iit 1 unlocked maths
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Dear Student,
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Document Number : SRMGW - IIT-JEE - MATH-01
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Preface
About the book
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IIT - MATHS
SET - 1
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INDEX
1. BASIC MATHEMATICS...................................................................................
2. ALGEBRA PROGRESSION .............................................................................
3. BASIC TRIGONOMETRY ..............................................................................
4. TRIGONOMETRY EQUATIONS ...................................................................
5. INVERSE TRIGONOMETRY FUNCTION...................................................
6. PROPERTIES OF TRIANGLE .......................................................................
7. CO - ORDINATE GEOMENTRY ...................................................................
2
54
132
164
186
210
232
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IIT- MATHS
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1 BASIC MATHEMATICS
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BASIC TRIGONOMETRY
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Number System
(i) Natural numbers : N = {1, 2, 3, 4, . . . . . }
(ii) Whole numbers : W = {0, 1, 2, 3, 4, . . . . . }
(iii) Integers : Z or I = {. . . . . 3, 2, 1, 0, 1, 2, 3, . . . . .}
Natural numbers are also called positive integers (denoted by Z+or I+)Whole numbers are also called nonnegative integers.
The set of negative integers, Zor I= {. . .. . 3, 2, 1}.
The set of non positive integers is {, 3, 2, 1, 0}.
Zero is neither positive nor negative but it is nonpositive as well as nonnegative.
(iv)Rational numbers:Numbers of the form p/q where p, q Z and q 0 (because division by
zero is not defined). Q represents their set. All integers are rational numbers with q = 1
When q 1 and p, q have no common factor except 1, the rational numbers are calledfractions.
Rational numbers when represented in decimal form are either terminating or non
terminating but repeating.
e.g., 5/4 = 1.25 (terminating)
5/3 = 1.6666 . . . . . (non terminating but repeating)
(v) Irrational numbers: Numbers, which cannot be represented in qp
form.
In decimal representation, they are neither terminating nor repeating all surds fall into this
category
e.g., 2 , 3/115 , p, etc.
Note : p 22/7, 22/7 is only an approximate value of p in terms of rational numbers,
taken for convenience Actually p = 3.14159 . . . . .
(vi)Real numbers: All rational and irrational numbers taken together form the set of real
numbers, represented by R. This is the largest set in the real world of numbers.
Also note that
Integers which give an integer on division by 2 are called even integers otherwise they are
called odd integers.
Zero is considered as even number.
The set of natural numbers can be divided in two ways.
(i) Odd and even natural numbers.
(ii) Prime numbers (which are not divisible by any number except 1 and themselves) andcomposite numbers (which have some other factor apart from 1 and themselves).
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1 is neither prime nor composite
2 is the only even number which is prime
Set Theory
Basic Concept
Set: A set is a welldefined collection of objects or elements. Each element in a set is unique.Usually but not necessarily a set is denoted by a capital letter e.g. A, B, . . . . . U, V etc. and the elements
are enclosed between brackets {}, denoted by small letters a, b, . . . . . x, y etc. For example:
A = Set of all small English alphabets
= {a, b, c, . . . . . x, y, z}
B = Set of all positive integers less than or equal to 10
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
R = Set of real numbers
= {x : < x < }
The elements of a set can be discrete (e.g. set of all English alphabets) or continuous (e.g., set of
real numbers). The set may contain finite or infinite number of elements. A set may contain no elements
and such a set is called Void set or Null set or empty set and is denoted by (phi). The number of
elements of a set A is denoted as n(A) and hence n () = 0 as it contains no element.
Union of sets
Union of two or more sets is the set of all elements that belong to any of these sets. The symbol usedfor union of sets is
i.e., AB = Union of set A and set B = {x : x A or x B(or both)}
e.g. If A = {1, 2, 3, 4} and B = {2, 4, 5, 6} and C {1, 2, 6, 8}, then ABC = {1, 2, 3, 4, 5, 6, 8}
Intersection of sets
It is the set of all the elements, which are common to all the sets. The symbol used for intersection
of sets is i.e., AB = {x : xA and xB} e.g. If A = {1, 2, 3, 4} & B = {2, 4, 5, 6}and C = {1, 2, 6, 8},
then ABC = {2}.
Remember that n(A B) = n(A) + n(B) n (A B)
Difference of two sets
The difference of set A to B denoted as A B is the set of those elements that are in the set
A but not in the set B i.e., A B = {x : x A and xB}
Similarly B A = {x : x B and x A}. In general A B B A
e.g. If A = {a, b, c, d} and B = {b, c, e, f}, then A B = {a, d} and B A = {e. f}
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BASIC TRIGONOMETRY
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LOGARITHM
If a is a positive real number other then 1 and ab= c, then we write logac = b obviously c is positive.
For example log381 = 4 34= 81
Note
The expression logba is meaningful for a > 0 and for either 0 < b < 1. or b > 1
a = b alogb
logab = alog
blog
c
c
logba
1 log
ba
2 1b0ifaa0
1bif0aa
21
21
Brain Teaser 1 : log x2= 2logx, is it true or false?
Formulae
(i) loga|mn| = log
a|m| + log
a|n|
(ii) loganm
= loga|m| loga|n|
(iii) loga|mn| = n log
a|m|
(iv) logab = log
cb log
ac
(v) log ka N = k1
logaN
Modulus Function
Let x R, then the magnitude of x is called its absolute value and in general, denoted by |x| and
defined as |x| =
0x,x
0x,x
Note that x = 0 can be included either with positive values of x or with negative values of x. As we
know all real numbers can be plotted on the real number line, |x| in fact represents the distance of number
x from the origin, measured along the numberline. Thus |x| 0 secondly, any point x lying on the real
number line will have its coordinate as (x, 0). Thus its distance from the origin is 2x
Hence |x| = 2x . Thus we can define |x| as |x| = 2x
e.g if x = 2.5 then |x| = 2.5, if x = 3.8 then |x| = 3.8
Brain Teaser 2 : If x R then 2
1,
2
1,
2
1is
4x
x2
or not defined.
Basic Properties :
||x|| = |x|
|x| > a x > a or x < a if a R+and x Rif a R
|x| < a a < x < a if a R+and no solution if a R{0}
|x + y| |x| + |y|
|xy| = |x||y|
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|y||x|
yx
, y 0
Intervals
Intervals are basically subsets of R and are of very much importance in calculus as you will get to
know shortly. If there are two numbers a, b R such that a < b, we can define four types of intervals as
follows:
Open interval : (a, b) = {x : a < x < b} i.e., end points are not included.
Closed interval: [a, b] = {x : a x b} i.e., end points are also included. This is possible only when
both a and b are finite.
Openclosed Interval: (a, b] = {x : a < x b}
Closedopen interval:[a, b) = {x : a x < b}
The infinite intervals are defined as follows
(a, ) = {x : x > a}
[a, ) = {x : x a}
(, b] = {x : x b}
intervals are particularly important in solving inequalities or in finding domains etc.
Inequalities
The following are some very useful points to remember
a b either a < b or a = b.
a < b and b < c a < c.
a < b a + c < b + c cR
a < b a > b i.e., inequality sign reverses if both sides are multiplied by a negative number.
a < b and c < d a + c < b + d and a d < b c.
a < b ma < mb if m > 0 and ma > mb if m < 0.
0 < a < b ar< brif r > 0 and ar> brif r < 0.
a
1a 2 a > 0 and equality holds for a = 1.
a
1a 2 a < 0 and equality holds for a = 1.
If a1> b
1, a
2> b
2, a
3> b
3. . . . . , where a
i> 0, b
i> 0, i = 1, 2,
Then a1+ a
2+ a
3+ . . . > b
1+ b
2+ b
3+ . . . and a
1a
2a
3. . . > b
1b
2b
3. . .
If a > b, p and q are some positive integers, then following results are evident.
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BASIC TRIGONOMETRY
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a > b an> bn an< bnwhere n N
a > b al/q> bl/q ap/q> bp/q
Wavy Curve Method
In order to solve inequalities of the form
0
xQxP,0
xQxP , where P(x) and Q(x) are polynomials, we use the following method:
If x1and x
2(x
1< x
2) are two consecutive distinct roots of a polynomial equation, then within this
interval the polynomial itself takes on values having the same sign. Now find all the roots of the polyno-
mial equations P(x) = 0 and Q(x) = 0. Ignore the common roots and write
m321n321
x.....xxx
x.....xxxxf
xQ
xP
,
Where a1, a2, . . . . . an, b1, b2, . . . . . , bmare distinct real numbers. Then f(x) = 0 for x = a1,a
2, . . . . . , a
nand f(x) is not defined for x = b
1, b
2, . . . . . , b
mapart from these (m + n) real numbers
f(x) is either positive or negative. Now arrange a1, a
2, . . . . . , a
n, b
1, b
2, . . . . . , b
min an increasing order say
c1, c
2, c
3, c
4, c
5, . . . . . , c
m+n. Plot them on the real line. And draw a curve starting from right of c
m+nalong
the real line which alternately changes its position at these points. This curve is known as the wavy curve.
The intervals in which the curve is above the real line will be the intervals for which f(x) is positive
and intervals in which the curve is below the real line will be the intervals in which f(x) is negative.
Factor Theorem
Let p(x) be a polynomial of degree greater than or equal to 1 and a be a real number such that p(a)
= 0, then (x a) is a factor of p(x). Conversely, if (xa) is a factor of p(x), then p(a) = 0.
Remainder Theorem
Let p(x) be any polynomial of degree greater than or equal to one and a be any real number. If p(x)is divided by (xa), then the remainder is equal to p(a).
DETERMINANTS
Consider the equations a1x + b
1y = 0 and a
2x + b
2y = 0. These give
2
2
1
1
2
2
1
1
b
a
b
a
b
a
x
y
b
a
a
1b
2 a
2b
1= 0
We express this eliminant as 2211
ba
ba
= 0.
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IIT- MATHS
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The expression22
11
ba
bais called a determinant of order two, and equals a
1b
2 a
2b
1.
A determinant of order three consisting of 3 rows and 3 columns is written as
333
222
111
cba
cba
cba
and is equal to a1 3322
cb
cb
b1 3322
ca
ca
+ c1 3322
ba
ba
= a1(b
2c
3 c
2b
3) b
1(a
2c
3 c
2a
3) + c
1(a
2b
3 b
2a
3)
The numbers ai, b
i, c
i(i = 1, 2, 3) are called the elements of the determinant.
Function
In the study of natural phenomena and the solution of technical and mathematical problems, it is
necessary to consider the variation of one quantity as dependent on the variation of another. For example,
in studies of motion, the path traversed is regarded as a variable, which varies with time. Here we say that
the distance traversed is a function of time. The area of a circle, in terms of its radius R, is pR2. If R takes
on various numerical values, the area assumes different numerical values. So the variation of one variable
brings about a variation in the other. Hence area of the circle is a function of the radius R.
If to each value of variable x (within a certain range) there corresponds a unique value of another
variable y, then we say that y is a function of x, or, in functional notation y = f(x). The variable x is called
the independent variable or argument. And the variable y is called the dependent variable. The relation
between the variable x and y is called a functional relation. The letter f in the functional notation y = f(x)
indicates that some kind of operation must be performed on the value of x in order to obtain the values of y.
f(x)
y2
y1
x1
x2
x3 x
y = f(x)
f(x) L
C
B
A
x0 x
y = f(x)
y3
y2
y1
Fig (a) Fig (b)
These figures show the graph of two arbitrary curves. In the figure any line drawn parallel to y-
axis would meet the curve at only one point. That means each element of X would have one and only one
image. Thus the figure (a) would represent the graph of a function.
In the figure (b) certain line (e.g. line L) would meet the curve in more than one points (A, B and
C). Thus element x0of X would have three distinct images. Thus this curve will not represent a function.
The set of all possible values which the independent variable (here x) is permitted to take for a
given functional dependence to be defined is called the domain of definition or simply the domain of the
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BASIC TRIGONOMETRY
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function.
e.g. The function y = sin x is defined for all values of x. Therefore its domain of definition is the
infinite interval < x < .
The function y =1x
1
is defined for all x > 1 its domain is (1, ).
Elementary Functions:
(i) Constant function: y = c where c is a constant, defined for all real x.
(ii) Power function: xy
(a) a is positive integer. The function is defined in the infinite interval < x < .
(b) a is negative integer. The function is defined for all values of x except for x = 0.
(iii) General exponential function:y = ax, where a is positive not equal to unity. This function
is defined for all values of x.(iv)Logarithmic function:y = log
ax, a > 0 but a 1. This function is defined for all x > 0.
(v) Trigonometric function: y = sinx, y = cosx defined for all real x
y = tanx, y = secx, defined for R (2n + 1)2
.
y = cotx, y = cosecx, defined for R n , where nl
It must be noted that in all these function the variable x is expressed in radians. All these
function have a very important property that is Periodicity.
` Is sec2 tan2 = 1 valid for all R (real) ?
(vi) Algebraic function:
(a) Polynomial function: y = a0xn+ a
1xn1+ + a
n, where a
0, a
1 a
nare real constants
(a00) and n is a positive integer, called the polynomial of degree n.
e.g. y = ax + b, a 0 (a linear function)
y = ax2+ bx + c, a 0 (a quadratic function)
A polynomial function is defined for all real values of x.
(b) Rational Function y =m
1m1
m0
n1n
1n
0
b...xbxb
a...xaxa
e.g:y = a/x (inverse variation)
The rational function is defined for all values of x except for those where the denominatorbecomes zero.
(c) Irrational function e.g. y = 2
2
x51
xx2
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Differential Calculus
Let y = f(x) be a function. Putting the values of x in this relation, we obtain the corresponding
values of y. Suppose we start putting some values of x in increasing order. The respective values of y
that we obtain may turn out to be in increasing order, or in decreasing order, or they may remain constant,
or they may even have a mixed trend, depending upon the type of function.
Let us take two values of x: x1and x2(x1< x2). So, y1= f(x1) and y2= f(x2)
Then, the quantity
12
12
xx
yywill tell us the average rate of change of y w.r.t. x in the interval [x
1, x
2] .
Let y2> y
1
12
12
xx
yy
is positive Function is increasing on an average.
if y2< y
1
12
12
xx
yy
is negative Function is decreasing on an average.
If y2= y
1
12
12
xx
yy
is zero Function is constant on an average.
As you can see, if x1and x
2are sufficiently far apart, the quantity
12
12
xx
yycan not give the exact
idea of the variation of y w.r.t. x in the interval [x1, x
2]. it just provides an overall information. For
example if y2= y
1it does not necessarily mean that y is same for all x in the interval [x
1, x
2]. Thus, to
obtain a sufficiently accurate information, we have to choose x1and x
2 sufficiently close to each other.
This sufficiently close is the key word here. To know the rate of change of y w.r.t. x at x = x1, we take x2
very near to x1(as much as possible), i.e., 2x tends to x1and then calculate
12
12
xx
yy
. In the limiting case,
we say that x2nearly coincides with x
1and represent it as x
2 x
1. We use the notation
1xxdx
dy
for
12
12
xx
yy
as x2 x
1. dx means small change in x (near x = x
1) and dy means the corresponding change in y. We call
dx
dythe derivative or the differential coefficient of y w.r.t. x.
y2 y1
x2 x1
x1 x2
y
x
(You can understand it physically by taking x as time and y as displacement of a body,
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Thendx
dydenotes the magnitude of velocity).
dx
dyis also represented as f (x) or
dx
)x(df
Graphically, 1xxdx
dy
(i.e., dxdy
computed at x = x1) denotes the slope of the tangent to the curve y =
f(x) at x = x1
We will not here derive the formulae for (x) of various functions, but we give the results of the
derivations here,
Basic Differentiation Formulae
y = constant 0dx
dy y = tan1x 2
x1
1
dx
dy
y = xn 1nnxdx
dy y = cot1x 2x11
dx
dy
y = sinx xcosdx
dy y = cosec1x
1x|x|
1
dx
dy2
y = cosx xsindx
dy y = sec1x
1x|x|
1
dx
dy2
y = tanx xsecdxdy 2 y = ax alnadx
dy x
y = cotx xeccosdx
dy 2 y = ex xedx
dy
y = sin1x 2x1
1
dx
dy
y = log
ax alnx
1
dx
dy
y = cos1x 2x1
1
dx
dy
y = ln x
x
1
dx
dy
Some Important Theorems
The following are very important theorems, which can be applied directly.
Theorem 1:
If a function is of the form y = k f(x), where k is a constant, thendx
)x(dfk
dx
dy
Theorem 2:
The derivative of the sum or difference of a finite number of differentiable functions is equal
to the sum or difference of the derivatives of these functions.
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i.e., if y = u (x) + v (x) + w(x) then y= u(x) + v(x) + w(x).
Theorem 3
The derivative of the product of two differentiable functions is equal to the product of the
derivative of the first function with the second function plus the product of the first function
with the derivative of the second function: i.e., if y = uv, then y= uv + uv.
This formula can be extended for the derivatives of the product of any (finite) number of
functions.
Theorem 4
If y = )x(v)x(u
, then y = 2vuvvu
dx
dy
Theorem 5
If y = uv, where u and v are functions of x, then y= vuv-1u+ uvln u.v
Derivative of a Composite FunctionGiven a composite function y = f(x), i.e., a function represented by
y = F(u), u = f (x) or y = F[f(x)],
then y=dx
du
du
dF
dx
dy
This is called the chain rule. The rule can be extended to any number of composite function;
e.g. if
y = f(u(v)), then y=dx
dv
dv
du
du
df
dx
dy .
Parametric Representation of a Function and its Derivatives
We find the trajectory of a load dropped from an aeroplane moving horizontally with uniform
velocity v0at an altitude y
0. We take the co-ordinate system as shown and assume that the load is dropped
at the instant the aeroplane cuts the y-axis. Since the horizontal translation is uniform, the position of the
load at any time t, is given x = v0t, y = y
0
2
gt 2
X
(x, y)
v0
Y
Those two equations are called the parametric equations of the trajectory because the two vari-
ables x and y have been expressed in terms of the third variable t (parameter) i.e. two equations x = (t),
y = (t)
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where t assumes values that lie in a given interval (t1, t
2)
Then )t()t(
)dt/d(
)dt/d(
dt/dx
dt/dy
dx
dy
Second Derivative of a Function
The second derivative of y w.r.t. x is the function obtained by differentiatingdx
dyw.r.t. x.
It is represented as 22
dx
ydor y or f (x). e.g. If y = x5then
dx
dy= 5x4
So,
dx
dy
dx
d
dx
yd2
2
= )x5(dx
d 4 = 5.4 x3= 20x3
The acceleration a of a particle is the second derivative of the distance s (given as a function oftime).
i.e. if s = f(t) then v =dt
ds= f (t) and a = 2
2
dt
sd=
dt
dv= f (t)
THE BEHAVIOUR OF FUNCTIONS
The following behaviours of a function are important to study
Increasing and Decreasing Functions
(i) Increasing Functions
If y = f(x) and x2> x
1implies y
2> y
1for any x belonging to the interval [a, b], then y is said
to be an increasing function of x. (x) increases in [a, b] f (x) > 0 x in (a, b).
(ii) Decreasing Function
If x2> x
1y
2< y
1for any x belonging to [a, b], then y is said to be a decreasing function
of x.
f(x) decreases in [a, b] f (x) < 0 x in (a, b)
For a constant function, f (x) = 0For a non-decreasing function f (x) 0
For a non increasing function, f (x) 0
Maxima and Minima of Functions
A function f is said to have a maxima at x = x0if f(x) < f(x
0), x in the immediate neighbourhood
of x0.
Similarly, a function f is said to have a minima at x = x0 if f(x) > f(x
0), x in the immediate
neighbourhood of x0
We have used the word immediate here because a given function may have any number of high
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and low points. It is just like moving on an uneven surface (which has many bumps and depressions).
Mathematically, these bumps are called the points of local maxima and the depressions are called the
points of local minima. The highest of all the bumps is the global maxima and the lowest of all the
depressions is the global minima. We state here the preliminary methods only to find the maxima and
minima of functions.
(a) Finddxdy or f (x)
(b) Find the points at which it becomes zero. These points are called critical points.
To find the points of maxima and minima we resort to either of the following tests.
(a) First Derivative Test
Suppose x = x0is a critical point i.e., f (x
0) = 0.
If f (x) changes sign from positive to negative in the neighbourhood of x = x0
Maxima at x0
If f (x) changes sign from negative to positive in the neighbourhood of x = x0
Minima at x0
(b) Second Derivative Test
(i) Find 22
dx
ydor f(x)
(ii) Compute the value of f (x) at the critical pointsIf it is positive Minima at those values of x.
If it is negative maxima at those values of x.
If the function is defined in an interval [a, b], then to find the maxima and minimum
values i.e., global maxima and global minimum of the function in that interval we com
pare the values of the function (i.e., y) at all the critical points and also the end points (i.e.,
y = f(a) and f(b)). Then the largest among them gives the global maximum values and
smallest gives the global minimum values.
Integral Calculus: The Antiderivative of Function
A function F(x) is called the antiderivative of the function f(x) on the interval [a, b] if, at all points
of the interval f(x) = F(x).
For example, the antiderivative of the function f(x) = x2 is3x3
, as'
3
x3
= x2. The function
13
xand2
3
x 33 are also antiderivatives of f(x) = x2. Infact, C
3x3
, where C is an arbitrary constant, is
the antiderivative of x
2
. So if a function f(x) possesses an anti-derivative F(x), then it possesses infinitelymany antiderivatives, all of them being contained in the expression F(x) + C, where C is a constant.
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If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is called the indefinite
integral of the function f(x) and is denoted by the symbol f(x) dx. Thus, by definition f(x) dx = F(x) +
C, if F(x) = f(x). If a function f(x) is continuous on an interval [a, b], then this function has an antideriva-
tive. The process of finding the antiderivative of a function f(x) is called integration. Two different inte-
grals of a function differ by a constant.
Standard Elementary IntegralsIn the following integrals, C stands for an arbitrary constant.
)1n(,c1n
xdxx
1nn sec x dx = ln |sec x + tan x| + c
1n(C1n
))x(f(dx)x(f))x(f(
1nn ) 22 xa
dx=
a
1tan1 c
a
x
c|x|lndxx1
cxtan
x1
dx 12
cedxe xx c
a
xsin
xa
dx 122 orcos1 a
x+ C
sin x dx = cosx + c 2x1dx
= sin1 x + c
cos x dx = sinx + c 1xx
dx2 = sec
1x + c or cosec1x + c
sec2x dx = tanx + c
cosec2x dx = cotx + c
tan x dx = ln|cosx| + c = ln |secx| + c
cot x dx = ln |sin x| + c = ln |cosecx| + c
e.g. (i)
12
1x
dxx
12
1
2
1
+ c =3
2. x3/2+ c (ii)
c
x1
c12
xdxxdx
x
1 1222
The following points are to be noted:
dxx1
lnx + c if x is positive = ln (x) + c if x is negative becausedxd
(ln (x)) =x
1
(
1) =x
1 C|x|lndxx
1
2x1
1dx = sin1x or cos1x. It does not mean that sin1x = cos1x.
The only legitimate conclusion is that they differ by some constant. In fact sin1x (cos1x)
= sin1x + cos1x = /2.
If a is a constant, then a f(x) dx = a )x(f dx
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[f(x) g(x)] dx, = f(x) dx g(x) dx
Methods of Integration
(i) Integration by Substitution
This method consists of expressing the integral f(x) dx, where x is the independent variable, in
terms of another integral where some other, say t, is the independent variable; x and t being connected
by the relation x =f (t). i.e., f(x)dx = f[j (t)] j (t) dt. This method is useful only when a relation x = (t)
can be so selected that the new integrand f(x)dt
dxis of a form whose integral is known
(ii) Integration by parts
dx)x()x(f)ii()i(
= f (x)
dxdx)x(dx
dfdx)x(
Integral of the product of two function = first function integral of secondintegral of (derivative
of first integral of second).Definite Integral
The difference in the values of an integral of a function f(x) for two assigned values of the inde-
pendent variable x, say a, b, is called the definite integral of f(x) over the interval (a, b) and is denoted by
b
a
.dx)x(f Thus b
a
),a(F)b(Fdx)x(f where F(x) is the antiderivative of f(x). Or, we write
).a(F)b(F|dx)x(F|dx)x(f ba
b
a
a is called the lower limit and b the upper limit of integration.
Note:
b
a
a
b
dx)x(fdx)x(f
b
a
c
a
b
c
dx)x(fdx)x(fdx)x(f
where c is any point inside or outside the interval (a, b). Geometrically definite integral represents area under curve.
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Ex amp l e 1 :
In a DABC, the medians AD, BE and CF pass through G.
(a) If BG = 6, what is BE ?
(b) If FG = 4, what is GC ?
So l u t i o n :
(a) We have, BG =3
2BE
6 =3
2BE BE = 9
(b) We have,
GC = 2 FGGC = 2 4 = 8
Ex amp l e 2 :
Triangles ABC and DBC are on the same base BC with A, D on opposite sides of lne BC, suchthat ar. (DABC) = ar. (DDBC). Show that BC bisects AD.
So l u t i o n :
Since Ds ABC and DBC are equal in area and have a common side BC.
Therefore the altitudes corresponding to BC are equal
i.e.,AE = DF
Now, in Ds AEO and DFO,
We have
1 = 2 (vertically opposite angles)
AEO = DFO (90 each)
and AE = DF
DAEO @ DDFO (by A.A.S)
A
C
D
O F
2
1
BE
AO = OD
SOLVED SUBJECTIVE EXAMPLES
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BC bisects AD
Ex amp l e 3 :
Solve |x2 3x 4| = x23x 4
So l u t i o n :
We know |x| = x when x 0
So, x2 3x 4 0
(x 4) (x + 1) 0 x 4 or x 1.
Ex amp l e 4 :
Solve 184x3= (54 2 )3x4
So l u t i o n :
Given equation is 184x3= 254( )3x4
Taking log on both the sides, we get(4x 3)log 18 = (3x 4) log(18.3 2 ) (since 3 2 = 18 )
or, (4x 3) log 18 = (3x 4) log (18)3/2or, 4x 3 = (3x 4)2
3
or, 8x 6 = 9x 12, or x = 6
Ex amp l e 5 :
Solve for x if log3x + log
9(x2) + log
27(x3) = 3
So l u t i o n :
log3x + log
9(x2) + log
27(x3) = 3
33log3xlog3
3log2
xlog2
3log
xlog 33log
xlog3 logx = log3
x = 3
Ex amp l e 6 :
If r be the ratio of the roots of the equation ax2
+ bx + c = 0, show that ac
b
r
)1r( 22
So l u t i o n :
Let a and ra be the roots of the equation ax2+ bx + c = 0
So, a + ra =a
b a(1 + r) =
a
ba = )1r(a
b
(1)
Also, a ra =a
cra2=
a
c
a
c
)1r(a
b.r
22
2
[Using (1)]
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BASIC TRIGONOMETRY
19
ac
b
r
)1r( 22
Ex amp l e 7 :
If x is real, prove that 3x2 5x + 4 is always positive
So l u t i o n :
3x2 5x + 4 = 3
3
4x
3
5x2
= 036
23
6
5x3
36
2548
6
5x3
22
since square of real number is always non-negative.
Ex amp l e 8 :
Find the area of the largest circle that can be inscribed in a square of side 14 c.m.
So l u t i o n :
BC = 14 c.m.
radius of circle = 72
14 c.m.
A
D C
B7
7
Now the area of circle = 2r = 2)7( = 49 c.m2
Ex amp l e 9 :
If f(x) = 22
x1
x
, x R, find the range of f(x)
So l u t i o n :
f(x) = 22
x1
x
= 2
2
x1
11x
= 2
x1
11
Clearly f(x) [0, 1)
Ex amp l e 10 :
If x = 2 ln cot t and y = tan t + cot t, finddx
dy
So l u t i o n :
Sincedt/dx
dt/dy
dx
dy
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Now t2sin4
tsintcos
2
tcot
teccos2
dt
dx 2
Also 4t2sin
t2cos
tcostsin
tcostsinteccostsec
dt
dy222
2222
Hence t2sin t2cos4 t2sint2sin t2cos4dxdy 2 = cot 2t
Ex amp l e 11 :
xcosxsin
xcosxsin22
33
dx
So l u t i o n :
xcosxsin
xcosxsin
22
33
dx = dx
xcosxsin
xcosdx
xcosxsin
xsin
22
3
22
3
= tanx secx dx + cotx cosecx dx = secx cosec x + c
Ex amp l e 12 :
Evaluate: 2/1
02/32
1
dx)x1(
xsin
So l u t i o n :
Let x = sinq dx = cos q dq and q = sin1
x; when x = 0, sinq = 0When x = 2/1 , sinq = p/4
2
1
0
4
0
4
0
232/32
1
dseccos
dcosdx
x1
xsin
=
4/
0
4/0 dtan.1tan (Integrating by parts) = 2ln
2
1
4
Ex amp l e 13 :
Solve : log| x |
| x | = 0
So l u t i o n :
We have
log| x |
| x | = 0
|x| = 1
but | x | 1 (being in base of logarithm)
x
Ex amp l e 14 :
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The maximum and minimum value of f (x) = 2x3 24x + 107 in the interval [1, 3].
So l u t i o n :
We have f(x) = 2x3 24x + 107
So, f (x) = 6x2 24
Now, f (x) = 0 6x2
24 = 0 x = 2Butx = 2 [1, 3]
So x = 2 is the only stationary point.
Now, (1) = 2 14 + 107 = 85, f(2) = 2(2)3 24 (2) + 107 = 75
And, (3) = 2 (3)3 24 3 + 107 = 89.
Hence, the maximum value of f (x) is 89 which attains at x = 3 and the minimum value is 75which is attained at x = 2.
Ex amp l e 15 :
Find area bounded by y = cosx, xcos , y-axis and x =2
.
So l u t i o n :
The represented area =/ 2
0
cosxdx
/ 2
0sin x 1 0
= 1 sp. unit.
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1. Find the solution to the inequality )4x()5x()1x()3x()3x( 52
> 0
2. Solve for x
(a) |x 4| > 7 (b) |x| > x
3. (a) Solve for x: log2x > 3
(b) Which is greater: log23 or log1/25
4. If the roots of (1 + m) x2 2 (1 + 3m) x + (1 + 8m) = 0 are equal, then find the value of m.
5. For every x R, prove that 2x2 6x + 9 is always positive.
6. Differentiate the following with respect to x.
(i) sinx + cosx
(ii) xlogx(iii) Differentiate sin2x + cos2x with respect to x
7. If y = acos t + sin t , then prove that2
22
d yy 0
dt
8. (a) Find domain of the definition of the following:
(i) x (ii) elog x(b) Draw the graph of the following:
(i) f(x) = x3 (ii) f(x) = logex
9. Find the intervals of increase and decrease of (x 3) (x + 1).
10. Evaluate : (i) 2(x 2x 1) dx . (ii)2x(2xe ) dx
(ii) dx)xcosx(sin2/
0
LEVEL - I
REVIEW YOUR CONCEPTS
SECTION - I
SUBJECTIVE
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1. Find the solution common to both the inequalities 0)4x4x(
|4x|)2x3x()1x(72
523
& 1 < |x 3| < 5
2. Solve for x
(i) |x 3| + |x + 2| = 3 (ii) (a) Solve for x, 57x2
2x
3. If a2+ 4b2= 12 ab, then prove that log (a + 2b) = )2log4bloga(log21
.
4. Find Domain of definition (i)1x3x
sin
(ii) loge
)4x()2x()3x()1x(
5. Differentiate cos (4x3
3x) with respect to x.
6. If for the function h, given by h(x) = kx2+ 7x 4, h(5) = 97, find k.
7. Find the intervals of increase and decrease of the function y = cosx,
x2
8. Integrate the following
2
0
2 dxxcos.xsin .
9. In the figure given below, find the value of angle P
36 54
A
B D
R P
C
24
10. In the figure given below, find the value of x
P
B
T
Ax4
5
LEVEL - II
BRUSH YOUR CONCEPTS
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1. Solve: log0.3
(x 1) < log0.09
(x 1).
2. Find the set of all solution of the equation 2|y|= 12 1y
3. Evaluate: 3log7log7log5log 5353 7537
4. Evaluate:7
1log 5
3
10
15
log 0.1
.
5. Find domain of definition (i) f(x) =2x3x
)3x(log2
2
(ii) f(x) =
|x|2
1|x|
(iii)
4
xx5logy
2
10 (iv)x1x1
2x2x
y
6. Evaluate: 2q
5cos x 3sin x dxcos x
.
7. Find the intervals of decrease and increase of (x + 2) ex.
8. In the figure given below, ABCD is square and triangles BCX and DYC are equilateral triangle.Find the value of y.
A
B
D C
x
y
9. (i) In the figure given below. Find QSR
S
R
Q
50O T P
LEVEL - III
CHECK YOUR SKILLS
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BASIC TRIGONOMETRY
25
(ii) In the given figure, O is the centre of the circle. If OCA = 26, then find ODB
O
A
C B
D
10. ABC is a triangle in which BC is produced to D. CA is produced to E, DCA = 108 and EAB= 124. Then find ABC.
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1. If log16x + log4x + log2x = 14, then x =(a) 16 (b) 32
(c) 64 (d) none
2. If | 4 3x | 2
1then x is equal to
(a)
2
3,
6
7(b)
2
3,
6
7
(c)
23,
67 (d) none of these
3. The product of all the roots of the equation x2- |x| 6 = 0 is
(a) 9 (b ) 6
(c) 9 (d) 36
4. The value of p for which (x 1) is a factor of x3+ (p + 1)2 x2 10 is given by
(a) 4, 2 (b) 2, 4(c) 2, 4 (d) none of these
5. The domain of definition of the function f(x) =1
x x is
(a) R (b) (0,)
(c) (, 0) (d) none of these
6. The differential coefficient of f(x) = log (logx) with respect to x is
(a) xlogx
(b)x
xlog
(c) (xlogx)1 (d) xlogx
7. The function f(x) = tanx x,
(a) always increases (b) always decreases
(c) never decreases (d) some times increases and some times decreases
8. The function f(x) = x3 3x is
(a) increasing on (, 1] [1, ) and decreasing on (1, 1)
SECTION - II
OBJECTIVE
LEVEL - I
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(b) decreasing on (, 1] [1, ) and increasing on (1, 1)
(c) increasing on (0, ) and decreasing on (, 0)
(d) decreasing on (0, ) and increasing on ( , 0)
9. The minimum value of 2 22( 3) 27x is
(a) 227 (b) 2(c) 1 (d) none of these
10. The maximum value of x3 3x in the interval [0, 2] is
(a) 2 (b) 0
(c) 2 (d) 1
11. Evaluate
/ 4
0
sin
cos 3 3cos
xx
dxx x
(a)2
1(b)
6
1
(c)8
1(d)
12
1
12. If 10log 3 = 0.477, the number of digits in 340is
(a) 18 (b) 19
(c) 20 (d) 21
13. The interior and its adjacent exterior angle of a triangle are in the ratio 1 : 2. What is the sum of
the other two angles of the triangle?
(a) 112 (b) 110
(c) 120 (d) 90
14. In the figure PAQ is a tangent of the circle with centre O at a point A if OBA = 32. The value
of x and y is
A
P
Q
C
xB
y
O
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(a) 30, 50 (b) both 40
(c) both 58 (d) 30, 60
15. The difference between the interior and exterior angles of a regular polygon is 132. The numberof sides of the polygon is
(a) 12 (b) 8
(c) 15 (d) 20
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1. If A = log2log
2log
4256 + 22log 2, then A equals to
(a) 2 (b) 3(c) 5 (d) 7
2. If logkA . log
5k = 3, then A =
(a) 53 (b) k3
(c) 12 (d) 243
3. If the product of the roots of the equation 22 log5 4 0x x is 8, then l is
(a) 22 (b) 2 2(c) 3 (d) none of these
4. The number of real roots of the equation (x 1)2
+ (x 2)2
+ (x 3)2
= 0(a) 3 (b) 2(c) 1 (d) 0
5. If a and b are roots of the equation x2+ x + 1 = 0 then a2+ b2=(a) 1 (b) 2(c) 1 (d) 3
6. The domain of definition of the function f (x) =2 9
log
x
x
(a) R (b) (1, )(c) (0, 1) (1, (d) [1, )
7. If x = a cos3, y = a sin3,2
dy1+ =
dx
(a) tan2 (b) sec2(c) sec (d) |sec|
8. The function y = x3
3x2
+ 6x 17(a) increases everywhere(b) decreases everywhere(c) increases for positive x and decreases for negative x(d) increases for negative x and decreases for positive x
9. The largest value of 2x3 3x2 12x + 5 for 2 x 4 occurs at x =(a) 2 (b) 1(c) 2 (d) 4
10. The greatest value of f(x) = cos (xex+ 7x2 3x), x[ 1, ) is(a) 1 (b) 1(c) 0 (d) none of these
LEVEL - II
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11. 21
x
x
edx
e
(a) cot1ex+c (b) tan1ex+ c(c) tanex+ c (d) sin1ex+ c
12. Eva luate/ 4
2
0
tanx
x dx
(a)2
3 (b)
62
(c) 14
(d) 5
2
13. If the area of parallelogram ABCD is 32 sq. cm. Area of DAMN is equal to
D M
N
A B
C
(a) 8 cm2 (b) 2 cm2
(c) 4 cm2
(d) none of these
14. In the given figure if AX = 5 cm, XD = 7 cm, CX = 10 cm find BX
A B
C O
DP
T
x
(a) 3 cm (b) 3.5 cm(c) 4 cm (d) 4.5 cm
15. If A, B, C are three consecutive points on the arc of a semicircle such that the angles subtended by
the chords AB and AC at the centre O is 90 and 100 respectively. Then the value of angle BAC
is equal to
(a) 5 (b) 10(c) 20 (d) 30
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1. Find the solution to the inequality2 5
( 3) ( 3) ( 1) 0( 5)( 4)
x x xx x
Solution:
2 5(x 3) (x 3) (x 1)0
(x 5) (x 4)
By wavy curve method,
+ ++
5 3 1 3 4
x ( , 5) ( 3 1) (4 )
2. Solve for x
(a) |x 4| > 7 (b) |x| > x
Solution:
(a) |x 4| > 7
x 4 > 7 or x 4 < 7
x > 11 or x < 3
3 11
x ( , 3) (11, )
(b) |x| > x
Case I : x > 0,
x > x which is not possible.
Case II: x < 0
x < x 2x < 0 x < 0
x( , 0) or R
3. (a) Solve for x: log2x > 3
(b) Which is greater: log23 or log
1/25
Solution:
(a) log2x > 3 x > 23 [ base is greater than 1]
SUBJECTIVE SOLUTIONS
CHECK YOUR SKILLS
LEVEL - I (CBSE Level)
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x > 8 x(8, )
(b) log23 or log
1/25
log1/25 = log25 < 0. [ log1/ax = logax]
log23 is greater than log1/25.
4. If the roots of (1 + m) x2 2 (1 + 3m) x + (1 + 8m) = 0 are equal then find the value of m.
Solution:
Given equation is
(1 + m) x2 2(1 + 3m) x + (1 + 8m) = 0
roots are equal , then D = 0
4 (1 + 3m)2 4 (1 + m) (1 + 8m) = 0
1 + 9m2 + 6m 1 9m 8m2= 0
m2 3m = 0
m (m 3) = 0
m = 0, 3.
5. For every x R, prove that 2x2 6x + 9 is always positive.
Solution:
Let f(x) = 2x2 6x + 9 = 2(x2 3x + 9/2)
= 2( x 3/2)2 + 9/4 > 0 [ square of real number is always nonnegative]
Hence f(x) is always positive.
6. Differentiate with respect to x(i) sinx + cosx
(ii) xlogx
Solution:
(i) Let y = sinx + cosx
dy
cos x sin xdx
d
dx
(sinx + cosx) = cosx sinx
(ii) Let y = x logx
dy d dx (log x) log x (x)
dx dx dx =
1x log x1
x
d(xlogx)
dx= 1 + logx
8. (a) Find domain
(i) x (ii) log ex
(b) Draw graph
(i) f(x) = x3 (ii) f(x) = logex
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Solution:
(a) (i) Let f(x) = x
f(x) is real for all x 0
Df = [0, ]
(ii) Let f(x) = elog x
f(x) is real for all x 1
Df= [1, )
(b) Let f(x) = x3
It is odd function so graph will be symmetric about the origin
x
y
(ii) DfR
(iii) R fR
(iv) f (x) = 3x2
f (x) = 0 x = 0
(v) f(x) = 6x
(vi)ve, x 0
f (x) 6xve, x 0
(ii) Let f(x) = logx
(i) f (x) =1
x, f(x) = 2
1
x
(ii) Df R {0}
y
x
y = logx
(1, 0)
(iii) Rf R
9. Differentiate with respect to x
(i) sin2x + cos2x
(ii) x = a (cost + log tan2t
), y = a sint.
Solution:
(i) Let y = sin2x + cos2x
2dy d d(sin x)
dx dx dx (sinx) +
2d d(cos x) (cos x)dx dx
= 2sinx cosx + 2cosx (sinx) = 2sinx cosx 2sinx cosx = 0
2 2d (sin x cos x) 0
dx
d
(1) 0
dx
[ d
dx
(constant) = 0]
(ii) x = a (cost + log tant
2), y = asint
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Diff. w.r.to t
dx
adt
21 t 1sin t sec
t 2 2tan2
,dy
acostdt
= a[ sint + cosect]
dy cos tdx cos ect sin t
2
cos t.sin t1 sin t
= 2cost.sint
cos t= tant
10. Find the intervals of increase and decrease of (x 3) (x + 1).
Solution:
Let f(x) = (x 3) (x + 1) = x2 2x 3
f (x) 2x 2 = 2(x 1)
for increasing, f (x) 0
2(x 1) 0 x 1
x [1, )
for decreasing, f (x) 0
x ( ,1]
11. Find the point of local maxima and minima of the following(i) x2+ 3x (ii) logex + x
So l u t i o n :
Let f(x) = x2+ 3x = x(x + 3)
f (x) = 2x + 3, for local maxima or local minima we must have, f (x) = 0
3
x2
f (x) = 2 > 0 3/2
f(x) is local minima at 3x2
(ii) f(x) = logx + x
1f (x) 1
x
for local maxima or local minima, we must have f (x) 0
1
1 0x
x = 1
21f (x) 0x
f(x) is local maxima at x = 1.
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12. Integrate the following
(i) x2+ 2x + 1 (ii) 22 xxe
(iii)/2
0
(sin cos )x
x x dx
(iv)2
1
x dx
Solution:
(i) Let I = 2(x 2x 1) dx 2x dx 2 x dx 1.dx
3 2x x2 x C
3 2
=3
2x x x C,3
where C is integral constant.
(ii) Let I = 2x2x e dxPut x2= t2xdx = dt
tI e dt = et+ C =
2xe C, where C is integral constant.
(iii)
/ 2
0
I (sin x cos x) dx
=
/ 2 / 2
0 0sin xdx cos x dx
= / 2 / 20 0[sinx]cosx
= cos cos 0 sin sin 02 2
= [0 1] + [1 0] = 2
(iv) Let2
1
I | x |dx
Let f(x) = |x| = x, x 0x, x 0
0 2
1 0
I f (x)dx f (x) dx
0 2
1 0
xdx xdx
0 22 2
1 0
x x
2 2
1 1[0 1] (4 0)
2 2
5
2
13. In the figure given below. Find QSR
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S
R
Q
50O T P
Solution:
QSR + PRQ + QPR + QOR = 360
90 + 90 + 50 + QOR = 360 QOR = 130
QSR =1
2 QOR = 65
14. In the given figure, O is the centre of the circle. If OCA = 26, then find ODB
O
A
C B
D
Solution:
OCA = 26 = DBA
OB = OD
OBD = ODB = 26 O
A
C B
26 26
D
15. ABC is a triangle in which BC is produced to D. CA is produced to E, DCA = 108 and EAB= 124. Then find ABC.
Solution:
DCA = 108 ACB = 72
EAB = 124
E
B C
A
56D
124
108 BAC = 56 ABC = 180 (72 + 56) = 62
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1. Find the solution common to both the inequalities
0)4x4x(
|4x|)2x3x()1x(72
523
& 1 < |x 3| < 5
Solution:
Case I : x < 43 5 5
14
(x 1) (x 2) (x 1) (x 4)0
(x 2)
3 5 5
14
(x 1) (x 2) (x 1) (x 4)0
(x 2)
4
+ + +
2 1 1
x ( , 4) Case II : x > 4
4
+ + +
2 1 1
Similarly, x ( 4 2) ( 1,1) x ( , 4) ( 4, 2) ( 1,1) ... (i)
And 1 < | x 3| < 5Case I : x < 3
1 < x + 3 < 5 2 < x < 2 2 < x < 2 x (2, 2)Case II: x > 31 < x 3 < 5 4 < x < 8
x (4,8)
x ( 2, 2) (4,8) ... (ii)from (i) & (ii)
4 2 1 2 4 8
x ( 1,1)
2. Solve for x
(i) |x 3| + |x + 2| = 3 (ii) 57x2
2x
Solution:
(i) |x 3| + |x + 2| = 3Case I:
BRUSH UP YOUR CONCEPTS
LEVEL - I
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x < 2x + 3 x 2 = 32x = 2 x = 1, which is impossibleCase II:
2 x < 3x + 3 + x + 2 = 3, which is not possible
Case III:x 3x 3 + x + 2 = 32x = 4x = 2, which is impossible
x
(ii)x 2
52x 7
x 2 5(2x 7) x 2 10x 350 0
2x 7 2x 7
9x 37
02x 7
(9x + 37) (2x + 7) > 0
37/9
+ +
7/2
37 7
x ( , ) ( , )9 2
3. If a2+ 4b2= 12 ab, then prove that log (a + 2b) = 1 (log log 4log 2)2
a b
Solution:
a2+ 4b2= 12aba2+ 2a2b + 4b2 = 16 ab(a + 2b)2= 16abtaking log on both the sides,2 log (a + 2b) = log16 + log a + logb
log (a + 2b) =1
2
(loga + logb + 4log2)
4. Find Domain (i)3
sin1
x
x
(ii) loge
( 1)( 3)( 2)( 4)x x
x x
Solution:
(i) Let f(x) =x 3
sinx 1
f(x) is defined for
x 3
0x 1
(x 1) (x + 3) 0
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+3
x
1
x ( ,3] (1, )
(ii) Let f(x) =e
(x 1) (x 3)log
(x 2)(x 4)
f(x) is defined for
(x 1) (x 3)
0(x 2)(x 4)
+ + +
31
42
x ( , 3) ( 2,1) (4, )
5. Differentiate with respect to x
(i) 2sin
(1 )
x x
x
+ log 2(1 )x
(ii) (sinx)x
(iii) cos (4x33x)Solution:
(i) Let2
2
xsinxy log 1 x
1 x
2
2 2 2 2
dy (1 x ) (sin x x cos x) (2x)x sin x 1 (x 2x)
dx (1 x ) 1 x 2 1 x
2 2 2
2 2(1 x ) x cos x (1 x )sin x 2x sin x 1(1 x )2 1 x
2 2
2 2 2
(1 x ) x cos x (1 x )sin x 1
(1 x ) 1 x
(ii) Let y = (sinx)x. Taking log on both the sides, we getlogy = x log sinx
1 dy x cos x. log sin x
y dx sin x = x cotx + log sinx
dydx
= (sinx)x [x cotx + log sinx]
x xd {(sin x) } (sin x ) [x cot x log sin x]dx
(iii) Let y = cos(4x3 3x)
dy
dx= sin (4x3 3x) [12x2 3]
d
dx (cos 4x3 3x) = 3 (1 4x2) sin (4x3 3x)
6. If for the1 function h, given by h(x) = kx2+ 7x 4, h(5) = 97, find k.Solution:
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h(x) = kx2+ 7x 4
h (x) 2kx + 7
h (5) = 10k + 7
97 = 10k + 7 k = 9 [ h (5) = 97]
7. (i) Find the intervals of increase and decrease of the function y = cosx,
2
x
(ii) Find the point of local maxima & minima of the function, f (x) = x1
x .
Solution:
(i) Let f(x) = cosx, x2
f (x) = sinxf(x) increases, if f (x) 0 sinx 0 sinx 0
O/2x ,02
f(x) decreases, if f (x) 0
sinx 0 sinx 0
x [0, ]
O /2
(ii) 1
f (x) xx
21
f (x) 1x
3
2f (x)
x
For local maximum or local minimum we must have f (x) 0
1 21
0x
x = 1
at x = 1, f (x) 1 0
Hence local minima at x = 1
at x = 1, f (x) 1 0 Hence local maxima at x = 1.
8. Integrate the following (i)
2
0
2 dxxcos.xsin (ii) 8
13/2
x
dxx
e3
.
Solution:
(i) Let I =
/ 22
0
sin x.cos x dx
put cosx = t =0 2
1
t dt sinx dx = dt when x = 0, t = 1, when x = , t 02
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=
11 32
00
1tt dt33
(ii) Let I =
38 x
2 / 31
edx
x put x1/3 = t 2 /3
1 1. dx dt
3 x when x = 8, t = 2 ,
x = 1, t = 1 =2
t t 2
1
3 e dt 3(e ) , = 3(e2 e) = 3e (e 1)
9. In the figure given below, find the value of angle P
36 54
A
B D
R P
C
24
So l u t i o n :
ACD = ABC + BAC = 36 + 24 = 60 P = RCD + RDC = 60 + 52 = 104.
10. In the figure given below, find the value of x
P
B
T
Ax4
5
Solution:
PT2 = PA.PB
PA =5 5 25
4 4
AB = PA PB
x =25 9
44 4
units
Hence x =9
4units.
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1. Solve, log0.3
(x 1) < log0.09
(x 1)Solution:
log
0.3 (x 1) < log
0.09(x 1)
log0.3 (x 1) 1 x > 2
x (2, )
2. Find the set of all solution of the equation 12 2 1y y Solution:
2|y| = 2y1+ 1Case I: y < 0 2y= 2y1+ 1
1 =2y
y2 22
(2y)2 + 2.2y 2 = 0
a2 + 2a 2 = 0 , where a = 2y
2 4 8 2 2 3a
2 2
a = 1 3 , 1 + 3
a = 1 3 , which is not possible
a = 3 1
2y= 3 1
log22y = log
2( 3 1 )
y = log2( 3 1 )
Case II: y < 02y= 2y1 + 1
2.2y = 2y + 2 2y= 2 y = 1
set of solution 1, log2 ( 3 1)
3. Evaluate: 3 5 3 55 7 7 37log 3log 5log 7 log
Solution:
Let y =5 7 7 3
3 5 3 5log log log log7 3 5 7
= 3 5 3 5log 5 log 7 log 5 log 77 3 7 3 [ b blog c log aa c ]
LEVEL - III
CHECK YOUR SKILLS
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= 0Hence the result.
4. Evaluate: 5 0.17 101 1
log log3 2
Solution:
Let 7 101 1
3 log 5 log 0.1y 5
5
1/ 3log 7
10
15
log 10
= (7 + 1)1/3 = (8)1/3 = 2
Hence the result.
5. Find domain (i) f(x) = (ii) f(x)1
2
x
x
Solution:
(i) f(x) = 22log (x 3)
x 3x 2
Let g(x) = log2(x + 3) and h(x) = x2+ 3x + 2
g(x) is defined forx + 3 > 0 x > 3 Dg= (3, )And h(x) is defined for x2+ 3x + 2 0 (x + 2) (x + 1) 0 Dh = R {2, 1} Df = Dg Dh = (3, ) R {2, 1} = (3, ) {2, 1}
(ii) f(x) =| x | 1
2 | x |
f(x) is defined for
2
| x | 1 (| x | 1) (2 | x |)0 0
2 | x | (2 | x |)
(2 |x|) (|x| 1) 0 2 |x| > 0 or |x| 1 0 |x| < 2 or |x| 1 2 < x < 2 or x 1 or x 1
Df = (2, 1] [1, 2)
6. Find the domain
(i)
4xx5
logy2
10 (ii) 3/1)x(sinecxcosy
(iii)x1x1
2x2x
y
Solution:
(i)2
105x xy log f (x)
4
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f(x) is defined for
log10
25x x 0
4
2
25x x 1 5x x 4 04
x2
5x + 4 0 (x 1) (x 4) 0either x 1 or x 4
Df= [1, 4]
(ii) Let f(x) = cosecx + (sinx)1/3
Let g(x) = cosecx and h(x) = (sinx)1/3
g(x) is defined forcosecx > 0D
g= (2n , (2n + 1) ).
And h(x) is defined for R
Df = Dg Dh = (2n , (2n + 1) )
(iii) Letx 2 1 x
f(x)x 2 1 x
Let g(x) =x 2
x 2
and h(x) =1 x
1 x
g(x) is defined for
x 20
x 2
either x 2 0
or x + 2 < 0x 2 or x + 2 < 0
Dg = ( , 2) [2, )
h(x) is defined for
1 x x 10 0
1 x x 1
either 1 x 0 or 1 + x > 0
x 1 or x > 1
Dn = (1, 1] Df = Dg Dh =
7. Integrate the following
(a) 5 cos x 3 sinx +xcos
q2 (b)
4
02
dxxcos
xsin32
Solution:
(a) Let I = 2a
(5cos x 3sin x )dxcos x 25 cos x dx 3 sin xdx 9 sec x dx
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= 5 sinx + 3cosx + 9 tanx + C,where C is integral constant.
(b) Let/ 4
20
2 3sin xI dx
cos x
=
/ 42
02 sec xdx 3 tan xsec x dx
= / 4 / 40 02[tan x] 3[sec x]
= 2[1 0] + 3 [ 2 1] = 3 2 1
8. Differentiate 2 1xe w.r.t. x.
Solution:
Let y = 2x 1e
2x 1dy e
dx
21
2 x 1 (2x 0)
2x 1
2
x.e
x 1
9. (a) Find the intervals of decrease and increase of (x + 2) ex.(b) Find the greatest and least values of the following functions on the intervals
y = 3x4+ 6x2 1 (2 x 2)
y = 1x3x23x 2
3
(1 x 5)
Solution:
(a) Let f(x) = (x + 2) ex
f (x) = ex
(x + 2) + ex
= ex
(x + 1)for decreasing f (x) 0
ex (x + 1) 0 ex (x + 1) 0 ( ex 0] x + 1 > 0
x > 1 x [ 1, )
for increasing , f (x) 0
ex(x + 1) 0ex (x + 1) 0
x + 1
0 [ ex
0]
x 1
x ( , 1] (b) (i) y = 3x4 + 6x2 1, 2 x 2
3dy 12 x 12x
dx
22 2
2
d y36x 12 12(x 1) 0
dx
dy
0dx 12x (x2 1) = 0 x = 0, 1 2least value y = 2, at x = 0
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greatest value y = 25 at x = 2
(ii)3x
y3
2x2+ 3x + 1 , 1 x 5
2dy x 4x 3dx
2
2
d y2x 4 2(x 2)
dx
dy 0dx
x2 4x + 3 = 0
(x 1) (x 3) = 0 x = 1, 3, 5
at x = 1,2
2
d y2 0
dx
maximum at x = 1
maximum value of y =7
3
at x = 3,2
2d y 2 0dx
minimum at x = 5
minimum value of23
y3
10. In the figure given below, ABCD is square and triangles BCX and DYC are equilateraltriangle. Find the value of y.
AB
D C
x
y
So l u t i o n :
BCX and DYC are equilateral
AB = BC = CD = DA = CX = DY = CY = BX CBX = 60 ABX = 150 BAX = 15
DAY = 75 = DYAA [ AD DY]
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Illustration 1:
If A = {a, b, c} and B = {b, c, d} then evaluate AB, AB, A B and B ASolution:
AB = {x : x A or x B} = {a, b, c, d}AB = {x : x A and x B} = {b, c}A B = {x : x A and x B} = {a}
B A = {x : x B and x A} = {d}
Illustration 2:
Find the logarithms of 0.0625 to the base 2.Solution:
Suppose 2x= 0.0625 = 421
16
1
or 2x= 24 or x = 4
Illustration 3:
Solve the equation a2xb3x= c5, where a, b, c R+
Solution:
Equation is a2x.b3x= c5
Taking log on both sides, we have2x log a + 3x logb = 5 logc or x (2loga + 3 logb) = 5logc
x = blog3alog2
clog5
Illustration 4:
Solve for x, log1/2
(x 2) > 2.
Solution:
log1/2
(x 2) > log1/2
4
1 0 < x 2 <
4
12 < x <
4
9
So x 49,2
Illustration 5:
Solve |2x 1| < 3.Solution:
|2x 1| < 3 3 < 2x 1 < 3 2 < 2x < 4 1 < x < 2.
Illustration 6:
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If f(x) =
7x1x5x2x3x
then find x such that
(i) f(x) > 0. (ii) f(x) < 0.Solution:
Given f(x) =
7x1x5x2x3x
Illustration 7:
If xx5x
= 24,, then find the value of x.
Solution:
Here, xx5x
= 24 x2 5x = 24
x2 8x + 3x 24 = 0 (x 8) (x + 3) = 0 x = 3, 8
Illustration 8:
Find the value of x and y when xy32
= 4 and 27
12
yx
Solution:
Since, 4y3x24xy32
(1)
and 27y2x
27
12yx (2)
Solving (1) and (2), we get x =2
5, y = 3.
Illustration 9:
Find the derivative of y =x
1)x(x3 3/1
Solution:Here we have y = u + v + w,
where u = 3/1xv,x3 and w =x
1
Hence we can use theorem 2
23/211
13
11
2
1
x
1
)x(
131
x3
21
x.1x31
x21
3dxdy
Illustration 10:
Find the derivative of y = (a + x) exw.r.t. x.Solution:
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Using theorem 3
)xa(dx
dee
dx
d)xa(
dx
dy xx
= )1xa(e1.ee)xa( xxx
Illustration 11:
Find the derivative of y =xa
xa
w.r.t. x
Solution:
Illustration 12:
Differentiate the following w.r.t x.(i) y = xx
(ii) y =2x)x(sin
Solution:(i) y = xx
)xln1(x1.xlnx1.x.xdx
dy xx1x
(ii)2x)x(siny
dx
dxxsinln)x(sin)x(sin
dx
d)x(sinx
dx
dy 2x1x2 22
= x2 .x2.xsinln)x(sinxcos)x(sin22 x1x
Illustration 13:
Find the derivative of the following functions w.r.t. x(i) y = sinx2
(ii) y = (lnx)3
(iii) y = sin (lnx)3
(iv) y = cos1(lnx)Solution:
(i) y = sin (x)2. Let u = x2y = sin u
x2.ucos)x(dxd)u(sinduddxdy2 2x cosx2
(ii) y = (ln x)3. Let u = ln x y = u3dx
dy= 22
3
)x(lnx
3
x
1u3
dx
du
du
)u(d
(iii) y = sin (lnx)3 Let u = ln x, v = u3 y = sin v
)x(lndx
d
du
)u(d
dv
)v(sind
dx
du
du
dv
dv
dy
dx
dy 3 = cos v. ])x[(lncos
x
)x(ln3
x
1u3 3
22
(iv) Let u = ln x = cos1u
222 )x(ln1x
1
x
1
u1
1
dx
du
u1
1
dx
dy
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Illustration 14:
The function y of x is given by, x = a cos t, y = a sin t. Find the derivative of y w.r.t. x.Solution:
tcottsin
tcos
)tcosa(
)tsina(
dx
dy
If we want to compute dx
dy
at a particular t, say t = 4
, then 14cotdx
dy
4/x
Illustration 15:
Find 22
dx
yd, where y = sin2x
Solution:
dx
dy= 2sinxcosx = sin2x
Illustration 16:
Find the interval of increase and decrease of the function y = x4.Solution:
y = x4y= 4x3
For x > 0, y> 0 the function increases in (0, ).For x < 0, y< 0 the function decreases in (, 0)
Illustration 17:
Separate the intervals in which f(x) = 2x3 15x2+ 36x + 1 is increasing or decreasing
Solution:
We have f (x) = 6x2 30x + 36 = 6 (x 2) (x 3)Thus for x < 2, f (x) > 0over 2 < x < 3, f(x) < 0 and for x > 3, f(x) > 0Hence the given function is increasing in (, 2) and (3, ), and decreasing in (2, 3)Is tanx always increasing Rx ?
Illustration 18:
Test y = 1x4for maximum and minimum
Solution:Here y= 4x3= 0 for x = 0
x = 0 is the critical point.
Now y= 12 x2= 0 at x = 0
It is thus impossible to determine the character of the critical point by means of the sign of thesecond derivative. Thus we investigate the character of the given function in an interval containing pointx = 0.
For x < 0, y> 0 the function is increasing for x < 0
For x 0, y< 0 the function is decreasing for x > 0.
Consequently, at x = 0, the function has a maximum i.e., yx = 0= 1
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Illustration 19:
Determine the maximum and minimum of the function y = x3 3x + 3 on the interval [3,2
3].
Solution:
For the given function, y= 3x2 3
For the critical points, 3x2 3 = 0 x = 1Then y= 6x > 0 at x = 1, y < 0 at x = 1
Hence there is maximum at x = 1 at which y = 1 + 3 + 3 = 5
Also there is minimum at x = 1 at which y = 13 + 3 = 1
Now at x = 3, y = 27 + 9 + 3 = 15
and at x = 3/2, y = 15/8
Hence the minimum value of the given function is 15 at x = 3 and the maximum value is 5 at
x = 1. It should be noted that the values are actually the largest and smallest values of the functionin the given interval.
Illustration 20:
Evaluate:
(i) (a0+ a
1x + a
2x2) dx (ii)
dxex
2xcos x
(iii) dxx1x
2
2
(iv) dx1x
x2
4
Solution:
(i) (a0+ a1x + a2x2
) dx = a0dx + a1x dx + a2x2
dx= a
0x + a
1 c
3x
a2
x 32
2
(ii)
dxedx
x
12dxxcosdxe
x
2xcos xx = sin x + 2 log |x| ex+ c
(iii)
222
2
2
2
x1
dxdxdx
x1
11dx
x1
11xdx
x1
x= x tan1x + c
(iv)
dx
x1
11xdx
x1
11xdx
1x
x2
22
4
2
4
= cxtanx
3
x
x1
dxdxdxx 1
3
22
Illustration 21:
Integrate the following w.r.t. x.(i) sin2x cosx
(ii) 83
x1
x
Solution:
(i) Let sin x = t cos x dx = dt
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c3xsin
c3t
dttdxxcosxsin33
22
(ii) Let x4= t = 4 x3dx = dt x3dx = dt
dt)t1(41
dxx1
x28
3
= cxtan4
1cttan
4
1
t1
dt
4
1 4112
Illustration 22:
Evaluate
dxxa
122
Solution:
Let x = a sin dx = a cosda2 x2= a2cos2
cax
sind.1cosa
dcosa
xa
dx 122
Note:
a cbxsinb1)cbx(adx 122
Illustration 23:
Evaluate
(i) dxxex (ii) dxxsin 1
Solution:
(i) let f(x) = x, (x) = ex cexedxe.1e.xdxxe xxxxx
(ii) Let f(x) = sin1x. (x) = 1
dxxx1
1x.xsindx1.xsindxxsin
2
111
= x sin1x cx1xsinxdx)x1(x 212/12
Note: dx)x(fedx)x(fedx))x(f)x(f(e xxx
= f(x) ex ce)x(fdx)x(fedxe)x(f xxx
Illustration 24:
Evaluate :
2/
02
dxxsin1
xcos
Solution:
Let sin x = t cos x dx = dt
When x = 0, t = 0, x =
/2, t = 1
2/
0
1
0
10
122
)t(tant1
dt
xsin1
xdxcos= tan11 tan10 =
4
0 = /4
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2 ARITHMETIC PROGESSION
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SQUENCE AND SERIES
A succession of numbers a1, a
2, a
3..., a
n, ... formed, according to some definite rule, is called a
sequence.
ARITHMETIC PROGESSION (A.P.)
A sequence of numbers {an} is called an arithmetic progression, if there is a number d, such that d= a
n-a
n1for all n. d is called the common difference (C.D.) of the A.P.
(i) Useful Formulae
If a = first term, d = common difference and n is the number of terms, then
(a) nth term is denoted by tnand is given by
tn= a + (n 1) d.
(b) Sum of first n terms is denoted by Snand is given by
n nS [2a (n 1)d]2
or ( )2
n
nS a l , where l= last term in the series i.e., l= t
n = a + (n 1) d.
(c) Arithmetic mean A of any two numbers a and ba b
A2
.
(d) Sum of first n natural numebrs ( n )n(n 1)
n2
, where n N .
(e) Sum of squares of first n natural numbers ( 2n )2 ( 1) (2 1)
6
n n nn
(f) Sum of cubes of first n natural numbers 3( n )
23 n(n 1)n
2
(g)Middle term: If the number of terms is n, and
n is odd, thenth
n 1
2
term is the middle terms
n is even, then 2
thn
andth
n1
2
terms are middle terms.
(h) If terms are given in A.P., and their sum is known, then the terms must be picked
up in following way
For three terms in A.P., we choose them as (a d), a, (a + d)
For four terms in A.P. , we choose them as (a 3d), (a d), (a + d), (a + 3d)
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For five terms in A.P., we choose them as (a 2d), (a d), a, (a + d), (a + 2d) etc.
(ii) Useful Properties
(a) If tn= an + b, then the series so formed is an A.P.
(b) If Sn= an2+ bn + c, then series so formed is an A.P.
(c) If every term of an A.P. is increased or decreased by some quantity, the resulting terms` will also be in A.P.
(d) If every term of an A.P. is multiplied or divided by some non-zero quantity, the result
ing terms will also be in A.P.
(e) In an A.P. the sum of terms equidistant from the beginning and end is constant and
equal to sum of first and last terms.
(f) Sum and difference of corresponding terms of two A.P.s will form an A.P.
(g) If terms a1, a2, ..., an, an+1, ..., a2n+1are in A.P., then sum of these terms will be equal to(2n + 1)a
n+1.
(h) If terms a1, a
2, ..., a
2n1, a
2n are in A.P. The sum of these terms will be equal to
(2n)n n 1a a
2
.
GEOMETRIC PROGRESSION (G.P.)
A sequence of the numbers {an}, in which 1 0a , is called a geometric progression, if there is a
number 0r such thatn
n 1
a ra
for all n.r is called the common ratio (C.R.) of the G.P..
(i) Useful Formulae
If a = first term, r = common ratio and n is the number of terms, then
(a) nth term, denoted by tn, is given by t
n= arn1
(b) Sum of first n terms denoted by Snis given by
n
n
a(1 r )S
1 r
orna(r 1)
r 1
ora r
1 r
l
,
where lis the last term (the nth term) in the series, r 1 In case r = 1, Sn= na.
(c) Sum of infinite terms (S )a
S (for | r | 1)1 r
Note: When |r| 1, the series is divergent and so its sum is not possible.
(d) Geometric mean (G.M.)
G ab is the geometric mean of two positive numbers a and b.
(e) If terms are given in G.P. and their product is known, then the terms must be picked up
in the following way.
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For three terms in G.P., we choose them asa
, a , a r r
For four terms in G.P., we choose them as3
3
a a, , ar, ar
r r
For five terms in G.P., we choose them as2
2
a a, , a, ar, ar
r retc.
(ii) Useful Properties
(a) The product of the terms equidistant from the beginning and end is constant, and it isequal to the product of the first and the last term.
(b) If every term of a G.P. is multiplied or divided by the some non-zero quantity, the
resulting progression is a G.P.
(c) If a1, a
2, a
3... and b
1, b
2, b
3, ... be two G.P.s of common ratio r
1and r
2respectively, then
a1b
1, a
2b
2... and 31 2
1 2 3
aa a, ,
b b b... will also form a G.P. Common ratio will be r
1r
2and 1
2
r
r
respectively.
(d) If a1, a
2, a
3, ... be a G.P. of positive terms, then loga
1, loga
2, loga
3, ... will be an A.P. and
conversely.
HARMONIC PROGRESSION (H.P.)
(a) A sequence is said to be in harmonic progression, if and only if the reciprocal of its
terms form an arithmetic progression.
For example
1 1 1, ,2 4 6
... form an H.P., because 2, 4, 6, ... are in A.P..
(b) If a, b are first two terms of an H.P. then
n
1t
1 1 1(n 1)
a b a
(c) Harmonic mean H of any two numbers a and b
2 2abH1 1 a ba b
, where a, b are two non-zero numbers.
(d) If terms are given in H.P. then the terms could be picked up in the following way
(i) For three terms in H.P, we choose them as
1 1 1, ,
a d a a d
(ii) For four terms in H.P, we choose them as
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1 1 1 1, , ,
a 3d a d a d a 3d
(iii) For five terms in H.P, we choose them as1 1 1 1 1
, , , ,a 2d a d a a d a 2d
Insertion of Means Between two numbers
Let a and b be two given numbers.
(i) Arithmetic Means
If a, A1, A
2, ... A
n, b are in A.P., then A
1, A
2, ... A
nare called n A.M.s between a and b. If d is
the common difference, then
b = a + (n + 2 1) d d =b a
n 1
Ai= a + id = a + i
b a a(n 1 i) ib,
n 1 n 1
i = 1, 2, 3, ..., n
Note: The sum of n-A. M s, i.e., AA1+ A2+ ... + An=n
(a b)2
(ii) Geometric means
If a, G1, G
2... G
n, b are in G.P., then G
1, G
2... G
nare called n G.M.s between a and b. If r is the
common ratio, then
b = a.rn+1 r =1
(n 1)b
a
Gi= ari=
in 1 i i
n 1n 1 n 1
ba a .b ,
a
i = 1, 2, ..., n
Note: The product of n-G. M s i.e., G1
G2
... Gn
=
n
ab
(iii)Harmonic Means
If a, H1, H
2... H
n, b are in H.P., then H
1, H
2... H
nare called n H.M.s between a and b. If d is the
common difference of the corresponding A.P., then
1 1 a b(n 2 1) d d
b a ab(n 1)
i
1 1 1 a bid i
H a a ab(n 1)
ab(n 1), i 1, 2, 3, ..., n
b(n i 1) ia
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MEANS OF NUMBERS
Let a1, a
2, ... a
nbe n given numbers
(i) Arithmetic Means (A.M.) = 1 2 na a ... a
n
(ii) Geometric Means (G.M.) = (a1a2... an)1/n
(iii) Harmonic Mean (H.M.) =1 2 n
n
1/ a 1/ a ... 1/ a
If weights of a1, a
2, ... , a
nare w
1, w
2, ..., w
nrespectively, then their weighted arithmetic mean,
weighted geometric mean and weighted harmonic mean are respectively defined by
1 2 n 1 2 n
1w w w w w ...w1 1 2 2 n n
1 2 n1 2 n
a w a w ... a w(a .a ...a )
w w ... w
1 2 n
1 2 n
1 2 n
w w ... wand
w w w...
a a a
.
RELATION BETWEEN A, G AND H
If A, G and H are A.M., G.M. and H.M. of positive numbers a1, a
2... a
n( 1 2 na a ... a ) then
1 na H G A a ...(1)
Note:
(i) The equality at any place in (1) holds if and only if the numbers a1, a
2, ..., a
nare all equal
(ii) (1) is true for weighted means also
(iii) G2= AH, if n = 2.
ARITHMETIC MEAN OF mthPOWER
Let a1, a
2 . . . , a
n be n positive real numbers and let m be a real number, then
mm m m1 2 n 1 2 na a ... a a a ... a , if m R [0,1].
n n
However if m (0, 1), thenmm m m
1 2 n 1 2 na a ... a a a ... a
n n
Obviously ifmm m m
1 2 n 1 2 na a ... a a a ... am {0,1}, thenn n
ARITHMETIC-GEOMETRIC SERIES
A series whose each term is formed, by multiplying corresponding terms of an A.P. and a G.P., iscalled an Arithmetic-geometric series.
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e.g. 1 + 2x + 4x2 + 6x3+ ..... ; a + (a + d) r + (a + 2d)r2+ .....
(i) Summation of n terms of an Arithmetic-Geometric Series
Let Sn= a + (a + d) r + (a + 2d)r2+ ... + [a + (n 1)d] rn1, d 0 , r 1
Multiply by r and rewrite the series in the following way
rSn = ar + (a + d)r2+ (a + 2d)r3+ ... + [a + (n 2)d]rn1+ [a + (n 1)d ]rnon subtraction,S
n(1 r) = a + d(r + r2+ ... + rn1) [a + (n 1)d]rn
or,n 1
nn
dr(1 r )S (1 r) a [a (n 1)d].r
1 r
or,n 1
nn 2
a dr(1 r ) [a (n 1)d]S .r
1 r (1 r) 1 r
(ii) Summation of Infinite Series
If |r| < 1, then (n 1)rn, rn1 0, as n .
Thus S = S = 2a dr
1 r (1 r)
SUM OF MISCLENIOUS SERIES
(i) Defference Method
Let T1, T
2, T
3 ... T
nbe the trms of a sequence and let (T
2 T
1) =
1T , (T3 TT2) = 2T ... ,
(Tn Tn1) = n 1T .
Case I: If 1 2 n 1T , T ,....T are in A.P. then TTnis a quadratic in n. If 1 2 2 3T T , T T , ... are in
A.P., then Tnis a cubic in n.
Case II: If 1 2 n 1T , T ,....T are not in A.P., but if 1 2 n 1T , T ,..., T are in G.P., then TTn= arn+ b,
where r is the C.R. of the G.P.1 2 3T , T , T .....and a, b R.
Again if1 2 n 1T ,T ,...T are not in G.P. but 2 1 3 2 n 1 n 2T T ,T T ,...T T are in G.P., then TTn is of the
form arn+ bn + c; r is the C.R. of the G.P. 2 1 3 2 4 3T T ,T T T T ... and a, b, c R.
(ii) Vn Vn1Method
Let T1, T
2, T
3 , ... be the terms of a sequence. If there exists a sequence V
1, V
2, V
3... satisfying
Tk= V
k V
k1, k 1,
thenn n
n k k k 1 n 0k 1 k 1
S T (V V ) V V
.
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Multiple Choice Questions with Single Answer:
1. If a1, a
2, a
3... are in AP, then a
p, a
q, a
rare in AP, if p, q, r are in
(a) AP (b) GP
(c) HP (d) none of these
2. Let trdenote the rth term of an AP. If t
m=
1
mand n
1t =
n, then t
mn equal to
(a)
1
mn (b)
1 1
m n
1
(c) 1 (d) 0
3. If p, q, r, s N and they are four consecutive terms of an AP, then the pth, qth, rth, sth terms of
a GP are in
(a) AP (b) GP
(c) HP (d) none of these
4. Let a1, a
2, a
3, ... be in AP and a
p, a
q, a
rbe in GP. Then a
q: a
pis equal to
(a)r p
q p
(b)
q p
r q
(c)r q
q p
(d) none of these
5. If a, b, c are in G.P., then a + b, 2b, b + c are in
(a) A.P. (b) G.P.
(c) H.P. (d) none of these
6. If a, b, c, d are nonzero real numbers such that (a2+ b2+ c2) (b2+ c2+ d2)
(ab + bc + cd)2, then
a, b, c, d are in
(a) AP (b) GP(c) HP (d) none of these
OBJECTIVE
SECTION - I
LEVEL - I
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7. If 4a2+ 9b2+ 16c2= 2(3ab + 6bc + 4ca), where a, b, c are nonzero numbers, then a, b, c are in
(a) AP (b) GP
(c) HP (d) none of these
8 If a, b, c are in H.P., then c, c a, c b are in :
(a) A.P. (b) G.P.
(c) H.P. (d) none of these
9. Let S be the sum, P be the product and R be the sum of the reciprocals of n terms of a GP. Then
P2Rn: Snis equal to
(a) 1 : 1 (b) (common ratio)n : 1
(c) (first term)2: (common ratio)n (d) none of these
10. If a1, a
2,a
3are in AP, a
2, a
3, a
4are in GP and a
3, a
4, a
5are in HP, then a
1, a
3, a
5are in
(a) AP (b) GP
(c) HP (d) none of these
11. If x > 1, y > 1, z > 1 are three numbers in GP then1 1 1
, ,1+ ln x 1+ ln y 1+ ln z are in
(a) AP (b) HP(c) GP (d) none of these
12. If a, a1, a
2, a
3, ... a
2n1, b are in AP, a, b
1, b
2, b
3..., b
2n1, b are in GP and a, c
1, c
2, c
3, ... , c
2n1, b are
in HP, where a, b are positive, then the equation anx2 b
nx + c
n= 0 has its roots
(a) real and unequal (b) real and equal
(c) imaginary (d) none of these
13. The product of n positive numbers is unity. Then their sum is
(a) a positive integer (b) divisible by n
(c) equal to n +1
n(d) never less than n
14. If x > 0 and a is known positive number, then the least values of ax +a
xis
(a) a2 (b) a
(c) 2a (d) none of these
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15. If p, q, r be three positive real numbers, then the value of (p + q) (q + r) (r + p) is
(a) > 8 pqr (b) < 8 pqr
(c) 8 pqr (d) none of these
16. If a1, a
2, a
3.. a
nare in H.P., then
1 2 n
2 3 n 2 3 n 2 3 n 1
a a a, ,....
a a ....a a a ....a a a ....a
are in
(a) A.P. (b) G.P.
(c) H.P. (d) none of these
17. Sn = 1 + n
1 1 1 1..... ,
2 3 4 2 1
then
(a) S100
< 100 (b) S200
< 200
(c) S200
> 100 (d) S50
> 25
18. Given Sn
n
r rr 0 r 0
1 1,s .
2 2
If S Sn
1
1000 , then least value of n is
(a) 8 (b) 9
(c) 10 (d) 11
19. If x15 x13+ x11 x9+ x7 x5+ x3 x = 7,(x > 0), then
(a) x16is equal to 15 (b) x16is less than 15
(c) x16greater than 15 (d) none of these
20.n 1
1(n 1) (n 2) (n 3)......(n k)
is equal to
(a)1
(k 1) k 1 (b)1
k k
(c)1
(k 1) k (d)1
k
21. Maximum value of n for whichn n
14 1
11 n
2
is
(a) 4 (b) 5
(c) 6 (d) 7
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22. The nth term of the series1 7 1 20
2 1 1 .....2 13 9 23
is
(a)20
(5n 3) (b)20
(5n 3)
(b) 20 (5n + 3) (d) 220(5n 3)
23. If1 1 1 1
+ + +a a - b c c - b
= 0 and a c b then a, b, c are in
(a) A.P. (b) G.P.
(c) H.P. (d) None of these
24. If 2p + 3q + 4r = 15, the maximum value of p3q5r7will be
(a) 2180 (b)4 5
15
5 .3
2
(c)5 7
17
5 .7
2 .9(d) 2285
25. The minimum value of x4 21
x is
(a) 31/ 3
1
4
(b)1
2
(c)1/ 3
12
3
(d) 2
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Multiple Choice Questions with one or more than one correct Answers:
1. Total number of positive real values of x such that x, [x], {x} are in H.P, where [.] denotes the
greatest integer function and {.} denotes fraction part, is equal to
(a) zero (b) 1(c) 2 (d) none of these
2. In the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, ..... , where n consecutive terms have the
value n, the 1025th term is
(a) 210 (b) 211
(c) 29 (d) 28
3. Sr denotes the sum of the first r terms of an AP. Then S3n: (S2n Sn) is(a) n (b) 3n
(c) 3 (d) independent of n
4. a, b, c are distinct real numbers such that a, b, c are in A.P. and a2, b2, c2are in H.P. then
(a) 2b2= ac (b) 4b2= ac
(c) 2b2= ac (d) 4b2= ac
5. If x1, x2, ... xnare in H.P. then
n-1
r r+1r=1 x x is equal to
(a) (n 1)x1x
n(b) nx
1x
n
(c) (n + 1) x1x
n(d) none of these
6. If ax = by = czand x, y, z are in GP then logc b is equal to
(a) logba (b) log
ab
(c) z/y (d) none of these
7. The value ofn
r=1
1
a + rx + a + (r -1)x is
(a)n
a a nx (b)a nx a
x
(c)n( a nx a)
x
(d) none of these
8. Sides a, b, c of a triangle are in G.P. If5c 3b a
ln , ln and lna 5c 3b
are in A.P., then triangle
must be
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(a) Isosceles (b) Equilateral(c)