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Chief Advisor
Rashmi Krishnan, IAS
Director, SCERT
Guidance
Dr. Anita Setia, Additional Director, SCERT
Dr. Pratibha Sharma
Joint Director, SCERT-cum-state pedagogy coordinator
Academic Co-ordinator
Dr. Anil Kumar Teotia
Sr. Lecturer, DIET Dilshad Garden
Mr. Sanjay Kumar
Lecturer, SCERT
Contributors
Dr. Anup Rajput
Dr. Anil Kumar Teotia
Dr. Satyavir Singh
Mr. D.R. Sharma
Dr. R.P. Singh
Mr. Sanjeev Kumar
Mr. Ashutosh Kr. Agarwal
Associate Professor, NCERT
Sr. Lecturer, DIET Dilshad Garden
Principal, SN Inter College, Pilana
Vice Principal, Navodaya Vidhalya, Mungeshpur
Lecturer, RPVV, Gandhi Nagar
Lecturer, RPVV, Raj Niwas Marg
Lecturer, RPVV, Nand Nagri
Editors
Dr. Anil Kumar Teotia
Sr. Lecturer, DIET Dilshad Garden
Dr. Kusum Bhatia
Sr. Lecturer, DIET, Pitampura
Publication Officer
Mr. Mukesh Yadav
Published by : State Council of Educational Research & Training, New Delhi and printed at
Educational Stores, S-5, Bsr. Road Ind. Area, Ghaziabad (U.P.)
Contents
S.No.
Chapter Name
Page No.
Syllabus for Class-XI
5
Syllabus for Class-XII
8
Question Paper-Delhi (2012)
11
Question Paper-Outside Delhi (2012)
23
Question Paper-Outside India (2012)
26
1.
Relations and Functions
30
2.
Inverse Trigonometric Functions
35
3.
Matrices and Determinants
42
4.
Continuity and Differentiability
49
5.
Application of Derivative
61
6.
Integrals
71
7.
Application of Integrals
81
8.
Differential Equations
84
9.
Vectors
91
10.
Three Dimensional Geometry
99
11.
Probability
107
12.
Linear Programming
114
MATHEMATICS (Code No 041)
The Syllabus in the subject of Mathematics has undergone changes from time to time in accordance
with growth of the subject and emerging needs of the society. Senior Secondary stage is a launching
stage from where the students go either for higher academic education in Mathematics or for professional
courses like engineering, physical and Bioscience, commerce or computer applications. The present
revised syllabus has been designed in accordance with National Curriculum Frame work 2005 and as
per guidelines given in Focus Group on Teaching of Mathematics 2005 which is to meet the emerging
needs of all categories of students. Motivating the topics from real life situations and other subject
areas, greater emphasis has been laid on application of various concepts.
Objectives
The broad objectives of teaching Mathematics at senior school stage intend to help the pupil:
to acquire knowledge and critical understanding, particularly by way of motivation and visualization,
of basic concepts, terms, principles, symbols and mastery of underlying processes and skills.
to feel the flow of reasons while proving a result or solving a problem.
to apply the knowledge and skills acquired to solve problems and wherever possible, by more
than one method.
to develop positive attitude to think, analyze and articulate logically.
to develop interest in the subject by participating in related competitions.
to acquaint students with different aspects of mathematics used in daily life.
to develop an interest in students to study mathematics as a discipline.
to develop awareness of the need for national integration, protection of environment, observance
of small family norms, removal of social barriers, elimination of sex biases.
to develop reverence and respect towards great Mathematicians for their contributions to the field
of Mathematics.
Syllabus for Class-XI
Units
I.
II.
III.
IV.
V.
VI.
Marks
Sets and Functions
Algebra
Coordinate Geometry
Calculus
Mathematical Reasoning
Statistics and Probability
29
37
13
06
03
12
Total
100
UNIT I : SETS AND FUNCTIONS
1. Sets :
(Periods 12)
Sets and their representations. Empty set. Finite and Infinite sets. Equal sets. Subsets. Subsets of the
set of real numbers especially intervals (with notations). Power set. Universal set. Venn diagrams. Union
and Intersection of sets. Difference of sets. Complement of a set. Properties of complement sets.
2. Relations and Functions :
(Periods 14)
Ordered pairs, Cartesian product of sets. Number of elements in the cartesian product of two finite
sets. Cartesian product of the reals with itself (upto R × R × R).
Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. Function as a
special kind of relation from one set to another. Pictorial representation of a function, domain,
co-domain & range of a function. Real valued functions of the real variable, domain and range
of these functions, constant, identity, polynomial, rational, modulus, signum and greatest integer
functions with their graphs. Sum, difference, product and quotients of functions.
3. Trigonometric Functions :
(Periods 15)
Positive and negative angles. Measuring angles in radians and in degrees and conversion from
one measure to another. Definition of trigonometric functions with the help of unit circle. Truth
of the identity sin2 x + cos2 x = 1, for all x. Signs of trigonometric functions and sketch of their
graphs. Expressing sin (x ± y) and cos (x ± y) in terms of sin x, sin y, cos x & cos y. Deducing
the identities like following :
5
tan x tan y
cot x cot y 1
, cot( x y )
,
1 tan x tan y
cot y cot x
x y
x y
x y
x y
cos
, cos x cos y 2 cos
cos
,
sin x sin y = 2sin
2
2
2
2
x y
x y
x y
x y
sin
, cos x cos y
2sin
sin
.
sin x sin y = 2 cos
2
2
2
2
Identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x. General solution of trigonometric
equations of the type sin = sin , cos = cos and tan = tan . Proof and simple application
of sine and cosine formulae.
tan( x
y) =
UNIT 2 : ALGEBRA
1. Principle of Mathematical Induction :
(Periods 6)
Processes of the proof by induction, motivating the application of the method by looking at natural
numbers as the least inductive subset of real numbers. The principle of mathematical induction
and simple applications.
2. Complex Numbers and Quadratic Equations :
(Periods 10)
Need for complex numbers, especially,
1 , to be motivated by inability to solve every some
of the quadratic equations. Algebraic properties of complex numbers. Argand plane and polar
representation of complex numbers. Statement of Fundamental Theorem of Algebra, solution of
quadratic equations in the complex number system. Square-root of a complex number.
3. Linear Inequalities :
(Periods 10)
Linear inequalities. Algebraic solutions of linear inequalities in one variable and their representation
on the number line. Graphical solution of linear inequalities in two variables. Solution of system
of linear inequalities in two variables-graphically.
4. Permutations and Combinations :
(Periods 12)
Fundamental principle of counting. Factorial n. (n!) Permutations and combinations, derivation
of formulae and their connections, simple applications.
5. Binomial Theorem :
(Periods 8)
History, statement and proof of the binomial theorem for positive integral indices. Pascal’s triangle,
general and middle term in binomial expansion, simple applications.
6. Sequence and Series :
(Periods 10)
Sequence and Series. Arithmetic progression (A.P.), arithmetic mean (A.M.). Geometric progression
(G.P.), general term of a G.P., sum of n terms of a G.P., Arithmetic and geometric series, infinite
G.P. and its sum, geometric mean (G.M), relation between A.M. and G.M. Sum to n terms of the
special series :
n
n
k,
k 1
n
2
k 3.
k , and
k 1
k 1
6
UNIT 3 : COORDINATE GEOMETRY
1. Straight Lines :
(Periods 9)
Brief recall of 2D from earlier classes. Shifting of origin. Slope of a line and angle between two
lines. Various forms of equations of a line : parallel to axes, point-slope from, slope-intercept form,
two-point form, intercepts form and normal form. General equation of a line. Equation of family
of lines passing through the point of intersection of two lines. Distance of a point from a line.
2. Conic Sections :
(Periods 12)
Sections of a cone : circle, ellipse, parabola, hyperbola, a point, a straight line and pair of intersecting
lines as a degenerated case of a conic section. Standard equations and simple properties of parabola,
ellipse and hyperbola. Standard equations of a circle.
3. Introduction to Three-dimensional Geometry :
(Periods 8)
Coordinate axes and coordinate planes in three dimensions. Coordinates of a point. Distance between
two points and section formula.
UNIT 4 : CALCULUS
I.
Limits and Derivatives :
(Periods 18)
Limit of a function. Derivative of a function introduced as rate of change both as that of distance
x
function and its geometric meaning. lim log e (1 x) , lim e 1 . Definition of derivative, relate it
x 0
x 0
x
x
to slope of tangent of the curve, derivative of sum, difference, product and quotient of functions.
Derivatives of polynomial and trigonometric functions.
UNIT 5 : MATHEMATICAL REASONING
I.
Mathematical Reasoning :
(Periods 8)
Mathematically acceptable statements. Connecting words/phrases — consolidating the understanding
of “if and only if (necessary and sufficient) condition”, “implies”, “and/or”, “implied by”, “and”,
“or”, “there exists” and their use through variety of examples related to real life and Mathematics.
Validating the statements involving the connecting words-difference between contradiction, converse
and contrapositive.
UNIT 6 : STATISTICS AND PROBABILITY
1. Statistics :
(Periods 10)
Measure of dispersion; mean deviation, variance and standard deviation of ungrouped/ grouped
data. Analysis of frequency distributions with equal means but different variances.
2. Probability :
(Periods 10)
Random experiments : outcomes, sample spaces (set representation). Events : occurrence of events,
‘not’, ‘and’ & ‘or’ events, exhaustive events, mutually exclusive events. Axiomatic (set theoretic)
probability, connections with the theories of earlier classes. Probability of an event, probability
of ‘not’, ‘and’ & ‘or’ events.
7
Syllabus for Class-XII
Units
1.
II.
III.
IV.
V.
VI.
Marks
Relations and Functions
Algebra
Calculus
Vectors And Three-dimensional Geometry
Linear Programming
Probability
10
13
44
17
06
10
Total
100
UNIT-I: RELATIONS AND FUNCTIONS
1. Relations and Functions :
(10 Periods)
Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto
functions, composite functions, inverse of a function, Binary operations.
2.
Inverse Trigonometric Functions:
(12 Periods)
Definition, range, domain, principal value branches. Graphs of inverse trigonometric functions.
Elementary properties of inverse trigonometric functions.
UNIT-11: ALGEBRA
1. Matrices:
(18Periods)
Concept, notation, order, equality, types of matrices, zero matrix, transpose of a matrix, symmetric
and skew symmetric matrices. Addition, multiplication and scalar multiplication of matrices, simple
properties of addition, multiplication and scalar multiplication. Non-commutativity of multiplication
of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square
matrices of order 2). Concept of elementary row and column operations. Invertible matrices and
proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).
8
2. Determinants:
(20 Periods)
Determinant of a square matrix (up to 3 x 3 matrices), properties of determinants, minors, cofactors
and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square
matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples,
solving system of linear equations in two or three variables (having unique solution) using inverse
of a matrix.
UNIT-III: CALCULUS
1. Continuity and Differentiability :
(18 Periods)
Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inverse
trigonometric functions, derivative of implicit function. Concept of exponential and logarithmic
functions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, Derivative
of functions expressed in parametric forms. Second order derivatives. Rolle’s and Lagrange’s Mean
Value Theorems (without proof) and their geometric interpretations.
2. Application of Derivatives:
(10 Periods)
Applications of derivatives : rate of change of bodies, increasing/decreasing functions, tangents
and normals, use of derivatives in approximation, maxima and minima (first derivative test motivated
geometrically and second derivative test given as a provable tool). Simple problems (that illustrate
basic principles and understanding of the subject as well as real-life situations).
3. Integrals:
dx
x
2
a
2
(20 Periods)
dx
,
x
dx
,
ax bx c
2
( px q )
dx
ax 2 bx c
2
a
2
dx
,
a
2
x2
,
dx
ax
2
and
bx c
( px q ) ax 2 bx c dx
be evaluated.
Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof). Basic
properties of definite integrals and evaluation of definite integrals.
4. Applications of the integrals :
(10 Periods)
Applications in finding the area under simple curves, especially lines, areas of circles/parabolas/
ellipses (in standard form only), area between the two above said curves (the region should be
clearly identifiable).
5. Differential Equations :
(10 Periods)
Definition, order and degree, general and particular solutions of a differential equation. Formation
of differential equation whose general solution is given. Solution of differential equations by method
9
of separation of variables, homogeneous differential equations of first order and first degree. Solutions
of linear differential equation of the type :
dy
dx
Py
Q, where P and Q are functions of x or constants.
dx
dy
Px
Q, where P and Q are function of y or constants.
UNIT-IV: VECTORS AND THREE-DIMENSIONAL GEOMETRY
1. Vectors:
2.
(12Periods)
Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios
of vectors. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of
a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector
by a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) product
of vectors, projection of a vector on a line. Vector (cross) product of vector, scalar triple product.
Three - dimensional Geometry:
(12 Periods)
Direction consines and direction ratios of a line joining two points. Cartesian and vector equation
of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation
of a plane. Angle between (i) two lines, (ii) two planes. (iii) a line and a plane. Distance of a
point from a plane.
UNIT-V: LINEAR PROGRAMMING
1. Linear Programming :
(12 Periods)
Introduction, definition of related terminology such as constraints, objective function, optimization,
different types of linear programming (L.P) problems, mathematical formulation of L.P. problems,
graphical method of solution for problems in two variables, feasible and infeasible regions, feasible
and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).
UNIT- VI: PROBABILITY
Probability:
(18 Periods)
Multiplication theorem on probability. Conditional probability, independent events, total probability,
Baye’s theorem, Random variable and its probability distribution, mean and variance of random
variable. Repeated independent (Bernoulli) trials and Binomial distribution.
1 0
Question Paper-Delhi (2012)
General Instructions:
(i) All questions are compulsory.
(ii) The question paper consists of 29 questions divided into three Sections A, B and C, Section
A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four
marks each and Section C comprises of 7 questions of six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact
requirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 4 questions of four
marks each and 2 questions of six marks each. You have to attempt only one of the alternatives
in all such questions.
(v) Use of calculators is not permitted.
SECTION-A
Questions numbers 1 to 10 carry 1 mark each.
Q1. If a line has direction ratios 2, –1, –2, then what are its direction cosines?
1
Q2. Find ‘ ’ when the projection of a
1
Q3. Find the sum of the vectors a
i
j 4k on b
i 2j k, b
2i 6 j 3k is 4 units.
2i 4 j 5k and c i 6 j 7k .
1
3
Q4. Evaluate :
1
dx
x
2
1
Q5. Evaluate
(1 x) x dx.
1
5 3 8
Q6. If
Q7. If
= 2 0 1 , write the minor of the element a23.
1 2 3
2 3
5 7
1
2
Q8. Simplify : cos
3
4
cos
sin
1
4 6
, write the value of x.
9 x
sin
cos
Q9. Write the principal value of cos
sin
1
1
2
sin
cos
2sin
1
cos
sin
1
1
.
2
Q10. Let * be a ‘binary’ operation on N given by a * b = LCM (a, b) for all a, b
1 1
1
1
N. Find 5 * 7.
1
SECTION-B
Question numbers 11 to 22 carry 4 mark each.
Q11. If (cos x)y = (cos y)x, find
dy
.
dx
4
OR
dy sin 2 (a y )
.
dx
sin a
Q12. How many times must a man toss a fair coin, so that the probability of having at least one head
is more than 80%?
4
Q13. Find the Vector and Cartesian equations of the line passing through the point (1, 2, –4) and
If sin y = x sin (a + y), prove that
perpendicular to the two lines
x 8
3
y 19
16
z 10
x 15
and
7
3
y 29
8
z 5
.
5
Q14. If a, b, c are three vectors such that a = 5, b = 12 and c = 13, and a b c
value of a . b b . c c . a.
4
0, find the
4
Q15. Solve the following differential equation :
4
dy
2 xy y 2 0.
dx
Q16. Find the particular solution of the following differential equation.
4
2x2
dy
dx
1 x2
Q17. Evluate :
y2
x 2 y 2 , given that y = 1 where x = 0.
sin x sin 2 x sin 3x dx
OR
Evaluate :
2
dx
(1 x)(1 x 2 )
Q18. Find the point on the curve y = x3 – 11x + 5 at which the equation of tangent is y = x – 11.
OR
Using differentials, find the approximate value of
49.5 .
4
Q19. If y = (tan 1 x) 2 , show that
4
d2y
dy
( x 1)
2 x( x 2 1)
2.
2
dx
dx
Q20. Using properties of determinants, prove that
6
2
2
b c
q r
y
z
a
p
x
c a
a b
r p
p q
z
x
x
y
2b
c
q
r
y
z
1 2
Q21. Prove that tan
1
Prove that sin
1
cos x
1 sin x
OR
8
17
sin
4
1
3
5
x
,x
2
, .
2 2
cos
6
36
.
85
1
Q22. Let A = R – {3} and B = R – {1}. Consider the function f : A
Show that f is one-one and onto and hence find f –1.
B defined by f ( x)
x 2
.
x 3
6
SECTION-C
Questions numbers 23 to 29 carry 6 mark each.
Q23. Find the equation of the plane determined by the points A(3, –1, 2), B (5, 2, 4) and C(–1, –1, 6)
and hence find the distance between the plane and the point P(6, 5, 9).
6
Q24. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars
(not residing in hostel). Previous year results report that 30% of all students who reside in hostel
attain ‘A’ grade and 20% of day scholars attain ‘A’ grade in their annual examination. At the
end of the year, one student is chosen at random from the college and he has an ‘A’ grade, what
is the probability that the student is a hostlier?
6
Q25. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours
on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine
B to produce a package of bolts. He earns a profit of ` 17.50 per package on nuts and ` 7
per package of bolts. How many packages of each should be produced each day so as to maximize
his profits if he operates his machines for at the most 12 hours a day? Form the above as a
linear programming problem and solve it graphically.
6
4
Q26. Prove that
( tan x
0
cot x ) dx
2.
6
2
OR
3
(2 x 2
Evaluate
5 x ) dx as a limit of a sum.
1
Q27. Using the method of integration, find the area of the region bounded by the lines 3x – 2y + 1 = 0,
2x + 3y – 21 = 0 and x – 5y + 9 = 0.
6
Q28. Show that the height of a closed right circular cylinder of given surface and maximum volume,
is equal to the diameter of its base.
6
Q29. Using matrices, solve the following system of linear equations :
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12
OR
Using elementary operations, find the inverse of the following matrix :
1 1 2
1 2 3
3 1 1
1 3
Marking Scheme
Class-XII
Mathematics (March 2012)
Q.No.
Value Pooints/Solution
65/1//1
Marks.
SECTION-A
1-10
1.
2 1 2
, ,
3 3 3
6. M 2,3
7
2.
= 5
7. 13
3.
4j k
4. log
8.
1 0
0 1
9.
3
2
2
3
5.
3 32
x
2
10. 35.
2 52
x
5
c
1×10 = 10
SECTION-B
11.
(cos x) y
(cos y ) x
y log cos x
1/2
x log cos y
( sin x)
dy
( sin y ) dy
log cos x.
x
cos x
dx
cos y dx
dy
(log cos x x tan y )
log cos y y tan x
dx
y.
log cos y
1+1
1
log cos y y tan x
dy
=
log cos x x tan y
dx
1/2
OR
sin y
dy
dx
x
x sin( a
cos y
sin(a y )
cos y x cos(a
sin y
sin(a y )
dy
dx
y)
sin(a
dy
dx
dy
dx
x cos( a
y)
dy
sin( a
dx
1
1
y)
sin(a y )
sin y
cos y
.cos(a
sin(a y )
sin 2 (a y )
y ) cos y cos(a
y)
y ) sin y
y)
sin 2 (a y )
sin a
1 4
1
1
12.
Let the coin be tossed n times
80
100
P(getting at least one heat) >
8
10
1 P(0)
1
C0
2
n = 3.
0
1
2
n
13.
8
10
P(o) 1
n
1
1
or n
5
2
2
10
1
1
5
1
or
5
1
2n
5
1
1
Let the vector equation of required line be a
a
b
than
a = i 2 j 4k
and
b = (3i 16 j 7k ) (3i 8 j 5k )
1
= 24i 36 j 72k
1
Vector equation of line is
r = (i 2 j 4k )
(24i 36 j 72k )
or
14.
r = (i 2 j 4k )
x 1 y 2 z 4
and cartesian from is
2
3
6
2
a b c 0
(a b c) 0
a
2
b
or | a |2
2
c
2
| b |2
2(a.b b.c c.a )
| c |2
15.
2 x2
dy
2 xy
dx
Putting
v x
2
2x
y
y
x
dv
dx
dv
v2
y2
dy
dx
2 xy y
2x2
v so that y = vx and
1 2
v
2
v
x
dx
x
log x c
2
v
y
dv
dx
1
1/2
1
0
1
1
(25 144 169)
2
0
}
1
0
2( a.b b.c c.a )
a.b b.c c.a =
(2i 3 j 6k )
2
dy
dx
169 .
2
y2
x2
y
x
1
1/2
2
v x
1 2
v
2
dv
dx
1
1/2
log x c
1
2x
log x c
1
1 5
16.
dy
dx
1 x2
y2
dy
1 y2
tan 1 y
tan y
17.
I =
(1 x 2 )(1 y 2 )
½
(1 x 2 )dx
x
x = 0, y = 1
1
x2 y2
x
1
x3
c
3
c = /4
x3
3
4
or
sin x sin 2 x sin 3 xdx
1
1
y
tan
4
x
x3
3
1
2sin 3 x sin x.sin 2 xdx
2
=
1
(cos 2 x cos 4 x) sin 2 xdx
2
=
1
1
sin 4 xdx
2 cos 4 x sin 2 xdx
4
4
1
(sin 2 x cos 2 x cos 4 x sin 2 x) dx
2
½
½
1
=
1
1
cos 4 x
(sin 6 x sin 2 x) dx
16
4
1
=
1
cos 4 x
16
1
2
(1 x)(1 x 2 )
2
A
1 x
1
1
cos 6 x
cos 2 x c
24
8
OR
Bx C
1 x2
½
A(1 x 2 ) (Bx C)(1 x)
1½
0 = A – B, B – C = 0 A + C = 2
2
dx =
(1 x)(1 x 2 )
=
18.
½
1
1 x
dx
log |1 x |
A= B = C = 1
x 1
dx
x2 1
1 2
( x 1) tan 1 x c
2
Slope of tangent, y = x – 11 is 1
dy
y x 3 11x 5
3 x 2 11
dx
1½
½
½
If the point is (x1, y1) then 3 x12 11 1
x1
x1 = 2 then y1
2 then y1 19
8 22 5
½
9 and if x1
2
Since (–2, 19) do not lie on the tangent y = x – 11
Required point is (2, – 9)
1 6
1
1
½
½
OR
Let y =
y
x
dy
x
dx
y
x
1
2 x
x
. x
1
(0.5)
2 49
49.5
19.
y
7
(tan 1 x) 2
(1 x 2 )
dy
dx
(1 x 2 )
d2y
dx 2
x
1
28
½
x
x
x
Putting x = 49 and
49
y
x
1
x = 0.5 we get
1
49.5
½
7.0357
1
dy
dx
2 tan 1 x
1
1 x2
2 tan 1 x
2 x.
dy
dx
2
1 x2
d2y
dy
(1 x ) . 2 2 x(1 x 2 )
2.
dx
dx
Using R1
R1 + R2 + R3 we get
2 2
20.
2(a b c) 2( p q r ) 2( x y z )
c a
r p
z x
LHS =
a b
p q
x y
a b c
= 2 c a
a b
p q r
x
r p
p q
a b c
= 2
b
c
p q r
q
r
a
p
x
= 2b
c
q
r
y
z
Using
= RHS
1 7
y
z
x
x
1
z
1
x
y
y z
y
z
R1
R2
R3
Using R2
R3
R1 + R2 + R3
– R2
– R3
R2 – R 1
R3 – R 1
1
1
21.
tan
1
2sin
= tan
=
4
sin
cos x
1 sin x
1
tan
1 cos
x
cos
4 2
4
x
2 cos 2
4 2
x
2
1
1
x
2
x
2
tan
1
tan
x
2
4
1+1
x
2
1
OR
Writing sin
1
LHS = tan
Getting tan
22.
1
8
17
1
tan
1
8
and sin
15
8
3
tan 1
15
4
77
36
cos
1
tan
1
36
85
f ( x2 )
1
3
5
tan
8
15
3
4
8 3
1
,
15 4
1
3
4
tan
1
1
77
36
1
A and f ( x1 )
x2 2
x1 x2 2 x2 3x1 x1 x2 2 x1 3x2
x2 3
x1 = x2
Hence f is 1 – 1
x 2
y f ( x)
y
xy 3 y x 2
Let y B,
x 3
3y 2
or x =
y 1
3y 2
3
x A
Since y 1 and
y 1
Hence f is ONTO
3y 2
1
and f ( y )
y 1
Let x1 , x2
x1 2
x1 3
1+1
½
1
½
1
1
SECTION-C
23.
Normal to the plane is n
½
AB BC
n =
i
2
6
j k
3 2
3 2
1 8
12i 16 j 12k
1½
Equation of plane is
r.(12i 16 j 12k ) = (3i
= 76
or
r.(3i 4 j 3k ) = 19
j 2k ).(12i 16 j 12k )
2
or 3x – 4y + 3z – 19 = 0 }
Distance of plane from the point P(6, 5, 9) is
18 20 27 19
d =
24.
6
34
9 16 9
Let E1 : selected student is a hostlier
E2 : selected student is a day scholar
A : selected student attain ‘A’ grade in exam. }
P(E1) =
60
,
100
P(E2) =
40
100
30
20
, P(A/E2) =
100
100
P(E1 ) . P(A/E1 )
P(E1/A) =
P(E1 ) . P(A/E1 ) + P(E 2 ) . P(A/E 2 )
P(A/E1) =
1
1
1
1
60
25.
. 30
9
100
100
=
40 20 13
60
30
.
100 100 100 . 100
Let x package of nuts and y package of bolts be produced each day
LPP is
maximise P = 17.5x + 7y
2
x + 3y
12
3x + y
12
correct graph
x 0, y 0 }
vertices of feasible region are A(0, 4), B (3, 3), C (4, 0)
Profit is maximum at B(3, 3)
i.e. 3 package of nuts and 3 package of bolts
1+1
1
subject to
1 9
2
1
4
26.
4
( tan x
I =
cot x )dx
0
0
Putting sin x cos x
sin x cos x
dx
sin x cos x
t , to get (cos x sin x) dx
sin x cos x =
and
0
I =
2
1
dt
1 t
1
1
dt
1 t2
2
1
=
2.[sin 1 t ]0 1
=
2(sin 1 0 sin 1 ( 1)
2
1+1
2.
1
2
OR
3
(2 x 2
I =
1
where
5 x ) dx = lim h[ f (1)
h 0
f (1 h)
2
f ( x) = 2 x
f (1 2h) ...
f (1 n 1h)]
2
or nh = 2.
n
5 x and h
1
f(1) = 7
f(1 + h) = 2(1 h) 2 5(1 h)
f (1 2h) = 2(1 2h) 2
7 9h 2 2
5(1 2h)
7 18h 222 h 2
f (1 3h) = 2(1 3h) 2 5(1 3h)
7 27 h 2.32 h 2
2
...........
f (1 (n 1)h) = 7 9(n 1)h 2.( n 1) 2 h 2
h 7 n 9h
I = lim
h 0
27.
2h 2 .
n(n 1)(2n 1)
6
9
1
nh(nh h)
nh(nh h)(2nh h)
2
3
7 nh
= lim
h 0
= 14 18
n(n 1)
2
16
3
112
3
6
0 correct figure : 1
1½
6
1
1
1
(3 x 1)dx
(21 2 x) dx
( x 9) dx
area of ( ABC) =
21
33
51
2 0
1
1
Let AB be 3 x 2 y 1 0, BC be 2 x 3 y 21 0 and AC be x 5 y 9
Solving to get A(1, 2), B (3, 5) and C(6, 3)
3
1
1
3
1
(3 x 1) 2
=
12
V=
=
3
( x 9) 2
10
6
1½
1
½
13
sq. U.
2
½
2
Surface area A = 2 rh 2 r
A 2 r2
h=
2 r
28.
6
25
2
= 7 12
=
1
(21 2 x) 2
12
r 2h
(Given)
½
...(1)
A 2 r2
2 r
r2
1
1
.[Ar 2 r 3 ]
2
½
dv
1
2
= [A 6 r ]
2
dr
dv
=0=
dr
29.
6 r2
A
4 r2
2 rh
1
2 rh 2 r 2
1
4
2
5
2
1
3
a11 = 7,
a21 = 1,
a31 = –3,
1
h = 2r = diameter
1
h
1
1
d 2v
[ 12 r ] 0
2 =
2
dr
Given equations can be written as
1
3
2r will give max. volume.
7
x
y = 5 or AX = B
12
z
a12 = –19
a22 = – 1
a32 = 11
1
1
a13 = – 11
a23 = – 1
a33 = 72
A
–1
x
1
=
4
1
y =
4
z
7
19
1
3
1 11
11
7
1
1
19
11
½
7
3
7
2
1 11
1 7
5
12
1
3
x = 2, y = 1, z = 3.
1½
2 1
OR
Let A =
1 1 2
1 2 3
3
c1
c2
1 0 0
2 3 =A 0 1 0
0 0 1
1 1
1 1 2
Writing
1 1
1
3
0 1 0
1 2
1 3 = A 1 0 0
0 0 1
3 1
1
2
1
1 0
c2
c2 + c1
c3
c3 –2c1
2 3
1 4
c1
c1 + 2c3
1
0
c2
c2 + 2c3
0
1
1 2
1
½
0 1
0
1 = A 1 1
0 0
1
2
0
1 = A
0
1
0 0
1
1
0
1
0
3
2
3
2
2
1
0
1
1
c3
c3 + –c2
0 1 0 = A
1 2 1
3
2
3
2
5
3
c1
c1 + c3
1 0 0
c2
c2 + 2c3
0 1 0 = A
0 0 1
1
8
1
7
1
5
5
4
3
A–1 =
1
1
8
1
7
1
5
5
4
3
2 2
½
½
1
1
QuestionPaper-OutsideDelhi(2012)
General Instructions:
(i) All questions are compulsory.
(ii) The question paper consists of 29 questions divided into three Sections A, B and C, Section
A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four
marks each and Section C comprises of 7 questions of six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact
requirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 4 questions of four
marks each and 2 questions of six marks each. You have to attempt only one of the alternatives
in all such questions.
(v) Use of calculators is not permitted.
SECTION-A
Questions numbers 1 to 10 carry 1 mark each.
Q1. The binary operation * : R × R
R is defined as a * b = 2a + b. Find (2 * 3) * 4.
Q2. Find the principal value of tan 1 3 sec 1 ( 2) .
Q3. Find the value of x + y from the following equation :
2
x
7
5
y 3
3
1
4
2
7 6
15 14
3 4
1 2 1
Q4. If A = 1 2 and B =
, then find AT – BT.
1 2 3
0 1
Q5. Let A be a square matrix of order 3 × 3. Write the value of |2A|, where |A| = 4.
T
2
4 x 2 dx
Q6. Evaluate :
0
x
Q7. Given e (tan x 1) sec xdx
e x f ( x) c.
Write f(x) satisfying the above.
Q8. Write the value of (i
j ).k i. j .
Q9. Find the scalar components of the vector AB with initial point A(2, 1) and terminal point B (–5, 7).
Q10. Find the distance of the plane 3x – 4y + 12z = 3 from the origin.
2 3
SECTION-B
Questions numbers 11 to 22 carry 4 mark each.
Q11. Prove the following :
cos sin –1
3
3
cot 1
5
2
6
5 13
Q12. Using properties of determinants, show that
b c
b
c
a
a
c a
b
c
a b
Q13. Show that f : N
f(x) =
4abc
N, given by
x 1, if x is odd
x 1, if x is even
is both one-one and onto.
OR
Consider the binary operations * : R × R R and o : R × R R defined as a * b = |a – b| and a ob
= a for all a, b R. Show that ‘*’ is commutative but not associative, ‘o’ is associative but not
commutative.
Q14. If x =
1
a sin t , y
a cos
1
t
, show that
dy
dx
y
.
x
OR
Differentiate tan
1
1 x2 1
with respect to x.
x
d 2x d 2 y
d2y
,
and
.
Q15. If x = a (cos t + t sin t) and y = a (sin t – t cos t), 0 < t <
2
dt 2 dt 2
dx 2
Q16. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground,
away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot
of the ladder is 4 m away from the wall?
, find
2
| x3
Q17. Evaluate :
x | dx
1
OR
Evaluate :
x sin x
dx
2
x
1
cos
0
Q18. Form the differential equation of the family of circles in the second quadrant and touching the
coordinate axes.
2 4
OR
Find the particular solution of the differential equation
dy
x( x 2 1)
1; y 0 when x = 2.
dx
Q19. Solve the following differential equation :
( (1 x 2 )dy 2 xy dx
cot x dx; x
0
Q20. Let a = i 4 j 2k , b = 3i 2 j 7k and c = 2i
j 4k . Find a vector p which is perpendicular
to both a and b and p . c 18.
Q21. Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses
the XY-plane.
Q22. Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of 52 cards.
Find the mean and variance of the number of red cards.
SECTION-C
Questions numbers 23 to 29 carry 6 mark each.
Q23. Using matrices, solve the following system of equations :
2 x 3 y 3z
5, x 2y
z
4 , 3x
y 2z
3.
Q24. Prove that the radius of the right circular cylinder of greatest curved surface area which can be
inscribed in a given cone is half of that of the cone.
OR
An open box with a square base is to be made out of a given quantity of cardboard of area c2 square
units. Show that the maximum volume of the box is
Q25. Evaluate :
x sin 1 x
1 x2
c3
cubic units.
6 3
dx
OR
x2 1
dx
( x 1) 2 ( x 3)
Q26. Find the area of the region {(x, y) : x2 + y2
Evaluate :
4, x + y
2}.
x 1 y 2 z 3
x 1 y 2 z 3
and
are perpendicular, find the value of k and
3
2k
2
k
1
5
hence find the equation of plane containing these lines.
Q28. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin 3 times and notes the number of
heads. If she gets 1, 2, 3 or 4 she tosses a coin once and notes whether a head or tail is obtained. If she
obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Q29. A dietician wishes to mix two types of foods in such a way that the vitamin contents of the mixture
contains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin
A and 1 unit/kg of vitamin C while Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin
C. It costs ` 5 per kg to purchase Food I and ` 7 per kg to purchase Food II. Determine the minimum
cost of such a mixture. Formulate the above as a LPP and solve it graphically.
Q27. If the lines
2 5
QuestionPaper-OutsideIndia(2012)
General Instructions:
(i) All questions are compulsory.
(ii) The question paper consists of 29 questions divided into three Sections A, B and C, Section
A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four
marks each and Section C comprises of 7 questions of six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact
requirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 4 questions of four
marks each and 2 questions of six marks each. You have to attempt only one of the alternatives
in all such questions.
(v) Use of calculators is not permitted.
SECTION-A
Questions numbers 1 to 10 carry 1 mark each.
Q1. If the binary operation * on the set Z of integers is defined by a * b = a + b – 5, then write the identity
element for the operation * in Z.
Q2. Write the value of cot (tan–1 a + cot–1 a).
Q3. If A is a square matrix such that A2 = A, then write the value of (I + A)2 – 3A.
Q4. If x
2
3
1
1
y
10
, write the value of x.
5
Q5. Write the value of the following determinant :
102 18 36
1
17
3
3
4
6
x 1 x
e dx
x2
Q6. If
f ( x )e x
c , then write the value of f(x).
a
Q7. If 3 x 2 dx
8 , write the value of ‘a’.
0
Q8. Write the value of (i
j) . k ( j k ) . i .
Q9. Write the value of the area of the parallelogram determined by the vectors 2i and 3 j .
Q10. Write the direction cosines of a line parallel to z-axis.
2 6
SECTION-B
Questions numbers 11 to 22 carry 4 mark each.
4x 3
,x
6x 4
Q11. If f ( x)
Q12. Prove that : sin
2
, show that fof(x) = x for all x
3
63
65
1
sin
1
5
13
cos
2
. What is the inverse of f ?
3
3
5
1
OR
Sovle for x :
2 tan 1 (sin x)
tan 1 (2sec x), x
2
Q13. Using properties of determinants, prove that
a
a b
a b c
2a 3a 2b 4a 3b 2c
3a 6a 3b 10a 6b 3c
Q14. If x m y n
y ) m n , prove that
(x
1
ea cos x , 1 x
Q15. If y
(1 x 2 )
a3
d2y
dx 2
x
dy
dx
dy
dx
y
.
x
1 show that
a2 y
0.
OR
If x 1 y
y 1 x
Q16. Show that y
1 x 1, x
0,
log(1 x)
2x
,x
2 x
y , then prove that
dy
dx
1
.
(1 x) 2
1 , is an increasing function of x throughout its domain.
OR
Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.
x 2 tan 1 x dx
Q17. Evaluate :
OR
3x 1
dx
( x 2) 2
Q18. Solve the following differential equation :
Evaluate :
e
2 x
x
y dx
x dy
1, x
0.
2 7
Q19. Solve the following differential equation :
3e x tan ydx (2 e x ) sec 2 y dy
= 3i 4 j 5k and
Q20. If
to
and
2i
0 , given that when x = 0, y
j 4k , then express
4
.
in the form
1
2
, where
1
is parallel
is perpendicualr to
2
Q21. Find the vector and cartesian equations of the line passing through the point P(1, 2, 3) and parallel to
the planes r . (i
j 2k ) 5 and r . (3i
j k ) 6.
Q22. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability
distribution of the number of successes and hence find its mean.
SECTION-C
Questions numbers 23 to 29 carry 6 mark each.
Q23. Using matrices, solve the following system of equations :
x
y
z
4; 2 x
y 3z
0; x
y
z
2
OR
If A
1
3
15
1
6
5
2
1
5 and B =
2
1
1
0
2
3
2
2
0 , find (AB)–1.
1
Q24. Show that the altutude of the right circular cone of maximum volume that can be inscribed in a
4R
.
sphere of radius R is
3
Q25. Find the area of the region in the first quadrant enclosed by x-axis, the line x
x2
y2
3 y and the circle
4.
3
Q26. Evaluate : ( x 2
x ) dx as a limit of a sum.
1
OR
4
Evaluate :
0
cos 2 x
dx
cos 2 x 4sin 2 x
Q27. Find the vector equation of the plane passing through the points (2, 1, – 1) and (–1, 3, 4) and
perpendicular to the plane x – 2y + 4z = 10. Also show that the plane thus obtained contains the line
r
i 3 j 4k
(3i 2 j 5k ).
2 8
Q28. A company produces soft drinks that has a contract which requires that a minimum of 80 units of the
chemical A and 60 units of the chemical B go into each bottle of the drink. The chemicals are
available in prepared mix packets from two different suppliers. Supplier S had a packet of mix of 4
units of A and 2 units of B that costs ` 10. The supplier T has a packet of mix of 1 unit of A and 1 unit
of B that costs ` 4. How many packets of mixes from S and T should the company purchase to
honour the contract requirement and yet minimize cost ? Make a LPP and solve graphically.
Q29. In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the
students in the college are girls. A student is selected at random from the college and is found to be
taller than 1.75 metres. Find the probability that the selected student is a girl.
2 9
1
Relations and Functions
Teaching-Learning Points
ð
l
Let A and B are two non empty sets then a relation from set A to set B is defined as R = {(a.b) : a A
and b B}. If (a.b) R, we say that a is related to b under the relation R and we write as a R b.
l
R R
l
A relation R in a set A is a subset of A × A.
Types of relations :
ð
ð
A B.
(i) empty relation : R
CA A
(ii) Universal relation R = A × A
(iii) Reflexive relation : ( a, a ) R
A.
(iv) Symmetric relation : If ( a, b) R (b, a )
(v) Transitive relation : If ( a, b)
l
l
l
l
l
l
l
R
a, b
A.
R and
( a, c ) R
a , b, c A .
A relation R is set A is said to be equivalence relation. If R is reflexive, symmetric and transitive.
Let R is an equivalence relation is set A and R divides A into mutually disjoint subset A called partitions
or subdivisions of A subsfying the conditions :
(i) all element of Ai are related to each other,
i.
(ii) no element of Ai is related to any element of Aj, it j
(iii) UAi = A and Ai
Aj = , j.
Type of Functions :
(i) one-one (or injective) function : Let F : A
B, then for every x1, x2
A, f(x1) = f(x2)
x1 = x2.
B, then for every y B, there exists an element
(ii) onto (or surjective function) : Let F : A
x
A such that f(x) = y.
(iii) A function which is not one-one is called many-one function.
A function which is not onto is called into function.
A function which is both one-one and onto is called a bijective function.
Let A be a finite set then an injective function F : A A is subjective and conversely.
Let F : A B and g : B
C be two functions. Then the composition of F and g, denoted as gof is
defined as the function gof : A C given by g of (x) = g [f(x)]
x A
3 0
l
l
l
l
l
Composition of functions need not to be commutative and associative.
If F : A B and g : B C be one-one (or on to) functions, then gof : A c is also one-one (or on to)
but converse is not true.
A function F : A B is said to be invertible if there exists another function g : B
A such that
gof = IA and fog = IB. The function g is called the inverse of the function F.
A function F : A B is said to be invertible if and only if F is one-one and onto (i.e. bijective).
If F : A B and g : B C are invertible functions, then gof : A C is also invertible and (gof)–1 =
F–1 og–1.
Binary operations :
l
l
l
l
l
A binary operation * on a set A is a function * A × A A we denoted * (a, b) by a * b.
A binary operation * on a set A is called commutative if a * b = b * a a, b A.
A binary operation * on a set A is said to be associative if a * (b * b) * c a, b, c A.
The element e A, if it exists, is called identity element for binary operation * if a * e = a = e * a
a A.
The element a A is said to be invertible with respect to the binary operation * if there exile b A
such that a * b = e = b * a. The element b is called morse of a and is denoted as a–1.
Question for Practice
Evaluate the following Integrals
Very Short Answer Type Questions (1 Mark)
Q1. Let R be a relation on A defined as R = {(a, b) A × A : a is a husband of b} can we say R is
symmetric? Explain your answer.
Q2. Let A = {a, b, c} and R is a relation on A given by R = {(a, a), (a, b), (a, c), (b, a), (c, c)}. Is R
symmetric? Give reasons.
Q3. Let R = {(a, b), (c, d), (e, f )}, write R–1.
Q4. Let L be the set of are straight lines in a given plane and R = {(x, y) : x y x, y L}. Can we say
that R is transitive? Give reasons.
Q5. The relation R in a set A = {x : x z and 0 x 12} is given by R = {(a, b) : |a – b| is a multiple if 4}
is an equivalence relation. Find the equivalence class related to {3}.
Q6. Let R1 be the relation on R defined as R = {(a, b) : a b2}. Can we say that R is reflexive? Give
reasons.
Q7. Let R {(a, b) : a, b Z (Integers) and |a – b| 5}. Can we say that R is transitive? Give reason.
Q8. If A = {2, 3, 4, 5}, then write the relation R on A, where R = {(a, b) : a + b = 6}.
Q9. If A {1, 2}, and B = {a, b, c}, then what is the number of relations on A × B?
3 1
Q10. State reason for the relation R in the set {1, 2, 3} given by R {(1, 2), (2, 1)} not to be transitive.
Q11. If f is invertible function, find the inverse of f(x) =
3x 2
.
5
R, find fog (x).
Q13. Write the inverse of the function f(x) = 5x + 7, x R.
Q14. Show that f : R R defined as f(x) = x + 1 is not one-one.
Q15. Show that the function f : N N defined by f(x) = 3x is not an onto function.
Q12. If f(x) = x + 7 and g(x) = x – 7, x
2
Q16. Let * be a binary operation on Z defiand by a * b = 2a + b – 3, find 3 * 4.
Q17. Let * be a binary operation on N defined by a * b = a2 + b and O be a binary operation on N defined
by aob = 3a – b find (2 * 1) 02.
Q18. let * be a binary operation on R defined by a * b = a – b. Show * is not commutative on R.
Q19. Let * be a binary operation on N given by a * b = l.c.m (a, b), a, b
N find (2 * 3) * 6.
Q20. Can we say that division is a binary operation on R? Give reasons.
Q21. Show that * : R × R
R given by a * b = a + 2b is not associative.
Q22. Explain that addition operation on N does not have any identity.
Q23. What is inverse of the element 2 for addition operation on R?
Q24. Let * be the binary operation on N given by l.c.m (a : b) find the identify element for * on N.
Q25. Let * be the binary operation on N defined by a * b = HCF (a, b). Deos there exist identify element
for * on N?
Short Answer Type Questions (4 Marks)
Q26. Show that f : N
{
N givne by
f(x) = x + 1 if x is odd
x – 1 is x is even, is bijective
Q27. Let * be a binary operation on the set A = {0, 1, 2, 3, 4, 5} as
{
a * b = a + b if a + b < 6
a + b – 6 if a + b
6,
Show that O is the identify element for this operation and each element a of the set is invertible with
6 – a being the inverse of a.
Q28. Let N be the set of all natural numbers and R be a relation on N × N, defined by (a, b) R (c, d)
ad = bc (a, b), (c, d) N × N. Show that R is an equivalence relation.
Q29. Let f : R
R be defined by f(x) = 3x + 2. Show that f is invertible. Find f : R
R.
Q30. Let * be a binary operation on N × N defined by (a, b) * (c, d) = (a + c, b + d). Show that * is
commutative as well as associative. Find the identity element for * on N × N if any.
Q31. Let T is a set of all triangles in a plane and R be a relation as R : T
T}. Show that R is an equivalence relation.
2
3 2
T = {( 1,
2
):
1
2
1
,
Q32. Let * be the binary operation on Q (Rational numbers) defiend by a * b = |a – b|, show that
(i) * is commutative
(ii) * is not associative
(iii) * does not have identity element
Q33. Show that f : R R defined by f(x) = x3 – 1; is invertible. Find f (x).
A is defined by f(x) = 35xx 74 and g : A B is defined by g(x) = 57xx 31 , then
3 and B = R – 7 .
and gof = I , where A = R –
Q34. Show that if f : B
Fog = IA
5
5
Q, given by F(x) =
2x 3
is not a bijective function.
x 3
B
Q35. Show that the function F : Q – {3}
Answers
Very Short Answer (1 Mark)
1.
2.
3.
4.
5.
No, if is a husband of b, then b being a female can not be husband of anybody.
No, because (a, c) R but (c, a ) R.
R–1 = {(b, a), (d, c), (f, e)}
No, If x y & y z x || z.
{3, 7, 11}.
6.
1
No, example
3
7.
No, Let a = 5, b = 10, c = 12, then (a, b)
8.
R = {(2, 4), (3,3), (4, 2)}
1
3
2
.
R, (b, c)
9. 64
10. (1,1) R.
1
11. f ( x)
12. x
13.
16.
19.
20.
21.
R but (a, c) R.
5x 2
3
x 7
5
7
17. 13
6
No, because Number divided by 0 does not belong to R.
Let a = 2, b = 5, c = 8, (a * b) * c = (2 + 2 × 5) * 8 = 12 * 8
= 12 2 8
28 and a * (b * c)
2 * (5*)
2 * (5 2 8)
= 2 * 21 2 2 21 44 .
3 3
22. Because O + number = Number but O does not belong to N.
23. –2
24. 1
25. No
Very Short Answer (4 Mark)
29. f –1 = (x) =
33.
35.
x 2
3
30. Identity does not exist
( x 1)
f(x ) = f(x ) x = x is one-one.
Let y codomain then f(x) = y
3 3 y Q – {3} for some y Q
x=
2 y
Example 2 codomain but
3 3 2 = Not defined, does not belong to domain
=
2 2
f 1 ( x)
1
13
2
1
2
3 4
2
InverseTrigonometric
Functions
Teaching-Learning Points
l
The sine function is defined as
sin : R [–1, 1]
Which is not a one-one function over the whole domain and hence its inverse does not exist but if we
, then the sine function becomes a one-one and onto function and
2 2
[–1, 1] as sin :[1,1] , therefore we com define the inverse of the function sin : ,
!" 2 2
2 2
% #3$ , #$ & ,'% $ , 3$ & etc which may also be taken as range of
In fact there are other intervals also like '
(2 2 (2 2
% #$ $ &
the function sin . Corresponding to each interval we get branch of sin . The branch with range ' ,
(2 2
restrict the domain to
1
–1
l
–1
is called principal value branch similiary for other inverse trigonometric functions we have principal
value branches.
List of principal value branches and the domain of inverse trigonometric functions.
Functions
Domain
Range (Principal value Branch)
y = sin–1x
y = cos–1x
y = tan–1x
y = cot–1x
y = sec–1x
y = cosec–1x
+, - y - ,
)1 * x * 1
.1 / x / 1
2 x
2
0
x
1
1
y
2
0<y<
y
2
1
x
x
x
0 y01
2
x
2
0
y
y
2
1
0
3 5
2
y
0
2
l
Properties of inverse trigonometric functions :
1.
(i) sin 1 (sin x)
(ii)
(iii)
(iv)
(v)
(vi)
2.
4 x , x 9 7:; 562 , 62 8
sin(sin 3 x) 4 x , x <[=1,1]
cos 3 (cos x) 4 x , x >[0, ? ]
cos(cos 3 x) 4 x, x <[=1,1]
D BC , C EH
tan @ (tan x) A x, x F G
I 2 2J
tan(tan K x) L x, x M R.
O 1P
sin N R S Q cosec N x, |x| 1
T xU
3
(i)
1
1
1
1
1
1
V WZ 1 X[ Y secV x, |x| 1
\ x]
W 1X
tan V Z [ Y cot V x, x > 0
\ x]
sin 3 (^ x) 4 ^ sin 3 x, x M [–1, 1]
tan _ (` x) a ` tan _ x, x M R
cosec 3 (^ x) 4 ^ cosec 3 x, |x| 1
cos 3 (^ x) 4 b^ cos 3 x, x M [–1, 1]
secc (d x) e fd secc x, |x| 1
cot c (d x) e fd cot c x, x M R
h
sin g x i cos g x j , x k[l1,1]
2
n
tan m x o cot m x p , x q R.
2
s
cosec x t sec r x u , | x | 1
2
xv y
tan 3 x v tan 3 y 4 tan 3
, xy w 1
1 ^ xy
xy y
tan x x y tan x y z tan x
, xy { y1
1 | xy
(ii) cos
(iii)
3.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
4.
(i)
(ii)
(iii)
5.
(i)
(ii)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
–1
1
1
1
1
1
1
1
3 6
6.
} ~ tan } 2 x ,| x | 1
1 x
2x
2 tan x sin
,| x | 1
1 x
1 x
2 tan
x cos
,x 0
1 x
(i) 2 tan 1 x
(ii)
1
2
1
1
2
2
(iii)
1
1
2
Question for Practice
Evaluate the following Integrals
Very Short Answer Type Questions (1 Mark)
3 .
2
Q2. Write the principal value of cosec ( 2).
1 .
Q3. Write the principal value of cot
3
Q4. Write the principal value of tan ( 3) .
Q5. Write the principal value of sec ( 2) .
1 Q6. Write the principal value of cos ¡ ¢ .
£ 2¤
Q7. Show that sin ¥ x ¦ cos ¥ 1 § x .
© 1« x ª .
Q8. Show that cos ¨ x ¬ tan ¨
¯ x ®°
± x ²´ .
Q9. Show that tan x sin ³
µ 1 x ¶
± x ².
Q10. Show that sin
x tan
³
µ 1 x ´¶
Q1. Write the principal value of sin
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
1
1
2
3 7
· ¹ 2sin · 1 ¸ x .
2
º
Q12. Write sin (3 x » 4 x ) in the simplest form.
Q13. Write cos ¼ (4 x ½ 3 x) in the simplest form.
¾ ¿À .
Q14. Evaluate cosec cosec
Q11. Show that cos 1 x
1
1
3
1
3
–1
4
Q15. Evaluate cos
3
Q16. Show that cos 1 x
1
2
1
cos
2 cos
1
.
1 x
.
2
Q17. Write cos–1 (2 x 2 1) in the simplest form.
Q18. Write cos 1 (1 2 x 2 ) in the simplest form.
Q19. Write tan
1
1 cos x
,0
1 cos x
Q20. Show that sin 1 2 x 1 x 2
Q21. Evaluate : sin
sin
3
Q22. Evaluate : cos
1
cos
Q23. Find x, if tan 1 x
.
x
2sin 1 x .
1
2
1
.
sin
3
1
sin
2
3
.
/ 4.
Q24. Evaluate tan
1
tan
3
4
.
Q25. Evaluate cos
1
cos
7
6
.
Q26. Evaluate sin 1 (sin 2 3) .
Q27. Evaluate cosec –1 cosec
Q28. Evaluate cos
1
cos
3
4
.
5
.
3
3 8
Q29. Write tan
ÁÂ
x
1
a2
x2
,| x | a in the simplest form.
Q30. Find x, if cot 1 x tan 1 7
Q31. Find x, if sin 1 x
6
Q32. Find x, if 4 sin 1 x
1
Q35. Write sin
2
.
cos 1 x.
cos 1 x.
Q33. Find x, if tan 1 x 2 cot 1 x
Q34. Write sin
2
2x
1 x2
2
.
3
1 x 1 , in the simplest form.
2 x 1 x 2 in the simplest form.
Short Answer Questions Carrying 4 Marks each
Q36. Solve for x : tan 1 ( x 1) tan 1 ( x 1)
Q37. Solve for x : tan 1 2 x tan 1 3 x
1
1
1
Q38. If tan a tan b tan c
Q39. Solve for x : tan
1
Q40. Solve for x : tan
1
Q41. Solve for x : tan
1
Q42. Solve for x : tan
1
Q43. Solve for x : sin
1
4
x 1
x 1
tan
8
17
8
.
31
, prove that a + b + c = abc.
1
tan 1 x, x
2
x 1
x 1
1
.
1 x
1 x
2x
1 x2
tan
1
2x 1
2x 1
sin 1 x sin
Q44. Solve for x : tan 1 (2 x) tan 1 (3 x)
tan 1 7.
1 x2
2x
1
cot
tan
x 1
x
1
0.
1
.
2
tan
1
23
.
36
3
.
5
n
3
.
4
Q45. Solve for x : tan 1 ( x 1) tan 1 x tan 1 ( x 1) tan 1 3 x
3 9
0.
à ÄÉ 12 ÅÊ Æ cosà ÄÉ 4 ÅÊ Æ tan à ÄÉ 63ÅÊ Ç È.
Ë 13Ì
Ë 5Ì
Ë 16 Ì
1
1
1
1 Î
Q47. Prove that tan Í Ï tan Í Ï tan Í Ï tan Í Ð .
3
5
7
8 4
1
1
4
Q48. Prove that 2 tan Ñ Ò tan Ñ Ó tan Ñ .
Q46. Prove that sin
1
1
1
1
1
1
1
1
1
1
5
8
7
Ô 3 Õ sin Ô 8 Ö sin Ô 77 .
5
17
85
5
7
Ø 253ÙÝ .
Q50. Prove that sin ×
sin ×
cos × Ü
Ú
Û
Þ 325 ß
13
25
á 1 ã x â å tan à çá 2 x èâ æ ä .
Q51. Prove that tan à ç
é 1ã x ê 2
é 2 x èê
ì 4í
ì 3í
ì 27 í
Q52. Prove that cos ë ñ ò î tan ë ñ ò ï tan ë ñ ò .
ó 5ô
ó 5ô
ó 11 ô
ö 63÷
ö 1÷
ö 3÷
Q53. Prove that cos õ ú û ø 2 tan õ ú û ù sin õ ú û .
ü 65ý
ü 5ý
ü 5ý
1
2 1
ÿ 3ð
Q54. Prove that tan þ
tan þ cos þ .
Q49. Prove that sin
1
1
1
1
1
1
2
1
1
2
1
1
1
1
1
1
1
1
4
1
9
2
5
5
1
1 tan 1 tan 1 .
12 70 99 4
Q55. Prove that : 2 tan 1 cos x x
.
1 sin x 4 2
Q56. Prove that : tan 1 12 3
sin 1 ! "
"
$
# 5$
13
Q57. Prove that cos 1 !
#
ìñ
ó
Q58. Prove that cos 2 tan
56 .
# 65 "$
sin 1 !
ë 1 íò ï sin ìñ 4 tan ë 1íò .
ó 3ô
7ô
1
1
1
1 &
Q59. Prove that : 2 tan %1 ' cosec %1 5 2 ' 2 tan %1 ( .
5
8 4
3
17 *
Q60. Prove that : 2sin )1 + tan )1 , .
5
31 4
4 0
Answers
1.
5.
14.
19.
24.
-
2.
3
31
4
4
x
2
./
28. 8 3
32.
1
2
-
4.
./
3
15. 1
1
17. 2 cos 6 x
18. 2sin 31 x
21. 1
22. 7
23.
26. 8 3
27. 8 4
91 : x ;
29. sin <> =?
a
30. 7
31.
33.
34. 2 tan @1 x
A1
35. 2 sin x
1
4
37.
41. x = 1
42.
36. B8,
4
20
3
1
13. 3cos 3 x
25.
4
3.
12. 3sin 21 x
6.
45
./
3
51
6
3
1
6
C3 4
,
8 3
39.
1
3
43.
77
85
1 1
45. 0, , E
2 2
4 1
4
3
2
40. x = 2
44.
D1
6
,1
3
MatricesandDeterminants
Teaching-Learning Points
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A matrix is an ordered rectangular array (arrangement) of numbers and enclosed by capital bracket
[ ]. These numbers are called elements of the matrix. Matrix is denoted by capital letters of the
English alphabet and its elements are denoted by small letters.
In a matrix horizontal lines of numbers are called rows of the matrix and vertical lines are called
columns of the matrix.
A matrix having m rows and n columns is called a matrix of order m by n, written as m × n. In general
a matrix of order m × n is written as:
F a11 a12
Ha
H 21 a22
H a31 a32
A= H
H
H ai ai
H
HJ am am
1
1
l
l
l
2
2
a13
a23
a33
a1 j
a2 j
a3 j
ai 3
am 3
ai j
am j
a1n G
a2 n I
I
a3n I
I
I
ain I
I
amn IK
(OR)
A = [aij]m × n where 1 L i L m, 1 L J L n, m, n M N
Types of matrices : A matrix A = [aij]m × n is said to be a:
(i) Row matix if m = 1
(ii) Column matrix if n = 1
(iii) Zero/Null matrix if each of its elements is zero.
(iv) Square matrix if m = n
(v) Diagonal matrix, if m = n and aij = 0 when i N J
(vi) Scalar matrix, if m = n, aij = 0 when i N j and aij = k when i = J
(vii) Unit/identity matrix if m = n, aij = 0 when i ¹ j
Two matrices A = [aij]m × n and B = [bij] p × q are said to be equal (i.e. A = B) if m = p, n = q and aij = bij
for all i, j.
Scalar multiplication of a matrix.
Let
A = [aij]m × n and k be any scalar,
then
KA = K[aij]m × n = [Kaij]m × n.
i.e. to multiply a matrix by a scalar multiply each element of the matrix by the scalar.
4 2
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Addition of matrices:
Let
A = [aij]m × n and B = [bij]m × n, then
A + B = [aij + bij]m × n
Properties of matrix addition.
(i) A + B = B + A (commutative law)
(ii) (A + B) + C = A + (B + C) [Associative law]
(iii)A + O = A = O + A, where O is Nuel matrix (existence of additive identity)
(iv) A + (–A) = O = (–A) + A [existence of additive inverse]
A – B = A + (–B).
Multiplication of matrices.
Let A = [aij]m × n and B = [bij]p × q be two matrices, then the product of A and B (i.e. AB) is defined if n
= p and it is a matrix of order m × q.
Let A = [aij]m × n and B = [bij]n × p, then
A B = C (say) = [cij]m × b, where cij is obtained by taking ith row of A and jth column of B, multiplying
their corresponding elements and taking sum of these produce.
(i) In general, AB O BA.
(ii) (AB)C = A(BC)
(iii) A.I = IA where I is identity matrix.
(iv) A(B + C) = AB + AC (or) (A + B) C = AC + BC. (Distributive Law)
Transpose of a matrix A is obtained by interchanging its rows and columns. It is denoted by A 1.
Properties of transpose:
(i) (KA)1 = KA1
(ii) (A + B)1 = A1 + B1
(iii) AB)1 = B1A1
(iv) (A1)1 = A.
A square matrix is called, a symmetric matrix if A1 = A and a skew symmetric matrix if A1 = –A.
For any square matrix A, A + A1 is always symmetric matrix and A – A1 is always skew symmetric
matrix.
Every square matrix can be expressed as a sum of symmetric and skew symmetric matrix.
i.e.
l
P A R A1 Q P A – A1 Q
R
2 TV SU 2 TV
A= S
U
Elementary operations (transformations) of a matrix.
(i) Interchange of any two rows (or two columns)
i.e. Ri W Rj (or) Ci W Cj
(ii) Multiplication of the elements of any row (or any column) by a non zero number i.e.
i.e. Ri X KRi (or) Ci X KCi
(iii) Addition of the elements of any row (or any column) to the corresponding elements of any other
row (or column) multiplied by any non zero number i.e.
i.e. Ri X Ri + KRj (or) Ci = Ci + KCi
4 3
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A square matrix A is said to be invertible if there exists another square matrix B of the same order such
that AB = I = BA, then B is called the inverse of A and is denoted by A–1.
Properties of inverse of matrix.
(i) AB = I Y B = A–1 and A = B–1
(ii) (A–1)–1 = A
(iii) (AB)–1 = B–1A–1
(iv) (A–1)1 = (A1)–1
For finding the inverse of a square matrix by using.
(i) elementary row operations (transformations) we write A = IA.
(ii) elementary column operations (transformations) we write A = AI.
A number which is associated to a square matrix A = [aij]n × n is called a determinant of the matrix A and
it is denoted as |A|.
Determinant ca be expanded by using any row or column.
Properties of determinants:
(i) |A1| = |A|.
(ii) Ri Z Rj (or) Ci Z Cj Y |A| = –|A|
(iii) If Ri is identical to Rj or Ci is identical to Cj Y |A| = 0.
(iv)
ka1 kb1 kc1
a1 b1 c1
a2 b2 c2 = k a2 b2 c2 = k|A|
a3 b3 c3
a3 b3 c3
In general
l
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l
n
kA n [ n = k A n\ n
(v) Ri ] Ri + kRj (or) Ci = Ci + kCj
Y |A| remains unchanged
(vi) If some or all elements of a row (or column) of a determinant are expressed as the sum two or
more terms, then the determinant can be expressed as a sum of two or more determinants.
|AB| = |A| |B|
1
A
A square matrix is said to be singular if |A| = 0 and non singular if |A| ^ 0
|A–1| =
Using determinants Area if a _ABC with vertices A(x1 y1), B(x2 y2), C(x3 y3) is given by
x1
1
x2
2
x3
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y1 1
y2 2
y3 3
Minor of an element aij of a determinants is the determinant obtained by deleting the ith row and yth
column in which elements aij lies. It is denoted by Mij.
Cofactor of an element aij of a determinant is denoted as Aij and is defined as Aij = (–1)i + y Mij
Adjoint of a square matrix A = [aij]n × n is defined as the transpose of the matrix [Aij]n × n where Aij is the
cofactor of the element aij and it is denoted as adj.A
4 4
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A(adjA) = (adjA) A = |A| I
|adjA| = |A|n–1, where A is a square matrix of order n.
A square matrix A is invertible if and only |A| ` 0 if A is a non singular matrix
adjA
A–1 =
.
|A|
If A and B are non singular matrices of the same order then AB and BA are also non singular matrices
of the same order.
A system of linear equations in three variables can be expressed in a matrix equation.
For example,
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
a
where
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b a1 b1 c1 c b x c h d1 i
da b c e d y e jd k
d 2 2 2 e d e = j 2 k a AX = B
df a3 b3 c3 eg df z eg jl d3 km
n a1 b1 c1 o
p
q
A = p a2 b2 c2 q , X =
pr a3 b3 c3 qs
txu
v yw
v w, B =
xv z yw
z d1 {
|d }
| 2}
|~ d3 }
By solving the matrix equation AX = B or X = A–1B we get solution of given system of equations.
If |A| ` 0, then the system of equations has unique solution and hence it is consistent.
If |A| = 0 and (adjA) B ` 0 (zero matrix), then the system of equations has no solution and hence it is
inconsistent.
If |A| = 0 and (adjA) B = 0 (zero matrix), then the system of equations may be either consistent or
inconsistent according as the system has either infinitely many solutions or no solution.
Question for Practice
Very short answer questions carrying one mark each:
1.
construct a 2 × 2 matrix A = [aij], where aiJ 3i J
2.
a b 2 6 2
If
=
, find the values of a and b.
a 5 8
5
3.
2 –1 10
If a b = , find a and b.
3 1 5
4.
If a matrix has 12 elements, then how many possible orders it can have.
4 5
5.
cos x – sin x
¢
If A =
,0<x<
and A + A1 = I, then find the value of x.
sin
x
cos
x
2
¡
6.
£3 4 ¤
If ¥
¦ = P + © where P is symmetric and © is skew symmetric matrix, find ©.
§ 1 –1¨
7.
8.
cos ª – sin ª
sin ª
cos ª
x 3 4 1
Find the value of x if
«
1 2 3 2
Evaluate
1
3
5
9.
What is the value of the determinant 2 6 10
11 13 15
1 0 4
10. Find cofactor of a23 in 3 5 –1
0 1 2
1 2 3
11. Find the minor of a12 in the determinant 2 3 4
–3 2 –1
12. Let A be a square matrix of order 3 and |A| = 5, find the value of |2A|.
13. Let A be a square matrix of order 3 × 3 and |A| = 2, then find |adjA|.
14. If A is invertible matrix of order 2 and |A| = –11, then find |A –1|.
¬1 2
15. If a singular matrix A is given by ®
¯ , then find the value of x.
° x 4±
Short answer questions carrying 4 marks each:
16. Express the following matrix as a sum of a symmetric and a skew symmetric matrix.
² 1 3 5³
´ –6 8 3µ
´
µ
´¶ –4 6 5µ·
¸ 3 –2 ¹
17. Given that A = º
» and I =
¼ 4 –2 ½
¾1 0¿
2
À 0 1 Á , find the real number k such that A – kA + 2I = 0.
Â
Ã
Ä2 1 Å
18. Given that A = ÆÆ –1 3 ÇÇ and B =
ÈÆ 4 0ÇÉ
Ê –3 1 2 Ë
1
1 1
Ì 0 4 5 Í , verify that (AB) = B A .
Î
Ï
¸ 2 Ð 3¹
19. If A = º
» show that A2 – 6A + 17 I = 0. Hence find A–1.
¼3 4 ½
4 6
x –6
20. Show that x = 2 is one of the roots of the equation 2 –3 x
–3 2 x
21. Using properties of determinants show that:
–1
x – 3 = 0. Find other roots also.
xÑ2
1 a a2
1 b b 2 = (a – b) (b – c) (c – a)
1 c c2
bÒc
22.
cÒa
qÒr rÒ p
yÒz zÒx
aÒb
c
2x
= (x + y + z)3
y – z – x 2y
2z
z–x– y
2y
2z
b–c
a
24.
b
pÒq = 2 p q r
xÒ y
x y z
x – y – z 2x
23.
a
cÑb
aÑc b
c – a = (a + b + c) (a2 + b2 + c2)
a–b aÑb c
1
x
2
25. x 1
x
x2
x = (1 – x3)2.
x2 1
Long Answer type questions carrying 6 marks each:
Ó 2 –3 3 Ô
26. Find the inverse of the matrix using elementary transformations: Õ 2
2 3Ö
Õ
Ö
×Õ3 –2 2 ÖØ
27. Solve the following system of linear equations using inverse of matrix.
2x + 3y + 4z = 8, 3x + y – z = 2, 4x – y – 5z = –9.
Ù 2 –3 5Ú
28. If A = Û 3 2 –4 Ü , find A–1. Hence solve the following system of linear equations:
Û
Ü
ÝÛ 1 1 –2 ÜÞ
2x – 3y + 5z = 11
3x + 2y – 4z = –5
x + y – 2z = –3
4 7
ß 1 1 2à
29. If A = á 2 –1 3â , find A–1 and hence solve the following system of linear equations:
á
â
áã 5 –1 –1âä
[Hint : (A1)–1 = (A–1)1
x + 2y + 5z = 10
x – y – z = –2
2x + 3y – z = –11
å 1 –1 2 æ
30. If A = ç 0 2 –3è and B =
ç
è
çé 3 –2 4 èê
ë –2 0 1ì
í 9 2 –3î . Find AB and hence solve the following system of equation.
í
î
íï 6 1 –2 îñ
x – y + 2z = 1
[Hint : AB = I ò A–1 = B]
2y – 3z = 1
3x – 2y + 4z = 2
Answers
1.
ó2 1 ô
õ5 4 ö
÷
ø
5.
ù
9.
0
2. a = 8, b = –2
6.
3
10. –1
13. 4
16.
1 ú 0 3û
2 üþ –3 0 ýÿ
14.
ð1
–3 2
1 2
–1
11
3. a = 3, b = –4
4. 6
7. 1
8. 1
11. 10
12. 40
15. 2
–3 2 1 2
92 92
ð0
8
9 2 –9 2 0
–3 2 9 2 5 –9 2 3 2 0 19. A–1 =
18. 1
1 4 3
17 3 2 20. 1, –3.
26.
– 2
–1
2
5
0
3 5
5 15
5 1 5 –2
0
5
27. x = 1, y = –2, z = 3.
ß 4 –1 5à
1
á17 –11 1â x = –1, y = –2, z = –3.
29. A–1 =
â
27 á
áã 3
6 –3âä
4 8
28.
0
–2
–1
1
–2
9 –23 x = 1, y = –2, z = 3.
5 –13
30. x = 0, y = 5, z = 3.
4
ContinuityandDifferentiability
Teaching Learning points
l
Continuity and discontinuity of function: A function y = f(x) is said to be continuous in an interval
If for every value of x in that interval y exist. If we plot the points, the graph is drawn without lifting
the pencil.
l
Continuity and discontinuity of a function at a point: A function f(x) is said to be continuous at a
point ‘a’ of its domain If
f(x),
f(x),
lim1
lim
and
f(a) exist
x a
x a
f(x)
= f(x)
lim
x a
= f(a)
lim
x a
A function f(x) said to be descontinuous at x = a If it is not continous at x = a
l
Let f(x) and g(x) are two real Continous functions at point x = a, then
(i) f(x) + g(x) is also continuous at x = a
(ii) f(x) · g(x) is also continuous at x = a
(iii)
l
f ( x)
is also continuous at x = a, provided g(a)
g ( x)
0
(iv) λ · f(x) is also continuous at x = a, where λ is any scalar
(v) |f(x)| is also continuous at x = a
(vi) fog or gof is also continuous at x = a
Before doing Exercise, the students must know the following facts
(i) Absolute value function is continuous for all real values of ‘x’.
(ii) A Polynomial function f(x) is continuous Α x R
(iii) Any constant function f(x) = c and identity function f(x) = x is continuous A x
(iv) Every Rational function
R
f ( x)
is continuous for all values of x in the Domain, where the points
g ( x)
at which g(x) is zero are not included in the Domain.
(v) The Trigonometric functions sin x, cos x, are continuous function every where.
(vi) f(x) = [x] is not continuous at any integral value of ‘x’.
4 9
l
l
The process of finding derivative of a function is called differentiation.
Let f(x) be any real valued function defined on an open interval (a, b), then f(x) is said to be differentiable
at x = e (a, b) if = lim
h
0
f (c h )
h
f (c )
lim
0
h
f (c h )
h
f (c )
a finite number..
l
A function f(x) is said to be differentiable in an interval (a, b) if it is differentiable at every point of
(a, b).
l
Chain Rule: If y = f(u) and u = g(x), then
l
l
l
l
l
dy
dx
dy du
!
du dx
This process also known as Derivative of function of a function.
Derivative of implicit function: When the variable x and y are related in such a way that y cannot be
expressed as a function of ‘x’ in a easier way, then the function is known as implicit function.
So the derivative from implicit function is obtained by differentiating directly w.r.t. the suitable variable.
While differentiating inverse trigonometric functions, first express it in simplest form by using suitable
substitution and then differentiating the simplest form.
Some important substitutions:
Expression
Substitutions
(i)
a2
x2
x = a sin θ or x = a cos
(ii)
a2
x2
x = a tan
"
or x = a cot
(iii)
a2 a2
x = a sec
"
or x = a cosec
(iv)
a# x
x = a cos
"
(v)
a x
or
a x
a x
a x
x = a cos2
"
"
"
"
Logarithmic differentiation:
(i) This process is used when function is given in a complicated form
(ii) When
y = [f(x)]g(x), then
dy
d
= [ f ( x)]g ( x ) $ [ g ( x) $ log{ f ( x)}]
dx
dx
Derivative of function in Parametric forms: If y = f(x) be a function in which x and y are the
variables, when the variables x and y are the functions of third variable ‘t’ i.e., x = u(t) and y = v(t) then
this form is called parametric form and ‘t’ is called the parameter.
In order to find
dy
, we use the following:
dx
dy
dy
= dt
dx
dx
dt
5 0
l
l
l
Roll’s theorem: If f(x) be a real valued function defined in [a, b] such that
(i) f(x) is continuous in [a, b]
(ii) Derivable in (a, b)
(iii) f(a) = f(b)
Then, there exists at least one c% (a, b)
Such that
f &(c) = 0
Geometrical interpretation:
*in the interval (a, b) there must exist at least one point where the tangent is parallel to x-axis.
Lagrange’s mean value theorem: If ‘f’ be a function such that
(i) f(x) is defined in [a, b]
(ii) f(x) is continous in [a, b]
(iii) Derivable in (a, b)
Then there exists at least one c% (a, b)
Such that
l
f &(c) =
f (b) ' f ( a )
b'a
Geometrically:
In the interval (a, b) there must exist at least one point ‘c’ where the tangent is parallel to the chord
joining the end points.
5 1
Question for Practice
Very Short Answer Type Questions (1 Mark)
Q1.
Q2.
Q3.
Q4.
Q5.
If y = (1 + x1/3) (1 – x1/3) (1 + x2/3) find dy/dx
If y = tan–1 (cot x) find dy/dx
f(x) = 53x find f ((2)
f(x) = tan–1 x + tan–1 1/x find f ((x) x > 0
f(x) = log10 x find f ((x)
Q6. If y )
x
x
*
1
then find dy/dx
x
Q7. If y = ex then what is the value of
,
1 + d 2 y dy
- y/
2
.2
a 0 dx
dx
1
dy
at (1, 1)
dx
Q9. If f(x) = [x] write points where f(x) is not differentiable
Q10. For what value of λ the f(x) = sin (λx) is continous every where
Q8. x2/3 + y2/3 = 2 find
Short Answer Type Questions (4 Marks)
Q1.
f(x) =
2
4
4
4
4
4
4
4
9
1 3 cos 4 x
x2
a
x50
x60
x
16 8 x
x70
34
For what value of ‘a’ f(x) is continuous at x = 0
Q2.
f(x) =
:
sin(a ; 1) x ; sin x
<
<
x
<
c
<
<
x ; bx 2 ? x
<
<
b x3
A
x=0
x>0
x@0
Determine the value of a, b and c for which the function f(x) may be continuous at x = 0
5 2
Q3. Examine the following function for continuity at x = 0 where n B 1
f(x) =
C n
E x sin
F
E
0
H
1
x
xD0
xG0
Q4. Determine the value of constants ‘a’ and ‘b’ such that the function defined as is continuous at x = 4
f(x) =
J
M
M
M
M
M
M
P
xI4
xI4
Ka
If x L 4
aKb
xI4
Kb
xI4
xN4
xO4
Q5. Show that the function f(x) defind by f(x) is continuous at x = 0
f(x) =
Q sin x
R cos x
T
T x
2
T
T
T 4(1 V 1 V x )
T
x
X
xS0
xU0
W
Q6. If x = a(Y – sin Y) y = a(1 – cos Y)
Find
d2y
at
dx 2
Z
[\
2
Q7. Diff. w.r.t. ‘x’
5x _
2a
c1 b 6 x d
^
tan ]1 `
Q8. If xmyn = (x + y)m+n
Prove that
Q9.
1 f x2
dy
dx
j
g
dy
dx
e
y
x
1f y2
h
a ( x f y ) , then show that
1i y2
1 i x2
Q10. If y k log[ x l 1 l x 2 ] prove that
( x 2 m 1)
d2y
dy
mx
2
dx
dx
\
0
Q11. y = ( x m x 2 n 1) m
Prove that (x2 – 1)y2 + xy1 = m2y
5 3
0
Q12. If x
d2y
o1
p
tan q log yr show that (1 x 2 ) 2
sa
t
dx
Q13. If x = log t and y =
d2y
1
Prove that
dx 2
t
(2 x a )
dy
dx
dy
dx
0
0
Q14. If y = (cos–1 x)2 Prove that (1 – x2)y2 – xy1 = 2
3
Q15. If x = a sin u
Q16. If y = sin
1
d2y
y = b cos u find
at x
dx 2
3
w 5x
y
y
{
12 1 x 2
13
x
z
z
|
v
4
find dy/dx
Q17. Verify the applicablity of rolle’s theorem for the following
1. f(x) = sin 2x
[0, }/2]
2/3
[–1, 1]
2. f(x) = x
[1, 2]
3. f(x) = (x – 1)(x – 2)2
Q18. Verity Lagrange’s mean value theorem for the following functions in the given interval and find c of
this theorem
[1, 3]
f(n) = x3 – 5x2 – 3x
f(n) = (x – 1) (x – 2) (x – 3) [0, 4]
1
f(n) = x
[1, 3]
x
Q19. Find
dy
If y
dx
sin 1 ( x 1 x
~
1 x
1 x
1
Q20. If y = sin 2 tan
dy
Prove that dx
Q21. Differentiate x x
x 1 x2 )
x
1 x2
2
2
3
( x 3) x w.r.t. ‘x’ for x > 3
Q22. Differentiate xxcosx + (xcosx)x w.r.t. x
Q23. Differentiate tan
1
1 x2
1 x2
1 x2
1 x2
w.r.t. tan
Q24. Differentiate sin2 x w.r.t. esin x
Q25. f(n) =
x
2x
3 x 2
at x = 2
5 x2
Show that f is not differentiable at x = 2
5 4
1
2x
1 x 2
Q26. If
x2
a2
y2
b2
1 Prove that
d2y
dx 2
b4
a2 y3
Q27. Differentiate the following w.r.t. ‘x’
(i)
The electric potenti
( x 3)( x 2 1)
3x 2 4 x 5
Q28. If (x – a)2 + (x – b)2 = c2 for some c > 0. Prove that
1
2 32
dy
dx
is a constant independent of a and b
d2y
dx 2
Q29. Find
Q30. If y
dy
If yx + xy + xx = 1010
dx
x
x y show that
dy
dx
y log y ( x log x log y 1)
(1 x log y ) x log x
Q31. Discuss the differentiability and continuity of
f(x) = |x – 1| + |x – 2| at x = 1 and x = 2
Q32. f(x) =
x
2 x
1 x 2
3 x ,2
Show that f(x) is continuous at x = 2 but not differentiable at x = 2
Q33. If the f(x) =
3ax b
11
5ax ¡ 2b
£
x 1
If x 1
x ¢1
is continuous at x = 1 find value of a and b
dy
Q34. If sin y = x sin (a + y). Prove that
dx
¥
sin 2 (a ¤ y )
sin a
Q35. If y = sin log x
Prove that x 2
©
Q36. If y = tan ¨1
¯
d2y
dy
¦x
¦ y§0
2
dx
dx
dy
1 « sin x « 1 ¬ sin x ª
® Prove that
dx
1 « sin x ¬ 1 ¬ sin x °
5 5
²
±1
2
Answers
1.
4.
4 13
x
3
2. –1
1
1
´
log10 x
8. –1
0
7. 0
10. ¶
3. 3log 5
56
6. –2x –1
5.
µ
³
9. for x
µ
I (I for integets)
R
Hints 2 Solutions of 4 Marks Questions
1.
Now f (0) = a
LHL
f(x) =
Lt
x½ 0
1 cos 4 x
x2
2sin 2 2 x
x2
Lt
· sin 2 x ¸
º
» 2x ¼
Lt
x½0–
Lt
x½0–
x½ 0–
[x ¾ 0– ¿ 2x ¾ 0–]
8(1)2 = 8
RHL
x
f(x) =
16
Lt
x ½ 0+
x ( 16
x 4)
=
16
x 16
Lt
2
8¹
x ( 16
x
Lt
( 16
x
4)( 16
4)
x
4)
Lt
x ½ 0+
16
4
x
x ½ 0+
x
4 8
Lt
x À 0+
x À 0+
f(x) = f(x) = 8
Lt
Lt
x ½ 0+
x ½ 0+
Á
2.
f(x) is continuous at x = 0 only. If a = 8
Here f(0) = c
LHL
f(n) =
Lt
xÀ 0 –
=
sin( a 1) x sin x
x
Lt –
xÀ 0
sin(a 1) x(a 1)
x(a 1)
Lt
sin( a 1) x
x
Lt –
sin x
x
xÀ 0
sin x
x
Lt
xÀ 0–
xÀ 0 –
[x ¾ 0– ¿ (a + 1)x ¾ 0–]
a+1+1= a+2
5 6
RHL
f(n) =
Lt
x à 0+
=
x [ 1 bx 1
bx x
x ( 1 bx 1)
1 bx 1
Â
bx x
1 6x 1
Lt
1 bx 1
bx  1 bx 1)
Lt
x à 0+
Lt
x à 0+
1
1 bx 1
Lt+
x à 0+
1
2
xÃ0
Now f is continuous at x = 0. If Lt+ f(x) = f(0) = f(x)
xÃ0
i.e.,
4.
a+2= c=
d2y
=
dx 2
cosec 2 È 2 É
=
1 cosec2 Ê 2
2a 2sin 2 Ê 2
1
cosec4 Ê 2
4a
=
1
cosec 4 Ò 4
4a
1
Ó4
4a
2
tan
1
2
1 dÈ
2 dÈ
Ô 3x 2 x Õ
Ö1 3 x Ø 2 x ×
Ù
Ú
tan–1 3x + tan–1 2x
Differentiate w.r.t. (x)
3
2
2
1 9 x 1 4 x2
8.
b Ä R – {0}
(RHL is independent of b)
a = 1 b = –1
x = a(Å – sin Å) y = a(1 – cos Å)
Differentiating both w.r.t. ‘Å’ we get
dy
dy
= a( 1 – cos Å)
= a sin Å
dÆ
dÆ
dy
dy / d Ç
a sin Ç
2sin Ç 2 cos Ç 2
=
dx
dx / d Ç a (1 cos Ç)
2sin 2 Ç 2
Differentiating w.r.t. ‘x’
Ì d yÍ
2 Ï
Î
Ð dx Ñ 0 Ë
7.
1
3
1
⇒a=
c=
2
2
2
xmyn = (x + y)m+n
taking logarithm both side
m log x + n log y = (m + n) log (x + y)
m
x
dy á n
dx ãå y
n dy
(m n) Û dy Ü
=
1
y dx
x y Ýß dx Þà
m n
m nâ
ä =
x y
x yæ
m
x
5 7
cot Ç 2
1
1
cosec 2 È 2
2
a (1 cos È)
1
a
dy ç nx
dx éë
9.
1 x2
ny my ny è
mx nx mx my
ê =
y( x y)
x( x y )
ì
dy
y
=
dx
x
1 y2
a( x
y)
x = cos A y = cos B
sin A + sin B = a(cos A – cos B)
í
2 sin ï
ò
A Bî
í A Bî
ô A Bõ
ô A Bõ
sin ö
ñ cos ï
ñ = 2 a sin ö
÷
÷
ò 2 ó
ø 2 ù
ø 2 ù
2 ó
ú
cot ü
þ
A Bû
ý = a
2 ÿ
A – B = 2 cot–1a
cos–1x – cos–1y = 2 cot–1a
Differentiate w.r.t. (x)
ð
1
1 x
1
dy
= 0
1 y dx
2
2
dy
=
dx
1 y2
1 x2
1
x = tan log y
a
a
tan 1 x
a tan–1x = log y ð e
12. Hint:
1
dy
aea tan x
=
dx
1 x2
dy
(1 x 2 )
= ay
dx
ay
1 x2
Again Differentiate w.r.t. (x)
(1 x 2 )
(1 x 2 )
d2y
dx 2
d2y
dx 2
ady
dy
2x =
dx
dx
(2 x a )
dy
= 0
dx
15. 4 2 b .
3a 2
16.
y = sin
1
5x
12 1 x 2 13
x = sin t t = sin–1 x
5 8
12
cos t 13
5 = r cos d 12 = r sin d r = 13
y = sin–1
y = sin
1
5sin t
13
r cos d sin t r sin d cos t 13
r
sin(t d ) t d
13
1
1 12
y = sin x tan
5
dy
1
=
dx
1 x2
= sin–1
17. (ii) f(x) = x2/3 [–1, 1]
clearly f(x) has definite and unique value for each x [–1, 1]
f(x) is continuous in [–1, 1]
2 13
which does not exist for x = 0 (–1, 1)
x
3
Hence Roll’s theorem is not applicable
1
18. (ii) f(x) = x
[1,3]
x
f(x) is defined in [1, 3]
f(x) is rational function such that denominator is not zero
for any value in [1, 3]
f(x) is continuous function in [1, 3]
1
f (x) = 1 2 which exist in (1, 3)
x
Hence all the condition of LMV are verified. Hence there exist atleast one value ‘c’ c (1, 3) such that
f(x) =
f (b) f ( a )
ba
10
2
1
1 2 = 3
c
3 2
f (c) =
only possible value
19.
c=
c
!
3 but " 3 # (1,3)
3 $ (1,3)
y = sin %1 ( x 1 x x 1 x 2 )
y = sin &1 ( x 1 ' ( x ) 2
%1
y = sin x sin
%1
'
2
(1
x 1 ' x 2 ) using sin–1 x – sin–1 y = sin )- x 1 + y
x
Differentiate w.r.t. (x)
dy
=
dx
1
1/ x
2
/
1
1/ x
0
20. Hint: Put x = cos 1
5 9
1
2 x
,
y 1 + x2 *
.
21. x x
2
3
2x
4
6
2
3
3
2 x log x 5 ( x 3) x
2
2 x
4
6x
2
3
2 x log | x 3 |5
3
7
22. x
[(1 log x) cos x x sin x log x] ( x cos x) x [log( x cos x) x tan x 1]
24. Hint :
y = sin2x
z = ecos x
Diff w.r.t. ‘x’
dy
dz
cos x
= 2sin cos x
= e 8 (sin x)
dx
dx
2sin x cos x
dy
dz
dy dx
=
=
9 0
cos x
sin x e
dz
dx
dz dx
2 cos x
=
ecos x
x
7
x cos x
1 ( x 3)( x 2 1) : 1
27.
2 3x 2 4 x 5 <> x 3
2x
x
2
6x 4 ;
1 3 x 4 x 5 =?
2
28. Given
( x a ) 2 ( y b) 2 = c2
Diff. twice w.r.t. (x)
2( x a) 2( y b) y1 = 0 @ ( x a) ( y b) y1
1 ( y b) y2
...(i)
0
2
1
y1 ( y1 0) = 0 @ ( y b)
Put that value of (y – b) from (iii) to (ii)
y1 (1 y12 )
x–a=
y2
Now put the value of (iii) and (iv) in (i) we get
y12 (1 y12 ) 2 (1 y12 ) 2
= c2
y22
y22
(1 y )
y2
y x log y yx y 1 x x (1 log x)
29.
x y x 1 x y log x
30. Hint :
f(x) = | x 1| | x 2 |
f(x) =
f(x) =
( x 1) ( x 2)
x 1
( x C 1) C ( x C 2) 1 D x E 2
x C1G x C 2
xH2
x L1
M J2 x K 3
N
P
N
R
1
2x J 3
1O x L 2
xQ2
Ans. f(x) is continuous at x = 1 & x = 2 lut not
Differentiable at x = 1 and x = 2
33. a = 3
b=2
6 0
...(iii)
...(iv)
(1 y12 )3 2
= c
y2
A
B
F
B
I
...(ii)
ApplicationofDerivative
5
Teaching-Learning Points
l
l
Concept of derivative originated from the study of rate of change of one quantity with respect to the
other. So, the notion of derivatives has a wide range of application in basic sciences, engineering,
economics and other field. In this chapter we shall learn, derivative as a Rate measure, tengents and
normal, increasing & decreasing function, maxima and minima of function and approximation.
Derivative as a Rate Measure : If a quantity ‘y’ varies with another quantity x, satisfying some rule
dy
x = x0 represents the rate of change of y w.r.t ‘x’ at x = x0
dx
Exp. Area of circle depends upon radius
A = Sr2
dy
= 2Sr will represent rate of change of area w.r.t radius
dr
If two variables x and y varying w.r.t another vareable. If y = f(t) and x = g(t) by chain rule
y = f(x) then
l
dy
dy dt dx
=
T0
dx
dx dt dt
Hence rate of change of y w.r.t ‘x’ can be calculated by using rate of change of both y and x with
respect to t.
Increasing and Decreasing functions–
(i) Teacher must explain the pre-idea of increasing and decreasing functions graphically.
(ii) If f (x) be a real valued continuous function defined on (a, b) is said to be increasing function
on (a, b) if
x1, x2 U(a, b) such that x1 < x2 V f(x1) < f(x2)
OR
x U(a, b)
f W(x) > 0
6 1
if f(x) be real valued continuous function defined on (a, b) is said to be decreasing in (a, b)
if x1, x2 X (a, b) such that x1 < x2 Y f(x1) > f(x2)
OR
f \( x) Z 0
l
x [ ( a, b)
Tangents and Normals :
(i) If we use the Geometrical meaning of the derivative, then f ](x) at (x1, y1) represents the slope of
the tangent at the point where f(x) is continuous and differentiable.
(ii) Slope of the Normal to the curve y = f(x) at point (x1, y1) is given by
1
^ dy _
`
a
b dx c ( x , y
1
1)
(iii) Since the tangent to the curve y = f(x) at point (x1, y1) is a straight line. Hence tangent will also be
(x x )
represented by an equation. So equation of tangent is given by y
given by y
y1 h
1
i dy j
k
l
m dx n ( x , y
1
l
( x x1 )
1)
(iv) Tangent and Normal to any curve y = f(x) at a given point are the lines passing the point and
perpendicular to each other.
Approximations :
Let f : D o R, D p R, R set of real number
and
y = f(x)
qy = f(x + qx) – f(x)
and also
Y
l
y1 e
dy d 1
and normal is
f
dx g ( x1 , y1 )
dy
= f \( x)
dx
r dy s
qy = u
v .tx
w dx x
if dx y qx then dy y qy
Hence we get the approximate value of qy
Maximum and minium :
Before attempting the exercise the students must know the following facts :
(i) | ax b | z 0 x { R , where a, b any real number
(ii)
1 | sin( ax b) | 1, x } R
(iii) 0 ~ | sin( ax b) | ~ 1,
(iv)
a | a sin(x) a,
x [R
x R and a 0
(v) (ax b) 2 0 x R
First derivative test – let f(x) be a differentiable function defined on Interval I and x = C X I, then
(a) x = c is said to be a point of local maxima, if
(i) f ](x) = 0, and
6 2
(ii) f
(x) changes sign from (+) ive to (–) ive as increases through c i.e., f
(x) < 0 at every point
sufficiently close to and the left of C, and f
(x) < 0 at every point sufficiently close to and to the
right of C.
(b) x = c is called the point of local minima, if
(i) f
(c) = 0 and
(ii) f
(x) changes sign from (–)ive to (+)ive as x increases through c i.e., f
(x) < 0 at every point
sufficiently close to and the left of c and f
(x) > 0 at every point sufficiently close to and to the
right to c.
(c) f
(x) = 0 and f
(x) does not change signs as x increases through c i.e., f
(x) has the same sign in the
complete neighbourbood of c, then point c is neither a point of local maxima and nor a point of local
minima. Such point is said to be a point of inflexion.
l
Second derivative test :
Theorem let f he a real valued function having second derivative at c such that
(i) f
(c) = 0 and f
(c) > 0 then f has a local minimum value at c
(ii) f
(c) = 0 and f
(c) < 0 then f has a local maximum value at c
(iii) f
(c) = 0 and f
(c) = 0 test fail
Question for Practice
Very Short Answer Type Questions (1 Mark)
Q1. Find the rate of change of area of circle with respect to its radius r when r = 3cm.
Q2. If the radius of a circle is increasing at the rate of .7 cm/sec at what rate its circumference is increasing.
Q3. What is the point on curve y = 3x2 –1 at which slope of tangent is 3.
3x 2
tangents makes 60° with x axis.
2
Q5. What is the slope of the normal to the curve
y = a sin3 at = /4
x = a cos3
Q4. At what point on curve y
Q6.
Q7.
Q8.
Q9.
Q10.
What is the point on the curve y x 2 2 x 3 at which tangent is || to x-axis.
What is maximum value of |sin 2x + 3|
What is the minimum value of |4 sin 2x + 3|
What is the max value of sin x + cos x
Find an angle which increases twice as fast as its sine.
Short Answer Type Questions (4 Mark)
Q1. A man of height 2 m walks at a uniform speed of 5 km/hr away from a lamp post which is 6 m high.
Find the rate at which shadow increases.
Q2. The two equal sides of an isoceles triangles with fixed base b are decreasing at the rate of 3cm/sec.
How fast is the area decreasing when two equal sides are equal to the base.
6 3
Q3. Water is leaking from a conical funnel at the rate of 5cm3/sec. If the radius of the base of the funnel
is 10 cm and its height is 20 cm. Find the rate at which the water level is dropping when it is 5 cm
away from the top.
Q4. Water is dripping out from a conical funnel at the uniform rate of 4 cm3/sec through a tiny hole at the
vertex in the bottom. When the slant height of the water is 3 cm. find the rate of decrease in slant
height of the water, given that the vertical, angle of the cone is 120°.
Q5. Find the points on the curve 9y2 = x3 where the normal to the curve makes equal intercepts on its
coordinate axis.
x y
Q6. Show that the line 1 touches the curve y be x / a at a pt where is crosses y-axis.
a b
x2 y 2
Q7. Find the points on the curve
1 at which tangents are
9 16
(i) parallel to x-axis
(ii) parallel to y-axis
Q8. A kite is 120 m height and 130 m of string is out. If the kite is moving horizontally at the rate of
5.2 m/s. Find the rate at which string is being paid out at that instant.
Q9. Using differential find approximate values of
(i) .037
(ii) .0037
Q10. Find the intervals in which the following function are strictly increasing or strictly decreasing:
(i) f(x) = 20 9 x 6 x 2 x 3
(ii) f ( x) x 3 12 x 2 36 x 17
(iii) f ( x)
3 4 4 3
36
x x 3x 2
x 11
10
5
5
(iv) f ( x) x 4 2 x 2
(v) f ( x) ( x 2)3 ( x 1) 2
(vi) f ( x) ( x 2)e x
(vii) f ( x) x 2 ( x 2) 2
(viii) f ( x) log(1 x)
x
x 1
1 x
(ix) f ( x) sin x ¡ cos x (0, 2¢ )
(x) f ( x ) £ sin 3 x (0, ¤ / 2)
Q11. Find the least value of ‘a’ so that the function f(x) = x2 + ax + 1 is strictly increasing on (1, 2).
Q12. Find the equation of tangent lines to the curve y ¥ 4 x3 ¦ 3 x § 5 which are perpendicular to the line
9 y ¨ x ¨ 3 © 0.
Q13. Show that the curves x y 2 xy k cut orthogonally if 8k2 = 1
Q14. If the tangents to the curve y ª x3 « ax « b at P(1, –6) 11 to the line y – x = 5. Find the values of a
and b.
Q15. A particle moves along the curve 6 y x 3 2 . Find the points on the curve at which y coordinate is
changing 8 times as fast as x coordinate.
6 4
Q16. Find the absolute maximum and munimum value of f ( x) ¬ 2 x 3 9 x 2 ® 12 x 5 [0,3]
Q17. Find the local maximum and local minimum of f ( x) ± sin 2 x ¯ x
¯°
2
²
x²
°
2
4
4
Q18. Find the local maximum and local minimum f ( x) ³ sin x ´ cos x 0 µ x µ ¶
2
Q19. If y ³ a log | x | ´bx ´ x has extreme value at x = –1 and x = 2 find a and b.
Q20. If y ¸
ax · b
has a turning point (2, –1) find the value of a and b.
( x · 1)( x · 4)
Long Answer Type Questions (6 Mark)
Q1. A rectangle is inscribed in a semicircle of radius r with one of its side on the diameter of semicircle.
Find the dimensions of the rectangle so that its area is maximum. Also find the max area
Q2. A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of
the window is 10 m. Find the dimensions of the window to admit maximum light through the whole
opening.
Q3. A wire of length 36 m is cut into two pieces. One of the piece is turned into the form of a square and
other in the form of equilateral triangle. Find the length of each piece so that sum of areas of two
figures be minimum.
Q4. Given the perimeters of circle and square, show that sum of area is least when side of square is
double the radius of circle.
Q5. Find the point on the parabola y2 = 4x which is nearest to the point (2, –8).
Q6. Prove that semivertical angle of right circular cone of maximum volume and of given slant height is
Q7.
Q8.
Q9.
Q10.
Q11.
tan ¹1 2 .
Prove that of all rectangles with given area, the square has the smallest perimeter.
Prove that all the rectangles with given perimeter, the square has largest area.
Prove that the perimeter of a right angled triangle of given hypotenuse is maximum when the triangles
is isosceles.
Show that of all rectangles inscribed in a given fixed circle. The square has the maximum area.
An open box with a square box is to be made out of a given quantity of sheet of area c2. Show that
c3
.
6 3
Q12. If the lengths of three sides of a trapezium other than base are equal to 10 cm each, then find the area
of the trapezium when it is maximum.
Q13. A point on the hypotenuse of right angles triangle is at a distance a and b from the sides. Show that
the minimum length of the hypotenuse is (a a 3 º b 2 3 )3 2 .
max volume of the box is
Q14. Show that the volume of the largest cone that can be inscribed in a sphere of radius R is
volume of sphere.
6 5
8
of the
27
Q15. Show that maximum volume of cylinder which can be inscribed in a cone of height ‘h’ and semivertical
4
3
2
»h tan d.
‘d’ is
27
Q16. Find the area of greatest isosceles triangle that can be inscribed in a given ellipse having its vertex
coincident with one extremity of major axis.
Q17. Show that the volume of the greatest cylinder which can be inscribed in a cone of height h and
4
3
¼h .
semivertical angle 45° is
27
Q18. Show that the right circular cone of least curved surface area and given voluem has an altitude equal
to
2 times the radius of base.
Q19. Show that a cylindrical vessel of given volume has the least surface area when its height is twice its
radius.
Q20. Find the equation of the line through the point (3, 4) which cuts from the first quadrant a triangle of
minimum area.
Answers
Answer (1 Mark)
1.
6 ½ cm2/sec
2. 1.4 ½ cm/se
3. x ¾
5.
1
6. (1, 2)
7. 4
9.
10.
2
dÃ
d
= 2 sin Ä
dt
dt
dÃ
dÅ
= 2 cos Å
dt
dt
1
cos Æ =
2
Æ = ½/3
Answer and Hints of 4 Mark Question
1.
2.
5
km/hr
2
Let ABC is isoceles Ç
AB = AC
At any time
AS = AC = a
6 6
1
y ¾ ¿1 4
2
4.
À
Á 1,
Â
8. 1
3
2
Èb
2
2
AD = AB – BD = a É Ê
2
2
2
Ë2
A=
1
BC Ì AD
2
A=
1 b 4a 2 Í b 2
2
2
Î
b
4a 2 Í b 2
4
b 1
da
dA
2
2 Ï1 2
= 4 Ð 2 (4a Ñ b ) 4 2a dt
dt
when a = b
bÓb
dA
Ô
dt
2
2
Ò
ab
da
4a 2 Ñ b 2 dt
(Õ3) Ô Õ 3b cm 2 /s
4b Õ b
–ve sign indicate the area of isosceles Ö is decreasig at rate of
3 cm 2 /sec
3.
4
cm/sec
45×
4.
V he the volume of cone & l be the slant height of water at any time
V=
dl
dV
when
dt
dt
9y2 = x3
5.
3
(l sin 60Ù) 2 l cos 60Ù Ú
4 cm 3 /s
Û
Find
Ø
Ø
8
l3
lÛ3
Diff w.r.t. ‘x’ we get
9 Ì 2y
dy
= 3x2
dx
Ü
dy
dx
Ý
x2
6y
scape of normal to curve at x, y, = 2 1
6y
x1
Normal to the curve makes equal intercepts on coordinate axes its slope will be = ± 1
6 y1
x12 = ± 1
(x1, y1) lies on curve (1) 9 y12
ß
9á
ã
Þ
x13
2
x12 à
3
= x1
â
6ä
Ü
when x1 = 0 y1 = 0 when x1 = 4 y1 ç
x14
å
4 x13
æ
8
as normal makes equal intercept it cannot passing through
3
origin
8é
ë 4, 8 3
ì 3í
(i) for no real pt tangent is || to x-axis
(ii) at (± 3, 0) tangent is || to y-axis
2 cm/sec.
Ans. èê 4,
7.
8.
x1 å 0, 4
6 7
9.
(i) Hint : Let
(ii) Hint : Let
10. (i) f is strictly
f is strictly
(ii) f is strictly
f is strictly
(iii) f is strictly
f is strictly
(iv) f is strictly
f is strictly
x = .04
Ans. .1925 approximate
îx = – .003
x = .0036 îx = .0001
Ans. .0608 approximate
increasing in (1, 3)
decreasing in (– ï, 1) ñ (3, ï)
decreasing in (2, 6)
increasing in (– ï 2) ñ (6, ï)
decreasing in (– ï 2) ñ (1, 3)
increasing in (–2, 1) ñ (3, ï)
increasing in (–1, 0) ñ (1, ï)
decreasing in (– ï –1) ñ (0, 1)
ò
(v) f is strictly decreasing in ô 1,
ö
f is strictly increasing in (
(vi) f
f
(vii) f
f
(viii) f
f
is
is
is
is
is
is
strictly
strictly
strictly
strictly
strictly
strictly
7ó
õ
5÷
ú
ø7
ù
1) û ü , úý
þ5
ÿ
increasing in (– ï –1)
decreasing in (–1, ï)
increasing in (0, 1) ñ (2, ï)
decreasing in (– ï, 0) ñ (1, 2)
decreasing in (–1, 0)
increasing in (0, ï)
(ix) f is strictly increasing in
f is strictly decreasing in
(x) f is strictly decreasing in
f is strictly increasing in
0,
5ð
ð
, 2 ð
4 4
5 ,
4 4 , 6 2
0, 6
11. a = – 2
12. 9x – y – 3 = 9x – y + 13 = 0
14. Hints : Slpe of tangent = slope of line
dy
= 3x2 + a
dx
dy dx ! (1,
slop of line = 1
= 3+a=1a=–2
6)
Curve passes through (1, –6) 6 " 13
a b
#
b" 5
Ans. a = – 2 b = – 5
6 8
15. (4, 11)
$31
%
& $4,
'
3
16. pt of maxima is 3 and absolute max value is 4
pt of maxima is 0 and absolute man value is – 5
17. f has local min at
()
local min value
6
-
*
3
2
,
+
2
3 .
/
6
2 6
1
18. f has local min at
and min value is
2
4
f has local max at
f has local max at
local max value
0
and mix volue is 1
2
31
1
and min valueis
4
2
f has local min value at
21
19. a = 2 b =
20. y =
2
ax 3 b
( x 3 1)( x 3 y )
4
ax 3 b
x 3 5x 5 4
2
Domain x 6 R – {1, 4}
( x 2 7 5 x 8 4)a 7 (ax 7 b)(2 x 7 5)
dy
=
( x 2 7 5 x 8 4) 2
dx
dy
= 0 and x = 2 we get b = 0
dx
trining pt (2 – 1) lies on curve also
–1 =
2a 9 b
4 9 10 : 4
;
Ans. a = 1
2a – b = 2
b=0
Answer and Hints of Six Marks Questions
<BO
C= =
OC = r/radius of semicircle
Area of rectangle = AB × BC
AB = 20B = 2r cos=
A = r2 2sin= cos= = r2 sin 2=
BC = r sin =
1.
Let
2.
Radius of semi circle is
are
20
4
?
m and
10
side of rectangle
4 >
10
cm
4 ?
6 9
3.
Let the length of the piece bent in the form of a square be x cm, then the length of the piece bent in the
form of equilateral @ is 36 – x cm. Let s be the combined are of two figure
S=
B
D
E
x
4
2
C
2
3 B 36 A x
D
4 E 3
d5
= 0
dt
144 3
x=
9F 4 3
Now find
Ans.
12. 75 3 cm 2
13.
AC = AD + DC
l = a sec G + b cosec G
dl
= a sec G tan G – b cosec G cot G
dH
a sin I b cos I
J
cos I
sin 2 I
dl
= 0
dH
b
Lb
tan M N O
tan G =
Pa
a
K
AC = l
0
13
3
R
d 2l
a[sec3 S T sec S tan 2 S] T b[cosec3S T cos ecS cot 2 S] U 0 for 0 V S V
2 =
2
dQ
l = a[1 X tan
2
12
Y]
2
12
X b[1 X cot Y ]
W
Z a [1 X
\
23
23
23 12
23
23
23 12
l = a [a ] b ] ] b [a ] b ]
20. eq of line
3 4
a
a b
x y
_
a b
b1 c
`1
3b + 4a = ab c b =
dA
=0
da
49
ad3
ah3
A is munimum at a = 6 b = 8
7 0
12
W
X b [1 X
\
( a 2 3 ] b 2 3 )3 2
It passes through (3, 4)
1
1 a e 49 2a 2
f
A = Area of @ = ab f
2
2 ag3 ag3
find
^
b2 3
a2 3
a2 3
b2 3
12
6
Integrals
Teaching Points
l
Antiderivative or Primitives : If
d
[ f ( x)]
dx
i( x ),
the f(x) is called antideivative of f(x). As we
d
[ f ( x) j c] k l( x) so f(x) + c is also antiderivative of f(x), which depends on C (constant)
dx
as C may attain infinitely many values, therefore antderivative of a function is not unique.
know that
Now f(x) + c is called the indefinite integral of f(x) w.r.t ‘x’ which is written as
l
d q
f ( x ) dx = .f(x) and
s
ds r
Rules of Integration :
2.
v. f1 ( x) w xf 2 ( x)
dx y v f1 ( x)dx w x f 2 ( x)dx where a and b are constants.
Fundamental Integration Formulae
n
x n z1
| c, n } ~ 1
n |1
l
x dx {
l
1
dx t log x
x
l
d
[ f ( x)]dx t f ( x )
dx
k . f ( x ) dx u k f ( x )dx where k is any constant
1.
l
f ( x) o C
and C is known as constant of integration.
Integration : The process of finding integral is called integration. Thus differentiation and integration
are inverse process.
p
l
m( x ) dx n
c
e x dx e x c
ax
x
a dx
c
log ca
l
sin xdx
cos x c
l
cos xdx n sin x o c
l
sec 2 xdx tan x c
l
cosec2 xdx cot x c
l
sec x tan xdx sec x c
7 1
cosec
xc
l
cosec x cot xdx
l
tan xdx
log | cos x |
l
sec xdx
log | sec x tan x | c
l
cosec x cot xdx
1
x
c
4 2
x
log | cosec x cot x | c log tan c
2
dx sin 1 x c
2
l
l
f (ax ¥ b)dx
l
cos
1
F (ax ¥ b)
¥ c, a ¦ 0
a
F ( x ) § c i.e. if the integral of function of x is known, then if in place of x we have linear
f ( x )dx
If
l
log tan
xc
1 x
1
dx tan 1 x c ¡ cot 1 x c
2
1 x
1
dx sec ¢1 x £ c ¤ cosec –1 x £ c
x 2
x ¤1
l
l
log | sec x | c
c
function of x, the integral is of same form but it is divided by coefficient of x.
Following are some substitutions which are useful in evaluating integrals
Expression
Substitutions
a 2 ¨ x 2 or a 2 ¨ x 2
x a tan © / a cot ©
a 2 ª x 2 or a 2 ª x 2
x a sin © / a cos ©
x 2 ª a 2 or x 2 ª a 2
x
a sec « / a cosec «
a
a cos 2©
a¬x
ax
or
ax
a¬ x
Special Integrals
1
dx
x®a
log
2
2
2a
x ®a
x¯a
1
x
2
a
2
³a
2
dx
1
± x²
tan °1 ´ µ ³ c
a
¶a·
2
dx
1
a¸x
log
2a
a¹x
1
¹x
1
x
2
a
2
»a
2
1
x
ºa
2
1
x
2
dx
¸c
log x º x 2 » a 2
ºc
x¾
Á ¿c
 aÃ
½
2
Ä
¯c
dx sin ¼1 À
dx
log x º x 2 º a 2
ºc
7 2
Use of Partial Fractions
l
In some cases we find integral using partial fractions of the following types.
px Å q
(ax Å b)(cx Å d )
A
B
Å
(where Dr. can be factorised into linear factors)
(ax Å b) (cx Å d )
px Æ q
(ax Æ b)(cx Æ d ) 2
A
B
Æ
(cx Æ b) ( x Æ d ) 2
Æ
C
(where Dr. can be factorised into repeated linear factors)
cx Æ d
px 2 Ç qx Ç r
A
bx Ç c
Ç
2
2
(ax Ç b)(cx Ç dx Ç e) ax Ç b cx Ç dx Ç e
When cx2 + dx + e can not fractorised further.
When we have to evaluate the integral of the type.
e x [ f ( x ) È f É( x )]dx Ê e x f ( x ) È c
l
Some special Integrals
x 2 Ë a 2 dx =
x 2
x Ì a2
2
Ì
a2
log x Í x 2 Í a 2
2
a 2 Ë x 2 dx =
x 2
a Ñ x2
2
Ò
a2
ÏxÐ
sin Î1 Ó Ô Ò c
2
ÕaÖ
Íc
x 2
a2
a × x 2 × log x × a 2 × x 2 × c
2
2
Definite Integral : Let f(x) be a continous function defined on the closed integral [a, b] and F(x) be an
antiderivative of f(x) then
a 2 Æ x 2 dx =
l
b
f ( x)dx = [F( x )]ba
Ø
F(b) Ù F( a )
a
l
Definite Integral as a limit of sum
b
h[ f (a ) Û f (a Û h) Û f (a Û 2h) Û ... Û f (a Û n Ü 1h)]
f ( x)dx = hLt
Ú0
a
l
where nh = b – a
Properties of definite Integral
b
a
f ( x)dx Ý Þ f ( x )dx
P(1) :
a
b
b
a
f ( x)dx ß f (t )dt
P(2) :
a
b
b
c
b
f ( x)dx ß f ( x )dx à f ( x )dx
P(3) :
a
a
b
b
c
f ( x)dx á f (a â b ã x )dx
P(4) :
a
a
7 3
a
a
f (a ä x)dx
f ( x)dx
P(5) :
0
2a
a
a
f ( x)dx å f (2a æ x )dx
f ( x)dx
P(6) :
0
a
0
0
a
2a
f ( x)dx
P(7) :
0
ç
é2
é 0
é
ê0
f ( x)dx if f (2a è x)
f ( x)
if f (2a è x)
è
f ( x)
a
a
f ( x)dx
P(8) :
ëa
ì
î2
î 0
î
ï0
f ( x)dx if f (í x)
f ( x)
if f (í x)
í
f ( x)
Question for Practice
Evaluate the following Integrals
Very Short Answer Type Questions (1 Mark)
cos 6 x ñ x
dx
3 x 2 ñ sin 6 x
1
dx
x(2 ò 3log x)
1.
3.
1
5.
x ó x ô1
x4 ò 1
dx
x2 ò 1
x
dx
( x ò 1) 2
7.
9.
dx
2.
sin 2 xdx
4.
x 3 sin( x 4 ) dx
6.
e x ø log x ÷
8.
1
dx
1 ü sin x
10.
15.
x 99 cos 4 x dx
ý1
1
þ x ÿ dx
0
2
| x | dx
1
1ð x dx
1 x log 12.
0
1
13.
ú
| sin x | dx
14.
0
1
e
x
1ö
ù dx
xû
1
32
11.
õ
dx
ñ1
7 4
Short Answer Type Questions (4 Mark)
1 dx
x2 16.
e x log x 18.
sin( x a )
dx
sin( x a )
20.
sin 2 x
dx
2
2
a sin x b 2 cos 2 x
1
26.
x (log x ) 2 3log x 4
dx
x 1 x x 2 dx
28.
1
30.
x( x 9 1)
1
32.
x1 2
x1 3
dx
dx
34.
( x 2 # 1)( x 2 # 4)
dx
( x 2 # 3)( x 2 $ 5)
36.
ex '
& 2 % sin 2 x
( 1 % cos 2 x )
19.
1
dx
sin( x a ) cos( x b)
21.
tan x cot x dx
24.
sin 3 x cos5 x dx
1
sec x 1dx
22.
17.
dx
23.
x
dx
x x2 1
25.
sin x cos x
dx
sin(2 x)
27.
5x 2
dx
3x 2 x 1
29.
x2
dx
x2 5x 3
31.
x2 x 2
dx
( x 1)( x 2)
33.
a x .tan !1 (a x )
dx
a 2 x "1
2
1
3
4
*4
.a
39.
x sin 2 x
dx
sin x - cos 4 x
4
32
| x cos 1x | dx
41.
0
3
3
1
x sin x
dx
1 cos 2 x
0 4
1
| x 2 2 2 x | dx
42.
dx
x 5 sin( x 3 ) dx
37.
0
a/x
dx
a0 x
5
(sin x) (cos x) 4
,2
| x | + | x + 3 | + | x + 6 | dx
a
40.
4
35.
0
38.
cos3 x.cos( x a )
ecos x
dx
44.
ecos x 6 e 5 cos x
0
43.
45.
71
9 1 8 sin x :
< dx
> 1 = sin x ?
log ;
7 5
dx
Long Answer Type Questions (6 Mark)
tan x dx
46.
tan @ A tan 3 @
d@
1 A tan 3 @
47.
D2
B
x tan x
dx
48.
sec x C tan x
0
log(cos x)dx
49.
0
1
50.
log(1 E x)
dx
1 E x2
0
2sin 2G H cos G
dG
6 H cos 2 G H 4sin G
52.
dx
sin x(3 F 2 cos x)
51.
5
53. Evaluate ( x 2 I 3 x)dx as limit of sum.
2
2
54. Evaluate ( x 2 J x J 1)dx as limit of sum.
1
3
55. Evaluate (3 x 2 K e 2 x )dx as limit of sum.
0
Answers
1.
1
log | 3 x 2 L sin 6 x | L c
6
2.
x sin 2 x
M
Nc
2
2
3.
1
log | 2 O 3log x | O c
3
4.
P
5.
7.
9.
3
S
2 R 32
2
V x T ( x T 1) W U c
3
x 3 Z x [ 2 tan Y1 ( x) [ c
1
log | x ^ 1| ^
^c
( x ^ 1)
6. e x log x X c
8. tan x \ tan x sec x ] c
10. 0
1
2
13. 1
11.
15.
12. 0
14. 4
` log |1 a e
6
_x
b
x
16. e elog x d
| ac
g
1c
Fc
x fh
8
cos x cos x
j
jc
6
8
1
sin( x m a)
19. cos(a m b) .log cos( x m b)
17.
1
cos( x 4 ) Q c
4
i
18. x cos 2a k sin 2a log | sin( x l a ) | k c
nc
20.
1
(a
2
pb
2
)
log | a 2 sin 2 x o b 2 cos 2 x | o c
7 6
21.
23.
q2
sin a
cos a q sin a tan x r c
22.
2
| 2 x ~ 1}
1
tan {1
~c
3
3
28.
2 log x 3
2
26. log
1
3
3
2 2
(1 x x )
x9 « 1
1
log
30.
x9
9
12
¯x
32. 6 ³
µ
3
±
1 2 x 1
2
¢ 1 x x
£¡
¥
¦
2§ 4
x1 3
2
¬c
±
(log x) 2 3log x 4
5 1 2 x 1
sin ¡
¢¤ c
¥
8
5 ¦¨
©c
31. x 2 log | x 1| ®4 log | x 2 | c
°
¶
34. x ½
1
4 3
»
tan ¹1 ¾
À
x ¼ 27
xº 5
log
¿½
3Á 8 5
x½ 5
1
39.
Æ
Æ
47.
36. e x tan x à c
38. 29
2
40. aÇ
42. 2
2
4
45. 0
46.
Âc
1
[Ä x 3 cos ( x 3 ) Å sin ( x 3 )] Å c
3
8
5È É 2
41.
2È 2
43.
c
x1 6 ± log | x1 6 ² 1|´ ± c
1
[tan ·1 (a x )]2 ¸ c
2 log a
35. 4(tan x) 4
37.
sc
5
11
3 x 1
log | 3 x 2 2 x 1|
tan 1
c
6
3 2
2
5
19
2 x © 5 ª 13
2
log
29. x ª log | x © 5 x © 3 | ©
2
2 13
2 x © 5 © 13
33.
cos 2 x s cos x
2 sin 1 (sin x cos x)
c
24.
25. sin 1 (sin x cos x) c
27.
t 2 cos x s 1u
v log w
xs
y
z
2
44.
Ê
2
1
1
tan x Ì 2 tan x Í 1
Î tan x Ì 1 Ï
tan Ë1 Ð
log
ÑÍ
Ò 2 tan x Ó 2 2
2
tan x Í 2 tan x Í 1
Õ1
Íc
1
1
Ö 2 tan x Õ 1×
log |1 Ø tan x | Ø log | tan 2 x Õ tan x Ø 1| Ø
tan Ô1 Ù
Ú Øc
Û
3
6
3
3 Ü
7 7
½c
48.
50.
ÞÝ
ß
Ý á à 1â
ã2
ä
ç
log 2
8
åæ
49.
51.
2
log 2
1
1
2
log |1 è cos x | è log |1 é cos x | é log | 3 é 2 cos x | é c
10
2
5
52. 2 log | sin 2 ë ì 4sin ë í 5 | í7 tan ê1 (sin ë ì 2) í c
141
53.
2
29
54.
6
55.
(53 î e6 )
2
Hints
19.
1
cos(a ï b)
dx
cos(a ï b) sin( x ï a ) cos( x ï b)
=
1
cos[( x ñ b) ñ ( x ñ a )]
dx
cos(a ñ b) sin( x ñ a) cos( x ñ b)
=
1
[cot( x ò a ) ó tan( x ò b)]dx
cos(a ò b)
1 ô cos x
dx
cos x
22.
sin x
=
cos x õ cos 2 x
1
35.
3
4
(sin x) (cos x)
5
4
dx
Put cos x = t
dx
Dividing num. and den. by cos2x, we get
3 5 8
ö
÷
sec 2 xdx
ó
2ù
=
ø
3
4 4 4
ú
û
(tan x) 4
Now put tan x = t
ü
39.
2
I=
0
þ
I=
I=
x sin 2 x
dx
sin x ý cos 4 x
4
ð
ÿ
4
4
2
0
2
0
sin 2 x
dx
4
sin x cos 4 x
a
Using prop. f ( x)dx
0
2 tan x sec 2 x
dx
tan 4 x 1
Put tan2x = t
7 8
a
f (a x)dx 0
ax
dx
a x
a
40. I =
a
ax
a
I=
a2 x2
a
a
I=
a
adx
a x
a
2
a
dx
2
dx
a
a
f ( x)dx 2 f ( x)dx if f ( x) f ( x) a
0
0
if f ( x) f ( x) 0
a2 x2
0
a x2
a
dx
I = 2a
x
2
I = a.
32
12
| x cos x | dx
41.
32
( x cos x)dx 0
0
( x cos x)dx
12
Use integral by parts to evaluate ( x cos x)dx.
Put tanx = t2
tan x dx
46. I =
2t 2
I=
t4 1
dt
t2 1
t2 1
dt 4
dt
t4 1
t 1
I=
1!1 t2
1"1 t2
dt ! 2
dt
I=
2
1
!
t
1
t
2
t ! 2
t
1#1 t2
I=
2
% 1&
') t $ (* # 2
t
Put t +
1
t
dt #
a
1$1 t2
dt
(t # 1 t ) 2 $ 2
and
t,
1
b
t
da
db
- 2
2
a - ( 2)
b . ( 2) 2
I=
2
log(1 / x)
dx
1 / x2
0
1
50. I =
Put x = tan0
7 9
14
I=
log(1 2 tan 3)d 3
0
4
4
I=
a
7
Using
property
f ( x)dx
:
0
<
log 2d 5 6 I
0
I=
>
8
a
0
log 2.
dx
sin x(3 ? 2 cos x)
51.
=
sin xdx
(1 @ cos x)(3 ? 2 cos x)
=
1
dt
(t A 1)(t B 1)(3 B 2t )
Put cos x = t
2
5
( x 2 C 3 x)dx
52.
2
b
As
a
n
f ( x)dx = Lt h
h 0
5
F
2
( x 2 G 3 x ) dx = Lt h
h 0
f (a E rh), where h
r 1
n
bDa
n
[10 H 7rh H r 2 .h 2 ]
r 1
n
n
I n
J
2
r2 M
= Lt h L 10 K 7 h r K h
h 0
r 1
r 1
Nr 1
O
Q
= Lt h S10n P 7h
h 0 U
W
= Lt h Y10nh ?
h 0 [
=
n(n P 1)
n(n P 1)(2n P 1) R
P h2
TV
2
6
7(nh)(nh ? h) (nh)(nh ? h)(2nh ? h) X
?
Z\
2
6
141
2
8 0
8
f (a 9 x)dx ;
=
ApplicationofIntegrals
Teaching Points
l
The area bouned by the curve y = F(x) above the x-axis and
between the lines x = a, x = b is given by
b
b
ydx =
a
l
F ( x)dx
a
If the curve between the lines x = a, x = b lies below the x-axis,
then the required area is given
b
b
(] y )dx =
a
l
b
ydx ^ F ( x )dx
a
a
The area bounded by curve y = F(x), x-axis and between lines
x = a, x = b to given by
b
F ( x)dx
a
l
The area bounded by the curve x = F(y) above the y-axis and the
lines y = c, y = d is given by
d
d
xdy =
c
l
F ( y ) dy
c
If the curve between the lines y = c, y = d lies below the y-axis
(to the left of y-axis) then the area is given by
d
d
d
(_ x ) dy = ` xdy a F ( y )dy
c
c
l
c
The area bounded by curve x = F(y), y-axis and between lines
y = c and y = d is gives by
d
| F ( y ) | dy
c
l
If 0 £ g(x) £ f(x), the area of region bounded between curves
and ordinates x = a and x = b is given by
b
[ F ( x ) b g ( x )]dx
=
a
8 1
7
l
When we find the area bounded by the curves y = f(x)
and y = g(x) and after drawing the graphs the shaded
region is of such type.
We find the x-cordinate of their point of inter-sections.
Let for point A and B the values of x are a and b. Then
b
Required Area = [ f ( x) c g ( x)]dx
a
Note : If the power of ‘x’ is even in the given curve then the graph of the curve is symmetric about y-axis.
If equation of curve contains only even power of ‘y’ then the graph is symmertic about x-axis. If curve
contains even power in both ‘x’ and ‘y’ then graph is symmetric about both axis.
Question for Practice
2
2
Q1. Find the common area bounded by the circles x 2 d y 2 e 4 and ( x f 2) g y h 4 .
Q2. Using integration find area of region bounded by the triangle whsoe vertices are
(a) (–1, 0), (1, 3) and (3, 2)
(b) (–2, 2) (0, 5) and (3, 2)
Q3. Using integration find the area bounded by the lines.
(i) x + 2y = 2,
y–x=1
and
2x + y – 7 = 0
(ii) y = 4x + 5,
y=5–x
and
4y – x = 5.
Q4. Find the area of the region i( x, y ) :| x f 1| k y k 5 f x 2 j .
Q5. Find the area of the region bounded by
y2 = x
and
line x + y = 2
Q6. Find the area enclosed by the curve y = sin x between x = 0 and x = 3l/2 and x-axis.
Q7. Find the area bounded by semi circle y =
25 m x 2 and x-axis.
Q8. Find area of region given by [( x, y ) : x 2 n y n | x |] .
Q9.
Q10.
Q11.
Q12.
Q13.
Q14.
x2 y 2
o p 1 and straight line 2x + 3y = 6.
Find area of smaller region bounded by ellipse
9
4
Find the area of region bounded by the curve x2 = 4y and line x = 4y – 2.
Using integration find the area of region in first quadrant enclosed by x-axis the line x q 3 y and
the circle x2 + y2 = 4.
Draw a rough sketch of the region [( x, y ) : x 2 r y 2 s 4 s x r y ] and find its area.
Find the smaller of two areas bounded by the curve y = |x| and x2 + y2 = 8.
Find the area lying above x-axis and included between the circle x2 + y2 = 8x and the parabola y2 = 4x.
8 2
Q15. Using integration find the area enclosed by the curve y = cos x, y = sin x and x-axis in the interval
u tv
wy 0, 2 xz .
6
Q16. Sketch the graph y = |x – 5|. Evaluate | x { 5 | dx .
0
Answers
1.
} |
~
8 3 2 3 sq units
3.
(a) 6 sq units (b)
5.
9
sq units
2
8.
1
sq units
3
11.
2. (a) 4 sq units, (b) 2 sq units
5 1 2
1 1 1
4. sin sin sq units
5
5 2
2
25
6. 3 sq units
7.
sq units
2
15
sq units
2
9.
sq units
3
8
14.
(8 3 ) sq units
3
3
( 2) sq units
2
13. 2 sq units
15. (2 2) sq units
16. 5 sq units
2
1
4 ( x 2) 2 dx 4 x 2 dx ¡ 0
¢
1
Required area = 2
2
4.
Required area =
1
5 ¤ x dx ¤ (¤ x ¥ 1)dx ¤ ( x ¤ 1)dx
£1
1
5.
Required area =
2
2
£1
¦2
4
¦2
y 2 dy
8
14. Required area = 2 xdx ¨
0
1
1
(2 § y ) dy §
9
sq units
8
12. ( – 2) sq units
Hints
1.
10.
16 © ( x © 4) 2 dx
4
8 3
DifferentialEquations
8
Teaching Points
Definition. A differential equation is an equation which involves unknown functions and their derivatives
w.r.t. one or more independent variables.
An ordinary differential equation is an equation which involves only one independent variable and
derivatives w.r.t. that independent variable.
In this unit, we shall be dealing only with ordinary differential equations.
Order and Degree of a Differential Equation
The order of a differential equation is the order of the highest order derivative occuring in the differential
equation and its degree is the index of the highest order derivative which appears in the differential equation
after making it free from negative and fractional powers so far as the derivatives are concerned.
Linear Differential Equation
Definition. A differential equation ª( x, y ¬, y ¬¬,..., y ( n ) ) « 0 is called linear iff the function is a linear
(n)
function of the variables y, y ¬, y ¬¬,..., y i.e. iff the dependent variable and its derivatives occur only in the
first degree and are not multiplied together. Otherwise, it is called non-linear.
Thus a general linear differential equation of order n may be written as
dny
d n ®1 y
d n® 2 y
P0 n ¯ P1 n ®1 ¯ P2 n ® 2 ¯ ... ¯ Pn y ° Q
...(1)
dx
dx
dx
where P0, P1, P2,..., Pn and Q are functions of x only (not containing y) and P0 ± 0.
The linear differential equation (1) is called homogeneous iff Q is the zero function, and it is called
non-homogeneous iff Q is a non-zero function.
Note that a linear differential equation is always of the first degree but every differential equation of
the first degree need not be linear.
Formation of a differential equation.
Consider the general equation (primitive) ª( x, y, c1 , c2 ,..., cn ) « 0 containing n independent arbitrary
constants (called parameters) c1 , c2 , c3 ,..., cn apart from x and y..
To obtain the differential equation which is satisfied by ²( x, y, c1 , c2 , c3 ,..., cn ) ³ 0,
we differentiate this equation n times w.r.t.x and eliminate n arbitrary constants c1 , c2 ,..., cn from the
n + 1 equations; n equations obtained on differentiation and the given equation. In this way, we obtain an
ordinary differential equation
8 4
(i) of an order equal to the number of independent arbitrary constants in the given equation,
(ii) consistent with the given equation i.e., satisfied by the given equation and
(iii) free from arbitrary constants.
Solution of a diferential Equation
Definition. A solution of a differential equation is a relation between the variables, which satisfies the
given differential equation.
Let the equation involving x,y and n independent arbitrary constants be
´( x, y, c1 , c2 ,..., cn ) = 0
...(i)
and the differential equation obtained from (i) be
µ
¹
f · x, y ,
dy d 2 y
dn y¶
, 2 ,..., n ¸ = 0
dx dx
dx º
...(iii)
then (i) is called the most general solution (or complete primitive or complete solution) of (ii).
Thus complete solution of an ordinary differential equation of order n contains n independent arbitrary
constants.
The most general solution of an ordinary differential equation of first order contains one arbitrary
constant and of second order contains two arbitrary constants and so on.
Definition. Any solution obtained from the general solution of a differential equation by giving
particular values to some or all the arbitrary constants is called a particular solution or particular primitive.
Equation with Variables Separable.
An equation whose varibles are separable can be put into the form
f1 ( x)dx » f 2 ( y )dy = 0
Integrating, the general solution of this equation is
f1 ( x)dx ¼ f 2 ( y )dy = C, where C is an arbitrary constant.
Reducible to Variables Separable.
Type 1. To solve an equation of the type
dy
= f(ax + by + c), put ax + by + c = v and it will be reducible
dx
to variables separable.
Type 2. Homogeneous Equation
A differential equation of the form
dy f ( x, y )
½
where f(x, y) and g(x, y) are both homogeneous
dx g ( x, y )
functions of the same degree in x and y i.e. and equation of the form
differentiable equation.
dy
¾ y¿
À F Á Â is called a homogeneous
à xÄ
dx
n Å yÆ
A homogeneous function of degree n in x and y is a function which can be written as x Ç É Ê , n È N . .
Ë xÌ
To solve homogeneous differential equation:
8 5
dy
dy
dv
and on substituting these values of y and
in the given differential
Í vÎ x
dx
dx
dx
equation it will be reducible to variables. separable.
put y = v x, then
Note. Just as the differential equation of the form
dy
Ï F ( y x) can be solved by the substitution
dx
y = y x, the differential equation
Ð xÑ
dx
Ò G Ó Ô , which is also homogeneous, can be solved by the substitution x = vy..
dy
Õ yÖ
Linear Differential Equation (of first order)
Definition. A linear differential equation of the first order in y is an equation of the form.
dy
× Py Ï Q , where both P and Q are functions of x only (not containing y).
dx
If Q ¹ 0, then the above equation is called non-homogeneous linear differential equation and if Q = 0,
then it is called a homogeneous linear differential equation.
A linear differential equation of the first order in x is an equation of the form
dx
Ø P1 x Ù Q1 , where
dy
both P1 and Q1 are functions of y only (not containing x).
Solution of Linear Differential Equation.
(i)
Pdx
If dy Ú Py Û Q , where P and Q are functions of x only, then it has eÜ as integrating factor and its
dx
solution is given by y.eß
(ii) If
Pdx
Pdx
à Ý Q.eß Þ dx á C .
â
ã
P dy
dx
Ø P1 x Ù Q1 , where P1 and Q1 are functions of y only, then it has eä 1 as I.F and its solution is
dy
given by x.eå
P1dy
æ (Q1.eå
P1dy
)dy ç C .
Question for Practice
Evaluate the following Integrals
Very Short Answer Type Questions (1 Mark)
Q1. What is integrating factor of ( x log x).
dy
è y é 2 log x .
dx
8 6
d2y
ê 9y ë 0 .
dx 2
Q3. Show that ( x 2 ì xy )dy í ( x 2 ì y 2 )dx is a homogeueous differential equation.
Q2. Show that y = 4 sin 3x is a solution of differential equation
3
d 3 y î d 2 y ï dy
Q4. Write the order and degree of differential equation
ñ
ñ ñ 4 y ò sin x .
dx 2 óõ dx 2 ôö dx
2
÷
dy ø
Q5. Write order and degree of differential equation ý1 û ùÿ úð þ
dx þ
ý
32
d2y
ü 2 .
dx
Short Answer Type Questions (4 Marks)
Q6. Solve the differential equation.
( x y x y )dx ( x y x y )dy 0
Q7. Show that y = a cos(log x) + b sin(log x) is a solution of the differential equation
d2y
dy
êx êyë0
x
2
dx
dx
2
Q8. (i) Form the differential equation of the family of circles touching y-axis at (0,0).
(ii) Form the differential equation of family of parabolas having vertex at (0,0) and axis along the (i)
positive y-axis (ii) + ve x-axis.
(iii) Form differential equation of all circles passing through origin and whose centre lie on x-axis.
dy x y
is homogeneous and solve it.
dx x y
Q10. Show that the differential equation :
(x2 – 2xy) dy + (2y2 – 3xy + x2) dx = 0 is homogeneous and solve it.
Q11. Solve the following differential equations :
Q9. Show that the differential equation
dy
y cos 2 x.
dx
dy
(ii) sin x
+y cos x = 2 sin2 x cos x
dx
Q12. Solve each of the following differential equations :
dy
2 dy 2 y (i) y x
dx
dx (i)
(ii) cos y dx ñ (1 ñ 2e x ) sin y dy ò 0
(iii) x 1 y 2 dy y 1 x 2 dx 0.
(iv)
(1 x 2 )(1 y 2 ) dy xy dx 0.
(v) ( xy 2 ê x)dx ê ( yx 2 ê y )dy ë 0; y (0) ë 1.
dy
3
3
x
y sin x cos x xy e .
(vi)
dx
8 7
Q13. Solve the following differential equations :
(i) xy dy y dx x 2
!
#
(ii) y ( x cos *,
.
y 2 dx.
y$
!
# y$ "
# y$
# y$ "
+ % y sin * + ) dx & x ( y sin * + & x cos * + ) dy ' 0.
, x-/
, x, x-/
x.
(iii) x 2 dy 0 y ( x 0 y )dx 1 0 given that y = 1 when x = 1.
dy
(iv) x 2 ay 3 x 4 1, x 5 0
dx
(v) ( x 3 6 3 xy 2 )dx 7 ( y 3 6 3 x 2 y )dy.
Q14. Solve the following differential equations:
(i) x 2 y dx 6 ( x 3 8 y 3 )dy 7 0
(ii) y dx 9 ( x : 2 y 2 )dy ; 0
(iii) ( x 2 6 y 2 )dx 8 2 xy dy 7 0, y (1) 7 1.
(iv)
<
@
B
y sin
<
=
x=
x
A dx > @ x sin ? y A dy
yC
y
B
C
(v)
dy
dx
F
y
D yE
G tan H I
J xK
x
(vi)
dy
dx
7
2 xy
x 8 y2
(vii)
dy
dx
M
e xL y N x 2e y
2
(viii) ( x O 3 y 2 )dy P y dx
Answers
1.
log x
4. order = 3. degree = 1
5.
order = 2, degree = 2
1
6. x R x 2 S y 2
8.
2
2
(i) x T y U 2 xy
9.
tan [1 `
y]
a
b xc
\
^
dy
dx
V
0 (ii) (a)
1
log( x 2 _ y 2 ) _ c
2
dy
dx
W
x
dy
(b)
2a
dx
10.
X
c
2
2
2a (c) ( x Y y ) Z 2 xy
y
d log x e c
x
8 8
Q
dy
dx
X
0
1
2 3
x
11. (i) y f (g cos 2 x h 2sin 2 x) h ce (ii) y sin x f sin x h c
5
3
12. (i) cy i ( x j 2)(1 k 2 y )
(ii) (e x l 2) sec y m c
1n y2
p
1q y2 q 1
1
log
(iv) 2
1q y2 s 1
r
(iii)
1 n x2
o
c
1 q x2
q
1q y2
sc
(v) ( x 2 t 1)( y 2 t 1) u 2
1
1
(vi) log y v w cos3 x x cos 5 x x xe x w e x x c
3
5
cos 2 x
(vii) log | tan y | y
zc
y
13. (i) y { x 2 { y 2
}
(ii) xy cos
(iii) y
(iv) y f
14. (i)
|
c2 x4
k
2x
3x 2
1
x
1g a
(v) x 2 y 2
x
y~
x
2
g
1
a
h cx
a
( x 2 y 2 )2 c 2
3
3 y3
log |
y|c
(ii) x 2 y 2 cy
(iii) x 2 y 2
2x
(iv) y cecos( x y )
(v) sin
y
x
[Hint : Put
x
= v]
y
cx
(vi) c( x 2 y 2 ) y
(vii)
e
y
ex
x3
3
c
(viii) x 3 y 2 cy
8 9
Hints
6.
( x y x y )dx ( x y x y )dy 0
x¢ y ¢ x£ y
dy
=
dx
¡
x¢ y £ x£ y
Put y = vx
¡
dxv
=
dx
v¤
xdv
1 ¥ 1 ¦ v2
=
dx
v
x
¡
v
¡
Put 1 ª v 2
1¨ v
«t
2
¡
dv
1 ¥ 1 ¦ v2
=
dx
v
dv =
1
dx
x
t dt
=
t t2
1
dx
x
© (1 ¨ v
2
)
¦v
§
2
2
¬
¡
1¥ v ¥ 1¦ v
1¥ v ¦ 1¦ v
v¤
® log |1 ¯ t |
1
1 ± 1 ² v2
= by | x | ° log | c |
= cx
1
¡
12. (i)
¡
¡
¡
x ³ x2 ´ y 2
= c
µ 2 dy ¶
dy
= 2 ·¹ y ± ¸º
dx
dx
1
1
dx =
dy
(2 » x)
y (1 ¼ 2 y )
y¬x
1
(2 ¥ x)
dx =
½1
À ¿
Ãy
2 ¾
dy
(1 Â 2 y ) ÁÄ
cy = (2 + x) (1 – 2y)
9 0
1¥ v ¥ 1¦ v
1¥ v ¦ 1¦ v
9
Vectors
Teaching Learning Point
l
A quantity that has magnitude as well as direction is called is called a vector.
l
A directed line segment represents a vector and is denoted by AB or a .
l
Position vector of a point P(x, y, z) w.r.t. origin 0(0,0,0) is denoted by OP , where OP = xiˆ È yjˆ È zkˆ
Å
Æ
Ç
Ç
É
l
l
l
and OP = x 2 Ê y 2 Ê z 2
The angles Ë, Ì, Í made rÎ with positive direction of X, Y and Z-axies respectivily are called
direction angles and cosines of these angles are called direction cosines of rÎ usually denoted
as l = cos Ë, m = cosÌ and n = cos Í where l2 + m2 + n2 = 1.
a b c
Ï
Ï .
The numbers a, b, c, propotional to l, m and n are called direction ratios i.e.
l m n
Ð
Æ
If two vector a and b are represented in magnitude and direction by the sides of a triangle taken
Ò
Ò
in order, then their sum a Ñ b is represented in magnitude and direction by the third side of triangle
taken in opposite order. This is called triangle law of addition of vectors.
l
If aÆ is any vector and
Ó
is a scalar, then
Æ
l
l
Æ
a is a vector, collinear with aÆ and
2
Ô
Õ
a .
1
Ð
If a and b be the position vectors of points A and B, and C is any point dividing AB internally
Æ
Æ
mb Ø na
. If C divides AB in m : n externally,,
mØ n
mb – na
.
m–n
Ð
Æ
Ù Ù
Ü Ü
Ü Ü
Scalar Product (Dot Product) of two vectors a and b is denoted by a.b and is defined as a.b Ú a b cos Û ,
then C =
Æ
l
=
Ö
in m : n, then position vector c of C is given as c =
l
Õ
Ôa
Any vector aÆ can be written is a = a aˆ where â is a unit vector in the direction of â .
If A(x1, y1, z1) and B(x2, y2, z2) be any two points in space, then AB = (x2 – x1) iˆ + (y2 – y1)
ˆj × ( z – z )kˆ.
Æ
l
Ó
Ð
where Ý is the angle between a and b (0 Þ Ý Þ )
à
à
Dot product of two vectors is commutative i.e., aà.b ß b .aà
Ð
Ù
Ð
Ù
á á
l
Æ
ââ
Ù
ã2
For unlike parallel vectors a and b , a.b = a b , so a.a = a
l
Æ
Ù
For unlike parallel vectors a and b , a.b = – a b
ä ä
9 1
è è
å
è
è
0 æ a å 0 or b
å
è
è
0 or a ç b .
l
If a.b
l
iˆ. ˆj é ˆj.kˆ é kˆ.iˆ é 0, iˆ.iˆ é ˆj. ˆj é kˆ.kˆ é 1
l
ì
If a ê a1iˆ ë a2 ˆj ë a3kˆ and b
ì
ê b1iˆ ë b2 ˆj ë b3 kˆ ,
l
í
a1a2 î b1b2 î c1c2 .
ó ó
ò
ñ
ï ï
then a.b
Projection of a vector a on b =
a.b
ó
b
ü
ö ü ÷
a.b ˆ
ø ü ù b.
ø b ù
ú
û
õ
ô
l
Projection of a vector a along b =
l
Cross product (Vector product) of two vectors a and b is denoted as a b and is defined as
õ
ô
þ
þ
ÿ
a b = aÿ b sinð ˆ , where
ñ
ý
ý
ò
ð is the angle between a and b (0 ð ) and ˆ is a unit
ò
ñ
ñ
ò
vector perpendicular to both a and b such that a , b and ˆ form a right handed system.
ò
ñ
l
Vector product of two vectors a and b is not commutative i.e., a b
l
If a b
l
a.a O, So iˆ iˆ O ˆj
l
If a ê a1iˆ ë a2 ˆj ë a3kˆ and b
O a O or b O , or a b.
If
ê b1iˆ ë b2 ˆj ë b3 kˆ ,
l
l
l
then
kˆ
a3
b3
ð is angle between two non zero vector a and b , then
sinð =
l
–b a .
ˆj kˆ kˆ.
iˆ ˆj
a b = a1 a2
b1 b2
l
b a , but a b
a b
a b
a b is the area of that parallelogram where adjacent sides are vectors a and b .
1
a b is the area of that parallelorgam whose diagonals are a and b .
2
1
1
1
If a , b and c forms a triangle, then area of the triangle =
c a.
a b =
b c =
2
2
2
1
If a , b and c are position vectors of vertices of a triangle, then area of triangle = a b b c c a
2
Scalar triple product of vectors
The scalar triple product of three vectors
a b c 9 2
Geometrical Interpretation : ( a b ).c represents the volume of parallelepiped whose coterminous
edges are represented by a b and c .
Scalar triple product of three vectors remains unchanged as long as their cyclic order remains
unchanged.
( a b ).c = (b c ).a = (c a ).b
or
[ a b c ] = [b c a ] = [ c a b ]
Scalar triple product changes its sign but not magnitude, when the cyclic order of vectors is changed.
[a b c ] = – [a c b ]
Scalar triple product vanishes if any two of its vector are equal.
Scalar triple product vanishes if any two if its vector are parallel or collinear.
Necessary and sufficient condition for three non zero non collinear vectors a b and c to be coplanar
is that [ a b c ] = 0
Scalar triple Product in terms of component
Let
then
= a a1iˆ a2 ˆj a3kˆ b
a1 a2
[ a b c ] = b1 b2
c1 c2
b1iˆ b2 ˆj b3kˆ c c1iˆ c2 ˆj c3kˆ
a3
b3
c3
Question for Practice
Very Short Answer Type Questions (1 Mark)
1.
2.
3.
4.
Write direction cosines of vector 2iˆ ˆj 2kˆ
For what value of ‘x’ the vectors
2iˆ – 3 ˆj 4kˆ and xiˆ – 6 ˆj 8kˆ are collinear
Write the projection of the vector iˆ – ˆj on the vector iˆ ˆj
Write the angle between two vectors a and b with magnitudes
a.b 6
3 and 2 respectively having
6.
If a.a 0 and a.b 0 = 0 then what can be concluded about vector b
What is the cosine of the angle which the vector 2iˆ ˆj kˆ makes with z axis
7.
Write a vector of Magnitude 3 units in the direction of vector iˆ ˆj – kˆ
5.
9 3
8.
If a = 2iˆ ˆj 3kˆ and b = 6iˆ ˆj 9kˆ and a b find the value of
9.
For what value of
a = 2iˆ ˆj – kˆ and b = iˆ ˆj kˆ are prependicular to each other
10. If a is a unit vector such that a i
11.
a = 2 b = 7, a b
ˆj find a.i
! 3iˆ " 2 ˆj " 6kˆ write the angle between a and b
12. If a , b represent Diagonal of rhombus, then write the value of a.b
13. Write unit vector in the direction of a # b where a $ iˆ % ˆj – kˆ , b
& iˆ ' ˆj ' 2kˆ
14. Write a – b , If two vector a and b are such that
a = 2, b = 3, a.b = 4
15. If a and b are unit vectors such that a b is also a unit vector. Find the angle between a
and b
16. What is the value of x If for unit vector a
( x – a ).( x ( a ) ) 15
17. Write the value of iˆ.( ˆj kˆ) * ˆj.(( kˆ iˆ) * kˆ.( ˆj i )
18. Find the value of so that the vectors
a = 2iˆ – 3 ˆj kˆ and b = iˆ 2 ˆj – 3kˆ and c + ˆj , -kˆ are coplanar
19. Find the value [iˆ ˆj kˆ]
20. Find the volume of parallelepiped whose coterminous edges are represented by the vectors
a = iˆ 2 ˆj 3kˆ and b = 3iˆ 7 ˆj – 4kˆ and c . iˆ 5 ˆj 3kˆ
Short Answer Type Questions (4 Marks)
1.
If a , b , c are unit vectors such that a. b
prove that a =
2.
/ a.c / 0 and the angle between b and c is 0 6 then
1 2 (b c )
Find a vector of magnitude 5 units, prependicular to each of the vectors (a * b ) and (a – b ) where
a 2 iˆ 3 ˆj 3 kˆ , b
4 iˆ 5 2 ˆj 5 3kˆ
. a.c
3.
If a , b , c are vectors such that a , b
4.
5.
a b 6 a c and a 0 then prove that b 7 c
If a 2 iˆ 3 ˆj 3 kˆ and b 8 ˆj – kˆ find a vector c such that a c 9 b and a.c : 3
Using vectors find the area of triangle with vertices A(1, 1, 2) B(2, 3, 5) and (1, 5, 5)
6.
7.
If a , b , c are the position vectors of the vertices A, B, C of ABC respectively, find an expression
for the area of ABC and Hence deduce the condition for the points A, B, C to be collinear
Show that the area of || gm having Diagonals 3iˆ * ˆj – 2kˆ and iˆ – 3 ˆj * 4kˆ is 5 3 sq unit.
9 4
8.
Let a , b , c be three vectors of magnitude 3, 4, 5 units respectively. If each of these is
; to the
sum of other two vectors, then find a < b < c
9.
If a , b and c are three vectors such that a = b = c > 0
a = 3 b = 5 c = 7 find the angle between a and b
10. If a and b are unit vectors and
@ 1
(i) sin A aˆ – b
2 2
B 1
(ii) cos C aˆ D b
2 2
11. If a unit vector a makes
4
and
? is the angle between them, then show that
3
with x axis and y axis respectively and an obtuse angle
?
? and the components of a along axis.
12. If a and b are unit vectors such that 2a – 4b and 10a E 8b are perpendicular to each other..
with z axis, then find
Find the angle between a and b
13. Find a vector a such that
a.(iˆ F ˆj ) G 2 a.(iˆ – ˆj ) H 3 a.kˆ I 0
14. If a makes equal angles with the coordinate axes and has magnitude 3, then find the angle between
a and each of three coordinate axes.
15. A girl walks 4 km west wards, then she walks 3 km is a direction 30° east of north and stops.
Determine the girls displacement from her initial point of departure
16. Prove the following
(i) [ a J b , b J c , c J a ] K 2 [ a b c ]
(ii) [ a – b , b – c , c – a ] H 0
17. Show that the points with position vectors
6iˆ – 7 ˆj , 16iˆ –19 ˆj – 4kˆ, 3iˆ – 6kˆ and 2iˆ – 5 ˆj L 10kˆ are coplanar..
18. If a b
M c d , a c M b d show that a – d is || to b – c
19. Let a N iˆ – ˆj b
20. If a
P b P a Q b P 1 then find a – b
21. Prove a b
22. If a
N 3iˆ – kˆ c N 7iˆ – kˆ find a vector d which is ; to both a and b and c .d O 1
P2
2
b
=
a.a
a.b
a.b
b .b
P5
a b
P 8 find a.b
23. Let u , v and w be vectors such that u +v R w S o
If u
T 3 v T 4 w T 5 find u.v U v.w U w.u
9 5
V
V V
V
X X X X X X X
24. In a Regular hexagon ABCDEF if AB W a BC W b then express CD, DE, EF, FA, AC, AD, AE, and
Y
CE is terms of a and b
25. Given a Z 3iˆ – ˆj and b
] ]
]
[ 2iˆ \ ˆj – 3kˆ express b ^ b1 _ b2 where b , is parallel to a and b2 is ` a
Answers
1 1 1
, ,
3 3 3
1.
3.
0
5.
b
7.
9.
3(iˆ c ˆj – kˆ)
d = –1
2. x = –4
4.
aa
4
6. cosb =
1
2
8. –3
10. 0
12. 0
11.
6
1 ˆ
13.
(2i e 2 ˆj e kˆ)
3
5
14.
15.
16. 4
2
17. 1
19. 1
20. 39 cubic unit
18.
d = –1
Answer of four marks question
1.
Given a.b f 0 and a.c g 0
Also a is a unit vector
2.
3.
h is ` to both b and c
i=
b c
b c
1 1
But b c j b c sin j 1 1
j
6
2 2
5 ˆ
(– i k 2 ˆj – kˆ)
6
a.b l a.c m a.b n a.c l 0 m a.(b – c ) l o a o o or b – c p o or a q (b – c ) (1) but a
a b = a c
o (given)
h a b – a c r o h a (b – c ) s o a o o or b – c p o or a b – c (2) but a o
from (1) and (2) we get b
tc
( a u b – c and a b – c can not held simultaneously)
9 6
4.
6.
5
2
2
v iˆ w ˆj w kˆ
c
A
3
r
e
a
3
o
3
ABC =
f
5.
1
61
2
1 x
x
2 AB AC
C(c )
1
(b – a ) (c – a )
2
1
=
b –c –b a –a c ya a
2
=
1
a b zb c zc a
2
( a a { o) a b | –b a
A(a )
=
Point A, B, C to be collinear if area of
7.
Hint Area of 11gm
8.
5 2
9.
ab c 0
1
d1 d 2
2
d 1 & d 2 are the vectors along diagonal
a b –c
(a b ).(a b ) (– c ).(– c )
a
a
2
2
b 2a.b
c 2
2
b 2 a b cos
c 2
2
32 + 52 + 2 × 3 × 5 cos = 72
11.
ABC = 0 a b } b c } c a ~ 0
=
2
,
3
= 60°
1 1
1
i, j , k
2
2 2
12. angle between a and b is 120°.
13. a
5ˆ 1
i– j
2
2
14. Let a x1iˆ y1 ˆj z1kˆ
a.iˆ
x
= 1
3
a iˆ
cos =
x1 = 3cos
y1 = 3cos z1 = 3cos
Similarly
a = 3
x12 y12 z12 = 3
3 3 cos 3
cos
1
3
9cos 2 9cos 2 9cos 2 = 3
cos1
1
3
9 7
B(b )
15.
In
OB = OA AB
OA = –4iˆ
AM = AB cos60°
MB = ABsin60°
AMB
3
2
AM =
MB =
3 3
2
AB = AM MB
3ˆ 3 3 ˆ
i
j
2
2
Girls displacement from initial point of departure
AB =
4i
3 3 3
i
j¡
2
2 ¢
1ˆ 1 ˆ 3 ˆ
i £ j£ k
4
4
4
22. 6
23. –25
5
2
i
3 3
j
2
19.
¥
§
©
¥
¬
¬
24. CD ¤ b – a DE ¦ –a EF ¨ –b FA ª a – b AC ¤ a « b AD ¦ 2b AE 2b – a CE b – 2a
25.
b1 =
®
b2 =
3ˆ 1 ˆ
i– j
2
2
1ˆ 3 ˆ
i ¯ j – 3kˆ
2
2
9 8
10
ThreeDimensionalGeometry
Teaching learning points
l
l
l
Distance between two given points P(x1, y1, z1) and Q(x2, y2, z2) is
PQ = ( x2 – x1 ) 2 ° ( y2 – y1 ) 2 ° ( z2 – z1 ) 2
Direction ratio of line joining the points (x1, y1, z1) and (x2, y2, z2) are x2 – x1, y2 – y1, z2 – z1
l m n
Let a, b, c be the direction ratio of a line whose direction cosines are l, m, n the
± ±
a b c
l 2 + m2 + n 2 = 1
l =
If line makes angles
a
²
a 2 ³ b2 ³ c2
m =
²
b
a 2 ³ b2 ³ c2
n =
²
c
a 2 ³ b2 ³ c2
´, µ, ¶ with coordinate axes then l = cos´ m = cosµ n = cos¶
l
cos2´ + cos2µ + cos2¶ = 1
l
Vector equation of a straight line passing through a fixed point with the position vector a and
|| to given vector b
l
l
l
r = a · ¸b where ¹ is parameter and r = xiˆ º yjˆ º zkˆ
Cartesian equation of straight line passing through a fixed point (x1, y1, z1) having direction ratio
x – x1 y – y1 z – z1
(a, b, c) is given by
»
»
a
b
c
x – x1 y – y1 z – z1
The coordinates of any point on the line
¼
¼
¼ ½ are (x1 + a¹, y1 + b¹, z1 + c¹)
a
b
c
where ¹ ¾R
Angle between two lines whose direction rahos are a1, b1, c1 and (a2, b2, c2) is given by
a1a2 À b1b2 À c1c2
cos¿ =
a12 À b12 À c12 a22 À b22 À c22
If lines are perpendicular then a1a2 + b1b2 + c1c2 = 0
If lines are parallel then
l
a1
a2
Angle between tow lines : r
Á
b1
b2
Á
c1
c2
 a1 à Äm1 and r  a2 à Åm2 is given as cos¿ =
lines are perpendicular if m1.m2 = 0. Lines are parallel m1 Æ Ç m2
99
m1.m2
m1 m2
and so, two
l
Skew lines : Lines in space, which are neither parallel, nor intersecting are called skew lines,
such pair of lines are non-coplanar.
l
Shortest distance : r
them is given by
È a1 É Êb1 and r È a2 É Ëb2 are two skew lines, then distance ‘d’ between
d =
l
l
l
l
Î a1 Ï Ðb and r Ñ a2 Ò Ób are two parallel lines, then
(a2 – a1 ) Ô b
b
Plane : A plane is uniquely determined if any one of the following is known:
(i) The normal to the plane and its distnace from origin.
(ii) It passes through a given point and is perpendicular to a given direction.
(iii) It passes through three given non collinear points.
Equation of a plance at a distance ‘p’ from origin and normal vector n is given by r .nˆ = p or
lx + my + nz = p.
General equation of plane passing through a point a and having normal vector to plane as n
is ( r – a ).n Õ 0. Corresponding Cartesian form is a(x – x1) + b(y – y1) + c(z – z1) = 0, where
a, b, c are direction ratios of normal to plane.
General equation of plane which cuts off intercepts a, b and c on x, y and z-axis respectively
is
l
b1 Í b2
Distance ‘d’ between parallel line : if r
distance d, between them is given by
d =
l
( a2 – a1 ) Ì (b1 Í b2 )
x y z
Ö Ö
a b c
× 1.
Equation of plane passing through three non-collinear points a , b , c is (r – a ) Ú Ø(b – a ) Û (c – a ) Ù Ü 0.
Ý
Þ
x – x1
y – y1
z – z1
Corresponding Cartesian from is x2 – x1
x3 – x1
y2 – y1
z2 – z1
y3 – y1
z3 – z1
ß0
where (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) are co-ordinates of known point.
l
Angle between two planes is cosà =
cosà =
n1.n2
or
n1 n2
a1a2 á b1b2 á c1c2
a12 á b12 á c12 a22 á b22 á c22
where n1 & n2 are vectors normal to planes or a1, b1, c1 and a2, b2, c2 are dr’s of normal to planes.
100
l
Condition of Coplanarity of two lines
â a1 ã äb1 and r å a2 æ çb2 be two lines then these lines are coplanar if (a 2 è a1 ).(b1 é b 2 ) ê 0
and equation of plane containing them is ( r ë a1 ).(b1 ì b 2 ) í 0 or ( r – a2 ) î (b1 ï b2 ) ñ 0
Let r
l
Distance of point a from plane r .n = d is
Distance =
l
Angle
aòn – d
n
ó between line r ô a õ öb and plane r ÷ n = d is given as
sinó =
b øn
b n
A line is parallel to plane of b ù n = 0.
Question for Practice
Very Short Answer Type Questions (1 Mark)
Q1. Write intercept cut off by the plane 2x + y – z = 12 on x axis.
x–5 yú4 z –6
û
û
3
7
2
Q3. Write the direction cosine of line joining (1, 0, 0) and (0, 1, 1)
Q4. What are the direction cosine of a line which makes equal angles with coordinate axes.
Q5. What is the cosine of the angle which vector 2iˆ ü ˆj ü kˆ makes with y axis
Q6. What is the distance of the following plane from origin 2x – y + 2z + 1 = 0.
Q7. Write the distance of the point (a, b, c) from x axis.
Q2. Write the vector equation of a line
x – 2 y –1 z – 3
its perpendicular to tbe plane 3x – y – 2z = 7.
þ
þ
9
–6
ÿ
Q9. If a line makes ð, , and with the x axis, y axis and z axis respectively. Find the value of
sin2ð + sin2 + sin2.
Q8. For what value of
ý the line
Q10. Find the coordinate of the pt where line
x–3 y –5 z –2
crosses the yz plane.
2
–3
5
x – 2 2y – 5 z 1
.
2
–3
1
Q12. What is the equation of plane parallel to XOY plane and passing through (3, –4, 8)
Q11. Write the direction ratio of the line
Q13. Write the direction cosines of the perpendicular from the origin to the plane r (iˆ ü ˆj ü kˆ) = 10.
Q14. Write the direction ratio of the normal of plane 2x + y + z = 7.
101
Q15. Write the value of for which the plane 2x – 4y + 3z = 7 and x + 2y + z = 18 are to
each other.
Q16. Write the vector eq. of plane 2x + y + z = 7.
x –1 y –1 z – 2
and passes through (0, 0, –1)
2
3
4
Q18. What is the equation of line passes through (1, 1, 1) and to the plane 2x + y + z = 7.
Q19. If a line makes an angle 60°, 30°, 90° with the positive direction of x, y, z axis respectively
then write the direction cosines of line.
Q20. What is the equation of a plane that cut the coordinates axis (a, 0, 0), (0, b, 0) and (0, 0, c).
Q17. Write the equation of line || to the line
Q21. Write the angle between line
x – 2 y 1 z – 3
and the plane 3x + 4y + z + 5 = 0.
3
–1
2
Short Answer Type Questions (4 Marks)
Q1. Find the equation of the line passing through the points (1, 2, –1) and (3, –1, 2). At what point
it meet yz plane.
Q2. Find the equation of the line passing through the point (–1, 3, –2) and perpendicular to the line
x y z
x 2 y –1 z 1
and
1 2 3
–3
2
5
x –1 y – 2 z – 3
x–2 y –4 z –5
Q3. Find the shortest distance between the lines
and
2
3
4
3
4
5
Q4. Show that the four points (0, –1, 0) (2, 1, –1) (1, 1, 1) and (3, 3, 0) are coplanar & also find
the equation of plane containing these point.
Q5. A variable plane which remains a constant distance 3P from the origin cuts the coordinate axis
A, B and C show that the locus of the centroid of ABC is x–2 + y–2 + z–2 = P–2
Q6. Find the equation of the plane passing through pt (2, 3, 4) and || to the plane 5x – 6y + 7z = 3.
Q7. Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular
to the plane 2x + 6y + 6z = 1.
Q8. Find the equation of plane passing through origin and perpendicular to each of the plane x +
2y – z = 1 and 3x – 4y + z = 5.
Q9. Find the distance between two parallel planes
2x – y + 3z + 4 = 0
6x – 3y + 8z – 3 = 0
Q10. Find the equation of plane which contains the line of intersection of the plane r (iˆ 2 ˆj 3kˆ) 4
and r (2iˆ ˆj kˆ) –5 and which it to the plane r (5iˆ 3 ˆj – 6kˆ) –8 .
Q11. Find the distance of the point A(–1, –5, –10) from the point of intersection of the line r =
(2iˆ – ˆj 2kˆ) + (3iˆ 4 ˆj 2kˆ) and the plane r (iˆ – ˆj kˆ) 5
Q12. Find the distance of the point (1, –2, 3) from the plane x – y + z = 5 measured along a line
x y
z
parallel to 2 3 –6
x 2 2 y 3 3z 4
Q13. Find the distance of the point A(–2, 3, –4) from the line
parallel to the
3
4
5
plane 4x + 12y – 3z + 1 = 0.
102
x 2 y 1 z – 3
at a distance of 5 units from the point (1, 3, 3).
3
2
2
x–3 y –3 z
Q15. Find the equations of the two lines through the origin which intersect the line
2
1
1
Q14. Find a points on the line
at angle of
3
.
x 1 y – 3 z 2
x y–7 z7
and are coplanar. Also find the equation
–3
2
1
1
–3
2
of plane containing them.
Q17. Show that the line r (2iˆ – 2 ˆj ! 3kˆ) + " (iˆ – ˆj # 4kˆ) is || to the plane r.(i $ 5 j $ k ) % 5 . Also
Q16. Show that the lines
find the distance between them.
Q18. Find the value of & so that the lines
to each other.
1 – x 7 y – 14 5 z – 10
7 – 7x y – 5 6 – z
and
are +
'
'
)
)
3
2(
11
3*
1
5
Q19. Show that the plane whose vector equation is r , (iˆ - 2 ˆj – kˆ) = 3 contains the line whose vector
equation is r . (iˆ / ˆj ) / 0 (2iˆ / ˆj / 4kˆ) .
Q20. If the point (1, 1, P) and (–3, 0, 1) be equidistant from plane r 1 (3iˆ 2 4 ˆj –12kˆ) 2 13 3 0 find value of P.
Long Answer Type Questions (6 Marks)
Q1. A line makes 4, 5, 6, 7 with the four diagonals of a cube prove that cos24 + cos25 + cos26
4
+ cos27 = .
3
1
Q2. Show that angles between any two diagonals of cube is cos–1 .
3
Q3. If l1, m1, n1 and l2, m2, n2 be the direction cosines of two mutually prependicular lines. Show
that direction cosines of the line + to both of them are (m1n2 – m2n1) (n1p2 – n2p1) (l1m2 – l2m1).
Q4. Find the foot of the perpendicular from the point (0, 2, 3) on the line
x 3 y –1 z 4
.
5
2
3
Also find the length of perpendicular.
x y –1 z – 2
)
)
1
2
3
x – 4 y 3 z 1
x –1 y 8 1 z 8 10
Q6. Prove that the lines
and
intersect. Also find the cordinates
9
9
1
–4
7
2
–3
8
of their point of intersection.
Q7. Find the shortest distance between the lines
r : (1 - ; )iˆ - (2 – ; ) ˆj - (1 - ; ) kˆ
r : 2(1 - < )iˆ – (1 – < ) ˆj - (–1 - 2< ) kˆ
Q5. Find the image of the point (1, 6, 3) on the line
Q8. Find the length and the foot of the prependicular from the point (7, 14, 5) to the plane 2x +
4y – z = 2.
Q9. Find the image of the point (1, 3, 4) on the plane 2x – y + z + 3 = 0.
103
Q10. Find the equation of a plane passing through the points (0, 0, 0) and (3, –1, 2) and || to the
x – 4 y = 3 z =1
>
>
.
line
1
–4
7
Q11. Find the equation of the plane passing through the point (0, 7, –7) and containing the line
x ?1 y – 3 z ? 2
.
@
@
–3
2
1
Answers
Very Short Answer (1 Mark)
1.
2. r A (5iˆ – 4 ˆj B 6kˆ) B C (3iˆ B 7 ˆj B 2kˆ)
6
3.
–1 1 1
,
,
3 3 3
5.
cos K L
b2 M c2
7.
9.
1
2
T 1
3
,
1 1 U
, W
3 3Y
15. Z = 2
17.
1
3
8. –3
N 19 –11O
,
Q
2 2 S
4, –3, 2
13. V
X
6.
1
1
1 E
,F
,F
H
3
3
3J
10. P 0
R
2
11.
D
4. GI F
x y z ]1
^ ^
2 3
1
` 3 1 a
19. b
, ,0
d 2 2 ce
12. z = 8
14. (2, 1, 1)
16. r .(2iˆ [ ˆj [ kˆ) \ 7
18.
x –1 y –1 z –1
_
_
2
1
1
20.
x y z
f f _1
a b c
g 7 h
j
2 91 l
–1
21. sin i
k
Short Answer (4 Mark)
1.
m 7 –5 n
oq 0, 2 , 2 pr
2.
x ?1 y – 3 z ? 2
@
@
2
–7
4
104
6.
1
6
5x – 6y + 7z = 20
8.
x + 2y + 5z = 0
3.
4. 4x – 3y + 2z = 3
10. r .(33iˆ s 45 ˆj s 50kˆ) = 41
7. 3x + 4y – 5z = 9
5
9.
14
11. 13
11. 1
13.
17
units
2
12. (–2, –1, 3) and (4, 3, 7)
15.
x y
z
x
y
z
and
t t
u u
1 2 –1
–1 1 –2
x–3 y –3 z
t
t tv
2
1
1
Any pt on line (2w + 3, w + 3, w)
DR of op (2w + 3 – 0, w + 3 – 0, w – 0)
Hint : Give line
line op makes
cos
x
3
x
3
with line pq
2(3 y 2z ) y 1(3 y z ) y z { 1
=
2 y1 y1
2
2
2
(3 y 2z ) y (3 y z ) y z
2
2
w = –1, or –2 eq of requred line
2
|
6} ~ 9
6 6} 2 ~ 18} ~ 18
x–0 y–0 z–0
x–0
y–0
z–0
and
1 – 0 2 – 0 –1 – 0
–1 – 0 1 – 0 –2 – 0
x y
z
x
y
z
t t ,
u u
1 2 –1 –1 1 –2
10
unit
3 3
7
20. P = 1,
3
16. x + y + z = 0
17.
18. w = 7
Answer (6 Mark)
3.
Let l, m, n be the direction cosines of the line to each one of given line, then
ll1 + mm1 + nn1 = 0
ll2 + mm2 + nn2 = 0
l
m
n
=
=
=
m1n2 – m2 n1
n1l2 – n2l1
l1m2 – l2 m1
(m1n2 – m2 n1 ) 2 = sin = sin
2
= 1
Hence direction cosines of the line are
(m1n2 – m2n1), (n1l2 = n2l1), (l1m2 – l2m1)
105
l 2 m2 n2
(m1n2 – m2 n1 ) 2
=
1
sin
4.
5.
Gen pt on the line
5
– 3, 2
+ 1, 3
– 4
for some value of
the coordinate of A
(5
– 3, 2
+ 1, 3
– 4)
(1)
DR of PN (5
– 3 – 0, 2
+ 1 – 2, 3
– 4 – 3)
PN is to given line
5(5
– 3) + 2(2
– 1) + 3(3
– 7) = 0
= 1
Putting
= 1 in (1)
We get (2, 3 – 1) as foot of
2
2
2
length of = PN = (2 – 0) (3 – 2) (–1 – 3) = 21 unit
Fing foot of ^ by using above method (mentioned in Q No 4) we get (1, 3, 5) since B pt is
image of pt A P is the mid pt of AB
1
1
2
= 1
6.
7.
8.
6
2
3
=0
3
2
5
= 7
Give line will intersect at (5, –7, 6)
3 2
unit
2
Eq. of line PN (line PN is to plane)
x – 7 y –14 z – 5
2
4
–1
Gen pt on line 2
+ 7, 4
+ 4, –
+ 5
If N is the foot of Gen pt must satisfies the eq. of plane 2(2
+7) + 4(4
+ 14) – (–
+ 5) = 2
= –3
pt N is (1, 2, 8)
9.
Also find length of = PN = (7 – 1) 2 (14 – 2)5(5 – 8) 2 = 3 21
(–3, 5, 2)
Hint : Find foot of by using above method.
(Mentioned in question No 8.
N is the mid pt of PM
1
1
3
4
2
2
= –3
=5
10. x – 19y – 11z = 0
11. x + y + z = 0
4
2
3
= 2
106
11
Probability
Teaching points
l
Conditional probability: If A and B are two events associated with any random experiment, then
P(A/B) repersents the probability of occurence of event-A knowing that event B has already occurred
P( A B)
, P( B) 0
P( B)
P(B) 0, means that the events should not be impossible
P(A B) = P(A and B) = P(B) × P(A/B)
Similarly
P(A B C) = P(A) × P(B/A) × P(C/AB)
l Multiplication Therem on Probability:
If the event A and B are associated with any random experiment and the occurrence of one depends on the other
then
P(A B) = P(A) × P(B/A) where P(A) 0
l
when the occurrence of one does not depends on the other then these events are said to be independent
events.
Here
P(A/B) = P(A) or P(B/A) = P(B)
P(A B) = P(A) × P(B)
l
Theorem on total probability: If E1, E2, E3,...En be a partition of sample space and E1, E2 ... En all has
non-zero probability. A be any event associated with sample space S, then
P(A) = P(E1). P(A/El) + P(E2). P(A/E2)+...+ P(En). P(A/En)
l
Baye’s theorem: Let S be the sample space and E1, E2...En be n mutually exclusive and exhaustive
events associated with a random experiment. If A is any event which occurs with E1, or E2 or ... En,
then
P( Ei ) P( A Ei )
P(Ei/A) = n
P( Ei ) P( A Ei )
P(A/B) =
i 1
l
l
Random variable: It is real valued function whose domain is the sample space of a random experiment
and whose range is a subset of real numbers.
Probability distribution: It is a system of numbers of random variable (x), such that
X2
X3
Xn
X:
X1
P(X1):
P(X1)
P(X2)
P(X3)...
P(Xn)
n
Where
P(X1) > 0 and
1
P( X i ) 1
107
l
Mean or expectation of a random variables (x) denoted by E(x)
n
E(x) = =
l
P( X i ) 1
i 1
Variance of X denoted by var (x) or a ¡x2 and
var(x) = ¡x2 =
n
¢
( X i £ ¤ )2 P( X i ) ¥
i 1
l
l
l
¢
X i P ( X i ) ¥ (¤ ) 2
l 1
The non-negative number ¦ x § var( x) is called standard deviation of random variable X.
Bernouli Trials: Trials of random experiment are called Bernouli trials if
(i) No. of trials are finite.
(ii) Trials are independent
(iii) Each trial has exactly two outcomes either success or failure,
(iv) Probability of success remains same in each trial.
Binomial Distribution:
P(X = r) =
l
n
n
Cr q n ¨ r p r , where r = 0, 1, 2,...n
p = prob. of success
q = prob. of Failure
n = total no. of trials
r = value of random variable.
Recurrence formula for Binomal Distribution:
P(r + 1) =
n©r p
. p (r ).
r ª1 q
Questions for Practice
Very Short Answer Type Questions (1 Mark)
Q1. Three coins are tossed once. What is the probability of getting at least one head?
Q2. If P(A) = 0.3, P(B) = 0.5 and P(A/B) = 0.4, then find P(B/A)?
Q3. A policeman tries three bullets on a dacoit. The probability that the dacoit will be killed by one bullet
is 0.7. What is the probability that dacoit is still alive?
Q4. What is the probability that a leap year will have 53 Sundays?
Q5. If A and B are independent events with P(A) = 0.3 and P(B) = 0.5 then what is P (A « B)?
1
1
Q6. Two students A and B solve a problem independently with probabilities
and
respectively..
3
4
What is the probability that problem is not solved?
108
Q7. Two balls are drawn one by one without replacement from a bag containing 10 red and 5 black balls.
What is the probability that both balls are black?
Q8. The random variable X has a probability distribution is
x
0
1
2
P(x)
K
2K
3K
What is the value of K.
Q9. Given P(A) = 1/4, P(B) = 2/3 and P(A ¬ B) = 3/4. Are the events independent?
Q10. A and B are two events such that P(A) = 0.3 and P( A B ) ® 0.8 . If A and B are independent events.
Find P(B)?
Short Answer Type Questions (4 Marks)
Q11. Two cards are drawn without replacement from a well shuffled pack of 52 cards. What is the probability
that one card is a red queen and other is a king of black colour?
Q12. Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls one by
one without replacement. Find the probability that both the balls are of different colours.
Q13. The probability that a student A can solve a question is 6/7 and that another students B solving a
question 3/4. Assuming the two events "A can solve the question" and "B can solve that question" are
independent, find the probability that only one of them solves the question.
Q14. A problem in mathematics is given to three students whose chance of solving it are
Q15.
Q16.
Q17.
Q18.
Q19.
Q20.
Q21.
Q22.
1 1 1
, , . What is
2 3 4
the probability that the problem will be solved?
A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases they are
likely to contradict each other in stating the same fact.
There are two identical boxes containing respectively 4 white and 3 red balls, 3 white and 7 red balls.
A box is chosen at random and a ball is drawn from it. If the ball drawn is white, what is the probability
that it is from the first box?
A can hit a target 4 times in 5 shots, B can hit the target 3 times in 4 shots and C 2 times in 3 shots.
They fire a volley. What is the probability that atleast two shots hit?
A man takes a step forward with probability 0.4 and backward with probability 0.6 find the probability
that at the end of eleven steps he is one step away from the starting point.
A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one
without replacement, from the box. Find the probability distribution of the number of defective
bulbs.
A coin is biased so that the head is 3 times as likely to occurs as a tail if the coin is tossed twice, find
the probability distribution for the number of tails.
There are three coins, one is a two headed coin, another is a biased coin that comes up head 75% of
the time and third is an unsbiased coin. One of the three coins is chosen at random and tossed. If it
shows a head. What is the probability that it was a two headed-coin9
Two cards are darwn from a pack of 52 cards at random and kept out. Then one card is drawn from
the remaining 50 cards. Find the probability that it is an ace.
109
Q23. A random variable x has the following distribution:
X –2
–1
0
1
2
3
P(X) 0.l
K
0.2
2K
0.3
K
Find (i) the value of K, (ii) p(X ¯ 1), (iii) P(X ° 0).
Q24. Two dice are thrown. Find the probability that the number appeared have a sum 8 if it is known that
the second dice always exhibits 4.
Q25. A machine operates of all of its three components function. The probability that the first component
fails during the year is 0.14, the probability that the second component fails is 0.10 and the probability
that the third component fails is 0.05. What is the probability that the machine will fail during the
year?
Long Answer Type Questions (6 Marks)
Q26. A company has two plants to manufacture bicylces. The first plant manufactures 60% of the bicycles
and the second plant 40%. 80% of the bicyles are rated of standard quality at the first plant and 90%
of standard quality at the second plant. A bicycle is picked up at random and found to be of standard
quality. Find the probability that it comes from the second plant.
Q27. In a bolt factory, Machnies A, B and C and manufacture 60%, 25% and 15% of the total output
respectively. Of the total output 1 %, 2% and 1 % are defective bolts respectively manufactured by
machines A, B and C. A bolt is drawn at random form the total production and is found to be defective.
From which machine, the defective bolt is most likely to have been manufactured.
Q28. By examining the chest x-ray, the probability that T.B. is detected when a person is actually suffering
from it is 0.99. The probability that the doctor diagnosis incorrectly that a perosn has T.B. on the
basis of x-ray is 0.001. In a certain city. 1 in 1000 persons suffers from T.B. a person is sleeted at
random and is diagnosed to have T.B. what is the chance that he actually has T.B.?
Q29. A man is known to speak truth 3 out of 4 times. He throws a pair of dice and reports that sum of
numbers appeared is six. Find the probability that it is actually six.
Q30. Supose one of the three men A, B, C will be appointed as a vice chanceller of a university. The
respective probabilities of their appointment are 0.5,0.3,0. The probabilities that research facilities
will be enhanced by these people if they are appointed are 0.3, 0.7, and 0.8 respectively. If the
research facilities are enhanced. What is the probability that it was due to the appointment of C ?
Q31. A pair of dice is rolled twice. Let x denote the number of times, a total of 9 is obtained. Find the mean
and variance of the random variable X.
Q32. A bag contains 4 balls. Two balls are drawn at random and are found to be white. What is the
probability that in the bag all balls are white?
Q33. A letter is known to have come either from TATA NAGAR or KOLKATA. On the envelope, only two
consecutive letters TA are visible. What is the probability that letter has come from KOLKATA.
Q34. Two cards from a pack of cards are lost. From the remaining cards, a card is drawn and is found to
spade. Find the probability of the missing cards to be spades.
Q35. Bag A contains 3 red and 4 black balls and bag B contains 4 red and 5 black balls. Two balls are
transferred from bag A to bag B and then a ball is drawn from bag B. The ball so drawn is found to be
red in colour. Find the probability that the transferred ball were both black.
110
Answers
1.
7
8
2.
4.
2
7
5. 0.65
7.
2
21
8.
10.
2
7
11.
2
663
12.
13.
9
28
14.
3
4
15. 42%
16.
40
61
17.
5
6
18. 462 × (.24)5 = 0.3678
19.
2
3
3. (0.3)3
6.
1
6
9. Yes
X
0
1
2
3
P(X)
84
429
210
429
120
429
15
429
X
0
1
2
P(X)
9
16
6
16
1
16
20.
21.
4
9
22.
24.
1
6
25. .2467
27. Machine A.
1
2
1
13
71
91
23. (i) K =
26.
3
7
28.
110
221
29.
15
46
30.
4
13
31.
2 16
,
9 9
32.
3
5
33.
2
5
34.
22
425
35.
4
17
111
1
(ii) 0.6 (iii) 0.8
10
Hints
Require probability = 1 – P (balls are of same colour)
12.
±2 1 3 2 5 4 4 3²
³
³
³
= 1 ´ ³
¶14 13 14 13 14 13 14 13 µ·
71
91
18. Let X denote the number of steps taken forward
Required probability = P(x = 5) + P(x = 6)
¸
=
22. Events
¸
A1 =
A2 =
A3 =
B3 =
11
5
6
11
C6 (.6)5 (.4) 6
= C5 (.4) (.6)
drawn cards are both aces
drawn cards are both non aces
drawn cards are both one ace and one non ace
Ace is drawn from 50 cards
P(B) = P ( A1 ) P ( B / A1 ) P ( A2 ) P ( B / A2 ) P ( A3 ) P ( B / A3 )
¹ 4
= ¼¾
=
32. Let event
A1 =
A2 =
A3 =
B=
Required probability =
=
33. Let event
Let event
A1 =
A2 =
B=
P(A1) =
P(B/A1) =
Required prob. =
=
52
»
3 2º
» ½
51 50 ¿
3º
¹ 48 47 4 º ¹ 4 48
¼¾ 52 » 51 » 50 ½¿ ¼¾ 52 » 51 » 2 » 50 ½¿
1
13
the bag contains 2 white and 2 non white balls
the bag contains 3 white and 1 non white balls
the bag contains 4 white balls
two white balls are drawn from the bag.
P(A3/B)
1
À1
3
3
Á
[Use Baye’s thon]
1 1 1 1 1
À
À
À1 5
3 6 3 2 3
letter has come from TATA NAGAR
letter has come from KOLKATA
two consecutive letters visible are TA.
1
1
P(A2) =
2
2
1
1
P(B/A2) =
4
6
P(A2/B)
1 1
Â
2
2 6
Ã
1 1 1 1 5
Â
Â
2 4 2 6
112
35. Let event
P(A1) =
P(B/A2) =
A1 =
A2 =
A3 =
B=
3 2
Ä
7 6
6
11
balls transfered are both red
balls transfered are one red and one black
balls transfered are both back
Red ball is drown form bag B after transfer
P(A2) =
P(B/A2) =
3 4
Å Å2
7 6
5
11
P(A3) =
P(B/A3) =
113
4 3
Ä
7 6
4
11
12
LinearProgramming
Teaching-Learning Points
l
l
l
l
l
l
l
l
l
l
l
l
l
Linear programming is a technique to find optimal value (maximum or minimum) of a linear function
of a number of variables (say x and y) subject to a number of linear inequalities in the variables
involved and the variables take non negative values only.
A problem which seeks to maximise or minimise a linear function is called optimisation problem or
linear programming problem.
Linear function of the form z = ax + by where a, b are constants, which is to be maximized or minimized
is called objective function.
Linear inequalities in the variables of a linear programming problem are called constraints.
The common region determined by all the constraints including non negative constraints x, y Æ 0 of a
linear programming problem is called the feasible region. The region other than feasible region is
called infeasible region.
The points within and on the boundary of the feasible region represent feasible solutions. Any point
out the feasible region is called infeasible solution.
Any point in the feasible region that gives optimal value (maximum or minimum) of the objective
function is called an optimal solution.
Optimal value of an objective function must occurs at a corner point of the feasible region.
A feasible region is said to be bounded if it can be enclosed within a circle. Otherwise, it is called
unbounded.
If a feasible region is bounded, then the objective function has both a maximum and a minimum value
and each of these occurs at a corner point of the feasible region.
If the feasible region is unbouned, then a maximum or minimum value of a function may not exist.
However if it exists, if must occur at corner point of the feasible region.
After evaluating the objective function z = ax + by at each corner point, let M and m respetively
denote the largest and smallest values of these points and if
(i) Feasible region is bounded, then M and m are the maximum and minimum values of the function.
(ii) Feasible region is unbounded then M is the maximum value if z, provided the open half plane
determined by ax + by > M has no point in common with the feasible region, otherwise z has no
maximum value. Similiarly m is the minimum value of z, provided the open half plane determined
by ax + by < m has no point in common with the feasible region, otherwise z has no minimum
value.
If the obtimal solution of the objective function occures at two vertices (corners) of the feasible region,
then every point on the line segment joining these points will give the same optimal value.
114
Questions for Practice
Long Answer type Questions Carrying 6 Marks each
1.
A house wife wishes to mix two types of food x and y in such a way that the vitamin contents at the
x costs ` 60 per kg and
food y cost ` 80 per kg. Food x contains 3 units per kg of vitamin A and 5 units per kg of vitamin B
while food y contains 4 units per kg of vitamin A and 2 units per kg of vitamin B. Formulate is as a
linear programming problem to minimize the cost of the mixture and solve it graphically.
m
i
x
t
u
r
e
c
o
n
t
a
i
n
s
a
t
l
e
a
s
t
8
u
n
i
t
s
o
f
v
i
t
a
m
i
n
A
a
n
d
1
1
u
n
i
t
s
o
f
v
i
t
a
m
i
n
B
.
F
o
o
d
2.
A dealer has ` 15000 only for a purchase of rice and wheat. A bag of rice costs ` 1500 and bag of
wheat costs ` 1200. He has a storage capacity of ten bags only and the dealer gets a profit of ` 110 and
` 80 per bag of rice and wheat respectively. Formulate it as a linear programming problem to get the
maximum profit and solve it graphically.
3.
In a small scale industry a manufacture produces two types of book cases. The first type of book cose
requires 3 hours on machine A and 2 hours on machine B for completion. whereas the second type of
book case requires 3 hours on machine A and 3 hours on machine B. Machine A can run at most for 18
hours and machine B can run atmost for 14 hours per day. He earns a profit of ` 30 on each book case
of first type and ` 40 one each book case of 2nd type. How many book cases of each type should he
make everyday so as to have a maximum profit? Solve it as a linear programming problem.
4.
A machine producing either product A or B can produce A by using 2 units of chemicals and unit of a
compound and can produce B by using unit of chemicals and 2 units of compound. If only 800 units
of chemicals and 1000 units of the compound are available and the profits per unit of A and B are
respectively ` 30 and ` 20, then find the number of units of A and B to be produced so as to maximize
the total profit. Find the maximum profit also, by solving it as a linear programming problem.
5.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply
to 3 ration shops D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of
transportation per quintal from godowns to the shops are given below in the table:
Transportaion cost per quintal (in Rs)
From/To
A
B
D
6
4
E
3
2
F
2.50
3
How should the supplier be transported in order that the transportation cost is minimum? Also find the
minimum cost by solving it as a linear programming problem.
115
6.
A manufactures makes two types of toys A and B. Three machines are needed for this purpose and the
time required (in minutes) for each toy on the machines is given below :
Types of toys
Machines
I
II
III
A
12
18
6
B
6
0
9
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is
` 7.50 and that on each toy of type B as ` 5, find the number of toys of each type to be manufactured
in a day to get maximum profit. Solve it as a linear programming problem. Also find maximum profit.
Answers
1.
Í 1Î
8
Minimum cast = ` 160, at all points on the line segment joining ÇÉ , 0ÈÊ and Ï 2, Ð .
Ñ 2Ò
Ë3 Ì
2.
3.
Ten bags of rice only and maximum profit = ` 1100.
First type = 4, second type = 2, and maximum profit = ` 200.
4.
5.
6.
Maximum profit = ` 14000 for A = 200 units and B = 400 units.
From A : 10, 50, 40 units and from B : 50,0,0 units minimum cost = ` 10.
Type A toys = 15, type B toys = 30, maximum profit = ` 262.50.
116
