥ 1. This yields that
P ( x )≡ p ,a contradiction to the fact that P ( x )has degree atleast 1. Hence the result
follows.
3. Let F={E1 , E2 ,………, E s } be a family of sets with r elements each. Show that if
the intersection of anyr+1 ( not necessarily distinct) sets in F is non empty then
the intersection of all the sets in F is non empty.
Solution:- Again assume the contrary , namely that the intersection of all sets
in F is empty. Consider the set E1= {x1 , x2 , …… xr }Because none of x i for
i=1,2 , … ..r lies in the intersection of all the Ejs ( the intersection of all sets in F
being empty), it follows that for each i there we can find some set E jisuch that
x i E ji. Then we have
E1 ∩ E j1 ∩……∩ E jr=¿
Since, at the same time the intersection is included in E1 and does not contain
any of the elements ofE1. This contradicts the hypothesis and hence the result
follows.
Mathematical InductionWe have following variations of the induction principle:
Given P (n ) , a property depending on an integer n,
(i) If P (n0 ) is true for some integer n0 and
(ii) If for every k ≥ n0, P (k ) is true implies that P (k+1 ) is true
Then P (n ) is true for all n ≥ n0 .
Given P (n ) a property depending on an integer n,
(i) If P(n0) is true for some integer n0 and
(ii) If for n0 ≤ i ≤ k, P (i ) is true implies P (k+1 ) is true
Then P (n ) is true for all integers n ≥n0.
Given P (n ), a property depending on an integer n,
(i) If P(m),P(m+1),………,P(m+r) are true for some integer m and
(ii) If truth of P(k),P(k+1),………,P(k+r)for k ≥ m implies the truth of
P(k+r+1)
Then P(n), is true for all integers n ≥ m.
We can also state backwards induction as follows
Given P(n), a property depending on an integer n,
(i) If P(n0) is true for some integer n0 and
(ii) If for every m≤ k ≤ n0, P(k) is true implies that P(k-1) is true
Then P(n) is true for all n = m-1,m,……..,n0.
Examples1. Prove that for any n ≥ 1, a 2n×2n chess board with one 1×1 square removed can be tiled
by Triminos, the blocks consisting of three 1×1 squares put together to make L-shape, of
the type shown in fig. 1
Fig.1
Solution:-The result is obvious for n = 1 as after removing one square from the 2×2
square we are left with just one trinimo. Now suppose the result is true for n-1, i.e. for the
2n-1×2n-1 squares, consider the 2n×2n square with one square removed, the following
diagram shows such square with n = 4
Fig.2
Divide this square in 4 equal parts ( here the division is shown by red lines in the
fig.2).Clearly one of the four parts will have the removed square. Place one trinimo at the
centre in such a way that it cover one square each from the remaining three of the four
parts as shown in the fig.3
Fig.3
2. For any natural number N, prove that
√2√3√4 √………… ..√(N−1¿¿¿¿¿)√N ¿¿¿¿¿ ¿3(¿)
Solution:-Here we use backward induction we will prove the following inequality
√m√(m+1)√(m+2)√………… ..√(N−1¿¿¿¿¿)√N ¿¿¿¿¿ ¿m+1¿
For all m = 2,3,……N.
For K= N , (**) becomes √N < N+1 , which is clearly true for N > 1. Suppose the result is
true for k = m , for k = m-1 we need to prove
√(m−1)√m√(m+1)√………… ..√(N−1¿¿¿¿¿)√N ¿¿¿¿¿ ¿m
Using (**) we get
L.H.S. < √ (m−1 ) (m+1 ) < m , hence the result follows.
No square is negativeHere we use the fact that square of a real number is non negative.
Examples1. Let a be a real number, prove that 4a – a4 ≤ 3.
Solution: - The inequality is equivalent to
(a2 – 1)2 +2(a -1)2 0,
Which is clearly true for real values of a.
2. Determine whether there exists a one to one function f : R R with the property that for
all x ,f (x2 )−( f (x ))2≥ 14 .
Solution: - We will show that such a function does not exist. The idea is simple. We
look for two numbers that are equal to their squares, namely 0 and 1.
For x = 0, 1 we have f (x2 )=f (x ). We first put x=0 in the given equation and
obtain f (0 )−(f (0))2≥ 14 , this implies ( f (0 )−1
2)
2
≤ 0 so we conclude that
( f (0 )−12)
2
=0, i.e. f (0 )=12 similarly we conclude that f (1 )=1
2 = 12 so that f ( x ) is not
one to one.
Telescopic Sums and Products
There are sums which can be put in the form
∑i=0
n
(F (i+1)−F (i)¿)¿
And there are products which can be put in the form
∏k=1
n F (k+1)F(k )
Or
∏k=1
n F(k )F (k+1)
In both the cases the in between terms cancel and we compute the value of the
sum or the product whichever the case may be.
Examples1. Evaluate
∑0
n 1(k+1 )√k+ k √(k+1)
Solution:- the kth term in the sum is
1(k+1 )√k+ k √ (k+1 )
=(k+1 )√k−k √( k+1 )
k (k+1)=( 1√k− 1√k+1 )
So that the required sum becomes
∑0
n
( 1√k− 1√k+1 )=¿1− 1
√n+1¿
2. Prove that
∏n=2
∞
(1− 1n2 )=1
2
Solution:- for large values of N
∏n=2
N
(1− 1n2 )=∏n=2
N
(1−1n )∏n=2
N
(1+ 1n )=∏n=2
N
(n−1n )∏
n=2
N
( n+1n )=( 1
N )(N+12 )
Which 12 as N ∞.
Arranging Numbers in OrderWe illustrate this method by using following examples
Examples
1. Given 7 distinct positive integers that add upto100 Prove that some three of them add
upto at least 50.
Solution:- we assume that the seven numbers are a, b, c,…….., g with a < b < c
<…….< g. We will show that e + f+ g 50.
If e > 15 then we get e + f + g 16 + 17 + 18 = 51.
If e ≤ 15 , then we get a + b + c + d ≤ 14 + 13 + 12 + 11 = 50 and hence
e + f + g 100 – (a + b + c + d) 50.
2 Prove that among any 50 distinct positive integers strictly less than 100 there are two that
are coprime.
Solution:- Order the numbers as x1< x2<……….<x50. If in this sequence there are
two consecutive integers then they are coprime and we are done. Otherwise we have
x50≥ x1+2.49≥ 99. Sincex50<100, implies that x50=99 and x1=1 , x2=3 , x3=5 , ……so
on. Among these 3 and 5 are coprime.
Parity considerationsSome problems are solved by just looking at the parity( even and odd
property) of the numbers under consideration.
Examples1. There are 100 soldiers in a detachment, and every evening three of them are on
duty. Can it happen that after a certain period of time each soldier has shared duty
with every other soldier exactly once.
Solution: - Suppose that what ever required in the problem can happen. Fix
one of the soldier S1(say) and during that period of time S1 has been on duty m
times and as no person on duty with him is repeated the number of persons with
whom he has shared his duty is 2m which is even however he need to share duty
with 99 other soldiers once and as 99 is odd, this is not possible.
2. Forty five Points are chosen along line AB, all lying outside the segment AB.
Prove that the sum of the distances from these points to the point A is not equal to
the sum of distances of these points to the point B.
Solution:- for any point X lying out side segment AB, the difference
AX−BX=± AB.Let the given point be X1 , X2 ,………. X 45, then we obtain
∑i=1
45
A X i−¿∑i=1
45
B X i=∑i=1
45
(A X i−B X i¿)=∑i=1
45
± AB0¿¿
(As for the last sum to be zero we must have as many AB with positive sign as
are with negative sign. But the total number of terms is 45 which is odd. )
3. The product of 22 integers is 1. Show that their sum cannot be zero.
Solution:- Each of these integer must divide the product which is 1 so that the
values taken by these integers are ±1. To obtain the sum as zero we must have
equal number of +1’s as are -1’s. But then there are 11 -1’s, which is odd and the
product of these numbers must then be -1, a contradiction and hence the proof.
4. The numbers from 1 to 10 are written along a row . Can the signs + and – be
placed between them so that the value of the resulting expression is 0.
Solution:- Note that the sum of first 10 numbers is 45, which is an odd
number. Now denote by ‘a’ the sum of sum of numbers with positive sign and by
‘b’ the sum of numbers with negative sign. Clearly the value of the expression is a
- b If this value happens to be zero this will imply that a = b, and hence
45 = a + b = 2a, an even number, a contradiction. This proves the result.
5. Consider the chess board with two corner squares along one of the diagonals
removed as shown in the Fig.4 Can this be covered by 2×1 rectangles as shown
in Fig.5? If possible to cover 31 such rectangles are required.
Fig.4
Fig.5
Solution:- Answer is No. We justify our answer by the following argument.
As shown in the Diagram, there are 32 black squares and 30 white
squares. If we pair each black square with a white square in whichever way we
get 30 pairs and two black squares are left unpaired. Also when we place any
block of size 2×1 horizontally or vertically on the board it covers equal areas in
black as well as in white. So complete covering is not possible
Looking at InvariantsIn some of the problem we allow certain things or numbers to change, i.e. we replace a
set of numbers by another set of numbers etc. In order to solve such a problem we look out for
something which does not undergo any change during the whole process. This helps us in
solving the problem. The something which remains same during the process is called invariant.
We list below some examples.
Examples1. Suppose for a positive odd integer n, the numbers 1,2,3,………….2n are written on a
black board.. We pick any two numbers a and b erase them and write instead ,|a−b|.
Prove that an odd number will remain at the end.
q
p
C f b a Fig.6
Solution:- Suppose S is the sum of all numbers on the board at any stage. ( Here S is a
variable quantity). Initially S=1+2+3+………….+2 n=n (2n+1 ), which is an odd
integer. Each step reduces S by 2min (a ,b )which is an even number so that the parity of S
will remain odd. Hence S will also be odd at the end. But at the end we are left with only
one number, hence that number must be odd.
2. The numbers 1, 2, 3,…….., 20 are written on a blackboard. It is allowed to erase any two
numbers a and b and write the new numbera+b−1. What number will be on the
blackboard after 19 such operations?
Solution: - For any collection n of numbers on the black board, let
X=(∑ of thenumbers) – n. Suppose that we have transformed the collection of n
numbers to a collection of n−1 numbers as described. How would the value of X
change? If we have picked the numbers a and b and replaced these by a+b−1 and if we
denote the new value of X by Y then
Y=(∑ of the numbers ) – a−b+(a+b−1 )−(n−1 )=(∑ of the numbers ) – n=X . X is an
invariant. Initially the value of X=(1+2+… ..20 )−20=190. Therefore after 19 operations
only one number will be left and the value of X will again be 190. the number = X+1
=191.
Looking at Extreme casesWe illustrate this by the following examples
Examples1. A finite set of points in the plane has the property that any line through two them
passes through a third. Show that all the points are collinear. (Posed by Sylvester in
1893 and solved by T. Gallai in 1933).
Solution:- Suppose that the points are not collinear. Among all pairs ( p , L )
consisting of a line L joining two of the points of the set and a point p of the set
outside the line L, choose one such pair for which the distance d from p to L is least.
Fix that pair as ( p ,L ) again. (For all following details refer to fig.6)
Let f be the foot of the perpendicular from p to the line L there are at least three points a,b,c of the set on the line L. Hence at least two of these say ‘a’ and ‘b’ are on one side of ‘f’ as shown in Fig.6. Let ‘b’ be nearer to ‘f’ than ‘a’, here ‘b’ may coincide with ‘f’. then clearly the distance from ‘b’ to L❑ the line joining a and p is less than d , which contradicts the choice of the pair ( p , L ). Hence the proof.
Doing a problem in two different waysIn order to solve a problem, we use the method of approaching some other problem
related to it in two different ways and the results lead to the solution of the given problem.
Examples1. Prove that
2n=∑k=0
n
(nk) (1)
Solution:- here we give a proof without using binomial theorem.Let us count the number of subsets of the set S= {1, 2, 3, ……,n} consisting of n
elements. Clearly the number of subsets is same as number of ways of constructing a subset as different ways will give rise to distinct subsets. Consider any element say ‘k’ there are two ways of considering this element while forming a subset either to include this or to exclude this. Same is true for each of the n elements. So by multiplicative rule of permutations and combinations total number of ways of forming subsets and hence the total number of subsets of S is 2n.
Now we count the number of subsets having k elements for 0 ≤ k ≤ n, which is
clearly (nk), now by additive property the total number of subsets of S is ∑k=0
n
(nk ). Hence
(1) is established.2. Prove that there does not exist an equilateral triangle in the xy-cartisian plane whose
vertices have integral co-ordinates.Solution:- If there is such a triangle, with vertices (x1 , y1 ) , (x2 , y2 ) , ( x3 , y3 ) with x i and
y i integers. Now we compute the area of this triangle in two ways,Using determinants
Area of the ∆=12|x1 y1 1
x2 y2 1x3 y3 1|,
this is a rational number. On the other hand
Area of the ∆=14
side2
√3=(a rational number)√3,
Which √3 is a rational number, a contradiction, which proves the result. Relating two different situations
Here firstly we count the number of binary strings of length n with r zeros and n−rones. Which is equal to the number of permutations of n things where r are of one
kind and n – r of second kind. This number is(nr ).Now let us count the number of paths from A to C where along the following grid
horizontal left to right and vertical upwards motions are allowed.
To reach from A to C one need to make 6 vertical steps and 9 horizontal steps. (Total 15 steps) While constructing a path we write 1 if horizontal step is taken and we write 0 if vertical step is taken, this way we obtain a binary sequence of length 15 with 6 zeros there is unique such sequence associated to a given path. Also from a binary sequence of length 15 with 6 zeros we can construct an unique path. So that
Number of such paths = Number of binary sequences of length 15 with 6 zeros =(156 ).
Pigeon Hole PrincipleThis Principle states that if we have more than n+1 objects and n boxes and if we
distribute the objects into the boxes randomly or applying some rule, the conclusion is that there must be some box containing at least two objects
OrMore generally if we have more than nk+1 objects and n boxes and if we distribute the
objects into the boxes randomly or applying some rule, the conclusion is that there must be some box containing at leastk+1 objects.
Examples1. In any group of n persons there are two who have the same number of acquaintances.
Solution:- Each of the persons may have 0,1,2,3,……., n - 1 acquaintances. If any one of these has 0 acquaintances then none will have n−1 acquaintances and if any one of these has n - 1 acquaintances then none of them will have 0 acquaintance. So that the number of acquaintances for all n persons are coming from the set {0,1,2,3,
D
A Fig.7 B
C
……., n – 2} or {1,2,3,……., n – 1} We consider boxes numbered 0 to n -2 or 1 to n – 1,
as the case may be. Now we allot box numbered i to a person having i acquaintances. As the number of persons is n and number of boxes is n−1, there must be two associated to the same box so the result follows.
2. Let A be a subset consisting of 101 numbers chosen from. {1,2,3,……., 200} Show that there exists a , b A such that a|b.
Solution:- We make 100 boxes indexed by first 100 odd integers i.e. 1,3,5,…….,199. Now we pick any element say c of A and write it as c=2k c1, where c1is odd. We put this element in the box indexed c1, this is possible as c1 is odd and less than 200. Now there are 101 elements and 100 boxes one of the boxes must contain at least 2 elements. Let a=2k cand b=2l cbe two such elements in the box indexed c with k<l. Then clearly a|b
3. Prove that one of the numbers 1,11,111,1111,………. Is divisible by 2009.
Solution:- Let a i=111… .1⏟i׿¿
, The given sequence becomes a1 , a2 , a3 , ……… .., let
ri be the remainder obtained on dividing ai by 2009. Now as the sequence is infinite and the set of remainders consists of 2009 numbers ( 0 to 2008) there must be two numbers in the sequence a j∧ak , with j<k such that their remainders are same but
then ak−a j=11111 ……...0000⏟withk− j 1' s∧ j 0 ' s
, is divisible by 2009 and hence a j−k is divisible by
2009.
Symmetric considerationsWe illustrate this method by the following examples
Examples1. There are two piles of balls as shown in the fig.8. At each turn, a player may take
away as many balls as he chooses, but only from one of the piles. The winner is the player who takes away last ball. Who can win the game?. Explain the strategy.
Fig.8
Solution:- There are two cases(i) Heaps are of equal size.
(ii) Heaps are of unequal size.
In case (i) the second player can win and in the case (ii) first player can
win.
The strategy is that the wining player maintains the symmetry and the next
player is forced to disturb the symmetry. In the first case when heaps are of
equal sizes at the beginning the symmetry is already there so first player will
be disturbing the symmetry and the second player will be maintaining it so
second player is the winner. In the second case the first player will make
heaps equal by removing balls from the heap containing more balls so he is
maintaining the symmetry while the second player will be disturbing it. In the
figure second heap contains one ball less than the first heap in this case first
player can win if he adopts this strategy.
2. Two players take turns putting one rupee coins of equal size on a round table,
without piling one coin on top of other and also without overlapping. The player
who cannot place a coin is the looser (A player can place exactly one coin on his
turn). Who can win and with what strategy?
Solution:- In this game first player can win no matter how big the table may
be. To do so, he must place the first coin so that its centre coincides with the
center of the table. After this he replies to each move of the second player by
placing a coin in a position symmetric to the coin placed by the second player,
with respect to the centre of the table. Notice that in such a strategy the positions
of the two players are symmetric after each move of the first player. It follows
that if there is a possible turn for the second player, and then there is a possible
response for the first player, who will therefore win.
KENDRIYA VIDYALAYA SANGATHAN SAMPLE PAPER 1
Class – XISubject – Mathematics
Maximum Time allowed: 3 hrs. Maximum Marks: 100
General Instructions:1. All questions are compulsory.2. The question paper consist of 29 questions divided into four sections A, B, C and D.
Section A comprises of 4 questions of one mark each, section B comprises of 8 questions of two marks each section C comprises of 11 questions of four marks each and section D comprises of 6 questions of six marks each.
3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
4. There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 2 questions of six mark each. You have to attempt only one of the alternatives in all such questions.
5.Use of calculators is not permitted. You may ask for logarithmic tables, if required.
SECTION A
Q1. A function f:R→R is defined as f(x) = x2 , ∀ x∈R. Write its range.
Q2. Find the argument of the complex number z = –1+i.
Q3. Find the slope of the line which is perpendicular to the line y=5x+3.
Q4. Write the negation of the statement: “Australia is a continent.”
SECTION B
Q5. If A ∩ B= φ , show that A = A∩B and hence show that A ⊂ B .
Q6. Find the value of tanπ8 .
Q7. Express3 i3 + 6 i16 − 7 i29 + 4 i27in the form x + i y where x , y∈ R .
Q8. Which term of the series 1− 1
3+ 1
9− 1
27+ .. . .. .. . .. .. . is − 1
243?
Q9. Find the value of 91 /3 . 91 /9 .91/27 .. . ∞ . .
Q10. Evaluate limx→ 0
sin ax + bxax + sin bx .
Q11. (i) Write the contra positive of the statement: “If you are born in India, then you are a citizen of
India.” (ii) Write the converse of the statement: “If a number n is even, then n2 is even.”
Q12. A bag contains 4 red , 6 blue and 8 green balls. Two balls are drawn at random. Find the probability
that the balls drawn either both are blue or both are red.
SECTION C
Q13 .Let A , B and C be the sets such that A¿ B = A ∪C and A ∩ B = A∩ C . Show that B = C.
Q14. Find the domain and range of the function f ( x )= √x2−3 x + 2 .
Q15. If A + B + C =1800, prove that
sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = 4 sin B − C
2sin C − A
2sin A − B
2
Q16. Prove that
12× 5
+ 15× 8
+ 18 × 11
+ . .. .. . .. .. . .. + 1(3n −1 )(3n + 2)
= n6n + 4
, ∀ n∈ Nby principle
of mathematical induction.
Q17. Find the square root of –8 –6iOR
Convert the complex number
z= i−1
cos π3+isin π
3 in the polar form.
Q18. Solve the given system of inequalities graphically: 2 x+ y ≥ 4 , x+ y ≤3 , 2x−3 y≤ 6
Q19. A traffic policeman at point P(3,5) wants to reach the nearest point Q on the road along the line
5x + 6y = 106 as early as possible to catch a defaulter.(i) Locate the position of PQ(ii) Find the distance PQ.(iii) What value is depicted by follow the traffic rules?
ORLet the point P(-8,12) lies on the top of a Qutub Minar, Delhi. Find the image of the point on
the line 4x + 7y +13 = 0. Why is conservation of monuments important?
Q20. Find the coordinate of the foci , the vertices, the length of major axis, the minor axis, the eccentricity
and the length of latus rectum of the ellipse : 36 x2 + 4y2 = 144.
Q21. Verify that the points (3,–2,4), (1,0,–2) and (–1,2,–8) are collinear
Q22. Evaluate
limx→ π
6
√3 sin x − cos x
x − π6
OR Using the First principle of derivative find the derivative of tan √ x . Q23. Out of 100 students, two sections of 40 and 60 students are formed, if you and your
friends are among the 100 students . what is the probability that you both enter the same section? Why we need
friends in life?
SECTION D
Q24. In a survey of 100 persons, it was found that 28 read newspaper A, 30 read read newspaper B, 42 read
newspaper C, 8 read newspaper A and B, 10 read newspaper A and C, 5 read newspaper B and C and 3
read all the three newspapers. Find how many persons read none of the three newspapers and how many
read newspaper C only? What are the importance of reading newspaper?
Q25. Prove that :sin 3 A sin3 A + cos 3 A cos3 A =cos 3 2 A .OR
Prove that :4 cos120 cos 480 cos720= 1− 2 sin2180.
Q26. In how many ways can be the letter of the word PERMUTATIONS be arranged if the (i) words start with P and end with S?(ii) vowels are all together(iii) there are always 4 letters between P and S .
Q27.The 3rd , 4th and 5th terms in the expansion of (x + a)n are respectively 84, 280, and 560, find the value
of x , a, and n.
Q28. Find the sum to n terms of the series : 3 + 15 + 35 + 63+…………..OR
Find the sum to n terms of the series Sn= 12 + (12 + 22 ) + (12 + 22 + 32) +.. .. . .. .. . . Also
determine ∑
Sn
(n + 1 )2 .
Q29. Calculate the mean and the standard deviation for the following distributionMarks 20 –
3030 -
4040 -
5050 –
6060 -
7070 -
8080 –
90Numb
er of students
3 6 13 15 14 5 4
What is the importance of scoring more marks in the examination?
KENDRIYA VIDYALAYA SANGATHAN SAMPLE PAPER 2
Class – XISubject – Mathematics
TIME 3 Hrs M.M. 100General InstructionsAll questions are compulsory.1. The question paper consist of 29 questions divided into four sections A, B,C and D.
Section A comprises of 4 questions of one mark each, section B comprises of 8 questions of two marks each , section C comprises of 11 questions of four marks each and section D comprises 6 question of six marks each.
2. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
3. There is no overall choice. However, internal choice has been provided in 03 questions of four marks each and 02 questions of six mark each. You have to attempt only one of the alternatives in all such questions.4. Use of calculators is not permitted.
SECTION –A
Q1. If (x + 1, y – 2) = (3 , 1) find the value of x + y.
Q2. Find the multiplicative inverse of 2 – 3i.
Q3. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Q4. Write the statement “The Banana trees will bloom if it stays warm for a month” in the form “if-then”.
SECTION –B
Q5. Using properties of sets prove that : ( A∪B )'=A '∩B'
Q6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Q7. Convert the given complex number in polar form: – 1 + i.
Q8. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
Q9. Find the sum to n terms of the series 1
1× 2 + 12× 3 + 1
3× 4 + ……
Q10. Find the derivative of x4 (5 sin x – 3 cosx).
Q11.By method of contradiction, Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true .
Q12. If E and F are events such that P(E) =14 , P(F) =
12 , and P(E and F) =
18
find: (i) P(E or F), (ii)P(not E and not F).
SECTION –C
Q13. Let A, B and C be the sets such that A ∪ B = C and A ∩ B = φ . Show that A = C – B .
Q14.Find the domain of the function f(x) = 1
√x+¿ x∨¿¿OR
Let f = {(x , x2
1+x2 ): x∈R}be a function from R into R. Determine the range of f.
Q15. Solve: 2 cos2x + 3 sinx = 0OR
Prove that acos(B−C2 ) = (b + c)sin A
2 .Q16 .Prove the following by using the principle of mathematical induction for all n ∈ N:
12.5+ 1
5 .8+ 1
8 . 11+ . .. + 1
(3 n−1 ) (3 n+2 )= n
(6n+4 ).
Q17. Find the real value of θ such that 3+2i sinθ1−2 isinθ is purely imaginary.
Q18. A milk of 45% concentration is diluted at home by the seller by adding some water to it so that milk concentration is reduced between 25% and 30%. If 1125 litres of milk of 45% is available, how much water has been added? Which value system the seller is lacking?
Q19. If p is the length of perpendicular from the origin to the line whose intercepts on the
axes are a and b, then show that : 1
P2 = 1a2 + 1
b2 .
Q20. A rod of length 12 cm moves with its ends always touching the coordinate axes.
Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis
OR
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2 , 3).
Q21. If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.
Q22. Find the derivative of esin x
by first principle.
Q23. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) The student opted for NCC or NSS. (ii) The student has opted neither NCC nor NSS.(iii) The student has opted NSS but not NCC.
SECTION - D
Q24. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Q25. If a and b are the roots of x2 -3x + p = 0 and c and d are roots of x2 -12x + q = 0 , where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15.
OR Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …
Q26. Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
ORIf the coefficients of ar– 1, arandar+ 1 in the expansion of (1 + a)nare in arithmetic progression,
prove that n2 – n(4r + 1) + 4r2 – 2 = 0.
Q27. How many four- letter words can be formed using the letter of the word ‘INEFFECTIVE’ ?
Q28. Prove that : (1+cos π
8 )(1+cos 3 π8 )(1+cos 5 π
8 )(1+cos 7π8 )= 1
8
Q29. In a town of 10,000 families it was found that 40% families buy newspaper A, 20% buy newspaper B, 10% families buy newspaper C, 5% buy A and B,3% B and C,4% A and C. If 2% families buy all the three newspapers . Find
(a) The number of families which buy newspaper A only.(b) The number of families which buy none of A, B and C.(c)The number of families which buy only one newspaper. (d) Suggest two measures to inculcate reading habits.
KENDRIYA VIDYALAYA SANGATHAN SAMPLE PAPER 3
Class – XISubject – Mathematics
TIME 3 Hrs M.M. 100
General Instructions
1. All questions are compulsory.
2. The question paper consists of 29 questions divided in three sections A, B, C & D Section A comprises of 4 questions of one mark each, section B comprises of 8 questions of Two marks each and section C comprises of 11questions of Four marks each & section D comprises of 6questions of 6 marks each .
3. All questions in section A are to be answered in one word, one sentence or as per the exact requirement of the question.
4. There is no overall choice. However, internal choice has been provided in 03 questions of four marks each and 02 questions of six marks each .You have to attempt only one alternative in all such questions.
5. Use of calculators is not permitted. You may ask for logarithmic tables, if required.
SECTION – A [1 mark questions]
Q.1 Let f (x)= x+1 and g(x) = 2x–3 be two real functions. Find (f-g)(x)
Q.2 Evaluate (1+i)4(1+1i)
4
Q3. Find the distance of the point (−1 , 1) from the line: 12 x−5 y+82=0.
Q4.Write the negation of the statement, “Every natural number is greater than 0”.
SECTION – B [2 marks questions]
Q5. If A and B are two sets such that n(A - B) = 24, n(B - A) = 19 and n(AB) = 11. Find n(A) , n(B).
Q6. Solve the equation 2 cos2 x+3sin x=0
Q7. If x+iy = √ a+ibc+id
, then prove that ¿
Q8. How many term of G.P. 3,32,33,-------- are needed to give sum 120 ?
Q9 If AM and GM of roots of a quadratic equation are 8 and 5 respectively , then obtain there quadratic equation.
Q10. Find derivative ofx+cos x
tan x .
Q11. (a) Identify the quantifier in the statement, “There exists a number which is equal to its square”.
(b) Show that the following statement is true p: for every real number x, y if x = y , then 2x + a = 2y + a , a∈ z
Q12. Coefficient of Variation of two distributions are 70 and 60 and their Standard deviation
are 16 and 21 respectively. What are their Arithmetic Mean.
SECTION – C [4 marks questions]
Q13. Using properties of sets prove that : A−(B−C )=( A−B )∪(A ∩ C)
Q14. Find the domain and range of the function: f ( x )= 1
2−sin 3 x.
Q15. In a ABC, it is given that
2 cos Aa
+cos Bb+2 cosC
c= a
bc+ b
ca . Show that triangle is a right angled.
OR
Prove that :
cos Aa+cos B
b+cos C
c=a2+b2+c2
2abc .
Q16. Prove the following using principle of mathematical induction.
32 n+2– 8n – 9 is divisible by 8. OR
Using principle of mathematical induction prove that:
1.3+3.5+5.7+…………+(2n-1)(2n+1)= n(4 n2+6 n−1)3
.
Q17. Write (−3√2+i 3√2 ) in polar form.
Q18. Solve graphically : 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6.
Q19 . Find the foot of the perpendicular drawn from the point (-2, 3) on the line 12x – 5y + 13 = 0
Q20. Find the equation of an ellipse that satisfies the given conditions: (i) Vertices (0 , ± 7 ) (ii) Foci (0 , ± 3 )
OR Find the equation of the circle which passes through the points (20, 3),(19, 8) and (2, –9). Find its centre and radius.
Q21. Find the ratio in which the YZ- plane divides the line segment formed by joining the points(−2 , 4 ,7 )and (3 ,−5 ,8 ).
Q22. Find the derivative of √cos3 x from first principle.
Q23. In a class of 60 students, 30 opted for NCC , 28 opted for NSS and 24 opted For both NCC and NSS. If one of these students is selected at random , find the Probability that,
(i) The students has opted neither NCC nor NSS(ii) The students has opted NSS but not NCC
Do you think students should take part in NCC. Give reason.
SECTION – D[6 mark questions]
Q24. In a survey of 25 students, it was found that 15 had taken mathematics, 12 had taken physics and 11 had taken chemistry. 5 had taken mathematics and chemistry, 9 had taken mathematics and physics, 4 had taken physics and chemistry and 3 had taken all the threesubjects. Find the number of students who had
(i) At least one of the three subjects (ii) Only one of the subjects.
Q25. (a)Prove that :
tan( π4 +θ)+ tan( π4−θ)tan( π4 +θ)− tan( π
4−θ)
=cosec2θ
.
(b) Prove that : cos2 x+cos2(x+ π
3 )+cos2(x−π3 )=3
2 .
Q26. A mathematics question paper consists of 10 questions divided into two parts I and II, each containing 5 questions. A student is required to attempt 6 questions in all, taking at least 2 questions from each part. In how many ways can the student select the questions?
Q27. Show that the middle term in expansion of (1+x¿¿2 n is
1,3,5 ………….(2 n−1)n!
2n xn, where n is a positive integer.
OR
The coefficient of three consecutive term in the expansion of (1+x)n are in the 3:8:14, find n.
Q28.The Sum of two numbers is 6 times their geometric mean , show that numbers are in the Ratio (3+2√2 ): (3−2√2 ).
OR
Find the sum of the first n terms of the series : 3 + 7 + 13 + 21 + 31 + -----------------.
Q29. Find mean, variance and standard deviation for the following distribution:Classes 30-
4040-
5050-
6060-
7070-
8080-
9090-
100Frequency 3 7 12 15 8 3 2
KENDRIYA VIDYALAYA SANGATHAN SAMPLE PAPER 4
Class – XISubject – Mathematics
Time Allowed: 3 Hrs Maximum Marks: 100
1. All questions are compulsory.2. The question paper consist of 29 questions divided into four sections A, B C and D.
Section A comprises of 04 questions of one mark each, section B comprises of 08 questions of two marks each, section C comprises of 11 questions of four marks each and section D comprises of 06 questions of six marks each.
3. There is no overall choice. However, internal choice has been provided in some questions5. Use of calculators is not permitted. ___________________________________________________________________________
___Section A
Q1 . If f(x) = ax +b , where a and b are integers, f (–1) = –5 and f (3) = 3, then find the value of a and b.
Q2 . For a complex number z, what is the value of arg. z + arg.z, z ≠ 0?
Q3 . Find the intercept of the line 2x + 3y – 6 = 0 on the x-axis .
Q4 . Write the converse of the following statement :
“If a number n is even, then n2 is even”.
Section B
Q5 . Find the argument of 1
1−i .
Q6 . A and B are two sets such that n(A –B) = 14+ x , n(B–A) = 3x and n(AB) = x. If n(A) = n(B) , then find the value of x .
Q7 . Prove that:
sin 2θ1−cos2 θ
= cot θ
Q8 . Write the value of the tenth term of the sequence:
1(1) + 2(1 +2) +3 (1 +2+3) +4 ( 1+2+3+4) +………
Q9 Find the values of' k' for which – 27 , k , –
72 are in G.P.
Q10 If A, B, C are three mutually exclusive and exhaustive events of an experiment such that 3P(A) = 2P(B) = P(C), then find the value of P(A).
Q11 Find limx→ 0
1+cos 4 x1−cos6 x
Q12 Verify by method of contradiction that p = √ 3 is irrational.
Section C
Q13 Let R be a relation from Q to Q defined by R = {(a , b): a , b ∈ Q and a – b ∈ Z}. Show that
(i) (a , a) ∈ R for all a ∈ Q(ii) (a , b) ∈ R implies that (b, a) ∈ R(iii) (a , b) ∈ R and (b , c) ∈ R implies that (a , c) ∈R
Q14 Prove that :
sec 8x−1sec 4 x−1
=tan 8 xtan 2 x
OR
Find the general solution for : cos3x + cosx – cos2x = 0.
Q15 Prove by using the principle of mathematical induction that for all n∈N
102n-1+1 is divisible by 11.
OR Prove using PMI for all n∈N : (2n+7) < (n+3)2
Q16 Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that
A = B.
Q17 Find the equation of the line perpendicular to the line whose equation is 6x – 7y + 8 = 0 and that passes through the point of intersection of the two lines whose equations
are : 2x – 3y – 4 = 0 and 3x + 4y – 5 = 0.
Q18 How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain morethan 25% but less than 30% acid content?(i) What do you mean by acid rain ?(ii) What is the effect of acid rain on human health ?
Q19 Find the square root of – 15 – 8i.
Q20 Find the coordinates of foci, the vertices, the eccentricity and the length of latus rectum
of the hyperbola x2
16− y2
9=1.
OR
Find the equation of the circle passing through the points (2 , 3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.
Q21 Differentiate f(x) = 3−x
3+4 x with respect to x, by first principle
Q22 Determine the point in yz-plane which is equidistant from three points A (2, 0 3) ,B (0, 3, 2) and C (0, 0, 1).
Q23 Two students Anil and Ashima appeared in an examination. The probability that Anil willqualify the examination is 0.05 and that Ashima will qualify the examination is 0.10.
Theprobability that both will qualify the examination is 0.02. Find the probability that
(a)Both Anil and Ashima will not qualify the examination.(b) Atleast one of them will not qualify the examination
(c) Only one of them will qualify the examination.
Section D
Q24 In a survey of 60 people, it was found that 25 people read news paper H, 26 read news paper T, 26 read newspaper I, 9 read both H & I, 11 read both H & T, 8 read both T & I, 3 read all three newspapers. Find: (i) the number of people who read news paper H only (ii) the number of people who read news paper T only (iii) the number of people who read news paper I only (iv) the number of people who read at least one of the newspapers. What is the role of news papers in daily life ?
Q25 Find the mean and variance for the data :
xi 6 10 14 18 24 28 30fi 2 4 7 12 8 4 3
Q26 The first three terms in the binomial expansion of (a+b)n are given to be729,7290 and
30375 respectively. Find a, b and n.
Q27 ( i) Eighteen guests are to be seated , half on each side of a long table. Four particular guests desire to sit on a particular side and three others on other side of the table . Find the number of ways in which the seating arrangement can be made. (ii) If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary. Then what is the rank of the word RACHIT.
Q28 If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.
Q29In a ΔABC, prove that : (b−c ) cot A
2+( c−a ) cot B
2+(c−a ) cot C
2=0
KENDRIYA VIDYALAYA SANGATHAN SAMPLE PAPER 5
Class – XISubject – Mathematics
TIME: 3Hrs. MAX.MARKS:100
General Instructions:(a) All the questions are compulsory.(b) This question paper consists of 29 questions divided into four section A,B,C & D.
Section A comprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section C comprises of 06 questions of six marks each.
(c) All questions in section A are to be answered in one word, one sentence or as per the exact requirement of the question.
SECTION AQ1: Write the power set of the set A= {0, 1, 3} .
Q2: What is the value of:[ i4 n+1−i4 n−1
2 ].Q3: Find the equation straight line; intersect the y-axis at a distance of 2 units above the
origin and making an angle of 30o with positive direction of the x-axis.
Q4: Write contrapositive of the statements “If n is a natural number, then n is an integer”.
SECTION BQ5: Let A, B and C be the sets such that A∪B=A∪C And A∩ B=A ∩Cshow that
B= C
Q6: In any triangle ∆ ABC prove that a+b
c = cos ( A+B
2)
sin c2
Q7: find real value of such that, 3+2 isin ϑ1−2 isin ϑ
ispurelyimaginary .
Q8: How many terms of the G.P. 3 , 32
, 34
, ………….are needed to give the sum of3069512
.
Q9: Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that
S3 = 3(S2 – S1).
Q10: Evaluate limx →0
cos2 x−1cosx−1
Q11: (i) State the reason why the sentence “answer this question” is not a statement.(ii) Write the negation of the statement “New Delhi is the capital of India” using the
statement “ It is not the case”.
Q12: One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that
i) The card will be not a diamond.ii) A red colour
SECTION–CQ13: Solve : √2 sec x+ tan x=1 .
Q14: Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that
(a) Both Anil and Ashima will not qualify the examination.(b) Only one of them will qualify the examination.
Q15: (i) Show that A ∩ B = A ∩ C need not imply B = C. (ii) Using laws of set algebra, show that ( A∪B )∩( A∪B ' )=A .
Q16: Find the domain of the function: f ( x )= −3+x+ x3
(x3−2x¿¿2−5 x+6) (75−3 x2 )¿ .
Q17: Using the principle of mathematical induction Prove that (1 + x)n≥(1 + nx), for all natural
number n, where x > – 1.
OR
Prove by using the principle of mathematical induction for all n € N, 102n−1+1 is divisible
by 11.
Q18: Find the square root of a complex number 3+4iOR
Convert the complex number −16i+√3
into polar form.
Q19: Solve the system of inequalities: 2x + 3y ≥ 3, 3x + 4y ≤ 18, –7x + 4y ≤14, x – 6y ≤ 3, x ≥
0, y ≥ 0.
Q20: A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is
bisected at the point (1, 5). Obtain its equation.
Q21: If the length of the transverse axis along x-axis with centre at origin of a hyperbola is 7
and it passes through the point (5, –2).Then find equation of the hyperbola.
Q22: Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, – 8) is
divided by the YZ-plane.
Q23: Evaluate : limx→ π
6
2−√3 cosx−sinx(6 x−π )2 .
OR
Find the derivative of f(x) by first principle,f ( x )=2 x+3x−2 .
SECTION–DQ24: In a survey of 100 student ,the number of student studying the various languages were
found to be :English only 18, English but not Hindi 23 ,English and Sanskrit 8 ,English 26 ,Sanskrit and Hindi 8, Sanskrit 48 ,no language 24.Find:
i. How many students were studying Hindi?ii. How many students were studying English and Hindi?
iii. How many students were studying all three languages?Q25: Find the number of arrangements of the letters of the word INDEPENDENCE. In how
many of these arrangements,(i) do the words start with P
(ii) do all the vowels always occur together(iii) do the vowels never occur together(iv) do the words begin with I and end in P?
Q26: Show that : 2 sin2β + 4 cos(α + β) sinα sinβ + cos2 (α + β) = cos2α.
OR
In any Δ ABC , prove thata2+b2
a2+c2=1+cos ( A−B ) cosC1+cos ( A−C )cos B
Q27: Find the sum to n terms of the series : 3 × 12 + 5 × 22 + 7 × 32+ ...
Q28: Calculate the mean and variance for the following distribution Classes
30-40
40-50
50-60
60-70
70-80
80-90
90-100
Frequency
3 7 12 15 8 3 2
Q29: If the coefficients of ar– 1, arand ar+ 1 in the expansion of (1 + a)nare in arithmetic
progression, prove that : n2 – n(4r + 1) + 4r2 – 2 = 0.
OR
The first three terms in the binomial expansion of (a+b)nare given to be 729,7290 and 30375 respectively. Find a, b and n.
KENDRIYA VIDYALAYA SANGATHANSAMPLE PAPEER 1
Class – XII Time:3.00HrsSubject - Mathematics Maximum Marks – 100
General instructions – (i) All questions are compulsory.
(ii) The question paper consists of 29 questions divided in to four sections viz. A, B a C&D. Section A comprises of 4 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 11 questions of 4 marks each and section D comprises 6 questions six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions.
(v) Use of calculators is not permitted.
SECTION C [ 1 mark Questions]
1. Evaluate cos {π3−cos−1 (12 )}
2. If f(x) = x+7 and g(x) = x –7 where x∈R find fog(7)
3. If | x 218 x−5
|=| 6 218 7
| then find the value of x
4.If a→= i
¿
+2 j¿
−3 k¿
and b→=2 i
¿
+4 j¿
+9 k¿
find a unit vector parallel to a→+b→
SECTION C [ 2 marks Questions]
5.Find
dydx
if y= tan−1(sin x1+cos x ) .
6.Evaluate ∫ log x dx
7.Given that P( A )=0 .4 , P(B )=0.2 and P (A /B )=0 .2 , find P(A∩B ).
8.Ifcos−1 4
5+cos−1 12
13= tan−1 x
find the value of x.
9. Find the general solution of the differential equation
dydx+ y=1 , y≠1 .
10. IfA=[sin x −cos x
cos x sin x ] , find x if A+A '=I .
11. If a , b , c are mutually perpendicular unit vectors ,evaluate|a+b+ c|.
12. For the curve y = , find all the points at which tangent is parallel to x-axis.
SECTION C [ 4 marks Questions]
13. Using properties of determinants ,show that
|x y zx2 y2 z2
yz zx xy|
= (y – z)(z – x)(x – y) (yz + zx + xy).
14.Find when
15. Find the values of a and b such that the function defined by
f(x) = {
x−4|x−4|
+a , if x<4
a+b if x=4x−4|x−4|
+b if x>4 .
16.Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
ORShow that the right circular cylinder of given surface and maximum volume is such that its
height is equal to the diameter of the base.
17.Find the intervals in which the function f given by f ( x )=sin x+cos x , 0≤x≤2 π is strictly increasing or strictly decreasing.
18.Form the differential equation representing the family of ellipse having foci on x-axis and centre at the origin.
ORFind a particular solution of the differential equation satisfying the given condition:
(1+x2) dydx+2 xy= 1
1+x2; y=0 when x=1.
19.Let a= i+4 j+2 k , { b=3 i−2 j+7 k ¿ and c=2 i− j+4 k .Find a vector d which is
perpendicular to both a and b ,andc . d=15.
20. Given three identical boxes I, II and III, each containing two coins. In box I, coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?
21.Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of spades.
22.Find the shortest distance between the lines x−1
2= y+2
3= z−3−4
and x+3−1
= y−54= z−3
5
23.Evaluate :∫ dx
3+2sinx+cosxOR
Evaluate∫ 3 x+5√2x2−5 x+4
dx
SECTION D[ 4 marks Questions]
24. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.
25. Find the equation of the perpendicular drawn from the point (1, –2,3) to the plane2x – 3y + 4z + 9 =0.Also find the foot of the perpendicular.
OR
Find the equation of the plane through the line of intersection of the planes and which is perpendicular to the plane
26.Find the product
.Use the product to solve the system of equations
27.Find the area of the region in the first quadrant enclosed by x-axis, line and the
circle
28.∫0
π /2
log sin x dx
OR
Evaluate ∫−1
2
(7 x2−5 x+3 ) dxas a limit of sums
29.Consider . Show that f is invertible .Findthe inverse of f.
KENDRIYA VIDYALAYA SANGATHANSAMPLE PAPEER 2
Class – XII Time: 3.00 HrsSubject - Mathematics Maximum Marks – 100
General instructions – (vi) All questions are compulsory.
(vii) The question paper consists of 29 questions divided in to four sections viz. A, B a C &D. Section A comprises of 4 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 11 questions of 4 marks each and section D comprises 6 questions six marks each.
(viii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(ix) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions.
(x) Use of calculators is not permitted.
SECTION A[ 1 mark Questions]
1. If
|2 x 58 x
|
=
|6 −27 3
|
then find the value of x?
2. Find the projection of vector a
=2 i+3 j+2 k
on the vector b
=i+2 j
+3 k
3. If the function f:R→R
defined as f(x)=2x – 3 is invertible , then find f−1
.4. Let * be a binary operation defined by a*b=4a+3b –2. Find (3*5)*2
SECTION B[ 2 marks Questions]
5. Write in simplest form:tan−1(cos x _sin x
cos x+sin x ).
6. Express A=[ 6 7−2 1 ]
as a sum of symmetric and skew symmetric matrices.7. Differentiate ( xx )x w.r.t. x.8. Find a Point on the parabola y =(x-3)2 where the tangent is parallel to the line joining
(3,0) and (4,1).
9. (i) Write the degree of the differential equation: y=x
dydx+a√1+( dy
dx )2
(ii) write the integrating factor of x
dydx – y = x4–3x
10. Find µ, if the vector a
=i+3 j+ k , b=2 i− j− k
and c= μ j+ k
are coplanar.11. Find the probability of getting 5 exactly twice in 7 throws of a die.
12. Evaluate: ∫sin5 x . dx
SECTION C[ 4 marks Questions]
13. Prove that
|1+a2−b2 2 ab −2 b
2 ab 1−a2+b2 2 a2 b −2a 1−a2−b2
|
=(1+a2+b2)2
14. If x y=ex− y
, Prove that
dydx=log x(1+ log x )2
15. If y = (cot-1x)2, Prove thatd2 ydx2 ( x
2+1 )2+2x(x2+1)
dydx = 2
16. For the curve y = 4x3–2x5, find all the points at which the tangent passes through the origin.
OR
A ladder is leaning against a wall. The bottom of the ladder is pulled along the ground, , away from the wall, at the rate of 2m/sec. How fast is the height on the wall decreasing when the foot of ladder is 4 m away from the wall
17. Show that of all the rectangles inscribed in a given fixed circle, the square has the
maximum area.
18.
Evaluate :∫ sin( x+a )sin( x+b )
dx
OR
Evaluate: ∫ ex( 2+sin 2 x
1+cos2 x )dx
19. Find the particular solution of 2xy+y2–2x2
dydx =0, it is given that y = 2 when x=1.
OR
Solve dxdy+x cot y=2 y+ y2 cot y , ( y≠0 ) , given that y (π3 )=0 .
20. The probability of a shooter hitting a target is
34 . How many minimum number of
times he should fire so that the probability of hitting the target at least once is more than 0.99.21. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.
One ball is transferred from Bag I to Bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
22. If the three vectors a ,b and c are coplanar, prove that the vectors a+b ,b +c andc +a are also coplanar.
23. Show that the lines
x−12 =
y−23 =
z−34 and
x−45 =
y−12 =z intersect. Find their
point of intersection.SECTION D[ 6 marks Questions]
24. Consider f : R+→ [−5 ,∞ ¿defined by f(x) = 9x2+6x-5. Show that f is invertible. Find the inverse of f.
25. Evaluate :∫0
π xsin x1+cos2 x dx OR
Evaluate: ∫0
π xdxa2 cos2 x+b2 sin2 x dx
26. Using integration compute the area of the region bounded by the triangle whose vertices are (4 , 1), (6 , 6), and (8 , 4).
27. Find the equation of the plane passing through the line of intersection of the planes
2x+y-z=3, 5x-3y+ 4z+9=0 and parallel to the line
x−12= y−3
4= z−5
5 . ORFind the length and the foot of the perpendicular from the point P (7,14,5) to the plane
2x+4y-z=2. Also, find the image of the point P in the plane.
28. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 cost Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
29. Use the product[1 −1 20 2 −33 −2 4 ] [
−2 0 19 2 −36 1 −2 ] to solve the system of equations
x-y+2z=1, 2y-3z=1, 3x-2y +4z=2.
KENDRIYA VIDYALAYA SANGATHANSAMPLE PAPEER 3
Class – XII Time:3.00HrsSubject - Mathematics Maximum Marks – 100
General instructions – (xi) All questions are compulsory.
(xii) The question paper consists of 29 questions divided in to four sections viz. A, B a C&D. Section A comprises of 4 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 11 questions of 4 marks each and section D comprises 6 questions six marks each.
(xiii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(xiv) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions.
(xv) Use of calculators is not permitted.
SECTION C [ 1 mark Questions]
1. If | x 218 x
|=| 6 218 6
| then find the value of x.
2. If the binary operation * on the set of integers Z, is defined by a *b = a + 3b2 , then find the value of 2 * 4.
3. Find gof if f ( x )=|x| and g ( x )=|5 x+1|.4. Write the value of , if two vectors are such that .
SECTION B[ 2 marks Questions]
5. Prove thattan−1 1−x
1+ x=1
2tan−1 x
.
6. Write the ad joint of the Matrix [2 −14 3 ] .
7. Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm ?
8. If sin y = x sin (a + y), show that dydx=
sin2 (a+ y )sin a .
9. Evaluate: ∫ xex
(1+x )2dx
.
10. Write the order & degree of the differential equation
(1+( dydx )
2)32
d2 ydx2
=k
11. Find the value of k if k i¿
−6 j¿
−2 k¿
, −i¿
+4 j¿
+3k¿
and −8 i¿
− j¿
+3 k¿
are coplanar.
12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that the youngest is a girl.
SECTION C[ 4 marks Questions]
13. Show that
|x+1 x+2 x+ax+2 x+3 x+bx+3 x+4 x+c
|=0 where a, b, c are in Arithmetic Progression
14. Determine the value of k so that the function:
f ( x )={k . cos2xπ−4 x
, if x≠ π4
5 , if x= π4
is continuous at x=π4
15. If y = (x+√ x2+1)m ,then show that (x2+1 ) d
2 yd x2 +x dy
dx−m2 y=0.
16. Determine the intervals in which the function f(x) = ( x−1 ) ( x+2 )2 is strictly increasing or decreasing. At what points are the tangents of the graph of the function parallel to x-axis?
17. Ramesh appears for an interview for two posts A and B for which selection is independent. The probability of his selection for post A is 1/6 and for post B is 1/7. He prepared well for the two posts by getting all the possible information. What is the probability that he is selected for at least one of the post? Which values in life he is representing ?
18. If a , b , and c are three unit vectors such that a . b=a .c = 0 and the angle between
b∧c is π6 , prove that a=± 2(b× c) .
19. Find whether the lines r=¿) + λ (2 i+ j) and r=(2 i− j) +µ(i+ j−k ) intersect or not.If intersecting , find their points of intersection.
20. Find the particular solution of the differential equation:
[ x sin2( yx )− y ]dx+ xdy=0 , given y= π
4when x=1
OR
Solve the differential equation (1+x2 ) dydx−2 xy = (1+x2¿ (2+ x2)
21. Evaluate∫3x−2( x+1)2 ( x+3 )
dx
OR
Evaluate ∫0
π x tanxsec x+tan x
dx
22. Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
23. If two cards are drawn at random from a deck of 52 cards and X is the number of aces obtained then write the probability distribution for random variable X and find the mean.
ORIn answering a question on a MCQ test with 4 choices per question, a student knows the
answer, guesses or copies the answer. Let ½ be the probability that he knows the answer , ¼ be the probability that guesses the answer and ¼ that he copies it. Assuming that a student who copies the answer will be correct with the probability 1/4, what is the probability that the student knows the answer given that he answered it correctly.
SECTION D[ 6 marks Questions]
24. Let A= {1, 2, 3… 9} and R be the relation in A × A defined by (a, b) R(c, d) if a+d=b+c for a, b, c, d ∈ A.Prove that R is an equivalence relation also obtain the equivalence class [(2,5)].
OR
If f:R→R given by the function f(x) = x + 1x , show that f is invertible . find f-1.
25. Evaluate ∫0
π2
x sinx cosxsin4 x+cos4 x
dx
OR
Evaluate ∫0
2
(x¿¿2+ x+1)dx¿ as the limit of a sum
26. Find the area bounded by the curves x2 + y2 = 4 and (x – 2)2 + y2 = 4 using integration.
27. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6)28. A toy company manufactures two types of dolls, A and B. Market tests and available
resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of 12 and 16 per doll respectively on ₹ ₹dolls A and B. Formulate the above as a linear programming problem for maximum profit and solve it graphically.
29. Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award Rs. x each, Rs. y each and Rs. z each for the three respective values to 3, 2 and 1 students respectively with a total award money of
Rs.1,600. School B wants to spend Rs.2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is Rs.900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
KENDRIYA VIDYALAYA SANGATHANSAMPLE PAPEER 4
Class – XII Time:3.00HrsSubject - Mathematics Maximum Marks – 100
General instructions – (xvi) All questions are compulsory.
(xvii) The question paper consists of 29 questions divided in to four sections viz. A, B a C&D. Section A comprises of 4 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 11 questions of 4 marks each and section D comprises 6 questions six marks each.
(xviii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(xix) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions.
(xx) Use of calculators is not permitted.
SECTION C [ 1 mark Questions]1. Check whether the relation R in R defined by R = {(a, b) : a ≤ b3} is transitive.2. If A = {1 , 2, 3} and B = {3, 0, 5}, find the number of bijections from A to B.
3. Let A be a square matrix of order 3 × 3, then find | kA|.
4. If a→= i
¿
+3 j¿
+7 k¿
and b→=7 i
¿
− j¿
+8k¿
find the projection of a→
on b→
SECTION C [ 2 marks Questions]
5. If yx=x2 , find dy
dx.
6. Evaluate ∫ x
1+x tan xdx
7. 10% of the bulbs produced in a factory are of red colour and 2% are red and defective. If one bulb is picked up at random, determine the probability of its being defective if it is red.
8. If 2 tan−1 1
2+tan−1 1
7=tan−1 x
, find the value of x.9. Find the differential equation of the family of curves y = Ae2x + B.e–2x.
10. IfA is a square matrix such that A2=I , then find the value of (A−I )3+( A+ I )3−7 A .
11. If a→
×b→
=c→
×d→
and a→
×c→
=b→
×d→
, show that (a→−d→)×(b
→−c→)=0.
12. Using differentials, find the approximate value of f(2.01), where f(x) = 4x3 + 5x2 + 2.
SECTION C [ 4 marks Questions]
13. Using properties of determinants , show that
|b+c a+b ac+a b+c ba+b c+a c
|= a3 + b3 + c3 – 3abc
14.
If x=a(cosθ+ logtan θ2 ) , y=a sin θ , find d2 y
dx2 at θ= π4
.
15. Find the relationship between a and b so that the function defined by
f(x) = {ax+1 , if x≤3bx+3 ,if x>3 is continuous at x = 3.
16. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
17. Find the value of p for when the curves x2 = 9p (9 – y) and x2 = p (y + 1) cut each other at right angles.
OR
Find the intervals in which the function f ( x )=3 x4−4 x3−12 x2+5 is (a) strictly increasing (b) strictly decreasing
18. Solve the differential equation : ( x2− y2 )dx+2 xydy=0. .
OR
Solve the following differential equation :
cos2 x dydx+ y=tan x
19. If the vectors a→
,b→
, c→
are coplanar then show that a→
+b→
, b→
+c→
and c→
+a→
are also coplanar.
20. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?
21. The random variable X has a probability distribution P(X) of the following
form, where k is some number : P(X) = { k , if x=02 k , if x=13 k , if x=20 , otherwise
(a) Determine the value of k. (b) Find P (X < 2), P (X ¿ 2), P(X ¿ 2)
22. Find the shortest distance between the lines
r→
= ( i+2 j+ k )+λ ( i− j+k )andr→
= (2 i− j−k )+μ (2 i+ j+2 k )
23.Evaluate : ∫ x2+2x+8( x−1 )(x−2)
dx
OR
Evaluate : ∫cosx . cos2x . cos3x . dx
SECTION D[ 4 marks Questions]
24. A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximize the profit?
25. Find the image of the point P(3, 2, 1) in the plane 2x – y + z + 1 = 0.OR
Find the equation of the plane through the line of intersection of the planes
r→
.(2 i+6 j )+12=0 andr→
.(3 i− j+4 k )=0 which are at unit distance from origin. 26. The management committee of a residential colony decided to award some of its
members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.
27. Using integration compute the area of the region bounded by the lines x + 2y = 2, y – x =1, and 2x + y = 7.
28 . Evaluate : ∫0
π/2xsinx .cosxsin4 x+cos4 x
dx
OR
Evaluate : ∫−1
2
|x3−x| dx
29. If A = Q×Q and ‘*’ be a binary operation defined by (a , b) * (c , d) = (ac , b + ad), for (a, b), (c, d)∈ A . Then with respect to ‘*’ on A (i) examine whether ‘*’ is associative, (ii) find the identity element in A, (iii) find the invertible elements of A.
KENDRIYA VIDYALAYA SANGATHANSAMPLE PAPEER 5
Class – XII Time:3.00HrsSubject - Mathematics Maximum Marks – 100
General instructions – (xxi) All questions are compulsory.
(xxii) The question paper consists of 29 questions divided in to four sections viz. A, B a C&D. Section A comprises of 4 questions of 1 mark each, Section B comprises of 8 questions of 2 marks each, Section C comprises of 11 questions of 4 marks each and section D comprises 6 questions six marks each.
(xxiii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
(xxiv) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all such questions.
(xxv) Use of calculators is not permitted.
SECTION C [ 1 mark Questions]1. Give an example of a relation on set A = { a, b, c} which is reflexive & symmetric but not transitive.
2. If f : R→R is given by f ( x )=(5− x5 )15 , then find fof ( x ) .
3. Find the minor of element 6 in the determinant
Δ=|1 2 34 5 67 8 9
|.
4.If a→= i
¿
+2 j¿
−3 k¿
and b→=2 i
¿
+4 j¿
+9k¿
find a unit vector parallel to a→+b→
SECTION C [ 2 marks Questions]
5.If sin y=x cos (a+ y ) , find dy
dx.
6.Evaluate ∫( x3−1 )13 x5 dx .
7.If P (A) = 0.4, P (B) = p and P (A ∪ B) = 0.7 find the value of p, if A and B are independents events.
8.Find the value of sin(2 tan−1 2
3 )+cos ( tan−1√3 )
9. Find the general solution of the differential equation
dydx= y
x.
10. If
A=[−123 ] , B= [−2 −1 4 ] find the relation between ( AB )′ and (BA )′ .
11. If │a→
│=√2 , │b→
│=√2and angle between a→
and b→
is 120°, find a→
.b→
.
12. Using differentials, find the approximate value of √49 . 5 .
SECTION C [ 4 marks Questions]
13. Using properties of determinants ,show that
|a2+1 ab ac
ab b2+1 bcca cb c2+1
|=1+a2+b2+c2 .
14.If √1−x2 +√1− y2=a ( x− y ) , show that dy
dx=√ 1− y2
1−x2
15.Show that the function f defined by f(x) = {3 x−2 , 0<x≤1 ¿ {2x2−x , 1<x ≤ 2 ¿¿¿¿
is continuous at x = 2 but not differentiable .
16.A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?
17.Prove that the curves x = y2 and xy = k intersect at right angles if 8k2 = 1.
OR
Find the intervals in which the function f(x) =32 x4 – 4x3 – 45x2 + 51 is
(a) strictly increasing.(b) strictly decreasing.
18. Solve the differential equation : (3 xy− y 2)dx+( x2+xy )dy=0 .OR
Solve the following differential equation : x dy
dx+ y=x log x ; x≠0
19. Show that the four points A, B, C and D with position vectors 4 i+5 j+ k , − j−k ,
3 i+9 j+4 k and 4 (− i+ j+k ) respectively are coplanar.
20. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability
that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4 What is the probability that the student knows the answer given that he answered it correctly?
21.Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of spades.
22. Find the equation of a line parallel to r→
= ( i + 2 j + 3k ) + λ (2 i + 3 j + 4k ) and passing
through 2 i + 4 j + 5k . Also find the S.D. between these lines.
23.Evaluate :∫ x2 tan−1 x . dxOR
Evaluate : ∫3x−2( x+1)2 ( x+3 )
dx
SECTION D[ 6 marks Questions]
24. A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid at least 270 kg of potash and at most 310 kg of chlorine.If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
kg per bagBra
nd PBran
d QNitrogen
Phosphoric acidPotash
Chlorine
313
1.5
3.52
1.52
25. Find the distance of the point P (6, 5, 9) from the plane determined by the points A(3, – 1, 2), B(5, 2, 4) andC (–1, – 1, 6).
ORFind the equation of the plane passing through the point P(1, 1, 1) and containing the line
r→
= (–3 i + j + 5k ) + λ (3 i – j – 5k ). Also, show that the plane contains the line
r→
= (– i +2 j + 5k ) + μ ( i – 2 j – 5k ).26.10 students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers, the second group has honest and law abiding students and the third group contains vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 13, while the combined strength of first and second group is four times that of the third group. Using matrix method, find the number of students in each group. Apart from the values, hard work, honesty and respect for law, vigilance and obedience, suggest one more value, which in your opinion, the school should consider for awards.
27.Using integration compute the area of the region bounded by the triangle whose vertices are (4 , 1), (6 , 6), and (8 , 4).
28 . Evaluate: ∫0
a
sin−1√ xa+ x
dx
OR
Evaluate∫1
4
( x2−x ) dxas a limit of sums
29.Given a non-empty set X, consider the operation : P(X) × P(X) ∗ →P(X) given by A B = A ∗ ¿ B, ∀A, B in P(X), where P(X) is the power set of X. Then (i) Is ‘ ’ commutative ? ∗(ii) Is ‘ ’ associative ?∗ (iii) Find the identity element in P(X) w.r.t. ‘ ’∗(iv) Find all invertible elements of P(X).
