120 2016 1. Sets, Relations & Functions DAY 1 l
DAY 21 l
Sets, Venn Diagram Algebra
DAY 2 l
Types of Relation, Equivalence Relation
DAY 3 l
One-one, Into and Onto Functions
Solutions of Radii of Inscribed & Circumscribed Circles of Regular Polygon
Domain-Range, odd-Even and Periodic Functions
DAY 5 l
Inverse Function, Composition of Function and Binary Operations
DAY 6 l
Practice Test of Sets, Relations and Functions
2. Matrix and Determinants DAY 7 l
Matrices, Algebra of Matrices, Transpose of Matrix
DAY 8 l
Symmetric & Skew-symmetric Matrices
DAY 9 l
Transformation of a Matrix, Inverse of Matrix
DAY 10 l
Determinants & Properties of Determinants, Minor and Cofactors
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Heights and Distances
DAY 23 l
Practice Test of Trigonometry
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Adjoint of Matrix, Solution of Linear Equations by Cramer’s Rule
DAY 12 l
Practice Test of Matrix and Determinants
3. Trigonometry DAY 13 l
Trigonometric Basics, Trigonometric Ratios, Ratios of Compound Angles
DAY 14 l
Maximum-Minimum Values
DAY 15 l
Trigonometric Periodic Functions
DAY 16 l
Trigonometric Equations
DAY 17 l
Transformation Formulae
DAY 18 l
Inverse Trigonometric Functions
As most of the part of first 14 days serves as the basis for gripping the subject, thus it is advisable not to move on day 15th topics until you develop confidence in these topics.
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DAY 25 l
Inverse Hyperbolic Functions
DAY 20 l
Solutions of Triangle
Continuity
DAY 26 l
Differentiability
DAY 27 l
Differentiation Formulae, Differentiation by Substitution
DAY 28 l
Higher Order Derivatives and Partial Derivative
DAY 29 l
DAY 19 l
Limits
Mean Value Theorems, Maxima-Minima of Functions of One Variable
DAY 30 l
Maxima-Minima & Tangents & Normals
Definite Integration by Substitution Parts and Partial Functions
DAY 35 l
Properties of Summation of Series of Definite Integration
DAY 36 l
Definite Integration by Gamma Function, Leibnitz’s Rule, Walli’s Formulae
DAY 37 l
Area of Curves
several application based questions such as in finding the sum of a given series, while Leibnitz rule are useful for differentiation under integrationsign.Topics based on application of definite integration are very important for direct questions,while all topics of day 41 and 42 are important for direct as well as application based questions.Random and planned testing is required to judge the exact learning. For topics underlined above learning with direct solved examples will be beneficial. 5. Mathematical Induction DAY 44 l
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Volume and Surface Area of Solids
DAY 39 l
Concept of Ordinary Differential Equations and their Order and Degree
DAY 40 l
Formation of Differential Equations and Solution of Differential Equations
DAY 41
Homogeneous Differential Equations, applications and Standard Results, of Solutions
DAY 42 l
Linear Differential Equations Application and Standard Results of Solutions
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DAY 46 l
MENTORS VOICE Topic 1 days 24, 25 and 26 are important for JEE as generally 1 question is asked every year,however, this portion of syllabus series as the basic need to grip the differential calculus. Topic 1 day 29 is important and has wide application such as in finding the roots of a polynomial within the given interval. Topics 1 and 2 day 30 are important from board as well as JEE point of view. Aspirants are advised to focus on these topics. Topics 1 and 2 of day 32 are very helpful in gripping the basics of integral calculus. The topics related to definite integral of days 34, 35 and 36 are very important for JEE and has
Binomial Theorem and its Simple Applications
DAY 47 l
Properties of Binomial coefficients
DAY 48 l
Properties of Multinomial Theorem
DAY 49 l
Practice Test of Binomial
7. Permutations & Combinations DAY 50 l
Test of Calculus
Practice Test of Mathematical Induction
6. Binomial Theorem
DAY 43 l
Mathematical Induction and its Applications
DAY 45
DAY 38
4. Calculus DAY 24
Fundamental of Definite Integral
DAY 34
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DAY 11
Integration by Substitution Parts & Partial Functions
DAY 33
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MENTORS VOICE Topic 1, day 1 is important from board point of view. Topic 2 day 1, venn diagram is very helpful in solving practical problems based on sets,however, this portion of syllabus is very helpful to understand the concept of probability,as well.In JEE usually 1 question is asked from topics 1 & 2 of day 4.Topic 4 day 8 is useful in solving practical problems based on application of matrix. Topics 1 & 2 days 18 and 19 are very important for JEE specially graph of Trigonometric and inverse trigonometric functions for solving questions in very short span of time.Topics 2 & 3 days 20 and 21 and topic 1 day 22 is also important specially for finding incentre and excenter of in circle and escribed circle and in finding the height of tower or building in different situations.
Indefinite Integrals Fundamental
DAY 32
DAY 22
DAY 4 l
DAY 31
Permutation & Simple Applications
DAY 51 l
Combination as Selection
DAY 52 l
Practice of Permutation & Combination
8. Sequence & Series DAY 53 l
Concept of AP,GP & HP
DAY 54 l
Concept of AM, GM and HM
DAY 55 l
Relations between AP,GP & HP
DAY 56 l
Relation between AM and GM
DAY 57 l
Sum of Special Series
DAY 58 l
Logarithmic
DAY 73 l
DAY 59 l
Exponential Series
DAY 60 l
Practice of Sequence & Series
DAY 74 l
MENTORS VOICE Topic 44 is less important from JEE point of view but important for divisibility based problems. Topics of binomial theorem, till day 49, are important and has several application based problems.It also helps in solving the problems based on combinations. Topics on days 50 and 51 are important as they have wide application and are useful to solve problems of probability. Topic after day 60th belongs to class XIth and need to recall more than once. All the topics till days 55 and 56 have problems related to nth,sum to n terms and also sum of infinite term are important to be asked as direct and application based questions. All the topics till day 58 are required essentially for basics of logarithm and exponential functions. Students are advised to focus on some special series like arithmeticogeometric series and telescopicsum. Use random testing for topics till day 60th for self analysis and use planned tests in two sections, i.e. (a) Day 1st to 23rd (b) Day 34th to 60th for proper assessment and to gain confidence.
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Cartesian System of Rectangular Coordinates
DAY 62 l
Straight Line
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Circle & its Equation
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Chord of Contact of Tangent Pole
DAY 66 l
Polar System of Circles
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Parabola & its Equation to Tangent
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Director Circle, Pole & Polar, Length of Tangent & Normal
DAY 69 l
The Ellipse
DAY 70 l
Hyperbola, Chord , Tangent
DAY 71 l
Asymptote , Conjugates, Rectangular Hyperbola
DAY 72 l
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Practice Test of Coordinate Geometry
Practice of Coordinate Geometry Based on day 61 to day 77.
conics. In three dimensional geometry, skew-lines, shortest distance between them, intersections of lines and planes, coplanar lines. Students are advised to focus on these topics. Topics 1 and 2 of day 80 are important and have direct and application based questions. All the topics till day 82 have wide application in Mathematics as well as in Physics. Topic 2 days 87 and 88, topic 1 day 89 and topic 1 days 89 and 91 are very important as every year 1 question is asked in JEE from them of direct as well as application based nature.Begin this complete section everyday with randomized check through random test to analysis your points of emphasis for your preparation. DAY 94
10. Vector Algebra DAY 79 l
Vector & Scalars, Algebra of Vectors in 2D & 3D
DAY 80 l
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Scalar & Vector Triple Products
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11. Statistics & Probability DAY 83
Mean, Median & Mode of Grouped Data
DAY 84 l
Mean, Median and Mode of ungrouped Data
DAY 85 l
Standard Deviation, Variance & Mean Deviation for Grouped Data
DAY 86 l
Standard Deviation, Variance and Mean Deviation for Ungrouped data
DAY 87 l
Practice Test of Statistics
DAY 88 l
Probability of an Event Addition & Multiplication Theorem
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Probability Distribution
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Bernoulli’s Trial
DAY 99 l
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Triangle Inequality of Complex Numbers
DAY 102 l
Square Root
DAY 103 l
Quadratic Equation in Real & Complex Numbers System & their Solution
DAY 104 l
Relation between Roots & Coefficients
DAY 105 l
Nature of Roots & Formation of Quadratic Equations with Given Roots
DAY 106 l
MENTORS VOICE All the topics till day 75th are very important as 1 or 2 questions are asked in JEE specially director circle, pole and polar, length of tangent normal related to different
Algebra of Complex Numbers, Modulus & Argument
DAY 101
Binomial Distribution Practice Test of Probability
Complex Numbers Representation in a Plane Argand Diagram
DAY 100
DAY 92 DAY 93
Practice Test
13. Complex Numbers & Quadratic Equations
Baye’s Theorem
DAY 91
Feasibles & Infeasible Region & Optimal Feasible Solutions
DAY 98
DAY 89 DAY 90
Mathematical & Graphical Solution for Problems in Two Variables
DAY 97
DAY 82
Test of 3D & Vector Algebra
Types of LPP
DAY 96
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Linear Programming Introduction, Objective Function, Optimization
DAY 95
Scalar & Vector Products
DAY 81
14. Mathematical Reasoning DAY 108 l
Practice of Complex Numbers & Quadratic Equations
DAY 107 l
Test of Algebra
Mathematical Reasoning Logic & Truth Table
DAY 109 l
Tautology, Contradiction, Converse & Contrapositive Quantifiers
DAY 110 l
Unit Test of Calculus
DAY 111 l
Unit Test of Trigonometry
DAY 112 l
Unit Test of Coordinate
DAY 113 l
12. Linear Programming l
DAY 67 DAY 68
Equation of Coplanar Lines
DAY 78
Tangent & Normal to a Circle
DAY 65
Equation of Intersection of a Line and Plane
DAY 77
DAY 63 DAY 64
Equation of a Line & Plane in Different Forms
DAY 76
9. Coordinate Geometry l
Skew-lines, Shortest Distance between Them & its Equation
DAY 75
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DAY 61
Coordinate of a Point in Space, Direction Ratios and Direction Cosines , Angle between Two Intersecting Lines
Unit Test of Vector & 3D Geometry
DAY 114 l
Unit Test of Probability & Mathematical Reasoning
DAY 115 l
Practice Set ( Full Syllabus )
DAY 116 l
Practice Set ( Full Syllabus)
DAY 117 l
Practice Set (Full Syllabus)
DAY 118 l
Practice Set (Full Syllabus)
DAY 119 l
Practice Set (Full Syllabus)
DAY 120 l
Revision of Important Points
MENTORS VOICE This last segment of your preparation is important as considerable number of questions is asked in JEE generally from the topics properties of modulus and argument triangle inequality rotation of complex numbers. All the topics till day 105 have direct and application based questions. Topic of days 108 and 109 is important from JEE Mains point of view as every year 1 question is asked. Concepts of these topics are useful in boolean algebra as well as in computer organisation. Try to complete the schedule of last 20 days in 17 days (approx) as some loose days (days with lesser syllabus) are given in between. In the last 3 days use randomized testing of complete syllabus to filter your weaker portion and utilize these days to plug your weak points left. For gaining the confidence after wards use 2-3 simulator papers of full syllabus and try to complete them approximately 15 min before the stipulated time, so that you can achieve the same under examination conditions of stress within the stipulated time.
Revision The Ultimate Tool “Toppers are in the habit of doing the things differently.” Here surely the word differently has the art of revision or reconciliation of learning in its basis especially for the toppers of entrance examinations. Revision plays a prominent role in preparation of school, college or other entrance exams. Once syllabus is covered, revision is a mandatory requirement for better scoring, even, if one had not covered 100% of syllabus but only 70% or so, still he has to revise what syllabus has been covered.
How to Revise?
Special Mantras
In a generalised methodology to be adopted during revision following steps are followed: Step 1 Read your notes and seek answers to questions of your own. Try to be as active in your reading as possible. Your activeness can be in the form of talking to yourself, walking around the room etc., Try to recollect what you have learned Step 2 When you feel that you have understood and can remember what you have read. Then close your source of information (copy, book, notes etc.,) Step 3 Try to recall the contents you were reading so far. You may pen down the same roughly and quickly on a new waste paper (if you are going for the first revision). This will help in filtering the remembered and non-remembered portions. Step 4 Check the new notes with the old content. You may find certain points at which your memory is washed out. These points only require attention again. Try to answer a previous question paper, i.e. past papers to see your progress once you revise. Step 5 You may recheck your memory in random manner after a certain period of time too. The washed out memory points may increase or decrease. Repeated attention on these points can kill them then- n - there and boost up your confidence. Step 6 Repetition of the same procedure for the same content many a times reduces the quantity of content requiring attention.
Remember, revision is not something that starts in the end of academic session rather it is what you can do on daily basis; in fact you can initiate it right from the very first day of your learning.
For Maths and Stats Revision Then best way as everyone says, practice-practice-practice. But some more tips to consider l
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Formulae; make a note of them on a plain paper and fix them before your bed or sort so as to view them frequently and memorise them. Because if you remember the formulae possibly the problem solving is half done. In mathematical problems you need to understand the flow of steps towards the answer but not mug-up the steps. Because, if you forget even a single step you may stop solving it. If you try to flow steps towards answer you can find one way on your own and also remember the step better. Mathematical logics are developed with more and more exposure, so try to utilize more and more of your study time in the development of logics. This ensures higher success rate. This not only builds your confidence but also trains your mind to solve the problems faster.
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Skills @ Effective Revision While going for revisions keep following points in mind It is always better to prepare a revision schedule on daily basis and after making the schedule, it is must to possess the will to abide by that. It is not necessary to revise large number of chapters every day. Revision must be slow and steady. It must be in the order to get firm grip on all types of questions asked on that topic. Many times, we tend to miss out on writing some points when the teacher speaks. It is best to pen them down when they are still fresh in the memory. Revise as many times as possible the entire subjects. There is a huge difference between revision and repetition i.e. revision is not going through all the notes over and over again. The specialized feature of revision is lessening of time in each and every attempt i.e. time consumed in revising the same content every time is less than the previous one. Just imagine a potential aspirant can revise the complete content of Physics syllabus in about two and a half hours in his 4th or 5th revision cycle. Decrease in the time of revision is possible only when you have a proper analysis of 'what to revise' and 'what not to revise'. Such an analysis can be done on the basis of 'what you know' and 'what is whipped off from your memory after learning'. Every time when you go for only those parts which are whipped off, you find a lesser content to revise. This, in turn, certainly requires lesser time to reconcile. Understand the concept and don't just mug. This is a big mistake made by many school going students till the mind maturity. Understanding the concept shows you the actual sweetness of the subject and makes you perceive even more of the subject with enthusiasm. The difference between mugging and understanding is that, mugging is a memorising aspect while understanding is realsing the facts about the subject or topic with its related prior knowledge. Use same notes or books or study material for all the revisions because the page of the book also strikes in memory during answering the questions asked. There is no substitute for hard preparation and excitement and enthusiasm in revision. n
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Last but not least you are the best to know your situation, analyze it thoroughly and work out the best method for your revision. Think for yourself, you are mature enough to organize your own study plan. Use your discretion to judge what will work for you and what will not.
@CLASS XI SYLLABUS
J
Final Touch
Complex Numbers A number of the form x + iy, where x, y Î R and i = -1, is called a complex number. If z = x + iy, then x is called real part of z and denoted as Re(z), whereas y is called imaginary part of z and denoted as Im(z).
Properties of Conjugate of a Complex Number If z is a complex number, then following are the properties of conjugate of complex number (z) = z l
MATHS MAXIMA n
n
If (a + ib ) (< or > ) c + id is not defined, i.e. order relation in complex number is not defined (comparision of complex number is only defined when imaginary part is zero). a ´ b = ab is valid. If atleast one of a and b is non-negative. If a and b both are negative, then a ´ b = - ab .
Integral Power of Iota (i) As we know that square of every real number is positive, hence solution of the equation x2 = - 9 is not possible in real number system. i = -1 is known as iota which is introduced in complex number system. Also, i =
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z + z = 2Re(z), z - z = 2 i Im(z) z + z = 0 Û z is purely imaginary; z - z = 0 Û z is purely real.
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zz = [Re(z)]2 + [Im(z)]2
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(z1 + z2 ) = z1 + z2
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(z1 - z2 ) = z1 - z2
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(z1 z2 ) = z1 z2
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æ z1 ç çz è 2
The sum of four consecutive powers of i is zero i.e. i n + i n + 1 + i n + 2 + i n + 3 = 0.
Conjugate of Complex Number Complex numbers z = a + ib and z = a - ib are called conjugate to each other (here, complex conjugate are obtained by just changing the sign of i). z+ z Also, Re(z) = =a 2 z-z Im (z) = =b 2i
z2 ¹ 0
Modulus and Argument of Complex Number If complex number z = x + iy, then modulus of complex number z is |z|, which is the distance of point ‘z’ from the origin and inclination of OP with positive X-axis is called argument of the complex number. Hence, |z| =
-1, i2 = - 1, i3 = - i, i4 = 1 i4 n = 1, i4 n + 1 = i, i4 n + 2 = - 1 and i4 n + 3 = - i
ö z1 ÷= ; ÷ z 2 ø
and
x 2 + y2
arg (z) = tan-1
y + 2 kp , k Î I x
Note The complex number z represents a point, where as | z| is distance of this point from origin. Y P (x, y) |z |
Note Complex number may be defined as an ordered pair of real numbers written as ( x, y). A complex number is purely real, if Im( z) = 0 and it is purely imaginary, if Re( z) = 0.
y
q O
x
X
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Properties of Modulus l
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|z| = 0 Û z = 0 |z1 z2 | = |z1||z2|
|z| = |z| = |- z | = |- z| ½z1½ |z1| ½ ½= , if z2 ¹ 0 ½z2½ |z2|
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If z = a , then maximum and minimum values of z + c + id is (a + c 2 + d 2 ) and (a - c 2 + d 2 ).
|z1 + z2|2 = |z1|2 + |z2|2 + z1 z2 + z1 z2 |z1 |z1 |z1 |z1
+ + -
z2|2 z2|2 z2|2 z2|2
Trigonometrical or Polar Form of a Complex Number
= |z1|2 + |z2|2 + 2 Re(z1 z2 ) = |z1 |2 + |z2|2 - 2 Re(z1 z2 ) + |z1 - z2|2 = 2 (|z1|2 + |z2|2 ) = |z1|2 + |z2|2 + 2|z1||z2| cos(q1 - q2 ) = |z1|2 + |z2|2 - 2|z1||z2| cos(q1 - q2 )
Let z = a + ib, then the polar form of z is r(cos q + i sin q) i.e. a + ib = r (cos q + i sin q), where b r = z and q = tan- 1 + 2 kp, k Î I a Note Since e iq = cos q + i sin q \
Properties of Argument
This is the Euler’s form of the complex numbers.
æz ö arg çç 1 ÷÷ = arg (z1 ) - arg (z2 ) + 2 kp, k = 0 or 1 or - 1 è z2 ø z arg æç ö÷ = 2 arg(z) + 2 kp, k = 0 or 1 or - 1 èzø
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Product and Division of Two Complex Numbers Let two complex numbers be z1 = eiq and z2 = eif Now,
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=
Square Roots of a Complex Number Let z = a + ib, then square root of z é z + a |z| - a ù i.e. z =± ê + i ú for b >0 2 2 ú êë û
z q
a
é =± ê êë
Angle between two lines = q1 - q2 z3
q2
q2
n
q1
æ z - z3 Angle between line joining z 1, z 2 , z 3 and z 4 is q = argçç 4 è z2 - z1 D(z4)
ö ÷ ÷ ø
n
B(z2) C(z3)
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z - aù ú for b < 0 2 úû
A(z1)
é 1+ i ù The square roots of i = ± ê ë 2 úû é 1- i ù The square roots of - i = ± ê ë 2 úû
Cube Roots of Unity Let
q
2
-i
MATHS MAXIMA
z1
(z2–z1)
z +a
z2
q 1– q 2
(z3–z1)
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log z = log z + i arg (z)
Þ
Locus of z, if arg ( z - a ) = q (fixed) and a >0 is a ray.
q1
[cos(q - f) + i sin (q - f)]
Þ log z = log z + log eiq Þ log z = log z + iq
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z2
[Q a + ib = z e iq ]
q
O
z1
Let z = a + ib, then log z = log (a + ib) = log ( z eiq )
Locus of z, if arg( z ) = q is ray excluding origin. z
n
f)
Logarithm of a Complex Number
Facts Based on Arguments n
z2 ei( q +
z1 . z2 = z1 z2 [cos (q + f) + i sin (q + f)] z eiq z z1 = 1 if = 1 ei( q - f ) z2 z2 e z2
and
arg(z) = - arg(z) If arg(z) = 0 Þ z is real. arg (z1 z2 ) = arg(z1 ) + arg(z2 ) + 2 kp, k = 0 or - 1 or 1
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z1 . z2 = z1 eiq ´ z2 eif z1 . z2 = z1
arg(z ) = n arg(z) + 2 kp, k =0, - 1 or 1 æz ö æz ö If arg çç 2 ÷÷ = q, then arg çç 1 ÷÷ = 2 kp - q, where k Î I z è z2 ø è 1ø n
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a + ib = r(cos q + i sin q ), a + ib = r × e iq
\
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1 = z Þ z3 - 1 = 0 Þ (z - 1) (z2 + z + 1) = 0 -1 + i 3 -1 - i 3 z = 1, , 2 2
Note If second root represented by w, then third root is w 2 . Therefore, cube roots of unity are 1, w, w 2 .
Properties of Cube Roots of Unity
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n n n n n n n n
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x + x + 1 = (x - w) (x - w ) x 2 - x + 1 = (x + w) (x + w2 ) x 2 + xy + y 2 = (x - yw) (x - yw2 ) x 2 - xy + y 2 = (x + wy ) (x + yw2 ) x 2 + y 2 = (x + iy ) (x - iy ) x 3 + y 3 = (x + y ) (x + yw) (x + yw2 ) x 3 - y 3 = (x - y ) (x - yw) (x - yw2 ) x 2 + y 2 + z 2 - xy - yz - zx = (x + yw + yw2 )( x + yw2 + zw) or (xw + yw2 + z ) (xw2 + yw + z ) or (xw + y + zw2 ) (xw2 + y + zw) x 3 + y 3 + z 3 - 3 xyz = (x + y + z ) (x + wy + w2 z ) (x + w2y + wz ) 2
then the nth roots of unity are 1, a, a2 , a3 , ..., a n - 1 which are in GP.
n
n
1 + a + a 2 + a 3 + ... + a n - 1 = 0 ì- 1, n is even 1 ´ a ´ a 2 ´ a 3 ´ ... ´ a n - 1 = í î 1, n is odd
The points represented by the nth roots of unity are located at the vertices of a regular polygon of n sides inscribed in a unit circle having centre at the origin, one vertex on the positive real axis.
ì 3, when n is a multiple of 3 1+ w + w = í î 0, when n is not a multiple of 3 2n
De-Moivre’s Theorem (i) If n is an integer, then (cos q + i sin q) n = cos nq + i sin nq (ii) If n is a rational number, then (cos nq + i sin nq) is one of the value of (cos q + i sin q) n
A3(a2)
3)
a A 4(
2
n
A2(a)
2p n 2p n
Use of Complex Numbers in Coordinate Geometry Addition of Two Complex Numbers Sum of complex numbers can be interpreted in terms of geometry as a vector. Sum of two complex numbers z1 and z2 is a diagonal of parallelogram formed by the vectors corresponds to complex numbers z1 and z2 .
|z2|
|z1|
P (z1)
Difference of two complex numbers z1 and z2 can be considered as sum of the complex numbers z1 and -z2 . Q (z2)
P (z1) – 1 |z
O
| z2
R (z1 – z2)
nth Roots of Any Complex Number
The equation x n = 1 has n roots which are called the nth roots of unity. x n = 1 Þ x n = cos 2 kp + i sin 2 kp \ x = (cos 2 kp + i sin 2 kp)1 / n
R (z1+z2)
Q (z2)
Difference of Complex Numbers
( q1 + q 2 + q 3 + ... + q n )
nth Roots of Unity
A1(1)
An(an–1)
Note If z = (cos q1 + i sin q1 ) (cos q 2 + i sin q 2 ) (cos q 3 + i sin q 3 ) ... (cos q n + i sin q n ), then z = cos( q1 + q 2 + q 3 + ... q n ) + i sin
If z = r (cos q + i sin q) and n is a positive integer, then æ æ 2 kp + q ö æ 2 kp + q ö ö z1 / n = r1 / n ç cos ç ÷ + i sin ç ÷÷ n n è ø è øø è where, k = 0, 1, 2, 3, ..., (n - 1)
2p 2p + i sin , n n
|
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where, k = 0, 1, 2, ..., n - 1. Let a = cos
z2
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1 + w + w2 = 0 w3 = 1 w3 n = 1, w3 n + 1 = w, w3 n + 2 = w2 w = w2 and (w)2 = w Each complex cube root of unity is the square of the other. Each complex cube root of unity is the reciprocal of 1 1 the other. i.e. w = 2 and w2 = w w Cube roots of – 1 are - 1, - w and - w2 . Cube roots of unity represent vertices of equilateral triangle on the argand plane.
1+
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2 kp 2 kp + i sin n n
|z
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x = cos
Q¢ (–z2)
Multiplication of Complex Numbers When two complex numbers z1 = r1 eiq 1 and z2 = r2 eiq2 are represented by the vectors OP and OQ, respectively. Then, product z1 with z2 rotate z1 anti-clockwise by an angle q2 (q2 >0) or clockwise when q2 <0. The modulus of the complex number z1 z2 is z1 z2 . \
z1 z2 = r1 r2 e
Y
R (z1 z2)
Q (z2) q1+q2 q2 q2 q1
i( q 1 + q2 )
O
P (z1) X
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æ z - z1 ö ÷÷ . (a) AB is inclined to CD at the angle arg çç 2 è z4 - z3 ø æ z - z1 ö p ÷÷ = ± . (b) If CD is inclined at 90° to AB, then arg çç 2 z z 2 3 ø è 4
Division of Two Complex Numbers When two complex numbers are z1 = r1 eiq1 and z2 = r2 eiq2 represented by the vectors OP and OQ, respectively. Then, division of z1 by z2 rotate the vector z1 in clockwise by an angle q2 and modulus of the z complex number is 1 . z2
Y
P (z1) Q (z2)
q2 q1
z1 R z 2
q2
(c) If z 1 and z 2 are fixed complex numbers, then the locus of a point z
q1 – q2
X
O
Triangle Inequalities
z1 r1 i( q 1 - q2 ) = e z2 r2
\
By the concept of multiplication or division an important concept of rotation is generated which helps the students to find many problem in coordinate geometry.
Coni’s Theorem (Rotation Theorem) If z1 , z2 , z3 are the vertices of DABC described in counter clockwise sense, then
Y
OQ = OP OQ OP ia = e OQ OP
C (z3)
n
Q (z3 – z1) A (z1) a
z3 - z1 z - z1 ia = 2 e |z3 - z1| |z2 - z1|
B (z2)
X
O
If
z3 - z1 = z2 - z1
Then,
z3 - z1 = (z2 - z1 ) ei a
n
i.e. if we have to find the new position of complex number when z1 rotated by a angle, then new complex number is z2 = z1 eia .
REMEMBER (i) Q (zeia)
Y
P (z) a q1 O
X
Y¢
(a) ze is the complex number whose modulus is r and argument q + a . (b) Multiplication by e - ia to z rotates the vector OP in clockwise sense through an angle a. (ii) If z 1, z 2 , z 3 and z 4 are the affixes of the points A, B, C and D, respectively in the argand plane.
8
n
P (z2 – z1)
z - z1 ia z3 - z1 e = 3 z2 - z1 z2 - z1
ia
It is an important concept to find maximum and minimum values. If z1 and z2 are two complex numbers, then (i) z1 + z2 £ z1 + z2 , i.e. sum of the two sides is always greater than third side. (ii) z1 + z2 ³ z1 – z2 , i.e. difference between two sides of triangle is always less than third side. (iii) z1 - z2 £ z1 + z2 £ z1 + z2
SOME IMPORTANT RESULTS
ia
X¢
æ z - z1 ö p ÷÷ = ± . satisfying arg çç 2 è z - z2 ø
n
Distance between A(z 1 ) and B (z 2 ) is given by AB = z 2 - z 1 The points P (z ) divides the line segment joining A(z 1 ) and B (z 2 ) in the mz + nz 1 ratio m : n . (a) For internal division, z = 2 m+n mz 2 - nz 1 (b) For external division, z = m- n Let ABC be a triangle with vertices A(z 1 ), B (z 2 ) and C (z 3 ), then 1 (a) Centroid G (z ) of the triangle is given by z = (z 1 + z 2 + z 3 ) 3 az + bz 2 + cz 3 (b) Incentre of I (z ) of the triangle is given by z = 1 , a +b + c where a, b and c are sides of triangle. Area of DABC with vertices A(z 1 ), B (z 2 ), C (z 3 ) is given by z1 z1 1 1 D = z2 z2 1 . 2 z3 z3 1
For an equilateral triangle z 12 + z 22 + z 32 = z 2 z 3 + z 3 z 1 + z 1z 2 n The general equation of straight line is az + az + b = 0, where a is complex number, b is real number. n If z 1 and z 2 are two fixed points, then z - z 1 = z - z 2 represents perpendicular bisector of the line segment joining A(z 1 ) and B (z 2 ) . n The equation of a circle with centre z 0 and radius r is z - z 0 = r or zz - z 0 z - z 0 z + z 0 z 0 - r 2 = 0. n General equation of circle is zz + az + az + b = 0, where a is complex number, b is real number. n If z 1 and z 2 are two fixed points and k > 0, k ¹ 1 is a real number, then z - z1 = k represents a circle. For k = 1, it represent perpendicular z - z2 n
bisector of the segment joining A(z 1 ), B (z 2 ). The diameter form of circle is (z - z 1 ) (z - z 2 ) + (z - z 2 ) (z - z 1 ) = 0 where A(z 1 ) and B (z 2 ) are end points of diameter.
MASTER p p 12. The amplitude of sin + i æç1 - cos ö÷ is
1. The square root of x + - x4 - x2 - 1 is
2p (a) 5
(b) -4
3. The conjugate of (a)
8+ i 5
(b)
(c) -4i
(a) cos q
2 + 3i 5
(c) 6 - 8i
50
(c) y
(d) 1
(b) parabola (d) None of these
C = {z :Re((1 - i)z) = 2 } be three sets of complex numbers. Then, the number of elements in the set A Ç B Ç C is (c) 2
(d) ¥
10. If w is a complex root of the equation z = 1, then 3
w+ w (a) 0
9 27 æ1 3 ö + + ... ÷ ç + + è 2 8 32 128 ø
(b) 1
is (c) -2
(d) -1
11. If z1 , z2 , z3 are complex numbers such that |z1| = |z2| = |z3| = (a) equal to 1 (c) greater than 1
(d) 4, 0
1 1 1 + + = 1, then |z1 + z2 + z3| is z1 z2 z3 (b) less than 1 (d) equal to 3
(b) 2( a 3 - b 3 - c 3 ) (d) a 3 - b 3 - c 3
= 0, then the value of
50
åz
r =1
(b) 25
r
(c) 75
1 is -1 (d) -25
17. Let w be the solution of x - 1 = 0 with Im(w) >0. If a = 2 3
with b and c satisfying é1 9 7 ù [abc] ê2 8 7 ú = [0 0 0] ê ú êë7 3 7 úû 3 1 3 then the value of a + b + c is equal to w w w (a) -2
(b) 2
(c) 3
(d) -3
1 - zi , then |w| = 1 implies that, in 18. If z = x + iy and w = z-i the complex plane (a) z lies on the imaginary axis(b) z lies on the real axis (c) z lies on the unit circle (d) None of these
9. Let A = {z:Im(z) ³ 1}, B = {z :|z - 2 - i|= 3} and
(b) 1
r
(a) -85
2z + 1 is -z, then the locus of the 8. If the imaginary part of iz + 1 point represented by z is a
(a) 0
(c) 6, 1
r =0
a+ i , x, y, a Î R, then ay - x is equal to a- i
(a) straight line (c) circle
(b) 6, 0
(a) 2( a 3 + b 3 + c 3 ) (c) a 3 + b 3 + c 3 - 3abc
å(z)
(b) right angle isosceles (d) obtuse-angled isosceles
(b) x
(a) 4, 1
16. If zr , r = 1, 2, 3, ...., 50 are the roots of the equation
z1 - z3 1 - i 3 are vertices of a triangle which is = z2 - z3 2
(a) a
(d) sin2 q
(d) 6 + 8i
6. The complex numbers z1 , z2 and z3 satisfying
7. If x + iy =
(c) sin q
where a and b are complex cube roots of unity, then xyz is
(b) 2 - 3i > 3 + 2 i (d) None of these
(a) of zero area (c) equilateral
(b) cos2 q
2 - 3i 5
(d)
5. Which of the following is correct? (a) 2 + 3i > 3 + 2 i (c) 3 + 4i > 2 + 3i
p 10
15. If x = a + b + c, y = aa + bb + c and z = ab + ba + c, (c)
(b) 8 - 6i
(d)
14. If|z + 4| £ 3, then greatest and least values of|z + 1|are (d) 4
4. (1 - i)(1 + 2 i)(1 - 3 i) is equal to (a) 1 - 6i
5ø
p (c) 15
equal to
2 + 3i is 1 + 2i
8-i 5
p (b) 5
13. If sin6 q = 32 cos5 qsin q - 32 cos3 qsin q + 3 x, then x is
2. The value of (1 + i)4 is (a) 4i
è
5
(a) ( x 2 + x + 1) + i ( x 2 - x + 1) 1 (b) ± { x 2 + x + 1 + i x 2 - x + 1} 2 1 (c) ( x 2 - 1 + i x 2 + 1) 2 (d) None of the above
19. The complex number z satisfying z + |z| = 1 + 7 i, then the value of |z|2 equals (a) 625
(b) 169
(c) 49
(d) 25
20. Given that the equation z2 + (p + iq)z + (r + is) = 0, where p, q, r and s are real and non-zero has a real root, then (a) pqr = r 2 + p2 s (c) qrs = p2 + s 2q
(b) prs = q 2 + r 2 p (d) pqs = s 2 + q 2 r
21. Let zr = (1 £ r £ 4) be complex numbers such that |z r | = r + 1 and |30 z1 + 20 z2 + 15 z3 + 12 z4 | = K|z1 z2 z3 + z2 z3 z4 + z3 z4 z1 + z4 z1 z2|, then the value of K is equal to (a) 20
(b) 24
(c) 30
(d) 10
9
22. If |z - 3| = min{|z - 1|,|z - 5|}, then Re(z) equals 5 (b) 2
(a) 2
7 (c) 2
32. If w = e
(d) 6
-n (1 - a )2 -2 n (c) 1- a
-n 1- a -2 n (d) (1 - a )2 (b)
(a)
plane, equidistant from the roots of the equation (z + 1)4 = 16 z4 ? æ -1 ö (b) ç , 0 ÷ è 3 ø 2 ö æ (d) ç 0, ÷ 5ø è
(a) (0, 0) æ1 ö (c) ç , 0 ÷ è3 ø
2 3
(c)| z| = 1
(d)| z| =
3 4
27. z1 and z2 are the roots of 3 z2 + 3 z + b = 0, if O(0), (z1 ), (z2 ) form an equilateral triangle, then b equals (b) 2
(c) 3
(d) 4
28. If z and w are two complex numbers simultaneously satisfying the equations z + w = 0 and z w 3
5
2
-4
= 1, then
29. If a = ei( 2 p / n ) , then (11 - a)(11 - a2 )...(11 - a n - 1 ) is equal to 11n - 1 - 1 11
(c)
11n - 1 - 1 10
(d)
11n - 1 10
30. The length of perpendicular from p(z0 ) to the line az + a z + b = 0 is | az0 + az0 + b| | a| | az0 + az0 + b| (c) | a|
(a)
| az0 + az0 + b| 2| a| | az0 + az0 + b| (d) 2| a| (b)
31. A man walks a distance of 3 units from the origin towards the North-East (N45° E) direction. From there, he walks a distance of 4 units towards the North-West (N45° W) direction to reach a point P. Then, the position of P in the argand plane is (a) 3 e ip/ 4 + 4i (c) ( 4 + 3i ) e ip / 4
10
(b) ( 3 - 4i ) e ip / 4 (d) ( 3 + 4i ) e ip / 4
(a) a
(d) 1
(b) b
(c) a - b
(d) a + b
34. If z1 and z1 represent adjacent vertices of a regular
(b) 16
(a) Re( z) = 1, Im( z) = 2 (c) Re( z) + Im( z) = 0
36. If log
3
(c) 24
(d) 32
-1, can be
(b) Re( z) = 1, 0 £ Im( z) £ 1 (d) None of these
æ |z|2 -|z|+1 ö ÷÷ > 2, then the locus of z is çç è 2 + |z| ø
(a)| z| = 5 (c)| z| > 5
(b)| z|< 5 (d) None of these
37. Reflection of the line (2 + 3 i)z + (2 - 3 i) z = 0 on the real axis is (a) (2 (b) (2 (c) (2 (d) (2
(a) z and w both are purely real (b) z is purely real and w is purely imaginary (c) w is purely real and z is purely imaginary (d) z and w both are imaginary
(b)
(c) 9
1 1 ù 1 2a , then the 33. If x = 2 + 5 i and 2 éê + + = ë1!9 ! 3 !7 !úû 5 !5 ! b! 3 2 value of (x - 5 x + 33 x - 19) is equal to
1 35. sin-1 ìí (z - 1)üý, where z is non-real and i = îi þ the angle of triangle, if
26. If 8 iz3 + 12 z2 - 18 z + 27 i = 0, then (b)| z| =
(b) 6
(a) 8
(b) 2 + 1 (d) None of these
(a) less than 1 (c) 2 - 1
|x|2 +|y|2 +|z|2 is |a|2 +|b|2 +|c|2
polygon of n sides whose centre is origin and if Im(z1 ) = 2 - 1, then n is equal to Re(z1 )
25. If |z|< 2 - 1, then |z2 + 2 z cos a| is
(a) 11n - 1
a + b + c = x, a + bw + cw2 = y , a + bw2 + cw = z
(a) 3
24. Which of the following represents a point in an argand
(a) 1
and a, b, c, x, y, z be non-zero complex
Then, the value of
terms equal to
3 2
2p 3
numbers such that
23. If a is the nth root of unity, then 1 + 2 a + 3 a2 + ... to n
(a)| z| =
i
+ + -
3i ) z - (2 + 3i ) z = 0 3i ) z - (2 - 3i ) z = 0 3i ) z + (2 - 3i ) z = 0 3i ) z + (2 + 3i ) z = 0
38. Let complex numbers a and 1 / a lies on
circles
(x - x0 )2 + (y - y0 )2 = r2 and (x - x0 )2 + (y - y0 )2 = 4 r2 , respectively. If z0 = x0 + iy0 satisfies the equation 2|z0|2 = r2 + 2, then|a|is equal to (a)
1 2
39. If w =
(b)
1 2
(c)
1 7
(d)
1 3
z and |w| = 1, then z lies on z - i/3
(a) parabola (c) a circle
(b) a straight line (d) an ellipse
40. If the roots of the equation z4 + az3 + (12 + 9 i) z2 + bz = 0 are the vertices of a square, then |a|2 + |b|2 is equal to (a) 270
(b) 280
(c) 290
41. If |z| = max{|z - 1|,|z + 1|}, then 1 2 (c)| z + z | = 1
(a)| z + z | =
(b) z + z = 1 (d) z Î f
(d) None
42. If x2 + x + 1 = 0, then the numerical value of
46. Let a and b be real and z be complex number, if 2
2
2
2
æ x + 1 ö + æ x2 + 1 ö + æ x3 + 1 ö + ... + æ x27 + 1 ö ç ÷ ç ÷ ç ÷ ç ÷ xø x2 ø x3 ø x27 ø è è è è is (a) 54 (c) 27
(b) 36 (d) 18
æ 1 + i tan a ö
n
z2 + az + b = 0 has two distinct roots on the line Re(z) = 1, then it is necessary that (a) b Î ( -1, 0) (c) b Î (1, ¥ )
p 2
æ 1 + i tan na ö
simplified, reduces to
equal to (a) tan q
(a) 4 (c) 8
real axis on the complex plane is (b) ( a + a )2 - 2 r
(c) ( a + a ) + r
(d) ( a + a )2 - 4r
æ z - z1 ö ö æ ÷ (b) arg çç ÷÷ = arg çç 3 ÷ z z 1 ø ø è è 2 æ z3 - z1 ö æ z3 ö ÷ (d) arg çç ÷ = 2 arg çç ÷ z2 ÷ø è è z2 - z1 ø
(b) 6 (d) More than 8
that z3 + iw z2 = (1 + iw) z1 , where w(¹ 1) be a cube root of ö æ z3 - z1 ÷ is equal to unity, then argçç ÷ z + z 2 z 3 1 ø è 2
of the following is true? z3 z2
(d) i tan q / 2
49. Let A(z1 ), B(z2 ) and C(z3 ) be the vertices of D ABC such
45. If z1 = 2 + 3 i, z2 = 3 - 2 i and z3 = - 1 - 2 3 i, then which æ (a) arg çç è æ (c) arg çç è
(c) tan q / 2
z z + = 1 is z z
44. Intercept made by the circle zz + az + az + r = 0 on the (a) ( a + a ) - r
(b) i tan q
48. Number of complex numbers z such that |z| = 1 and
(b) 2 sin na (d) None of these
2
2 z
47. If |z - 1| = 1 and arg z = q(z ¹ 0) and 0 < q < , then 1 - is
÷÷, when ÷÷ - çç 43. The expression çç è 1 - i tan na ø è 1 - i tan a ø (a) 0 (c) 2 cos na
(b)|b| = 1 (d) b Î ( 0, 1)
ö æz ö ÷÷ = arg çç 2 ÷÷ ø è z1 ø æ z - z1 ö z3 ö 1 ÷ ÷ = arg çç 3 ÷ z2 ÷ø 2 è z2 - z1 ø z3 z2
(a)
p 6
(b)
p 3
(c)
p 12
(d)
p 4
50. If a is non-real and a = 5 1, then the value of 2|1 +
a + a 2 + a -2 - a -1 |
(a) 4
is equal to
(b) 2
(c) 1
(d) None
Answers 1. 11. 21. 31. 41.
(b) (a) (c) (d) (d)
2. 12. 22. 32. 42.
(b) (d) (a) (a) (b)
3. 13. 23. 33. 43.
(a) (d) (b) (b) (a)
4. 14. 24. 34. 44.
(c) (b) (c) (b) (d)
5. 15. 25. 35. 45.
(d) (c) (a) (b) (a)
6. 16. 26. 36. 46.
(c) (d) (a) (c) (c)
7. 17. 27. 37. 47.
(d) (a) (a) (d) (b)
8. 18. 28. 38. 48.
(a) (b) (a) (c) (c)
9. 19. 29. 39. 49.
(b) (a) (d) (b) (c)
10. 20. 30. 40. 50.
(d) (d) (b) (c) (a)
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AIPMT 2016 11
Formula at a Glance 15. If z1 , z2 , z3 are the vertices of D ABC described in counter
1. If n is an integer, then i 4 n = 1, i 4 n + 1 = i , i 4 n + 2 = - 1 and i 4 n + 3 = - i
clockwise sense, then
2. The sum of four consecutive powers of i is zero, 16. Triangle Inequality
i.e. i n + i n + 1 + i n + 2 + i n + 3 = 0
(a) | z1 + z2| £| z1| + | z2| (b) | z1 + z2| ³| z1| - | z2| (c) | z1| - | z2| £| z1 + z2| £| z1| + | z2|
3. If z is a complex number, then its conjugate is z. \ Re( z) =
z+ z z- z , Im( z) = 2 2i
4. The modulus and argument of a complex number z = x + iy is | z| =
æ yö x 2 + y 2 and arg( z) = tan-1 ç ÷ + 2 kp. è xø
( a + c 2 + d 2 ) and ( a - c 2 + d 2 ).
6. The polar form of complex number a + ib is ( r cos q + i sin q), b + 2 kp. a
7. If z1 = cos q + i sin q, z2 = cos f + i sin f are two complex numbers, then (a) z1 z2 = | z1|| z2|[cos( q + f) + i sin( q + f)] | z| z (b) 1 = 1 [cos( q - f) + i sin( q - f)] z2 | z2| log z = log| z| + i arg ( z) roots
of
a
complex
number
z = a + ib
is
é | z| + a | z| - a ù ± ê + i ú for b > 0 2 2 û ë é | z| + a | z| - a ù and ± ê -i ú for b < 0 2 2 û ë
(a) For internal division, z =
mz2 + nz1 m+ n
(b) For external division, z =
mz2 - nz1 m-n
19. Let ABC be a triangle with vertices A( z1 ), B( z2 ) and C ( z3 ), then (a) Centroid G ( z) of the triangle is given by 1 z = ( z1 + z2 + z3 ) 3 (b) Incentre of I( z) of the triangle is given by az + bz2 + cz3 , where a, b, c are sides of triangle. z= 1 a+ b+c
where a is complex number, b is real number.
22. If z1 and z2 are two fixed points, then | z - z1| = | z - z2| represents perpendicular bisector of the line segment joining A( z1 ) and B( z2 ).
12. De-Moivre’s theorem
23. The equation of a circle, with centre z0 and radius r is
(a) If n in an integer, then (cos q + i sin q)n = cos nq + i sin nq (b) If n is a rational number, then (cos nq + i sin nq) is one of the value of (cos q + i sin q)n
13. The nth roots of any complex number z = r(cos q + i sin q) is
(a) 1 + a + a 2 + ... + a n - 1 = 0 ì -1, n is even (b) 1 ´ a ´ a 2 ´ ... ´ a n - 1 = í î 1, n is odd
z1 1 z2 1 z3 1
21. The general equation of straight line is az + az + b = 0 ,
ì 3, when n is a multiple of 3
14. If 1, a, a 2 , …, a n -1 are nth roots of unity, then
z1 1 z2 2 z3
z12 + z22 + z32 = z2 z3 + z3 z1 + z1 z2
11. 1 + w n + w 2 n = í î 0, when n is not a multiple of 3
é æ 2 kp + q ö æ 2 kp + q ö ù z1 / n = r1 / n êcos ç ÷ + i sinç ÷ ú, n n è ø è øû ë k = 0, 1, 2, 3, ..., ( n - 1)
D =
For an equilateral triangle
10. Cube roots of unity are 1, w, w 2 .
12
18. The point P( z) divides the line segment joining A( z1 ) and
20. Area of D ABC with vertices A( z1 ), B( z2 ), C ( z3 ) is given by
8. Logarithm of a complex number z is 9. Square
17. Distance between A( z1 ) and B( z2 ) is given by AB = | z2 - z1| B( z2 ) in the ratio m : n .
5. If| z| = a, then maximum and minimum value of| z + c + id| is
where r = | z| and q = tan-1
z3 - z1 | z3 - z1| ia = e z2 - z1 | z2 - z1|
where
| z - z0| = r or z z - z0 z - z0 z + z0 z0 - r 2 = 0
24. General equation of circle is zz + az + az + b = 0, where a is complex number, b is real number.
25. If z1 and z2 are two fixed points and k > 0, k ¹ 1 is a real number, then
| z - z1| = k represents a circle. For k = 1, it | z - z2|
represent perpendicular bisector of the segment joining A( z1 ), B( z2 ).
26. The
diameter form of circle is ( z - z1 )( z - z2 ) + ( z - z2 )( z - z1 ) = 0 , where A( z1 ) and B( z2 ) are end points of diameter.
1. The shortest distance between the line y = x + 1 and the curve x = y2 is (a)
3 2 8
(b)
2 3 8
(c)
3 2 5
(d)
2. Let f(x) satisfy the requirement of Lagrange’s Mean 1 for all 2
x Î[0, 2], then
(a) 0 (c) 13l
3. The area enclosed between the curves y = x3 and (a)
(b)
5 4
(c)
5 12
(d)
12 5
4. If f(x) = sin-1 (cos x), then the value of f (10) + f ¢ (10) is equal to (a) 11 -
7p 2
(b)
7p - 11 2
(c)
5p - 11 2
(d) None of these
5. One ticket is selected at random from 50 tickets numbered 00, 01, 02, …, 49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals 1 (a) 14 5 (c) 14
1 (b) 7 1 (d) 50
(a) 2, 4
3 /4
ö ÷ ÷ ø
1/3
æ d2 y ö = çç 2 ÷÷ è dx ø
(b) 2, 3
are, respectively (c) 6, 4
(d) 6, 9
7. The number of terms common between the series 1 + 2 + 4 + 8 + ... + to 100 terms and 1 + 4 + 7 + 10 + K + to 100 terms is (a) 6
28
(b) 4
(b) 13 (d) None of these
10. Number of solutions of the equation z3 +
(c) 5
(a) 2 (c) 6
(d) None of these
3(z)2 = 0, |z|
(b) 3 (d) 5
11. If the distances from the origin of the centres of three
circles x2 + y2 + 2 lk x - c2 = 0 (K = 1, 2, 3) are in GP, then the lengths of the tangents drawn to them from any point of the circle x2 + y2 = c2 are in
(a) AP (c) HP
(b) GP (d) None of these
12. Let a = $i - $j, b = $j - k$ , c = k$ - $i. If d$ is a unit vector such $ = 0 = [b c d $ ], then d $ equals that a × d (a) ±
6. The order and degree of the differential equation 2 æ dy ö ç1 + æç ÷ ç è dx ø è
(b) 7 / 3 (d) 5 / 3
where z is a complex number is
x is [in square units]
5 3
(a) 5 / 4 (c) 7 / 4
9. Let P(n) = 12 n + 125 n - 1 , then the least divisor of P(n) is
(a) f( x ) £ 2 (b)| f( x )| £ 1 (c) f( x ) = 2 x (d) f( x ) = 3 for atleast one x in [0, 2]
y=
y2 x2 + 2 = 1, (b < 4), so 16 b that the fourth vertex S of parallelogram PQRS lies on the circumcircle of D PQR, then the eccentricity of the ellipse is circle x2 + y2 = 25 to the ellipse
3 4
value theorem in [0, 2]. If f(0) = 0 and |f ¢(x)| £
8. If tangents PQ and PR are drawn from a point on the
$i + $j - 2k$
6 $i + $j + k$ (c) ± 3
$i + $j - k$
(b) ±
3
(d) ± k$
13. Ten apples are distributed at random among 6 persons. The probability that atleast one of them will receive none is 14
(a)
6 143
(b)
(c)
137 143
(d)
C4 C5
15 14
C5 C4
15
14. The
function
f(x) = e
x
3
-3x + 2
f : (-¥, - 1] ® (0, e5 ),
2
ò1 / 2
by
(b) many-one and into (d) one-one and into
(b) 1
(c) 0
(d) 101/2
equally in two groups. Teams of each group will play a match against each other. From each group 3 top teams will qualify for the next round. In this round, each team will play against others once. Four top teams of this round will qualify for the semifinal round, where each teams of this round will go to final round, where they will play the best of three matches. The minimum number of matches in the world cup will be (b) 53
(c) 38
(b) a 2 - b 2 = c 2 (d) None of these n
16. In world cup of cricket there are 12 teams divided
(a)54
(a) a 2 + b 2 > c 2 (c) a 2 + b 2 = c 2
23. The total number of terms which are dependent on the
1 1 cos ec101 æç x - ö÷ dx is equal to x xø è
(a) 1/4
22. If in a DABC, tan A / 2 and tan B / 2 are the roots of the equation 6 x2 - 5 x + 1 = 0, then
, is
(a) many-one and onto (c) one-one and onto
15.
defined
(d) 40
17. If the mean deviation of the numbers 1, 1 + d, 1 + 2 d, …,
1 value of x in the expansion of æç x2 - 2 + 2 ö÷ is equal to x ø è (a) 2 n + 1
(d) n + 1
(c) 2 n
(b) n [ x ]+|x|
- 2) x(e , where [×] [x] + |x| integer function, then lim f(x) is
24. If f(x) = (a) -1
x ®0
(b) 0
represents greatest
(c) 1
(d) Does not exist
25. Let a, b and c be three real numbers satisfying é1 9 7 ù [a b c] ê8 2 7 ú = [0 0 0] ú ê êë7 3 7 úû
1 + 100d from their mean is 255, then d is equal to
If point P(a, b, c) lies on the plane 2 x + y + z = 1, then the value of 7a + b + c is
(a) 10.0
(a) 0
(b) 20.0
(c) 10.1
(d) 20.2
then the value of n is (b) any even integer (d) any integer
19. Statement I ~(p « ~ q) is equivalent to p « q. Statement II ~(p « ~ q) is a tautology. (a) Statement I is true, Statement II is true Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true Statement I adj(adj A) = A Statement II |adj A| = |A| (a) Statement I is true, Statement II is true Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true
21. The product of all values of x satisfying the equation é æ 2 x2 + 10|x|+ 4 ö ù ÷÷ ú = sin-1 ê cos çç 2 è x + 5|x|+ 3 ø û ë é æ 2 - 18 |x|ö ù p ÷÷ ú + is cot ê cot -1 çç è 9 |x| ø û 2 ë (b) - 9
a2 x2 - y2 = 4 intersect at right angles. Then, the equation of the circle through the points of intersection of two conics is (a) x2 + y2 = 5 (b) 5 ( x2 + y2 ) - 3 x - 4 y = 0 2 2 (c) 5 ( x + y ) + 3 x + 4 y = 0 (d) x2 + y2 = 25
27. If tanq1 , tanq2 , tanq3 are the real roots of the equation x3 - (a + 1)x2 + (b - a)x - b = 0, where q1 + q2 + q3 Î(0, p), then q1 + q2 + q3 is equal to
(a) p / 2
(c) - 3
(d) - 1
(b) p / 4
(c) 3 p / 4
(d) p
28. Suppose A, B, C are defined as
A = a2 b + ab2 - a2 c - ac2 , B = b2 c + bc2 - a2 b - ab2 and C = a2 c + ac2 - b2 c - bc2 , where a > b > c >0 and the equation Ax2 + Bx + C = 0 has equal roots, then a, b, c are in (a) AP
20. Let A be a 2 ´ 2 matrix.
(a) 9
(d) 6
2
26. The ellipse 4 x + 9 y = 36 and the hyperbola
a a + 1 a -1 a+1 b+1 c -1 -b b + 1 b - 1 + a -1 b -1 c + 1 = 0, c c -1 c + 1 (-1)n + 2 a (-1)n + 1 b (-1)n c
(a) zero (c) any odd integer
(c) 7 2
18. Let a, b, c be such that b(a + c) ¹ 0 If
(b) 12
(b) GP
(c) HP
(d) AGP
29. Statement I The plane 5 x + 2 z - 8 = 0 contains the lines 2 x - y + z - 3 = 0 and 3 x + y + z = 5 and is perpendicular to 2 x - y - 5 z - 3 = 0. Statement II The plane 3 x + y + z = 5 meets the line x- 1 = y + 1 = z - 1 at the point (1, 1, 1). (a) Statement I is true, Statement II is true Statement II is a correct explanation for Statement I (b) Statement I is true, Statement II is true Statement II is not a correct explanation for Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true
30. For x >1 and y = log x which one of the following is not true? (a) x - 1 > y
(b) x 2 - 1 > y
(c) y > x - 1
(d)
x -1
29
Answer with Explanations 1. (a) Given line is …(i) x - y + 1= 0 Let P( x, y) be any point on the parabola. Then, perpendicular distance of P from Eq. (i), x - y + 1 y2 - y + 1 P= = [Q x = y 2 ] 2 2 2
2
1ö 1 æ 1ö 3 æ çy - ÷ + 1çy- ÷ + 2ø 4 è 2ø 4 è P= = 2 2 1 1 P is minimum, when y - = 0 Þ y = 2 2 3 3 2 Pmin = = 8 4 2 Lagrange’s mean value theorem in [0, 2]. So, it also satisfies in [0, x]. Consequently, there exists c Î ( 0, x ) such that f ( x ) - f ( 0) f ¢ (c ) = x-0
Þ Þ Þ Þ
3. (c) y =
f ¢(c ) =
f( x ) x 1ù é êëQ| f ¢( x )| £ 2 úû
[Q x ³ 0] [Q x Î ( 0, 2 )\| x| < 2]
x or y 2 = x ( y ³ 0) and y = x 3.
We get points of intersection (0, 0) and (1, 1). 1
ò0
1
é x 3/ 2 x4 ù ( x - x 3 ) dx = ê ú 4 û0 ë3 /2
x
y2 =
A (1, 1)
y=x3 O
=
5 sq unit 12
4. (a) f(10) = sin-1 cos 10 æp ö = sin-1 sin ç - 10 ÷ è2 ø pö æ -1 = - sin sinç 10 - ÷ 2ø è pö æ -1 = - sin sinç 3p - 10 + ÷ 2ø è 7p p æ ö = - ç 3p + - 10 ÷ = 10 2 2 è ø
30
=
- sin x |sin x|
- sin10 =1 |sin10 |
So, f(10) + f ¢(10) = 11 -
7p 2
04, 05, 06, 07, 08, 09, 10, 20, 30, 40. Number of such tickets = 14 There is only one ticket numbered as 08, sum of whose digits is 8. \ Total number of cases = 14 Favourable number of cases = 1 1 \ Probability = 14
6. (a) The given differential equation can be written as ( 3/ 4 ) ´ 12
2 é æ dy ö ù ê1 + ç ÷ ú è dx ø úû êë
9
f( x ) 1 = | f ¢(c )| £ x 2 | x| | f( x )| £ 2 x | f( x )| £ 2 | f( x )| £ 1
\ Required area =
Þ f ¢(10) =
- sin x 1 - cos 2 x
5. (a) Tickets having product of digits as zero are 00, 01, 02, 03,
2. (b) Let x Î ( 0, 2 ). Since, f( x) satisfies the requirements of
Þ
f ¢( x ) =
Þ
(1/ 3) ´ 12
æ d 2y ö = çç 2 ÷÷ è dx ø
2 é æ d 2y ö æ dy ö ù ê1 + ç ÷ ú = çç 2 ÷÷ dx è ø êë úû è dx ø
4
Here, we see that the order of the highest derivative is 2 and the degree is 4.
7. (c) For GP, t n = 2 n - 1 for AP t m = 1 + ( m - 1)3 = 3m - 2 They would be common, if 2 n - 1 = 3m - 2 3m or 2n - 2 + 1 = £ 150 2 Þ n £ 9, m £ 100 Þ ( n = 1, m = 1), ( n = 3, m = 2 ) ( n = 5, m = 6), ( n = 7, m = 22 ) ( n = 9, m = 86) and for n = 2, 4, 6, 8, is a fraction, which is not feasible.
8. (c) A cyclic parallelogram will be a rectangle or a square. So, ÐQPR = 90° Therefore, P lies on the director circle of the ellipse x2 y2 + 2 =1 16 b Hence, x 2 + y 2 = 25 is the director circle of x2 y2 + 2 = 1. 16 b Then, 16 + b 2 = 25 or b 2 = 9 9 Now, a 2(1 - e 2 ) = 9 Þ 1 - e 2 = 16 7 7 Þ e2 = Þ e = 16 4
9. (b) P( n) = 12 n + 125n - 1 Þ P(1) = 12 1 + 1251 - 1 = 12 + 1 = 13 So, the least divisor is 13.
Now, the number of integral solutions of
10. (d) Given, 3( z)2 z + =0 | z|
x1 + x2 + x3 + x4 + x5 + x6 = 10
3
Such that, x1 ³ 1, x2 ³ 1, …, x6 ³ 1
z = re iq
Let 3 i 3q
Þ
re
+ 3re
- i 2q
gives n( E ) and the number of integral solutions of x1 + x2 + ... + x5 + x6 = 10
=0
Since, r cannot be zero, so r 2e i 5q = - 3
Such that x1 ³ 0, x2 ³ 0, …, x6 ³ 0 gives n( s ).
Which will hold for r = 3 and five distinct values of q. Thus, there are five solutions.
\ The required probability = 1 -
11. (b) Let ( a, b ) be any point on the locus of the circle. \ Now,
Now,
x 2 + y2 = c 2 a2 + b2 = c2 S1 : x 2 + y 2 + 2 l1 x - c 2 = 0 S 2 : x 2 + y 2 + 2 l2 x - c 2 = 0 S 3 : x 2 + y 2 + 2 l3 x - c 2 = 0 |TS1| =
Þ …(i)
a 2 + b 2 + 2 l1a - c 2
2 l1a , 2 l2a , 2 l3a are in GP.
Þ
|TS1|,|TS 2|,|TS 3| are in GP. $ (a) Let d = x $i + y$j + zk$ Since, d$ is a unit vector, where
Also, Þ
2
2
2
x + y + z =1 a × d$ = 0 Þ x - y = 0
…(i)
x= y
…(ii)
13. (c) Here, Probability that atleast one of them will receive none = 1 - Probability of each receiving atleast one n( E ) Required probability = 1 n( s )
C5 137 = C 5 143
15
14. (d) We have, f( x) = e x
3
- 3x + 2
g ( x) = x 3 - 3 x + 2 g ¢( x ) = 3 x 2 - 3 = 3( x 2 - 1) Þ g ¢( x ) ³ 0 for x Î ( - ¥, - 1] \ g ( x ) is an increasing function. \ f( x ) is one-one. Now, the range of f( x ) is [0, e 4 ] but the codomain is (0, e 5). \ f( x ) is an into function. 2 1 1ö æ 15. (c) I = ò1/ 2 cosec 101 ç x - ÷dx x xø è 1 -1 Let = t Þ 2 dx = dt x x 1/ 2 1 æ1 ö I=-ò cosec 101 ç - t ÷ dt \ 2 t èt ø
Þ
0 1 -1 -1 0 1 = 0 x y z
Þ x+ y + z = 0 Þ 2x + z = 0 Þ z = -2x From Eqs.(i), (ii) and (iii), we get x 2 + y 2 + z2 = 1 2 Þ x + x2 + 4 x2 = 1 Þ 6 x2 = 1 1 Þ x=± 6 1 $ $ $ d=± ( i + j - 2 k$ ) \ 6
P( E ) = 1 -
=-
$ =0 Also, [b c d] Þ
C6 - 1
Let Þ
[using Eq. (i)] = 2 l1a |TS 2| = 2 l2a |TS 3| = 2 l3a Now, distances from the origin of the centres of these circles are in GP. \ l1, l2, l3 are in GP. Þ 2 l1a, 2 l2a, 2 l3a are in GP.
12.
C6 - 1
10 + 6 - 1
9
Similarly,
Þ
10 - 1
2
1
æ
ò1/ 2 t cosec 101 çè t
1ö - ÷ dt tø
I=-I
Þ
2I = 0
Þ
I=0
16. (b) The number of matches in the first round 6
…(iii)
C 2 + 6C 2 = 15 + 15 = 30 The number of matches in the next round = 6C 2 = 15 The number of matches in the semi-final round = 4 C 2 = 6 \ The required number of matches = 30 + 15 + 6 + 2 = 53 (for best of three atleast two matches are played) S| x - x| …(i) 17. (c) Mean deviation = n For given data with 101 terms ( x = 101), 1 + 1 + d + 1 + 2d + K + 1 + 100 d x= 101 101 + d(1 + 2 + ... + 100) = 101 101 d [(100 ´ 101) / 2 ] = + = 1 + 50 d 101 101 Now,
S| x - x | = |1 - (1 + 50 d )| + |1 + d - (1 + 50 d )| + |1 + 2d - (1 + 50 d )| + ... + |(1 + 100 d ) - (1 + 50 d )|
31
S| x - x | = 50 d + 49 d + 48 d + ... + 0 d + ... + 4 ad + 50d æ 50 ´ 51 ö = 2 (1 + 2 + 3 + ... + 50)d = 2 ´ ç ÷ d = ( 50 ´ 51)d è 2 ø
adj(adj)
From Eq. (i), 50 ´ 51 d 101 æ 50 ´ 51 ö 255 = ç ÷d è 101 ø
Mean deviation =
Þ
20. (b) If
d = 101 .
18. (c) Let a+ 1 b+ 1 c -1 a a+ 1 a-1 D = - b b + 1 b - 1 + ( -1)n a - 1 b - 1 c + 1 = 0 a c c c -1 c + 1 -b [Taking ( -1)n common from the last row of the second determinant] Interchanging rows and columns in the second determinant, then apply C 3 « C 2 and C1 « C 2, we get
q
pÚq
T
T
T
T
T
T
T
F
T
F
F
F
p
pÙq
p®q
p«q
F
T
T
F
T
F
F
F
F
F
T
T
p - cos -1 2
é æ 2 x 2 + 10 | x| + 4 ö ù ÷ú = êcos çç 2 ÷ êë è x + 5 | x| + 3 ø úû é æ 2 - 18 | x| ö ù p ÷ú + cot êcot -1 çç ÷ è 9 | x| ø û 2 ë
p 2 x 2 + 10 | x| + 4 2 - 18 | x| p = + 2 9 | x| 2 x 2 + 5 | x| + 3
é 2( x 2 + 5 | x| + 3) - 2 ù 2 -2 Þ -ê ú= 2 9 | x| x | x | + 5 + 3 ë û 2 2 = -2 + 2 = -2 x + 5 | x| + 3 9 | x|
p
q
~q
p « ~q
T
T
F
F
T
T
Þ x 2 - 4 | x| + 3 = 0 Þ | x| = 1, 3 Þ x = 1, - 1, 3, - 3 \ Product = 9 A B 5 A B 1 22. (c) tan + tan = and tan tan = 2 2 6 2 2 6 A B 5 tan + tan Bö æA 2 2 = 6 \ tanç + ÷ = =1 A B æ 1ö 2ø è2 1 - tan tan ç1 - ÷ 2 2 6ø è A B p p + = Þ A+ B= 2 2 2 4 p Þ ÐC = 2 Þ c 2 = a2 + b 2
T
F
T
T
F
F
23. (c) ç x 2 - 2 +
F
T
F
T
F
F
F
F
T
F
T
T
Now, (i) p ® q is false (F) if and only if p is true (T) and q is false (F) in all other cases it is true (T). (ii) p « q is true (T) if and only if both p and q are true (T) or p and q are both false (F). (iii) Tautology : The compound statement involving p and q as component statements is called a tautology, if this statement is true (T) for all values of p and q. (iv) Fallacy : The compound statement involving p and q as component statements is called a fallacy, if it is false (F) for all values of p and q. ~ (p «~ q)
p«q
From the last two columns ~ ( p «~ q ) º p « q Thus, statement I is correct. But ~ ( p «~ q ) is not a tautology because all values are not true. Hence, statement II is false.
32
- bù a úû
21. (a) We have,
Þ
19. (c) Consider the following table.
bù éd , adj A = ê d úû ë -c bù = A. d úû
Thus, for a second order matrix, adj(adjA) = A Also, adj (adj A) = | A|n - 2 A = A [Q n = 2] Thus, statement I is correct. Also, for a square matrix of order n, |adj A| = | A |n - 1 For a second order matrix A, i.e. when n = 2 |adj A | = | A |2 - 1 = | A | Statement II is correct. But statement II cannot be a correct explanation for statement I.
a a+ 1 a-1 a a+ 1 a-1 D = - b b + 1 b - 1 + ( -1)n - b b + 1 b - 1 c c -1 c + 1 c c -1 c + 1 Now, D = 0, if n is any odd integer.
éa A=ê ëc éa A=ê ëc
æ è
n
1 ö 1 4 2 n ÷ = 2n ( x - 2 x + 1) x2 ø x =
( x 2 - 1)2n
x 2n Total number of terms that are dependent on x is equal to number of terms in the expansion of ( x 2 - 1)2n that have degree of x different from 2 n, which is given by (2 n + 1) - 1 = 2 n.
24. (d) We have,
28.
x(e -1- x - 2 ) lim f( x ) = lim - 1- x x ®0 x ®0
Þ and
Þ \
25.
[Q [ x ] = - 1 and| x| = - x for -1 < x < 0] 0 ´ (e -1 - 2 ) =0 lim f( x ) = -1 x ®0x æe - 2 ö ÷ lim f( x ) = lim x çç ÷ x ®0 x ®0+ è x ø [Q [ x ] = 0 and| x| = x for 0 £ x < 1] lim f( x ) = lim e x - 2 = 1 - 2 = - 1
x ®0
+
x ®0
lim f( x ) does not exist.
x ®0
é1 9 7 ù (d) [a b c ] ê 8 2 7 ú = [0 0 0] ú ê êë7 3 7 úû a + 8b + 7c = 0 9a + 2 b + 3 c = 0 7 a + 7 b + 7c = 0 Þ a = 1, b = 6, c = - 7 Clearly, P( a, b, c ) lies on the plane 2 x + y + z = 1 \ 7a + b + c = 7 + 6 - 7 = 6 Y¢
P(x1, y1)
X
X¢
Y
Since, ellipse and hyperbola intersect orthogonally, they are confocal. Hence, a = 2 (equation foci) Let the point of intersection in the first quadrant be P( x1, y1 ) P lies on both the curves. Therefore, 4 x12 + 9 y12 = 36 and 4 x12 - y12 = 4 Adding these two results,we get 8( x12 + y12 ) = 40 Þ x12 + y12 = 5 or r= 5 Hence, the equation of circle is x 2 + y2 = 5
27. (b) tan q1 + tan q 2 + tan q 3 = ( a + 1)
\
S tan q1 tan q 2 = ( b - a ) tan q1 tan q 2 tan q 3 = b S tan q1 - P tan q1 tan( q1 + q 2 + q 3 ) = 1 - S tan q1 tan q 2 =
Þ
q1 + q 2 + q 3 =
a + 1- b a + 1- b = =1 1 - ( b - a) 1 - b + a p 4
B = b(c - a )( a + b + c ) C = c( a - b )( a + b + c ) Now, Ax 2 + Bx + C = 0 Þ ( a + b + c ){ a( b - c ) x 2 + b(c - a ) x + c( a - b )} = 0 Given that, roots are equal. Hence, D = 0 Þ b 2(c - a )2 - 4ac( b - c )( a - b ) = 0 Þ
b 2c 2 - 2 ab 2c + b 2a 2 - 4a 2bc + 4acb 2 + 4a 2c 2 - 4abc 2 = 0
Þ
b c + b a + 4a c + 2 ab 2c - 4a 2bc - 4abc 2 = 0
Þ
( bc + ab - 2 ac )2 = 0
Þ
bc + ab - 2 ac = 0
Þ
Þ
26. (a)
(c) A = a( b - c )( a + b + c )
Þ
2 2
2 2
2 2
bc + ab = 2 ac 1 1 2 [dividing throughout by abc] + = a c b
Hence, a, b and c are in HP.
29. (c) The equation of the family of planes containing the lines 2 x - y + z - 3 = 0, 3 x + y + z = 5 is 2 x - y + z - 3 + l ( 3 x + y + z - 5) = 0 For l = 1, this reduces to 5 x + 2 z - 8 = 0 So, the plane 5 x + 2 z - 8 = 0 contains the given line. Also, 2 ´ 5 - 1 ´ 0 - 5 ´ 2 = 0 So, the plane 5 x + 2 z - 8 = 0 is perpendicular to 2 x - y - 5z - 3 = 0 Hence, Statement I is true. The coordinates of any point on line x-1 y+ 1 z-1 are ( r + 1, r - 1, r + 1). = = 1 1 1 If this point lies on the plane 3 x + y + z = 5. 2 Then, 3r + 3 + r - 1 + r + 1 = 5 Þ r = 5 æ 7 -3 7 ö Thus, the line meets the plane at ç , , ÷. è5 5 5ø So, Statement II is not true.
30. (c) Let Then,
f( x ) = log( x ) - ( x - 1) f ¢( x ) =
1- x 1 - 1= x x
Clearly , f ¢( x ) < 0 for x > 1 Þ f( x ) is decreasing function for x > 1 Þ f( x ) < f(1) for x > 1 Þ log x - ( x - 1) < 0 x > 1 Þ log x < x - 1 for x > 1 But x 2 - 1 > x - 1 for x > 1 \ x 2 - 1 > x - 1 and log x < x - 1 Þ log x < x 2 - 1 Similarly, it can be proved that x-1 < log x x x-1
[Q y = log x]
[Q log x = y]
33
Questions to Measure Your Problem Solving Skills 1. If (x2 + x + 2)2 - (a - 3) (x2 + x + 1) 2
2
8. If (3 + x2008 + x2009 )2010 = a0 + a1 x + a2 x2 + ... + anx n,
2
(x + x + 2) + (a - 4) (x + x + 1) = 0 has atleast one root, then æ 19 ö æ 19 ù (a) a Î ç 5, ÷ (b) a Î ç 5, ú è 3ø è 3û 4
æ3 ö (c) a Î ç , ¥ ÷ è4 ø 3
é3 ö (d) a Î ê , ¥ ÷ ë4 ø
2
then the value of 1 1 1 1 a0 - a1 - a2 + a3 - a4 - a5 + a6 - ... is 2 2 2 2 (a) 3 2010
(b) 1
(c) 2 2010
(d) None
2. If the equation z + a1 z + a2 z + a3 z + a4 = 0,
9. Let p, q be chosen one-by-one from the set
where a1 , a2 , a3 , a4 are real coefficients different from zero, has a purely imaginary root, then the a a a expression 3 + 4 1 has the value a1 a2 a2 a3
{1, 2 , 3, 2, e, p} with replacement. Now, a circle is drawn taking (p, q) as its centre, then the probability that at the most two rational points exist on the circle is
(a) -2
(b) 0
(c) 1
(d) 2
3. In a geometric series, the first term is a and common ratio is r. If Sn denotes the sum of the n terms and Un =
n
å Sn, then rSn + (1 - r) Un equals
n =1
(a) 0
(b) n
(c) na
(d) nar
4. In a certain test, there are n questions. In the test, n-i
students gave wrong answers to atleast i 2 questions, where i = 1, 2, 3, ..., n. If the total number of wrong answers given is 2047, then n is equal to (a) 10
(b) 11
(c) 12
(d) 13
5. A vector r has length 21 and direction ratios are proportional to 2, - 3, 6. If r makes an acute angle with X-axis, then r is equal to
(a) 6 $i - 9 $j - 18 k$ (c) 6 $i + 9 $j - 18 k$
(b) 6 $i - 9 $j + 18 k$
(d) None of these
6. If b > 1, sin t > 0, cos t > 0 and logb (sin t ) = x , then logb (cos t ) is equal to 1 (a) log b (1 - b2x ) 2 (c) Both (a) and (b)
(b) log b 1 - b
2x
(d) None of these
7. Let {D1 , D2 , D3 ..., Dn} be the set of third-order determinants that can be made with the distinct non-zero real numbers a1 , a2 ,..., a9 , then n
(a)
å Di = 1
i=1
34
n
(b)
å Di = 0
i=1
(c) Di = D j , " i, j (d) None
(a) 2/3
(b) 7/8
(c) 8/9
(d) 9/10
1 2ù 1+ x 10. If A = éê ú and f(x) = 1 - x , then f( A) is ë2 1 û é -1 -1ù (a) ê ú ë -1 -1û
é1 1ù (b) ê ú ë1 1û
é2 2 ù (c) ê ú ë2 2 û
é -2 (d) ê ë -2
-2 ù -2 úû
11. If A = {(x, y); y = 4 , x ¹ 0} and
x B = {(x, y); x2 + y2 = 8, x, y Î R}, then
(a) (b) (c) (d)
AÇB=f A Ç B contains one point only A Ç B contains two points only A Ç B contains four points only
12. For n Î N, 32 n+ 2 - 23 n - 9 is when divided by 64, the remainder is equal to (a) 0
(b) 1
(c) 2
(d) 3 2
2
x -x 13. The function f : R ® R defined by f (x) = e 2 - e 2 , is
ex + e- x
(a) one-one but not onto (c) one-one and onto
(b) many-one but onto (d) neither one-one nor onto
14. Statement I If a and b are positive real numbers and [.] denotes the greatest integer function, then x é bù b = . lim ê x ûú a x ® 0+ a ë {x} Statement II lim = 0, where {x} denotes the x ® ¥ x fractional part of x.
(a) Statement I is true, Statement II is true ; Statement II is a correct explanation for Statement I (b) Statement I is true; Statement II is true ; Statement II is not a correct explanation for Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true
15. The line joining the incentre to the circumcentre of
23. If l r (x) means log log log ...x, the log being repeated r times, then 2 3 r -1 ò [x l (x) l (x) l (x).... l (x)] dx is equal to
(a) l r + 1( x ) + c (b)
æ cos B + cos C - 1ö (b) tan -1 ç ÷ è sin C + sin B ø
æ cos B + cos C - 1ö (c) tan -1 ç ÷ è sin C - sin B ø
(d) None of these
x cos a + y sin a = p, p Î R+ . If these lines and the line x sin a - y cos a = 0 are concurrent, then
(b) different centroid (d) None of these
17. The equation of the curve whose tangent at any point (x, y) makes an angle tan (2 x + 3 y) with X-axis and which passes through (1 , 2) is (a) 6 x + 9 y + 2 = 26 e 3 ( x - 1)
(b) 6 x - 9 y + 2 = 26 e 3 ( x - 1)
(c) 6 x + 9 y - 2 = 26 e 3 ( x - 1)
(d) 6 x - 9 y - 2 = 26 e 3 ( x - 1)
(d) - 3 2
(b) 2 {| x| - | x - 1|}
ì 0 for x < 0 and for x > 1 (c) í î 4 (2 x - 1) for 0 < x < 1
ì 0 for x < 0 (d) í î 4 (2 x - 1) for x > 0
(d) a 2 - 8 ab + b2 = 0
+ x
(b) more than 10 -7 (d) None of these
(a) less than 10 (c) less than 10 -8
27. Let x1 , x2 , ..., x n be n observations and let wi = lx i + k for i = 12 , , ..., n, where l and k are constants. If the mean of x i is 48 and their standard deviation is 12, the mean of wi ’s is 55 and standard deviation of wi ’s is 15, the values of l and K should be
(a) l = 1.25, k = - 5 (c) l = 2 .5, k = - 5
(b) l = - 1.25, k = 5 (d) l = 2 .5, k = 5
28. Let Ar , r = 1, 2, 3 ... be the points on the number line
19. If f (x) = {|x| - |x - 1|}2 , then f ¢ (x) equals (a) 0 for all x
(b) a 2 + 2 ab - b2 = 0
(c) a 2 - 4 ab + b2 = 0 10 1
on the curve such that its abscissa is increasing at the rate of 4 units/s . Then, the rate of increase of the projection of SP on x + y = 1, when P is at (4 ,4) is (c) - 2
(a) a 2 - 6 ab + b2 = 0
-7
18. Let S be the focus of y2 = 4 x and a point P be moving
(b) - 1
(d) None of these
19 26. The absolute value of ò cos x8 dx is
-1
2
(b) a 2 + b2 = 2
(c) 2 ( a 2 + b2 ) = 1
which pass through P and touch both the coordinates axes cut at right angles, then
the ratio 1:2. The three lines AP , BQ and CR enclose a D XYZ. The D ABC and D XYZ have
(a)
(a) a 2 + b2 = 1
25. P is a point (a, b) in the first quadrant. If the two circles
16. P, Q, R divide the sides BC, CA and AB of a D ABC in (a) the same centroid (c) same area
such that OA1 , OA2 , OA3 ,.... are in GP where O is the origin and the common ratio of the GP be a positive proper fraction. Let Mr be the middle point of the line ¥
segment Ar Ar + 1. Then, the value of
20. From a point p (x ¢ , y ¢ , z ¢), a plane is drawn at right r5 (b) 2 x¢ y¢ z ¢
r4 (c) x¢ y¢
y2 x2 = 1 whose centre C be such that CP is a2 b2 perpendicular to CQ , a < b. Then, the value of 1 1 is + CP2 CQ2 (a)
b2 - a 2 2 ab
(b)
1 a2
+
1
(c)
b2
2 ab b2 - a 2
(d)
1 a2
-
1
OA1 (OA1 + OA2 ) 2 (OA1 + OA2 ) OA1 (c) 2(OA1 + OA2 )
OA1 (OA1 - OA2 ) 2 (OA1 + OA2 )
(d) ¥
29. The area of the region bounded by the parabola (y - 2)2 = (x - 1), the tangent to the parabola at the point (2, 3) and the X-axis is (a) 3 (c) 9
(b) 6 (d) 12
30. The largest term of the sequence < an > given by
b2
an =
û
n2 , n Î N is n3 + 200
49 543 1 (c) 52 (a)
æ pö greatest integer function, then f‘ ç5 ÷ is equal to è 2ø (a) 0
(b)
(a)
22. If f (x) = sin éê p [x] - x5 ùú, |1 < x < 2, and []× denotes the ë2
is equal
to
(d) None
21. Let two points P and Q lie on the hyperbola
åOMr
r =1
angles to OP to meet coordinate axes at A, B and C. Then, the area of the D ABC is r5 (a) x¢ y¢ z ¢
(d) None
24. Line ax + by + p = 0 makes angle p / 4 with
DABC is inclined to the side BC at an angle æ cos B - cos C + 1ö (a) tan -1 ç ÷ è sin C - sin B ø
l r + 1 ( x) + c (c) l r ( x ) + c r +1
(b) 5(p / 2 ) 4 / 5 (c) - 5 (p / 2 )4 / 5 (d) None of these
(b)
8 89
(d) None of these
Answers 1. (b) 11. (c) 21. (d)
2. (c) 12. (a) 22. (c)
3. (c) 13. (d) 23. (a)
4. (b) 14. (a) 24. (b)
5. (b) 15. (c) 25. (c)
6. (c) 16. (a) 26. (a)
7. (b) 17. (a) 27. (b)
8. (c) 18. (c) 28. (b)
9. (c) 19. (c) 29. (c)
10. (a) 20. (b) 30. (a)
35
T E G TAR
E E J0 1 6 2
Comprehensive Simulator Test Series for JEE Main & Advanced
JEE ADVANCED Questions to Measure Your Problem Solving Skills Paper I
Single Integer Answer Type 1. The number of integral value of a; a Î(6, 100) for which the equation [tan x]2 + tan x - a = 0 has real roots, where [.] denotes greatest integer function, is
2. P is a point on positive X-axis, Q is a point on the positive Y-axis and ‘O’ is the origin. If the line passing through P and Q is tangent to the curve y = 3 - x2 , then the minimum area of the DOPQ, is
3. OPQR is a square and M, N are the mid-points of the sides PQ and QR, respectively. If the ratio of the areas of the square and the DOMN is l : 6, then l /4 is equal to
4. If A is a square matrix of order n such that |adj(adj A)| =| A|9 , then the value of n can be
5. A region S in complex plane is S = {x + iy, - 1 £ x £ 1 and -1 £ y £ 1}. A complex number z = x + iy is chosen uniformly at random from S. If the 3 probability P that the complex number (1 + i) × z is 4 m also in S can be expressed in lowest form , then n find the sum of digits in (m + n). 1
6. The value of ò [ x[1 + sin p x] + 1] dx is (where, [.] is -1
the greatest integer function)
36
A B = x1, sin = x2 , 2 2 2016 2015 æ x1 ö æ x3 ö A B -ç ÷ = 0, cos = x3 and cos = x4 with ç ÷ 2 2 è x2 ø è x4 ø then the length of AC is n(n - 1) (n - 4)n The value of nn - (n - 2)n + … to 2! n (n + 1) terms = n !(k) , then k is
7. In D ABC, if BC is unity sin
8.
One or More than One Option Correct Type 9. If a function satisfies
(x - y) f(x + y) - (x + y) f(x - y) = 2(x2 y - y3 ), "x, y, Î R and f(1) = 2, then (a) f( x ) must be polynomial function (b) f( 3) = 12 (c) f( 0) = 0 (d) f( x ) may not be differentiable
10. Let unit vectors a and b be perpendicular and unit vector c be inclined at an angle q to both a and b. If c = aa + bb + g (a ´ b), then (a) a = b
(b) 1 - 2 a2 = g 2
1 + cos 2 q (c) a = 2
(d) a2 - b 2 = g 2
2
11. The
roots of equation x5 - 40x4 + ax3 + bx2 + gx + d = 0 are real and in GP. If the sum of their reciprocal is 10, then d can be (a) -32
(b) -
1 32
(c) 32
(d)
1 32
12. If the normal at P to the rectangular hyperbola
x2 - y2 = 4 meets the axes in G and g and C is the centre of the hyperbola, then
(a) PG = PC (c) PG = Pg
(b) Pg = PC (d) Gg = 2 PC
æ xdx + ydy ö ÷= è xdy - ydx ø
13. The solution of ç
a2 - x2 - y2 is x2 + y2
y ü ì x 2 + y 2 = a siní æç tan-1 ö÷ + c ý xø þ îè ì æ -1 y ö ü 2 2 (b) x + y = a cos í ç tan ÷ + cý xø îè þ y ì x 2 + y 2 = a taní æçsin-1 ö÷ + xø îè 1 ì (d) y = x tanísin-1 æç x 2 + y 2 ö÷ èa ø î
3 46 ö (a) æç - , ÷ è 5 5ø (c) ( 6, 4)
Column II. Column I
ü cý þ ü + cý þ
The maximum value of æ 7 - 5 ( x 2 + 3)ö sec -1 ç ÷ is 2 è 2( x + 2 ) ø
p.
p 4
B.
The minimum value of 5ö 1ö æ æ cosec -1 ç 3 x 2 + ÷ + sec -1 ç 3 x 2 + ÷ is è è 4ø 4ø
q.
2p 3
C.
Points of non-differentiability of the function f ( x ) = min ì p ö p öü æ 3p ö æ æ ÷ ÷, cot ç x + ÷ ý, "x Î ç 0, í tan ç x + è 2 ø è è 12 ø 12 ø þ î
r.
7p 6
s. x2 y2 =1 8 1
p 6
å[ r (n n - 1C r - 1 - r nC r - 1) + (2r + 1)nC r ], then
r =1
(a) f( n) = n2 - 1
(b) f( n) = ( n + 1)2 - 1
10
(c)
10
å f(n) = 495
(d)
n=1
å f(n) = 374
n=1
is/are
15. If there are three square matrices A, B, C of same n
order satisfying the equation A2 = A -1, B = A2 and ( n - 2)
C = A2 are true?
D.
of q such that sum of intercepts on axes made by this tangent is maximum is
(b) ( B + C ) ( B - C ) = 0 (d) None of these
16. If f(x) = cos -1 x + cos -1 æç + x è2
2 p (a) f æç ö÷ = è 3ø 3 1 p (c) f æç ö÷ = è 3ø 3
17. The coefficient of x50 in
1 ö 3 - 3x2 ÷ , then ø 2
2 2 p (b) f æç ö÷ = 2 cos -1 æç ö÷ è 3ø è 3ø 3 1 1 p (d) f æç ö÷ = 2 cos -1 æç ö÷ è 3ø è 3ø 3 100
å100C
K =0
K
20. Match the statements of Column I with values of Column II. Column I
(a) Number of ways in which 50 identical books can be distributed in 100 students, if each student can get atmost one book (b) Number of ways in which 100 different white balls and 50 red balls can be arranged in a circle, if no two red balls are together (c) Number of dissimilar terms in ( x1 + x2 + x3 + .... + x50 )51 2 × 6 × 10 × 14...198 (d) ( 50)!
18. Point M moved on the circle (x - 4)2 + (y - 8)2 = 20, then it broke away from it and moving along a tangent to the circle cuts the X-axis at the point
Column II
A.
Out of four machines, exactly two are faulty, they are tested one-by-one in a random order till both faulty machines are identified, then the probability that only two tests are needed, is
p.
1 2
B.
If m is selected at random from set {12 , , ...,10} and the probability that the quadratic equation 2 x 2 + 2 mx + m + 1 = 0 has real roots is k, then value of 10 k is
q.
12 17
C.
If three points are lying in a plane, then what is the probability that a triangle will be formed by joining them?
r.
4 5
D.
A letter is known to have come either from LONDON or CLIFTON; on the postmark only the two consecutive letters ON are legible. The probability that it came from LONDON, is
s.
1 3
(x - 2)100 - K × 3K is
also equal to
Tangent is drawn to hyperbola
æ pö at (2 2 sec q, tan q); q Î ç 0, ÷. The value è 2ø
, then which of the following statements
(a)| B - C| = 0 (c) B must be equal to C
Column II
A.
n
14. If f(n) =
2 44 ö (b) æç - , ÷ è 5 5ø (d) ( 3, 5)
Matching Type 19. Match the statements of Column I with values of
(a)
(c)
(-2, 0). The coordinates of a point on the circle at which the moving point broke away are
37
Paper II Single Integer Answer Type 1. Let f be a function defined on R (the set of all real
2.
numbers) such that f ¢ (x) = 2010 (x - 2009) (x - 2010)2 (x - 2011)3 (x - 2012)4 for all x Î R. If g is a function defined on R with values in the interval (0, ¥) such that f(x) = (g(x)) for all x Î R, then the number of points in R at which g has local maximum, is é 1ù Let f :[0, 1] ® ê0, ú be a function such that f(x) is a ë 2û polynomial of 2nd degree, satisfy the following conditions 1 (a) f(0) = 0 (b) has a maximum value of at x = 1. If A 2 is the area bounded by y = f(x), y = f -1 (x) and the line 2x + 2y - 3 = 0, in Ist quadrant, then value of 24A is
3. ABC is a triangle and P, Q and R are points on the sides BC, CA and AB respectively dividing them in the ratio 1 : 4, 3 : 2 and 3 : 7, respectively. The point S divides AB in the ratio 1 : 3, if the ratio of ( AP + BQ + CR) : CS = m : n, where m and n are coprime. Then, the value of m + n is
7. A sequence a1, a2 , a3 ,..., an of real numbers is such that a1 = 0,| a2| = | a1 + 1|,| a3| = | a2 + 1|,…, | an| = | an - 1 + 1, | where the arithmetic mean of l a1, a2 , a3 ,..., an cannot be less than - , then the m value of l + m is
8. The set of natural number N partitioned into arrays of rows and columns in the form of matrices as é6 7 8ù é2 3ù ê 9 10 11ú ,..., Mn = æç ö÷ , = M1 = (1), M2 = ê M 3 è ø ë4 5ûú ê12 13 14ú ë û and so on, where the sum of the elements of the diagonal in M6 is k, then sum of digits of k is
One or More than One Option Correct Type 9. If the vectors b = (tan a, -1, 2 sin a / 2) and æ ö ç ÷ -3 ÷ ç are orthogonal and a c = tan a, tan a, ç a÷ sin ç ÷ è 2ø vector a = (13 , ,sin 2a) makes an obtuse angle with the Z-axis, then the value of a is
4. With respect to a particular question on a multiple
(a) a = ( 4n + 1) p - tan-1 2
choice test (having 4 alternatives with only 1 correct) a student knows the answer and therefore can eliminate 3 of the 4 choices from consideration with probability 2/3, can eliminate 2 of the 4 choices from consideration with probability 1/6, can eliminate 1 choice from consideration with probability 1/9 and can eliminate none with probability 1/18. If the student knows the answer, he answers correctly otherwise he guesses from among the choices not eliminated. If the answer given by the student was found correct, then the probability a that he knew the answer is , where a and b are b relatively prime. Then, the value of (a + b) /317 is
(c) a = ( 4n + 1) p + tan 2
5. If the equation of the curve on the reflection of the
6.
38
(x - 4)2 (y - 3)2 ellipse + = 1 about the line 16 9 2 2 x - y - z = 0 is 16x + 9y + ax - 36y + b = 0, then unit digit of (a + b) is n n C If lim å r r = e - x, then x is n® ¥ ( + 3) n r r =0
-1
(b) a = ( 4n + 2 ) p - tan-1 2
(d) a = ( 4n + 2 ) p + tan-1 2
10. Given a real valued function f such that
ì tan2 x , if x > 0 ï 2 2 ï (x - [ x] ) f(x) = í 1 , if x =0 ï 1 ï {x} cot {x} , if x < 0 î where, [ x] is the integral part and {x} is the fractional part of x, then (a) lim f( x ) = 1 x®0 -1
(c) cot
(b) lim 2
æ ö ç lim - f( x )÷ = 1 èx ® 0 ø
x ® 0-
f( x ) = cot 1
(d) f is continuous at x = 0
11. Let PM be the perpendicular from the point (1, 2, 3) to XY-plane. If OP makes an angle q with the positive direction of Z-axis and OM makes an angle f with the positive direction of X-axis, where O is the origin, then (when q and f are acute) (a) tan q =
5 3
(c) tan f = 2
2 14 1 (d) cos q cos f = 14 (b) sin q sin f =
12. If f(x) = ae2 x + be x + cx satisfies the conditions f(0) = - 1, f ¢(log 2) = 31, ò
log 4 0
[ f(x) - cx] dx =
39 , then 2
(a) a = 5 (b) b = - 6 (c) c = 2 (d) a = 3
13. If the roots of the equation x3 + ax2 + bx - 1 = 0 are
Linked Comprehension Type Passage I
Directions (Q. Nos. 17-18) A function f ( x ) having the following properties (i) f(x) is continuous except at x = 3 (ii) f(x) is differentiable except at x = - 2 and x = 3 (iii) f(0) = 0, lim f(x) ® - ¥, lim f(x) = 3, lim f(x) = 0
(a) a - 2 b = 0 (b) b Î ( 3, ¥) (c) one root is 2 (d) one root is greater than 1 and other root is less than 1
17. Maximum possible number of solutions of f(x) =| x| is
14. If k1 and k2 (k1 < k2 ) are the points of discontinuity of 1 , then the 1- x value of k for which the points (k1, k2 ) and (k, k2 ) lie on the same side of x + 2y - 3 = 0, is the function f [ f{f(x)}], where f(x) =
the parabola y2 = 2px, such that it touches the directrix of the parabola. Then, the point of intersection of the circle and the parabola is p (a) æç , pö÷ è2 ø
p (b) æç , - pö÷ è2 ø
-p (c) æç , è 2
-p (d) æç ,è 2
2x
xe
2x
1 + e2 x
dx = f(x) 1+ e
(a) (b) (c) (d)
(c) g ( x ) =
Directions (Q. Nos. 19-20) f ( x ) is continuous and differentiable function. Given f ( x ) takes values of the form ± I, where I denotes set of whole numbers, whenever x = a or b ; otherwise f ( x ) takes real values. 3 Also, f ( c ) = - and |f ( a )| £ |f ( b)| 2 Y
X¢
1 - log g(x) + c, then 2
b
c
X
O a Y¢
19. The number of rational values that f(a) + f(b) + f(c)
1+ e 1+ e
(d) 4
differentiable for all x, if f ¢ ( 0) = 0 continuous but not differentiable at two points, if f ¢ ( 0) = 0 continuous but not differentiable at one point, if f ¢ ( 0) = 0 discontinuous at two points, if f ¢ ( 0) = 0
pö÷ ø
(a) f( x ) = x - 1 (b) g ( x ) =
(c) 3
Passage II
15. Consider a circle with centre lying on the focus of
pö÷ ø
(b) 1
18. Graph of the function y = f(-| x|) is
(b)
(c) -1
16. If ò
(a) 2
1 2 (d) 2
(a) -4
x ®0
x®¥
x ®3
(iv) f ¢ (x) > 0, "x Î (-¥, - 2) È (3, ¥) and f ¢(x) £ 0, "x Î (-2, 3) (v) f ¢¢(x) > 0, "x Î (-¥, - 2) È (-2, 0) and f ¢¢(x) < 0, "x Î (0, 3) È (3, ¥)
increasing GP, then
2x
2x
can take is /are
-1
(a) 4
+1
(b) 2
(c) 3
(d) 5
20. The number of values that [ f(a)]2 + [ f(b)]2 + [ f(c)]2 can
2 + e 2x - 2 1 + e 2x
take is
e2x
(a) 5 (c) 3
(d) f( x ) = x
(b) 2 (d) 7
Answers Paper I 1. (7) 2. (4) 3. (4) 11. (a,c) 12. (a,b,c,d) 13. (a,d) 19. A ® q, B ® s, C ® q, r, s, D ® p
4. (4) 5. (5) 6. (2) 14. (b,c) 15. (a,b,c) 16. (a,d) 20. A ® s, B ® r, C ® p, D ® q
7. (1) 17. (a,d)
8. (2) 18. (b, c)
9. (a,b,c)
10. (a,b,c)
Paper II 1. (1) 11. (b,c)
2. (5) 12. (a,b)
3. (7) 13. (b,d)
4. (1) 14. (b,c)
5. (2) 15. (a,b)
6. (2) 16. (a,b,c)
7. (3) 17. (c)
8. (9) 18. (d)
9. (a,b) 19. (c)
10. (c) 20. (a)
39
Continuity and Differentiability
Limits Limit of a Function
Right Hand Limit Y
Let f(x) be defined on an open interval about x0 , except possible at x0 itself. If f(x) gets arbitrarily close to L for all x sufficiently close to x0 , we say that f approaches the limit L as x approaches x0 and we write lim f(x) = L.
The right hand limit of f(x) as x tends to ‘a’ exist and is equal to l1 , if x approaches ‘a’ through values greater than ‘a’, then f(x) approaches a definite unique real number l1 ,
L
i.e. x0
O
lim f(x) = l1
Y
x ® a+
X l1
x ® x0
Classical Definition of Limit Let f(x) be defined on an open interval about x0 except possible at x0 itself and let L be a real number lim f(x) = L means that x ® x0 for each real e > 0 there exist a real d > 0 such that for all x with 0 < |x - x0 | < d, we have |f(x) - L| < e
X¢
Y¢
Note Evaluate RHL
or simbolically,
X
a x=a+h
O
To evaluate lim f( x ) x ® a+
(i) Put x = a + h in f( x ), then lim f( x ) reduces to lim f( a + h)
" e > 0, d > 0
x ® a+
h® 0
(ii) Simplify the limit as h ® 0
|f(x) - L| < e, where |x - x0 | < d. Y
Left Hand Limit The left hand limit as x ® a exist and is equal to l2 , if x approaches ‘a’ through values less than a, then f(x) approaches a definite unique real number l2 .
Î L Î d
d x0
X
Caution Normally students have the perception that limit should be a finite number. But it is not really, so it is quite possible that f(x) has infinite limit as x ® a. If lim f(x) = ¥, it would simply x ® a mean that function has tendency to assume a very large 1 positive values in the neighbourhood of x = a . e.g. lim =¥ x ® 0 |x|
Neighbourhood (nbd ) of a Point Let ‘x 0 ’ be a real number and let d be a positive real number. Then, the set of all real numbers lying between x 0 - d and x 0 + d is called the neighbourhood of x 0 of radius d and is denoted by Nd( x 0 ). Thus, Nd( x 0 ) = ( x 0 - d, x 0 + d) = {x ÎR | x 0 - d < x < x 0 + d} The set ( x 0 - d, x 0 ) is called the left nbd of x 0 and the set (x 0 , x 0 + d) is called the right nbd of x 0 .
l2
i.e. lim- f(x) = l2
a–h a
x ® a
Note Evaluate LHL To evaluate lim f( x ) x ® a-
(i) Put x = a - h in f ( x ), then lim - f ( x ) reduces to lim f (a - h ) h®0
x®a
(ii) Simplify the limit as h ® 0
Existence of Limit Limit of a function f(x) exists, if lim - f(x) and lim+ f(x) exist x® a x® a and both are equal. Thus, lim f(x) exists Û lim- f(x) = lim+ f(x) x® a
x® a
x® a
Reasons for Non-existence of the Limit Here, student should aware about reasons for non-existence of limit. From the discussion, so far it is clear that lim f(x) will x® a not exist due to any of these three reasons (i) f(x) is not defined in the neighbourhood of x = a. e.g. Consider, lim sec-1 (sin x). x ® p /2
53
Indeterminate Forms
(ii) f(x) does not have a unique tendency. 1 e.g. Consider, lim cos æç ö÷ . x® ¥ èx ø (iii) Left and right tendencies of f(x) are not same. e.g. Consider, 1 lim{x}, where {} x ®1
x+1 -1 0
denotes fractional part.
x
x-1
In finding the limit of a function f(x) at x = a, if a unique value cannot be assigned to f(a), then f(x) is said to be indeterminate at x = a A list of indeterminate forms is given below 1 / ¥2 0 ¥ 0 0 (i) 1 = 0´¥= = = ¥2 1 / ¥1 0 1/ ¥ 0 (ii) (iii) (iv) (v)
1
Note For the existence of the limit at x = a, f( x ) need not be defined at x = a. However, if f( a ) exists limit need not exist or even, if it exists, then its limit need not be equal to f( a ).
L’ HOSPITAL’S RULE
REMEMBER n
(i) If lim f ( x ) g ( x ) exists, then we have following cases x ®a
(a) Both lim f (x ) and lim g (x ) exist. x ®a
(b)
y = 00 Þ log y = log(00 ) = 0 ´ log(0) = 0 ´ ¥ y = ¥0 Þ log y = log(¥0 ) = 0 ´ log(¥) = 0 ´ ¥ y = 1¥ Þ log y = log(1¥ ) = ¥ ´ log(1) = ¥ ´ 0 ¥1 - ¥2 is also an indeterminate form as the ¥1 and ¥2 does not necessarily approach to the same infinity.
x ®a
lim f (x ) exists and lim g (x ) does not exist.
x ®a
n
x ®a
(c) Both lim f (x ) and lim g (x ) do not exist. x ®a
x ®a
(ii) If lim [f ( x ) + g ( x )] exists, then we can have the x ®a
n
f (x ) 0 has an indeterminancy at x = 0 of the type or If g (x ) 0 f (x ) f ¢ (x ) , where f ¢ is called the derivative of f. lim = lim x ® a g (x ) x ® a g ¢ (x ) f ¢( x ) 0 too has an indeterminancy at x = a of the type or If g ¢(x ) 0 f ¢(x ) f ¢¢(x ) lim = lim x ® a g ¢ (x ) x ® a g ¢¢( x ) This can be done continued till we find a determinate result.
¥ , then ¥
¥ , then ¥
following cases
(a) If lim f (x ) exists, then lim g (x ) must exist. x ®a
x ®a
(b) Both lim f (x ) and lim g (x ) do not exist. x ®a
x ®a
Evaluation of Algebraic Limits
Algebra of Limits Let lim f(x) = l and lim g(x) = m. If l and m exist, then x® a
Evaluation of Limits (i)
x® a
(a) Factorisation Method f(x) Consider, lim x ® a g(x)
(i) lim(f ± g)(x) = lim f(x) ± lim g(x) = l ± m x® a
x® a
x® a
(ii) lim(fg)(x) = lim f(x) lim g(x) = lm x® a
x® a
x® a
If by substituting x = a,
lim f(x)
æf ö l (iii) lim ç ÷(x) = x ® a = , provided m ¹ 0 x ® aè g ø lim g(x) m x® a
(iv) lim k f(x) = k × lim f(x), where k is constant. x® a
x® a
(v) lim|f(x)| = |lim f(x)| = |l| x® a
x® a
(vi) lim(f(x)) x® a
g( x )
lim g( x )
= lim f(x)x ® a x® a
= lm
(vii) lim fog(x) = f(lim g(x)) = f(m), only if f is x® a
x® a
continuous at g (x) = m . In particular, (a) lim log f(x) = log(lim f(x)) = log l x® a
x® a
lim f ( x )
(b) lim ef ( x ) = ex ® a x® a
54
= el
0 Form 0
l
l
f(x) 0 reduces to the form . Then, we first g(x) 0
factorise the f(x) and g(x), then cancel out the common factor to evaluate the limit. (b) Rationalisation Method Rationalisation is followed when we have fractional powers (like 1/2, 1/3 etc) on expression in numerator or denominator or in both. After rationalisation the terms are factorised which on cancellation gives the result. (c) Based on Standard Formula x n - an To evaluate the limit of the type lim we use following x® a x - a formulas x n - an lim = nan -1 , where n is a rational number x® a x - a x m - am m m- n lim n = a , where m and n are rational numbers x ® a x - an n
(ii) Algebraic Function of ¥ Type
¥ (a) form Consider the function ax . First we should know ¥ the limiting values of ax (a > 0) as X ® ¥. See the graphs of these functions. f(x)=a x When 0 S a<1 Y
O
Y 1 X
f(x)=a x When a=1
Y
X
O
f(x)=a x When a>1
O
If lim f(x) = 0, then x ® a
(i) lim
x ® a
(ii) lim cos f(x) = 1 x ® a
(iii) lim
x ® a
X
sin f(x) tan f(x) = lim =1 x ® a f(x) f(x) sin-1 f(x) tan-1 f(x) = lim =1 x ® a f(x) f(x)
Evaluation of Logarithmic Limits In this section, we shall deal with the problems based on expansion of logarithmic series which is given below.
Now, see the graph for ax when a > 1. This graph appears to touch X-axis in the negative side of X-axis and thereafter it increases rapidly. That is why because lim ax ® 0, again you will also x ® -¥
find the result, lim ax ® ¥. x® ¥ a >1 ì¥, if ï Thus, we have lim ax = í1, if a =1 x® ¥ ï0, if 0 £ a < 1 î Note This type of problem are solved by taking the highest power of the terms tending to infinity as common numerator and denominator. That is after they are cancelled and the rest output is the result or (apply L’ Hospital’s rule).
(b) ¥ - ¥ form Such problems are simplified (generally ¥ rationalised) first, thereafter they generally acquire ¥ form.
log(1 + x) = x -
x2 x3 + + ... ¥, where -1 £ x £ 1 2 3
And it should be noted that expansion is true only, if the base is e. To evaluate the logarithmic limit, we use the following formula log(1 + x) lim =1 x ®0 x
Evaluation of Exponential Limits (a) Based on Series Expansion ex = 1 + x +
x2 x3 + + ... ¥ 2! 3!
To evaluate the exponential limit, we use the following result ex - 1 ax - 1 (i) lim (ii) lim =1 = log ea x ®0 x ®0 x x
(b) Evaluation of Exponential Limits of the Form 1 ¥ To evaluate the exponential form 1¥ , we use the following results Result If lim f(x) = lim g(x) = 0, then
If m, n are positive integers and a 0 , b 0 ¹ 0 are non-zero real numbers. Then, a x m + a 1x m - 1 + K+ a m - 1x + a m lim 0 n x ® ¥ b x + b x n - 1 + K+ b 0 1 n - 1x + b n ì0, m < n ïa 0 ï ,m =n = í b0 ï¥, m > n when a 0 b 0 > 0 ï î-¥, m > n when a 0 b 0 < 0
Evaluation of Trigonometric Limits To evaluate limits involving trigonometrical functions we have to follow the following steps. Step I Reduce the given expression in terms of sin q and cos q. Step II If the expression obtained in step I is sum or difference of two terms, then remove the positive or negative signs in between two terms. Step III Arrange the terms and use the standard result given below (which is applicable).
x® a
x® a
lim
f( x )
lim{1 + f(x)}1 / g( x ) = ex ® a g( x )
x® a
When lim f(x) = 1 and lim g(x) = ¥ x® a
x® a
Then, lim{f(x)}
g( x )
x® a
lim [ f ( x ) - 1 ]g( x )
= lim[1 + f(x) - 1]g( x ) = ex ® a x® a
REMEMBER (i) lim(1 + x)1 / x = e x ®0
(iii) lim(1 + lx)1 / x = el x ®0
x
1 (ii) lim æç1 + ö÷ = e x ® ¥è xø 1/ x
l (iv) lim æç1 + ö÷ x ® ¥è xø
= el
Evaluation of Limits of the Form 0 0 When lim f(x) ¹ 1 but f(x) is positive in the neighbourhood of x = a.
x® a
In this case, we write {f(x)}g( x ) = elog e { f ( x )} Þ
g( x )
lim g( x ) log e f ( x )
lim[f(x)]g( x ) = ex ® a
x® a
55
Some Standard Limits
This is illustrated in figure. Y
(i) If p ( x ) is a polynomial, then lim p ( x ) = p (a ). x ®a
(ii) lim ( 1 + x ) 1/ x = e x ®0
(1 + x ) n - 1 =n x ln ( 1 + x ) (v) lim =1 x ®0 x
(iii) lim
x ®0
y=h(x)
ex - 1 =1 x ®0 x sin-1 x tan-1 x (vi) lim = 1 = lim x ®0 x ®0 x x sin x tan x (vii) lim = lim = lim cos x = 1 (where, x is in radians) x ®0 x x ®0 x x ®0 x ax - 1 1ö æ (viii) lim ç 1 + ÷ = e (ix) lim = ln (a ), a ÎR + x ® ¥è x ®0 xø x x m - am m m - n log ( 1 + x ) (x) lim n (xi) lim a = loga e, a > 0, ¹ 1 = a x ® a x - an x ®0 x n
(iv) lim
If lim f ( x ) = 0, then the following results will be holding true: x ®a
sinf ( x ) tanf ( x ) = lim = lim cos f ( x ) = 1 x ® a f (x ) x ®a f (x ) -1 -1 sin f ( x ) tan f ( x ) (xiii) lim = lim =1 x ®a x ®a f (x ) f (x ) bf ( x ) - 1 (xiv) lim = In b ( b > 0 ) (xv) lim ( 1 + f ( x )) 1/ f ( x ) = e x ®a x ®a f (x )
y=g(x) y=f(x) O
Following are some of the frequently used series expansions:
sin x = x -
x3 x5 x7 + +K 3! 5! 7!
cos x = 1 -
x2 x4 x6 + +K 2! 4! 6!
tan x = x +
y( x )
r 1 å f æçè n ö÷ø = x® ¥ n r = f( x )
(i) S is replaced by ò (iii)
(ii)
x2 + K , a Î R+ 2! n(n - 1) 2 n(n - 1)(n - 2) 3 (1 + x)n = 1 + nx + x + x 2! 3!
b
òa f(x)dx,
+ K, n Î R.| x| < 1 x2 x3 ln (1 + x) = x + - K , -1 < x £ 1 2 3
r is replaced by x n
(iv) To obtain lower and upper limits we have a = lim
n® ¥
n® ¥
y(x) n
f(x) n
The value, so obtained is the required sum of the given series.
Use of Newton-Leibnitz’s Formula in Evaluating the Limits Let us consider the definite integral I(x) =
y( x )
òf( x ) f(t)dt
Newton-Leibnitz formula states that d d d (I(x)) = f(y(x))ìí y(x)üý - f {f(x)}ìí f(x)üý dx þ þ î dx î dx
Sandwich Theorem for Evaluating Limits Let f, g and h be real functions such that f(x) £ g(x) £ h(x) for all x in the common domain of definition. For some real number a, if lim f(x) = l = lim h(x), then lim g(x) = l. x ® a
56
x ® a
x ® a
x2 x3 + +K 2! 3!
ax = 1 + (ln a) x + (ln a)2
1 is replaced by dx n
and b = lim
x3 2x5 + +K 15 3
ex = 1 + x +
Summation of Series Using Definite Integral as the Limit The expression of the form, lim
X
Series Expansion
(xii) lim
x ®a
a
VITEEE 2016 VIT University will conduct VIT Engineering Entrance Exam (VITEEE) 2016 for granting admissions to various undergraduate engineering programmes. Admissions to its two campuses, Chennai and Vellore, will be purely on the basis of the marks secured in the VITEEE 2016.
VITEEE 2016 Exam Pattern The Paper will have 125 questions in total with 40 questions each in Physics, Chemistry, Maths/ Biology and 5 questions in English. No negative marking will be there in the question paper. All questions will be of objective type and as per the exam pattern. Each question will be followed by four alternative answers and the candidates have to select the right option. Candidates will be given two and half hours to complete the online test. There will be no negative marking for wrong answers.
VITEEE 2016 Important Dates Complete schedule of VITEEE 2016 will be announced by the university later. In the meantime, aspirants can check here other important dates at official website.
Best Practice SHOTS 1.
The value of lim
x ®1 y®0
y3 as (x, y) ® (1, 0) along the x 3 - y2 - 1
6.
lim
The value of lim
x ® a
(a) 2 (c) 4
7.
æ x2 + 1 ö If lim ç - ax - b÷ = 0, then find the values of a x ® ¥è x + 1 ø and b.
4.
(b) 1, 1 (d) None of these
1 × n + 2 × (n - 1) + 3 × (n - 2) + K + n ×1 The value of lim n® ¥ 12 + 22 + K + n2 is
5.
lim
x ®0
9.
n n n + 2 + K+ 2 + 12 n + 22 n + n2
p 2
(b)
is equal to
p 3
(c)
p 4
ö ÷ is equal to ø (d)
p 6
x é bù (a ¹ 0) (where, [.] denotes the a êë x úû greatest integer function) is equal to The value of the lim
x ®0
(b) b x2
10. lim ò0 x ®0
(b) - 1 1 (d) 2
(c) 0
æ
(a) a
x tan 2 x - 2 x tan x is (1 - cos 2 x)2
(a) 1
lim
x
(b) e -1 (d) e -2
ç n ® ¥ è n2 (a)
(b) - 1 1 (d) 2
(a) 1 1 (c) 2
log p The value of lim æç tan æç + log x ö÷ ö÷ x ® 1è è4 øø
(a) e (c) e 2
8.
(a) 1, - 1 (c) - 1, 1
(b) 3 (d) None of these 1
(b) cos 3 a (d) cot a
(a) - sin a (c) sin3 a
3.
non-zero number is
cos x - cos a is cot x - cot a
3
xn
x ®0
(b) - 1 (d) does not exist
(a) 1 (c) 0
x3 2 is a finite
cos2 x - cos x - ex cos x + ex -
line y = x - 1 is
2.
The integer n for which the
cos t 2 dt x sin x
(a) 1
(c)
b a
(d) 1 -
b a
is equal to
(b) 2
(c) - 1
(d) 0
Continuity A continuous function is a function whose graph can be drawn without lifting the pen from the paper. Graphically, it could be stated as shown in figure (i). When the function f(x) is continuous at a point x = a, we mean that at point (a, f(a)) the graph of the function has no hole or gaps. Here, lim f(x) = 2 and f(1) = 2 x ®1
Thus, lim f(x) = f(1), hence f(x) is continuous. Y
f(x)
f(x)
2
O
x ® a
Caution
2
X
O
x=1
lim f( x ) = 2
x ®1
(ii)
x ® a
i.e. LHL = RHL = value of the function at x = a .
(1)
x=1
Continuity of a Function A function f(x) is said to be continuous at x = a, if lim- f(x) = lim+ f(x) = f(a).
x ®1
Y
Note The graph of the function is said to be continuous at x = a, if while moving along the graph of the function and in crossing over the point at x = a either from left to right or from right to left one does not have to lift his pen.
X
It should be noted that continuity of a function is the property of interval and is meaningful at x = a only if the function has a graph in the immediate neighbourhood of x = a, not necessarily at x = a. Hence, it should not be mislead that continuity of a function is taken only in its domain.
(i)
57
Discontinuity of a Function
Continuity in Interval (i) Continuity in an Open Interval A function f(x) is said to be continuous in an open interval (a, b), if it is continuous at each point of (a, b). (ii) Continuity in a Closed Interval A function f(x) is said to be continuous in a closed interval [a, b], if (a) f(x) is continuous from right at x = a, i.e. lim+ f(x) = f(a).
If f(x) is not continuous at x = a, then we say that f(x) is discontinuous at x = a. f(x) will be discontinuous at x = a in any of the following cases. (i) lim- f(x) and lim+ f(x) exist and are not equal. x ® a
x ® a
Y
f(x)
l1
x ® a
(b) f(x) is continuous from left at x = b i.e. lim- f(x) = f(b).
l2
x ® b
(c) f(x) is continuous at each point of the interval (a, b). O
Discontinuity (Graphical Approach)
(ii) lim- f(x) and x ® a
In case one has to lift his pen when moving through the graph of a function, is said to have a break or discontinuous at x = a. Different types of situation, which may come up at x = a along the graph, can be shown below. Y
X
x=a
lim f(x) exist and are equal but not
x ® a+
equal to f(a).
Y f(x) f(a)
Y
L
O
X O
x=a Continuous at x=a
X
(iii) f(a) is not defined
X O
a
Y
x=a Discontinuous at x=a
Y
X x=a X O
x=a Discontinuous at x=a
Y
(iv) Atleast one of the limits does not exist.
Types of Discontinuity
Y
There are two types of discontinuity.
(i) Removable Discontinuity Here, lim f(x) necessarily exists, but either is not equal to f(a) x® a
X O
x=a Discontinuous at x=a
X O
x=a Discontinuous at x=a
Y
O
58
x=a Discontinuous at x=a
X
or f(a) is not defined. In this case, therefore it is possible to redefine the function in such a manner that lim f(x) = f(a) and x® a thus making function continuous. These discontinuities can be further classified as (a) Missing Point Discontinuity Y (x - 1)(9 - x2 ) e.g. Let f(x) = , 9 (x - 1) 0 clearly f(1) ® form. 0 \ f(x) has missing point -3 O 1 discontinuity can be shown as
3
X
(b) Isolated Point Discontinuity
ì0, if x Î I e.g. Let f(x) = [x] + [- x] Þ f(x) = í î-1, if x Ï I
Theorems on Continuity (i) Sum, difference, product and quotient of two continuous functions are always continuous function. f(x) is continuous at x = a only if g(a) ¹ 0 However, h(x) = g(x)
where, x = Integer, has isolated point discontinuity, can be shown as Y
-4 -3 -2 -1 O
1
2
3
4
5
(ii) If f(x) is continuous and g(x) is discontinuous at x = a, then product function f(x) = f(x) × g(x) is not necessarily be discontinuous at x = a. (iii) If f(x) and g(x) both are discontinuous at x = a, then the product function f(x) = f(x)g(x) is not necessarily be discontinuous at x = a.
X
-1
(ii) Non-removable Discontinuity If lim f(x) does not exist and therefore it is not possible to x® a
redefine the function in any manner to make it continuous. Such discontinuities can further be classified into three types. (a) Finite type (Both limits finite and unequal) Consider, the function f(x) = [x], greatest integer function. As shown in figure the graph has jump of discontinuity at all integral values of x.
Continuity of Composite Function If the function u = f(x) is continuous at the point x = a and the function y = g(u) is continuous at the point x = f(a), then the composite function y = gof(x) = g(f(x)) is continuous at that point.
MATHS MAXIMA
Y
n
2
n
1 n
-2 -1
O
1
2
3
X
1 2
n
(b) Infinite type (Atleast one of the two limits is infinite or vertical asymptote) Consider, the function f(x) = tan x.
If f is continuous on a closed interval [a , b ], then it is bounded. A continuous function whose domain is some closed interval must have its range also in the closed interval. If f (a ) and f (b ) possess opposite signs, then there exists atleast one solution of the equation f (x ) = 0 in the open interval (a, b) provided f is continuous in [a , b ]. If f is continuous on [a , b ], then f -1 is also continuous.
Intermediate Value Theorem If f is continuous on [a, b] and f(a) ¹ f(b), then for any value c Î(f(a), f(b)), there is atleast one number x0 in (a, b) for which f(x0 ) = c.
Y
Y -2p -3p/2 -p -p/2
0 p/2
Y
p 3p/2 2p
X f(a) f(b)
p 3p Here, the function is not defined at points ± , ± and 2 2 near these points, the function becomes both arbitrarily large and small. Since, the function is not defined at these points, it cannot be continuous. (c) Oscillatory (limit oscillate between two finite p 1 quantities) f(x) = sin . When x ® 0, ® ± ¥ and x x sin(® ± ¥) can take any value between -1 to 1 or we can say when x ® 0, f(x) oscillates between -1 and 1 as shown in figure. Y
p y=sin x
1 X¢
–1 1 –1 Y¢
c f(a) b O
x=a x0 x=b
X
x1 O f(b)
(i)
x3 a
x2
X
(ii)
Y Y f(b) f(b)
c a
f(a)
O X
c
O a (iii)
b
X
b
X
f(a) (iv)
From fig. (iii) and fig. (iv), it is clear that continuity in interval [a, b] is essential for the validity of this theorem.
59
Best Practice SHOTS 11.
ln(1 + ax) - ln(1 - bx) is not defined x at x = 0. The value which should be assigned to f at x = 0, so that it is continuous at x = 0, is
The function f(x) =
(a) a - b
12.
(b) a + b
(c) ln a - ln b
1 ìï 2 -1 , x ¹ 2 is continuous from right f(x) = í{x + e2 - x } , x =2 ïî K
(d) ln a + ln b
at x = 2 , then K is equal to (a) 0
one of the following statement on f(x) is true. That is f(x) is
(c)
17.
ìx2 cos e1 / x , when x ¹ 0. Then, f(x) is Let f(x) = í 1 , when x = 0. î (a) discontinuous at x = 0 (b) continuous but not differentiable at x = 0 (c) differentiable at x = 0 (d) lim f( x ) exist
a2 - ax + x2 a+ x -
18.
19.
a2 + ax + x2
15.
(b) a1/ 2
(c) - a1/ 2
(b) 5
(c) 6
(d) None of these
ìïx m sinæ 1 ö , x ¹ 0 ç ÷ If f(x) = í is continuous at x = 0, then èx ø , x =0 ïî 0 (b) m Î ( - ¥, 0) (d) m Î ( - ¥, 1)
The set of points of discontinuity of the function 1 is f(x) = log|x|
If f(x) =
(b) {-1, 1} (d) None of these
loge (1 + x2 tan x) , x ¹ 0 is to be continuous at sin x3
(a) 1 1 (c) 2
(d) - a 3/ 2
Let f(x) = [x3 - 3], where [x] is the greater integer function. Then, the number of points in the interval (1, 2), where function is discontinuous is (a) 4
1 2
x = 0, then f(0) must be defined as
a- x
becomes continuous for all x, is given by (a) a 3/ 2
1 4
(a) {0} (c) { -1, 0, 1}
The value of f(0), so that the function f(x) =
(b) -
(a) m Î ( 0, ¥) (c) m Î (1, ¥)
x ®0
14.
The function f(x) defined by
1 Give f(0) = 0 and f(x) = for x ¹ 0. Then, only (1 - e-1 / x )
(a) continuous at x = 0 (b) not continuous at x = 0 (c) both continuous and differentiable at x = 0 (d) not defined at x = 0
13.
16.
(d) 7
20.
(b) 0 (d) - 1 2
Let f(x) = degree of (ux + u2 + 2u + 3). Then, at x = 2 f(x) is (a) continuous (c) discontinuous
(b) differentiable (d) None of these
Differentiability f(x) - f(c) f(c + h) - f(c) is called the right = lim h®0 (x - c) h hand derivative of f(x) at x = c and is denoted by f ¢(c + ) or Rf ¢(c).
Let f(x) be a real valued function defined on an open interval (a, b), where c Î(a, b).Then, f(x) is said to be differentiable or f(x) - f(c) derivable at x = c, if and only if lim exists finitely. x ® c (x - c)
While lim+
This limit is called the derivative or differentiable coefficient of the function f(x) at x = c and is denoted by f ¢(c) or Df(c) or d (f(x))x = c dx f(x) - f(c) f(c - h) - f(c) lim = lim h®0 x ® c - (x - c) -h
If Lf ¢(c) ¹ Rf ¢(c), we say that f(x) is not differentiable at x = c.
is called the left hand derivative of f(x) at x = c and is denoted by f ¢(c - ) or Lf ¢(c).
60
x® c
Differentiability in a Set (i) A function f(x) defined on an open interval (a, b) is said to be differentiable or derivable in open interval (a, b), if it is differentiable at each point of (a, b). (ii) A function f(x) defined on [a, b] is said to be differentiable or derivable at the end points a and b, if it is differentiable from the right at a and from the left at b. In other words,
lim
x ® a+
f(x) - f(a) f(x) - f(b) and lim both exist. x® b x-a x-b
Y
Y
If f is derivable in the open interval (a, b) and also the end points a and b, then f is said to be derivable in the closed interval [a, b]. Note A function f is said to be differentiable function, if it is differentiable at every point of its domain.
Relation between Continuity and Differentiability If a function is differentiable at a point, then it is necessarily continuous at that point. But converse is not necessarily true or f(x) is differentiable at x = c Þ f(x) is continuous at x = c but converse is not necessarily true. e.g. The function f(x) = |x| is continuous at x = 0 but it is not differentiable at x = 0, as shown in figure given below Which shows we have sharp edge at x = 0. Hence, function is Y
O
X O a Continuous and differentiable
a
X
Continuous but not differentiable
Y O
O
a Neither continuous differentiable
X nor
f(x)=|x|
Theorems on Differentiability O
X
not differentiable but continuous at x = 0. Note
If f ( x ) is differentiable, then its graph must be smooth, i.e. there should be no break or corner. Thus, for a function f ( x ) n n n
Differentiable Þ Continuous Continuous Þ May or may not be differentiable Not continuous Þ Not differentiable
How can a function fail to be differentiable
(i) If f(x) and g(x) are both derivable at x = a, f(x) are also derivable at then f(x) ± g(x), f(x) × g(x) and g(x) f(x) ] x = a, [only if g(a) ¹ 0 for g(x) (ii) If f(x) is derivable at x = a and g(x) is not differentiable at x = a, then f(x) ± g(x) will not be derivable at x = a. (iii) If both f(x) and g(x) are non-derivable, then nothing can be said about sum/difference/ product functions. (iv) If f(x) is derivable at x = a and f(a) = 0 and g(x) is continuous at x = a. Then, the product function h(x) = f(x) × g(x) will be derivable at x = a. Note If y = f( x ) is differentiable at x = a, then it is not necessarily that the derivative is continuous at x = a.
The function f(x) is said to be non-differentiable at x = a, if (i) both Rf ¢(a) and Lf ¢(a) exist but are not equal. (ii) either or both Rf ¢(a) and Lf ¢(a) are not finite. (iii) either or both Rf ¢(a) and Lf ¢(a) do not exist.
MATHES MAXIMA If f ( x ) is a function such that RHD = f (a + ) = l and LHD = f (a - ) = m.
Case IIf l = m = some finite value, then the function f (x ) is differentiable as well as continuous. Case II If l ¹ m but have some finite value, then the function f (x ) will not be differentiable but it will be continuous. Case III If atleast one of the l or m is infinite, then the function is non-differentiable but we cannot say about continuity of f (x ).
Some Standard Differentiable Function (i) Exponential function ax , a > 0 is differentiable at each x Î R. (ii) The logarithmic function is differentiable at each point in its domain. (iii) Absolute value functions are always continuous throughout but not differentiable at their critical points. e.g. |x - a| is continuous but not differentiable at x = a . (iv) Every polynomial function is differentiable at each x Î R. (v) Every constant function is differentiable at each x Î R. (vi) Trigonometric and inverse trigonometric functions are differentiable in their respective domain. (vii) The composition of differentiable functions is differentiable.
61
Best Practice SHOTS 21.
ìïx p cos æ 1 ö , x ¹ 0 ç ÷ If f(x) = í is differentiable at x = 0, èx ø ïî x =0 0, then (a) p < 0
22.
(b) 0 < p < 1
(b) a = 1, b = 2 (d) a = 4, b = 5
25.
29.
x + 1), then
If f(x) = min{x, x2 , x3 }, then (a) f( x ) is everywhere differentiable (b) f ¢( x ) > 0 for x > 1 (c) f( x ) is not differentiable at three points but continuous for all x Î R (d) f( x ) is not differentiable for two values of x
30.
Let a function y = f(x) be defined as x = 2t -|t |, y = t 2 + t|t |, where t Î R . Then, f(x) is
The set of points where the function f(x) = |x - 1| ex is differentiable, is
If f(x) = x ( x +
(a) f( x ) is continuous but not differentiable at x = 0 (b) f( x ) is differentiable at x = 0 (c) f( x ) is not differentiable at x = 0 (d) None of the above
If f(x) = x5 sgn(x), then at x = 0, f(x) is
(a) continuous and differentiable in [-1,1] (b) continuous but not differentiable in [-1,1] (c) continuous in [-1, 1] and differentiable in ( -1,1) only (d) None of the above
26.
28.
p 3p ö If f(x) = |x| + |sin x|, x Î æç - , ÷, then f is è 2 2 ø
(a) continuous and differentiable (b) continuous but not differentiable (c) differentiable but not continuous (d) neither continuous nor differentiable
Let f(x) = min{x3 , x4 } for all x Î R. Then, (a) f( x ) is continuous for all x (b) f( x ) is differentiable for all x (c) f( x ) is not differentiable at two points (d) None of the above
(d) p > 1
(a) continuous, " x Î R - { 0} (b) continuous and differentiable everywhere (c) no where differentiable (d) differentiable everywhere except at x = 0
24.
27.
If f(x) = ae|x| + b|x|2 , a, b Î R and f(x) is differentiable at x = 0. Then, a and b are (a) a = 0, b Î R (c) b = 0, a Î R
23.
(c) p = 1
(b) R - {1} (d) R - { 0}
(a) R (c) R - { - 1}
If
f(x) = |x| - 1,
then
points, where
f(x) is
not
differentiable, is/are (a) 0, ± 1
31.
(b) ± 1
(c) 0
(d) 1
The function f(x) = max {(1 - x), (1 + x), 2} x Î(-¥, ¥) is (a) continuous at all points (b) differentiable at all points (c) differentiable at all points except at x = 1 and x = - 1 (d) None of the above
Differentiation The rate of change of quantity ‘y’ with respect to another quantity x is called derivative or differential coefficient of y dy with respect to x and is denoted by . dx
Rules for Differentiation (i) If f(x) and g(x) are differentiable functions and a ÎR, then d d d [f(x) ± g(x)] = [f(x)] ± [g(x)] dx dx dx d d and [af(x)] = a f(x). dx dx (ii) Product rule If f(x) and g(x) are differentiable functions, d d d then [f(x) × g(x)] = g(x) [f(x)] + f(x) [g(x)] dx dx dx
62
d / dx {f ( x ) × g ( x ) × h ( x )} = f ( x ) × g ( x ) × h ¢ ( x ) + f (x ) × g ¢ (x ) × h (x ) + f ¢ (x ) × g (x ) × h (x ) (fg ) ¢ h + ( gh ) ¢f + ( hf ) ¢ g = 2 d d g(x) [f(x)] - f(x) [g(x)] d é f(x) ù dx dx (iii) Quotient rule = dx êë g(x) úû [g(x)]2 dy dy du (iv) Chain rule If y = h(u) and u = f(x), then = × dx du dx This rule can be extended as follows dy dy du dv If y = u(x), u = v(x), v = f(x) then = × × dx du dv dx Note If at all points of a certain interval, f ¢ ( x ) = 0, then the function f( x ) preserves a constant value within this interval. This function is said to be constant function.
Derivative of Some Standard Functions (i) The trigonometric functions have the following derivatives d d (sin x) = cos x, (cos x) = - sin x dx dx d d (tan x) = sec2 x, (cot x) = - cosec2 x dx dx d d (sec x) = sec x tan x, (cosecx) = - cosecx cot x dx dx d n (ii) (x ) = nx n - 1 , if f(x) is a differentiable function then dx d using chain rule [f(x)n] = n[f(x)]n - 1 f ¢(x). dx d 1 (iii) (log x) = , if f(x) is a differentiable function then dx x d 1 [log f(x)] = f ¢(x). dx f(x) d x (iv) (a ) = ax log a. dx d x d f( x ) In particular, (e ) = ex and (e ) = ef ( x )f ¢(x). dx dx d d 1 (v) for -1 < x < 1. At (sin-1 x) = (cos -1 x) = dx dx 1 - x2 the points x = ± 1, sin-1 x and cos -1 x are not differentiable. d d 1 (vi) for all x Î R . (tan-1 x) = (cot -1 x) = dx dx 1 + x2 d d 1 for |x| > 1. (vii) (sec-1 x) = (cosec-1 x) = dx dx |x| x2 - 1
Step I Differentiate each term of f (x, y) = 0 dy Step II Collect the terms containing , and transpose other dx terms to the right side. Step III Divide both side of the equation obtained in step II by dy dy coefficient of to get . dx dx
¶f dy ¶ If f ( x , y ) = 0 is an implicit function, then =- x ¶f dx ¶y ¶f ¶f represents partial derivative of ‘f’ with respect to x and where, ¶x ¶y represents partial derivative of ‘f’ w.r.t. y .
Differentiation of Parametric Functions If x = f(t ), y = g(t ), where t is parameter, then d æ dy ö g(t ) ç ÷ dy è dt ø dt g¢(t ) = = = d dx æ dx ö f(t ) f ¢(t ) ç ÷ è dt ø dt
Differentiation of a Function with Respect to Another Function If u = f(x) and v = g(x), then in order to find derivative of f(x) du f ¢(x) . = dv g ¢(x)
with respect to g(x). We use following formula
Differentiation Using Substitution
Logarithmic Differentiation
In order to find differential coefficients of complicated expression involving inverse trigonometric functions some substitutions are very helpful, which are listed below.
The process of taking logarithms on both sides and then differentiating them is called logarithmic differentiation. It is useful in the following cases: (i) If the given function consists of three or more factors which are function of x. (ii) If the given function is of the form [f(x)]h( x ). h (x)h2 (x) (iii) If the given function is of the form 1 . h3 (x)h4 (x)
S.No.
Function
Substitution
1.
a - x
2
x = a sin q or acos q
2.
a2 + x2
x = a tan q or acot q
3.
x2 - a2
x = a sec q or a cosec q
4.
a + x and a - x
x = a cos2 q
5. 6. 7.
2
a sin x + b cos x x - a and b - x 2 ax - x
2
If y = {f ( x )}h ( x ) , then
a = r cos a, b = r sin a 2
2
x = asin q + b cos q
dy = h ( x ){f ( x )}h ( x ) - 1 f ¢( x ) + {f ( x )}h ( x ) log f ( x ) ´ h ¢( x ) dx
x = a(1 - cos q)
Differentiation of Implicit functions If f(x, y) = 0 be given implicit function, in which y is not expressible explicitly in term of x, then following steps are dy very useful to find . dx
REMEMBER dy y 2f ¢ ( x ) = dx f ( x ){1 - y logf ( x )} dy f ¢ ( x ) 1 = × = 2f ( x ), then dx g ¢ (y ) 1 + {f ( x )}2
(i) If y = f ( x ) {f ( x )} K¥ , then (ii) If e g (y ) - e - g (y )
63
1 - g (x ) 1 + g (x ) dy g ¢ (x ) , then × = 1 + g (x ) 1 - g (x ) dx [ 1 - g ( x )] 2 dy f ¢ (x ) (iv) y = f ( x ) + f ( x ) + f ( x ) + K ¥ , then = dx 2y - 1 dy f ¢ ( x )log( x ) (v) If {f ( x )}g (y ) = ef ( x ) - g (y ) , then = dx g ¢ (y ) × {1 + logf ( x )}2 (iii) If y =
(vi) If {f ( x )}g (y ) = {g (y )}f ( x ) , then dy g (y ) f ¢ ( x ) é f ( x )log g (y ) - g (y ) ù = × dx f ( x ) g ¢ (y ) êë g (y )logf ( x ) - f ( x ) úû
Differentiation of Determinant To differentiate a determinant, we differentiate one row (or column) at a time, keeping others unchanged. f(x) g(x) e.g. If D(x) = , then u(x) v(x) f ¢(x) g¢(x) f(x) g(x) d [D(x)] = + u(x) v(x) u¢(x) v ¢(x) dx Note Similar results hold for the differentiation of determinants of higher order.
Best Practice SHOTS 32.
dy is equal to dx y ( x log y - y) x (log y - y)
If x y = y x ,
y ( x log y + y) (a) (b) (c) (d) None x ( y log x - x ) y (log x + x ) x ( y log x - x )
33.
If y =
1 x2 + a2 +
x2 + b2
,
dy is equal to dx
é ù x 1 1 + ê ú 2 2 2 2 2 a -b ê x + a x + b úû ë é ù 1 1 1 (b) 2 ê ú 2 2 a - b 2 ê x2 + a2 x + b úû ë é ù x 1 1 (c) 2 ê ú 2 a - b ê x2 + a2 x 2 + b 2 úû ë é ù 1 1 (d) 2 ê ú 2 2 2 2 2 a -b ê x + a + x + b ú ë û
(a)
34.
If y = ax
36.
y2 log y
(b) 7/8 x
x ... ¥
(c) 1
, then x (1 - y log x log y) (b)
y log y
(c) y 2 log y
64
at x =
38.
p is equal to 2
8 p2 + 4
39.
(c) -
(b) 0
(b) - 1
dy is dx (d) y log y
(b) 1 for 2 < x < 3 (d) None of these 1
(b) - 1
(c) x
dy is dx (d)
x
If f(x), g(x) and h(x) are three polynomials of degree 2, f(x) g(x) h(x) then f(x) = f ¢(x) g¢(x) h¢(x) f ¢¢(x) g¢¢(x) h¢¢(x) (a) Constant polynomial (c) Polynomial of degree
42.
1
If y = (1 + x 4 ) (1 + x 2 ) (1 - x 4 ), then (a) 1
41.
(d) 2
If f(x) =|x -1| and g(x) = f(f(f(x))), then for x > 2, g ¢(x) is equal to
1
(d) None of these
(d) 1
dy at x = 1 is dx
(c) 1
(a) 1 for all x > 2 (c) -1 for 2 < x < 3
40.
dy 2x , then dx 1 + x2
8 p2 + 4
If 2 x + 2 y = 2 x + y , then the value of (a) 0
æ2 x + 3 ö dy If f(x) = sin (log x) and y = f ç is equal to ÷, then dx è 3 - 2x ø ì æ2 x + 3 öü 1 (a) cos ílog ç ÷ý 2 9 - 4x è 3 - 2 x øþ î ì æ2 x + 3 öü 12 (b) cos x ílog ç ÷ý 9 - 4 x2 è 3 - 2 x øþ î ì æ 3 - 2 x öü 12 (c) cos ílog ç ÷ý 2 9 - 4x è2 x + 3 øþ î
ì æ2 x + 3 öü 12 cos ílog ç ÷ý 2 9 + 4x è 3 - 2 x øþ î
If y = (logcos x sin x)(logsin x cos x) + sin-1
(a)
dy 1 If 5 f(x) + 3 f æç ö÷ = x + 2 and y = xf(x), then at x = 1 is èx ø dx equal to
(a)
37.
2
(a) 14
35.
(d)
(b) Polynomial of degree 2 (d) None of these
If ai , bi Î N for i = 1, 2, 3 then coefficient of x in the determinant (1 + x)a1 b1 (1 + x)a2 b1 (1 + x)a3 b1 (a) 0 (c) 2
(1 + x)a1 b2 (1 + x)a2 b2 (1 + x)a3 b2 (b) 1 (d) 3
(1 + x)a1 b3 (1 + x)a2 b3 is (1 + x)a3 b3
Higher Order Derivative Mean Value Theorems Let y = f(x) be a differentiable function such that z = f ¢(x) is also differentiable. Then, the second derivative of y = f(x) is d2 y denoted by y2 (x), f ¢¢(x) or 2 and is defined by dx d2 y d æ dy ö = ç ÷ = f ¢¢(x) dx è dx ø dx2 Similarly, we can define d3 y d4 y d æ d2 y ö ( iv ) = ç 2 ÷ = f ¢¢¢(x) and f (x) = 3 dx è dx ø dx dx4 or in general,
dny d æ dn - 1 y ö = ç ÷ for any n Î N. n dx è dx n - 1 ø dx
REMEMBER np dn [sin(ax + b )] = a n sin æç + ax + b ö÷ è 2 ø dx n n dn p n (ii) n [cos(ax + b )] = a cos æç + ax + b ö÷ è 2 ø dx m! dn m n m -n (iii) n (ax + b ) = a (ax + b ) ( m - n )! dx
(i)
( -1) n - 1 ( n - 1)a n dn ax b [log( + )] = dx n (ax + b ) n n d (v) (a) [e ax sin( bx + c )] = r n e ax sin( bx + c + nf) dx n
(iv)
Note Most of the functions like, polynomial, trigonometric, exponential and logarithmic functions are differentiable at each point of their domain.
(b)
Second Order Derivative of Parametric Functions If x = f(t ) and y = y(t ), then dy y¢(t ) d d , where y¢(t ) = = {y(t )} and f¢(t ) = {f(t )} dx f¢(t ) dt dt and
d2 y d æ dy ö d ì y¢(t ) ü = ç ÷= í ý dx è dx ø dx î f¢(t ) þ dx2 d dt
ì y¢(t ) ü dt í ý× î f¢(t ) þ dx
d2 y d = dt dx2
ì y¢(t ) ü dt í ý× î f¢(t ) þ dx
= Þ
Leibnitz Theorem for nth Derivatives of Product of Two Functions Let u(x) and v(x) be functions possessing derivatives upto nth order. Then,
(uv)n = u n(x)v(x) + nC1u n - 1 (x)v1 (x) + K
d n ax [e cos(bx + c )] = r n e ax cos(bx + c + nf ) dx n where, r = a2 + b2 æbö and f = tan -1ç ÷ èa ø
Mean Value Theorems Rolle’s Theorem Let f be a real valued function defined in the closed interval [a, b], such that (i) f is continuous in the closed interval [a, b]. (ii) f(x) is differentiable in the open interval (a, b). (iii) f(a) = f(b) Then, there is some point c in the open interval (a, b), such that f ¢(c) = 0. Geometrically Under the assumptions of Rolle’s theorem, the graph of f(x) starts at point (a,0) and ends at point (b, 0) as shown in figures. Y Y
+ nC ku n - k (x)v k (x) + K + nC nu(x)v n(x) k
d u(x) [k th derivative of the function u(x)], dx k dk v(x), 1 £ k £ n 1 £ k £ n and v k (x) = dx k where, uk (x) =
nth derivatives of some elementary functions dn m m!x m - n dn m , if n £ m and (x ) = 0, if n > m. (x ) = n (m - n)! dx dx n
O
a
b
X
O
a
b
X
dn np ö d n np ö (sin x) = sinæç x + (cos x) = cos æç x + ÷, ÷ n è è 2 ø dx n 2 ø dx
The conclusion is that there is atleast one point c between a and b, such that the tangent to the graph at (c, f(c)) is parallel to the X-axis.
dn (emx ) = mn emx dx n
Between any two roots of a polynomial f(x), there is always a root of its derivative f ¢(x).
65
Lagrange’s Mean Value Theorem
Geometrically Any chord of the curve y = f(x), there is a point on the graph, where the tangent is parallel to this chord.
Let f be a real function, continuous on the closed interval [a, b] and differentiable in the open interval (a, b). Then, there is atleast one point c in the open interval (a, b), such that f(b) - f(a) f ¢(c) = b- a
Y
Note In the particular case, where f( a ) = f( b ) , the expression f( b ) - f( a ) becomes zero. Thus, when f( a ) = f( b ), f ¢(c ) = 0 for b-a
O
a
X
b
c
some c in ( a, b ). Thus, Rolle's theorem becomes a particular case of the mean value theorem.
Best Practice SHOTS 43.
If y = ( A + Bx) emx + (m - 1)-2 ex , then d2 y dy - 2m + m2 y is equal to dx dx2 (b) e mx
(a) e x
44.
48.
(a) (0, 1) (c) ( - 1, 0)
(d) e(1 - m)x
(c) e - mx
If y = at 2 + 2 bt + c and t = ax2 + 2 bx + c, then
d3 y dx3
49.
equals to (a) 24a 2 ( at + b ) (c) 24a ( at + b )2
45.
(a) 1
46.
(b) 2
n
(c) 2
n-1
50.
(d) 0
d y dy 1 is If x = tan æç log y ö÷, then (1 + x2 ) 2 - (a - 2 x) èa ø dx dx equal to (b) 1
(c) 2
(d) 2 x
2 æ x ö 3 d y If y = x log ç is equal to ÷ , then x dx2 è a + bx ø
dy (a) æç x - yö÷ è dx ø dy (c) æç - yö÷ è dx ø
2
2
(b)
If the equation ax2 + bx + c = 0 has two positive and real root, then the equation ax2 + (b + 6 a)x + (c + 3 b) = 0 has
If f(x) = 3 x2 + 5 x + 7 satisfies the Lagrange mean value theorem in the interval [1, 3], then the value of c is (a) 1 (c) 3
51.
(b) 2 (d) None of these
The value of c in lagrange’s mean value theorem for the p 5p ù function f(x) = log sinx in the interval é , ëê 6 6 ûú p 4 2p (b) 3 p (c) 2 (d) None of the above (a)
a2 x ( a + bx )2
dy (d) æç x + yö÷ è dx ø
(b) [0, 2] (d) ( - 2, 0)
(a) atleast one positive roots (b) exactly one positive roots (c) exactly two positive roots (d) None of the above
2
(a) 0
47.
(b) 24a ( ax + b )2 (d) 24a 2 ( ax + b )
If f(x) = x n , then the value of f ¢(1) f ¢¢(1) f ¢¢¢(1) (- 1)n f n (1) is f(1) + + + ... + 1! 2! 3! n!
If 2 a + 3 b + 6 c = 0, then the equation ax2 + bx + c = 0 has atleast one root lies in the interval
2
Answers 1. (c)
2. (c)
3. (a)
4. (d)
5. (d)
6. (c)
7. (c)
8. (c)
9. (c)
10. (a)
11. (b)
12. (b)
13. (a)
14. (c)
15. (c)
16. (c)
17. (a)
18. (c)
19. (a)
20. (a)
21. (d)
22. (a)
23. (d)
24. (a)
25. (a)
26. (b)
27. (a)
28. (c)
29. (c)
30. (a)
31. (c)
32. (a)
33. (c)
34. (b)
35. (c)
36. (b)
37. (c)
38. (b)
39. (c)
40. (b)
41. (a)
42. (a)
43. (a)
44. (d)
45. (d)
46. (a)
47. (d)
48. (a)
49. (a)
50. (b)
51. (c)
66
Formula at a Glance (ii) When lim f( x ) = 1 and lim g ( x ) = ¥
1. Let lim f( x) = l and lim g ( x) = m. If l and m exist, then x®a
(i) lim( f ± g )( x ) = lim f( x ) ± lim g ( x ) = l ± m x®a
x®a
x®a
x®a
x®a
Then, lim{ f( x )}
x®a
g(x )
x®a
= lim [1 + f( x ) - 1]g ( x ) x®a
(ii) lim( fg )( x ) = lim f( x ) lim g ( x ) = lm x®a
x®a
lim [ f( x ) - 1]g ( x )
= e x ®a
x®a
lim f( x ) l æfö (iii) lim ç ÷( x ) = x ® a = , provided m ¹ 0 x ® aè g ø lim g ( x ) m
6. When lim f( x) ¹ 1 but f( x) is positive in the neighbourhood of x®a
x = a. In this case, we write { f( x )} g ( x ) = eloge { f( x )}
x®a
(iv) lim k f( x ) = k × lim f( x ), where k is constant. (v) lim| f( x )| = | lim f( x )| = | l| x®a
(vi) lim( f( x )) x®a
= lim f( x )
lim g ( x )
=l
x ®a
x®a
m
(vii) lim fog ( x ) = f( lim g ( x )) = f( m), only if f is continuous at x®a
x®a
g ( x) = m In particular, (a) lim log f( x ) = log( lim f( x )) = log l x®a
(b) lim e
x®a
f( x )
x®a
=e
lim f( x )
x ®a
= el
(viii) If lim f( x ) = + ¥ or -¥, then lim
1
x ® a f( x )
x®a
= 0.
(ix) If f( x ) £ g ( x ) for every x in the nbd of a, then lim f( x ) £ lim g ( x ). x®a
x®a
2. L’ Hospital’s Rule f( x ) 0 has an indeterminancy at x = 0 of the type or g ( x) 0 ¥ f( x ) f ¢( x) , then lim , where f ¢ is called the = lim x ® a g ( x) x ® a g ¢( x) ¥ derivative of f. f ¢( x ) 0 too has an indeterminancy at x = a of the type (ii) If g ¢( x ) 0 ¥ f ¢( x ) f ¢¢( x ) or , then lim = lim x ® a g ¢( x) x ® a g ¢¢( x ) ¥ (i) If
3.
4.
This can be done continued till we find a determinate result. a>1 ì ¥, ï lim a x = í 1, a=1 x ®¥ ï 0, 0 £ a < 1 î lim
x ®¥
a 0 x m + a1 x m-1 + ... + am-1 x + am b 0 x n + a1 x n-1 + ... + an-1 x + an m n when a0b 0 > 0 ï î - ¥, m > n when a0b 0 < 0
5.
(i) If lim f( x ) = lim g ( x ) = 0, then x®a
x®a
lim
f( x )
x®a
(i) If p( x ) is a polynomial, then lim p( x ) = p( a ).
7.
x®a
g(x )
lim g ( x )loge f( x )
lim[f( x )]g ( x ) = e x ® a
Þ
x®a
x®a
g( x )
x®a
(ii) lim(1 + x )1/ x = e x®0 (1 + x )n - 1 (iiii) lim =n x®0 x ex - 1 (iv) lim =1 x®0 x ln (1 + x ) (v) lim =1 x®0 x -1 sin x tan-1 x (vi) lim = 1 = lim x®0 x®0 x x sin x tan x (vii) lim = lim = lim cos x = 1 (where, x®0 x x®0 x x®0 radians)
x
is
in
x
1 (viii) lim æç1 + ö÷ = e x ®¥è xø ax - 1 = ln ( a ), a Î R + x®0 x xm - am m m - n (x) lim n = a x ® a x - an n
(ix) lim
(xi) lim
x®0
log a(1 + x ) = log a e, a > 0, ¹ 1 x
If lim f( x ) = 0, then the following results will be hold. x®a
(xii) lim
x®a
sin f( x ) tan f( x ) = lim = lim cos f( x ) = 1 x ® a f( x ) x®a f( x )
(xiii) lim
sin-1 f( x ) tan-1 f( x ) = lim =1 x®a f( x ) f( x )
(xiv) lim
b f( x ) - 1 = In b ( b > 0) f( x )
x®a
x®a
(xv) lim(1 + f( x ))1/ f( x ) = e
8. If
x®a
f( x ) £ g ( x ) £ h( x )
and
lim f( x ) = lim h( x ) = l,
x ®a
x ®a
then
lim g ( x ) = l.
x ®a
9. Following are some of the frequently used series expansions: x3 + 3! 2 x (ii) cos x = 1 + 2! (i) sin x = x -
x5 x7 +K 5! 7 ! 4 6 x x +K 4! 6!
lim{1 + f( x )}1/ g ( x ) = e x ® a g ( x )
x®a
67
x3 2 x5 + +K 3 15 x2 x3 (iv) e x = 1 + x + + +K 2! 3! x2 (v) a x = 1 + (ln a ) x + (ln a )2 + K, a Î R + 2! n( n - 1) 2 n( n - 1)( n - 2 ) 3 (vi) (1 + x )n = 1 + nx + x + x 2! 3! + K, n Î R.| x| < 1 x2 x3 (vii) ln (1 + x ) = x + - K, - 1 < x £ 1 2 3
(ii) If f( x ) is continuous and g ( x ) is discontinuous at x = a, then product function f ( x ) = f( x ) × g ( x ) is not necessarily be discontinuous at x = a. (iii) If f( x ) and g( x ) both are discontinuous at x = a, then the product function, f x = f( x ) g ( x ) is not necessarily be discontinuous at x = a .
(iii) tan x = x +
10.
(i) Sum, difference, product and quotient of two continuous functions is always a continuous function. f( x ) However h( x ) = is continuous at x = a only, if g ( x) g ( a ) ¹ 0.
11. If f( x) is differentiable, then its graph must be smooth i.e. there should be no break or corner. Thus, for a function f( x ) (i) Differentiable Þ Continuous (ii) Continuous Þ May or may not be differentiable (iii) Not continuous Þ Not differentiable
12. The function f(x) is said to be non-differentiable at x = a, if (i) both Rf ¢( a ) and Lf ¢( a ) exist but are not equal. (ii) either or both Rf ¢( a ) and Lf ¢( a ) are not finite. (iii) either or both Rf ¢( a ) and Lf ¢( a ) do not exist.
Continuity and Differentiability of Different Functions
13. Types of functions
Curve
Identity function
f( x ) = x
Domain and range Domiain = R, Range = ] - ¥, ¥[ x
Exponential function
f( x ) = a , a > 0, a ¹ 1
Domain = R, Range =] 0, ¥ [
Logarithmic function
f( x ) = log a x; x, a > 0 and a ¹ 1
Domain = ( 0, ¥), Range = R
Root function
f( x ) =
Domain = [0, ¥), Range = [0, ¥)
Greatest integer function
f( x ) = [ x ]
Domain = R, Range = l
Least integer function
f( x ) = ( x )
Domain = R, Range = l
Fractional part function
f( x ) = { x} = x - [ x ]
Domain = R, Range = [0, 1)
Signum function
| x| f( x ) = x ì -1, x < 0 ï = í 0, x = 0 ï 1, x > 0 î
Domain = R, Range = { -1, 0, 1}
Constant function
f( x ) = c
Domain = R, Range = {c}, where c ® constant
Polynomial function
Domain = R f( x ) = a0 x n + a1 x n - 1 + a2 x n - 2 + K + an
14.
68
Continuity and differentiability
x
(i) If f( x ) and g ( x ) are differentiable functions and a Î R, d d d and then [f( x ) ± g ( x )] = [f( x )] ± [g ( x )] dx dx dx d d [af( x )] = a f( x ). dx dx (ii) Product rule If f( x ) and g ( x ) are differentiable functions, d d d then [f( x ) × g ( x )] = g ( x ) [f( x )] + f( x ) [g ( x )]. dx dx dx d d g ( x ) [f( x )] - f( x ) [g ( x )] d é f( x ) ù dx dx (iii) Quotient rule ê . = dx ë g ( x )úû [g ( x )]2 dy dy du (iv) Chain rule If y = h(u ) and u = f( x ), then . = × dx du dx
15.
Continuous and differentiable in their domain Continuous and differentiable in ( 0, ¥) Other than integral value it is continuous and differentiable Continuous and differentiable everywhere except at x = 0
Continuous and differentiable everywhere (i) The trigonometric functions have the following derivatives d d (sin x ) = cos x, (cos x ) = - sin x dx dx d d (tan x ) = sec 2 x, (cot x ) = - cosec 2 x dx dx d d (sec x ) = sec x tan x, (cosecx ) = - cosec x cot x dx dx d (ii) ( x n ) = nx n - 1, if f( x ) is a differentiable function then dx d using chain rule [f( x )n ] = n[f( x )]n - 1 f ¢ ( x ). dx
(ii) If e g ( y ) - e - g ( y ) = 2 f( x ), then 1 dy f ¢( x) = × dx g ¢ ( y) 1 + { f( x )} 2
d 1 (log x ) = , if f( x ) is a differentiable function then dx x d 1 [log f( x )] = f ¢ ( x ). dx f( x ) d (iv) ( a x ) = a x log a . dx d d In particular, (e x ) = e x and (e f( x ) ) = e f( x )f ¢ ( x ). dx dx 1 d d for -1 < x < 1. (v) (sin-1 x ) = - (cos -1 x ) = dx dx 1 - x2 (iii)
At the points x = ± 1, sin-1 x and cos -1 x are differentiable. 1 d d for all x Î R. (vi) (tan-1 x ) = - (cot -1 x ) = dx dx 1 + x2 (vii)
(vi) y =
f( x ) +
order. Then,
Function
(uv )n = u n( x )v( x ) + nC1u n - 1( x )v1( x ) + K where, u k ( x ) = 1 £ k £ n.
1.
a2 - x2
x = a sin q or acos q
2.
a2 + x2
x = a tan q oracot q
3.
x2 - a2
x = a sec q or a cosec q
a + x and a - x a sin x + b cos x
+ nC ku n - k ( x )v k ( x ) + K + nC nu( x )v n( x )
Substitution
23.
x = a cos2 q a = r cos a, b = r sin a 2
2
6.
x - a and b - x
x = asin q + b cos q
7.
2 ax - x 2
x = a (1 - cos q)
17. If f( x, y) = 0 be an implicit function, then
¶f dy = - ¶x ¶f dx ¶y
18. If x = f(t ), y = g (t ), where t is parameter, then
dy g ¢ (t ) = dx f ¢ (t )
du f ¢( x) = dv g ¢( x)
20. To differentiate a determinant, we differentiate one row (or column) at a time, keeping others unchanged. f( x ) g ( x ) e.g. If D( x ) = , then u( x ) v ( x ) f ¢( x) g ¢( x) f( x ) g ( x ) d [D( x )] = + u( x ) v ( x ) u ¢( x) v ¢( x) dx (i) If y =
f( x ){ f( x )} K¥ ,
dy y 2f ¢ ( x ) then = dx f( x ){1 - y log f( x )}
dk u( x ) [kth derivative of the function u( x )], dx k
(i)
dn m m! x m - n dn , if n £ m and n ( x m ) = 0, if n > m . (x ) = n ( m - n)! dx dx
(ii)
dn np ö (sin x ) = sinæç x + ÷, è 2 ø dx n
(iii)
dn np ö (cos x ) = cos æç x + ÷ è 2 ø dx n
(iv)
d n mx (e ) = mn e mx dx n
(v)
dn np [sin( ax + b )] = a n sinæç + ax + b ö÷ è2 ø dx n
(vi)
dn np [cos( ax + b )] = a n cos æç + ax + b ö÷ è2 ø dx n
(vii)
dn m! ( ax + b )m = a n( ax + b )m - n ( m - n)! dx n
(viii)
( -1)n - 1( n - 1)a n dn [log( ax + b )] = dx n ( ax + b )n
19. If u = f( x) and v = g ( x), then derivative of f( x) w.r.t. g ( x) is
21.
dy f ¢ ( x )log( x ) = dx g ¢ ( y) × {1 + log f( x )} 2
22. Let u( x) and v( x) be functions possessing derivatives upto nth
expression involving inverse trigonometric functions some substitutions are very helpful, which are listed below.
4.
dy f ¢( x) = dx 2 y - 1
dy g ( y) f ¢ ( x ) é f( x )log g ( y) - g ( y)ù = × dx f( x ) g ¢ ( y) êë g ( y)log f( x ) - f( x ) úû
for| x| > 1.
5.
f( x ) + K ¥ , then
(vi) If { f( x )} g ( y ) = { g ( y)} f( x ), then
16. In order to find differential coefficients of complicated
S. No.
f( x ) +
(v) If { f( x )} g ( y ) = e f( x ) - g ( y ), then
not
1 d d (sec -1 x ) = - (cosec -1 x ) = dx dx | x| x 2 - 1
1 + g ( x) 1 - g ( x) dy g ¢( x) , then = × dx [1 - g ( x )]2 1 + g ( x ) 1 - g ( x)
(iii) If y =
(ix) (a) (b)
d n ax [e sin( bx + c )] = r ne ax sin( bx + c + nf) dx n d n ax [e cos( bx + c )] = r ne ax cos( bx + c + nf) dx n a2 + b 2
where,
r=
and
b f = tan-1 æç ö÷ èaø
69
Here, we are introducing some important shortcut methods on relations and functions to save your precious time in tedious calculations based problems asked in IIT JEE and various engineering entrance exams. Use these shortcuts, to practice questions and be acquainted with it. Relation and function are very important chapter for JEE, point of view. This chapter is very conceptual. The following tips help you to crack the some particular type of problems in a very short span of time. So, students are advised to learn these tips and practice the examples based on each tip.
Sol.
(c) Given, A = { x : x is a natural number and x £ 10} \
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} n( A ) = 10 \ Number of reflexive relation = 2n
2
- n
2
= 2 (10)
- 10
= 2 100 - 10 = 2 90
4. Let A be a finite set containing n elements. Then, total 2
number of binary operations on A is nn .
1. If A and B are any two non-empty sets having n elements in common, then A ´ B and B ´ A have n2 elements in common.
Example If A = { x : x is first five even multiple of 3}, then the number of binary operations on A is
Example A and B are two sets having 3 elements in common. If n( A) = 8 and n[( A ´ B) Ç ( B ´ A)] is equal to (a) 9 Sol.
(b) 3
(c) 16
n( B) = 6,
(a) 5 25 (c) 510
then
(d) 20
Sol.
2. Let A and B be two non-empty finite sets consisting of m and n elements respectively, then total number of relation from A to B is 2mn .
2
5. Let A be a finite set containing n elements, then the total number of commutative binary operation on A is é n ( n + 1) ù ê ú. 2 ë û
then the total number of relation from A to B is Sol.
(b) 24
(c) 2 24
(d) None
number of commutative binary operation on set A is
B = {7, 8, 9, 10, 11, 12} n( A ) = 4 , n( B) = 6 \ Total number of relation from A to B = 2 4 ´ 6 = 2 24
relations from A to A is 2
Sol.
.
then the number of reflexive relation from set A to itself is (b) 2 65
(c) 2 90
(d) Given, A = { x : x Î N, 3 x - 5 £ 25} \
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} n( A ) = 10
Example If A = { x : x is a natural number and x £ 10},
(a) 2 100
(b) (10)100 (d) 55
(a) 50 (c) (100)100
3. If a set A has n elements, then number of reflexive
84
Example If A = { x : x Î N , 3 x - 5 £ 25}, then the total
(c) Given, A = { 3, 5, 7, 9}
n2 - n
A = { 6, 12, 18, 24, 30} n( A ) = 5
\ Total number of binary operation on A is ( 5)5 = ( 5)25
Example If A = {3, 5, 7, 9 } and B = {7, 8, 9, 10, 11, 12}, (a) 2 10
(a) Given, A = { x : x is first five even multiple of 3} \
(a) Since, A and B have 3 elements in common. \ n[( A ´ B) Ç ( B ´ A )] = 3 2 = 9
(b) 25 5 (d) None of these
(d) None
éQ 3 x - 5 £ 25ù ê x £ 10 úû ë
Total number of commutative binary operations 10(10 + 1) = 2 = 55
6. If a set A contains m elements and another set B
Sol.
Example If A = {2, 3, 5, 7 } and B = { 4, 8, 12, 16, 20}, then the total number of functions from A to B is (a) 225 Sol.
(b) 125
(c) Given,
(c) 625
(d) 256
A = {2, 3, 5, 7}, B = { 4, 8, 12, 16, 20}
n( A ) = 4, n( B) = 5 Total number of functions = ( 5)4 = 625
7. Let f : A ® B be a function such that A and B are finite
9. If A and B are finite sets and f : A ® B is a bijection, then A and B have the same number of elements. If A has n elements, then the number of bijections from A to B is n!. Note The term bijection means one-one and onto.
sets having m and n elements respectively (where, n > m). The number of one-one functions is ì nP , n ³ m ü n( n - 1)( n - 2) K ( n - m + 1) = í m ý n < mþ î 0,
Example If X = {1, 2, 3, 4, 5} and Y = { a, b, c, d, e, f } and f : X ® Y , then the total number of one-one functions is (a) 620 Sol.
(b) 240
(c) Given, X = {1, 2, 3, 4, 5}, Here,
(c) 720
Example If X = {1, 2, 3, 4 }, then the number of one-one onto mappings f : X ® X is (a) 25 Sol.
(d) 7776
(c) 36
(d) Given, X = {1, 2, 3, 4},
(d) 24
n( X ) = 4
10. If f ( x ) is periodic with period T, then kf ( ax + b) is periodic with period
6! = 6 ! = 720 1!
8. If a set A has m elements and another set B has n (i) Total number of constant functions = n
(a) 2 p Sol.
(ii) Total number of many-one functions ìnm - nPm , n ³ m =í m n
f : A ® B, then the total number of (i) constant function (ii) many-one function (iii) onto function (iv) into function, are respectively (b) 5, 505, 0, 625 (d) None of these
(c)
p 4
(d)
p 6
(d) Since, tan x is periodic with period p. p \ Period of tan 6 x is . 6
respectively, then we have h( x ) = f1( x ) + f2 ( x ) has period as ìLCM of {T1 , T2 }, if h( x )is not an even functions. ï =í 1 LCM of {T1 , T2 }, if f1( x )and f2 ( x ) îï 2
ì n m - nC1( n - 1) m + nC2 ( n - 2 ) m - nC3 ( n - 3 ) m + K , n < m ïï n! n=m =í 0, n >m ï ïî
Example If A = {2, 4, 6, 8 } and B = { a, e, i, o, u} and
(b) p
11. If f1( x ), f2( x ) are periodic functions with periods T1,, T2
(iii) Total number onto functions
(iv) Total number of into functions ìï nC ( n - 1)m - nC ( n - 2)m + nC ( n - 3)m - K , n £ m 2 3 =í 1 nm , n>m îï
T , where a, b, k Î R and a, k ¹ 0 | a|
Example The period of tan 6x is
elements and a function f : A ® B is defined, then
(a) 4, 200, 325, 625 (c) 5, 0, 225, 325
(b) 30
Total number of bijection from X to X = 4 ! = 24
Y = { a, b, c, d , e, f}
n( X ) = 5, n( Y ) = 6 n( Y ) > n( X )
Total number of one-one functions = 6P5 =
(b) Given, A = {2, 4, 6, 8}, n( A ) = 4( m) and B = { a, e, i , o, u}, n( B) = 5( n) Here, n( B) > n( A ) (i) Total number of constant functions is 5. (ii) Total number of many-one functions 5! = 625 – 120 = 505 = nm - nPm = ( 5)4 - 5P4 = 625 1! (iii) Total number of onto functions = 0 [Q n > m, i.e. 5 > 4] (iv) Total number of into functions = ( 5)4 = 625 [Q n > m]
contains n elements, then total number of functions = ( n)m .
are complementary pairwise comparable functions.
x Example The period of f ( x) = tan 3 x + sinæç ö÷ is è3 ø
(a) 2 p Sol.
(b) 2 p
2
2 (c) p 3
(d) 6p
(d) Since, period of tan x is p p \ Period of tan 3 x is½ ½. Period of sin x is 2 p. ½3½ x 3 \ Period of sin is½2 p ´ ½ = | 6p| 3 ½ 1½ p 6p is 6p. Thus, LCM of and 3 1 Thus, f( x ) is periodic with period 6p.
85
Example The period of f ( x) = sin4 x + cos4 x is (a) p Sol.
(b) 2 p
p (c) 2
(b) We know that, sin x has period 2 p. Then, 5 sin x + 4 is also periodic with period 2 p.
Sol.
(d) None
(c) sin4 x and cos 4 x both has period p. But f( x ) is an even function and sin x,cos x are complementary. 1 p Hence, f( x ) has period = [LCM of ( p, p )] = 2 2 p Thus, period of f( x ) is . 2
12. If f ( x ) is periodic with period T, then
13. sinn x, cos n x, sec n x and cosec nx are periodic with period 2p, when n is odd or fraction and periodic with period p, if n is even
Example The period of cos x is (a) p
i.e. if a constant is added, subtracted, multiplied or divided in a periodic function, its period remains the same.
Example The period of cot3 x is (a)
(b) 2 p p (d) 2
(c) 3p
p 2
(d)
p 4
14. The period of tann x, cot n x is p when n is even or odd
Example The period of f ( x) = 5 sin x + 4 is (a) p
(c)
(b) Since, period of cos x is 2 p. 1 is fraction. \ Period of cos x is 2 p as n = 2
Sol.
(a) C × f ( x ) is periodic with period T (b) f ( x + C ) is periodic with period T (c) f ( x ) ± C is periodic with period T (C is constant)
(b) 2 p
p 4
(b)
p 2
(c)
p 3
(d) p
(d) Since, period of cot x is p. \ Period of cot 3 x is p.
Sol.
ANSWERS OF BOARD EXAM CORNER (NOVEMER ISSUE) 1. Given a * b = HCF ( a, b ), a, b Î N Value of 20 * 6 = HCF ( 20, 6) = 2 3 2. Let cos -1 æç - ö÷ = q, where q Î[ 0, p ] è 5ø 9 4 3 = Þ cos q = - Þ sin q = 1 25 5 5 æ æ 3öö Þ sin ç 2cos -1 ç - ÷ ÷ Þ sin 2q è 5øø è 4 æ 3ö 24 Þ 2 ´ ´ ç- ÷ Þ 5 è 5ø 5
3. We know that,|adj A| = | A|n -1 , where n is order Þ 625 = | A|3 - 1 Þ | A| = 625 = 25 Þ | A¢| = 25 é 1 0 0ù é xù é 1 ù é xù é 1 ù 4. ê0 1 0ú êy ú = ê- 1ú Þ êy ú = ê- 1ú ê úê ú ê ú ê ú ê ú êë0 0 1úû êëz úû êë 0 úû êëz úû êë 0 úû Þ x = 1, y = - 1, z = 0 ( Q Equality of matrix) Þ x + y + z Þ 1 + ( - 1) + 0 Þ 0 5. f ( x) = [ x], - 4 < x < 4 f ( x) would be discontinuous at every integral point in the given interval. So, f ( x) is discontinuous at seven point. 1 6. Consider f ( x) = x Þ f ¢( x) = 2 x Let x = 0.0036 and Dx = 0.0001 Now, f ( x + Dx) @ f ( x) + Dx f ¢( x) 1 ´ Dx Þ x + Dx @ x + Dx ´ 2 x 0.0001 Þ 0.0036 + 0.0001 = 0.0036 + 2 0.0036 Þ 0.0037 @ 0.06083 7. Since, R is reflexive symmetric and transitive. Hence, R is an equivalence relation. 8. By LHS sin [cot -1 {cos (tan -1 x)}]
86
é ì ï = sin êcot -1 ícos ê îï ë é æ x2 = sin êsin -1 ç ç x2 ê è ë
æ çcos -1 ç è + 1 ö÷ù ú= + 2 ÷øúû
öüïù ÷ ú ý x + 1 ÷øþïúû 1
2
x2 + 1 x2 + 2
or
a a = q Þ cos q = b b æp q ö æp q ö By LHS tan ç + ÷ + tan ç - ÷ è4 2ø è4 2ø p q p q tan + tan tan - tan 4 2 + 4 2 = p q p q 1 - tan × tan 1 + tan × tan 4 2 4 2 2 2 2b = = = = RHS cos q a/ b a
Let
cos -1
9. Let cos -1 x = a, cos -1 y = b and cos -1 z = g Þ cos a = x, cos b = g and cos g = z From question a + b + g = p Þ a + b=p - g Þ cos ( a + b) = cos (p - g ) Þ cos a cos b - sin a sin b = - cos g Þ x × y - 1 - cos 2 a 1 - cos 2 b = - z Þ xy - 1 - x2 × 1 - y 2 = - z Þ x2 + y 2 + z 2 + 2 xyz = 1 or Let x = sin q Þ q = sin -1 x p then sin -1 ( 1 - sin q) - 2q = 2 p Þ sin -1 ( 1 - sin q) = + 2q 2 æp ö Þ 1 - sin q = sin -1 ç + 2q ÷ è2 ø After simplification put sin q = x we get, x = 0
- a+ b - a+c 3b c-b a-c b -c 3c Applying operation C1 ® (C1 + C 2 + C 3 ) a+ b +c - a+ b - a+c = a+ b +c - 3b c-b a+ b +c b -c 3c 3a
10. Let D = a - b
Taking out common factor ( a + b + c ) from C1 and again apply properties, we get the required result. - bc b 2 + bc c 2 + bc 11. Let D = a2 + ac - ac c 2 + ac 2 2 a + ab b + ab - ab Applying R1 ® aR1 , R2 ® bR2 , R3 ® cR3 - abc ab 2 + abc ac 2 + abc 1 D= - abc a2 b + abc bc 2 + abc abc 2 2 - abc a c + abc b c + abc Taking out common factors a, b, c from C1 , C 2 , C 3 respectively. Apply the properties we get the required result. a2 + 1 ab ac 12. By LHS let D = ab b 2 + 1 bc ac bc c2 + 1 Applying operation C1 ® aC1 , C 2 ® bC 2 and C 3 ® cC 3 a( a2 + 1) ab 2 ac 2 1 = a2 b b( b 2 + 1) bc 2 abc a2c b 2c c(c 2 + 1) Taking out common factors a, b, c from R1 , R2 , R3 , respectively and again apply properties, we get result. é5 4 3ù é xù é11000ù 13. (i) ê4 3 5ú êy ú = ê10700ú ê úê ú ê ú êë 1 1 1úû êëz úû êë 200 úû
Þ 5 x + 4y + 3z = 4 x + 3y + 5z = 5 4 (ii) Let A = 4 3 1 1
11000 10700, x + y + z = 2700 3 5 = - 3¹ 0 1
A is invertible, so equations have a unique solutions. (iii) Any values of three variables x, y, z which proper and logical will be consider correct. 14. The given function f ( x) is continuous at x = 1 lim f ( x) = lim f ( x) = f ( 1)
x ®1-
x ®1+
Now, LHL lim f ( x) = lim f ( 1 + h) = 3a + b x ®1-
h®0
RHL = lim f ( x) = lim f ( 1 + h) = 5a - 2b x ®1+
h®0
After simplification, we have a = 3, b = 2
15. Given, x = asin Þ x=
1 sin -1 t , a2
-1
t
, y = acos
y=
t
1 cos -1 t a2
Differentiating w.r.t. t, we get
16. Given,
-1
dy y =dx x
y = (tan -1 x)2
…(i)
On differentiating both sides w.r.t. x, we get 1
Þ y1 = 2 tan -1 x ×
1 + x2
Þ ( 1 + x2 )y1 = 2 tan -1 x On squaring both sides, we get Þ ( 1 + x2 )2 y12 = 4(tan -1 x)2 Again differentiating both sides, we get Þ ( 1 + x2 )2 y 2 + 2 x( 1 + x2 )y1 = 2
17. Given, f ( x) = x2 - 4, for x Î[ 2, 4] Clearly f ( x) is continuous in & differentiable in (2, 4). So, there must be exists some c Î( 2, 4). f ( 4) - f ( 2) Such that f ¢(c ) = ( 4 - 2) Therefore, c = 6 Î ( 2, 4) Hence, mean value theorem is verified. 18. Given, y = ( x)cos x + (cos x)sin x Let u = xcos x , v = (cos x)sin x , y = u + v Differentiating w.r.t. x, we get dy du dv …(i) = + dx dx dx Now, find du / dx, dv / dx and put in Eq. (i), we get dy ì cos x ü = xcos x í - (log x)sin xý dx î x þ + (cos x)sin x {- sin x tan x + cos x × log (cos x)}
19. Given equation of line is 5y - 15 x = 13 Here, m=3 \ Slope of the given line is 3. 1 Slope of tangent = - & equation of the curve. 3 …(i) y = x2 - 2 x + 7 dy = 2x - 2 Þ dx 1 5 217 and y = \ 2x - 2 = - Þ x = 3 6 36 So, the point on the given curve at which tangent is perpendicular to given line is æ 5 217 ö ç , ÷ and equation of the tangent is è 6 36 ø Þ 12 x + 36y - 227 = 0 20. Given, function f : R+ ® [ - 5, ¥)
defined by f ( x) = 9 x2 + 6 x - 5 Let y be any arbitrary element of ( - 5, ¥) y+6-1 y = 9 x2 + 6 x - 5 Þ x = 3 Therefore, f is onto and range of f = [ - 5, ¥) let us define g : ( - 5, ¥) ® R y+6-1 such that g ( y ) = 3 Now, ( gof )( x) = g ( f ( x)) = g ( 9 x2 + 6 x - 5) æ y + 6 - 1ö ÷=y and ( fog )( y ) = f ( g ( y )) = f ç ÷ ç 3 ø è Therefore, gof = I R + and fog = I[ - 5, ¥) Hence, f is invertible and the inverse of f, if given y+6-1 by f -1 ( y ) = g ( y ) = 3 21. The given equations are x - y + 2z = 1, 2y - 3z = 1, 3z - 2y + 4z = 2 é1 - 1 2 ù é xù é 1ù Let A = ê0 2 - 3ú, X = êy ú and B = ê 1ú ê ú ê ú ê ú êë3 - 2 4 úû êëz úû êë2úû then, the given system in matrix form is AX = B Simplify the matrix equation and use 1ù é 1 0 0ù é 1 -1 2ù é-2 0 ê0 2 -3ú ê 9 2 -3ú = ê0 1 0ú ê úê ú ê ú êë3 -2 4úû êë 6 1 -2úû êë0 0 1úû we get x = 0, y = 5, z = 3 1 1 1 22. Putting = u, = v and = w x y z 2u + 3v + 10w = 4, 4u - 6v + 5w = 1 6u + 9v - 20w = 2 Simplify and then x = 2, y = 3, z = 6 23. Let the first, second and third numbers be x, y, z respectively. Then, according to question, …(i) x+ y + z = 6 …(ii) x + 2z = 7 …(iii) 3 x + y + z = 12 é 1 1 1ù é xù é6ù Let A = ê 1 0 2ú, X = êy ú and B = ê 7 ú ê ú ê ú ê ú êë3 1 1úû êëz úû êë12úû Then, the given system in matrix form is AX = B Now simplify and then x = 3, y = 1, z = 2 24. We have, f ( x) = sin x + cos x f ¢( x) = cos x - sin x Now, f ¢( x) = 0 gives sin x = cos x p 5p as 0 £ x £ 2p Þ tan x = 1 Þ x = , 4 4 0
p 4
5p 4
Þ
2 x = 10 - r(p + 2) pr
25. Let radius of semi-circle = r \ One side of rectangle = 2r & let other side = x \ P (Perimeter of figure) = 10 m (given) 1 Þ 2 x + 2r + ( 2pr ) = 10 2
r
D
x
x
D
D 2r Let A be area of figure A = Area of semi-circle + Area of rectangle 1 A = (pr 2 ) + x ´ 2r 2 1 A = pr 2 + r ´ {10 - r(p + 2)} 2 1 = (pr 2 ) + 10r - r 2p - 2r 2 2 pr 2 …(i) A = 10r - 2r 2 2 Differentiate Eq. (ii) twice and use concept of maxima and minima. 20 10 Hence, length = , Breadth = p+4 p+4 or Let P be a point on the hypotenuse AC of right angled D ABC. A q
P
A
L
A C
q
B M Such that PL ^ AB, PL = a, PM ^ BC , PM = b Let ÐAPL = ÐACB = q AP = a sec q, PC = b cosec q Let l be the length of the hypotenuse. then, l = AP + PC p …(i) Þ l = asec q + b cosec q, 0 < q < 2 Differentiate Eq. (i) use concept of maxima and 1/ 3 æbö and the minima, l is minima at q = ç ÷ a è ø 2/ 3 2/ 3 3 minimum value is ( a +b ) .
26. Let AB be the line passing through P( 3, 4). Equation of any line through P( 3, 4) is y - 4 = m( x - 3)
…(i)
B
2p
é pö Þ f is strictly increasing in the intervals ê0, ÷ ë 4ø æ 5p ù and ç , 2p ú. è 4 û æ p 5p ö Also, f ¢( x) < 0, if x Î ç , ÷ è4 4 ø æ p 5p ö or f is strictly decreasing in ç , ÷. è4 4 ø
r
D
…(i)
P (3, 4)
O
A
X
4 ö æ This meets X-axis ( y = 0) at A ç 3 - , 0 ÷ m ø è and Y-axis ( x = 0) at B( 0, 4 - 3m) 4ö 1æ \ Area of DOAB is A = ç 3 - ÷( 4 - 3m) …(i) 2è mø Differentiate Eq. (i) twice and use concept of maxima and minima. 4 A is minimum, when m = - and the equation 3 of line is 4 x + 3y = 24.
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Quizzer (No.14) 1. If x is so small that x3 and higher power of x may be 3 éæ ù xö 3 /2 ê ç1 + ÷ - (1 + x) ú 2ø êè úû may be neglected, then - ë (1 - x)1/ 2
approximated as 3 2 x 8 3 (c) - x 2 + 1 8
(a) -
(b) (d)
3 2 x + 3x 8
minimum value of the expression æ (a4 + 3a2 + 1) (b4 + 5b2 + 1)(c4 + 7c2 + 1) ö is not ç ÷ a2 b2 c2 è ø divisible by the prime number (b) 2 (d) 11
3. Ramesh has x children by his first wife. Selina has (x + 1) children by her first husband. They marry and have children of their own. The whole family has 24 children. Assuming that two children of the same parents do not fight, then the maximum possible number of fights that can takes place is (a) 201 (c) 180
(b) 181 (d) 191
x 3 ï x ï x+ 3 í x x+ ï 3 x + x ... ¥ ïî
4. The value of lim ì x ®¥
(a) 1 (c) 3
(b) 2 (d) None of these
1ù é 2ù é f(x) = [ x] + ê x + ú + ê x + ú is discontinuous for 3 3û ë û ë x Î(0, 3) is (since,[ × ] denotes greatest integer function) (a) 7
(b) 8
(c) 10
(d) 6
6. The base AB of a DABC are fixed and the vertex C
3 2 x ×x + x 2
2. If a, b, c are non-zero real numbers, then the
(a) 3 (c) 5
5. Number of points, where
lies on a fixed circle of radius r. Lines through A and B are drawn to intersect CB and CA respectively at E and F such that CE : EB = 1 : 2 and CF : FA = 1 : 2. If the point of intersection P of these lines lies on the median through AB for all positions of AB, then the locus of P is (a) a circle of radius 2 r (b) a rectangular hyperbola (c) a parabola of latusrectum 4r r (d) a circle of radius 2
7. If 1, a, a2 , a3 , ..., an - 1 are n, nth roots of unity, then the
of (2011 - a)(2011 - a2 )(2011 - a3 )
K (2011 - a n - 1) is (2010)n - 1 2011 (2010)n + 1 (c) 2011
(2011)n + 1 2010 (2011)n - 1 (d) 2010
(a)
(b)
ìï tan[ x] , x ¹0 , g(x) = í x ïî 1, x =0 log2 ( x + 3 ) , then in [0, 1] Lagrange’s h(x) = {x}, k(x) = 5
8. Given ü is ï ï ý ï ïþ
value
æ1 ö f(x) = 4 - ç - x ÷ è2 ø
2 /3
mean value theorem is applicable to the (where, [×] and {×} represent greatest integer function and fractional part function, respectively.) (a) g
(b) k
(c) h
(d) f
95
9. Consider the function defined implicitly by the 2
equation y - 2ye
sin -1 x
2
2 sin -1 x
+ x - 1 + [ x] + e
=0
(where, [ x] denotes the greatest integer function). Then, the area of the region bounded by the curve and the line x = - 1 is æp ö (a) ç - 1÷ sq unit è2 ø
æp ö (b) ç + 1÷ sq unit è2 ø
(c) ( p + 1) sq unit
(d) ( p - 1) sq unit
10. Given a function ‘g’ which has a derivative g¢(x) for every real x and which satisfies g¢ (0) = 2 and g(x + y) = e y g(x) + e x g(y) for all x and y. Then, the range of the function g(x) is (a) (2, ¥) (c) [e, ¥)
é 2 ö (b) ê - , ¥÷ ë e ø (d) None of these
KNOWLEDGE Coefficient Quizzer (No. 14)
Winner of Knowledge Coefficient Quizzer (No. 13) (November Issue) Kavita Sharma (Gonda)
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Mathematics Spectrum Arihant Media Promoters C/O Arihant Prakashan Kalindi, TP Nagar, Meerut (UP)-250002
