pedagogical issues in mathematics at sr. sec. level for lecturer 2012
TRANSCRIPT
Chief AdvisorRashmi Krishnan, IAS
Director, SCERT
GuidanceDr. Anita Setia, Additional Director, SCERT
Dr. Pratibha SharmaJoint Director, SCERT-cum-state pedagogy coordinator
Academic Co-ordinatorDr. Anil Kumar Teotia
Sr. Lecturer, DIET Dilshad Garden
Mr. Sanjay KumarLecturer, SCERT
ContributorsDr. Anup Rajput Associate Professor, NCERTDr. Anil Kumar Teotia Sr. Lecturer, DIET Dilshad GardenDr. Satyavir Singh Principal, SN Inter College, PilanaMr. D.R. Sharma Vice Principal, Navodaya Vidhalya, MungeshpurDr. R.P. Singh Lecturer, RPVV, Gandhi NagarMr. Sanjeev Kumar Lecturer, RPVV, Raj Niwas MargMr. Ashutosh Kr. Agarwal Lecturer, RPVV, Nand Nagri
EditorsDr. Anil Kumar Teotia
Sr. Lecturer, DIET Dilshad Garden
Dr. Kusum BhatiaSr. Lecturer, DIET, Pitampura
Publication OfficerMr. Mukesh Yadav
Published by : State Council of Educational Research & Training, New Delhi and printed atEducational 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 accordancewith growth of the subject and emerging needs of the society. Senior Secondary stage is a launchingstage from where the students go either for higher academic education in Mathematics or for professionalcourses like engineering, physical and Bioscience, commerce or computer applications. The presentrevised syllabus has been designed in accordance with National Curriculum Frame work 2005 and asper guidelines given in Focus Group on Teaching of Mathematics 2005 which is to meet the emergingneeds of all categories of students. Motivating the topics from real life situations and other subjectareas, 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 morethan 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, observanceof small family norms, removal of social barriers, elimination of sex biases.
to develop reverence and respect towards great Mathematicians for their contributions to the fieldof Mathematics.
Syllabus for Class-XI
Units Marks
I. Sets and Functions 29
II. Algebra 37
III. Coordinate Geometry 13
IV. Calculus 06
V. Mathematical Reasoning 03
VI. Statistics and Probability 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 theset of real numbers especially intervals (with notations). Power set. Universal set.Venn diagrams. Unionand 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 finitesets. Cartesian product of the reals with itself (uptoR × R × R).
Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. Function as aspecial 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 rangeof these functions, constant, identity, polynomial, rational, modulus, signum and greatest integerfunctions 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 fromone measure to another. Definition of trigonometric functions with the help of unit circle. Truthof the identity sin2 x + cos2 x = 1, for all x. Signs of trigonometric functions and sketch of theirgraphs. Expressing sin (x ± y) and cos (x ± y) in terms of sinx, sin y, cosx & cos y. Deducingthe identities like following :
5
6
tan( )x y =tan tan cot cot 1
,cot( ) ,1 tan tan cot cot
x y x yx y
x y y x
sin sinx y = 2sin cos ,cos cos 2cos cos ,2 2 2 2
x y x y x y x yx y
sin sinx y = 2cos sin ,cos cos 2sin sin .2 2 2 2
x y x y x y x yx y
Identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x. General solution of trigonometricequations of the type sin = sin , cos = cos and tan = tan . Proof and simple applicationof sine and cosine formulae.
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 naturalnumbers as the least inductive subset of real numbers. The principle of mathematical inductionand 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 polarrepresentation of complex numbers. Statement of Fundamental Theorem of Algebra, solution ofquadratic 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 representationon the number line. Graphical solution of linear inequalities in two variables. Solution of systemof linear inequalities in two variables-graphically.
4. Permutations and Combinations : (Periods 12)
Fundamental principle of counting. Factorialn. (n!) Permutations and combinations, derivationof 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 ofn terms of a G.P., Arithmetic and geometric series, infiniteG.P. and its sum, geometric mean (G.M), relation between A.M. and G.M. Sum to n terms of thespecial series :
2 3
1 1 1
, , and .n n n
k k k
k k k
7
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 twolines. 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 familyof 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 intersectinglines 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 betweentwo 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
function and its geometric meaning.0 0
log (1 ) 1lim , lim
xe
x x
x e
x x. Definition of derivative, relate it
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 understandingof “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, converseand contrapositive.
UNIT 6 : STATISTICS AND PROBABILITY
1. Statistics : (Periods 10)
Measure of dispersion; mean deviation, variance and standard deviation of ungrouped/ groupeddata. 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, probabilityof ‘not’, ‘and’ & ‘or’ events.
8
Units Marks
1. Relations and Functions 10
II. Algebra 13
III. Calculus 44
IV. Vectors And Three-dimensional Geometry 17
V. Linear Programming 06
VI. Probability 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).
Syllabus for Class-XII
9
2. Determinants: (20 Periods)
Determinant of a square matrix (up to 3 x 3 matrices), properties of determinants, minors, cofactorsand applications of determinants in finding the area of a triangle. Adjoint and inverse of a squarematrix. 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 inverseof a matrix.
UNIT-III: CALCULUS
1. Continuity and Differentiability : (18 Periods)
Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inversetrigonometric functions, derivative of implicit function. Concept of exponential and logarithmicfunctions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, Derivativeof 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, tangentsand normals, use of derivatives in approximation, maxima and minima (first derivative test motivatedgeometrically and second derivative test given as a provable tool). Simple problems (that illustratebasic principles and understanding of the subject as well as real-life situations).
3. Integrals: (20 Periods)
2 2 2 2 2 2, , ,
dx dx dx
x a x a a x
2 2,
dx dx
ax bx c ax bx c
22
( )and ( )
px qdx px q ax bx c dx
ax bx c
be evaluated.
Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof). Basicproperties 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 beclearly identifiable).
5. Differential Equations : (10 Periods)
Definition, order and degree, general and particular solutions of a differential equation. Formationof differential equation whose general solution is given. Solution of differential equations by method
1 0
of separation of variables, homogeneous differential equations of first order and first degree. Solutionsof linear differential equation of the type :
P Q,dy
ydx
where P and Q are functions ofx or constants.
P Q,dx
xdy
where P and Q are function ofy or constants.
UNIT-IV: VECTORS AND THREE-DIMENSIONAL GEOMETRY
1. Vectors: (12Periods)
Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratiosof vectors. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector ofa point, negative of a vector, components of a vector, addition of vectors, multiplication of a vectorby a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) productof vectors, projection of a vector on a line. Vector (cross) product of vector, scalar triple product.
2. Three - dimensional Geometry: (12 Periods)
Direction consines and direction ratios of a line joining two points. Cartesian and vector equationof a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equationof a plane. Angle between (i) two lines, (ii) two planes. (iii) a line and a plane. Distance of apoint 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, feasibleand 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 1
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 4a i j k on 2 6 3b i j k is 4 units. 1
Q3. Find the sum of the vectors 2a i j k , 2 4 5b i j k and 6 7 .c i j k 1
Q4. Evaluate :3
2
1dx
x1
Q5. Evaluate (1 ) .x x dx 1
Q6. If =
5 3 8
2 0 1
1 2 3, write the minor of the element a
23. 1
Q7. If2 3 1 3 4 6
5 7 2 4 9 x, write the value ofx. 1
Q8. Simplify :cos sin sin cos
cos sinsin cos cos sin 1
Q9. Write the principal value of 1 11 1cos 2sin
2 2. 1
Q10. Let * be a ‘binary’ operation on N given by a * b = LCM (a, b) for all a, b N. Find 5 * 7. 1
Question Paper-Delhi (2012)
1 2
SECTION-B
Question numbers 11 to 22 carry 4 mark each.
Q11. If (cos x)y = (cos y)x, find .dy
dx4
OR
If sin y = x sin (a + y), prove that2sin ( )
.sin
dy a y
dx aQ12. 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
perpendicular to the two lines8 19 10
3 16 7
x y z and
15 29 5.
3 8 5
x y z4
Q14. If , ,a b c are three vectors such thata = 5, b = 12 and c = 13, and 0,a b c find the
value of . . . .a b b c c a 4
Q15. Solve the following differential equation : 4
2 22 2 0.dy
x xy ydx
Q16. Find the particular solution of the following differential equation. 4
2 2 2 21dy
x y x ydx
, given thaty = 1 wherex = 0.
Q17. Evluate : sin sin 2 sin3x x x dx
OR
Evaluate : 2
2
(1 )(1 )dx
x x
Q18. Find the point on the curvey = x3 – 11x + 5 at which the equation of tangent isy = x – 11.
OR
Using differentials, find the approximate value of49.5. 4
Q19. If y = 1 2(tan )x , show that 4
22 2 2
2( 1) 2 ( 1) 2d y dy
x x xdx dx
.
Q20. Using properties of determinants, prove that 6
2
b c q r y z a p x
c a r p z x b q y
a b p q x y c r z
1 3
Q21. Prove that1 cos
tan , , .1 sin 4 2 2 2
x xx
x6
OR
Prove that 1 1 18 3 36sin sin cos .
17 5 85
Q22. Let A = R – {3} and B = R – {1}. Consider the function f : A B defined by2
( )3
xf x
x.
Show thatf is one-one and onto and hence findf –1. 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). 6Q24. 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 hostelattain ‘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? 6Q25. 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 machineB to produce a package of bolts. He earns a profit of` 17.50 per package on nuts and` 7per package of bolts. How many packages of each should be produced each day so as to maximizehis profits if he operates his machines for at the most 12 hours a day? Form the above as alinear programming problem and solve it graphically. 6
Q26. Prove that4
0
( tan cot ) 2.2
x x dx 6
OR
Evaluate3
2
1
(2 5 )x x dx as a limit of a sum.
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 = 73x + 4y – 5z = – 5
2x – y + 3z = 12OR
Using elementary operations, find the inverse of the following matrix :
1 1 2
1 2 3
3 1 1
1 4
Q.No. Value Pooints/Solution 65/1//1 Marks.
SECTION-A
1-10 1.2 1 2
, ,3 3 3
2. = 5 3. 4 j k 4.3
log2 5.
3 2 5 23 2
2 5x x c
6. 2,3M 7 7. 13 8.1 0
0 1 9.2
3 10. 35. 1×10 = 10
SECTION-B
11. (cos ) (cos ) log cos log cosy xx y y x x y 1/2
( sin ) ( sin ). log cos . log cos
cos cos
x dy y dyy x x y
x dx y dx 1+1
(log cos tan ) log cos tandy
x x y y y xdx
1
dy
dx =
log cos tan
log cos tan
y y x
x x y 1/2
OR
sin sin( ) cos cos( ) sin( )dy dy
y x a y y x a y a ydx dx
1
sin( )
cos cos( )
dy a y
dx y x a y 1
sin sin( )sinsin( ) cos .cos( )
sin( )
y dy a yx
ya y dx y a ya y
1
2 2sin ( ) sin ( )
sin( )cos cos( )sin sin
dy a y a y
dx a y y a y y a1
Marking SchemeClass-XII
Mathematics (March 2012)
1 5
12. Let the coin be tossedn times
P(getting at least one heat) >80
1001
8 8 2 11 P(0) P( ) 1
10 10 10 5o 1
0
0
1 1 1 1 1or or 2 5
2 2 5 2 5
nn n
nC 1
n = 3. 1
13. Let the vector equation of required line bea a b
than a = 2 4i j k
and b = (3 16 7 ) (3 8 5 )i j k i j k 1
= 24 36 72i j k 1
Vector equation of line is
r = ( 2 4 ) (24 36 72 )i j k i j k
}or r = ( 2 4 ) (2 3 6 )i j k i j k 1
and cartesian from is1 2 4
2 3 6
x y z1
14. 20 ( ) 0a b c a b c 1/2
2 2 22( . . . ) 0a b c a b b c c a 1
or 2 2 2| | | | | | 2( . . . ) 0a b c a b b c c a 1
. . .a b b c c a =1
(25 144 169) 1692 . 1
15.
2
2 22 2
2
222 2 0
2 2
y ydy dy xy y x xx xy ydx dx x
1/2
Puttingy
vx
so thaty = vx anddy dv
v xdx dx
1
2 21 1
2 2
dv dvv x v v x v
dx dx1/2
2
22 log
dv dxx c
v x v1
2 2log
log
x xx c y
y x c 1
1 6
16.2 2 2 2 2 21 (1 )(1 )
dyx y x y x y
dx½
22 (1 )
1
dyx dx
y 1
31tan
3
xy x c 1
x = 0, y = 1 c = /4 1
31tan
3 4
xy x or
3
tan4 3
xy x ½
17. I =1
sin sin 2 sin 3 2sin 3 sin .sin 22
x x xdx x x xdx ½
=1 1
(cos 2 cos 4 )sin 2 (sin 2 cos 2 cos 4 sin 2 )2 2
x x xdx x x x x dx ½
=1 1
sin 4 2cos 4 sin 24 4
xdx x xdx 1
=1 1
cos 4 (sin 6 sin 2 )16 4
x x x dx 1
=1 1 1
cos 4 cos6 cos 216 24 8
x x x c 1
OR
2 2
2 A B C
(1 )(1 ) 1 1
x
x x x x ½
22 A(1 ) (B C)(1 )x x x 1½
0 = A – B, B – C = 0 A + C = 2 A = B = C = 1
2
2
(1 )(1 )dx
x x = 2
1 1
1 1
xdx dx
x x½
=2 11
log |1 | ( 1) tan2
x x x c 1½
18. Slope of tangent,y = x – 11 is 1 ½
3 211 5 3 11dy
y x x xdx
½
If the point is (x1, y
1) then 2
1 13 11 1 2x x 1
x1 = 2 then 1 8 22 5 9y and if 1 2x then 1 19y 1
Since (–2, 19) do not lie on the tangent y = x – 11 ½Required point is (2, – 9) ½
1 7
OR
Let y = x y y x x ½
dyy x x x
dx
1.
2x x x x
x1
Putting x = 49 and x = 0.5 we get 1
149 (0.5) 49.5
2 49½
149.5 7 7.0357
281
19.1 2 1
2
1(tan ) 2 tan
1
dyy x x
dx x
2 1(1 ) 2 tandy
x xdx
22
2 2
2(1 ) 2 .
1
d y dyx x
dx dx x
22 2 2
2(1 ) . 2 (1 ) 2.d y dy
x x xdx dx
20. Using R1
R1 + R
2 + R
3 we get
LHS =
2( ) 2( ) 2( )a b c p q r x y z
c a r p z x
a b p q x y
1
= 2
a b c p q r x y z
c a r p z x
a b p q x y
1
= 2
a b c p q r x y z
b q y
c r z
1
= 2
a p x
b q y
c r z
1
Using R2
R2 – R
1
R3
R3 – R
1
Using R1
R1 + R
2 + R
3
= RHS R2
– R2
R3
– R3
1 8
21. 1 1
sincos 2
tan tan1 sin
1 cos2
xx
xx
1
=1 1
2
2sin cos4 2 4 2
tan tan tan4 2
2cos4 2
x xx
x1+1
=4 2
x1
OR
Writing1 18 8
sin tan17 15 and
1 13 3sin tan
5 4 1
LHS =1 1 1 1
8 38 3 7715 4tan tan tan tan
8 315 4 361 ,15 4
1+1
Getting 1 177 36tan cos
36 851
22. Let 1 2, Ax x and 1 2( ) ( )f x f x ½
1 2
1 2
2 2
3 3
x x
x x 1 2 2 1 1 2 1 22 3 2 3x x x x x x x x
x1 = x
21
Hence f is 1 – 1
Let y B,2
( ) 3 23
xy f x y xy y x
x
or x =3 2
1
y
y ½
Since y 1 and3 2
3 A1
yx
yHence f is ONTO 1
and1 3 2( )
1
yf y
y 1
SECTION-C
23. Normal to the plane is AB BCn ½
n = 2 3 2 12 16 12
6 3 2
i j k
i j k 1½
1 9
Equation of plane is
.(12 16 12 )r i j k = (3 2 ).(12 16 12 )i j k i j k= 76 2
or .(3 4 3 )r i j k = 19 or 3x – 4y + 3z – 19 = 0 }
Distance of plane from the point P(6, 5, 9) is
d =18 20 27 19 6
9 16 9 342
24. Let E1 : selected student is a hostlier
E2
: selected student is a day scholar}
1A : selected student attain ‘A’ grade in exam.
P(E1) =
60
100, P(E
2) =
40
1001
P(A/E1) =
30
100, P(A/E
2) =
20
1001
P(E1/A) = 1 1
1 1 2 2
P(E ) . P(A/E )
P(E ) . P(A/E ) + P(E ) . P(A/E )1
=60 30. 9100 100
40 20 1360 30. .100 100 100 100
1+1
25. Let x package of nuts andy package of bolts be produced each dayLPP is maximise P = 17.5x + 7y 1
subject to x + 3y 12
3x + y 12}
2x 0, y 0 correct graph
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
2 0
26. I =
4 4
0 0
sin cos( tan cot )
sin cos
x xx x dx dx
x x1
Putting sin cos ,x x t to get (cos sin )x x dx dt 1
and sin cosx x =21
2
t1
I =0
21
21
dt
t= 1 0
12.[sin ]t 1+1
=1 12(sin 0 sin ( 1) 2.
21
OR
I =3
2
1
(2 5 )x x dx =0
lim [ (1) (1 ) (1 2 ) ... (1 1 )]h
h f f h f h f n h
where ( )f x = 22 5x x and2
hn
or nh = 2. 1
f(1) = 7
f(1 + h) = 2 22(1 ) 5(1 ) 7 9 2h h h
(1 2 )f h = 2 2 22(1 2 ) 5(1 2 ) 7 18 22h h h h 2
(1 3 )f h = 2 2 22(1 3 ) 5(1 3 ) 7 27 2.3h h h h
...........
(1 ( 1) )f n h = 2 27 9( 1) 2.( 1)n h n h
I =2
0
( 1) ( 1)(2 1)lim 7 9 2 .
2 6h
n n n n nh n h h 1
=0
9 1lim 7 ( ) ( )(2 )
2 3hnh nh nh h nh nh h nh h 1
=16 112
14 183 3
1
27. Let AB be 3 2 1 0,x y BC be2 3 21 0x y and AC be 5 9 0x y correct figure : 1
Solving to get A(1, 2), B (3, 5) and C(6, 3) 1½
area of ( ABC) =3 6 6
1 3 1
1 1 1(3 1) (21 2 ) ( 9)
2 3 5x dx x dx x dx 1
2 1
=
6 63 2 22
1 3 1
1 (21 2 ) ( 9)(3 1)
12 12 10
x xx 1½
=25
7 122
½
=13
2 sq. U. ½
28. Surface area A =22 2rh r (Given) ½
h =2A 2
2
r
r...(1) 1
V =2
2 2 A 2
2
rr h r
r 1
=31
.[A 2 ]2
r r ½
dv
dr =
21[A 6 ]
2r 1
dv
dr = 0 = 2 26 A 2 2r rh r 1
24 2r rh h = 2r = diameter 1
2
2
d v
dr =
1[ 12 ] 0 2
2r h r will give max. volume. 1
29. Given equations can be written as
1 1 2
3 4 5
2 1 3
x
y
z
=
7
5
12 or AX = B 1
a11
= 7, a12
= –19 a13
= – 11
a21
= 1, a22
= – 1 a23
= – 1
a31
= –3, a32
= 11 a33
= 72
–1A =
7 1 31
19 1 114
11 1 7
½
x
y
z
=
7 1 3 7 21
19 1 11 5 14
11 1 7 12 3
x = 2, y = 1,z = 3. 1½
2 2
OR
Let A =
1 1 2
1 2 3
3 1 1
Writing
1 1 2
1 2 3
3 1 1
= AA
1 0 0
0 1 0
0 0 1
1
c1
c2
1 1 2
2 1 3
1 3 1
= A
0 1 0
1 0 0
0 0 1
½
c2
c2+ c
1
1 0 0
2 3 1
1 4 1
= A
0 1 0
1 1 2
0 0 1
1
c3
c3 –2c
1
c1
c1+ 2c
3
1 0 0
0 1 1
1 2 1
= A
0 1 0
3 3 2
2 2 1
½
c2
c2+ 2c
3
c3
c3+ –c
2
1 0 0
0 1 0
1 2 1
= A
0 1 1
3 3 5
2 2 3½
c1
c1+ c
3
c2
c2+ 2c
3
1 0 0
0 1 0
0 0 1
= A
1 1 1
8 7 5
5 4 3
1
A–1 =
1 1 1
8 7 5
5 4 31
2 3
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 fourmarks 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 exactrequirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 4 questions of fourmarks each and 2 questions of six marks each. You have to attempt only one of the alternativesin 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 asa * b = 2a + b. Find (2 * 3) * 4.
Q2. Find the principal value of 1 1tan 3 sec ( 2).
Q3. Find the value ofx + y from the following equation :
5 3 4 7 62
7 3 1 2 15 14
x
y
Q4. If A T =
3 4
1 2
0 1
and B =1 2 1
1 2 3, then find AAT – BT.
Q5. Let A be a square matrix of order 3 × 3. Write the value of |2A|, where |A| = 4.
Q6. Evaluate :2
2
0
4 x dx
Q7. Given (tan 1)sec ( ) .x xe x xdx e 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 vectorAB 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.
Question Paper-Outside Delhi (2012)
2 4
SECTION-B
Questions numbers 11 to 22 carry 4 mark each.Q11. Prove the following :
–1 13 3 6cos sin cot
5 2 5 13
Q12. Using properties of determinants, show that
4
b c a a
b c a b abc
c c a b
Q13. Show thatf : N N, given by
f(x) =1, if is odd
1, if is even
x x
x x
is both one-one and onto.OR
Consider the binary operations * : R × R R and o : R × R R defined asa * b = |a – b| anda ob= a for all a, b R. Show that ‘*’ is commutative but not associative, ‘o’ is associative but not
commutative.
Q14. If x = 1 1sin cos,t ta y a , show that .dy y
dx x
OR
Differentiate2
1 1 1tan
x
x with respect tox.
Q15. If x = a (cost + t sin t) andy = a (sin t – t cost), 0 < t < ,2
find2 2
2 2,d x d y
dt dt and
2
2 .d y
dx
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 footof the ladder is 4 m away from the wall?
Q17. Evaluate :2
3
1
| |x x dx
OR
Evaluate :2
0
sin
1 cos
x xdx
x
Q18. Form the differential equation of the family of circles in the second quadrant and touching thecoordinate axes.
2 5
ORFind the particular solution of the differential equation
2( 1) 1; 0dy
x x ydx
whenx = 2.
Q19. Solve the following differential equation :
( 2(1 ) 2 cot ; 0x dy xy dx x dx x
Q20. Let a = 4 2 ,i j k b = 3 2 7i j k andc = 2 4i j k . Find a vectorp which is perpendicular
to both a andb and . 18.p c
Q21. Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crossesthe 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 3 3 5x y z , 2 4x y z , 3 2 3x y z .
Q24. Prove that the radius of the right circular cylinder of greatest curved surface area which can beinscribed in a given cone is half of that of the cone.
ORAn open box with a square base is to be made out of a given quantity of cardboard of areac2 square
units. Show that the maximum volume of the box is3
6 3
c cubic units.
Q25. Evaluate :1
2
sin
1
x xdx
xOR
Evaluate :2
2
1
( 1) ( 3)
xdx
x xQ26. Find the area of the region {(x, y) : x2 + y2 4, x + y 2}.
Q27. If the lines1 2 3
3 2 2
x y z
k and
1 2 3
1 5
x y z
k are perpendicular, find the value ofk and
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 sheobtained 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 mixturecontains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitaminA and 1 unit/kg of vitamin C while Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitaminC. It costs 5 per kg to purchase Food I and` 7 per kg to purchase Food II. Determine the minimumcost of such a mixture. Formulate the above as a LPP and solve it graphically.
2 6
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 fourmarks 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 exactrequirement of the question.
(iv) There is no overall choice. However, internal choice has been provided in 4 questions of fourmarks each and 2 questions of six marks each. You have to attempt only one of the alternativesin 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 bya * 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. If2 1 10
3 1 5x y , write the value ofx.
Q5. Write the value of the following determinant :
102 18 36
1 3 4
17 3 6
Q6. If 2
1( )x xx
e dx f x e cx
, then write the value off(x).
Q7. If 2
0
3 8a
x dx , write the value of ‘a’.
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 and3 j .Q10. Write the direction cosines of a line parallel to z-axis.
Question Paper-Outside India (2012)
2 7
SECTION-B
Questions numbers 11 to 22 carry 4 mark each.
Q11. If4 3 2
( ) ,6 4 3
xf x x
x, show that fof(x) = x for all
2
3x . What is the inverse off ?
Q12. Prove that : 1 1 163 5 3sin sin cos
65 13 5OR
Sovle forx :
1 12 tan (sin ) tan (2sec ),2
x x x
Q13. Using properties of determinants, prove that
32 3 2 4 3 2
3 6 3 10 6 3
a a b a b c
a a b a b c a
a a b a b c
Q14. If ( )m n m nx y x y , prove that .dy y
dx x
Q15. If1cos , 1 1a xy e x show that
22 2
2(1 ) 0.d y dy
x x a ydx dx
OR
If 1 1 0,x y y x 1 1,x x y, then prove that 2
1.
(1 )
dy
dx x
Q16. Show that2
log(1 ) , 12
xy x x
x, is an increasing function ofx throughout its domain.
OR
Find the equation of the normal at the point (am2, am3) for the curveay2 = x3.
Q17. Evaluate : 2 1tanx x dx
OR
Evaluate : 2
3 1
( 2)
xdx
xQ18. Solve the following differential equation :
2
1, 0xe y dx
xdyx x .
2 8
Q19. Solve the following differential equation :
23 tan (2 )sec 0x xe ydx e y dy , given that whenx = 0, .4
y
Q20. If = 3 4 5i j k and 2 4i j k , then express in the form 1 2 , where 1 is parallel
to and 2 is perpendicualr to
Q21. Find the vector and cartesian equations of the line passing through the point P(1, 2, 3) and parallel to
the planes . ( 2 ) 5r i j k and . (3 ) 6.r i j k
Q22. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probabilitydistribution 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 :
4; 2 3 0; 2x y z x y z x y z
OR
If 1
3 1 1
A 15 6 5
5 2 2
and B =
1 2 2
1 3 0
0 2 1, find (AB)–1.
Q24. Show that the altutude of the right circular cone of maximum volume that can be inscribed in a
sphere of radius R is4R
.3
Q25. Find the area of the region in the first quadrant enclosed by x-axis, the line 3x y and the circle
2 2 4.x y
Q26. Evaluate :3
2
1
( )x x dx as a limit of a sum.
OR
Evaluate :
4 2
2 20
cos
cos 4sin
xdx
x x
Q27. Find the vector equation of the plane passing through the points (2, 1, – 1) and (–1, 3, 4) and
perpendicular to the planex – 2y + 4z = 10. Also show that the plane thus obtained contains the line
3 4 (3 2 5 ).r i j k i j k
2 9
Q28. A company produces soft drinks that has a contract which requires that a minimum of 80 units of thechemical A and 60 units of the chemical B go into each bottle of the drink. The chemicals areavailable in prepared mix packets from two different suppliers. Supplier S had a packet of mix of 4units 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 unitof B that costs 4. How many packets of mixes from S and T should the company purchase tohonour 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 thestudents in the college are girls. A student is selected at random from the college and is found to betaller than 1.75 metres. Find the probability that the selected student is a girl.
3 0
Teaching-Learning Points
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) : að Aandbð B}. If ( a.b) ð R, we say that a is related tob under the relation R and we write asa R b.
l R R A B.
l A relation R in a set A is a subset of A × A.
Types of relations :
(i) empty relation : R CA A
(ii ) Universal relationR = A × A
(iii ) Reflexive relation :( , ) A.a a R
(iv) Symmetric relation : If( , ) ( , )a b R b a R ,a b A.
(v) Transitive relation : If ( , )a b Rand
( , ) , ,a c R a b c A.
l A relation R is set A is said to be equivalence relation. If R is reflexive, symmetric and transitive.
l Let R is an equivalence relation is set A and R divides A into mutually disjoint subset A called partitionsor subdivisions of A subsfying the conditions :
(i) all element ofAi are related to each other, i.
(ii ) no element ofAi is related to any element ofA
j, it j
(iii ) UAi = A and A
iAj = , j.
l Type of Functions :
(i) one-one (or injective) function : Let F : A B, then for everyx1, x
2A, f(x
1) = f(x
2)
x1 = x
2.
(ii ) onto (or surjective function) : Let F : A B, then for everyy B, there exists an elementx A such thatf(x) = y.
(iii ) A function which is not one-one is called many-one function.l A function which is not onto is called into function.
l A function which is both one-one and onto is called a bijective function.
l Let A be a finite set then an injective function F : A A is subjective and conversely.
l Let F : A B and g : B C be two functions. Then the composition of F and g, denoted as gof isdefined as the function gof : A C given by g of (x) = g [f(x)]
x A
Relations and Functions1
3 1
l Composition of functions need not to be commutative and associative.
l 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.
l A function F : A� B is said to be invertible if there exists another function g : B� A such that
gof = IA and fog = I
B. The function g is called the inverse of the function F.
l A function F : A� B is said to be invertible if and only if F is one-one and onto (i.e. bijective).
l 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 A binary operation * on a set A is a function * A × A� A we denoted * (a, b) by a * b.
l A binary operation * on a set A is called commutative ifa * b = b * a a, b A.
l A binary operation * on a set A is said to be associative ifa * (b * b) * c a, b, c A.
l The elemente A, if it exists, is called identity element for binary operation * if a * e = a = e * aa A.
l The elementa A is said to be invertible with respect to the binary operation * if there exileb Asuch thata * b = e = b * a. The elementb is called morse ofa and is denoted asa–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 ofb} can we say R issymmetric? 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 Rsymmetric? 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 saythat 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 2
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 off(x) =3 2
5
x�.
Q12. If f(x) = x + 7 andg(x) = x – 7, x� R, find fog (x).
Q13. Write the inverse of the functionf(x) = 5x + 7,x� R.
Q14. Show thatf : R� R defined asf(x) = x2 + 1 is not one-one.
Q15. Show that the functionf : N� N defined byf(x) = 3x is not an onto function.
Q16. Let * be a binary operation on Z defiand bya * b = 2a + b – 3, find 3 * 4.
Q17. Let * be a binary operation on N defined bya * b = a2 + b and O be a binary operation on N definedby aob = 3a – b find (2 * 1) 02.
Q18. let * be a binary operation on R defined bya * b = a – b. Show * is not commutative on R.
Q19. Let * be a binary operation on N given bya * 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 bya * 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 bya * b = HCF (a, b). Deos there exist identify elementfor * on N?
Short Answer Type Questions (4 Marks)
Q26. Show thatf : 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 with6 – a being the inverse ofa.
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 byf(x) = 3x + 2. Show thatf is invertible. Findf : R R.
Q30. Let * be a binary operation on N × N defined by (a, b) * (c, d) = (a + c, b + d). Show that * iscommutative 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 = {(1,
2) :
1 2 1,
2T}. Show that R is an equivalence relation.
3 3
Q32. Let * be the binary operation on Q (Rational numbers) defiend bya * b = |a – b|, show that
(i) * is commutative
(ii ) * is not associative
(iii ) * does not have identity element
Q33. Show thatf : R� R defined byf(x) = x3 – 1; is invertible. Find f (x).
Q34. Show that if f : B� A is defined byf(x) =3 4
5 7
x
x
�
� and g : A� B is defined byg(x) =
7 1
5 3
x
x
�
, then
Fog = IA and gof = I
B, where A = R –
3
5
and B = R –7
5 .
Q35. Show that the function F : Q – {3} Q, given by F(x) =2 3
3
x
x is not a bijective function.
AnswersVery Short Answer (1 Mark)
1. No, if is a husband of b, then b being a female can not be husband of anybody.
2. No, because (a, c) R but ( , ) .c a R
3. R–1 = {(b, a), (d, c), (f, e)}
4. No, If x y & y z x ||z.
5. {3, 7, 11}.
6. No, example2
1 1
3 3.
7. No, Leta = 5,b = 10,c = 12, then (a, b) R, (b, c) R but ( , ) .a c R
8. R = {(2,4), (3,3), (4,2)} 9. 64
10. (1,1) .R 11.1 5 2( )
3
xf x
12. x 13.7
5
x
16. 7 17. 13
19. 6
20. No, because Number divided by 0 does not belong to R.
21. Leta = 2,b = 5,c = 8, (a * b) * c = (2 + 2 × 5) * 8 = 12 * 8
= 12 2 8 28 and *( * ) 2 *(5*) 2*(5 2 8)a b c
= 2*21 2 2 21 44.
3 4
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) =2
3
x�30. Identity does not exist
33. 1 1 3( ) ( 1)f x x� �
35. f(x1) = f(x
2)� x
1 = x
2� is one-one.
Let y� codomain thenf(x) = y
� x =3 3
Q –{3} for some Q2
yy
y
� �� �
�
Example 2� codomain but
=3 3 2
2 2
� �
� = Not defined, does not belong to domain
3 5
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
restrict the domain to ,2 2
�� �� ���
then the sine function becomes a one-one and onto function and
therefore we com define the inverse of the function sin : ,2 2
�� �� ���
� [–1, 1] as 1sin :[ 1,1] ,2 2
� �� �� �� !
"
In fact there are other intervals also like3 3
, , ,2 2 2 2
# $ #$ $ $% & % &' '( (
etc which may also be taken as range of
the function sin–1. Corresponding to each interval we get branch of sin–1. The branch with range ,2 2
#$ $% &'(
is called principal value branch similiary for other inverse trigonometric functions we have principalvalue branches.
l List of principal value branches and the domain of inverse trigonometric functions.
Functions Domain Range (Principal value Branch)
y = sin–1x 1 1x) * * 22y
+, ,- -
y = cos–1x 1 1x. / / 0 y0 0 1
y = tan–1x x22 2
y
y = cot–1x x 0 < y <
y = sec–1x1
1
x
x
2
02
y
y
y = cosec–1x1
1
x
x
02
02
y
y
Inverse TrigonometricFunctions
2
3 6
l Properties of inverse trigonometric functions :
1. (i) 1sin (sin )x x3 4 , ,2 2
x56 67 8
9:;
(ii ) 1sin(sin )x x3 4 , [ 1,1]x< =
(iii ) 1cos (cos )x x3 4 , [0, ]x> ?
(iv) 1cos(cos ) ,x x3 4 [ 1,1]x< =
(v) 1tan (tan ) ,x x@ A ,2 2
xBC CD E
FG HI J
(vi) 1tan(tan ) ,x xK L x M R.
2. (i) 1 11sin cosec ,x
xN NO P
QR ST U |x| 1
(ii ) 1 11cos sec ,x
xV VW X
YZ [\ ] |x| 1
(iii ) 1 11tan cot ,x
xV VW X
YZ [\ ]x > 0
3. (i) 1 1sin ( ) sin ,x x3 3^ 4 ^ x M [–1, 1]
(ii ) 1 1tan ( ) tan ,x x_ _` a ` x M R
(iii ) 1 1cosec ( ) cosec ,x x3 3^ 4 ^ |x| 1
(iv) 1 1cos ( ) cos ,x x3 3^ 4 b ^ x M [–1, 1]
(v) 1 1sec ( ) sec ,x xc cd e f d |x| 1
(vi) 1 1cot ( ) cot ,x xc cd e f d x M R
4. (i)1 1sin cos , [ 1,1]
2x x xg g hi j k l
(ii )1 1tan cot , R.
2x x xm m no p q
(iii ) –1 1cosec sec , | | 12
x x xr st u
5. (i) 1 1 1tan tan tan , 11
x yx y xy
xy3 3 3 vv 4 w
^
(ii ) 1 1 1tan tan tan , 11
x yx y xy
xyx x x yy z { y
|
3 7
6. (i) 1 12
22 tan tan ,| | 1
1
xx x
x} }~ �
�
(ii ) 1 12
22 tan sin ,| | 1
1
xx x
x� �� �
�
(iii )2
1 12
12 tan cos , 0
1
xx x
x� � ��
�
Question for Practice
Evaluate the following Integrals
Very Short Answer Type Questions (1 Mark)
Q1. Write the principal value of 1 3sin .
2� � �� �� �
Q2. Write the principal value of 1cosec ( 2).� �
Q3. Write the principal value of 1 1cot
3� � ��� �� �
.
Q4. Write the principal value of 1tan ( 3)� � .
Q5. Write the principal value of 1sec ( 2)� � .
Q6. Write the principal value of 1 1cos
2� � ¡ ¢£ ¤
.
Q7. Show that 1 1 2sin cos 1x x¥ ¥¦ § .
Q8. Show that2
1 1 1cos tan
xx
x¨ ¨ © ª«¬ ®
¯ °.
Q9. Show that 1 1
2tan sin
1
xx
x
� � ± ²� ³ ´µ ¶�
.
Q10. Show that 1 1
2sin tan
1
xx
x
� � ± ²� ³ ´µ ¶�
.
3 8
Q11. Show that 1 1 1cos 2sin
2
xx· · ¸¹ .
Q12. Write 1 3sin (3 4 )x xº » in the simplest form.
Q13. Write 1 3cos (4 3 )x x¼ ½ in the simplest form.
Q14. Evaluate –1cosec cosec4
¾ ¿À.
Q15. Evaluate 1 1cos cos
3 2.
Q16. Show that 1 1 1cos 2cos
2
xx .
Q17. Write cos–1 2(2 1)x in the simplest form.
Q18. Write 1 2cos (1 2 )x in the simplest form.
Q19. Write 1 1 costan , 0
1 cos
xx
x.
Q20. Show that 1 2 1sin 2 1 2sinx x x.
Q21. Evaluate : 1 1sin sin
3 2.
Q22. Evaluate : 1 1 2cos cos sin sin
3 3.
Q23. Find x, if 1tan / 4.x
Q24. Evaluate 1 3tan tan
4.
Q25. Evaluate 1 7cos cos
6.
Q26. Evaluate 1sin (sin 2 3).
Q27. Evaluate –1 3cosec cosec
4.
Q28. Evaluate 1 5cos cos .
3
3 9
Q29. Write 1
2 2tan ,| |
xx a
a x
Á Â in the simplest form.
Q30. Find x, if 1 1cot tan 72
x .
Q31. Find x, if 1 1sin cos .6
x x
Q32. Find x, if 1 14sin cos .x x
Q33. Find x, if 1 1 2tan 2cot .
3x x
Q34. Write 12
2sin 1 1
1
xx
x, in the simplest form.
Q35. Write 2 2sin 2 1x x in the simplest form.
Short Answer Questions Carrying 4 Marks each
Q36. Solve forx : 1 1 1 8tan ( 1) tan ( 1) tan .
31x x
Q37. Solve forx : 1 1tan 2 tan 3 .4
x x
Q38. If 1 1 1tan tan tana b c , prove thata + b + c = abc.
Q39. Solve forx : 1 11 1tan tan , 0.
1 2
xx x
x
Q40. Solve forx : 1 1 11 1tan tan tan 7.
1
x x
x x
Q41. Solve forx :2
1 12
2 1tan cot
1 2 2
x x
x x.
Q42. Solve forx : 1 1 11 2 1 23tan tan tan
1 2 1 36
x x
x x.
Q43. Solve forx : 1 1 18 3sin sin sin
17 5x .
Q44. Solve forx : 1 1 3tan (2 ) tan (3 ) .
4x x n
Q45. Solve forx : 1 1 1 1tan ( 1) tan tan ( 1) tan 3 0.x x x x
4 0
Q46. Prove that 1 1 112 4 63sin cos tan .
13 5 16Ã Ã ÃÄ Å Ä Å Ä Å
Æ Æ Ç ÈÉ Ê É Ê É ÊË Ì Ë Ì Ë Ì
Q47. Prove that 1 1 1 11 1 1 1tan tan tan tan .
3 5 7 8 4Í Í Í Í ÎÏ Ï Ï Ð
Q48. Prove that 1 1 11 1 42 tan tan tan .
5 8 7Ñ Ñ ÑÒ Ó
Q49. Prove that 1 1 13 8 77sin sin sin .
5 17 85Ô Ô ÔÕ Ö
Q50. Prove that 1 1 15 7 253sin sin cos .
13 25 325× × × Ø ÙÚ Û Ü ÝÞ ß
Q51. Prove that2
1 12
1 2tan tan .
2 1 2
x x
x xà àá âã äá â
å æç èç è é êãé ê
Q52. Prove that 1 1 14 3 27cos tan tan .
5 5 11ë ë ëì í ì í ì í
î ïñ ò ñ ò ñ òó ô ó ô ó ô
Q53. Prove that 1 1 163 1 3cos 2 tan sin .
65 5 5õ õ õö ÷ ö ÷ ö ÷
ø ùú û ú û ú ûü ý ü ý ü ý
Q54. Prove that 1 1 11 2 1 3tan tan cos .
4 9 2 5þ þ þ ÿ ð
� � � �� �
Q55. Prove that : 1 1 15 1 12 tan tan tan .
12 70 99 4� � � �� � �
� � � � �� � � � � �
Q56. Prove that : 1 costan .
1 sin 4 2
x x
x� �� �
� �� �� ��
Q57. Prove that 1 1 112 3 56cos sin sin .
13 5 65� � �� � � � � �
� ! " ! " ! "# $ # $ # $
Q58. Prove that 1 11 1cos 2 tan sin 4 tan .
7 3ë ëì í ì í
ïñ ò ñ òó ô ó ô
Q59. Prove that : 1 1 11 12 tan cosec 5 2 2 tan
5 8 4% % % &
' ' ( .
Q60. Prove that : 1 13 172sin tan .
5 31 4) ) *
+ ,
4 1
Answers1.
3
-2.
4
./3.
2
3
04.
3
./
5.3
4
16.
3
-12. 13sin x2 13. 13cos x3
14.4
4515. 1 17. 12cos x6 18. 12sin x3
19.2
x21. 1 22. 7 23.
4
-
24.4
./25.
5
6
126. 38 27. 48
28. 38 29.1sin
x
a9 : ;< => ? 30. 7 31.
3
2
32.1
233. 3 34. 12 tan x@ 35.
12sin xA
36.1
8,4
B 37.1
639.
1
340. x = 2
41. x = 1 42.3 4
,8 3
C43.
77
8544.
1,1
6
D
45.1 1
0, ,2 2E
4 2
Teaching-Learning Points
l 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 theEnglish alphabet and its elements are denoted by small letters.
l In a matrix horizontal lines of numbers are called rows of the matrix and vertical lines are calledcolumns of the matrix.
l A matrix havingm rows andn columns is called a matrix of orderm byn, written asm × n. In generala matrix of orderm × n is written as:
A =
1 2
1 2
11 12 13 1 1
21 22 23 2 2
31 32 33 3 3
3
3
n
n
n
i i i i in
m m m m mn
a a a a j a
a a a a j a
a a a a j a
a a a a j a
a a a a j a
F GH IH IH IH IH IH IH IH IJ K
(OR) A = [aij]
m × n where 1L i L m, 1L JL n, m, nM N
l Types of matrices : A matrix A = [aij]
m × n is said to bea:
(i) Row matix ifm = 1
(ii ) Column matrix ifn = 1
(iii ) Zero/Null matrix if each of its elements is zero.
(iv) Square matrix ifm = n
(v) Diagonal matrix, ifm = n andaij = 0 when iN J
(vi) Scalar matrix, ifm = n, aij = 0 wheni N j anda
ij = k wheni = J
(vii) Unit/identity matrix ifm = n, aij = 0 wheni ¹ j
l Two matrices A = [aij]
m × n and B = [b
ij] p × q are said to be equal (i.e. A = B) ifm = p, n = q anda
ij = b
ij
for all i, j.
l Scalar multiplication of a matrix.
Let A = [aij]
m × n andk be any scalar,
then KA = K[aij]
m × n = [Ka
ij]
m × n.
i.e. to multiply a matrix by a scalar multiply each element of the matrix by the scalar.
Matrices and Determinants3
4 3
l Addition of matrices:
Let A = [aij]
m × n and B = [b
ij]
m × n, then
A + B = [aij + b
ij]
m × n
l 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]
l A – B = A + (–B).
l Multiplication of matrices.
Let A = [aij]
m × n and B = [b
ij]
p × q be two matrices, then the product of A and B (i.e. AB) is defined ifn
= p and it is a matrix of orderm × q.
l Let A = [aij]
m × n and B = [b
ij]
n × p, then
A B = C (say) = [cij]
m × b, wherec
ij is obtained by takingith row of A andjth column of B, multiplying
their corresponding elements and taking sum of these produce.
(i) In general, ABO 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)
l Transpose of a matrix A is obtained by interchanging its rows and columns. It is denoted by A1.
l Properties of transpose:
(i) (KA) 1 = KA1 (ii ) (A + B)1 = A1 + B1
(iii ) AB)1 = B1A1 (iv) (A1)1 = A.
l A square matrix is called, a symmetric matrix if A1 = A and a skew symmetric matrix if A1 = –A.
l For any square matrix A, A + A1 is always symmetric matrix and A – A1 is always skew symmetricmatrix.
l Every square matrix can be expressed asa sum of symmetric and skew symmetric matrix.
i.e. A =1 1A A A – A
2 2
P Q P QRRS T S TU V U V
l Elementary operations (transformations) of a matrix.
(i) Interchange of any two rows (or two columns)
i.e. RiW R
j (or) C
iW C
j
(ii ) Multiplication of the elements of any row (or any column) by a non zero numberi.e.
i.e. RiX KR
i (or) C
iX KC
i
(iii ) Addition of the elements of any row (or any column) to the corresponding elements of any otherrow (or column) multiplied by any non zero numberi.e.
i.e. RiX R
i + KR
j(or) C
i = C
i + KC
i
4 4
l A square matrix A is said to be invertible if there exists another square matrix B of the same order suchthat AB = I = BA, then B is called the inverse of A and is denoted by A–1.
l 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
l 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.
l 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|.
l Determinant ca be expanded by using any row or column.
l Properties of determinants:
(i) |A1| = |A|.
(ii ) RiZ R
j(or) C
iZ C
jY |A| = –|A|
(iii ) If Ri is identical to R
j or C
i is identical to C
jY |A| = 0.
(iv)1 1 1
2 2 2
3 3 3
ka kb kc
a b c
a b c
= k1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
= k|A|
In general A n nk [ = Ann nk \
(v) Ri] R
i + kR
j (or) C
i = C
i + kC
j
Y |A| remains unchanged
(vi) If some or all elements of a row (or column) of a determinant are expressed as the sum two ormore terms, then the determinant can be expressed as a sum of two or more determinants.
l |AB| = |A| |B|
l |A–1| = 1
A
l A square matrix is said to be singular if |A| = 0 and non singular if |A|^ 0
l Using determinants Area if a_ABC with vertices A(x1y
1), B(x
2y
2), C(x
3y
3) is given by
1 1
2 2
3 3
11
22
3
x y
x y
x y
l Minor of an elementaij of a determinants is the determinant obtained by deleting theith row andyth
column in which elementsaij lies. It is denoted by M
ij.
l Cofactor of an elementaij of a determinant is denoted as A
ij and is defined as A
ij = (–1)i + y M
ij
l Adjoint of a square matrix A = [aij]
n × n is defined as the transpose of the matrix [A
ij]
n × n where A
ij is the
cofactor of the elementaij and it is denoted as adj.A
4 5
l A(adjA) = (adjA) A = |A| I
l |adjA| = |A|n–1, where A is a square matrix of ordern.
l A square matrix A is invertible if and only |A|` 0 if A is a non singular matrix
A–1 =A
.|A|
adj
l If A and B are non singular matrices of the same order then AB and BA are also non singular matricesof the same order.
l A system of linear equations in three variables can be expressed in a matrix equation.For example, a
1x + b
1y + c
1z = d
1
a2x + b
2y + c
2z = d
2
a3x + b
3y + c
3z = d
3
a
1 1 1
2 2 2
3 3 3
a b c x
a b c y
a b c z
b c b cd e d ed e d ed e d ef gf g
=1
2
3
d
d
d
h ij kj kj kl m
a AX = B
where A =
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
n op qp qp qr s
, X =
x
y
z
t uv wv wv wx y
, B =
1
2
3
d
d
d
z {| }| }| }~ �
l By solving the matrix equation AX = B or X = A–1B we get solution of given system of equations.
l If |A| ` 0, then the system of equations has unique solution and hence it is consistent.
l If |A| = 0 and (adjA) B 0 (zero matrix), then the system of equations has no solution and hence it isinconsistent.
l If |A| = 0 and (adjA) B = 0 (zero matrix), then the system of equations may be either consistent orinconsistent 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 J 3 Jia i� �
2. If2
5
a b
a
�� �� �� �
=6 2
5 8
� �� �� �
, find the values ofa andb.
3. If2 –1
3 1a b� � � �
�� � � �� � � �
=10
5
� �� �� �
, find a andb.
4. If a matrix has 12 elements, then how many possible orders it can have.
4 6
5. If A =cos – sin
sin cos
x x
x x
� �� � ¡
, 0 <x <2
¢ and A + AA1 = I, then find the value ofx.
6. If3 4
1 –1
£ ¤¥ ¦§ ¨
= P +© where P is symmetric and© is skew symmetric matrix, find©.
7. Evaluatecos – sin
sin cos
ª ª
ª ª
8. Find the value ofx if3 4 1
1 2 3 2
x«
9. What is the value of the determinant
1 3 5
2 6 10
11 13 15
10. Find cofactor ofa23
in
1 0 4
3 5 –1
0 1 2
11. Find the minor ofa12
in the determinant
1 2 3
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|.
15. If a singular matrix A is given by1 2
4x
¬ ® ¯° ±
, then find the value ofx.
Short answer questions carrying 4 marks each:16. Express the following matrix as a sum of a symmetric and a skew symmetric matrix.
1 3 5
–6 8 3
–4 6 5
² ³´ µ´ µ´ µ¶ ·
17. Given that A =3 –2
4 –2
¸ ¹º »¼ ½
and I =1 0
0 1
¾ ¿À ÁÂ Ã
, find the real numberk such that AA2 – kA + 2I = 0.
18. Given that A =
2 1
–1 3
4 0
Ä ÅÆ ÇÆ ÇÆ ÇÈ É
and B =–3 1 2
0 4 5
Ê ËÌ ÍÎ Ï
, verify that (AB)1 = B1A1.
19. If A =2 3
3 4
и ¹º »¼ ½
show that AA2 – 6A + 17 I = 0. Hence find A–1.
4 7
20. Show thatx = 2 is one of the roots of the equation
–6 –1
2 –3 – 3
–3 2 2
x
x x
x xÑ
= 0. Find other roots also.
21. Using properties of determinants show that:
2
2
2
1
1
1
a a
b b
c c = (a – b) (b – c) (c – a)
22.
b c c a a b
q r r p p q
y z z x x y
Ò Ò Ò
Ò Ò Ò
Ò Ò Ò
= 2
a b c
p q r
x y z
23.
– – 2 2
2 – – 2
2 2 – –
x y z x x
y y z x y
z z z x y
= (x + y + z)3
24.
–
–
–
a b c c b
a c b c a
a b a b c
Ñ
Ñ
Ñ
= (a + b + c) (a2 + b2 + c2)
25.
2
2
2
1
1
1
x x
x x
x x
= (1 – x3)2.
Long Answer type questions carrying 6 marks each:
26. Find the inverse of the matrix using elementary transformations:
2 –3 3
2 2 3
3 –2 2
Ó ÔÕ ÖÕ ÖÕ Ö× Ø
27. Solve the following system of linear equations using inverse of matrix.
2x + 3y + 4z = 8, 3x + y – z = 2, 4x – y – 5z = –9.
28. If A =
2 –3 5
3 2 –4
1 1 –2
Ù ÚÛ ÜÛ ÜÛ ÜÝ Þ
, find AA–1. Hence solve the following system of linear equations:
2x – 3y + 5z = 11
3x + 2y – 4z = –5
x + y – 2z = –3
4 8
29. If A =
1 1 2
2 –1 3
5 –1 –1
ß àá âá âá âã ä
, find AA–1 and hence solve the following system of linear equations:
x + 2y + 5z = 10 [Hint : (A1)–1 = (A–1)1
x – y – z = –2
2x + 3y – z = –11
30. If A =
1 –1 2
0 2 –3
3 –2 4
å æç èç èç èé ê
and B =
–2 0 1
9 2 –3
6 1 –2
ë ìí îí îí îï ñ
. Find AB and hence solve the following system of equation.
x – y + 2z = 1 [Hint : AB = I ò A–1 = B]2y – 3z = 1
3x – 2y + 4z = 2
Answers1.
2 1
5 4
ó ôõ ö÷ ø
2. a = 8,b = –2 3. a = 3,b = –4 4. 6
5.3
ù6.
0 31
–3 02
ú ûü ýþ ÿ
7. 1 8. 1
9. 0 10. –1 11. 10 12. 40
13. 4 14.–1
1115. 2
16.
1 –3 2 1 2 0 9 2 9 2
–3 2 8 9 2 –9 2 0 –3 2
1 2 9 2 5 –9 2 3 2 0
ð � ð �� � � ��� � � �� � � �� � � �
18. 1 19. A–1 =4 31
3 217
� �� � �
20. 1, –3.
26.
– 2 5 0 3 5
–1 5 1 5 0
2 5 1 5 – 2 5
�� �� �� �� �
27. x = 1,y = –2, z = 3. 28.
0 1 –2
–2 9 –23
–1 5 –13
� �� �� �� �� �
x = 1,y = –2, z = 3.
29. A–1 =
4 –1 51
17 –11 127
3 6 –3
ß àá âá âá âã ä
x = –1, y = –2, z = –3. 30. x = 0,y = 5,z = 3.
4 9
Teaching Learning points
l Continuity and discontinuity of function: A functiony = f(x) is said to be continuous in an intervalIf for every value ofx in that intervaly exist. If we plot the points, the graph is drawn without liftingthe pencil.
l Continuity and discontinuity of a function at a point: A function f(x) is said to be continuous at apoint ‘a’ of its domain If
f(x), f(x), f(a) exist
limx a� 1
limx a�
and f(x) = f(x) = f(a)
limx a�
limx a�
A function f(x) said to be descontinuous atx = a If it is not continous atx = a
l Let f(x) andg(x) are two real Continous functions at pointx = a, then
(i) f(x) + g(x) is also continuous atx = a
(ii ) f(x) · g(x) is also continuous atx = a
(iii )( )
( )
f x
g xis also continuous atx = a, providedg(a) � 0
(iv) λ · f(x) is also continuous atx = a, whereλ is any scalar
(v) |f(x)| is also continuous atx = a
(vi) fog or gof is also continuous atx = a
l 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 continuousΑ x � R
(iii) Any constant functionf(x) = c and identity functionf(x) = x is continuousA x � R
(iv) Every Rational function( )
( )
f x
g x is continuous for all values ofx in the Domain, where the points
at which g(x) is zero are not included in the Domain.
(v) The Trigonometric functions sinx, cos x, are continuous function every where.(vi) f(x) = [x] is not continuous at any integral value of ‘x’.
Continuity and Differentiability4
5 0
l The process of finding derivative of a function is called differentiation.
l Let f(x) be any real valued function defined on an open interval (a, b), thenf(x) is said to be differentiable
at x =� e (a, b) if =0 0
( ) ( ) ( ) ( )lim limh h
f c h f c f c h f c
h h 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) andu = g(x), thendy dy du
dx du dx!
This process also known as Derivative of function of a function.
l Derivative of implicit function: When the variablex andy are related in such a way thaty cannot beexpressed 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.
l While differentiating inverse trigonometric functions, first express it in simplest form by using suitablesubstitution and then differentiating the simplest form.
l Some important substitutions:
Expression Substitutions
(i) 2 2a x x = a sin θ or x = a cos "
(ii ) 2 2a x x = a tan " or x = a cot "
(iii ) 2 2a a x = a sec" or x = a cosec"
(iv) a x# x = a cos "
(v) ora x a x
a x a xx = a cos2 "
l Logarithmic differentiation:
(i) This process is used when function is given in a complicated form(ii ) When y = [f(x)]g(x), then
dy
dx = ( )[ ( )] [ ( ) log{ ( )}]g x d
f x g x f xdx$ $
l Derivative of function in Parametric forms: If y = f(x) be a function in whichx andy are thevariables, when the variablesx andy are the functions of third variable ‘t’ i.e., x = u(t) andy = v(t) thenthis form is called parametric form and ‘t’ is called the parameter.
In order to finddy
dx, we use the following:
dy
dx =
dy
dtdx
dt
5 1
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 onec% (a, b)
Such that f &(c) = 0
l Geometrical interpretation:
*in the interval (a, b) there must exist at least one point where the tangent is parallel tox-axis.
l 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 onec% (a, b)
Such that f &(c) =( ) ( )f b f a
b a
'
'
l 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 2
Question for Practice
Very Short Answer Type Questions (1 Mark)
Q1. If y = (1 +x1/3) (1 – x1/3) (1 +x2/3) find dy/dx
Q2. If y = tan–1 (cotx) find dy/dx
Q3. f(x) = 53x find f ((2)
Q4. f(x) = tan–1 x + tan–1 1/x find f ((x) x > 0
Q5. f(x) = log10
x find f ((x)
Q6. If1x
yx x
)*
then finddy/dx
Q7. If y = ex then what is the value of2
2
12
d y dyy
a dx dx
+ ,- -. /0 1
Q8. x2/3 + y2/3 = 2 finddy
dx at (1, 1)
Q9. If f(x) = [x] write points wheref(x) is not differentiableQ10. For what value ofλ thef(x) = sin (λx) is continous every where
Short Answer Type Questions (4 Marks)
Q1. f(x) =
2
1 cos 40
0
016 4
xx
xa x
xx
x
234 5
44 644
748 349
For what value of ‘a’ f(x) is continuous atx = 0
Q2. f(x) =2
3
sin( 1) sin0
0
0
a x xx
xc x
x bx xx
b x
:; ;< =
<< ><< ; ?
@<<A
Determine the value ofa, b andc for which the functionf(x) may be continuous atx = 0
5 3
Q3. Examine the following function for continuity atx = 0 wheren B 1
f(x) =1
sin 0
0 0
nx xx
x
CDE
FE GH
Q4. Determine the value of constants ‘a’ and ‘b’ such that the function defined as is continuous at x = 4
f(x) =
4If 4
4
4
44
4
xa x
x
a b x
xb x
x
IJK LM I
MM K NM
IM K OM IP
Q5. Show that the functionf(x) defind byf(x) is continuous atx = 0
f(x) =
sincos 0
2 0
4(1 1 )0
xx x
xx
x
x
QR ST
TUT
TV VT W
TX
Q6. If x = a(Y – sin Y) y = a(1 – cos Y)
Find2
2 at2
d y
dx
Z[ \
Q7. Diff. w.r.t. ‘x’
12
5tan
1 6
x
x] ^ _` abc d
Q8. If xmyn = (x + y)m+n
Prove thatdy y
dx xe
Q9. 2 21 1 ( )x y a x yf g f h f , then show that
2
2
1
1
dy y
dx x
ij
i
Q10. If 2log[ 1 ]y x xk l l prove that
22
2( 1) 0d y dy
x xdx dx
m m \
Q11. y = 2( 1)mx xm n
Prove that (x2 – 1)y2 + xy
1 = m2y
5 4
Q12. If1
tan logx ya
o pq rs t
show that2
22(1 ) (2 ) 0
d y dyx x a
dx dx
Q13. If x = log t andy =1
t Prove that
2
2 0d y dy
dx dx
Q14. If y = (cos–1 x)2 Prove that (1 – x2)y2 – xy
1 = 2
Q15. If x = a sin3u y = b cos3u find2
2 at4
d yx
dx
v
Q16. If y =2
1 5 12 1sin
13
x xw xy zy z{ |
find dy/dx
Q17. Verify the applicablity of rolle’s theorem for the following
1. f(x) = sin 2x [0, }/2]
2. f(x) = x2/3 [–1, 1]
3. f(x) = (x – 1)(x – 2)2 [1, 2]
Q18. Verity Lagrange’s mean value theorem for the following functions in the given interval and find c ofthis theorem
f(n) = x3 – 5x2 – 3x [1, 3]
f(n) = (x – 1) (x – 2) (x – 3) [0, 4]
f(n) =1
xx
[1, 3]
Q19. Find 1 2If sin ( 1 1 )dy
y x x x xdx
Q20. If y = 1 1sin 2 tan
1
x
x
~ �� �� �
Prove that 21
dy x
dx x
Q21. Differentiate2 23 ( 3)x xx x w.r.t. ‘x’ for x > 3
Q22. Differentiatexxcosx + (xcosx)x w.r.t. x
Q23. Differentiate2 2
1 122 2
1 1 2tan w.r.t. tan
11 1
x x x
xx x
� � � �� � � �
� �� �� �
Q24. Differentiate sin2 x w.r.t. esin x
Q25. f(n) =3 2
2 5 2
x x
x x
�� ��
atx = 2
Show thatf is not differentiable atx = 2
5 5
The electric potenti
Q26. If2 2 2 4
2 2 2 2 31 Prove thatx y d y b
a b dx a y
Q27. Differentiate the following w.r.t. ‘x’
(i)2
2
( 3)( 1)
3 4 5
x x
x x
Q28. If (x – a)2 + (x – b)2 = c2 for somec > 0. Prove that
3 22
2
2
1dy
dx
d y
dx
� �� �� �� �� �� �� � is a constant independent ofa andb
Q29. Finddy
dx If yx + xy + xx = 1010
Q30. Ifxyy x show that
log ( log log 1)
(1 log ) log
dy y y x x y
dx x y x x
Q31. Discuss the differentiability and continuity of
f(x) = |x – 1| + |x – 2| at x = 1 andx = 2
Q32. f(x) =1 2
2 3 ,2
x x
x x
���
Show thatf(x) is continuous atx = 2 but not differentiable atx = 2
Q33. If the f(x) =
3 1
11 If 1
5 2 1
ax b x
x
ax b x
��
� � ¡ ¢£
is continuous atx = 1 find value ofa andb
Q34. If sin y = x sin (a + y). Prove that2sin ( )
sin
dy a y
dx a
¤¥
Q35. If y = sin logx
Prove that2
22 0
d y dyx x y
dx dx¦ ¦ §
Q36. If y = 1 1 sin 1 sintan
1 sin 1 sin
x x
x x¨ © ª« « ¬ ®
« ¬ ¬¯ ° Prove that
1
2
dy
dx
±²
5 6
Answers1.
1 34
3x 2. –1 3. 3log 5 ³ 56
4. 0 5.1 1
log10 x´ 6. –2x –1
7. 0 8. –1 9. for x µ I (I for integets)10. ¶ µ R
Hints 2 Solutions of 4 Marks Questions
1. Now f (0) = a
LHL f(x) =22
2 2
1 cos 4 2sin 2 sin 28
2
x x x
x x x· ¸¹ º» ¼
0Lt
x½ –0Lt
x½ –0Lt
x½ –0Lt
x½
8(1)2 = 8 [x¾ 0– ¿ 2x¾ 0–]
RHL f(x) =( 16 4)
16 4 ( 16 4)( 16 4)
x x x
x x x
+0Lt
x½ +0Lt
x½ +0Lt
x½
( 16 4)
16 16
x x
x = 16 4 8x
+0Lt
xÀ +0Lt
xÀ
f(x) = f(x) = 8
+0Lt
x½ +0Lt
x½
Á f(x) is continuous atx = 0 only. Ifa = 8
2. Heref(0) = c
LHL f(n) =sin( 1) sin sin( 1) sina x x a x x
x x x
–0Lt
xÀ –0Lt
xÀ –0Lt
xÀ
=sin( 1) ( 1) sin
( 1)
a x a x
x a x
–0Lt
xÀ –0Lt
xÀ
a + 1 + 1 = a + 2 [x¾ 0– ¿ (a + 1)x¾ 0–]
5 7
RHL f(n) =[ 1 1 ( 1 1) 1 1 1 1
1 6 1 1 1)
x bx x bx bx bx
bx x bx x x bx bxÂ
Â
+0Lt
xà +0Lt
xà +0Lt
xà +0Lt
xÃ
=1 1
21 1bx
+0Lt
xÃ
Now f is continuous atx = 0. If+0
LtxÃ
f(x) = f(0) = f(x)
i.e., a + 2 = c =1
2⇒ a =
3
2c =
1
2bÄ R – {0}
(RHL is independent ofb)4. a = 1 b = –1
x = a(Å – sin Å) y = a(1 – cos Å)Differentiating both w.r.t. ‘Å’ we get
dy
dÆ = a( 1 – cos Å)
dy
dÆ = a sinÅ
dy
dx =
2
/ sin 2sin 2cos 2cot 2
/ (1 cos ) 2sin 2
dy d a
dx d a
Ç Ç Ç ÇÇ
Ç Ç ÇDifferentiating w.r.t. ‘x’
2
2
d y
dx = 2 21 1 1
cosec 2 cosec 22 2 (1 cos )
d
d a
ÈÈ É È
È È
=2
42
1 cosec 2 1cosec 2
2 2sin 2 4a a
ÊÊ
Ê
2
2
0 2
d y
dx Ë
Ì ÍÎ ÏÐ Ñ
= 41 1 1cosec 4 4
4 4a a aÒ Ó
7. 1 3 2tan
1 3 2
x x
x xÔ ÕÖ ×ØÙ Ú
tan–1 3x + tan–1 2x
Differentiate w.r.t. (x)
2 2
3 2
1 9 1 4x x8. xmyn = (x + y)m+n
taking logarithm both side
m log x + n log y = (m + n) log (x + y)
m n dy
x y dx =
( )1
m n dy
x y dxÛ ÜÝ Þß à
dy n m n
dx y x y
á âã äå æ
=m n m
x y x
5 8
( )
dy nx ny my ny
dx y x y
ç èé êë ì
=( )
mx nx mx my
x x y
dy
dx =
y
x
9. 2 21 1 ( )x y a x y
x = cosA y = cosB
sinA + sinB = a(cosA – cos B)
2 sin cos2 2
A B A Bí î í îï ñ ï ñò ó ò ó = 2 sin sin
2 2
A B A Ba
ô õ ô õö ÷ ö ÷ø ù ø ù
cot2
A Bú ûü ýþ ÿ
= a
ð A – B = 2 cot–1a
cos–1x – cos–1y = 2 cot–1a
Differentiate w.r.t. (x)
2 2
1 1
1 1
dy
dxx y = 0
dy
dx =
2
2
1
1
y
x
12. Hint: x =1
tan logya
� �� �� �
a tan–1x = log y ð1tana xe
dy
dx =
1tan
2 21 1
a xae ay
x x
2(1 )dy
xdx
= ay
Again Differentiate w.r.t. (x)2
22(1 ) 2
d y dyx x
dx dx� =
ady
dx2
22(1 ) (2 )
d y dyx x a
dx dx = 0
15.2
4 2
3
b
a.
16. y =2
1 5 12 1sin
13
x x� � � �
x = sin t t = sin–1 x
5 9
y = sin–15sin 12
cos13 13
tt
�� �� �
5 = r cosd 12 =r sind� r = 13
y = 1 cos sin sin cossin
13
r d t r d t� �� �� �
= sin–1 sin( )13
rt d t d�
y = 1 112sin tan
5x
dy
dx =
2
1
1 x17. (ii) f(x) = x2/3 [–1, 1]
clearlyf(x) has definite and unique value for eachx� [–1, 1]
f(x) is continuous in [–1, 1]
f(x) = 1 32
3x which does not exist forx = 0 (–1, 1)
Hence Roll’s theorem is not applicable
18. (ii ) f(x) =1
[1,3]xx
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]
f �(x) = 2
11
x� which exist in (1, 3)
Hence all the condition of LMV are verified. Hence there exist atleast one value ‘c’ c� (1, 3) such that
f �(c) =( ) ( )f b f a
b a
�
�
2
11
c� =
102
33 2
�
�3 but 3 (1,3)c ! " #
only possible value c = 3 (1,3)$
19. y = 1 2sin ( 1 1 )x x x x% � � �
y = 1 2 2sin ( 1 ( ) 1 )x x x x& ' ' ' using sin–1 x – sin–1 y = 1 2 2sin 1 1x y y x( ) *+ , +- .
y = 1 1sin sinx x% %�
Differentiate w.r.t. (x)
dy
dx =
2
1 1 1
1 21 x xx/ 0
//20. Hint: Putx = cos1
6 0
21.2 2
2 23 3
2 log ( 3) 2 log | 3 |3
x xx xx x x x x x
x x
2 3 2 34 5 4 56 7 6 7
22. cos [(1 log )cos sin log ] ( cos ) [log( cos ) tan 1]x x xx x x x x x x x x x x x24. Hint : y = sin2x z = ecosx
Diff w.r.t. ‘x’
dy
dx = 2sin cosx
dz
dx =
cos (sin )xe x8
dy
dz=
dy dx
dz dx = cos
2sin cos
sin x
x x
x e
dz
dx9 0
= cos
2cosx
x
e
27.2
2 2 2
1 ( 3)( 1) 1 2 6 4
2 3 4 5 3 1 3 4 5
x x x x
x x x x x x: ;< => ?
28. Given 2 2( ) ( )x a y b = c2 ...(i)Diff. twice w.r.t. (x)
12( ) 2( )x a y b y = 0@ 1( ) ( ) 0x a y b y ...(ii )
2 1 11 ( ) ( 0)y b y y y = 0@21
2
(1 )( )
yy b
y...(iii )
Put that value of (y – b) from (iii ) to (ii )
x – a =2
1 1
2
(1 )y y
y ...(iv)
Now put the value of (iii ) and (iv) in (i) we get2 2 2 2 21 1 1
2 22 2
(1 ) (1 )y y y
y y = c2
2 3 21
2
(1 )y
y = c
29.1
1
log (1 log )
log
x y x
x y
y y yx x x
x y x x30. Hint : f(x) = | 1| | 2 |x x
f(x) =
( 1) ( 2) 1
( 1) ( 2) 1 2
1 2 2
x x x
x x x
x x x
AB
C C C D EFB C G C HI
f(x) =
2 3 1
1 1 2
2 3 2
x x
x
x x
J K LMN
O LPN J QR
Ans. f(x) is continuous atx = 1 & x = 2 lut not
Differentiable atx = 1 andx = 2
33. a = 3 b = 2
6 1
Teaching-Learning Pointsl 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 andnormal, increasing & decreasing function, maxima and minima of function and approximation.
l Derivative as a Rate Measure : If a quantity ‘y’ varies with another quantity x, satisfying some rule
y = f(x) thendy
dxx = x
0 represents the rate of change ofy w.r.t ‘x’ at x = x
0
Exp. Area of circle depends upon radius
A = Sr2
dy
dr = 2Sr will represent rate of change of area w.r.t radius
If two variablesx andy varying w.r.t another vareable. Ify = f(t) andx = g(t) by chain rule
dy
dx =
dy dt
dx dt
dx
dtT 0
Hence rate of change ofy w.r.t ‘x’ can be calculated by using rate of change of both y andx withrespect tot.
l 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) ifx
1, x
2U(a, b) such thatx
1 < x
2V f(x
1) < f(x
2)
ORf W(x) > 0 x U(a, b)
Application of Derivative5
6 2
if f(x) be real valued continuous function defined on (a, b) is said to be decreasing in (a, b)if x
1, x
2X (a, b) such thatx
1 < x
2Y f(x
1) > f(x
2)
OR
( ) 0 ( , )f x x a bZ [\
l Tangents and Normals :
(i) If we use the Geometrical meaning of the derivative, thenf ](x) at (x1, y
1) represents the slope of
the tangent at the point wheref(x) is continuous and differentiable.
(ii ) Slope of the Normal to the curvey = f(x) at point (x1, y
1) is given by
1 1( , )
1
x y
dy
dx^ _` ab c
(iii ) Since the tangent to the curvey = f(x) at point (x1, y
1) is a straight line. Hence tangent will also be
represented by an equation. So equation of tangent is given by1
1 1
( )
1( , )
x x
x y
dyy y
dxd
e fg and normal is
given by
1 1
1 1
( , )
1( )
x y
y y x xdy
dx
hi jk lm n
(iv) Tangent and Normal to any curvey = f(x) at a given point are the lines passing the point andperpendicular to each other.
l 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 alsody
dx = ( )f x\
Y qy = .dy
xdx
r stu vw x
if dxy qx thendyy qy
Hence we get the approximate value ofqy
l Maximum and minium :
Before attempting the exercise the students must know the following facts :
(i) | | 0 Rax b xz { , wherea, b any real number
(ii ) 1 sin( ) 1,ax b x R| | }
(iii ) 0 | sin( ) | 1,ax b x R~ ~ [
(iv) | sin( ) , and 0a a x a x R a� � � � �
(v) 2( ) 0ax b x R� �
First derivative test – let f(x) be a differentiable function defined on Interval I andx = CX I, then
(a) x = c is said to be a point of local maxima, if
(i) f ](x) = 0, and
6 3
(ii ) f �(x) changes sign from (+) ive to (–) ive as increases through c i.e., f �(x) < 0 at every pointsufficiently close to and the left ofC, andf �(x) < 0 at every point sufficiently close to and to theright 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 throughc i.e., f �(x) < 0 at every point
sufficiently close to and the left ofc andf �(x) > 0 at every point sufficiently close to and to theright toc.
(c) f �(x) = 0 andf �(x) does not change signs asx increases throughc i.e., f �(x) has the same sign in thecomplete neighbourbood ofc, then pointc is neither a point of local maxima and nor a point of localminima. Such point is said to be a point of inflexion.
l Second derivative test :
Theorem letf he a real valued function having second derivative atc such that
(i) f �(c) = 0 andf ��(c) > 0 thenf has a local minimum value atc(ii ) f �(c) = 0 andf ��(c) < 0 thenf has a local maximum value atc
(iii ) f �(c) = 0 andf ��(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 radiusr whenr = 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 curvey = 3x2 –1 at which slope of tangent is 3.
Q4. At what point on curve23
2
xy � tangents makes 60° withx axis.
Q5. What is the slope of the normal to the curvex = a cos3 � y = a sin3� at� = �/4
Q6. What is the point on the curve 2 2 3y x x� at which tangent is || to x-axis.
Q7. What is maximum value of |sin 2x + 3|Q8. What is the minimum value of |4 sin 2x + 3|Q9. What is the max value of sinx + cosxQ10. 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 baseb are decreasing at the rate of 3cm/sec.How fast is the area decreasing when two equal sides are equal to the base.
6 4
Q3. Water is leaking from a conical funnel at the rate of 5cm3/sec. If the radius of the base of the funnelis 10 cm and its height is 20 cm. Find the rate at which the water level is dropping when it is 5 cmaway 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 thevertex in the bottom. When the slant height of the water is 3 cm. find the rate of decrease in slantheight 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 itscoordinate axis.
Q6. Show that the line 1x y
a b� � touches the curve /x ay be�� at a pt where is crossesy-axis.
Q7. Find the points on the curve2 2
19 16
x y� � at which tangents are
(i) parallel tox-axis (ii ) parallel toy-axisQ8. 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 ) .0037Q10. Find the intervals in which the following function are strictly increasing or strictly decreasing:
(i) f(x) = 2 320 9 6x x x� � �
(ii ) 3 2( ) 12 36 17f x x x x� � � �
(iii )4 3 23 4 36
( ) 3 1110 5 5
f x x x x x� � � � �
(iv) 4 2( ) 2f x x x� �
(v) 3 2( ) ( 2) ( 1)f x x x� � �
(vi) ( ) ( 2) xf x x e�� �
(vii) 2 2( ) ( 2)f x x x� �
(viii ) ( ) log(1 ) 11
xf x x x
x� � � � �
�
(ix) ( ) sin cos (0,2 )f x x x ¡ ¢
(x) ( ) sin 3 (0, / 2)f x x£ ¤
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 34 3 5y x x¥ ¦ § which are perpendicular to the line
9 3 0.y x¨ ¨ ©
Q13. Show that the curves 2x y xy k� � cut orthogonally if 8k2 = 1
Q14. If the tangents to the curve 3y x ax bª « « at P(1, –6) 11 to the line y – x = 5. Find the values ofa
andb.
Q15. A particle moves along the curve 36 2y x� � . Find the points on the curve at which y coordinate is
changing 8 times as fast asx coordinate.
6 5
Q16. Find the absolute maximum and munimum value of 3 2( ) 2 9 12 5 [0,3]f x x x x¬ ®
Q17. Find the local maximum and local minimum of( ) sin 22 2
f x x x x¯° °
± ¯ ² ²
Q18. Find the local maximum and local minimum 4 4( ) sin cos 0f x x x x³ ´ µ µ ¶
Q19. If 2log | |y a x bx x³ ´ ´ has extreme value atx = –1 and x = 2 finda andb.
Q20. If( 1)( 4)
ax by
x x
·¸
· · has a turning point (2, –1) find the value of a andb.
Long Answer Type Questions (6 Mark)
Q1. A rectangle is inscribed in a semicircle of radiusr 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 parabolay2 = 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 is1tan 2¹ .
Q7. Prove that of all rectangles with given area, the square has the smallest perimeter.
Q8. Prove that all the rectangles with given perimeter, the square has largest area.
Q9. Prove that the perimeter of a right angled triangle of given hypotenuse is maximum when the trianglesis isosceles.
Q10. Show that of all rectangles inscribed in a given fixed circle. The square has the maximum area.
Q11. An open box with a square box is to be made out of a given quantity of sheet of areac2. Show that
max volume of the box is3
.6 3
c
Q12. If the lengths of three sides of a trapezium other than base are equal to 10 cm each, then find the areaof 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 thatthe minimum length of the hypotenuse is 3 2 3 3 2( )aa bº .
Q14. Show that the volume of the largest cone that can be inscribed in a sphere of radius R is8
27 of the
volume of sphere.
6 6
Q15. Show that maximum volume of cylinder which can be inscribed in a cone of height ‘h’ and semivertical
‘d’ is3 24tan d.
27h»
Q16. Find the area of greatest isosceles triangle that can be inscribed in a given ellipse having its vertexcoincident with one extremity of major axis.
Q17. Show that the volume of the greatest cylinder which can be inscribed in a cone of heighth and
semivertical angle 45° is34.
27h¼
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 itsradius.
Q20. Find the equation of the line through the point (3, 4) which cuts from the first quadrant a triangle ofminimum area.
AnswersAnswer (1 Mark)
1. 6½ cm2/sec 2. 1.4½ cm/se 3.1
1 42
x y¾ ¾ ¿ 4.3
1,2
ÀÁÂ
5. 1 6. (1, 2) 7. 4 8. 1
9. 2
10.d
dt
à = 2 sin
d
dtÄ
d
dt
à = 2cos
d
dt
ÅÅ
cosÆ =1
2Æ = ½/3
Answer and Hints of 4 Mark Question
1.5
km/hr2
2. Let ABC is isocelesÇ
AB = AC
At any time AS = AC = a
6 7
AD2 = AB2 – BD2 =2
2
2
ba
ÈÉ ÊË
A =1
BC AD2
Ì
A =2 2
2 21 44
2 2 4
b a b ba b
ÍÎ Í
dA
dt =
2 2 1 2
2 2
1(4 ) 4 2
4 2 4
b da ab daa b a
dt dta b
ÏÐ Ñ ÒÑ
whena = b2
2 2( 3) 3 cm /s
4
dA b bb
dt b b
ÓÔ Õ Ô Õ
Õ
–ve sign indicate the area of isosceles Ö is decreasig at rate of 23 cm /sec
3.4
cm/sec45×
4. V he the volume of cone &l be the slant height of water at any time
V =2 3( sin 60 ) cos60
3 8l l l
Ø ØÙ Ù Ú
Finddl
dt when 34 cm /s 3
dVl
dtÛ Û
5. 9y2 = x3
Diff w.r.t. ‘x’ we get
9 2dy
ydx
Ì = 3x2 Ü
2
6
dy x
dx yÝ scape of normal to curve atx, y, = 1
21
6y
x
Normal to the curve makes equal intercepts on coordinate axes its slope will be = ± 1
121
6y
x = ± 1
(x1, y
1) lies on curve (1) 2 3
1 19y xÞ22
196
xß àá âã ä
= 31x Ü 4 3
1 1 14 0,4x x xå æ å
whenx1 = 0 y
1 = 0 whenx
1 = 4 1
8
3y ç as normal makes equal intercept it cannot passing through
origin
Ans. 84, 4, 8 3
3è éê ëì í
7. (i) for no real pt tangent is || tox-axis(ii ) at (± 3, 0) tangent is || toy-axis
8. 2 cm/sec.
6 8
9. (i) Hint : Let x = .04 îx = – .003 Ans. .1925 approximate(ii ) Hint : Let x = .0036 îx = .0001 Ans. .0608 approximate
10. (i) f is strictly increasing in (1, 3)f is strictly decreasing in (– ï, 1) ñ (3, ï)
(ii ) f is strictly decreasing in (2, 6)f is strictly increasing in (– ï 2) ñ (6, ï)
(iii ) f is strictly decreasing in (– ï 2) ñ (1, 3)f is strictly increasing in (–2, 1) ñ (3, ï)
(iv) f is strictly increasing in (–1, 0) ñ (1, ï)f is strictly decreasing in (– ï –1) ñ (0, 1)
(v) f is strictly decreasing in7
1,5
ò óô õö ÷
f is strictly increasing in7
( 1) ,5
ø ùú û úü ýþ ÿ
(vi) f is strictly increasing in (– ï –1)
f is strictly decreasing in (–1, ï)(vii) f is strictly increasing in (0, 1)ñ (2, ï)
f is strictly decreasing in (– ï, 0) ñ (1, 2)(viii ) f is strictly decreasing in (–1, 0)
f is strictly increasing in (0,ï)
(ix) f is strictly increasing in5
0, ,24 4
ð ð� � � �� ð� � � �� � � �
f is strictly decreasing in5
,4 4
� �� ��
(x) f is strictly decreasing in ,6 2
� �� �� �� �
f is strictly increasing in 0,6
�� �� �� �
11. a = – 2
12. 9x – y – 3 = 9x – y + 13 = 0
14. Hints : Slpe of tangent = slope of line
dy
dx = 3x2 + a slop of line = 1
(1, 6)
dy
dx� �� � !
= 3 +a = 1a = – 2
Curve passes through (1, –6) 36 1 5a b b" # "
Ans. a = – 2 b = – 5
6 9
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 at6
() local min value
3
2 2
* +,
f has local max at6
- local max value
3
2 6
./
18. f has local min at4
- and min value is
1
2
f has local max at2
0 and mix volue is 1
f has local min value at3
4
1 and min valueis
1
2
19. a = 2 b =1
2
2
20. y = 2( 1)( ) 5 4
ax b ax b
x x y x x
3 34
3 3 3 5Domainx 6 R – {1, 4}
dy
dx =
2
2 2
( 5 4) ( )(2 5)
( 5 4)
x x a ax b x
x x
7 8 7 7 7
7 8
dy
dx = 0 andx = 2 we getb = 0
trining pt (2 – 1) lies on curve also
–1 =2
4 10 4
a b9
9 :; 2a – b = 2
Ans. a = 1 b = 0
Answer and Hints of Six Marks Questions
1. Let <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=
2. Radius of semi circle is10
4 > side of rectangle
are20
4 ? m and
10
4 ? cm
7 0
3. Let the length of the piece bent in the form of a square bex cm, then the length of the piece bent in theform of equilateral@ is 36 – x cm. Let s be the combined are of two figure
S =2 2
3 36
4 4 3
x xAB BCD DE E
Now find5d
dt = 0
Ans. x =144 3
9 4 3F
12. 275 3 cm
13. AC = AD + DC
l = a secG + b cosecG
dl
dH = a secG tanG – b cosecG cotG AC = l
dl
dH = 0 2
sin cos0
cos sin
a bI IJ K
I I
tan3G =1 3
tanb b
a aL
M N OP
2
2
d l
dQ = 3 2 3 2[sec sec tan ] [cosec cos cot ] 0 for 0
2a b ec
RS T S S T S T S S U V S V
l =
1 2 1 22 3 2 32 1 2 2 1 2
2 3 2 3[1 tan ] [1 cot ] 1 1b a
a b a ba b
W WX Y X X Y Z X X X[ [
\ \
l = 2 3 2 3 2 3 1 2 2 3 2 3 2 3 1 2 2 3 2 3 3 2[ ] [ ] ( )a a b b a b a b] ] ] ^ ]
20. eq of line 1x y
a b_ ` It passes through (3, 4)
3 41
a ba b c 3b + 4a = abc b =
49
3ad
A = Area of@ =21 1 49 2
2 2 3 3
a aab
a a
ef f
g ga h 3
finddA
da = 0 A is munimum ata = 6 b = 8
7 1
Teaching Points
l Antiderivative or Primitives : If [ ( )] ( ),d
f x xdx
i the f(x) is called antideivative off(x). As we
know that [ ( ) ] ( )d
f x c xdx
j k l so f(x) + c is also antiderivative off(x), which depends on C (constant)
as C may attain infinitely many values, therefore antderivative of a function is not unique.
Now f(x) + c is called the indefinite integral off(x) w.r.t ‘x’ which is written as ( ) ( )x dx f x Cm n o
and C is known as constant of integration.l Integration : The process of finding integral is called integration. Thus differentiation and integration
are inverse process.
p ( )d
f x dxds
qr s = .f(x) and [ ( )] ( )
df x dx f x
dxt
l Rules of Integration :
1. . ( ) ( )k f x dx k f x dxu wherek is any constant
2. 1 2 1 2. ( ) ( ) ( ) ( )f x f x dx f x dx f x dxv wx y v wx wherea andb are constants.
Fundamental Integration Formulae
l
1
, 11
nn x
x dx c nn
z{ | } ~
|
l1
logdx x cx
t �
lx xe dx e c� �
l log
xx
ac
aa dx c� �
l sin cosxdx x c� � �
l cos sinxdx x cn o
l2sec tanxdx x c� �
l2cosec cotxdx x c� � �
l sec tan secx xdx x c� �
Integrals6
7 2
l cosec cot cosecx xdx x c� �
l tan log | cos | log | sec |xdx x c x c� � �
l sec log | sec tan | log tan4 2
xxdx x x c c
�� �� � � �� �� �
l cosec cot log | cosec cot | log tan2
xx xdx x x c c� � �
l1 1
2
1sin cos
1dx x c x c
x
� �� � ��
l1 1
2
1tan cot
1dx x c x c
x� � ¡
l1 –1
2
1sec cosec
1xdx x c x c
x
¢ £ ¤ £¤
l( )
( ) , 0F ax b
f ax b dx c aa
¥¥ ¥ ¦
If ( ) ( )f x dx F x c§ i.e. if the integral of function ofx is known, then if in place ofx we have linear
function ofx, the integral is of same form but it is divided by coefficient ofx.l Following are some substitutions which are useful in evaluating integrals
Expression Substitutions
2 2 2 2ora x a x¨ ¨ tan / cotx a a© ©
2 2 2 2ora x a xª ª sin / cosx a a© ©
2 2 2 2orx a x aª ª sec / cosecx a a« «
ora x a x
a x a x
¬
¬cos2a a ©
l Special Integrals
2 2
1log
2
dx x ac
x a a x a
®¯
® ¯
12 2
1 1tan
xdx c
x a a a° ± ²
³´ µ³ ¶ ·
2 2
1 1log
2
a xdx c
a x a a x
¸¸
¹ ¹
2 2
2 2
1logdx x x a c
x aº » º
»
1
2 2
1sin
xdx c
aa x
¼ ½ ¾¿À Á ÃÄ
2 2
2 2
1logdx x x a c
x aº º º
º
7 3
Use of Partial Fractionsl In some cases we find integral using partial fractions of the following types.
A B
( )( ) ( ) ( )
px q
ax b cx d ax b cx d
ÅÅ
Å Å Å Å (where Dr. can be factorised into linear factors)
2 2
A B C
( )( ) ( ) ( )
px q
ax b cx d cx b x d cx d
ÆÆ Æ
Æ Æ Æ Æ Æ (where Dr. can be factorised into repeated linear factors)
2
2 2
A
( )( )
px qx r bx c
ax b cx dx e ax b cx dx e
Ç Ç ÇÇ
Ç Ç Ç Ç Ç Ç
Whencx2 + dx + e can not fractorised further.When we have to evaluate the integral of the type.
[ ( ) ( )] ( )x xe f x f x dx e f x cÈ É Ê È
l Some special Integrals
2 2x a dxË =2
2 2 2 2log2 2
x ax a x x a cÌ Ì Í Í Í
2 2a x dxË =2
2 2 1sin2 2
x a xa x c
aÎ Ï Ð
Ñ Ò ÒÓ ÔÕ Ö
2 2a x dxÆ =2
2 2 2 2log2 2
x aa x x a x c× × × × ×
l Definite Integral : Let f(x) be a continous function defined on the closed integral [a, b] and F(x) be anantiderivative off(x) then
( )b
a
f x dx = [F( )] F( ) F( )bax b aØ Ù
l Definite Integral as a limit of sum
( )b
a
f x dx =0
[ ( ) ( ) ( 2 ) ... ( 1 )]hLt h f a f a h f a h f a n hÚ
Û Û Û Û Û Û Û Ü
wherenh = b – al Properties of definite Integral
P(1) : ( ) ( )b a
a b
f x dx f x dxÝ Þ
P(2) : ( ) ( )b a
a b
f x dx f t dtß
P(3) : ( ) ( ) ( )b c b
a a c
f x dx f x dx f x dxß à
P(4) : ( ) ( )b b
a a
f x dx f a b x dxá â ã
7 4
P(5) :0
( ) ( )a a
a
f x dx f a x dxä
P(6) :
2
0 0 0
( ) ( ) (2 )a a a
f x dx f x dx f a x dxå æ
P(7) :2
00
2 ( ) if (2 ) ( )( )
0 if (2 ) ( )
aa f x dx f a x f x
f x dx
f a x f x
çèé
éé è èê
P(8) : 0
2 ( ) if ( ) ( )( )
0 if ( ) ( )
aa
a
f x dx f x f xf x dx
f x f xë
ìíî
îî í íï
Question for Practice
Evaluate the following Integrals
Very Short Answer Type Questions (1 Mark)
1. 2
cos6
3 sin 6
x xdx
x x
ñ
ñ2. 2sin xdx
3.1
(2 3log )dx
x xò4. 3 4sin( )x x dx
5.1
1dx
x xó ô6.
1logxe x dx
xõ ö
÷ø ùú û
7.4
2
1
1
xdx
x
ò
ò8.
1
1 sindx
xü
9. 2( 1)
xdx
xò10.
199 4
1
cosx x dxý
11. þ ÿ3 2
0
x dx 12.
1
0
1log
1
xdx
x
ð� �� ��� �
13.1
1
| |x dx�
14.
2
0
| sin |x dx�
15.1
1xdx
e ñ
7 5
Short Answer Type Questions (4 Mark)
16. 2
1logxe x dx
x
�� � �
17. 3 5sin cosx x dx
18.sin( )
sin( )
x adx
x a
�
�19.
1
sin( )cos( )dx
x a x b� �
20. 2 2 2 2
sin 2
sin cos
xdx
a x b x�21. 3
1
cos .cos( )dx
x x a�
22. sec 1x dx� 23. 4 2 1
xdx
x x� �
24. tan cotx x dx� 25.sin cos
sin(2 )
x xdx
x
�
26.2
1
(log ) 3log 4dx
x x x� �27. 2
5 2
3 2 1
xdx
x x
�
� �
28. 21x x x dx� � 29.2
2 5 3
xdx
x x� �
30. 9
1
( 1)dx
x x �31.
2 2
( 1)( 2)
x xdx
x x
� �
� �
32. 1 2 1 3
1dx
x x 33.
1
2 1
. tan ( )x x
x
a adx
a
!
"
34.2 2
2 2
( 1)( 4)
( 3)( 5)
x xdx
x x
# #
# $35. 3 5
4 4
1
(sin ) (cos )dx
x x
36.2 sin 2
1 cos 2x x
e dxx
%&' %( )
37. 5 3sin( )x x dx
38.0
4
| | | 3 | | 6 |x x x dx*
+ + + + 39.
2
4 40
sin 2
sin cos
x xdx
x x
,
-
40.
a
a
a xdx
a x.
/
041.
3 2
0
| cos |x x dx1
42.3
2
1
| 2 |x x dx2 43. 20
sin
1 cos
x xdx
x
3
�
44.cos
cos cos0
x
x x
edx
e e
4
5645.
1
1
1 sinlog
1 sin
xdx
x7
89 :; <=> ?
7 6
Long Answer Type Questions (6 Mark)
46. tanx dx 47.3
3
tan tan
1 tand
@A @@
A @
48.0
tan
sec tan
x xdx
x x
B
C49.
2
0
log(cos )x dxD
50.
1
20
log(1 )
1
xdx
x
E
E51. sin (3 2cos )
dx
x xF
52. 2
2sin 2 cos
6 cos 4sind
GH GG
H GH G53. Evaluate
52
2
( 3 )x x dxI as limit of sum.
54. Evaluate2
2
1
( 1)x x dxJ J as limit of sum.
55. Evaluate3
2 2
0
(3 )xx e dxK as limit of sum.
Answers1.
21log | 3 sin 6 |
6x x cL L 2.
sin 2
2 2
x xcM N
3.1
log | 2 3log |3
x cO O 4.41
cos( )4
x cP Q
5.3 3
2 22
( 1)3
x x cR S
T T UV W 6. logxe x cX
7. 3 12 tan ( )x x x cYZ [ [ 8. tan tan secx x x c\ ]
9.1
log | 1|( 1)
x cx
^ ^ ^^
10. 0
11.1
212. 0
13. 1 14. 4
15. log |1 |xe c_` a a 16.1
logxe x cx
b cd Fe fg h
17.6 8cos cos
6 8
x xci j j 18. cos 2 sin 2 log | sin( ) |x a a x a ck l k
19.1 sin( )
.logcos( ) cos( )
x ac
a b x b
mn
m m20.
2 2 2 22 2
1log | sin cos |
( )a x b x c
a bo o
p
7 7
21.2
cos sin tansin
a a x ca
qq r 22.
22cos 1log cos cos
2
xx x c
st uv s s sw xy z
23.2
11 2 1tan
3 3
xc{ | }~
~� �� �24. 12 sin (sin cos )x x c� � �
25. 1sin (sin cos )x x c� � � 26. 22log 3log (log ) 3log 4
2
xx x c
�� �� � � �� �� �
27. 2 15 11 3 1log | 3 2 1| tan
6 3 2 2
xx x c� �� �� � � �� �� �
28.
32 2 12
1 1 2 1 5 2 1(1 ) 1 sin
3 2 4 8 5
x xx x x x c�� � � � �� � � �
� � ¡ ¢ ¡ ¢£ ¤¥ ¦ ¥ ¦§ ¨
29.25 19 2 5 13
log | 5 3 | log2 2 13 2 5 13
xx x x c
x
© ªª © © © ©
© ©
30.9
9
1 1log
9
xc
x
«¬ 31. 2log | 1| 4 log | 2 |x x x c ®
32.1 2 1 3
1 6 1 66 log | 1|3 2
x xx x c
¯ °± ± ± ² ±³ ´
µ ¶
33.1 21
[tan ( )]2 log
xa ca
· ¸ 34.11 27 5
tan log4 3 3 8 5 5
x xx c
x¹ º» ¼
½ ½ ½¾ ¿À Á ½
35.1
44(tan )x c 36. tanxe x cÃ
37. 3 3 31[ cos ( ) sin ( )]
3x x x cÄ Å Å 38. 29
39.2
8
Æ40. aÇ
41. 2
5 2
2
È É
È42. 2
43.2
4
Æ44.
2
Ê
45. 0
46.11 tan 1 1 tan 2 tan 1
tan log2 2 tan 2 2 tan 2 tan 1
x x xc
x x xË Ì Ì ÍÎ Ï
Í ÍÐ ÑÒ Ó Í Í
47.2 11 1 1 2 tan 1
log |1 tan | log | tan tan 1| tan3 6 3 3
xx x x cÔÕ ÕÖ ×
Ø Ø Õ Ø Ø ØÙ ÚÛ Ü
7 8
48. 12
ÝÞ ßÝ àá âã ä 49. log 2
2
åæ
50. log 28
ç51.
1 1 2log |1 cos | log |1 cos | log | 3 2cos |
10 2 5x x x cè è é é é é
52. 2 12log | sin 4sin 5 | 7 tan (sin 2)cêë ì ë í í ë ì í
53.141
254.
29
655.
6(53 )
2
eî
Hints
19.1 cos( )
cos( ) sin( )cos( )
a bdx
a b x a x b
ï
ï ï ï
=1 cos[( ) ( )]
cos( ) sin( )cos( )
x b x adx
a b x a x b
ñ ñ ñ
ñ ñ ñ
=1
[cot( ) tan( )]cos( )
x a x b dxa b
ò ó òò
22.1 cos
cos
xdx
x
ô
= 2
sin
cos cos
xdx
x xõPut cosx = t
35. 3 5
4 4
1
(sin ) (cos )
dx
x x
Dividing num. and den. by cos2x, we get
=2
3
4
sec
(tan )
xdx
x
3 5 82
4 4 4ö ÷
óø ùú û
Now put tanx = t
39. I =
2
4 40
sin 2
sin cos
x xdx
x x
ü
ý
þ I =
2
4 40
sin 2
4 sin cos
xdx
x x
ÿð� 0 0
Using prop. ( ) ( )a a
f x dx f a x dx� �
�� �� �
I =
2 2
40
2 tan sec
4 tan 1
x xdx
x
�
Put tan2x = t
7 9
40. I =a
a
a xdx
a x��
I = 2 2
a
a
a xdx
a x���
I = 2 2 2 2
a a
a a
adx xdx
a x a x� ��
� �
I = 2 20
2 0a dx
aa x
��
0
( ) 2 ( ) if ( ) ( )
0 if ( ) ( )
a a
a
f x dx f x dx f x f x
f x f x�
� ��� �
� �� �� �� �
I = a�.
41.
3 2 1 2 3 2
0 0 1 2
| cos | ( cos ) ( cos )x x dx x x dx x x dx� � � � �
Use integral by parts to evaluate( cos ) .x x dx�
46. I = tanx dx Put tanx = t2
I =2
4
2
1
tdt
t �
I =2 2
4 4
1 1
1 1
t tdt dt
t t
� �� �
I =2 2
2 22
2
1 1 1 11 1
t tdt dt
t ttt
! "!!!
I =2 2
2 2
1 1 1 1
( 1 ) 212
t tdt dt
t tt
t
# $## $% &$ #' () *
Put1
t at
+ and1
t bt
,
I = 2 2 2 2( 2) ( 2)
da db
a b-
- .
50. I =1
20
log(1 )
1
xdx
x
//
Putx = tan0
8 0
I =
4
0
log(1 tan )d1
2 3 3
I =
4
0
log 2 Id4
5 60 0
Using property ( ) ( )a a
f x dx f a x dx7 8
9: ;< =
I = log 2.8
>
51. sin (3 2cos )
dx
x x?
= 2
sin
(1 cos )(3 2cos )
xdx
x x@ ? Put cosx = t
=1
( 1)( 1)(3 2 )dt
t t tA B B
52.5
2
2
( 3 )x x dxC
As ( )b
a
f x dx =0
1
( ), wheren
hr
b aLt h f a rh h
n
DE
F5
2
2
( 3 )x x dxG =2 2
01
[10 7 . ]n
hr
Lt h rh r hH H
= 2 2
01 1 1
10 7n n n
hr r r
Lt h h r h rI JK KL MN O
= 2
0
( 1) ( 1)(2 1)10 7
2 6h
n n n n nLt h n h h
P P PQ RP PS TU V
=0
7( )( ) ( )( )(2 )10
2 6h
nh nh h nh nh h nh hLt h nh
? ? ?W X? ?Y Z[ \
=141
2
8 1
Teaching Points
l The area bouned by the curvey = F(x) above thex-axis andbetween the linesx = a, x = b is given by
b
a
ydx = ( )b
a
F x dx
l If the curve between the linesx = a, x = b lies below thex-axis,then the required area is given
( )b
a
y dx] = ( )b b
a a
ydx F x dx^
l The area bounded by curvey = F(x), x-axis and between linesx = a, x = b to given by
( )b
a
F x dx
l The area bounded by the curvex = F(y) above they-axis and thelinesy = c, y = d is given by
d
c
xdy = ( )d
c
F y dy
l If the curve between the lines y = c, y = d lies below the y-axis(to the left ofy-axis) then the area is given by
( )d
c
x dy_ = ( )d d
c c
xdy F y dy` a
l The area bounded by curvex = F(y), y-axis and between linesy = c andy = d is gives by
| ( ) |d
c
F y dy
l If 0 £ g(x) £ f(x), the area of region bounded between curvesand ordinatesx = a andx = b is given by
= [ ( ) ( )]b
a
F x g x dxb
Application of Integrals7
8 2
l When we find the area bounded by the curvesy = f(x)andy = g(x) and after drawing the graphs the shadedregion is of such type.We find thex-cordinate of their point of inter-sections.Let for point A and B the values ofx area andb. Then
Required Area = [ ( ) ( )]b
a
f x g x dxc
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 curvecontains even power in both ‘x’ and ‘y’ then graph is symmetric about both axis.
Question for Practice
Q1. Find the common area bounded by the circles2 2 4x yd e and 2 2( 2) 4x yf g h .
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 regioni j2( , ) :| 1| 5x y x y xf k k f .
Q5. Find the area of the region bounded by
y2 = x and line x + y = 2
Q6. Find the area enclosed by the curvey = sinx betweenx = 0 andx = 3l/2 andx-axis.
Q7. Find the area bounded by semi circley = 225 xm andx-axis.
Q8. Find area of region given by 2[( , ) : | |]x y x y xn n .
Q9. Find area of smaller region bounded by ellipse2 2
19 4
x yo p and straight line 2x + 3y = 6.
Q10. Find the area of region bounded by the curvex2 = 4y and linex = 4y – 2.
Q11. Using integration find the area of region in first quadrant enclosed by x-axis the line 3x yq and
the circlex2 + y2 = 4.
Q12. Draw a rough sketch of the region 2 2[( , ) : 4 ]x y x y x yr s s r and find its area.
Q13. Find the smaller of two areas bounded by the curvey = |x| andx2 + y2 = 8.
Q14. Find the area lying above x-axis and included between the circlex2 + y2 = 8x and the parabolay2 = 4x.
8 3
Q15. Using integration find the area enclosed by the curvey = cosx, y = sinx andx-axis in the interval
0,2
tu vw xy z .
Q16. Sketch the graphy = |x – 5|. Evaluate6
0
| 5 |x dx{ .
Answers1. 8 2 3
3
|} ~�� �� � sq units 2. (a) 4 sq units, (b) 2 sq units
3. (a) 6 sq units (b)15
2 sq units 4. 1 15 2 1 1
sin sin2 25 5
� �� �� �� � � �� �� �� � � �� �� � � �� �� � sq units
5.9
2 sq units 6. 3 sq units 7.
25
2 sq units
8.1
3 sq units 9.
3( 2)
2� � sq units 10.
9
8 sq units
11.3
� sq units 12. (� – 2) sq units 13. 2� sq units
14. 8(8 3 )
3� � sq units 15. (2 2)� sq units 16. 5 sq units
Hints
1. Required area =
1 22 2
0 1
2 4 ( 2) 4x dx x dx� �� �� � � �� � �¡ ¢
4. Required area =2 1 2
2
1 1 1
5 ( 1) ( 1)x dx x dx x dx£ £
¤ ¤ ¤ ¥ ¤ ¤
5. Required area =1 1
2
2 2
(2 )y dy y dy¦ ¦
§ §
14. Required area =4 8
2
0 4
2 16 ( 4)xdx x dx¨ © ©
8 4
Teaching Points
Definition. A differential equation is an equation which involves unknown functions and their derivativesw.r.t. one or more independent variables.
An ordinary differential equation is an equation which involves only one independent variable andderivatives 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 differentialequation and its degree is the index of the highest order derivative which appears in the differential equationafter making it free from negative and fractional powers so far as the derivatives are concerned.
Linear Differential Equation
Definition. A differential equation ( )( , , ,..., ) 0nx y y yª «¬ ¬¬ is called linear iff the function is a linear
function of the variables ( ), , ,..., ny y y y¬ ¬¬ i.e. iff the dependent variable and its derivatives occur only in thefirst degree and are not multiplied together. Otherwise, it is called non-linear.
Thus a general linear differential equation of ordern may be written as1 2
0 1 21 2 ... Pn n n
nn n n
d y d y d yP P P y Q
dx dx dx
® ®® ®¯ ¯ ¯ ¯ ° ...(1)
whereP0, P
1, P
2,..., P
nandQ are functions ofx only (not containingy) andP
0 ± 0.
The linear differential equation (1) is called homogeneous iff Q is the zero function, and it is callednon-homogeneous iff Q is a non-zero function.
Note that a linear differential equation is always of the first degree but every differential equation ofthe first degree need not be linear.
Formation of a differential equation.
Consider the general equation (primitive) 1 2( , , , ,..., ) 0nx y c c cª « containingn independent arbitrary
constants (called parameters)1 2 3, , ,..., nc c c c apart fromx andy..
To obtain the differential equation which is satisfied by 1 2 3( , , , , ,..., ) 0,nx y c c c c² ³we differentiate this equationn timesw.r.t.xand eliminaten arbitrary constants1 2, ,..., nc c c from the
n + 1 equations;n equations obtained on differentiation and the given equation. In this way, we obtain anordinary differential equation
Differential Equations8
8 5
(i) of an order equal to the number of independent arbitrary constants in the given equation,
(ii ) consistent with the given equationi.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 thegiven differential equation.
Let the equation involvingx,yandn independent arbitrary constants be
1 2( , , , ,..., )nx y c c c´ = 0 ...(i)
and the differential equation obtained from (i) be
2
2, , , ,...,n
n
dy d y d yf x y
dx dx dx
µ ¶· ¸¹ º = 0 ...(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 arbitraryconstants.
The most general solution of an ordinary differential equation of first order contains one arbitraryconstant and of second order contains two arbitrary constants and so on.
Definition. Any solution obtained from the general solution of a differential equation by givingparticular 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
1 2( ) ( )f x dx f y dy» = 0
Integrating, the general solution of this equation is
1 2( ) ( )f x dx f y dy¼ = C, where C is an arbitrary constant.
Reducible to Variables Separable.
Type 1. To solve an equation of the typedy
dx= f(ax+ by+ c), putax+ by+ c= vand it will be reducible
to variables separable.
Type 2. Homogeneous Equation
A differential equation of the form( , )
( , )
dy f x y
dx g x y½ wheref(x, y) andg(x, y) are both homogeneous
functions of the same degree in x and y i.e. and equation of the formdy y
Fdx x
¾ ¿À Á ÂÃ Ä is called a homogeneous
differentiable equation.
A homogeneous function of degreen in x andy is a function which can be written as , .n yx n N
xÅ ÆÇ ÈÉ ÊË Ì .
To solve homogeneous differential equation:
8 6
put y = v x, thendy dv
v xdx dx
Í Î and on substituting these values ofy anddy
dxin the given differential
equation it will be reducible to variables. separable.
Note. Just as the differential equation of the form ( )dy
F y xdx
Ï can be solved by the substitution
y = y x,the differential equation
dx xG
dy y
Ð ÑÒ Ó ÔÕ Ö , which is also homogeneous, can be solved by the substitutionx = vy..
Linear Differential Equation (of first order)
Definition. A linear differential equation of the first order in y is an equation of the form.
dyPy Q
dx× Ï , where bothP andQ are functions of x only (not containingy).
If Q ¹ 0, then the above equation is called non-homogeneous linear differential equation and ifQ = 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 1 1
dxPx Q
dyØ Ù , where
bothP1 andQ
1are functions ofy only (not containingx).
Solution of Linear Differential Equation.
(i) If dyPy Q
dxÚ Û , whereP andQ are functions ofx only, then it has
PdxeÜ as integrating factor and its
solution is given by . .Pdx Pdx
y e Q e dx CÝ Þß ßà áâ ã .
(ii ) If 1 1
dxPx Q
dyØ Ù , where P
1 and Q
1 are functions of y only, then it has 1Pdy
eä as I.F and its solution is
given by 1 1
1. ( . )Pdy Pdy
x e Q e dy Cå åæ ç .
Question for Practice
Evaluate the following Integrals
Very Short Answer Type Questions (1 Mark)
Q1. What is integrating factor of( log ). 2 logdy
x x y xdx
è é .
8 7
Q2. Show thaty = 4 sin 3x is a solution of differential equation2
2 9 0d y
ydx
ê ë .
Q3. Show that 2 2 2( ) ( )x xy dy x y dxì í ì is a homogeueous differential equation.
Q4. Write the order and degree of differential equation33 2
2 2 4 sind y d y dy
y xdx dx dx
î ïñ ñ ñ òó ôõ ö.
Q5. Write order and degree of differential equation3 22 2
21dy d y
dx dx
÷ øù úû üý þÿ ð� �ý þ� �
.
Short Answer Type Questions (4 Marks)
Q6. Solve the differential equation.
( ) ( ) 0x y x y dx x y x y dy� � � � � � � �
Q7. Show thaty = a cos(logx) + b sin(logx) is a solution of the differential equation
22
2 0d y dy
x x ydx dx
ê ê ë
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 ) + vex-axis.
(iii ) Form differential equation of all circles passing through origin and whose centre lie on x-axis.
Q9. Show that the differential equationdy x y
dx x y
��
is homogeneous and solve it.
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 :
(i) cos 2 .dy
y xdx
�
(ii ) sin xdy
dx+y cosx = 2 sin2 x cosx
Q12. Solve each of the following differential equations :
(i)22
dy dyy x y
dx dx�
� � �� �� �
(ii ) cos (1 2 )sin 0xy dx e y dy�ñ ñ ò(iii ) 2 21 1 0.x y dy y x dx� � � �
(iv) 2 2(1 )(1 ) 0.x y dy xy dx� � � �
(v) 2 2( ) ( ) 0; (0) 1.xy x dx yx y dy yê ê ê ë ë
(vi)3 3sin cos .xdy
y x x xy edx
� �
8 8
Q13. Solve the following differential equations :
(i) 2 2 .xy dy y dx x y dx� �
(ii ) cos sin sin cos 0.y y y y
y x y dx x y x dyx x x x
! " ! "# $ # $ # $ # $% & & '( ) ( )* + * + * + * +, - , - , - , -. ./ /
(iii ) 2 ( ) 0x dy y x y dx0 0 1 given thaty = 1 whenx = 1.
(iv) 1, 0dy
x ay x xdx
2 3 4 5
(v) 3 2 3 2( 3 ) ( 3 ) .x xy dx y x y dy6 7 6
Q14. Solve the following differential equations:
(i) 2 3 3( ) 0x y dx x y dy6 8 7
(ii ) 2( 2 ) 0y dx x y dy9 : ;
(iii ) 2 2( ) 2 0, (1) 1.x y dx xy dy y6 8 7 7
(iv) sin sinx x
y dx x y dyy y
< = < => ?@ A @ AB C B C
(v) tandy y y
dx x xD E
F G H IJ K
(vi) 2 2
2dy xy
dx x y7
8
(vii)2x y ydy
e x edx
LM N
(viii ) 2( 3 )x y dy y dxO P
Answers1. logx 4. order = 3. degree = 1
5. order = 2, degree = 2 6. 2 2
1c
x x yQ
R S
8. (i)2 2 2 0
dyx y xy
dxT U V (ii ) (a)
2
dy x
dx aW (b) 2
dya
dxX (c) 2 2( ) 2 0
dyx y xy
dxY Z X
9.1 2 21
tan log( )2
yx y c
x[ \ ]
^ _ _` ab c 10. logy
x cxd e
8 9
11. (i)1
( cos 2 2sin 2 )5
xy x x cef g h h (ii)32
sin sin3
y x x cf h
12. (i) ( 2)(1 2 )cy x yi j k
(ii ) ( 2)secxe y cl m
(iii ) 2 21 1x y cn o n p
(iv)
22 2
2
1 11log 1 1
2 1 1
yx y c
y
q qr q q q s
q s
(v) 2 2( 1)( 1) 2x yt t u
(vi) 31 1log cos cos5
3 5x xy x x xe e cv w x x w x
(vii)cos 2
log | tan |x
y cy
y z
13. (i)2
2 2 2 4y x y c x{ { |
(ii ) cosy
xy kx
} ~�� �� �
(iii ) 2
2
3 1
xy
x�
�
(iv)1
1ax
y cxa a
f g hg
(v) 2 2 2 2 2 2( )x y x y c� � �
14. (i)3
3 log | |3
xy c
y
�� �
(ii ) 22x y cy� �
(iii ) 2 2 2x y x� �
(iv) cos( )x yy ce� [Hint : Putx
y = v]
(v) siny
cxx
� ��� �� �
(vi) 2 2( )c x y y� �
(vii)3
3y x x
e e c�� � � �
(viii ) 23x y cy� �
9 0
Hints
6. ( ) ( ) 0x y x y dx x y x y dy� � � � � � �
¡dy
dx =
x y x y
x y x y
¢ ¢ £
¢ £ £
Puty = vx
dxvv
dx¤ =
1 1 1 1
1 1 1 1
v v v v
v v v v
¥ ¥ ¦ ¥ ¥ ¦§
¥ ¦ ¦ ¥ ¦ ¦
¡xdv
vdx
¤ =21 1 v
v
¥ ¦
¡dv
xdx
=2 21 1 v v
v
¥ ¦ ¦
¡ 2 21 (1 )
vdv
v v¨ © ¨ =
1dx
x
Put 2 21 v tª «
2
t dt
t t¬
=
1dx
x
¡ log |1 |t® ¯ = | | log | |by x c°
¡ 2
1
1 1 v± ² = cx
¡ 2 2
1
x x y³ ´ = c
12. (i)dy
y xdx
¬ =22
dyy
dxµ ¶
±· ¸¹ º
¡1
(2 )dx
x» =
1
(1 2 )dy
y y¼
¡1
(2 )dx
x¥ =
1 2
(1 2 )dy
y y
½ ¾¿À Á
ÂÃ Ä
¡ cy = (2 +x) (1 – 2y)
9 1
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 byABÅ
or aÆ .
l Position vector of a point P(x, y, z) w.r.t. origin 0(0,0,0) is denoted byOPÇ
, whereOPÇ
= ˆˆ ˆxi yj zkÈ È
and OPÉ
= 2 2 2x y zÊ Ê
l The anglesË, 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 = cosË, m = cosÌ and n = cos Í where l2 + m2 + n2 = 1.
l The numbersa, b, c, propotional tol, m and n are called direction ratiosi.e. .a b c
l m nÏ Ï
l If two vector aÆ
and bÐ
are represented in magnitude and direction by the sides of a triangle taken
in order, then their suma bÑÒÒ
is represented in magnitude and direction by the third side of triangletaken in opposite order. This is called triangle law of addition of vectors.
l If aÆ is any vector andÓ is a scalar, thenÓ a
Æ is a vector, collinear witha
Æ and aÔÕ
= aÔÕ
.
l Any vector aÆ can be written isa
Æ = ˆa a
Ö where a is a unit vector in the direction ofa .
l If A(x1, y
1, z
1) and B(x
2, y
2, z
2) be any two points in space, thenAB = (x
2 – x
1) i + (y
2 – y
1)
2 1ˆˆ ( – ) .j z z k×
l If aÆ
and bÐ
be the position vectors of points A and B, and C is any point dividing AB internally
in m : n, then position vectorcÆ
of C is given ascÆ
=mb na
m n
Ø
Ø. If C divides AB in m : n externally,,
then C =–
.–
mb na
m n
l Scalar Product (Dot Product) of two vectorsaÆ
andbÐ
is denoted by .a bÙÙ
and is defined as. cosa b a bÚ ÛÜ ÜÜ Ü
,
whereÝ is the angle betweenaÆ
andbÐ
(0Þ Ý Þ )l Dot product of two vectors is commutativei.e., . .a b b aß
à àà à
l For unlike parallel vectorsaÆ
and bÐ
, .a bÙÙ
= a báá
, so .a aâ â
=2
aã
l For unlike parallel vectorsaÆ
and bÐ
, .a bÙÙ
= – a bää
Vectors9
9 2
l If . 0 0 or 0 ora b a b a bå æ å å çè è èè è è
.
l ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ. . . 0, . . . 1i j j k k i i i j j k ké é é é é é
l If 1 2 3ˆˆ ˆa a i a j a kê ë ë
ì and 1 2 3
ˆˆ ˆb b i b j b kê ë ëì
, then 1 2 1 2 1 2. .a b a a b b c cí î îïï
l Projection of a vectorañ
on bò
= .a b
b
óó
ó
l Projection of a vectoraô
along bõ
= . ˆ.a b
bb
ö ÷ø ùø ùú û
üü
ü
l Cross product (Vector product) of two vectorsaô
and bõ
is denoted asa býý
and is defined as
a bþþ
= a bÿÿ sinð � , whereð is the angle betweena
ñ and b
ò (0 � ð � �) and � is a unit
vector perpendicular to bothañ
and bò
such thatañ
, bò
and � form a right handed system.
l Vector product of two vectorsañ
and bò
is not commutativei.e., a b b a� �� �
, but –a b b a�� �� �
.
l If O or , or .a b a O b O a b� � � �� � �� � �� � �
l ˆ ˆˆ ˆ ˆ ˆ. O, So O .a a i i j j k k
l If 1 2 3ˆˆ ˆa a i a j a kê ë ë and 1 2 3
ˆˆ ˆb b i b j b kê ë ë , then
a b = 1 2 3
1 2 3
ˆˆ ˆi j k
a a a
b b b
l If ð is angle between two non zero vectora and b , then
sinð =a b
a b
l a b is the area of that parallelogram where adjacent sides are vectors a and b .
l1
2a b is the area of that parallelorgam whose diagonals area and b .
l If ,a b and c forms a triangle, then area of the triangle =1
2a b =
1
2b c =
1.
2c a
l If ,a b andc are position vectors of vertices of a triangle, then area of triangle =1
2a b b c c a
Scalar triple product of vectors
The scalar triple product of three vectorsa b c� � �
9 3
Geometrical Interpretation : ( ).a b c represents the volume of parallelepiped whose coterminous
edges are represented bya 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 bor
[ ]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 andc to be coplanar
is that [ ]a b c = 0
Scalar triple Product in terms of component
Let = 1 2 3ˆˆ ˆa a i a j a k� � � 1 2 3
ˆˆ ˆb b i b j b k� � � 1 2 3ˆˆ ˆc c i c j c k� � �
then [ ]a b c =1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
Question for Practice
Very Short Answer Type Questions (1 Mark)
1. Write direction cosines of vector ˆˆ ˆ2 2i j k� �
2. For what value of ‘x’ the vectorsˆˆ ˆ2 – 3 4i j k� and ˆˆ ˆ– 6 8xi j k� are collinear
3. Write the projection of the vector ˆ–i j on the vectorˆ ˆi j�
4. Write the angle between two vectorsa and b with magnitudes 3 and 2 respectively having
. 6a b �
5. If . 0a a� and . 0a b� = 0 then what can be concluded about vectorb
6. What is the cosine of the angle which the vector ˆˆ ˆ2i j k� � makes with z axis
7. Write a vector of Magnitude 3 units in the direction of vector ˆˆ ˆ –i j k�
9 4
8. If a = ˆˆ ˆ2 3i j k� � and b = ˆˆ ˆ6 9i j k� � � and a b find the value of�
9. For what value of� a = ˆˆ ˆ2 –i j k� � and b = ˆˆ ˆi j k� � are prependicular to each other
10. If a is a unit vector such that ˆa i j find .a i
11. a = 2 b = 7, ˆˆ ˆ3 2 6a b i j k! " " write the angle betweena and b
12. If ,a b represent Diagonal of rhombus, then write the value of.a b
13. Write unit vector in the direction ofa b# where ˆˆ ˆ –a i j k$ % , ˆˆ ˆ 2b i j k& ' '
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 thata b is also a unit vector. Find the angle betweena
and b
16. What is the value ofx If for unit vector a
( – ).( ) 15x a x a( )
17. Write the value of ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ.( ) .(( ) .( )i j k j k i k j i* *
18. Find the value of� so that the vectors
a = ˆˆ ˆ2 – 3i j k� and b = ˆˆ ˆ2 – 3i j k� and ˆˆc j k+ , - are coplanar
19. Find the value ˆˆ ˆ[ ]i j k
20. Find the volume of parallelepiped whose coterminous edges are represented by the vectors
a = ˆˆ ˆ2 3i j k� � and b = ˆˆ ˆ3 7 – 4i j k� and ˆˆ ˆ5 3c i j k. � �
Short Answer Type Questions (4 Marks)
1. If , ,a b c are unit vectors such that. . 0a b a c/ / and the angle betweenb and c is 60 then
prove that a = 12( )b c
2. Find a vector of magnitude 5 units, prependicular to each of the vectors( )a b* and ( – )a b where
ˆˆ ˆa i j k2 3 3 , ˆˆ ˆ2 3b i j k4 5 5
3. If , ,a b c are vectors such that, .a b a c.
a b a c6 and 0a then prove thatb c7
4. If ˆˆ ˆa i j k2 3 3 and ˆˆ –b j k8 find a vector c such thata c b9 and . 3a c :5. Using vectors find the area of triangle with vertices A(1, 1, 2) B(2, 3, 5) and (1, 5, 5)
6. If , ,a b c are the position vectors of the vertices A, B, C ofABC respectively, find an expressionfor the area of ABC and Hence deduce the condition for the points A, B, C to be collinear
7. Show that the area of || gm having Diagonals ˆˆ ˆ3 – 2i j k* and ˆˆ ˆ– 3 4i j k* is 5 3 sq unit.
9 5
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 finda b c< <
9. If ,a b and c are three vectors such that 0a b c= = >
a = 3 b = 5 c = 7 find the angle betweena and b
10. If a and b are unit vectors and? is the angle between them, then show that
(i)1
ˆsin –2 2
a b@A
(ii )1
ˆcos2 2
a bBC D
11. If a unit vectora makes4
and3
with x axis andy axis respectively and an obtuse angle?
with z axis, then find? and the components ofa along axis.
12. If a and b are unit vectors such that2 – 4a b and 10 8a bE are perpendicular to each other..
Find the angle betweena and b
13. Find a vectora such that
ˆ ˆ.( ) 2a i jF G ˆ ˆ.( – ) 3a i j H ˆ. 0a k I
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 departure16. Prove the following
(i) [ , , ] 2 [ ]a b b c c a a b cJ J J K
(ii ) [ – , – , – ] 0a b b c c a H
17. Show that the points with position vectors
ˆ ˆˆ ˆ ˆ ˆ ˆ6 – 7 ,16 –19 – 4 , 3 – 6i j i j k i k and ˆˆ ˆ2 – 5 10i j kL are coplanar..
18. If ,a b c d a c b dM M show that –a d is || to –b c
19. Let ˆ ˆˆ ˆ ˆ ˆ– 3 – 7 –a i j b i k c i kN N N find a vector d which is ; to both a and b and . 1c d O
20. If 1a b a bP P Q P then find –a b
21. Prove 2a b =
. .
. .
a a a b
a b b b
22. If 2 5 8 find .a b a b a bP P P
23. Let ,u v and w be vectors such that+ ou v wR S
If 3 4 5u v wT T T find . . .u v v w wuU U
9 6
24. In a Regular hexagon ABCDEF ifAB BCa bV V V V
W W then expressCD,DE,EF,FA,AC,AD,AE,X X X X X X X
and
CEY
is terms of a and b
25. Given ˆ ˆ3 –a i jZ and ˆˆ ˆ2 – 3b i j k[ \ express 1 2b b b] ]
^ _ where ,b is parallel toa and 2b]
is` a
Answers
1.1 1 1
, ,3 3 3
2. x = –4
3. 0 4.4
5. b aa 6. cosb =1
27. ˆˆ ˆ3( – )i j kc 8. –3
9. d = –1 10. 0
11.6
12. 0
13.1 ˆˆ ˆ(2 2 )3
i j ke e 14. 5
15.2
16. 4
17. 1 18. d = –1
19. 120. 39 cubic unit
Answer of four marks question
1. Given . 0a b f and . 0a c g h is ` to both b and c
Also a is a unit vectori =b c
b c
But1 1
sin 1 16 2 2
b c b cj j j
2.5 ˆˆ ˆ(– 2 – )6
i j kk
3. . . . . 0 .( – ) oa b a c a b a c a b cl m n l m l oa o or – ob c p or ( – )a b cq (1) but oa (given)
a b = a c h – oa b a cr h ( – ) oa b c s oa o or – ob c p or –a b c (2) but oa
from (1) and (2) we getb ct ( –a b cu and –a b c can not held simultaneously)
9 7
4.5 2 2 ˆˆ ˆ3 3 3
c i j kv w w 5.1
612
6. A r e a o f ABC =1
AB AC2x x
=1
( – ) ( – )2
b a c a
=1
– – –2
b c b a a c a ay
=1
2a b b c c az z
( o)a a { –a b b a|
Point A, B, C to be collinear if area ofABC = 0 0a b b c c a} } ~
7. Hint Area of 11gm 1 2
1
2d d 1d & 2d are the vectors along diagonal
8. 5 2
9. 0a b c� � � � –a b c� �
� ( ).( ) (– ).(– )a b a b c c� � �
22 22 .a b a b c� � �
22 22 cosa b a b c� � � �
32 + 52 + 2 × 3 × 5 cos� = 72 � = 60°
11. � =2
3
�,
1 1 1, ,2 22
i j k�
12. angle betweena and b is 120°.
13.5 1ˆ –2 2
a i j�
14. Let 1 1 1ˆˆ ˆa x i y j z k� � �
cos� =ˆ.ˆ
a i
a i = 1
3
x
x1 = 3cos�
Similarly y1 = 3cos� z
1 = 3cos�
a = 3 2 2 21 1 1x y z� � = 3 �
2 2 29cos 9cos 9cos�� �� � = 3
3 3 cos 3� � �1
cos3
� � �1 1
cos3
�
A( )a
C( )c
B( )b
9 8
15. OB�
= OA AB� �� OA
� = ˆ–4i
In AMBAM = AB cos60° MB = ABsin60°
AM =3
2MB =
3 3
2
AB�
= AM MB� �
�
AB =3 3 3ˆ ˆ2 2
i j�
Girls displacement from initial point of departure
3 3 3 5 3 34
2 2 2 2i i j i j
� �� � � �¡¢
19.1 1 3 ˆˆ ˆ4 4 4
i j k£ £
22. 623. –25
24. CD –b a�
¤ DE –a¥
¦ EF –b§
¨ FA –a b©
ª AC a b�
¤ « AD 2b¥
¦ AE 2 –b a¬
CE – 2b a¬
25. 1b�
=3 1ˆ ˆ–2 2
i j
2b®
=1 3 ˆˆ ˆ – 32 2
i j k¯
99
Teaching learning points
l Distance between two given points P(x1, y
1, z
1) and Q(x
2, y
2, z
2) is
PQ = 2 2 22 1 2 1 2 1( – ) ( – ) ( – )x x y y z z° °
l Direction ratio of line joining the points (x1, y
1, z
1) and (x
2, y
2, z
2) are x
2 – x
1, y
2 – y
1, z
2 – z
1
l Let a, b, c be the direction ratio of a line whose direction cosines arel, m, n thel m n
a b c± ±
l2 + m2 + n2 = 1
l =2 2 2
a
a b c²
³ ³m =
2 2 2
b
a b c²
³ ³n =
2 2 2
c
a b c²
³ ³
If line makes angles , µ, ¶ with coordinate axes thenl = 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 vectora and
|| to given vectorb
r = a b· ¸ where¹ is parameter andr = ˆˆ ˆxi yj zkº º
l Cartesian equation of straight line passing through a fixed point (x1, y
1, z
1) having direction ratio
(a, b, c) is given by 1 1 1– – –x x y y z z
a b c» »
l The coordinates of any point on the line 1 1 1– – –x x y y z z
a b c¼ ¼ ¼ ½ are (x
1 + a¹, y
1 + b¹, z
1+ c¹)
where ¹ ¾Rl Angle between two lines whose direction rahos area
1, b
1, c
1 and (a
2, b
2, c
2) is given by
cos¿ = 1 2 1 2 1 2
2 2 2 2 2 21 1 1 2 2 2
a a b b c c
a b c a b c
À À
À À À À
If lines are perpendicular thena1a
2+ b
1b
2+ c
1c
2 = 0
If lines are parallel then 1 1 1
2 2 2
a b c
a b cÁ Á
l Angle between tow lines : 1 1r a mÂ Ã Ä and 2 2r a m à Šis given as cos¿ = 1 2
1 2
.m m
m m and so, two
lines are perpendicular if 1 2.m m = 0. Lines are parallel 1 2m mÆ Ç
Three Dimensional Geometry10
100
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 : 1 1r a bÈ É Ê and2 2r a bÈ É Ë are two skew lines, then distance ‘d’ between
them is given by
d =2 1 1 2
1 2
( – ) ( )a a b b
b b
Ì Í
Í
l Distance ‘d’ between parallel line : if 1r a bÎ Ï Ð and 2r a bÑ Ò Ó are two parallel lines, then
distanced, between them is given by
d =2 1( – )a a b
b
Ô
l 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.
l Equation of a plance at a distance ‘p’ from origin and normal vector n is given by ˆ.r n = p orlx + my + nz = p.
l General equation of plane passing through a pointa and having normal vector to plane asn
is ( – ). 0.r a n Õ Corresponding Cartesian form isa(x – x1) + b(y – y
1) + c(z – z
1) = 0, where
a, b, c are direction ratios of normal to plane.l General equation of plane which cuts off interceptsa, b and c on x, y and z-axis respectively
is 1.x y z
a b cÖ Ö ×
l Equation of plane passing through three non-collinear points, ,a b c is ( – ) ( – ) ( – ) 0.r a b a c aØ ÙÚ Û ÜÝ Þ
Corresponding Cartesian from is1 1 1
2 1 2 1 2 1
3 1 3 1 3 1
– – –
– – – 0
– – –
x x y y z z
x x y y z z
x x y y z z
ß
where (x1, y
1, z
1), (x
2, y
2, z
2) and (x
3, y
3, z
3) are co-ordinates of known point.
l Angle between two planes is cosà = 1 2
1 2
.n n
n n or
cosà = 1 2 1 2 1 2
2 2 2 2 2 21 1 1 2 2 2
a a b b c c
a b c a b c
á á
á á á á
where 1n & 2n are vectors normal to planes ora1, b
1, c
1 anda
2, b
2, c
2are dr’s of normal to planes.
101
l Condition of Coplanarity of two lines
Let 1 1r a bâ ã ä and 2 2r a bå æ ç be two lines then these lines are coplanar if2 1 21( ).( ) 0a a b bè é ê
and equation of plane containing them is 1 21( ).( ) 0r a b bë ì í or 2 1 2( – ) ( ) 0r a b bî ï ñ
l Distance of pointa from plane .r n = d is
Distance =–a n d
n
ò
l Angle ó between liner a bô õ ö and planer n÷ = d is given as
sinó =b n
b n
ø
A line is parallel to plane ofb nù = 0.
Question for Practice
Very Short Answer Type Questions (1 Mark)
Q1. Write intercept cut off by the plane 2x + y – z = 12 onx axis.
Q2. Write the vector equation of a line– 5 4 – 6
3 7 2
x y zúû û
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 ˆˆ ˆ2i j kü ü makes withy axisQ6. 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.
Q8. For what value ofý the line– 2 –1 – 3
9 –6
x y zþ þ
ÿ its perpendicular to tbe plane 3x – y – 2z = 7.
Q9. If a line makesð, �, � and with thex axis, y axis andz axis respectively. Find the value ofsin2ð + sin2� + sin2�.
Q10. Find the coordinate of the pt where line– 3 – 5 – 2
2 –3 5
x y z� � crosses theyz plane.
Q11. Write the direction ratio of the line– 2 2 – 5 1
2 –3 1
x y z�� � .
Q12. What is the equation of plane parallel to XOY plane and passing through (3, –4, 8)
Q13. Write the direction cosines of the perpendicular from the origin to the plane ˆˆ ˆ( )r i j k� ü ü = 10.Q14. Write the direction ratio of the normal of plane 2x + y + z = 7.
102
Q15. Write the value of� for which the plane 2x – 4y + 3z = 7 andx + 2y + �z = 18 are� toeach other.
Q16. Write the vector eq. of plane 2x + y + z = 7.
Q17. Write the equation of line || to the line–1 –1 – 2
2 3 4
x y z� � and passes through (0, 0, –1)
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 ofx, 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).
Q21. Write the angle between line– 2 1 – 3
3 –1 2
x y z and the plane 3x + 4y + z + 5 = 0.
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
1 2 3
x y z� � and
2 –1 1
–3 2 5
x y z� �
Q3. Find the shortest distance between the lines–1 – 2 – 3
2 3 4
x y z and
– 2 – 4 – 5
3 4 5
x y z
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 planex +
2y – z = 1 and 3x – 4y + z = 5.Q9. Find the distance between two parallel planes
2x – y + 3z + 4 = 06x – 3y + 8z – 3 = 0
Q10. Find the equation of plane which contains the line of intersection of the plane ˆˆ ˆ( 2 3 ) 4r i j k� � � �
and ˆˆ ˆ(2 ) –5r i j k� � � � and which it� to the plane ˆˆ ˆ(5 3 – 6 ) –8r i j k� � � .
Q11. Find the distance of the point A(–1, –5, –10) from the point of intersection of the line r =
ˆˆ ˆ(2 – 2 )i j k� + � ˆˆ ˆ(3 4 2 )i j k� � and the plane ˆˆ ˆ( – ) 5r i j k� � �
Q12. Find the distance of the point (1, –2, 3) from the plane x – y + z = 5 measured along a line
parallel to2 3 –6
x y z� �
Q13. Find the distance of the point A(–2, 3, –4) from the line2 2 3 3 4
3 4 5
x y z� � � parallel to the
plane 4x + 12y – 3z + 1 = 0.
103
Q14. Find a points on the line2 1 – 3
3 2 2
x y z� �� � at a distance of 5 units from the point (1, 3, 3).
Q15. Find the equations of the two lines through the origin which intersect the line– 3 – 3
2 1 1
x y z� �
at angle of3
�.
Q16. Show that the lines1 – 3 2
–3 2 1
x y z� �� � and
– 7 7
1 –3 2
x y z�� � are coplanar. Also find the equation
of plane containing them.
Q17. Show that the line ˆˆ ˆ(2 – 2 3 )r i j k ! + ˆˆ ˆ( – 4 )i j k" # is || to the plane .( 5 ) 5r i j k$ $ % . Also
find the distance between them.
Q18. Find the value of& so that the lines1– 7 –14 5 –10
3 2 11
x y z' '
( and
7 – 7 – 5 6 –
3 1 5
x y z) )
* are+
to each other.
Q19. Show that the plane whose vector equation is ˆˆ ˆ( 2 – )r i j k, - = 3 contains the line whose vector
equation is ˆˆ ˆ ˆ ˆ( ) (2 4 )r i j i j k. / / 0 / / .
Q20. If the point (1, 1, P) and (–3, 0, 1) be equidistant from plane ˆˆ ˆ(3 4 –12 ) 13 0r i j k1 2 2 3 find value of P.
Long Answer Type Questions (6 Marks)
Q1. A line makes4, 5, 6, 7 with the four diagonals of a cube prove that cos24 + cos25 + cos26
+ cos27 =4
3.
Q2. Show that angles between any two diagonals of cube is cos–11
3.
Q3. If l1, m
1, n
1 and l
2, m
2, n
2 be the direction cosines of two mutually prependicular lines. Show
that direction cosines of the line+ to both of them are (m1n
2 – m
2n
1) (n
1p
2 – n
2p
1) (l
1m
2 – l
2m
1).
Q4. Find the foot of the perpendicular from the point (0, 2, 3) on the line3 –1 4
5 2 3
x y z� �� � .
Also find the length of perpendicular.
Q5. Find the image of the point (1, 6, 3) on the line–1 – 2
1 2 3
x y z) )
Q6. Prove that the lines– 4 3 1
1 –4 7
x y z� �� � and
–1 1 10
2 –3 8
x y z8 89 9 intersect. Also find the cordinates
of their point of intersection.Q7. Find the shortest distance between the lines
ˆˆ ˆ(1 ) (2 – ) (1 )r i j k: - ; - ; - - ;
ˆˆ ˆ2(1 ) – (1– ) (–1 2 )r i j k: - < < - - <
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.
104
Q10. Find the equation of a plane passing through the points (0, 0, 0) and (3, –1, 2) and || to the
line– 4 3 1
1 –4 7
x y z= => > .
Q11. Find the equation of the plane passing through the point (0, 7, –7) and containing the line
1 – 3 2
–3 2 1
x y z? ?@ @ .
AnswersVery Short Answer (1 Mark)
1. 6 2. ˆ ˆˆ ˆ ˆ ˆ(5 – 4 6 ) (3 7 2 )r i j k i j kA B B C B B
3.–1 1 1
, ,3 3 3
4.1 1 1
, ,3 3 3
D EF F FG HI J
5.1
cos2
K L 6.1
3
7. 2 2b cM 8. –3
9. 2 10.19 –11
0 ,2 2
N OP QR S
11. 4, –3, 2 12. z = 8
13.1 1 1
, ,3 3 3
T UV WX Y
14. (2, 1, 1)
15. Z = 2 16. ˆˆ ˆ.(2 ) 7r i j k[ [ \
17.1
2 3 1
x y z]^ ^ 18.
–1 –1 –1
2 1 1
x y z_ _
19.3 1
, ,02 2
` ab cd e
20. 1x y z
a b cf f _
21. –1 7sin
2 91
g hi jk l
Short Answer (4 Mark)
1.7 –5
0, ,2 2
m no pq r 2.
1 – 3 2
2 –7 4
x y z? ?@ @
105
3.1
64. 4x – 3y + 2z = 3
6. 5x – 6y + 7z = 20 7. 3x + 4y – 5z = 9
8. x + 2y + 5z = 0 9.5
14
10. ˆˆ ˆ.(33 45 50 )r i j ks s = 41 11. 13
11. 1 13.17
units2
12. (–2, –1, 3) and (4, 3, 7)
15.1 2 –1
x y zt t and
–1 1 –2
x y zu u
Hint : Give line– 3 – 3
2 1 1
x y zt t t v
Any pt on line (2w + 3, w + 3, w)DR of op (2w + 3 – 0, w + 3 – 0, w – 0)
line op makes3
x with line pq
cos3
x = 2 2 2 2 2 2
2(3 2 ) 1(3 ) 1
2 1 1 (3 2 ) (3 )
y z y y z y z {
y y y z y y z y z| 2
6 9
6 6 18 18
} ~
} ~ } ~
w = –1, or –2 eq of requred line– 0 – 0 – 0
1– 0 2 – 0 –1– 0
x y z� � and
– 0 – 0 – 0
–1– 0 1– 0 –2 – 0
x y z� �
1 2 –1
x y zt t ,
–1 1 –2
x y zu u
16. x + y + z = 0 17.10
3 3 unit
18. w = 7 20. P = 1,7
3
Answer (6 Mark)
3. Let l, m, n be the direction cosines of the line� to each one of given line, thenll
1 + mm
1 + nn
1 = 0
ll2 + mm
2 + nn
2 = 0
1 2 2 1–
l
m n m n =
1 2 2 1–
m
n l n l =
1 2 2 1–
n
l m l m=
2 2 2
21 2 2 1( – )
l m n
m n m n
� � =
1
sin�
21 2 2 1( – )m n m n = sin� = sin
2
� = 1
Hence direction cosines of the line are(m
1n
2 – m
2n
1), (n
1l2 = n
2l1), (l
1m
2 – l
2m
1)
106
4. Gen pt on the line5� – 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 line5(5� – 3) + 2(2� – 1) + 3(3� – 7) = 0
� = 1Putting � = 1 in (1)We get (2, 3 – 1) as foot of �
length of� = PN = 2 2 2(2 – 0) (3 – 2) (–1– 3)� � = 21 unit
5. Fing foot of ^ by using above method (mentioned in Q No 4) we get (1, 3, 5) since B pt isimage of pt A P is the mid pt of AB
11
2
� ��
63
2
� ��
35
2
� ��
� = 1 � = 0 � = 7
6. Give line will intersect at (5, –7, 6)
7.3 2
2 unit
8. Eq. of line PN (line PN is� to plane)
– 7 –14 – 5
2 4 –1
x y z� � � �
Gen pt on line 2� + 7, 4� + 4, –� + 5If 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)
Also find length of� = PN = 2 2(7 –1) (14 – 2)5(5 – 8)� = 3 219. (–3, 5, 2)
Hint : Find foot of � by using above method.(Mentioned in question No 8.N is the mid pt of PM
11
2
� ��
34
2
� ��
43
2
� ��
� = –3 � = 5 � = 210. x – 19y – 11z = 011. x + y + z = 0
107
Teaching points
l Conditional probability: IfA andB are two events associated with any random experiment, thenP(A/B) repersents the probability of occurence of event-A knowing that event B has already occurred
P(A/B) =( )
, ( ) 0( )
P A BP B
P B
��
P(B)� 0, means that the events should not be impossible
P(A� B) = P(A andB) = 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 eventA andB are associated with any random experiment and the occurrence of one depends on the other
then P(A� B) = P(A) × P(B/A) whereP(A) � 0
l when the occurrence of one does not depends on the other then these events are said to be independentevents.
Here P(A/B) = P(A) or P(B/A) = P(B)
� P(A� B) = P(A) × P(B)
l Theorem on total probability: IfE1, E
2, E
3,...E
n be a partition of sample space andE
1, E
2 ... E
n all has
non-zero probability. A be any event associated with sample spaceS, then
P(A) = P(E1). P(A/E
l) + P(E
2). P(A/E
2)+...+P(E
n). P(A/E
n)
l Baye’s theorem: Let S be the sample space andE1, E
2...E
n ben mutually exclusive and exhaustive
events associated with a random experiment. IfA is any event which occurs withE1, or E
2 or ...E
n,
then
P(Ei/A) =
1
( ) ( )
( ) ( )
i in
i ii
P E P A E
P E P A E�
l Random variable: It is real valued function whose domain is the sample space of a random experimentand whose range is a subset of real numbers.
l Probability distribution: It is a system of numbers of random variable (x), such that
X: X1
X2
X3
Xn
P(X1): P(X
1) P(X
2) P(X
3)... P(X
n)
Where P(X1) > 0 and
1
( ) 1n
iP X��
�
Probability11
108
l Mean or expectation of a random variables (x) denoted byE(x)
E(x) = � =1
( ) 1n
ii
P X�
l Variance ofX denoted by var (x) or a¡x2 and
var(x) = ¡x2 = 2 2
1 1
( ) ( ) ( ) ( )n n
i i i ii l
X P X X P X¢ ¢
£ ¤ ¥ ¥ ¤
l The non-negative number var( )x x¦ § is called standard deviation of random variableX.
l 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.
l Binomial Distribution:
P(X = r) = n n r rrC q p¨ , wherer = 0, 1, 2,...n
p = prob. of success
q = prob. of Failure
n = total no. of trials
r = value of random variable.
l Recurrence formula for Binomal Distribution:
P(r + 1) = . ( ).1
n r pp r
r q
©
ª
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 andP(A/B) = 0.4, then findP(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)?
Q6. Two students A and B solve a problem independently with probabilities1
3 and
1
4 respectively..
What is the probability that problem is not solved?
109
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 isx 0 1 2
P(x) K 2K 3KWhat is the value ofK.
Q9. GivenP(A) = 1/4,P(B) = 2/3 andP(A¬ B) = 3/4. Are the events independent?
Q10. A andB are two events such thatP(A) = 0.3 and ( ) 0.8P A B ® . If AA andB are independent events.
FindP(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 probabilitythat 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 byone 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 studentsB solving aquestion 3/4. Assuming the two events "A can solve the question" and "B can solve that question" areindependent, 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 are1 1 1
, ,2 3 4
. What is
the probability that the problem will be solved?
Q15. A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases they arelikely to contradict each other in stating the same fact.
Q16. 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 probabilitythat it is from the first box?
Q17. 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?
Q18. A man takes a step forward with probability 0.4 and backward with probability 0.6 find the probabilitythat at the end of eleven steps he is one step away from the starting point.
Q19. A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by onewithout replacement, from the box. Find the probability distribution of the number of defectivebulbs.
Q20. A coin is biased so that the head is 3 times as likely to occurs as a tail if the coin is tossed twice, findthe probability distribution for the number of tails.
Q21. There are three coins, one is a two headed coin, another is a biased coin that comes up head 75% ofthe time and third is an unsbiased coin. One of the three coins is chosen at random and tossed. If itshows a head. What is the probability that it was a two headed-coin9
Q22. Two cards are darwn from a pack of 52 cards at random and kept out. Then one card is drawn fromthe remaining 50 cards. Find the probability that it is an ace.
110
Q23. A random variablex 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 ofK, (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 thatthe second dice always exhibits 4.
Q25. A machine operates of all of its three components function. The probability that the first componentfails during the year is 0.14, the probability that the second component fails is 0.10 and the probabilitythat the third component fails is 0.05. What is the probability that the machine will fail during theyear?
Long Answer Type Questions (6 Marks)
Q26. A company has two plants to manufacture bicylces. The first plant manufactures 60% of the bicyclesand 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 standardquality. Find the probability that it comes from the second plant.
Q27. In a bolt factory, MachniesA, B andC and manufacture 60%, 25% and 15% of the total outputrespectively. Of the total output 1 %, 2% and 1 % are defective bolts respectively manufactured bymachinesA, B andC. 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 sufferingfrom it is 0.99. The probability that the doctor diagnosis incorrectly that a perosn has T.B. on thebasis of x-ray is 0.001. In a certain city. 1 in 1000 persons suffers from T.B. a person is sleeted atrandom 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 ofnumbers 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. Therespective probabilities of their appointment are 0.5,0.3,0. The probabilities that research facilitieswill be enhanced by these people if they are appointed are 0.3, 0.7, and 0.8 respectively. If theresearch facilities are enhanced. What is the probability that it was due to the appointment ofC ?
Q31. A pair of dice is rolled twice. Let x denote the number of times, a total of 9 is obtained. Find the meanand 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 theprobability 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 twoconsecutive 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 tospade. 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 aretransferred from bag A to bag B and then a ball is drawn from bag B. The ball so drawn is found to bered in colour. Find the probability that the transferred ball were both black.
111
Answers
1.7
82.
2
33. (0.3)3
4.2
75. 0.65 6.
1
2
7.2
218.
1
69. Yes
10.2
711.
2
66312.
71
91
13.9
2814.
3
415. 42%
16.40
6117.
5
618. 462 × (.24)5 = 0.3678
19. X 0 1 2 3
P(X)84
429
210
429
120
429
15
429
20. X 0 1 2
P(X)9
16
6
16
1
16
21.4
922.
1
1323. (i) K =
1
10 (ii ) 0.6 (iii ) 0.8
24.1
625. .2467 26.
3
7
27. Machine A. 28.110
22129.
15
46
30.4
1331.
2 16,
9 932.
3
5
33.2
534.
22
42535.
4
17
112
Hints
12. Require probability =1 – P (balls are of same colour)
=2 1 3 2 5 4 4 3
114 13 14 13 14 13 14 13± ²
³ ³ ³ ³´ µ¶ ·
=71
9118. Let X denote the number of steps taken forward
¸ Required probability =P(x = 5) +P(x = 6)
=11 5 6 11 5 6
5 6(.4) (.6) (.6) (.4)C C
22. Events A1 = drawn cards are both aces
A2 = drawn cards are both non aces
A3 = drawn cards are both one ace and one non aceB
3 = Ace is drawn from 50 cards
¸ P(B) = 1 1 2 2 3 3( ) ( / ) ( ) ( / ) ( ) ( / )P A P B A P A P B A P A P B A
=4 3 2 48 47 4 4 48 3
252 51 50 52 51 50 52 51 50
¹ º ¹ º ¹ º» » » » » » »¼ ½ ¼ ½ ¼ ½¾ ¿ ¾ ¿ ¾ ¿
=1
1332. Let event A
1 = the bag contains 2 white and 2 non white balls
A2 = the bag contains 3 white and 1 non white balls
A3 = the bag contains 4 white balls
B = two white balls are drawn from the bag.Required probability =P(A
3/B)
=
11 33
1 1 1 1 1 513 6 3 2 3
ÀÁ
À À À
[Use Baye’s thon]
33. Let event A1 = letter has come from TATA NAGAR
Let event A2 = letter has come from KOLKATA
B = two consecutive letters visible are TA.
P(A1) =
1
2P(A
2) =
1
2
P(B/A1) =
1
4P(B/A
2) =
1
6Required prob. =P(A
2/B)
=
1 122 6
1 1 1 1 52 4 2 6
ÂÃ
 Â
113
35. Let event A1 = balls transfered are both red
A2 = balls transfered are one red and one black
A3 = balls transfered are both back
B = Red ball is drown form bag B after transfer
P(A1) =
3 2
7 6Ä P(A
2) =
3 42
7 6Å Å P(A
3) =
4 3
7 6Ä
P(B/A2) =
6
11P(B/A
2) =
5
11P(B/A
3) =
4
11
114
Teaching-Learning Points
l Linear programming is a technique to find optimal value (maximum or minimum) of a linear functionof a number of variables (sayx andy) subject to a number of linear inequalities in the variablesinvolved and the variables take non negative values only.
l A problem which seeks to maximise or minimise a linear function is called optimisation problem orlinear programming problem.
l Linear function of the formz =ax +by wherea, b are constants, which is to be maximized or minimizedis called objective function.
l Linear inequalities in the variables of a linear programming problem are called constraints.
l The common region determined by all the constraints including non negative constraintsx, y Æ 0 of alinear programming problem is called the feasible region. The region other than feasible region iscalled infeasible region.
l The points within and on the boundary of the feasible region represent feasible solutions. Any pointout the feasible region is called infeasible solution.
l Any point in the feasible region that gives optimal value (maximum or minimum) of the objectivefunction is called an optimal solution.
l Optimal value of an objective function must occurs at a corner point of the feasible region.
l A feasible region is said to be bounded if it can be enclosed within a circle. Otherwise, it is calledunbounded.
l If a feasible region is bounded, then the objective function has both a maximum and a minimum valueand each of these occurs at a corner point of the feasible region.
l 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.
l After evaluating the objective function z = ax + by at each corner point, let M and m respetivelydenote 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 ifz, provided the open half planedetermined byax + by > M has no point in common with the feasible region, otherwisez has nomaximum value. Similiarlym is the minimum value ofz, provided the open half plane determinedby ax + by < m has no point in common with the feasible region, otherwisez has no minimumvalue.
l 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.
Linear Programming12
115
Questions for Practice
Long Answer type Questions Carrying 6 Marks each
1. A house wife wishes to mix two types of foodx andy in such a way that the vitamin contents at the
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 x costs 60 per kg and
food y cost` 80 per kg. Foodx 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.
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.
116
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 manufacturedin a day to get maximum profit. Solve it as a linear programming problem. Also find maximum profit.
Answers1. Minimum cast = 160, at all points on the line segment joining8,0
3Ç ÈÉ ÊË Ì
and1
2,2
Í ÎÏ ÐÑ Ò
.
2. Ten bags of rice only and maximum profit =` 1100.
3. First type = 4, second type = 2, and maximum profit =` 200.
4. Maximum profit =` 14000 for A = 200 units and B = 400 units.
5. From A : 10, 50, 40 units and from B : 50,0,0 units minimum cost =` 10.
6. Type A toys = 15, type B toys = 30, maximum profit =` 262.50.