Download Maths Study Material – 2015-16

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Study Material
Class XII - Mathematics
2015-2016
2
KENDRIYA VIDYALAYA SANGATHAN
ZONAL INSTITUTE OF EDUCATION AND
TRAIING, MYSORE
Study Material
Class XII – Mathematics
2015-2016
Prepared/Revised by
Mr. S.K.Sivalingam
PGT (Maths), Faculty, KVS ZIET Mysore
Acknowledgement
श्री संतोष कुमार मल्ल, आईएएस
आयक्
ु त
Shri. Santosh Kumar Mall, IAS
Commissioner, KVS
श्री जी. के. श्रीवास्तव, आईएएस
अपर आयुक्त (प्रशासन)
Shri. G.K. Srivastava, IAS
Additional Commissioner (Administration)
श्री यू एन खवारे
अपर आयुक्त (शैक्षिक)
Shri. U N Khaware
Additional Commissioner (Academics)
डॉ. शचीकाांत
सांयुक्त आयुक्त (प्रशशिण)
Dr. Shachi Kant
Joint Commissioner (Training)
श्री ए. अरूमग
ु म
सांयुक्त आयुक्त (ववत्त)
Shri. M Arumugam
Joint Commissioner (Finance)
डा. वी. ववजयलक्ष्मी
संयक्
ु त आयक्
ु त (शिक्षा)
Dr. V. Vijayalakshmi
Joint Commissioner (Academics)
डॉ. ई. प्रभाकर
सांयुक्त आयुक्त (काशमिक)
Dr. E. Prabhakar
Joint Commissioner (Personnel)
श्री. एस. विजयकुमार
संयक्
ु त आयक्
ु त(प्रशासन)
Shri. S Vijaya kumar
Joint Commissioner (Admn)
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FOREWORD
The seven PGTs working as members of faculty at KVS,ZIET Mysore - Mr. K
Arumugam (Physics), Mr.S. Kalusivalingam (Maths), Mr. M Reddenna (Geo.), Mr. S.
Murugan (History), Mr. Hari Shankar (Hindi), Mr. Joseph Paul (Econ.) and Mr. U.P
Binoy (English) prepared Study Materials for Class XII for the academic year 20152016 in their respective subjects.
All these study materials focus on some select aspects, namely;
 Gist of lessons/chapter
 Marking scheme (CBSE)
 Important questions
 Solved Question papers with value point
 Tips for scoring well in the examination
The above mentioned seven members of faculty at ZIET Mysore have put in a lot of
efforts and prepared the materials in a period of two months. They deserve
commendation for their single-minded pursuit in bringing out these materials.
The teachers of these subjects namely, Physics, Mathematics, Geography, History,
Hindi, Economics and English may use the materials in the month of January &
February 2016 for Pre-Board Examination revision and practice purposes. It is hoped
that these materials will help the students perform better in the forthcoming Board
Examinations.
The teachers are requested to go through the materials thoroughly, and feel free
to send their opinions and suggestions for the improvement of these materials to
[email protected].
Dr. E.T ARASU
Deputy Commissioner & Director
KVS, ZIET Mysore
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PREFACE
It is a matter of great pleasure that after receiving encouragement from our Deputy Commissioner
DR. E.T Arasu, I now present the thoroughly revised latest edition of STUDY MATERIAL OF CLASS XII
MATHEMATICS based on the latest syllabus and revised question paper pattern to be followed from 2015
onwards.
THE SALIENT FEATURES OF THIS STUDY MATERIAL ARE AS FOLLOWS;
 It covers the syllabus given by CBSE for the class XII Mathematics
 Gist of each chapter
 Marking scheme given by CBSE ( 2015)
 Two sets of solved question papers with value points
 Two sets of model question papers -to be answered
 Tips for pre-examination and during examination
 The material can be used for the purpose of revision
 All the concepts of the subject have been included in the material
I am sure this material will serve the purpose of helping students perform better in the Board Examination.
However, suggestions and comments from the teachers and the students for the improvement of this
material will be highly appreciated.
Place: Mysore
Date: 17/12/2015
S.K.Sivalingam
Faculty ZIET Mysore
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Index
S.No.
Particulars
Page
No.
1
Marking Scheme
7
2
Gist of Chapters – Chapters 1 to 13
10
3
4
5
Important model questions –
Chapters 1 to 13
Solved Question Paper with value points
– Sample Paper 1
Solved Question Paper with value points
– Sample Paper 2
60
88
105
6
Model Question Paper 1
115
7
Model Question Paper 2
118
8
Tips for scoring well
122
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MARKING SCHEME
Course Structure
Unit Topic
Marks
I.
Relations and Functions
10
II.
Algebra
13
III.
Calculus
44
IV.
Vectors and 3-D Geometry 17
V.
Linear Programming
6
VI.
Probability
10
Total
100
Unit I: Relations and Functions
1. Relations and Functions
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
Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions. Elementary
properties of inverse trigonometric functions.
Unit II: Algebra
1. Matrices
Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix,
symmetric and skew symmetric matrices. Operation on matrices: Addition and multiplication and
multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication.
Noncommutativity 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).
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2. Determinants
Determinant of a square matrix (up to 3 x 3 matrices), properties of determinants, minors, co-factors and
applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix.
Consistency, inconsistency and number of solutions of system of linear equations by examples, solving
system of linear equations in two or three variables (having unique solution) using inverse of a matrix.
Unit III: Calculus
1. Continuity and Differentiability
Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inverse
trigonometric functions, derivative of implicit functions. Concept of exponential and logarithmic functions.
Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions
expressed in parametric forms. Second order derivatives. Rolle's and Lagrange's Mean Value Theorems
(without proof) and their geometric interpretation.
2. Applications of Derivatives
Applications of derivatives: rate of change of bodies, increasing/decreasing functions, tangents and normals,
use of derivatives in approximation, maxima and minima (first derivative test motivated geometrically and
second derivative test given as a provable tool). Simple problems (that illustrate basic principles and
understanding of the subject as well as real-life situations).
3. Integrals
Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by
partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on
them.
Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof). Basic properties of
definite integrals and evaluation of definite integrals.
4. Applications of the Integrals
Applications in finding the area under simple curves, especially lines, circles/parabolas/ellipses (in standard
form only), Area between any of the two above said curves (the region should be clearly identifiable).
5. Differential Equations
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Definition, order and degree, general and particular solutions of a differential equation. Formation of
differential equation whose general solution is given. Solution of differential equations by method of
separation of variables solutions of homogeneous differential equations of first order and first degree.
Solutions of linear differential equation of the type:
dy/dx + py = q, where p and q are functions of x or constants.
dx/dy + px = q, where p and q are functions of y or constants.
Unit IV: Vectors and Three-Dimensional Geometry
1. Vectors
Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector.
Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a
vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of
a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and
application of scalar (dot) product of vectors, vector (cross) product of vectors, scalar triple product of
vectors.
2. Three - dimensional Geometry
Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of
a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a
plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane.
Unit V: Linear Programming
1. Linear Programming
Introduction, 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 (bounded and unbounded), feasible
and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).
Unit VI: Probability
1. Probability
Conditional probability, multiplication theorem on 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.
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GIST OF CHAPTERS
Chapter1. Relations and Functions
Introduction:Any set of ordered pairs (x,y) is called a relation in x and y. Furthermore,
• The set of first components in the ordered pairs is called the domain of the relation.
• The set of second components in the ordered pairs is called the range of the relation.
Relation: - Let A and B be two sets. Then a relation R from set A to Set B is a subset of 𝐴 × 𝐵.
Types of relations: Empty relation: A relation R in a set A is called empty relation, if no element of A is related to any element
of A i
Universal relation :A relation R in a set A is called universal relation, if each element of A is related to every
element of A, i.e., R = A × A.
Equivalence relation.: A relation R in a set A is said to be an equivalence relation if R is reflexive, symmetric
and transitive
A relation R in a set A is called
(i) reflexive, if (a, a) ∈ R, for every a∈A,
(ii) symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b∈ A.
(iii) transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b, c∈ A.
Function:
A function is a relation between a set of inputs and a set of permissible outputs with the property that each
input is related to exactly one output. An example is the function that relates each real number x to its
square x2.
A function is a relation for which each value from the set the first components of the ordered pairs is
associated with exactly one value from the set of second components of the ordered pair.
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Types of functions:
One-one (or injective):
A function f : X Y is defined to be one-one (or injective), if the images
of distinct elements of X under f are distinct, i.e., for every a, b in X, f (a) = f (b)
implies a = b. Otherwise, f is called many-one.
Horizontal line test:
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To check the injectivity of the function, f (x) =2x. Draw a horizontal line such that this line cuts the graph
only at one place. Such types of functions are known as one-one functions.
In this case where the line cuts the graph of a function at more than one place, the functions are not oneone.
Onto (or surjective):
A function f : X Y is said to be onto (or surjective), if every element of Y is the image of some element of X
under f, i.e., for every y in Y, there exists an element x in X such that f (x) = y.
Bijective function:
A function f : X Y is said to be one-one and onto (or bijective), if f is both one-one and onto.
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Composition of Functions and Invertible Function
Let f : A B and g : B C be two functions. Then the composition of f and g, denoted by gof, is defined as the
function gof: A → C given by gof(x) = g(f (x)), for all x A.
Inverse of a Bijective Function
Let f: A → B be a function. If, for an arbitrary x ∈ A we have f(x) = y ∈ B, then the function, g: B → A, given
by g(y) = x, where y ∈ B and x ∈ A, is called the inverse function of f.
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Binary Operations
A binary operation ∗ on a set A is a function ∗ : A × A → A. We denote ∗ (a, b) by a ∗ b.
Binary Operation: A binary operation ‘*’ defined on set A is a function from A × A → A. * (a, b) is
denoted by a * b.
Binary operation * defined on set A is said to be commutative iff a * b = b * a.
Binary operation * defined on set A is called associative iff a *(b * c) = (a*b)*c
If * is Binary operation on A, then an element e in A is said to be the identity element iff a * e = e
* a for every a i n A
Identity element is unique.
If * is Binary operation on set A, then an element b is said to be inverse of a ∈ A iff a * b = b * a = e
Inverse of an element, if it exists, is unique.
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Chapter 2. Inverse Trigonometric Functions
Inverse trigonometric functions or cyclometric functions - are the so-called inverse functions of the
trigonometric functions, when their domain are restricted to principal value branch to make the
trigonometric functions bijective. The principal inverses are listed in the following table.
Range of usual
principal value
(radians)
Range of usual principal
value
(degrees)
Name
Usual notation
Definition
Domain of x for
real result
arcsine
y = sin-1 x
x = sin y
−1 ≤ x ≤ 1
−π/2 ≤ y ≤ π/2
−90° ≤ y ≤ 90°
arccosine
y = cos-1 x
x = cos y
−1 ≤ x ≤ 1
0≤y≤π
0° ≤ y ≤ 180°
arctangent
y = tan-1 x
x = tan y
all real numbers
−π/2 < y < π/2
−90° ≤ y ≤ 90°
arccotangent
y = cot-1 x
x = cot y
all real numbers
0<y<π
0° < y < 180°
arcsecant
y = sec-1 x
x = sec y
x ≤ −1 or 1 ≤ x
0 ≤ y < π/2 or π/2 <
y≤π
0° ≤ y < 90° or 90° < y ≤
180°
arccosecant
y = csc-1 x
x = csc y
x ≤ −1 or 1 ≤ x
−π/2 ≤ y < 0 or 0 < y
≤ π/2
−90° ≤ y ≤ 0° or 0° < y ≤
90°
Let us take a function
also write
,
,
. The inverse of this function is defined as
, provided the function is both one-one and onto).
(We may
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Very important:
does not mean
. This is just a notation given to write Inverse Trigonometric Function.
We can easily verify
function
etc. Further any
does not exist, as
is not an image. Hence, the
is not one-one and onto.
Now consider the function
defined by
can be visualized from the above graph. Hence,
. It is both one-one and onto which
exists.
In fact, if we restrict the domain of the sin function to any one of the intervals
,
.. etc., keeping the range
Then
,
domain and range.
,
Every branch of the graph of
,
the function becomes one-one and onto.
is a function whose domain is
,
,
and range is any one of the intervals
.. etc., [please recall that if
corresponding to branches
gives a one–one and onto function. The graph in the branch
exists,
and
,
interchange their
,
,
called the principle value
branch.
We can also justify our answer from the chapter Trigonometric Equations and General Values.
has infinitely many solutions,
called the principle value.
,
,
,
… etc.. The value
is
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We may say that
. Also value of
is called the principle value.
Similarly we restrict the domains of other trigonometric functions to make it invertible the table given below
shows the domain and the range of the six basic inverse trigonometric function.
The values of domain and range indicated for different trigonometric, inverse trigonometric functions can be observed
from the graphs of trigonometric functions and their corresponding inverse trigonometric functions.
We can draw the graph of a inverse function of a function by taking image on line y = x.
Let us apply this to draw graph of y =
2. he graph of the function y =
and y =
.
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3. The graph of the function y = tanx and y =
Note: Whenever no branch of an inverse function is mentioned, we consider the Principal value branch.
PROPERTIES OF INVERSE CIRCULAR FUNCTIONS
1. (a)
(b)
(c)
2. (a)
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(b)
(c)
3. (a)
(b)
(c)
4. (a)
(b)
(c)
5. (a)
(b)
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Chapter 3. Matrices
Definition: A matrix A is defined as an ordered rectangular array of numbers in m rows and n columns.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Row matrix: A matrix can have a single row A [ a11 a12 a13 … a1n]
Column matrix: - A matrix can have a single column A =
Zero or null matrix: – A matrix is called the zero or null matrix if all the entries are 0.
Square matrix: - A matrix for which horizontal and vertical dimensions are the same (i.e., an
matrix).
Diagonal matrix: - A square matrix A is called diagonal matrix if aij = 0 for .
Scalar matrix: - A diagonal matrix A is called the scalar matrix if all its diagonal elements are equal.
Identity matrix :– A diagonal matrix A is called the identity matrix if aij = 1 for i = j , it is denoted by In.
Upper triangular matrix: - A square matrix A is called upper triangular matrix if aij = 0 for
Lower triangular matrix :- A square matrix A is called lower triangular matrix if aij = 0 for
 Matrix operations
1. Definition: Two matrices A and B can be added or subtracted if and only if their dimensions are the
same (i.e. both matrices have the identical amount of rows and columns.
2. Addition
If A and B are matrices of the same type then the sum is a matrixC =
obtained by adding the corresponding elements aij+bij i.e. A+B = C if aij+bij =cij
3. Matrix addition is commutative , associative and distributive over multiplication
A+B=B+A
A + (B + C) = (A+ B) + C
A (B + C) = AB + AC
(A+B)C= AC + BC
Subtraction
If A and B are matrices of the same type then the difference is a matrix D = obtained by
subtracting the corresponding elements aij - bij i.e. A - B = C if aij - bij =dij
4.
Equal matrices –Two matrices are said to be equal if they have the same order and their
corresponding elements are also equal i.e. A = B if aij = bij for all I, j .
5.
Scalar multiplication- If A and B are matrices of the same order and k, m are scalars then, scalar
multiplication is defined as kA=[kaij].
6.
K(A+B) = KA + KB
7.
(m+n) A = mA+ nA
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8.
(mk)A = m(kA) =k(mA)
9.
Matrix multiplication
Definition: When the number of columns of the first matrix is the same as the number of rows in the
second matrix then matrix multiplication can be performed.
Let A and B . Then the product of A and B is the matrix C, which has dimensions mxp. The ijth element of
matrix C is found by multiplying the entries of the ith row of A with the corresponding entries in the jth
column of B and summing the n terms. The elements of C are:
Note: AxB is not equal to BxA
10.
Transpose of Matrices: The transpose of a matrix is found by exchanging rows for columns i.e. Matrix A =
(aij) and the transpose of A is:AT=(aji) where j is the column number and i is the row number of matrix A.
11.
Symmetric matrix. For a symmetric matrix A = AT i.e. aij = aji
Skew -symmetric matrix. For a skew-symmetric matrix AT = - A i.e. aji = - aij.
Properties
 Note that the diagonal elements of the skew symmetric matrix are 0.
 A + AT is a symmetric matrix.
 A - AT is a skew - symmetric matrix.
 Every square matrix A can be expressed as a sum of a symmetric P and skew symmetric Q matrices
10.
Elementary transformation - Following elementary row or column transformations can be applied to
a matrix
 Interchange of any two rows or columns.
 Multiplication of any row or column by any non-zero number(scalar)
 The addition to the elements of any row or column the scalar multiples of any other row or column
 Working rule to find A-1 by elementary transformations
a) Write A = InA, apply elementary row transformations to both the matrices A on LHS and In on RHS
till you get In= BA. Then B is the required A-1
b) Write A = AIn, apply elementary column transformations to both the matrices A on LHS and In on
RHS till you get In= AB. Then B is the required A-1
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NOTE : Apply only one kind of transformations (row or column ) in all the steps in one particular
answer.
Chapter 4 Determinants
Every square matrix A is associated with a number, called its determinant and it is denoted by det
(A) or |A| .
Only square matrices have determinants. The matrices which are not square do not have
determinants
(i) First Order Determinant If A = [a], then det (A) = |A| = a
(ii) Second Order Determinant
|A| = a11a22 – a21a12
(iii) Third Order Determinant
Evaluation of Determinant of Square Matrix of Order 3 by Sarrus Rule
then determinant can be formed by enlarging the matrix by adjoining the first two
columns on the right and draw lines as show below parallel and perpendicular to the diagonal.
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The value of the determinant, thus will be the sum of the product of element. in line parallel to the
diagonal minus the sum of the product of elements in line perpendicular to the line segment. Thus, Δ
= a11a22a33 + a12a23a31 + a13a21a32 – a13a22a31 – a11a23a32 – a12a21a33.
Properties of Determinants
(i) The value of the determinant remains unchanged, if rows are changed into columns and columns
are changed into rows e.g., |A’| = |A|
(ii) In a determinant, if any two rows (columns) are inter changed, then the value of the determinant is
multiplied by - 1.
(iii) If two rows (columns) of a square matrix A are proportional, then |A| = 0.
(iv) |B| = k |A|, where B is the matrix obtained from A, by multiplying one row (column) of A by k.
n
(v) |kA| = k |A|, where A is a matrix of order n x n.
(vi) If each element of a row (or column) of a determinant is the sum of two or more terms, then the
determinant can be expressed as the sum of two or more determinants, e.g.,
(vii) If the same multiple of the elements of any row (or column) of a determinant are added to the
corresponding elements of any other row (or column), then the value of the new determinant remains
unchanged, e.g.,
(viii) If each element of a row (or column) of a determinant is zero, then its value is zero. (ix) If any
two rows (columns) of a determinant are identical, then its value is zero.
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(x) If each element of row (column) of a determinant is expressed as a sum of two or more terms,
then the determinant can be expressed as the sum of two or more determinants.
Important Results on Determinants
(i)
|AB| = |A||B| , where A and B are square matrices of the same order.
(ii)
n
n
(ii) |A |=|A|
(iii) If A, B and C are square matrices of the same order such that ith column (or row) of A is the sum of i
th columns (or rows) of B and C and all other columns (or rows) of A, Band C are identical, then |A|
=|B| + |C|
(iv) |In| = 1,where In is identity matrix of order n
(v) |On| = 0, where On is a zero matrix of order n
(vi) If Δ(x) be a 3rd order determinant having polynomials as its elements.
(a) If Δ(a) has 2 rows (or columns) proportional, then (x – a) is a factor of Δ(x).
2
(b) If Δ(a) has 3 rows (or columns) proportional, then (x – a) is a factor of Δ(x).
(vii) A square matrix A is non-singular, if |A| ≠ 0 and singular, if |A| =0.
(viii) Determinant of a skew-symmetric matrix of odd order is zero and of even order is a non- zero
perfect square.
(ix) In general, |B + C| ≠ |B| + |C|
(x) Determinant of a diagonal matrix = Product of its diagonal elements
(xi) Determinant of a triangular matrix = Product of its diagonal elements
Minors and Cofactors
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then the minor Mij of the element aij is the determinant obtained by deleting the i
row and jth column.
i+j
The cofactor of the element aij is Cij = (- 1)
Mij
Adjoint of a Matrix - Adjoint of a matrix is the transpose of the matrix of cofactors of the give matrix, i.e.,
Properties of Minors and Cofactors
(i) The sum of the products of elements of any row (or column) of a determinant with the
cofactors of the corresponding elements of any other row (or column) is zero
(ii) The sum of the product of elements of any row (or column) of a determinant with the
cofactors of the corresponding elements of the same row (or column) is Δ
Applications of Determinant in Geometry
Let three points in a plane be A(x1, y1), B(x2, y2) and C(x3, y3), then Area =
= 1 / 2 [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
SOLVING SYSTEMS OF EQUATIONS USING INVERSE MATRIX METHOD
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DEFINITION: A system of linear equations is a set of equations with n equations and n unknowns, is of the
form of
The unknowns are denoted by x1,x2,...xn and the coefficients
(a's and b's above) are assumed to be given. In matrix form the system of equations above can be written
as:
which can be expressed in matrix equation as AX =B
By pre-multiplying both sides of this equation by A-1 gives:
A-1 (AX) = A-1 B
STEPS
i.
Evaluate . Refer the note given below.
ii.
Evaluate the cofactors of elements of A.
iii.
Form the adjoint of A as the matrix of cofactors
iv.
Calculate inverse of A and X.
NOTE - From the above it is clear that the existence of a solution depends on the value of the
determinant of A. There are three cases:
1. If the then the system is consistent with unique solution given by
2. If (A is singular) and adj A .B 0 then the solution does not exist. The system is inconsistent.
3. If (A is singular) and adj A .B = 0 then the system is consistent with infinitely many solutions.to find
these solutions put z = k in two of the equations and solve them by matrix method.
For homogeneous system of linear equations, AX = 0 (B = 0)
1. If the then the system is consistent with trivial solution x = 0, y = 0, z = 0
2. If (A is singular) and adj A .B 0 then the solution does not exist. The system is inconsistent.
3. If (A is singular) and adj A .B = 0 then the system is consistent with infinitely many solutions. to
find these solutions put z = k in two of the equations and solve them by matrix method.
27
Chapter 5 Continuity and Differentiability
Definition
A function f is said to be continuous at x = a if
A function is said to be continuous on the interval [a, b] if it is continuous at each point in the interval.
Definition of derivative
: If y = f(x) then y1 =
A function f is differentiable if it is continuous.
When LHD & RHD both exist and are equal then f is said to be derivable or differentiable.
Parametric differentiation: if y =f(t), x= g(t) then,
Logarithmic differentiation: If y = f(x)g(x) then take log on both the sides.
Write logy = g(x) log[f(x)]. Differentiate by applying suitable rule for differentiation.
If y is sum of two different exponential function u and v, i.e. y = u + v. Find by using logarithmic
differentiation separately then evaluate.
Higher Derivatives: The derivative of first derivative is called as the second derivative. It is denoted by y2 or
2
y’’ or d y2
dx
28
FORMULAE
2.
d n
x   nx n 1

dx
4.
d
 x  1
dx
d  1  1
 
dx  x  x 2
6.
d
dx
7.
d x
 a   a x log a
dx
8.
d x
e   ex
dx
9.
d
1
 log x  
dx
x
10.
d
 sin x   cos x
dx
1.
d
c  0
dx
3.
d  1

dx  x n
5.
 n
  n 1
 x
 x   21x
11.
d
 cos x    sin x
dx
12.
d
 tan x   sec2 x
dx
13.
d
 cos ec x    cos ec x cot x
dx
14.
d
 s ec x   sec x tan x
dx
15.
d
 cot x   cosec 2 x
dx
16.
d
1
sin 1 x  

dx
1  x2
17.
d
1
cos 1 x  

dx
1  x2
18.
d
1
tan 1 x  

dx
1  x2
19.
d
1
cot 1 x  

dx
1  x2
20.
d
1
sec 1 x  

dx
x x2 1
21.
d
1
cos ec 1 x  

dx
x x2 1
MEAN VALUE THEOREMS
29
ROLLE’S THEOREM
STATEMENT Let f be a real valued function defined on the closed interval [a, b] such that
1. It is continuous on the closed interval [a, b].
2. It is differentiable on the open interval (a, b).
3. f(a) = f(b).
Then, there exist a real number c
GEOMETRICAL INTERPRETATION OF ROLLE’S THEOREM
Let f(x) be a real valued function defined on [a, b] such that the curve y=f(x) is a continuous curve between
points A (a, f(a)) and B(b, f(b)) and it is possible to draw a unique tangent at every point on the curve y=f(x)
between A and B. Also, there exists at least one ordinates at the end points of the interval [a, b] are equal. Then
there exists at least one point (c, f(c)) lying between A and B on the curve. Let the curve is increasing at A. Now
f(a) = f(b) that means curve has to decrease from some point. Since it is continuous and differentiable so, it must
take smooth turn from somewhere (without forming edge) at that point the tangent parallel to X- Axis i.e. at
that point f’(x) = 0. Similarly it can be shown easily if starts decreasing at A then also there exist at least one
point in the given interval s=where f”(x) = 0. If it is neither increasing nor decreasing i.e. a parallel line to the Xaxis then, there will be infinite no. of points where f”(x) = 0.
Working Rule:
1.
We first check whether f(x) satisfies conditions of Rolle’s Theorem or not.
2.
Differentiate the function
3.
Equate the derivatives with Zero to get x.
4.
If x belongs to given interval then, Rolle’s Theorem verified.
The following rules may be helpful while solving the problem.
1.
A polynomial function is everywhere continuous and differentiable
30
2.
The exponential function, sine and cosine function are everywhere continuous and differentiable.
3.
Logarithmic function is continuous and differentiable in its domain.
4.
Tan x is not continuous
Chapter 6 Application of Derivatives
Motion in a straight Line
Suppose a particle is moving in a straight line
Take ‘O’ as origin in time ‘t’ the position of the particle be at A where OA= s and in time t+∆t the position of the particle be
at B where OB=s+∆s clearly the directed distance of the particle from ‘O’ is function of time ‘t’
s=f (t)
s+∆s=f (t+∆t)
Algorithm:
1. Solving any problem first, see what are given data.
2. What quantity is to be found?
3. Find the relation between the point no. (1) & (2).
4. Differentiate and calculate for the final result.
APPROXIMATIONS, DIFFERENTIALS AND ERRORS
 Absolute error  Relative error
 Percentage error - Approximation 1. Take the quantity given in the question as y +k = f(x + h)
2. Take a suitable value of x nearest to the given value.
31
3. Calculate y= f(x) at the assumed value of x.
4. Using differential calculate
5. find the approximate value of the quantity asked in the question as y +k , from the values of y and
evaluated in step 3 and 4.
Tangents and normals

Slope of the tangent to the curve y = f(x) at the point (x0,y0) is given by m.
 Equation of the tangent to the curve y = f(x) at the point (x0,y0) is (y - y0) = m (x – x0)
 Slope of the normal to the curve y = f(x) at the point (x0,y0) is given by -1 / m
 Equation of the normal to the curve y = f(x) at the point (x0,y0) is (y - y0) = ( - 1 /m) (x – x0)
 To curves y = f(x) and y = g(x) are orthogonal means their tangents are perpendicular to each other at the
point of contact
INCREASING AND DECREASING FUNCTION
STRICTLY INCREASING FUNCTION: A function f(x) is said to be a strictly increasing function on (a, b) if
x1<x2 => f(x1)<f(x2) for all x1, x2(a, b)
Thus, f(x) is strictly increasing on (a, b) if the values of f(x) increase with increase in the value of x.
Graphically, f(x) is increasing on (a, b) if the graph y=f(x) moves up as x moves to the right. The graph in fig is the graph of
strictly increasing on (a, b).
STRICTLY DECREASING FUNCTION: A function f(x) is said to be a strictly increasing function on (a, b) if
32
x1<x2 => f(x1)>f(x2) for all x1, x2(a, b)
Thus, f(x) is strictly increasing on (a, b) if the values of f(x) decrease with increase in the value of x. Graphically, f(x) is
decreasing on (a, b) if the graph y=f(x) moves down as x moves to the right. The graph in fig is the graph of strictly decreasing
on (a, b).
NECESSARY AND SUFFICIENT CONDITION FOR MONOTONICITY OF FUNCTION
In this section we intend to see how we can use derivative of a function to determine where it is increasing and where it is
decreasing.
Necessary condition: if f(x) is an increasing function on(a, b) then the tangent at every point on the curve y=f(x) makes an
acute angle with the positive direction of x- axis.
It is evident from fig that if f(x) is a decreasing function on (a, b), then tangent at every point on the curve y=f(x) makes and
obtuse angle with the positive direction of x- axis.
Thus, f’(x) > 0(<0) for all is the necessary condition for a function f(x) to be increasing (decreasing) on a given interval (a,
b). in other words, if it is given f(x) is increasing (decreasing on (a, b), then we can say that f’(x)>0(<0).
SUFFICIENT CONDITION
Theorem Let f be a differentiable real function defined on an open interval (a, b).
(a) If f’(x) > 0 for all , then f(x) increasing on (a, b)
(b) If f’(x) < 0 for all , then f(x) decreasing on (a, b)
ALGORITHM FOR FINDING THE INTERVAL IN WHICH A FUNCTION IS INCREASING OR DECREASING.
33
Let f(x) be a real function. To find the intervals of monotonicity of f(x) we proceed as follows:
Step 1 find f’(x).
Step II Put f’(x)>0 and solve this inequation.
For the values of x obtained in step II f(x) is increasing and for the remaining points in its domain it is decreasing.
Or
1. Calculate f’(x) =0 for critical points
2. Let c1, c2, c3 are the roots of f’(x) = 0.
3. Cut the no line at c1, c2, c3
4. Write the interval in table
5. Consider any point in the interval
6. See the sign of f’(x) in that interval and accordingly determine the function is increasing or decreasing in that
interval.
Maximum Let f(x) be the real function define on the interval I. then f(x)is said to have the maximum value in I, if there
exists a point a in I such that f(x)<f(a) for all x
in such a case, the number f(a) is called the maximum value of f(x) in the interval I
and the point a is called a point of maximum value of f in the interval I.
Minimum Let f(x) be the real function define on the interval I. then f(x) is said to have the minimum value in I, if there
exists a point a I such that
f(x)>f(a) for all x
In such a case, the number f(a) is called the minimum value of f(x) in the interval I and
the point a is called a point of minimum value of f in the interval I.
LOCAL MAXIMA AND LOCAL MINIMA
1.
2. Local maximum A function f(x) is said to attain a local maximum at x=a if there exists a neighborhood (a-,a+-)
of a such that
f(x)<f(a) for all x in (a-,a+)
f(x)-f(a)<0 for all x in (a-,a+)
In such a case, f(a) is called the local maximum value of f(x) at x=a.
3. Local minimum A function f(x) is said to attain a local minimum at x=a if there exists a
Neighborhood (a-, a+-) of a such that
f(x)>f(a) for all x in (a-,a+)
34
f(x)-f(a)>0 for all x in (a-,a+)
The value of the function at x= a i.e. f(a) is called the local minimum value of f(x) at x=a.
The points at which attains either the local maximum or local minimum values are called the extreme values of
f(x).
First Derivatives Test for finding the Local Extremum
Let y=f(x) be a function defined on the interval I and let f be derivable at an interior point c of I. Let f have an extreme
value at x=c,
Case I. Let f have a local maximum value at x=c, then f is an increasing function in the left nbd of x=c .
Thus, f’(x) changes continuously from +ve to –ve as increases through c.
Case II. Let f have a local minimum value at x=c, then f is an increasing function in the left nbd of x=c
f’(x) changes continuously from +ve to +ve as increases through c.
Working rule for first derivative test
Find the sign of f’(x) when x is slightly < c and when x is slightly >c.
If f’(x) changes sign from +ve to –ve as x increases through c, then f has a local maximum value at x=c.
If f’(x) changes sign from -ve to +ve as x increases through c, then f has a local minimum value at x=c.
Second Derivative Test for Finding Local Maxima and Local Minima
Suppose a quantity y depends on the another quantity x in a manner shown in fig. it becomes maximum at x1 and
minimum at x2. At these points the tangent to the curve is parallel to the x-axis and hence its slope is 0.
Maxima and minima by using second derivative test
35
Just before the maximum the slope is positive, at the minimum it is zero and just after the maximum it is negative.
Thus decreases at a maximum and hence the rate of change of is negative at a maximum.
Minima
Similarly, at a minimum the slope changes from negative to positive.
Hence with increases of x. the slope is increasing that means the rate of change of slope w.r.t x is positive.
Working Rule:
1.
Find f’(x).
2.
Solve f’(x) = 0 within the domain to get critical point let one of the value of x = c.
3.
Calculate f’’(x) at x = c.
4.
If f’’(c) > 0 then, f(x) is minimum at x = c, and if f’’(c) < 0, then f(x) is maximum at x= c.
Maxima
Similarly, at a maximum the slope changes from positive to negative.
Working Rule:
5.
Find f’(x).
6.
Solve f’(x) = 0 within the domain to get critical point let one of the value of x = c.
7.
Calculate f’’(x) at x = c.
8.
If f’’(c) < 0 then, f(x) is minimum at x = c, and if f’’(c) < 0, then f(x) is maximum at x= c.
Application of Maxima and Minima to problems
Working rule
(i)
In order to illustrate the problem, draw a diagram, if possible. Distinguish clearly between the variable and
constants.
(ii)
If y is the quantity to be maximized or minimized, express y in terms of a single independent variable with the
help of given data.
(iii)
Get dy/dx and d2y/dx2, then equate dy/dx=0 get the value of x.
(iv)
Determine the sign of d2y/dx2 at x and proceed.
36
Chapter 7 Integration
Integration is a way of adding slices to find the whole. Integration can be used to find areas,
volumes, central points and many useful things.
The Chain Rule tells us that derivative of g(f(x)) = g'(f(x))f'(x). But what about going the other
way around? What happens if you want to integrate g'(f(x))f'(x)? Well, that's what the
"reverse chain rule" is for. As you can see, a lot of integrals you'll run into can be solved this
way. It is also another way of doing substitution without having to substitute.
Trigonometric substitution
Another substitution technique where we substitute variables with trigonometric functions.
This allows us to leverage some trigonometric identities to simplify the expression into one
that it is easier to take the anti-derivative of.
Division and partial fraction expansion
When you're trying to integrate a rational expression, the techniques of partial fraction
expansion and algebraic long division can be *very* useful.
Integration by parts
When we wanted to take the derivative of f(x)g(x) in differential calculus, we used the
product rule. In this tutorial, we use the product rule to derive a powerful way to take the
anti-derivative of a class of functions--integration by parts.
If u and g are two functions of x then the integral of product of two functions = 1st function integral of the product of the derivative of 1st function and the integral of the 2nd function
Write the given integral where you identify the two functions u(x) and v(x) as the 1st
and 2nd function by the order
I – inverse trigonometric function
L – Logarithmic function
A – Algebraic function
T – Trigonometric function
E – Exponential function
37
Note that if you are given only one function, then set the second one to be the
constant function g(x)=1. integrate the given function by using the formula
DEFINITE INTEGRAL AS LIMIT OF A SUM
Definite integral is closely related to concepts like antiderivative and indefinite integrals. In
this section, we shall discuss this relationship in detail.
Definite integral consists of a function f(x) which is continuous in a closed interval [a, b] and
the meaning of definite integral is assumed to be in context of area covered by the function
f from (say) ‘a’ to ‘b’.
An alternative way of describing
is that the definite integral
is a
limiting case of the summation of an infinite series, provided f(x) is continuous on [a, b]
The converse is also true i.e., if we have an infinite series of the above form, it can be
expressed as a definite integral.
The Fundamental Theorem of Calculus justifies our procedure of evaluating an
antiderivative at the upper and lower limits of integration and taking the difference.
38
Fundamental Theorem of Calculus
Let f be continuous on [a b]. If F is any antiderivative for f on [a b], then
= F(b)−F(a)
Properties
Property 1:
Property 2:
Property 3:
Property4:
Property5:
Property6:
Property 7:
Property 8:
39
1.  x n dx 
x n 1
,  n  1
n 1
1
1
dx 
n
x
 n  1 xn1
1
4.  dx  log x
x
3.  dx  x
5.  a x dx 
2. 
ax
log a
6.  e x dx e x
7.  sin x dx   cos x
8.  cos x dx  sin x
9.  sec 2 x dx  tan x
10.  cos ec 2 x dx   cot x
11.  sec x tan x dx  sec x
12.  cos ec x cot x dx   cos ec x
13.  tan x dx  log sec x
14.  cot x dx  log sin x
15.  sec x dx  log sec x  tan x
16.  cos ecx dx  log cosec x  cot x
17. 
dx
1
 x
 tan 1  
2
a x
a
a
18. 
dx
1
ax
 log
2
a x
2a
ax
19. 
dx
1
xa
 log
2
x a
2a
xa
20. 
 x
 sin 1  
a
a2  x2
21. 
2
2
dx
x a
2
2
 log x  x 2  a 2
22. 
23.  a 2  x 2 dx 
x 2 2 a 2 1  x 
a  x  sin  
2
2
a
24.  x 2  a 2 dx 
x 2
a2
x  a 2  log x  x 2  a 2
2
2
25.  x 2  a 2 dx 
x 2
a2
x  a 2  log x  x 2  a 2
2
2
2
dx
dx
x a
2
2
 log x  x 2  a 2
40
Chapter 8. Application of Integration
A plane curve is a curve that lies in a single plane. A plane curve may be closed or open.
The area between the curve defined by a positive function and the x axis between
two values of y is called the definite integral of f between these values. The area of a
rectangle is the product of length and breadth the area under a curve will be generated. This
method is generalised to define a variety of integrals that do not describe area.
The standard approach deals with a more general class of functions by imagining that we
divide the interval between a and b into a large number of small strips and estimate the
area of each strip to be the product of its width and some value of f(x) for x within the strip.
The area will then be something like the sum of the areas of the strips. If we then let the
maximum strip length go to zero, we can hope to find the resulting sum of strip areas
approach the true area.
The standard way to do this is to let the i-th strip begin at xi-1 and end at xi; the area of that
strip is estimated as (xi-xi-1)f(x'i) with x'i anywhere in the strip.
A Riemann sum is a sum of the form just indicated: it is a sum over strips of the width of the
strip times a value of the f(x) within the strip. The function is said to be Riemann
integrable if the sum of the area of the strips approaches a constant independent of which
arguments are used within each strip to estimate the area of the strip, as maximum strip
width goes to zero.
The notation used above can be understood from this approach; we are summing the area
of the individual strips, which for a very small interval around x of size dx is estimated to
be f(x)dx, and summing this over all such strips. The integral sign represents the sum which
is not an ordinary sum, but the limit of ordinary sums as the size of the intervals goes to
zero.
41
Suppose we have a non-negative function f of the variable x, defined in some domain
that includes the interval [a, b] with a < b. If f is sufficiently well behaved, there is a well
defined area enclosed between the lines x = a, x = b, y = 0 and the curve y = f(x).
That area is called the definite integral of f dx between x = a and x = b (of course only for
those functions for which it makes sense).
It is usually written as
If c lies between a and b we obviously have
Area between two curves:There are actually two cases that we are going to be looking at.
In the first case we want to determine the area between y = f(x) and y = g(x) on the interval
[a,b]. We are also going to assume that
.
The second case is almost identical to the first case. Here we are going to determine the
area between x = f(y) and x =g(y) on the interval [c,d] with
.
In this case the formula is,
The area of a triangle is an extension of area of a plane curve.
42
Chapter 9 DIFFERENTIAL EQUATIONS
Definition: An equation containing an independent variable, a dependent variable and
differential coefficients of the dependent variable with respect to the independent variable
is called a differential equation DE.
The ORDER of a differential equation is the highest derivative that appears in the equation.
The DEGREE of a differential equation is the power or exponent of the highest derivative
that appears in the equation.
Unlike algebraic equations, the solutions of differential equations are functions and not just
numbers.
Initial value problem is one in which some initial conditions are given to solve a DE.
To form a DE from a given equation in x and y containing arbitrary constants (parameters) –
1. Differentiate the given equation as many times as the number of arbitrary
constants involved in it .
2. Eliminate the arbitrary constant from the equations of y, y’, y’’ etc.
A first order linear differential equation has the following form:
The general solution is given by
where
called the integrating factor. If an initial condition is given, use it to find the constant C.
Here are some practical steps to follow:
1. If the differential equation is given as
, where
2. Find the integrating factor .
3. Evaluate the integral
4. Write down the general solution .
, rewrite it in the form
43
5. If you are given an IVP, use the initial condition to find the constant C.
VARIABLE SEPARABLE FORM The differential equation of the form is called separable,
if f(x,y) = h(x) g(y); that is,
- - - (I)
In order to solve it, perform the following STEPS:
1. Solve the equation g(y) = 0, which gives the constant solutions of (I);
2. Rewrite the equation (I) as, and, then,
3. integrate
to obtain G(y) = H(x) + C
4. Write down all the solutions; the constant ones obtained from (1) and the ones given
in (3);
5. If you are given an IVP, use the initial condition to find the particular solution. Note
that it may happen that the particular solution is one of the constant solutions given
in (1). This is why Step 4 is important.
Homogeneous Differential Equations
 A function which satisfies for a given n is called a homogeneous function of order n.
 A differential equation is called a HOMOGENEOUS DIFFERENTIAL EQUATION if it can be
written in the form M(x,y)dx + N(x,y)dy = 0 where M and N are of the same degree.
 A first-order ordinary differential equation is said to be homogeneous if it can be written
in the form) Such equations can be solved by the change of variables which transforms
the equation into the separable equation
 Steps for Solving a Homogeneous Differential Equation
1. Rewrite the differential in homogeneous form or
2. Make the substitution y = vx or x = vy where v is a variable.
3. Then use the product rule to get
4. Substitute to rewrite the differential equation in terms of v and x or v and y only
5. Follow the steps for solving separable differential equations.
6. Re-substitute v = y/x or v = x / y.
7.
44
Chapter 10 VECTORS
 A quantity that has magnitude as well as direction is called a vector. It is denoted by
directed line segment, where A is the initial point and B is the terminal point . The
distance AB is called the magnitude denoted by and the vector is directed from A to
B.
 FIXED VECTOR is that vector whose initial point or tail is fixed. It is also known as
localised vector. For example, the initial point of a position vector is fixed at the
origin of the coordinate axes. So, position vector is a fixed or localised vector.
 FREE VECTOR is that vector whose initial point or tail is not fixed. It is also known as a
non-localised vector. For example, velocity vector of particle moving along a straight
line is a free vector.
Co-initial vectors: Vectors having the same initial point are called as co-initial vectors.
 Negative of a vector: A vector whose magnitude is the same of that of a given vector but in
the opposite direction is called as the negative of the given vector.
POSITION VECTOR gives the position of a point with reference to the origin of the
coordinate system. COLLINEAR VECTORS are those vectors that act either along the
same line or along parallel lines. These vectors may act either in the same direction
or in opposite directions.
vectors acting along the same direction.
PARALLEL VECTORS are two collinear
EQUAL VECTORS Two
vectors are said to be equal if they have the same magnitude and direction.
ADDITION OF VECTORS:
1. Triangle law of vectors for addition of two vectors. If two vectors can be represented
both in magnitude and direction by the two sides of a triangle taken in the same
order, then the resultant is represented completely, both in magnitude and
direction, by the third side of the triangle taken in the opposite order.
Corollary: 1) If three vectors are represented by the three sides of a triangle taken in order,
then their resultant is zero.
2) If the resultant of three vectors is zero, then these can be represented completely by the
three sides of a triangle taken in order.
2. Parallelogram law of vectors for addition of two vectors.If two vectors
are completely represented by the two sides OA and OB respectively of a
45
parallelogram. Then, according to the law of parallelogram of vectors, the diagonal
OC of the parallelogram will be resultant
, such that
MULTIPLICATION OPERATIONS FOR VECTORS:
1. MULTIPLICATION OF A VECTOR BY A SCALAR, When a vector . is multiplied by a real
number, say m, then we get another vector m. The magnitude of mis m times the
magnitude of . If m is positive, then the direction of m is the same as that of . If m is
negative, then the direction of m is opposite to that of .
2. SCALAR MULTIPLICATION OF TWO VECTORS to yield
a · b = |a| × |b| × cos(θ)where:
|a| is the magnitude (length) of vector a
|b| is the magnitude (length) of vector b
θ is the angle between a and b
It is important to remember that there are two different angles between a pair of
vectors, depending on the direction of rotation. However, only the smaller of the two is
taken in vector multiplication
Some Properties of the Dot Product
i.i = j.j = k.k = 1 and i.j = j.i = i.k = k.i = j.k = k.j = 0.
The dot product is commutative:
The dot product is distributive over vector addition:
The dot product is bilinear:
When multiplied by a scalar value, dot product satisfies:
Two non-zero vectors a and b are perpendicular if and only if a • b = 0.
The dot product does not obey the cancellation law: If a • b = a • c and a ≠ 0: then a
• (b − c) = 0:
If a is perpendicular to (b − c), we can have (b − c) ≠ 0 and therefore b ≠ c.
46
Vector product:
The Cross Product a × b of two vectors is another vector that is at right angles to both.
a × b = |a| |b| sin(θ) n




|a| is the magnitude (length) of vector a
|b| is the magnitude (length) of vector b
θ is the angle between a and b
n is the unit vector at right angles to both a and b
The cross product is anticommutative: a x b = - b x a
| a x b | is the area of the parallelogram formed by a and b
Distributive over addition. This means that a × (b + c) = a × b + a × c
For two parallel vectors a × b = 0
i × i = j × j = k × k = 0.
The cross product of a vector with itself is the null vector
If a × b = a × c and a ≠ 0 then: (a × b) − (a × c) = 0 and, by the distributive law above:
a × (b − c) = 0 Now, if a is parallel to (b − c), then even if a ≠ 0 it is possible that (b − c) ≠ 0
and therefore that b ≠ c.
SCALAR TRIPLE PRODUCT
It is defined for three vectors
in that order as the
scalar
. ( × )It denotes the volume of the
parallelopiped formed by taking a, b, c as the coterminus edges.
i.e. V = magnitude of × . = | × . |
The value of scalar triple product depends on the cyclic order of the vectors and is
independent of the position of the dot and cross. These may be interchange at pleasure.
However and anti-cyclic permutation of the vectors changes the value of triple product in sign
but not a magnitude.
47
Properties of Scalar Triple Product:
* If, then their scalar triple product is given by
 i.e. position of dot and cross can be interchanged without altering the product.

* , , in that order form a right handed system if . . > 0

i.e. scalar triple product is 0 if any two vectors are equal.

i.e. scalar triple product is 0 if any two vectors are parallel.

Show that
 If three vectors
are coplanar then
.
48
Chapter 11. Three Dimensional Geometry
In space, any point can be expressed as P(x, y, z). The equations of a line passing through a
given point and parallel to a given direction are given by
The equations of a line which passes through two given points are
.
.
Two lines in space may be parallel lines or intersecting lines or skew lines. Two parallel lines
or intersecting lines are in the same plane. If two lines do not intersect then there is no
angle between them.
Angle between two lines
Finding the angle between two lines in 2D is easy, just find the angle of each line with the xaxis from the slope of the line and take the difference. In 3D it is not so obvious, but it can
be shown (using the Cosine Rule) that the angle θ between two vectors a and b is given by:
Cos θ =
Unfortunately this gives poor accuracy for angles close to zero; for instance an angle of
1.00E-7 radians evaluates with an error exceeding 1%, and 1.00E-8 radians evaluates as
zero. A similar formula using the sine of the angle Sin θ =
has similar problems with
angles close to 90 degrees.
But combining the two gives Tan θ =
which is accurate for all angles, and since the
(|a||b|) values cancel out the computation time is similar to the other expressions.
49
Shortest distance between two skew lines:
The two key points two remember here are:
1. The shortest line between two lines is perpendicular to both
2. When two vectors are crossed, the result is a vector that is perpendicular to both
Thus the vector representing the shortest distance between AB and CD will be in the same
direction as (AB X CD), which can be written as a constant times (AB X CD). We can then
equate this vector to a general vector between two points on AB and CD, which can be
50
obtained from the vector equations of the two lines. Solving the three equations obtained
simultaneously, we can find the constant and the shortest distance.
Plane:
A plane is a flat (not curved) two dimensional space embedded in a higher number of
dimensions.
In two dimensional space there is only one plane that can be contained within it and that is
the whole 2D space.
Three dimensional space is a special case because there is a duality between points (which
can be represented by 3D vectors) and planes (also represented by 3D vectors). We can
visualise the vector representing a plane as a normal to the plane.
The angle between two planes is defined as the angle between their normals. It is given by
Coplanarity of four points:Coplanar points are three or more points which lie in the same plane where a plane is a flat
surface which extends without end in all directions. It's usually shown in math textbooks as
a 4-sided figure. You can see that points A, B, C and D are all coplanar points on a single
plane.
51
The concept of coplanar points may seem simple, but sometimes the questions about it may
become confusing. With a little bit of geometry knowledge and some real-world examples, it
can be mastered even the most challenging questions about coplanar points.
Any three points in 3-dimensional space determine a plane. This means that any group of
three points determines a plane, even if all the points don't look like they're located on the
same flat surface.
Let's look at another real life example. The tissue paper box is covered in leaves. Points have
been placed at the tips of four leaves and labeled W, X, Y and Z. From the first picture, we
can see that points X, Y and Z are coplanar. They form a triangle, and you can visualize that.
If you were to cut through the tissue box and pass through these points, you would have a
piece of the tissue box that would have a plane figure, a triangle, as its base.
Chapter 12 Linear Programming
52
Linear programming is the process of taking various linear inequalities relating to some
situation, and finding the "best" value obtainable under those conditions. A typical example
would be taking the limitations of materials and labor, and then determining the "best"
production levels for maximal profits under those conditions.
In "real life", linear programming is part of a very important area of mathematics called
"optimization techniques". This field of study (or at least the applied results of it) are used
every day in the organization and allocation of resources. These "real life" systems can have
dozens or hundreds of variables, or more. In algebra, though, you'll only work with the
simple (and graphable) two-variable linear case.
The general process for solving linear-programming exercises is to graph the inequalities
(called the "constraints") to form a walled-off area on the x,y-plane (called the "feasibility
region"). Then you figure out the coordinates of the corners of this feasibility region (that is,
you find the intersection points of the various pairs of lines), and test these corner points in
the formula (called the "optimization equation") for which you're trying to find the highest
or lowest value.
The Decision Variables: The variables in a linear program are a set of quantities that need to
be determined in order to solve the problem; i.e., the problem is solved when the best
values of the variables have been identified. The variables are sometimes called decision
variables because the problem is to decide what value each variable should take. Typically,
the variables represent the amount of a resource to use or the level of some activity.
The Objective Function: The objective of a linear programming problem will be to maximize
or to minimize some numerical value. This value may be the expected net present value of a
project or a forest property; or it may be the cost of a project;
The Constraints: Constraints define the possible values that the variables of a linear
programming problem may take. They typically represent resource constraints, or the
minimum or maximum level of some activity or condition.
Linear Programming Problem Formulation: We are not going to be concerned with the
question of how LP problems are solved. Instead, we will focus on problem formulation —
translating real-world problems into the mathematical equations of a linear program — and
interpreting the solutions to linear programs. We will let the computer solve the problems
for us. This section introduces you to the process of formulating linear programs. The basic
53
steps in formulation are: 1. Identify the decision variables; 2. Formulate the objective
function; and 3. Identify and formulate the constraints. 4. A trivial step, but one you should
not forget, is writing out the non-negativity constraints. The only way to learn how to
formulate linear programming problems is to do it.
The corner points are the vertices of the feasible region. Once you have the graph of the
system oflinear inequalities, then you can look at the graph and easily tell where the corner
points are. You may need to solve a system of linear equations to find some of the
coordinates of the points in the middle.
Chapter 13 Probability
Definition:
54
Two events are dependent if the outcome or occurrence of the first affects
the outcome or occurrence of the second so that the probability is changed.
The conditional probability of an event B in relationship to an event A is the probability that
event B occurs given that event A has already occurred. The notation for conditional
probability is P(B|A) [pronounced as probability of event B given A].
The notation used above does not mean that B is divided by A. It means the probability of
event B given that event A has already occurred. The notation used above does not mean
that B is divided by A. It means the probability of event B given that event A has already
occurred.
When two events, A and B, are dependent, the probability of both occurring is:
P(A and B) = P(A) · P(B|A)
Definition:
Two events, A and B, are independent if the fact that A occurs does not
affect the probability of B occurring.
Some other examples of independent events are:
 Landing on heads after tossing a coin AND rolling a 5 on a single 6-sided die.
 Choosing a marble from a jar AND landing on heads after tossing a coin.
 Choosing a 3 from a deck of cards, replacing it, AND then choosing an ace as the
second card.
 Rolling a 4 on a single 6-sided die, AND then rolling a 1 on a second roll of the die.
Multiplication Rule 1:
When two events, A and B, are independent, the probability of
both occurring is:
P(A and B) = P(A) · P(B)
Summary:
The probability of two or more independent events occurring in
sequence can be found by computing the probability of each
event separately, and then multiplying the results together.
55
Summary:
The probability of two or more independent events occurring in
sequence can be found by computing the probability of each
event separately, and then multiplying the results together.
Summary:
The probability of two or more independent events occurring in sequence
can be found by computing the probability of each event separately, and then multiplying
the results together.
Bayes’ theorem
Bayes' theorem (also known as Bayes' rule) is a useful tool for calculating conditional
probabilities. Bayes' theorem can be stated as follows:
Bayes' theorem. Let A1, A2, ... , An be a set of mutually exclusive events that together form
the sample space S. Let B be any event from the same sample space, such that P(B) > 0.
Then,
P( Ak ∩ B )
P( Ak | B ) =
P( A1 ∩ B ) + P( A2 ∩ B ) + . . . + P(
An ∩ B )
Bayes' theorem (as expressed above) can be intimidating. However, it really is easy to use.
The remainder of this lesson covers material that can help you understand when and how to
apply Bayes' theorem effectively.
When to Apply Bayes' Theorem
Part of the challenge in applying Bayes' theorem involves recognizing the types of problems
that warrant its use. You should consider Bayes' theorem when the following conditions
exist.
▪ The sample space is partitioned into a set of mutually exclusive events { A1, A2, . . . ,
An }.
▪ Within the sample space, there exists an event B, for which P(B) > 0.
56
▪ The analytical goal is to compute a conditional probability of the form: P( Ak | B ).
▪ You know at least one of the two sets of probabilities described below.
 P( Ak ∩ B ) for each Ak
 P( Ak ) and P( B | Ak ) for each Ak
Random Variable
When the value of a variable is determined by a chance event, that variable is called
a random variable.
Discrete vs. Continuous Random Variables
Random variables can be discrete or continuous.
▪ Discrete. Within a range of numbers, discrete variables can take on only certain
values. Suppose, for example, that we flip a coin and count the number of heads.
The number of heads will be a value between zero and plus infinity. Within that
range, though, the number of heads can be only certain values. For example, the
number of heads can only be a whole number, not a fraction. Therefore, the number
of heads is a discrete variable. And because the number of heads results from a
random process - flipping a coin - it is a discrete random variable.
▪ Continuous. Continuous variables, in contrast, can take on any value within a range
of values. For example, suppose we randomly select an individual from a population.
Then, we measure the age of that person. In theory, his/her age can take
on any value between zero and plus infinity, so age is a continuous variable. In this
example, the age of the person selected is determined by a chance event; so, in this
example, age is a continuous random variable.
Probability Distribution
A probability distribution is a table or an equation that links each possible value that
a random variable can assume with its probability of occurrence.
Discrete Probability Distributions
The probability distribution of a discrete random variable can always be represented by a
table. For example, suppose you flip a coin two times. This simple exercise can have four
possible outcomes: HH, HT, TH, and TT. Now, let the variable X represent the number of
57
heads that result from the coin flips. The variable X can take on the values 0, 1, or 2; and X is
a discrete random variable.
The table below shows the probabilities associated with each possible value of X. The
probability of getting 0 heads is 0.25; 1 head, 0.50; and 2 heads, 0.25. Thus, the table is an
example of a probability distribution for a discrete random variable.
Number of
Probability,
heads, x
P(x)
0
0.25
1
0.50
2
0.25
Note: Given a probability distribution, you can find cumulative probabilities. For example,
the probability of getting 1 or fewer heads [ P(X < 1) ] is P(X = 0) + P(X = 1), which is equal to
0.25 + 0.50 or 0.75.
Continuous Probability Distributions
The probability distribution of a continuous random variable is represented by an equation,
called the probability density function (pdf). All probability density functions satisfy the
following conditions:
▪ The random variable Y is a function of X; that is, y = f(x).
▪ The value of y is greater than or equal to zero for all values of x.
▪ The total area under the curve of the function is equal to one.
Mean and Variance
Just like variables from a data set, random variables are described by measures of central
tendency (like the mean) and measures of variability (like variance). This lesson shows how
to compute these measures for discrete random variables.
58
Mean of a Discrete Random Variable
The mean of the discrete random variable X is also called the expected value of X.
Notationally, the expected value of X is denoted by E(X). Use the following formula to
compute the mean of a discrete random variable.
E(X) = μx = Σ [ xi * P(xi) ]
where xi is the value of the random variable for outcome i, μx is the mean of random
variable X, and P(xi) is the probability that the random variable will be outcome i.
Variability of a Discrete Random Variable
The equation for computing the variance of a discrete random variable is shown below.
σ2 = Σ { [ xi - E(x) ]2 * P(xi) }
where xi is the value of the random variable for outcome i, P(xi) is the probability that the
random variable will be outcome i, E(x) is the expected value of the discrete random
variable x.
Binomial Distribution
To understand binomial distributions and binomial probability, it helps to understand
binomial experiments and some associated notation; so we cover those topics first.
Binomial Experiment
A binomial experiment is a experiment that has the following properties:
▪ The experiment consists of n repeated trials.
▪ Each trial can result in just two possible outcomes. We call one of these outcomes a
success and the other, a failure.
▪ The probability of success, denoted by P, is the same on every trial.
▪ The trials are independent that is, the outcome on one trial does not affect the
outcome on other trials.
Consider the following statistical experiment. You flip a coin 2 times and count the number
of times the coin lands on heads. This is a binomial experiment because:
▪ The experiment consists of repeated trials. We flip a coin 2 times.
▪ Each trial can result in just two possible outcomes - heads or tails.
59
▪ The probability of success is constant - 0.5 on every trial.
▪ The trials are independent; that is, getting heads on one trial does not affect
whether we get heads on other trials. Binomial Distribution
A binomial random variable is the number of successes x in n repeated trials of a binomial
experiment. The probability distribution of a binomial random variable is called a binomial
distribution.
The binomial distribution describes the behaviour of a count variable X if the following
conditions apply:
1: The number of observations n is fixed.
2: Each observation is independent.
3: Each observation represents one of two outcomes ("success" or "failure").
4: The probability of "success" p is the same for each outcome.
Mean and Variance of the Binomial Distribution
The binomial distribution for a random variable X with parameters n and p represents the
sum of n independent variables Z which may assume the values 0 or 1. If the probability that
each Z variable assumes the value 1 is equal to p, then the mean of each variable is equal
to 1*p + 0*(1-p) = p, and the variance is equal to p(1-p). By the addition properties for
independent random variables, the mean and variance of the binomial distribution are
equal to the sum of the means and variances of the n independent Z variables,
so
60
1. If f(x) = x + 7 and g(x) = x – 7 then find (f
g) (7).
2. If the function f is an invertible function find the inverse of f(x) =
3. If the function f is an invertible function defined as f(x) =
.
then write f -1(x).
4. Show that the relation R in the set of real numbers defined as R = {(a, b): a
} is neither
reflexive, symmetric nor transitive.
5. Let T be the set of all triangles in a plane with R as relation in T given by R = {(T1, T2): T1
T2}.
Show that R is an equivalence relation.
6. Let Z be the set of all integers and R be the relation on Z given by R = {(a, b): a,b Z, a – b is
divisible by 5}. Show that R is an equivalence relation.
7. If f: R
R defined by f(x) = (3 – x3)
then find (f
f) (x).
8. Show that the relation R in the set A = { 1, 2, 3, 4, 5} given by R = {(a, b):
is even} is an
equivalence relation.
9. Show that the relation S in the set A = { x : x Z,
} given by S = {(a,b): a,b Z,
is divisible by 4} is an equivalence relation.
10. Is the binary operation defined on N given by
for all a,b N, commutative?
ii) Is associative?
11. If the binary operation
12. Let
on Z is defined by
be the binary operation on N given by
then find
.
. Find
.
61
13. Let f:N
N defined by f(n) =
for all n N. Find whether the
function is bijective.
14. Let
be the binary operation on Q given by
15. What is the range of f(x) =
=2a + b - ab. Find 3 4.
?
defined by f(x) = 9x2 + 6x – 5. Show that f is invertible with
16. Consider f: R
.
17. Let A =
that
18. If f: R
and
be a binary operation on A defined by (a, b)
(c, d) = (a+c, b+d). Show
is commutative and associative. Also find the identity element for
R and g: R
R are given by f(x) = sin x and g(x) = 5 x2 find (
19. If f(x) = 27x3 and g(x) =
find (
) (x).
on A if any.
)(x).
62
1. Evaluate
2. Evaluate
3. Evaluate
4. Evaluate
5. Evaluate
6. Evaluate
7. Evaluate
8. Evaluate
9. Evaluate
10. Evaluate
11. What is the domain of sin-1 x?
12. Prove that tan-11 + tan-12 + tan-13 =
.
63
13. Prove that tan-1
+ tan-1
+ tan-1
14. Prove that tan-1
+ tan-1
=
+tan-1 =
.
.
15. Prove that
.
16. Prove that
.
17. Prove that tan-1x+tan-1
=
18. Prove that tan-1
.
.
19. Prove that
.
20. Prove that
21. Solve tan-12x + tan-13x =
.
.
22. Solve tan-1(x+1) + tan-1(x-1) = tan-1
23. Solve
24. Solve
.
.
25. Solve 2tan-1(Cos x) = tan-1(2 Cosecx)
26. Solve
.
64
27. Prove that
.
1. Find x and y if
2. Find x and y if
3. Find x if
4. Find y if
5. Find y if
6. Find y if
7. Using elementary transformations find the inverse of the matrix
.
8. Using elementary transformations find the inverse of the matrix
9. Using elementary transformations find the inverse of the matrix
.
.
65
10. Using elementary transformations find the inverse of the matrix
.
11. Using elementary transformations find the inverse of the matrix
.
12. Using elementary transformations find the inverse of the matrix
.
13. Solve 2x – y + z = 3, - x + 2y – z = 1 and x – y + 2z = 1.
14. Solve 3x – 2y +3z = 8, 2x + y – z =1 and 4x –3y+2z = 4.
15. Solve x + y + z = 4, 2x + y – 3z = -9 and 2x – y + z = -1.
16. Solve 2x – 3y + 5z = 11, 3x + 2y–4z = -5 and x+y-2z=-3.
17. Solve x + y + z = 6, x + 2z = 7 and 3x + y + z = 12.
18. Solve x+2y -3z=-4, 2x+ 3y +2z = 2 and 3x – 3y -4z = 11.
19. If A =
find A-1. Using A-1 solve
2x – 3y + 5z = 16, 3x + 2y – 4z = -4 and
find A-1. Using A-1 solve
2x + y + 3z = 9, x + 3y – z = 2and -2x + y
x + y - 2z = -3.
20. If A =
+ z = 7.
66
Differentiation of composite functions:
Differentiation by chain rule:
Find the derivative of the following functions wrt x:
1. (x2 – 3)5.

1 
2.  x2 



x3 
10
1

3.  x  
x

4.
5. 1  x2
6. 3x2  4
7.
1 x
1 x

8.
9. x2 x3  5
4 x3
11.
1  x2
x  3x
4
6

x2  1
x2  1
10. (x2+ 1)3 (x3 – 1)4.
12.
x 1  x 1
x 1  x 1
Differentiation of trigonometric functions
Find the derivative of the following functions wrt x:
1. Sin 5x
2. Cos (2x – 3)
3. Sin3 (5x + 4)
1  sin2x
5.
1  sin2x
1  tan3x
7.
1  tan3x
4. Tan104x
1  co s x
6.
1  cosx
dy
= y2 + y sin2x.
dx
2
 dy 
9. If y = a sin x + b cos x then prove that y2 + 
= a2 + b2.

 dx 
8. If y = x tan x then prove that x sin2x
10. If y =

1  sin2 x
dy
then prove that
+ sec2 ( -x) = 0.
dx
4
1  sin2 x
Differentiation of inverse trigonometric functions
Find the derivative of the following functions wrt x:
1. Sin–12x
2. Tan-15x
67
 2x 
4. sin1 


 1  x2 
3. Sin-1 1  x2

5. sin1 3x  4x3



1
6. sec1 

 2
 2x  1 
1

x 

x
7. co s1 
 x  1 

x
8. co s1
9. tan1  sec x  tan x 
 cos x 
10. tan1 

 1  sin x 
1  cos x
11. tan1
1  cos x
1  sin x
12. tan1
1  sin x
 1  x2 

14. tan1 
 1  x2 




1  a2 x2  1 
16. tan1 


ax


 a x
13. tan1 
 1  ax




1  x 
15. co t 1 

1  x 


x

17. tan1 


2
 1  x  1
1  x2
2

18. tan1  x 


x2  1 

 1 x  1 x 
 1  sin x  1  sin x 
19. tan1 
20. co t 1 
 1  x  1  x 
 1  sin x  1  sin x 






21. sin1  x 1  x  x 1  x2 
22. sin1 3x  4x3  cos1 4x3  3x


sin1 x
dy
 xy  1 .
23. If y =
then prove that 1  x2
dx
1  x2





Differentiation of log and exponential functions
Find the derivative of the following wrt x:
x2
1. e 
2. e2x 3
2
3. e 1 x
4. e 2x
5. esin x
6. esin x
e x  e x
8.
e x  e x
7. e x tan x
9. e x cos x

11. log x2  3
10. log sec x 


1 
12. log  x 

x



1  x2  1 

14. log


2
 1  x  1
13. log log x 
 1  cos 2 x 
15. log 

 1  cos 2 x 
17. log

xa xb
16. log

1  sin x
1  sin x
18. log cos ecx  cot x 

68
19. log s ecx  tan x 

x
 x 
20. log  s ec    tan   
2
 
 2 



21. log  x  x2  1 


ax
23. e
log  sin bx 

22. log  x 


x2  1 

Differentiation of Implicit functions
dy
Find
from the following:
dx
1.
x2  y2  25
2.
x2 y 2

1
a2 b2

3. x3  y3  3xy
4. x3  y3  3 x2y  xy2
5. x 4  y 4  4xy
6. x3y3  y  x
7.
x y 

a




dy
 xy  1  0.
8. If y 1  x2  log  1  x2  x  then prove that 1  x2
dx


dy
1

9. If x 1  y  y 1  x  0 then prove that
.
2
dx
1 x


dy
1  y2

 0.
10. If x 1  y2  y 1  x2  1 then prove that
dx
1  x2
dy
1  y2

 0.
1  x2  1  y2  a  x  y  then prove that
dx
1  x2
dy
12. If x + y = sin(x + y) then prove that
= - 1.
dx
dy
log x

.
13. If y log x = x – y then prove that
dx
1  log x 2
11. If
14.
x
x
x
x  ......
15.
sin x  sin x  sin x  sin x  ......
16.
log x  log x  log x  log x  ......
1
17. x 
1
x
x
1
x  .....
Logarithmic differentiation
Differentiate the following wrt x:
1. x x
2. xlog x
69
x
3. xsin x
4.  sin x 
cos x
5.  sin x 
sin x
6.  ta n x 
7. log x 
 
x
8. sin x x
2x
x
 2 x 
11.  x 
x2  1
x2  1
sin x
x
  sin x 
12.  x 
co s x
sin x
  co s x 
13.  sin x 
tan x
sin x
  tan x 
14.  sin x 
9. xcot x 
2x2  3
x2  x  2
10. x sin x cos x 
tan x
sin x
  tan x 
15.  sin x 
dy
.
dx
dy
y
x
17. If  x    y   0 then find
.
dx
16. If  x 
y
 y 
x
 c then find
dy
log x
y

18. If  x   e x  y then prove that
.
dx
1  log x 2
x y
dy
y
19. If x  e x then find
.
dx
dy
20. If x  y  x y then find
.
dx
21. x x
x ...
sin x
22. sin xsin x
...
Differentiation of parametric functions
dy
Find
from the following:
dx
1. x= at2, y = 2at
2.
x
2t
1  t2
, y 
.
2
1  t2
1t
3. x = a cos Ө, y= b sin Ө
4. x = a (1 – cos Ө), y = a (Ө + sin Ө)
5. x = a (cos Ө + Ө sin Ө), y = a (sin Ө – Ө cos Ө)
6. x = 2cos Ө – cos 2Ө, y = 2 sin Ө – sin 2Ө.
7.
Higher Derivatives
The derivative of the derivative of a function is called as the second derivative.
d2y
d  dy 
y '' or y or

 dx 
2
2
dx


dx
70
d3y
d  d2y  d3y
d  d2y 
y ''' or y or


3
dx3 dx  dx2  dx3 dx  dx2 
1. Find the second derivative of x2 + log x wrt x.
2. If y = ex (sin x + cos x) then prove that y2 – 2y1 + 2y = 0.
d2y
cos x
3. If y = sec x + tan x then prove that

2
dx
1  sin x 2
d2y

dx2
4. If y = cosec x + cot x then prove that
sin x
1  co s x 2
5. If y = a sin 2x + b cos 2x then prove that y  4y  0
2


6. If y = ta n1 x then prove that 1  x2 y  2xy  0
1
2

 then prove that 1  x2  y2  2x 1  x2  y1  2  0
2
8. If y =  si n1 x  then prove that 1  x2  y  xy  2  0
1
2
9. If y = esin x then prove that 1  x2  y  xy  y  0
1
2
2
10. If y = eta n x then prove that 1  x2  y  2 x 1  x2  y  y  0
1
2
7. If y = ta n1 x
2
2
1
1
11. If y = sin (log x) + 3 cos (log x) then prove that x2y  xy  y  0
1
2

12. If x = a (Ө – sin Ө), y = a (1 – cos Ө) then find y at Ө= .
2
2
71
1. Show that the rectangle of maximum area that can be inscribed in a circle is a square.
2. Verify LMVT for the function f(x) = x2 + 2x + 3 for [4, 6].
3. Prove that the curves x = y2 and xy = k intersect at right angles if 8k2=1.
4. Find the equation of the tangent to the curve y =
which is parallel to the line 4x–
2y+5=0
5. Find the intervals in which the function f(x) =
is increasing and decreasing.
6. The length x of a rectangle is decreasing at the rate of 5 cm / min and the width y is
increasing at the rate of 4 cm / min. when x = 8 cm y = 6 cm find the rate of change of
a)perimeter b) area of the rectangle.
7. Find the intervals in which the function f given by f(x) = sin x + cos x,
is strictly
increasing or strictly decreasing.
8.
Find the intervals in which the function f given by f(x) = sin x - cos x,
is strictly
increasing or strictly decreasing.
9. If the sum of the hypotenuse and a side of a right triangle is given, then show that the area
of the triangle is maximum when the angle between them is
.
72
10. A manufacturer sell x items at a price of Rs
each. The cost price of x items is Rs
. Find the number of items he should sell to earn maximum profit.
11. Find the equation of the tangent to the curve y =
which is parallel to the line 4x–
2y+5=0
12. Find the approximate value of f(2.01) where f(x) = 4x3+5x2+2.
13. Find the points on the curve y = x3 at which the slope of the tangent is equal to the y coordinate of the point.
14. Find the equations of the normal to the curve y = x3+2x+6 which are parallel to the line
x+14y+4= 0.
15. Find the equation of the tangent to the curve y =
at the point where it cuts
the x – axis.
16. Find the equations of the tangent and the normal to the curve x = 1 – cos
1. Evaluate
1
x2
 1  x3 dx
1 a. Evaluate
0

x sin x
dx
2. Evaluate 
1  cos 2 x
0
a
2a. Eva.
x tan x
0 secx cos ecx dx

4. Eva.
x tan x
 sec x  tan x dx
0

2
5. Eva.

0

a

3. Eva.

3a. Eva.
ax
dx
ax
x sin x
 1  cos
0
4a. Eva.
2
x
dx
x  cos 6 x
dx
2
 sin 6 x
 3x

tan x  cot x dx
dx
 1 x
2

5a.Eva. log sin x dx
0
2
,y=
at
73
2
6. Eva.
sec x
 3  tan xdx
1
7. If
  3x
2
6a. Eva.

x
e dx
5  4e x  e2 x
 2 x  k  dx = 0 then find the value of k.
0
8.
x  4  ex

Evaluate 
dx
3
 x  2

e cos x
dx
8 a .Evaluate  cos x
e  e  cos x
0

2
9. Evaluate
 2log sin x   log sin2x   dx
9a.Evaluate  sec2  7  x dx
0
1
2

1  log x  dx

cos x
dx
x
11a.Evaluate

12. Evaluate x sin1 x dx

12a.Evaluate

sec2 x
 x dx
13a.Evaluate
 2  sin x 3  4sin x dx

x4
dx
14a.Evaluate 
 x  1  x 2  1
2
10. Evaluate
11. Evaluate
13. Evaluate
x
1
14. Evaluate x co s x dx
2
10a.Evaluate

0
dx
1  x2
dx
5  4 x  2x2
sin x
dx
x
4
15. Evaluate
  x  1  x  2  x  4  dx
15a.Evaluate
1

x
dx
16. Evaluate  2
2
a cos x  b2 sin2 x
0
17. Evaluate
e2 x  e 2 x
 e2 x  e2 x dx
e x  sin4 x  4 
dx
18. Evaluate
1  cos4 x
cos x
dx
 x  log x
e2 x  e 2 x
dx
16a.Evaluate  2 x
e  e 2 x
17a.Evaluate

log x
dx
x
1  x2
dx
18a.Evaluate 
x 1  2 x 

3
19. Evaluate


sin x  cos x
dx
sin2 x
19a.Evaluate
6
3
20. Evaluate
  3x
2
 2x  dx as the limit of sum.
1
4
21. Evaluate
x
1
2
 x  dx as the limit of sum.
x
 x  sinx dx
74

2
22. Evaluate
 sin
5
xdx
22a.Evaluate
5x 2
1 x 2  4 x  3 dx
23a.Evaluate

2
2
23. Evaluate
x 2
  x  2 x  3 dx
1

24a.Evaluate
 x sin x dx
1
25. Evaluate x 2 1  x  dx

dx
3
2
0
1
2
0
3
2
24. Evaluate x cos  x dx
1
1 x
n
25a.Evaluate
0
x 2  3x
  x  1 x  2  dx
1. Find the area of the region included between y2= 4ax and x2 = 4ay, a > 0.
2. Find the area lying above the x- axis and included between x2 + y2 = 8x and y2 = 4x.
3. Find the area of the region enclosed by the curves y = x2 and y = x.
4. Find the area of the region enclosed by the curves y2 = x and y = x + 2.
5. Find the area of the region bounded by the curves y =
, x = - 3, x = 3 and y =0.
6. Find the area of the smaller region founded by the ellipse
and the line
.
7. Find the area of the region in the first quadrant enclosed by the x-axis, the line x =
y and
the circle x2 +y2 = 4.
8. Find the area bounded by the lines y = 4x + 5, y = 5 – x and 4y = x + 5.
9. Find the area of the region bounded by y2 = 4x and 4x2 + 4y2 = 9.
10. Find the area of the region enclosed between x2 + y2 = 4 and (x – 2)2 +y2 = 4.
11. Find the area of that part of the circle x2 +y2 = 16 which is exterior to the parabola y2= 6x.
75
12. Find the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y =7.
13. Find the area bounded by the lines 2x+ y = 4, 3x - 2y = 6 and x - 3y + 5 = 0.
14. Find the area bounded by the lines 4x - y + 5= 0, x + y – 5 = 0 and x - 4y + 5 = 0.
15. Find the area bounded by the curves x2 + y2 = 1 and (x – 1)2 +y2 = 1.
16. Prove that the curves y2= 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts.
17. Find the area of the region included between the parabola y2 = x and x+y = 2.
18. Find the area of the region included between the parabola x2 = 4y and x = 4y - 2.
19. Find the area of the region included between the parabola 4y = 3x2 and 3x - 2y+12 = 0.
20. Find the area of the region bounded by the curves
y2= 4ax and x2 = 4ay.
21. Find the area of the region
.
22. Find the area of the region
.
23. Find the area of the region
24. Find the area of the region
25. Find the area of the region
.
.
26. Find the area of the triangle ABC with vertices as A (- 1, 0), B(1, 3) and C(3, 2).
27. Find the area of the triangle ABC with vertices as A (4, 1), B (6, 6) and C (8, 4).
76
1. What is the degree of the DE
.
2. Form the DE representing the parabolas having the vertex at the origin and the axis along
the positive direction of x-axis.
3. Form the DE representing the family of ellipses whose foci on x- axis and the centre at the
origin.
4. Solve the following:
5. Solve :
6. Solve:
7. Solve
8. Solve
9. Solve
10. Solve
given that y=1 when x= 1.
given that y = 1 when x = 1.
77
11. Solve
12. Solve
, y= 0 when x = 1.
13. Solve
14. Solve
15. Solve
16. Solve
, y = 0 when x =
17. Solve
18. Solve
19. Solve
20. Solve
21. Solve
22. Solve
23. Solve
24. Solve
25. Solve
. Y = 1 when x = 0.
78
26. Solve
given that x = 1 and y =1.
27. For the DE
find the solution curve passing through the point (1, - 1).
1. Find a unit vector in the direction of a  3iˆ  2 ˆj  6kˆ .
2. Find a unit vector in the direction a  iˆ  2 ˆj of whose magnitude is 7.
3. Find the angle between a  iˆ  ˆj  kˆ and b  ˆi  ˆj  kˆ .
4. For what value of λ are the vectors a  2iˆ   ˆj  kˆ and b  iˆ  2 ˆj  3kˆ perpendicular to
each other.
5. If a  iˆ  ˆj  kˆ and b  ˆj  kˆ find a vector c such that a  b  c and a  b  3 .
6. If a  b  c  0 and a =3, b =5 & c =7 show that the angle between a & b is 60 .
7. If a = 3 , b = 2 and the angle between them is 60 then find a  b .
8. If ˆi  ˆj  kˆ , 2ˆi  5 ˆj , 3ˆi  2 ˆj  3kˆ & ˆi  6 ˆj  kˆ are the position vectors of A, B, C & D find the
angle between AB and CD .
9. Write the value of p for which a  3iˆ  2 ˆj  9kˆ and b  iˆ  pjˆ  3kˆ are parallel vectors.
10. If a  b  c  d and a  c  b  d then show that a  d b  c where a  d & b  c .
11. Write a unit vector in the direction of b  2iˆ  ˆj  2kˆ .
12. Write a unit vector in the direction of b  2iˆ  3 ˆj  6kˆ .
13. Find the angle between a & b with magnitudes 1 & 2 and a  b  3 .

 
14. Write the value of p for which 2ˆi  6 ˆj  14kˆ  ˆi   ˆj  7kˆ

 

 =0.
15. 2ˆi  6 ˆj  27kˆ  ˆi  3ˆj  pkˆ =0.
16. If p is a unit vector and  x  p    x  p  =80 then find x .
17. The scalar product of ˆi  ˆj  kˆ with the unit vector along the sum of 2ˆi  4 ˆj  5kˆ and
 ˆi  2 ˆj  3kˆ is equal to one. Find the value of  .
18. If a  3 , b  2 and a  b = 3 then find the angle between a & b .
79
19. If a , b & c are three vectors such that a  b  a  c and a  b  a  c , a  0then prove that
b c .
20. Write a vector of magnitude 15 units in the direction of ˆi  2 ˆj  2kˆ .
21. Find the position vector of a point R which divides the line hoining two points P and Q
whose position vectors 2a  b and a  3b externally in the ratio 1:2. Also show that P is
the midpoint of RQ.
22. Write a vector of magnitude 9 units in the direction of 2iˆ  ˆj  2kˆ .

 

23. Find  if 2ˆi  6 ˆj  14kˆ  ˆi   ˆj  7kˆ =0.
24. If a  iˆ  ˆj  kˆ , b  4iˆ  2 ˆj  3kˆ and c  iˆ  2 ˆj  kˆ find a vector of magnitude 6 units
which is parallel to 2a  b  3c .
25. Let a  iˆ  4 ˆj  2kˆ , b  3iˆ  2 ˆj  7kˆ and c  2iˆ  ˆj  4kˆ . Find a vector d which is
perpendicular to both a & b and c  d = 18.
26. If a & b are two vectors such that a  b  a  b then what is the angle between a & b .
27. Vectors a & b are such that a  3 , b 


2
and a  b is a unit vector. Write the angle
3
between a & b .
28. Find x if for a unit vector a ,  x  a    x  a  =0.


29. For what value of p, ˆi  ˆj  kˆ p is a unit vector ?
30. Let a  iˆ  4 ˆj  2kˆ , b  3iˆ  2 ˆj  7kˆ and c  2iˆ  ˆj  4kˆ . Find a vector d which is
perpendicular to both a & b and c  d = 1.
31. The scalar product of iˆ  2 ˆj  4kˆ with a unit vector along the sum of the vectors
iˆ  2 ˆj  3kˆ and  iˆ  4 ˆj  5kˆ is equal to one. Find the value of  .
80
1. Find the shortest distance between the lines
and
.
2. Find the point on the line
at a distance
from the point (1, 2,3).
3. Find the equation of the plane passing through the point (-1, -1, 2) and perpendicular to the
planes 2x + 3y – 3z = 2 and 5x – 4y + z = 6.
4. Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to the
planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
5. Find the equation of the plane passing through the point (3, 4, 1) and parallel to the line
.
6. If the equation of the line AB is
parallel to the line AB.
find the direction ratios of a line
81
7. Find the equation of the line passing through the point P (4, 6, 2) and the point of
intersection of the line
and the plane x + y – z = 8.
8. Find the distance of the point (- 2, 3, - 4) from the line
measured
parallel to the plane 4x + 12 y – 3z + 1 = 0.
9. Find the value of
so that the lines
and
are perpendicular to each other.
10. Find the value of
so that the lines
and
are
perpendicular to each other.
11. Find the value of
so that the lines
and
are
perpendicular to each other.
12. Find the direction cosines of the line equally inclined to the three co-ordinate axes.
13. Find the shortest distance between the lines
and
.
14. Find the shortest distance between the lines
and
15. Find the shortest distance between the lines
and
16. Find the equation of the plane determined by the points (3,-1, 2), (5, 2, 4) and (-1, -1, 6).
Also find the distance of the point (6, 5, 9) from the plane.
17. Find the direction cosines of the line passing through the points (-2, 4, -5) and (1, 2, 3).
18. Find the angle between the line
and the plane 10x + 2y – 11z = 3.
82
19. Show that the lines
and
are coplanar. Also find
te equation of the plane containing the lines.
20. Find the angle between the line
and the plane x + 2y + 2z – 5 = 0.
21. What is the cosine of the angle which the vector
makes with the y – axis?
22. Find the equation of the line passing through the points (0, 0, 0) & (3, -1, 2) and parallel to
the line
.
23. The points A (4, 5, 10), B(2, 3, 4), C (1, 2, -1) are three vertices of a parallelogram ABCD. Find
the vector equations of the sides AB & BC and also find the co-ordinates of the point D.
24. Find the vector equations of the lines
and
and
hence find the distance between them.
25. Write the distance of the plane 2x – y + 2z + 1 = 0 from the origin.
26. Find the points on the line
at a distance of 5 units from the point
(1,
3, 3).
27. Find the distance of the point (6, 5, 9) from the plane determined by the points
(3, -1,
2), (5, 2, 4) and (-1, -1, 6).
28. Find the co-ordinates of the foot of the perpendicular and the perpendicular distance of the
point (3, 2, 1) from the plane 2x – y + z + 1 = 0. Find also the image of the point in the plane.
29. Find the equation of the plane passing through (1, 1, 1) and containing the line
. Also show that the plane contains the line
.
30. Write the Cartesian equation of the line
31. Find the shortest distance between the lines
Hence write whether the lines intersect or not.
.
and
.
83
32. Find the co-ordinates of the point where the line through (3, -4, -5) & (2, 2, 1) crosses the
plane determined by (1, 2, 3), (2, 2, 1) & (-1, 3, 6).
33. Find the Cartesian and the vector equations of the plane passing through the points (0, 0, 0)
& (3, -1, 2) and parallel to the line
.
1. A factory owner purchases two types of machines, A and B for his factory. The requirements
and the limitations for the machines are as follows:
Machine
Area occupied
Labour force
Daily output (in units)
A
1000 m2
12 men
60
B
2000 m2
8 men
40
2.
He has maximum area of 9000 m2 available and 72 skilled labourers who can operate both
the machines. How many machines of each type should he buy to maximize the daily
output?
2. An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each
executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline
reserves at least 20 seats for executive class. However, at least 4 times as many passengers
prefer to travel by economy class than by the executive class. Determine how many tickets
of each type must be sold in order to maximize the profit for the airline. What is the
maximum profit?
3. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1and
F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food
F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units
of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find
the minimum cost for diet that consists of mixture of these two foods and also meets the
minimal nutritional requirements.
4. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs 5760 to
invest and has a space for atmost 20 items. A fan costs him Rs 360 and a sewing machine Rs
240. His expectation is that he can sell a fan at a profit of Rs 22 and a sewing machine at a
84
profit of Rs 18. Assuming that he can sell all the items that he can buy how should he invest
his money in order to maximize the profit? Formulate this as a LPP and solve it graphically.
5. One kind of cake requires 200g flour and 25g of fat, and another kind of cake requires 100g
of flour and 50g of fat. Find the maximum number of cakes which can be made from 5 kg of
flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in
making the cakes? Formulate this as a LPP and solve it graphically.
6. A small firm manufactures gold rings and chains. The total number of rings and chains
manufactured per day is atmost 24. It takes 1 hour to make a ring and 30 minutes to make a
chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300
and that on a chain is Rs 190 find the number of rings and chains that should be
manufactured per day so as to earn the maximum profit. Make it LPP and solve it
graphically.
7. One kind of cake requires 300g flour and 15g of fat, and another kind of cake requires 150g
of flour and 30g of fat. Find the maximum number of cakes which can be made from 7.5 kg
of flour and 600 g of fat assuming that there is no shortage of the other ingredients used in
making the cakes? Formulate this as a LPP and solve it graphically.
8. A factory makes two types of items A and B made of plywood. One piece of item A requires
5 minutes for cutting and 10 minutes for assembling. One piece of item B requires 8 minutes
for cutting and 8 minutes for assembling. There are 3 hours and 20 minutes available for
cutting and 4 hours for assembling. The profit on one piece of item A is Rs 5 and that on item
B is Rs 6. How many pieces of each type should the factory make so as to maximize the
profit? Make it as a LPP and solve it graphically.
85
1. A pair of dice is thrown 4 times. If getting a doublet is considered as a success, find the
probability distribution of number of successes.
2. A and B throw a pair of die turn by turn. The first to throw 9 is awarded a prize. If A starts
the game, show that the probability of A getting the prize is 9 / 17.
3. A man is known to speak 3 out of 4 times. He throws a die and report it is a 6. Find the
probability that it is actually a 6.
4. A man is known to speak 3 out of 5 times. He throws a die and report it is a number greater
than 4. Find the probability that it is actually a number greater than 4.
5. An insurance company insured 2000 scooter drivers, 3000 car drivers and 4000 truck drivers.
The probabilities of accidents are 0.04, 0.06 and 0.15 respectively. One of the insured
persons meets with an accident. What is the probability that he is a car driver?
6. An insurance company insured 2000 scooter drivers, 5000 car drivers and 7000 truck drivers.
The probabilities of accidents are 0.04, 0.05 and 0.15 respectively. One of the insured
persons meets with an accident. What is the probability that he is a car driver?
7. A die is thrown again and again until three sixes are obtained. Find the probability of
obtaining the third six in the sixth throw of the die.
8. Three bags contain balls as shown in the table below:
Bag
No. of White balls
No. of Black balls
No. of Red balls
I
1
2
3
II
2
1
1
86
III
4
3
2
A bag is chosen at random and two balls are drawn from it. They happen to be white and
red. What is the probability that they came from the III bag?
9. Two groups are competing for the position on the board of directors of a corporation. The
probabilities that the first and the second groups will win are 0.6 and 0.4 respectively.
Further, if the first group wins, the probability of introducing a new product is 0.7 and the
corresponding probability is 0.3 if the second group wins. Find the probability that the new
product introduced was by the second group.
10. Coloured balls are distributed in three bags as shown in the following table;
Bag
Colour of the ball
Black
White
Red
I
2
1
3
II
4
2
1
III
5
4
3
A bag is selected at random and then two balls are randomly drawn from the selected bag.
They happen to be white and red. What is the probability that they came from bag II?
11. There are three coins. One is two headed coin (having head on both faces), another is a
biased coin that comes up heads 25% of the time and third is an unbiased coin. One of the
three coins is chosen at random and tossed, it shows heads, what is the probability that it
was the two headed coin?
12. The probability that A hits a target is 1/3 and the probability that B hits the target is 2/5. If
each one of them shoots a target what is the probability that
a) the target is hit? b)
exactly one of them hits the target?
13. Two cards are drawn simultaneously (without replacement) from a well shuffled pack of 52
cards. Find the mean and the variance of the number of red cards.
14. A pair of dice is thrown 4 times. If getting a doublet is considered as a success find the mean
and the variance of the number of successes.
15. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random
with replacement. Find the probability distribution of the number of defective bulbs.
87
16. On a multiple choice examination with three possible answers for each of the five questions,
what is the probability that a candidate would get four or more correct answers just by
guessing?
17. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are
drawn and are found to be both clubs. Find the probability of the lost card being a club.
18. From a lot of 10 bulbs which include 3 defectives, a sample of 2 bulbs is drawn at random
with replacement. Find the probability distribution of the number of defective bulbs.
19. There are two bags, Bag I and Bag II. A bag contains 4 white and 4 red balls, bag II contains 3
white and 7 red balls. One of the two bags is selected at random and a ball is drawn from the
bag which is found to be white. Find the probability that the ball is drawn from the bag I.
20. A couple has two children. Find the probability that both are boys, if it is known that a) at
least one of the children is a boy. b) the elder child is a boy.
21. A bag contains 4 balls. Two balls are drawn at random, and are found to be white. What is
the probability that all balls are white?
22. An experiment succeeds twice as often as it fails. Find the probability that in the next six
trails there will be atleast 4 successes?
23. An urn contains 4 white and 3 red balls. Let X be the number of red balls in the random draw
of three balls. Find the mean and variance of X.
24. In answering a question on a multiple choice test, a student either knows the answer or
guesses. Let 3/5 be the probability that he knows the answer and 2/5 be the probability that
he guesses. Assuming that a student who guesses at the answer will be correct with
probability 1/3. What is the probability that the student knows the answer given that he
answered it correctly?
88
SAMPLE PAPER
CLASS: XII
MATHEMATICS
Max.Marks:100
Time Allowed: 3 hours
General Instructions:
1. All questions are compulsory
2. This question paper consists of 26 questions divided into three sections A, B and C.
Section A comprises of 06 questions of one mark each, section B comprises of 13
questions of four marks each and section C comprises of 07 questions of six marks
each
3. All questions in Section A are to be answered in one word, one sentence or as per
the exact requirement of the question
4. There is no overall choice. However, internal choice has been provided in 04
questions of four marks each and 02 questions of six marks each. You have to
attempt only one of the alternatives in all such questions.
5. Use of calculators is not permitted. You may ask for logarithmic tables, if required
SECTION - A
i  j
1. If A =[a ij ] where aij  
i  j




if i  j
, then construct a 2  3 matrix A
if i  j






2. If a  i  2 j  3 k and b  2 i  4 j  9 k find a unit vector parallel to a  b

3.
4.
5.
6.



Find the distance of the point (2,3,4) from the plane r . (3 i  6 j  2 k )  11
Find 𝜆 if (2𝑖̂ + 6𝑗̂ + 14𝑘̂ ) × (𝑖̂ − 𝜆𝑗̂ + 7𝑘̂) = 0
Form the differential equation of the family of curves y2 = 4ax.
Find the order and degree of the differential equation y´´´ + y2 + ex = 0
SECTION – B
7. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from
1
1
1
1
the origin, then 2  2  2  2
a
b
c
p
(OR)








Find the distance between the lines r  ( i  2 j  4 k )   (2 i  3 j  6 k ) and r 






(3 i  3 j  5 k )   (2 i  3 j  6 k )
8. Find the equation of the plane through the line of intersection of the planes
x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.
9. Find the probability distribution of number of doublets in three throws of
a pair of dice.
(P.T.O)
89
 1  x  1
10. Write the simplest form of tan 1 

x


(OR)
3
8
84
 cos 1
Prove that sin 1  sin 1
5
17
85
2
 x  2 if x  1
11. Discuss the continuity of the function f defined by f ( x)  
 x  2 if x  1
5
x
dx
12. Evaluate 
7x  x
2
dy
3a
2a
 5ax 
 2
 2
13. If y = tan 1  2
then show that
2
2 
dx a  9 x
a  4x2
 a  6x 
.
x 2  1
x 
dx
14. Evaluate  e
2
 x  1
(OR)
Evaluate

x

1
4
x 4
dx
x5
2
15. Evaluate  7 x  5dx as a limit of sum.
1
x 1 x  2 x  a
16. Show that x  2 x  3
x3 x4
x  b  0 where a, b, c are in Arithmetic Progression.
xc
 1 x2  1 x2 
1
tan
17. Differentiate

 with respect to x.
2
2
 1  x  1  x 
(OR)
d2y
If y = x + cot x then prove that sin x 2  2 y  2 x  0
dx
18. Find the matrix P satisfying the matrix equation
 2 1    3 2  1 2 
3 2 P  5  3  2  1 .

 
 

2
19. Using matrix method, solve the following:
x + y + z = 4, 2x – y + z = -1, 2x + y - 3z + 9 = 0
SECTION – C
20. If length of three sides of a trapezium other than base is equal to 10 cm each, then
find the area of the trapezium when it is maximum.
(P.T.O)
(OR)
90
Prove that the radius of the right circular cylinder of greatest curved
surface area which can be inscribed in a given cone is half of that of the cone.
21. Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2).
22. Find the image of the point (1,6,3) in the line
x y 1 z  2


1
2
3
23. Solve the differential equation
24. A merchant plans to sell two types of personal computers – a desktop model and a
portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates
that the total monthly demand of computers will not exceed 250 units. Determine
the number of units of each type of computers which the merchant should stock to
get maximum profit if he does not want to invest more than Rs 70 lakhs and if his
profit on the desktop model is Rs 4500 and on portable model is Rs 5000.
25. Let f : N  R be a function defined as f (x) = 4 x 2  12 x  15 . Show that f : N  S
where S is the range of f is invertible and find the inverse of f.
26. Given three identical boxes I, II and III, each containing two coins. In box I, coins
are gold , in box II, both are silver coins and in the box III, there is one gold and one
silver coin. A person choses a box at random and takes out a coin. If the coin is of
gold, what is the probability that the other coin in the box is also of gold?
(OR)
A man is known to speak truth 3 out of 4 times. He throws a die and reports
that it is a six. Find the probability that it is actually a six
***
91
ANSWER KEY
Q.NO
Value points
Marks
a11 =2, a12 = -1 a13 = -2 a 21 = 3 a 22 = 4 a 23 = -1
1
 2  1  2
Required matrix A = 

3 4  1 
2




1M

a  b = 3 i  6 j 6 k

1M


1  
Unit vector parallel to a  b is  (3 i  6 j  6 k )
9
3
4
5
6
7
Total
Marks


  
  

 2 i  3 j  4 k  .  2 i  3 j  4 k   11


Required distance = 
9  36  4
1M
=1
𝜆 = -3
2y´ – y = 0
Order = 3 and Degree = 1
1M
1M
1M
The equation of a plane having intercepts a, b, c with x, y, and z
axes respectively is given by,
1M
The distance (p) of the plane from the origin is given by,
1M
1M
4M
1M
92

OR

Since b1  b 2 , the lines are parallel



½M

a1  i  2 j  4 k




a2  3i  3 j  5 k



1M

b  2 i  3 j 6 k
4M
𝑏⃗ × ( ⃗⃗⃗⃗
𝑎2 − ⃗⃗⃗⃗
𝑎1 )
𝑑=|
|
|𝑏⃗|
1½ M
1M
293
=
7
8
The equation of plane passing through the point of intersection of
planes is
x + y + z - 1 +  (2 x + 3 y + 4 z - 5 ) = 0. …………..(1)
x (1 + 2  ) + Y (1 + 3  ) + z ( 1 + 4  ) -1 -5  =0.
1M
1M

n1  (1  2 )iˆ  (1  3 ) ˆj  (1  4 )kˆ

x - y + z = 0, n2  iˆ  ˆj  kˆ .
1M
 
Since the planes are perpendicular, n1.n2 = 0
1 + 2  -1 – 3  + 1 + 4  = 0 . 3  = -1 Therefore  = -
1
3
1
in equation (1) we get
3
1M
The equation of the required plane is x – z + 2 = 0.
By substituting the value of  = -
9
Let X denote the number of doublets. Possible doublets are
(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6)
1M
1M
½M
4M
½M
93
Clearly, X can take the value 0, 1, 2, or 3.
½M
½M
Similarly P(X=2)=P( two doublets and one non doublet)=
15
216
1
216
Required Probability distribution is
And P(X=3)=P(three doublets)=
X
P(X)
10
0
125
216
1
75
216
2
15
216
Let x -= tan 
 1  x 2  1
1 1  cos  
Then tan 1 
 = tan 
x
 sin  


 

2sin 2

2 
= tan 1 


 2sin cos 
2
2


=
3
1
216
1M
1M
4M
1M

2
1
 tan 1 x
2
½M
½M
94
OR
11
3
8
Let  sin 1 and  sin 1
5
17
4
15
 cos   ;cos  
5
17
now., cos(   )  cos  cos   sin  sin 
84

85
84
     cos 1
85
3
8
84
 sin 1  sin 1  cos 1
5
17
85
For x<1 or x>1 the given function is continuous being a polynomial
At x = 1,
LHL  lim f ( x)  lim x  2  1
x 1
x 1
RHL  lim f ( x)  lim x  2  3
x 1
1½M
1M
4M
1M
½M
1 M
1M
x 1
1M
4M
LHL not equal to RHL
For continuity at x= 1 , LHL should be equal to RHL and this
common value should be equal to f(1)
Hence the given function is discontinuous at x = 1 and x=1 is point
of discontinuity.
1M
12
b
We know that

a
5
I 
2
5
I 
2
1M
a
x
25 x
dx  
dx
7x  x
7  (2  5  x)  2  5  x
2
5
1M
7x
dx
7x  x
5
I I 
2
3
I
2
b
f ( x)dx   f (a  b  x)dx
4M
x  7x
5
dx   dx  x 2  5  2  3
7x  x
2
5
1½ M
½M
95
13
14
 5ax 


 5ax 
1
a2
y  tan 1  2

tan


2
2 
2
 a  6x 
 a  6x 


 a2

 3x 2 x 
 a a 
3x
2x
1
 tan 
 tan 1  tan 1

a
a
 1  3x  2 x 
a 
 a
dy
1
3
1
2
hence, 
 

2
2
dx
 3x  a
 2x  a
1  
1  
 a 
 a 
3a
2a
 2
 2
2
a  9x a  4x2
 x2 1  1  1  x
 e dx
I  
  x  12 


 x2 1
2  x
 e dx
 

  x  12  x  12 


 x 1
2  x
 e dx
 

 x  1  x  12 


x 1
take f ( x) 
x 1
2
Then f ' ( x) 
2
 x  1
½M
4M
1½ M
1M
1M
1M
1M
1M
4M
Now it is of the form  e x  f ( x)  f '( x)  dx
 x 1 
Hence I= e x f ( x)  c  e x 
c
 x 1 
OR
1M
1
1 4

1  3 
x
I    4  dx
x
1
3
put 1  3  1  x 3  t  4 dx  dt
x
x
1M
2M
4M
5
4
1 1
4
1
I   t 4 dt  1  3   c
3
15  x 
1M
96
15
a = -1; b= 2
ba 3
h
  nh  3
n
n
f ( x)  7 x  5
we know that
1M
 f ( x)dx  limh  f (a)  f (a  h)  f (a  2h)  ....  f (a  n  1h 
b
h 0
a
1M
now,   7 x  5  dx  limh  12  (7h  12)  (14h  12)  .....  (7(n  1)h  12 
2
1
4M
h 0


 lim h  7h(1  2  3  ......  n  1  12n 

h 0 
7

 lim h   nh  nh  h   12nh 
h 0
2

9

2
1M
1M
16
R3  R3  R2
1M
x 1 x  2 x  a
  x2 x3 xb
1
1
c b
1M
R2  R2  R1
x 1 x  2 x  a
 1
1
ba
1
1
c b
4M
Since a, b, c are in AP we have c-b= b-a
and hence second and third rows are identical.
Hence the value of the determinant is zero
x 1 x  2 x  a
i.e. x  2 x  3
x3 x4
17
1M
1M
xb  0
xc
x 2  cos  ,  cos1 x 2
 1  cos   1  cos 
y  tan 1 
 1  cos   1  cos 
1M



cos  sin 


1
2
2
  tan 




 cos  sin 
2
2

1M
97


1  tan 

2  tan 1  tan(    )
 tan 1 


4 2 
1  tan 
2


4M

1
 cos 1 x 2
4 2
1M
dy
1 1
x

*2 x 
dx
2 1  x4
1  x4
1M
( OR )
dy
 1 cos ec 2 x
dx
1M
d2y
2 cot x
 2 cos ec 2 x. cot x 
2
dx
sin 2 x
1M
4M
2
d y
 2 cot x  2( y  x)
dx 2
d2y
 sin 2 x 2  2 y  2 x  0
dx
 sin 2 x.
1M
1M
18
let APB=C
P=A-1CB-1
1M
2 1 
consider A= 

 3 2
1  2  1  2  1

A-1= 
1  3 2   3 2 
 3 2 
Consider B= 

 5  3
1M
1  3  2 3 2

B-1=  
1  5  3 5 3
15 
 25
P= 

 37  22
19
1M
4M
1M
1
Matrix form of system of linear equations AX  B  X  A B where
1
1
1
 x
4
2
1
 3
 z 
 9 
A = 2  1 1  X=  y  B=  1 so finding the inverse of matrices


 
 
A
1
2
2 4
1 
 8  5 1 
14
4 1  3
2M
4M
98
2   4    1
2 4
1 
1
Now X = A B  8  5 1    1 =  2 
 
14
4 1  3  9  3 
2M
So x =-1, y =2, and z = 3.
SECTION C
20
The required trapezium is as given in Fig below. Draw
perpendiculars DP and CQ on AB
1M
(fig)
𝛥APD and 𝛥 BQC
BQ= 𝑥 cm
DP  QC  100  x 2
A.area of trapezium   x  10  100  x 2
A '( x) 
1M
1M
2 x 2  10 x  100
100  x 2
A '( x)  0  x  5
A ''( x) 
2 x3  300 x  100
3
2 2
100  x 
30
0
75
Thus the area of the trapezium is maximum at x=5
Area= 75 3 cm2
Let OC = r be the radius of the cone and OA = h be its height. Let a
cylinder with radius OE = x inscribed in the given cone .The height
QE of the cylinder is given by
1M
1M
A ''( x) x 5 
OR
𝑄𝐸
𝑄𝐴
1M
1M
𝐸𝐶
= 𝑂𝐶 because 𝛥𝑄𝐸𝐶 ∼ 𝛥𝐴𝑂𝐶
 QE 
h( r  x )
r
1M
6M
S  S ( x) 
2 h
rx  x 2
r

99

2 h
 r  2x 
r
4 h
S ''( x) 
r
S '( x) 
S '( x)  0  x 
1½ M
1M
(fig)
½M
r
2
Now S ''( x)  0, x
r
is a point of maxima of S
2
Hence the result
x 
1M
100
21
BL and CM are drawn perpendicular to x-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)
1M
1M
(fig)
6M
Equation of line segment AB is
Equation of line segment BC is
1M
Equation of line segment AC is
1M
1M
Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units
1M
22
A(1,6,3) B (α,β,γ) be the image of A. L point on the line so AL=BL
1  6   3   
L
,
,

2
2 
 2
General point is L(  ,2  1,3  2 )
1M
1M
1M
101
Dr’s
AL (λ-1,2λ -5,3λ-1)
AL perpendicular line 1(λ-1)+2(2λ-5)+3(3λ-1)=0
 λ=1
Points are (1,3,5)
The eqn
 1
6
3

1
3
5
2
2
2
 ( ,  ,  )  (1,0,7)
23
1M
1M
6M
1M
The given differential equation is:
6M
1M
Integrating both sides of this equation, we get:
1M
1M
1M
Substituting these values in equation (1), we get:
1M
6M
This is the required general solution of the given differential
equation
1M
102
24
Let the merchant stock x desktop models and y portable models.
Therefore,
x ≥ 0 and y ≥ 0
The cost of a desktop model is Rs 25000 and of a portable model is
Rs 4000. However, the merchant can invest a maximum of Rs 70
lakhs.
1M
The monthly demand of computers will not exceed 250 units.
The profit on a desktop model is Rs 4500 and the profit on a
portable model is Rs 5000.
1M
Total profit, Z = 4500x + 5000y
Thus, the mathematical formulation of the given problem is
subject to the constraints,
1M
The feasible region determined by the system of constraints is as
follows.
The corner points are A (250, 0), B (200, 50), and C (0, 175).
The values of Z at these corner points are as follows.
Corner point
Z = 4500x + 5000y
2M
103
A(250, 0)
1125000
B(200, 50)
1150000
C(0, 175)
875000
Maximum
The maximum value of Z is 1150000 at (200, 50).
25
26
Thus, the merchant should stock 200 desktop models and 50
portable models to get the maximum profit of Rs 1150000.
1M
To prove f is 1-1
2M
f is onto
1 x  6
f-1=
for all x Є [-5,  )
3
2M
2M
Let E1, E2 and E3 be the events that boxes I, II and III are chosen,
respectively.
1
Then P(E1) = P(E2) = P(E3) =
3
Also, let A be the event that ‘the coin drawn is of gold’
Then P(A|E1) = P(a gold coin from bag I) =
2
1
2
6M
1½ M
1½ M
P(A|E2) = P(a gold coin from bag II) = 0
1
P(A|E3) = P(a gold coin from bag III) =
2
Now, the probability that the other coin in the box is of gold
= the probability that gold coin is drawn from the box I.
= P(E1|A)
By Bayes' theorem, we know that
6M
1½ M
Let E be the event that the man reports that six occurs in the
throwing of the die and let S1 be the event that six occurs and S2 be
the event that six does not occur.
6M
1½ M
104
OR
Then P(S1) = Probability that six occurs =
1
6
5
6
P(E|S1) = Probability that the man reports that six occurs when six
has
actually occurred on the die
3
= Probability that the man speaks the truth =
4
P(E|S2) = Probability that the man reports that six occurs when six
has
not actually occurred on the die
3 1
= Probability that the man does not speak the truth = 1  
4 4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has
occurred is
P(S2) = Probability that six does not occur =
1M
1M
1M
6M
actually a six
1½ M
Hence, the required probability is =
3
8
1½ M
105
MODEL QUESTION PAPER 2015-2016
Class XII Mathematics
Maximum marks:-100
Time-3hrs
BLUE PRINT
Form of questions
VSA
(1 M)
SA
(4 M)
LA
(6 M)
TOTAL
Unit
RELATIONS & FUNCTIONS
-------
1(4)
ALGEBRA
1(1)
3(12)*
CALCULUS
2(2)
16(24)***
3(18)*
11(44)
VECTORS AND 3-D GEOMETRY
3(3)
2(8)
1(6)*
6(17)
LINEAR PROGRAMMING
----
-----
1(6)
1(6)
PROBABLITY
----
1(4)
1(6)
2(10)
TOTAL
6(6)
13(52)
7(42)
26(100)
*:-Internal choices are given
1(6)
2(10)
4(13)
106
Sub: Mathematics
GENERAL INSTRUCTIONS:
Sample question paper CLASS XII 2015-16
TIME:3Hrs
Maximum Marks:100
 All questions are compulsory.
 The question paper consist of 26 questions divided into three sections A,B and Csection A comprises of 6 questions of 1 marks each ,section B comprises of 13
questions of 4 marks each and section C comprises of 7 questions of 6 marks each .
 All questions in section A are to be answered in one word
 There is no overall choice. However internal choice has been provided
Section A
𝑥 + 1 𝑥 − 1 4 −1
1. If |
|=|
| find the value of x.
𝑥−3 𝑥+2 1 3
2. Find∫ 𝑠𝑒𝑐 2 (7 − 4𝑥) 𝑑𝑥.
3. Find the slope of the tangent to the curve y=3𝑥 4 -4x at x=4.
4. Find a unit vector in the direction of sum of vectors 𝑎 = 𝑖̂ − 2𝑗̂ + 7𝑘̂ and 𝑏⃗ = 2𝑖̂ + 2𝑗̂ − 3 𝑘̂
5. For what value of 𝛽, 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 2𝑖̂ + 𝛽𝑗̂ + 𝑘̂ and 𝑖̂ − 2𝑗̂ + 3𝑘̂
perpendicular to each other
6. If 𝑎 = 3𝑖̂ − 2𝑗̂ 𝑎𝑛𝑑 𝑏⃗ = 𝑗̂ + 2𝑘̂ find |𝑎 + 2𝑏⃗|
Section B
8
−1
−1
7. If 0<x<1, solve tan ( 𝑥 + 1) +tan (𝑥 − 1)=tan−1 ( 31)
8. A school wants to award its students for the values of honesty, regularity and hard work
with a total cash award of Rs.6, 000.Three times the award money for hard work added to
that given for honesty amounts to Rs.11, 000.The award money given for honesty and hard
work together is double the one given for regularity .Represent the above situation
algebraically and find the award money for each using matrix method.
0 1 2
9. Using elementary transformation find the inverse of the following matrix A=[1 2 3]
3 1 1
1+𝑎
1
1
10. Using properties of determinants prove that | 1
1+𝑏
1 |=ab+bc+ca+abc
1
1
1+𝑐
OR
107
𝑏+𝑐
| 𝑏
𝑐
𝑑𝑦
𝑎
𝑐+𝑎
𝑐
𝑎
𝑏 |=4abc
𝑎+𝑏
𝑙𝑜𝑔𝑥
11. If 𝑥 𝑦 =𝑒 (𝑥−𝑦) then show that 𝑑𝑥 = (𝑙𝑜𝑔𝑥𝑒)2
OR
Differentiate with respect to x , (sin 𝑥)x+ (cos 𝑥) 𝑠𝑖𝑛𝑥
𝑑2 𝑦
12. If y=sin−1 𝑥 show that (1-x2) 𝑑𝑥 2 - x
𝑑𝑦
=0
𝑑𝑥
𝑑𝑦
13. Solve the differential equation sec 𝑥 𝑑𝑥 - y = sin 𝑥
(5𝑥−2)𝑑𝑥
14. ∫ (3𝑥 2 +2𝑥+1)
𝑂𝑅
∫
𝑠𝑖𝑛𝑥 𝑑𝑥
(1 − 𝑐𝑜𝑠𝑥)(2 − 𝑐𝑜𝑠)
15. A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a
six. Find the probability that it is actually a six.
𝑑𝑦
16 . Solve the following D.E ( 𝑥 2 +1) 𝑑𝑥 +2xy =√𝑥 2 + 4.
17.Form the differential equation representing the family of ellipses having foci on x-axis
and centre at the origin.
18.Find the shortest distance between the lines 𝑟 = 𝑖̂ + 2𝑗̂ + 3𝑘̂ + 𝜆(2𝑖̂ + 3𝑗̂ + 4𝑘̂ ) and
𝑟 = 2𝑖̂ + 4𝑗̂ + 5𝑘̂ + 𝜇(3𝑖̂ + 4𝑗̂ + 5𝑘̂ )
19. Find a unit vector perpendicular to the plane of triangle ABC where the vertices are
A(3,-1,2) B(1,-1,-3) and C(4,-3,1)
𝐒𝐞𝐜𝐭𝐢𝐨𝐧 𝐂
20. Let f:N->R be a function defined as f(x)=4x2+12x+15.Show that f:N->s where S is the range of
f is invertible .Find the inverse of f.
8
21. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 27
of the volume of the sphere.
OR
If the sum of hypotenuse and a side of a right angled triangle is given .show
𝜋
that the area of triangle is maximum when angle between them is 3 .
𝜋
22. Using properties of definite integral ,evaluate ∫0 log(1 + 𝑐𝑜𝑠𝑥) 𝑑𝑥
OR
Evaluate as limit of sums
+ 𝑒 2𝑥 )𝑑𝑥
23. Find the area of the region {(x,y):y2≤4x,4x2+4y2≤ 9}
24. Two cards are drawn successively without replacement from a well shuffled pack of 52 cards.
4
∫0 (𝑥
Find the mean, variance and standard deviation of the number of Kings?.
𝑥+2
25. Find the distance of the point (-2,3,-4) from the line
the plane 4x+12y- 3z+1=0
OR
3
=
2𝑦+3
4
=
3𝑧+4
5
measured parallel to
108
Find the equation of the plane through the intersection of the planes
𝑟.(𝑖̂ + 3𝑗̂) – 6 =0 and 𝑟.(3𝑖̂ − 𝑗̂ − 4𝑘̂) = 0,whose perpendicular
distance from the origin is unity.
26. An aero plane can carry a maximum of two hundred passengers. A profit of Rs. 1000 is
made on each executive class ticket and a profit of Rs 600 is made on each economic class
ticket. the airline reserves at least 20 seats for executive class . However, at least four times
as many passengers prefer to travel by economy class, than by executive class. Determine
how many tickets of each type must be sold in order to maximize the profit for the airline?
what is the profit? Form a LPP and solve graphically? Aero plane is the fastest means of
transportation and reduces travelling time, so what is the importance of time?
109
SCORING KEY
SECTION A
1. x=2
−1
2. 4 tan(7-4x)+c
3. 764
3
4
4. 𝑟̂ =5 𝑖̂ + 5 𝑘̂
(1)
(1)
(1)
(1)
5. 2
6. 5
(1)
(1)
5
SECTION B
−1
𝑥+1+𝑥−1
7. tan (1−(𝑥+1)(𝑥−1) )
2𝑥
= 8/31
+31x-8=0
X=1/4 x=-8(not possible)
(1)
8. let x,y and z be the awarded money for honesty ,regularity and hardwork
A.T.Q the system of equations are:
X+y+z =6000 , x+3z =11000, x-2y+z=0
Wkt AX = B
𝑥
1 1 1
6000
Where A = [1 0 3] , X =[𝑦] , B =[11000]
𝑧
1 −2 1
0
(1/2)
6 −3 3
1
Getting A-1 = 6 [ 2
0 −2]
−2 3 −1
1
(12 )
Applying X = A-1B to getting x=500,y=2000,z=3500
0 1 2
1 0 0
9. Write A = IA [1 2 3]
= [0 1 0] A
-----------------------------1/2
3 1 1
0 0 1
For R1 ↔ R2
----------------------------------------------------------- ½
For R3 → R3 - 3R1
--------------------------------------- 1/2
1 2 2
0 1 1
[0 1 2] = [1 0 0] A
3 1 1
0 −3 1
R1 → R1 - 2R2
1 0 −1
−2 1 0
[0 1
]
=
[
2
1
0 0] A -------------------------------------------------1/2
0 −5 −8
0 −3 1
R3 → R3 + 5R2
1 0 −1
−2 1 0
[0 1 2 ] = [ 1
0 0] A
0 0 2
5 −3 1
For R3 → 12R3 ------------------------------------------------------------------------------------------------------ 1/2
For R1 → R1 + R3 ---------------------------------------------------------1 /2
R2 → R2 - 2R3
2−𝑥 2
4x2
(1)
( 1)
(1)
(1)
110
1
2
1 0 0
[0 1 0] = [−4
5
0 0 1
2
1
2
1
2
5
2
−3
2
A-1 = [−4
3
1
2
3
1
2
−1] A --------------------------------------------------1/2
−3
2
1
2
1
2
−1]
--------------------------------------------------1/2
1
2
10.Taking out a,b,cfrom R1, R2, R3
1
+ 1 1/𝑎
1/𝑎
𝑎
abc 1/𝑏
1/𝑐
1
𝑏
+1
1/𝑐
1/𝑏
1
𝑐
(1)
+1
Applying R1→ R1+R2+R3
Applying C2→C2-C1 ,C3→ C3 – C1
Expand along R1 to get the answer
OR
Applying R1→R1+ R2+ R3
Taking 2 as common factor from R1
Apply R2→R2 –R1 , R3 →R3- R1
Apply R1→
R1+R2+R3
(1/2)
Expand along R1 to get the answer
(1/2)
11. Taking log on both sides
𝑦 𝑙𝑜𝑔𝑥 = (𝑥 − 𝑦) log 𝑒
𝑦(log 𝑥 + 𝑙𝑜𝑔𝑒 ) = 𝑥𝑙𝑜𝑔𝑒
𝑥
𝑦 = 𝑙𝑜𝑔𝑥+1
(1)
(1)
(1)
(1)
(1)
(1)
1
(loge=1)
½
Differentiating w.r.t x, using quotient form
𝑑𝑦
𝑙𝑜𝑔𝑥
=
𝑑𝑥
(𝑙𝑜𝑔𝑥+1)2
2
But loge =1
𝑑𝑦
𝑙𝑜𝑔𝑥
= (𝑙𝑜𝑔𝑥+𝑙𝑜𝑔𝑒)2
𝑑𝑥
=
𝑙𝑜𝑔𝑥
½
(𝑙𝑜𝑔𝑒𝑥)2
OR
u= .𝑠𝑖𝑛𝑥 𝑥
y= u+v
𝑑𝑦
=
𝑑𝑥
𝑑𝑢
𝑣 = 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥
𝑑𝑣
+ 𝑑𝑥
𝑑𝑥
½
𝑢 = 𝑠𝑖𝑛𝑥
𝑥
Taking log on both sides
log 𝑢 = 𝑥𝑙𝑜𝑔 𝑠𝑖𝑛𝑥
Differentiating & simplifying
𝑑𝑢
=𝑠𝑖𝑛𝑥 𝑥 (𝑥𝑐𝑜𝑡𝑥 + log 𝑠𝑖𝑛𝑥 )
𝑑𝑥
𝑑𝑣
Similarly 𝑑𝑥 = 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 (−𝑠𝑖𝑛𝑥 𝑡𝑎𝑛𝑥 + 𝑐𝑜𝑠𝑥 log 𝑐𝑜𝑠𝑥)
112
1 12
111
𝐻𝑒𝑛𝑐𝑒
12.
𝑑𝑦
= 𝑠𝑖𝑛𝑥 𝑥 (𝑥𝑐𝑜𝑡𝑥 + 𝑙𝑜𝑔 𝑠𝑖𝑛𝑥 ) + 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 (−𝑠𝑖𝑛𝑥 𝑡𝑎𝑛𝑥
𝑑𝑥
+ 𝑐𝑜𝑠𝑥 𝑙𝑜𝑔 𝑐𝑜𝑠𝑥)
½
y= sin−1 𝑥
𝑑𝑦
1
=
2
√1−𝑥
𝑑𝑥
1
𝑑𝑦
𝑥 2 𝑑𝑥
√1 −
=1
Differentiating w.r.t x
√1 − 𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦
𝑥
− 𝑑𝑥 √1−𝑥 2 =0
𝑑2 𝑦
2
𝑑𝑦
Hence (1 − 𝑥 2 ) 𝑑𝑥 2 − 𝑥 𝑑𝑥 = 0
13 Solution
P = - cos 𝑥 Q = sin 𝑥 cos 𝑥
I.F = 𝑒 − sin 𝑥
Solution is y 𝑒 − sin 𝑥 = ∫ sin 𝑥 cos 𝑥 𝑒 − sin 𝑥 dx
= ∫ 𝑡 𝑒 −𝑡 dt
=-𝑒 − sin 𝑥 ( sin 𝑥 + 1 ) +c
Y=-sin 𝑥 - 1 + c𝑒 sin 𝑥
1
sin 𝑥 = t
14.. 5𝑥 − 2 = 𝐴(6𝑥 + 2) + 𝐵
½
equating & finding value of A & B
A = 5/6 , B = -11/3
5
6𝑥+2
11
1
∴ 6 ∫ 3𝑥 2 +2𝑥+1 𝑑𝑥 − 3 ∫ 3𝑥 2 +2𝑥+1 𝑑𝑥
Integrating ,
3𝑥+1
= 5/6log|3𝑥 2 + 2𝑥 + 1| −11/3√2 tan−1 (
) +c
1
1+ 1
√2
½
OR
put cos 𝑥 = 𝑢 , differentiating , − sin 𝑥 𝑑𝑥 = 𝑑𝑢,
−1
𝐴
𝐵
+
,
(1−𝑢)(2−𝑢) (1−𝑈) (2−𝑈)
Solving & getting the value of A & B , A = -1 , B = 1 ,
∴ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 ,
𝑠𝑖𝑛𝑥
1−𝑠𝑖𝑛𝑥
∫ (1−𝑐𝑜𝑠𝑥)(2−cos 𝑥) 𝑑𝑥 = log | 2−𝑠𝑖𝑛𝑥 | +c
1
1
1+1
15. A : He report that it is 6
E1 : Getting Head
E1 :Not getting Head
For P(E1 ), P( E2 ), P(A/E1), P(A/E2)
For final answer
½ Mark for each
2 Marks
½ Mark
16,Solution
2𝑥𝑦
P=𝑥 2 +1
√𝑋 2+4
Q= 𝑋 2 +1
I.F = 1+𝑥 2
Solution is y(1+ 𝑥 2 ) = ∫
=
√𝑥 2+4
1+𝑥 2 dx
𝑥 2 +1
∫ √𝑥 2 +
4 dx
112
x 2
x  4  2 log x  x 2  4  c
2
(2)
17.Equation of the family of ellipses,
X2/a2+y2/b2 =1
𝑦 𝑑𝑦
Differentiating and get,𝑥 (𝑑𝑥 ) = -b2/a2
Diff ertiate and simplify
(1+1)
18. ⃗⃗⃗⃗
𝑎2 − ⃗⃗⃗⃗
𝑎1 = 𝑖̂ + 2𝑖̂ + 2𝑘̂
1/2)
𝑖̂ 𝑗̂ 𝑘̂
⃗⃗⃗
⃗⃗⃗⃗
𝑏1 𝑋 𝑏2 = |2 3 4| = −𝑖̂ + 2𝑗̂ − 𝑘̂
3 4 5
⃗⃗⃗
⃗⃗⃗⃗
|𝑏1 𝑥𝑏2 | = √6
(1/2)
(1)
(1)
(
(1)
⃗⃗⃗⃗1 𝑥𝑏
⃗⃗⃗⃗2 )
(𝑎
⃗⃗⃗⃗⃗2 −𝑎
⃗⃗⃗⃗⃗1 ).(𝑏
|
⃗⃗⃗⃗
⃗⃗⃗⃗
|𝑏1 𝑥𝑏2 |
Shortest distance = |
(1)
= 1/√6
19.Getting the equation of the plane ABC
Getting perpendicular vector
Getting the required unit vector
(1)
(2)
(1)
(1)
SECTION C
20 Let y be an arbitrary element of range f .
then y =4x2 + 12x +15 , for some x in N
 𝑥=
√𝑦−6 −3
2
, as y≥ 6
Let us define g: S -> N by g(y) =
√𝑦−6 −3
(1)
2
. gof(x)=g(f(x) = g(4x2 + 12x + 15) = g((2x+3)2 + 6)
=
21.
√(2𝑥+3)2 +6−6−3
2
=
2𝑥+3−3
2
(1)
=x
(1)
Similarly getting fog(y) = y
(2)
Hence gof = IN and fog =IS => f is invertible with f-1 = g
(1)
R2 = x2 + r2
1
V = 3 π r2h
1
= 3 π (R2 – x2)(R+x)
(1)
R
(1)
113
1
(R2
dV/dx = 3 π
–2R x –
dV/dx = 0 => x = R/3
3X2)
(1)
(1)
𝑑2 𝑣
< 0 for x= R/3
So V is maximum for x = R/3
1 4
V = 27 (3 π r3)
𝑑𝑥 2
(1)
(1)
OR
let h and x be the length of the hypotenuse and one side of a right triangle and y be the
length of the third side,
Let A be the area of the triangle
1
1
A = 2 X.Y = 2 X.√ℎ2 − 𝑥 2
(1)
1
Z = A2 = 4 ( k2x2 – 2kx3)
𝑑𝑧
𝑑𝑥
𝑑𝑧
(1)
1
= 4(2k2x – 6kx2)
(1)
𝑘
= 0 => 𝑥 = 3
(1)
𝑑𝑥
𝑑2 𝑧
< 0 for x = k/3
Z is maxima at x = k/3 and h = 2k/3
So Cosϴ = ½
ϴ = π/3
22. Ans: - - πlog2
OR
(1)
𝑑𝑥 2
(1)
15+𝑒 8
Ans:2
23. to draw the correct graph
To get the point of intersection
to get the area in terms of integration
1
9𝜋 9
ans =3√2 + 8 - 4 sin-1(1/3)
(1)
(1)
(2)
(2)
. 24 P(X=0) = 188/221
P(X=1) = 32/221
P(X=2) = 1/221
Mean = 34/221
Variance = 6800/2212
Standard deviation = 0.37
.25 Getting the equation of the line
Getting the general point
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
Getting the value of  = 2
(2)
B(4,5/2,2)
(1)
AB = 17/2
.26 Value points
Let the number of executive class tickets be x
Let the number of economic class tickets be y
Now the LPP is given by
(1)
marks
114
Maximize : Z = 1000x + 600y
Subject to constraints : x+y ≤ 200
x ≥ 20 , y ≥ 4x , x,y ≥ 0
1
the required feasible region is given below
y
A
B
C
. .
.
(2)
Hence ∆ ABC is the feasible region. The corner points of the region are
A(20,180) , B(40,160) and C(20,80)
Now we can find the profit Z at the various corner points
Corner points
Z = 1000x + 600y
A(20,180)
Rs 128000
B(40,160)
Rs 136000 (Maximum)
C(20,80)
Rs 68000
(1)
Conclusion the maximum profit is Rs 136000 and is achieved when 40
tickets of executive class and 160 tickets of economic class is sold.
(1)
Time is important as it does not wait for anyone.it is commonly said that time
and tide waits for none .we can never get back the lost time.
(1)
(for any other points related to importance of time )
-------------------------------------------------------
115
Model Question Paper 1
Time: 3 hours
Subject: Mathematics
M:M 100
________________________________________________________________________
General instructions:
I. All questions are compulsory
II.
The question paper consists of 26 questions divided into three sections A,B and C.
Section A comprises 6 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.
Section A
1. Write a example of matrices A and B of order 2 x 2 such that AB = BA
2. Write a differential equation for y  A cos x  B sin x where A and B are arbitrary
constants.
3. Find the product of the order and degree of the differential equation
2
 d 2 y   dy 
x  2      y 2  0
 dx   dx 
2




4. The magnitudes of a and b are 1 and 2 respectively. Find the angle between a and b if
it is given that a .b  1









5. Write the value of i .( j  k )  j .( k  i )  k .( i  j ) .
6. Find the direction ratios of the line 3x + 2 = 5 - 3y = 5z
SectionB
7. A school wants to award its students for the values of Honesty, Regularity and Hard
work with a total cash award of Rs 6,000. Three times the award money for Hard work added
to that given for honesty amounts to Rs 11,000. The award money given for Honesty and
Hard work together is double the one given for Regularity. Represent the above situation
algebraically and find the award money for each value, using matrix method. Apart from
these values, namely, Honesty, Regularity and Hard work, suggest one more value which
the school must include for awards.
116
1 1 
 3


8. Find the inverse of the matrix A    15 6  5  and hence show that A1 A  I
 5 2 2 


1 a a2
2
9. Using the properties of determinants, show that a 2 1 a  a 3  1
a a2 1

3ax  b

10. If the function f ( x )  11
 5ax - 2b


if x  1
if x  1 is Continuous at x = 1, find the values of a
if x  1
and b.
𝑑𝑦
11. If y=𝑠𝑖𝑛𝑥 𝑥 + 𝑐𝑜𝑠𝑥 𝑡𝑎𝑛𝑥 ,find 𝑑𝑥
( OR)
If x 1  y  y 1  x  0 , -1 < x < 1. Show that
dy
1

dx 1  x 2
12. .Find the intervals in which the function f(x) = 2x3 – 9x2 + 12x + 30 in increasing or
decreasing.
𝜋
1
13 . Evaluate∫02 3+2𝑐𝑜𝑠𝑥 𝑑𝑥
Evaluate

(OR)
2x  3
x 2  2x  5
dx
14. Evaluate
 x sin
15 Evaluate
 cos x (1  cos x)dx .
1
x dx
1  cos x
16. Find the equation of the plane passing through the intersection of the plane
2𝑥 + 3𝑦 − 𝑧 + 1 = 0 𝑎𝑛𝑑 𝑥 + 𝑦 − 2𝑧 + 3 = 0 and perpendicular to the plane
3𝑥 − 𝑦 − 2𝑧 − 4 = 0.
17. Show that the lines 𝑟 = 3𝑖̂ + 2𝑗̂ - 4𝑘̂ + 𝜆 (𝑖̂ + 2𝑗̂ +2𝑘̂) and 𝑟 = 5𝑖̂ - 2𝑗̂ + μ (3𝑖̂ +2𝑗̂ +6𝑘̂)
are intersecting. Hence find their point of intersection.
18. In a group of students, 200 attend coaching classes, 400 students attend school regularly
and 600 students study themselves with help of peers. The probability that a student
will succeed in life who attend coaching classes, attend school regularly and study
themselves with help of peers are 0.1, 0.2 and 0.5 respectively. One student is
selected who succeeded in life, what is the probability that he study himself with
help of peers.What type of study can be considered for the success in life and why?
1
1 
19. Prove that 2 tan 1    tan 1   
3
7 4
117
Section-C
20. Prove that the relation R on the set N x N defined by (a, b) R (c, d) <=> a + b = b + c,
for all (a, b), (c, d) N x N is an equivalence relation.
21. A wire of length 28m is to be cut into two pieces, one piece is bent into a circle and the
other into a square. What should be the length of the two pieces so that the
combined area of the square and the circle is minimum.
OR
A window is in the form of a rectangle surmounted by a semicircle. If the perimeter
of the window is 30 cm, find the dimensions of the window so that maximum
possible light is admitted.
22. A farmer has a piece of land. He wishes to divide equally in his two sons to maintain
peace and harmony in the family. If his land is denoted by area bounded by curve
y2=4x and x=4 and to divide the area equally he draws a line x=a , what is the value of
a? What is the importance of equality among the people?
dy

 y  x cos x  sin x if y   1 .
23. Solve the differential equation x
dx
2
24. A furniture dealer deals in only two items - tables and chairs. He has Rs. 10,000 to invest
and a space to store at most 60 pieces. A table costs him Rs. 500 and a chair costs
him Rs. 100. He can sell table at Rs. 550 and a chair at Rs. 115. Assume that he can
sell all the items that he buys. Formulate this problem as an LPP so that he
maximizes his profit.
25.
Find the image of the point (1,2,3) in the plane x+2y+4z=38
OR
Find the shortest distance between the lines r = i +2 j - 4k +(2i + 3j + 6k) and
r=3i+3j-5k+(-2i+3j+8k)
26. Three cards are drawn successively ,with replacement, from a well-shuffled pack of 52
cards. Find the probability distribution of number of aces. Also find mean and
variance of
the distribution.
118
Model Question Paper 2
Time: 3 hours
Subject: Mathematics
M:M 100
________________________________________________________________________
General instructions:
I. All questions are compulsory
II.
The question paper consists of 26 questions divided into three sections A,B and C.
Section A comprises 6 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.
Section A
1. Let A = 1, 2,3, 4 and R is a relation in A given by R =
1,3 ,  2, 4 ,  2,3 , 3,1 ,  4, 2
Is R symmetric? Give reason.

 1 
2. Evaluate: sin   sin1    
 2 
6
3. Find the value of
4. Evaluate

sin20
 cos 20
sin70
cos 70
.
dx
5  2 x  x2
5. Find the angle between the vectors i  j  k and i  j  k .
6. If the equation of a line is
x 3 y  2 z 5


, then find the direction cosines of the
1
2
4
line.
7. Prove that the function f defined by f(x) =
OR
3x  4
is invertible. Also find the inverse of f.
5x  7
119
Prove that the binary operation * defined by a * b =
for the
ab
is commutative. Verify
3
associative property.
a bc
2a
2a
3
2b
bca
2b   a  b  c  .
8. Prove that
2c
2c
c a b
Or
Prove that
.
 1
 1
 a 
 a   2b
9. Prove that tan   cos 1     tan   cos 1     .
 b 
 b  a
4 2
4 2
10. Differentiate wrt x:  sin x 
cos x
  cos x 
sin x
.
OR
If x  a cos    sin  and y  a sin    cos  , find
11. If y 1  x 2  log


d2y
.
dx2
1  x 2  x then find the value of 1  x 2 
dy
 xy  1 .
dx
12. Out of a group of 8 highly qualified doctors in a hospital, 6 are very kind and
cooperative with their patients and so are very popular, while the other two
remain reserved. For a health camp, three doctors are selected at random. Find
the probability distribution of the number of very popular doctors.
What values are expected from the doctors?
OR
A box of oranges is inspected by examining three randomly selected oranges drawn
without replacement. If all the three oranges are good, the box is approved for sale,
otherwise, it is rejected. Find the probability that a box containing 15 oranges out of
which 12 are good and 3 are bad ones will be approved for sale.
13. Evaluate
x2  1
  x  1  x  3dx
2
120
14. Evaluate


15. Evaluate
1
x sin x
1  x2
x dx
 1  cos
0
dx
2
x
16.Find the equation of the normal to the curve 2 x 2  y 2  14 which is
parallel to
the line x + 3y = 4.
17. Find the shortest distance between the lines
x 1 y  2 z  3


and
1
1
2
x 1 y 1 z 1


.
1
2
2
18. Let
and
perpendicular to both
19.
. Find a vector
and , and
which is
.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion,
4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg
rice is Rs70. Find the cost of each item per kg by using matrix method.
20. Show that the altitude of the right circular cone of maximum volume that can be
inscribed in a sphere of radius r is
.
OR
Find the point on the curve x 2 = 4y which is nearest to the point (-1, 2).
21. Find the area of the region enclosed between the circles x 2  y 2  1 and
 x  1
2
 y2  1 .
22. A doctor is to visit a patient. From past experience it is known that the
probability that he will come by train, bus, scooter or by car are 3/10, 1/5, 1/10,
2/5 respectively. The probabilities that he will be late are 1/ 4, 1/3 & 1/12 if he
comes by train, bus & scooter respectively but if he comes by car, he will not be late.
When he arrives, he is late. What is the probability that he has come by train?
OR
Three bags contain balls as shown in the table.
121
Bag
Number
Number of
Number of red
of white
black balls
balls
balls
I
1
2
3
II
2
1
1
III
4
3
2
A bag is chosen at random and 2 balls are drawn from it. They happen to be white
and red. What is the probability that they came from bag III?
23. A company manufactures two types of novelty souvenirs made of plywood.
Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for
assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes
each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours
of assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs.
How many souvenirs of each type should the company manufacture in order to
maximize the profit?
24. Find the equation of the plane passing through the point i  2j  3k perpendicular




to the planes r  i  j  k  3 and r  2i  3j  4k  0 .
25. Solve the Differential equation cos x

2
26. Evaluate  log sin x dx
0
dy
 y  sin x given that y (0) = 2.
dx
122
Tips for scoring well in the exam
Pre-exam preparations
Create a positive attitude. You may ask "how does creating a good attitude supposed to
boost up my mark?" Well, if you create a good attitude in studying and think positive things
about studying, it's likely that you will start liking a subject that you were really not fond of
before. You may also get marks for having a good attitude and for not disrupting a class.
Attend your classes regularly. If you attend classes daily instead of skipping them, you will
be there to listen to all the lessons the teachers are teaching and won't miss important days.
It's likely that you will remember or learn more if you are in class when the teacher is
teaching than if you skipped class and then take notes later on.
Set goals for yourself. Set goals like 'get good marks in math, science etc' for yourself and
then try your hardest to achieve those goals. If you fail once, don't give up, keep trying.
Do your homework daily. When you get homework, do it, don't leave it. If you're tempted
to watch your favourite TV show instead of do homework, think about it this way: Is your
favourite TV show going to help you get to a good university and possibly help you achieve
the job of your dreams? If the answer is yes, then by all means watch the show. Homework
is important because it gives you practice on the subject that you are learning. Always do
your homework.
Ask for help when you need it. If you have a question on something, ask for help, don't just
leave it behind. You can ask anybody who has knowledge on the subject that you need help
with like your parents, teacher, brother or sister. It will most definitely help you in tests and
quizzes.
Manage your time effectively. It will help you reduce anxiety and focus on studying. If you
have a test next week, start studying now. Try not to study at the last minute and cram the
night before. Try studying 1 or 2 hours daily and leave a half hour for homework. If you
study before you do homework, it will help you do your homework faster and help you
understand the subject better.
Always review. After school, review what you learned that day
Develop test smarts. This will really help and increase your confidence when taking exams if
you're familiar with the typical exam format, common errors to avoid, and know how the
concepts in a subject area usually tested.
Know your personal learning style. It will help you maximize your learning by using
effective study techniques, developing meaningful notes, and making the most efficient use
of your study time.
Linguistic learner: learns best by saying, hearing and seeing words; is good at
memorizing things such as dates, places, names and facts.
Logical/mathematical learner: learns best by categorizing, classifying and working with
abstract relationships; is good at mathematics, problem solving and reasoning.
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Spatial learner: learns best by visualizing, seeing, working with pictures, is good at
puzzles, imaging things, and reading maps and charts.
Musical learner: learns best by hearing, rhythm, melody, and music; is good at
remembering tones, rhythms and melodies, picking up sounds.
Bodily/kinesthetic learner: learns best by touching, moving, and processing knowledge
through bodily sensations; is good at physical activity.
Interpersonal learner: learns best by sharing, comparing, relating, cooperating; is good
at organizing, communicating, leading, and understanding others.
Intrapersonal learner: learns best by working alone, individualized projects, and selfpaced instruction.
During Exams Preparations
Do not panic
Have healthy food in time and sleep in time
Focus on the question paper and identify which questions can be answered correctly
Attempt all the questions
Answer the questions as per your knowledge and skills
Don’t think about those chapters which you have not remembered well. Feel
confident about those lessons which you have learned well. It will increase your
confidence and will remove your stress
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