Mathematics MCA11
Mathematics MCA11
Mathematics MCA11
by
K.S. Srinivasa
Retd. Principal &
Professor of Mathematics
Bangalore
Published by
Sharada Vikas Trust (R)
Bangalore
MCA 11 MATHEMATICS
Syllabus
1. Complex Trigonometry
Revision of Plane Trigonometry - trigonometric ratios, expressions for relation between allied angles and trigonometrical
ratios. Addition formulae for trigonometrical ratios and simple problems. Complex numbers and functions, definition,
properties, De Moivre's Theorem (without proof), Roots of a complex number, expansions of sin (nθ), cos (nθ) in
powers of sin θ & cos θ, addition formulae for any number of angles, simple problems.
2. Matrix Theory :
Review of the fundamentals. Solution of linear equations by Cramers' Rule and by Matrix method, Eigen values and
Eigen vectors, Cayley Hamilton's Theorem, Diagonalization of matrices, simple problems.
3. Algebraic Structures
Definition of a group, properties of groups, sub groups, permutation groups, simple problems, scalars & vectors, algebra
of vectors, scalar & vector products, scalar triple product, simple problems.
4. Differential Calculus
Limits, continuity and differentiability (definition only), standard derivatives, rules for differentiation, derivatives of
function of a function and parametric functions, problems. Successive differentiation, nth derivative of standard functions,
statement of Leibnitz's Theorem, problems, polar forms, angle between the radius vector and the tangent to a polar
curve, (no derivation) angle between curves, pedal equation, simple problems, indeterminate forms, L' Hospital's rule,
partial derivatives, definition and simple problems.
5. Integral Calculus
Introduction, standard integrals, integration by substitution and by parts, integration of rational, irrational and trigonometric
functions, definite integrals, properties (no proof), simple problems, reduction formulae and simple problems.
Text Books
1. Elementary Engineering Mathematics by Dr. B.S. Grewal, Khanna Publications
2. Higher Engineering Mathematics by B.S. Grewal, Khanna Publications
Reference Books
1. Differential Calculus by Shanti Narayan, Publishers S. Chand & Co.
2. Integral Calculus by Shanti Narayan, Publishers S. Chand & Co.
3. Modern Abstract Algebra by Shanti Narayan, Publishers S. Chand & Co.
CONTENTS
Page Nos.
1. Complex Trigonometry 01
2. Matrix Theory 27
3. Algebraic Structures 47
4. Differential Calculus 69
adjacent side BC
The ratio ie is defined as ‘cosine of θ ’ & written as cosθ.
hypotenuse AB
opposite side AC
The ratio ie is defined as ‘tangent of θ ’ & written as tanθ. θ 90º
adjacent side BC B C
sin è
It can be seen by the above definition that = tanè
cos è
1 AB
The reciprocal of sinθ ie ie is defined as ‘cosecant of θ ’ and written as cosecθ.
sin è AC
1 AB
The reciprocal of cosθ ie ie is defined as ‘secant of θ ’ and written as secθ.
cos è BC
BC
The Reciprocal of tanθ ie is defined as ‘cotangent of θ ’ and written as cotθ.
AC
AC 2 AB 2
divide by BC 2 , we have +1 = ie tan2θ + 1 = sec2θ
BC 2 BC 2
BC 2 AB 2
divide by AC 2 , then 1+ = ie 1 + cot2θ = cosec2θ
AC 2 AC 2
AD 3
cos 30° = =
AB 2
BD 1
tan30° = =
AD 3
2
also cosec 30° = 2, sec 30° = & cot 30° = 3
3
From the above results, it can seen that sin60º = cos30º, cos60º = sin30º and tan60º = cot30º.
AB C = 45° = B AC
Let AC = BC = 1 unit then AB = 2 units
2
AC 1 1
\ sin 45° = =
AB 2
BC 1
cos 45° = = B C
AB 2 1
AC 1
tan45° = = =1
BC 1
Note :- Trigonometric ratios of 30º, 45º and 60º are called ‘Standard Trigonometric’ ratios which are always useful,
hence these values have to be always remembered.
Let XOX' & YOY' be co-ordinate axes where O is the origin. Consider a circle of radius r with centre O. Let P be any point
on the cirlce whose co-ordinates are (x, y).
Then OM = x & PM = y Y
Let MO P = è
y x y
sin è = , cos è = , tanè =
r r x (–, +) (+, +) P (x, y)
r r x y
also cosecè = , sec è = , cot è =
y x y X' θ x
O M X
When MO P, satisfy 0º < θ < 90º the print P will be in first quadrant
of the circle, when it satisfy 90º < θ < 180º, P will in second quadrant, (–, –) (+, –)
when 180º < θ < 270º, P will be in third quadrant & finally when 270º
< θ < 360º the point P will be in fourth quadrant, because of these
positions, signs of the Trigonometric ratios changes.
Y'
In the I quadrant both x & y are +ve and r is always +ve.
Therefore sinθ, cosθ & tanθ are +ve, their reciprocals are also +ve.
Therefore sinθ & cosecθ are +ve cosθ, tanθ, secθ & cotθ are –ve.
Therefore, tanθ & cotθ are +ve and sinθ, cosθ, cosecθ & secθ are –ve.
Therefore, cosθ & secθ are +ve and sinθ, tanθ, cosecθ & cotθ are –ve.
Note :- The signs of the trigonometric ratios can be easily remembered with the help of the following diagram
As θ → 0, y → 0, x → r
0
\ sin 0° = =0
r
fig.1
4 KSOU Complex Trigonometry
P
r
cos 0° = =1
r
0 y
tan0° = =0
r
θ
O xM
If MO P, is very close to 90º as in fig. 2.
As θ → 90°, x → 0, y → r
r
\ sin 90° = =1
r fig.2
0
cos 90° = =0
r
r
tan90° = =∞
0 P
y
θ
If θ is very close to 180º as in fig. 3 x
M O
As θ → 180°, x → −r , y → 0
0
\ sin 180° = =0
r
−r
cos 180° = = −1 fig.3
r
0
tan180° = =0
r
When θ → 360° , x → r , y → 0
θ x M
0
\ sin 360° = =0 y
r
r
r P
cos 360° = =1
r
0
tan360° = =0
r fig.5
MCA 11 - Mathematics SVT 5
Note :- From the above derivations it can be seen that the values of sinθ & cosθ always be between –1 & 1. Whereas the
value of tan θ will be between −∞ & ∞ and hence the graphs of the trigonometric functions
y = sin x, y = cos x & y = tan x are as follows.
y = sin x
x
O 90° 180° 270° 360°
fig.6
y = cos x
x
O 90° 180° 270° 360°
fig.7
y = tanx
x
O 90° 180° 270° 360°
fig.8
6 KSOU Complex Trigonometry
Using the Trigonometric ratios of standard angles 30°, 45° & 60° and using the above rules for allied angles, we can
find the trigonometric ratios of other angles as follows.
3
Eg.1 sin120° = sin(90° + 30° ) = cos 30° =
2
3
or sin120° = sin(180° − 60° ) = sin 60° =
2
3
Eg.2 cos150° = cos(90° + 60° ) = − sin 60° = −
2
3
or cos150° = cos(180° − 30° ) = − cos 30° = −
2
1
Eg.3 sin 315° = sin(360° − 45° ) = − sin 45° = −
2
1
or sin 315° = sin(270° + 45° ) = − cos 45° = −
2
B
Radian Measure r r
Consider a circle of radius r with centre O. Let AB be a arc such that arc AB = r.
1 radian
Measure of the AO B is called a 'radian' denoted as 1C. O r A
Consider a circle of radius r with centre O. Let arc AB = r so that AO B is 1 radian.
Let C be a point on the circle such that AO C = 90°.
Since the angle subtended at the centre of a circle is proportional to the corresponding arc.
MCA 11 - Mathematics SVT 7
AO B arc AB C
=
AO C arc AC B
1 radian r 2
= = r
r
90° 1
(2πr ) π
4
O r A
1 1
Q AC = 4 Circumference = 4 ( 2πr )
∴ π radians = 180°
ie π = 180° imp. result.
π π π π
= 90°, = 45°, = 60°, = 30°.
2 4 3 6
Using the relation π = 180° measurement of an angle in degrees can be converted to radians and vice-versa.
Eg. (1) Qn. : Convert 40° to radians
π 2π
Solution : 40° = 40 × = radius
180 9
2π
Eg. (2) Qn. : Convert radians in to degrees
3
2π 180
Solution : × = 120°
3 π
Some Problems
1. Show that
(1 + cot A) 2 + (1 − cot A) 2 = 2 cosec2 A
Solution : LHS = 1 + cot 2 A + 2 cot A + 1 + cot 2 A − 2 cot A
= 2 + 2 cot 2 A
= 2(1 + cot2 A)
= 2 cosec2 A = RHS.
2. Show that
1 + sin A 1 − sin A
− = 4 sec A tan A
1 − sin A 1 + sin A
(1 + sin A) 2 − (1 − sin A) 2
Solution : LHS = (taking LCM)
(1 − sin A)(1 + sin A)
a 2 −1
3. If sec A + tan A = a, then prove that = sin A
a2 +1
4 sinθ + cos θ
4. If sinθ = , find the value of
5 tanθ − cotθ
4
Solution : If sinθ = then opposite side is 4 hypotenusis 5.
5
3 4
∴ adjacent side = 25 − 16 = 9 = 3 ∴ cos θ = & tanθ =
5 3
4 3 7
+
sinθ + cos θ 12
∴ = 5 5 = 5 =
tanθ − cotθ 4
−
3 7 5
3 4 12
3 π 5 cos θ + 8 tanθ
5. If tanθ = − , < θ < π find
4 2 8 secθ − 3 cosecθ
Solution : Since θ lies in II Quadrant sine is +ve cosine & tangant are –ve.
3 4
∴ sinθ = and cos θ = −
5 5
−4 −3
5 + 8
5 cos θ + 8 tanθ
=
5 4
∴
8 secθ − 3 cosecθ 5 5
8 − − 3
4 3
− 4 − 6 10 2
= = =
− 10 − 5 15 3
6. Prove that
tan(180° + θ ) sec(180° − θ ) cosec(90° + θ )
= sec 2 θ
sec(360° − θ ) cot(90° + θ ) sin(90° − θ )
MCA 11 - Mathematics SVT 9
7. Prove that
π
sin(π + θ ) cos(2π − θ ) cot − θ
2 = sinθ
π 3π
tan + θ cot + θ sin(−θ )
2 2
= cos θ tanθ
sinθ
= cos θ × = sinθ = RHS
cos θ
Addition Formulae
sin( A + B) = sin A cos B + cos A sin B
cos( A + B) = cos A cos B − sin A sin B
tan A + tan B
tan(A + B ) =
1 − tan A tan B
replacing B by –B
tan A − tan B
tan(A − B ) =
1 + tan A tan B
Using the above formulae we can find the trigonometric ratios of 15°, 75°, 105° etc.
1 3 1 1 3 +1
= × + × =
2 2 2 2 2 2
cos 75° = cos( 45° + 30°)
1 3 1 1 3 −1
= × − × =
2 2 2 2 2 2
sin 75° 3 +1
tan75° = =
cos 75° 3 −1
1 3 1 1 3 −1
= × − × =
2 2 2 2 2 2
1 3 1 1 3 +1
= × + × =
2 2 2 2 2 2
sin15° 3 −1
tan15° = =
cos15° 3 +1
3 1 1 1 3 +1
= × + × =
2 2 2 2 2 2
1 1 3 1 1− 3
= × − × =
2 2 2 2 2 2
3 +1
tan105° =
1− 3
Alternate method
sin15° = sin(90° − 75°) = cos 75°
3 −1
=
2 2
3 +1
=
2 2
3 +1
=
2 2
=
− ( 3 − 1) = 1 − 3
2 2 2 2
MCA 11 - Mathematics SVT 11
= 3 sinθ − 4 sin 3 θ
cos 3θ = cos(2θ + θ )
= 4 cos 3 θ − 3 cos θ
12 KSOU Complex Trigonometry
tan3θ = tan(2θ + θ )
tan2θ + tanθ
=
1 − tan2θ tanθ
2 tanθ
+ tanθ
= 1 − tan2 θ
2 tanθ
1− × tanθ
1 − tan2 θ
3 tanθ − tan3 θ
=
1 − 3 tan2 θ
Thus we have,
2 tanθ
tan2θ =
1 − tan2 θ
3 tanθ − tan3 θ
tan3θ =
1 − 3 tan2 θ
Problems
sin 3 A cos 3 A
1. Show that − =2
sin A cos A
2
2. If cos A = find sin2A & cos3A.
3
2 9−4 5
Solution : Given cos A = , sin A = =
3 3 3
MCA 11 - Mathematics SVT 13
5 2 4 5
sin 2 A = 2 sin A cos A = 2 × × =
3 3 9
cos 3 A = 4 cos 3 A − 3 cos A
8 2 32 32 − 54 − 22
= 4× − 3× = −2 = =
27 3 27 27 27
4 5 − 22
∴ sin 2 A = & cos 3 A =
9 27
3. Prove that
1 + sin 2θ
= tan2 ( 45° + θ )
−
1 sin 2 θ
1 + 2 sinθ cos θ
Solution : LHS =
1 − 2 sinθ cos θ
(cos θ + sinθ ) 2
=
(cos θ − sinθ ) 2
2
cos θ + sinθ
= (dividing Nr & Dr inside the bracket by cos θ )
cos θ − sinθ
2
1 + tanθ
=
1 − tanθ
= tan(45° + θ ) = RHS
1 − cos θ θ
4. Prove that = tan2 and hence prove that tan15° = 2 − 3
1 + cos θ 2
θ
1 − 1 − 2 sin 2
1 - cos θ 2
Solution : = using cos2 A formula
1 + cos θ θ
1 + 2 cos 2 − 1
2
θ θ
1 − 1 + 2 sin 2 2 sin 2
= 2 = 2 = tan2 θ
θ θ 2
1 + 2 cos 2 − 1 2 cos 2
2 2
1 − cos θ θ
∴ = tan2
1 + cos θ 2
3
1−
put θ = 30°. tan 15° =
2 1 − cos 30°
= 2 = 2− 3
1 + cos 30° 3 2+ 3
1+
2
(2 − 3 )(2 − 3 ) = (2 − 3 ) = (2 − 3 )
2
2
=
(2 + 3 )(2 − 3 ) 4 − 3
θ
∴ tan = 2 − 3
2
14 KSOU Complex Trigonometry
Complex Number
Definition : A number of the form x + iy where x ∈ R, y ∈ R & i = − 1 is defined as a Complex Number and usually denoted
as Z. x is called Real part & y is called Imaginary part. x – iy is called Conjugate. Complex number denoted as Z or z . A
complex number can be represented by a point on a plane by taking real part on x-axis & imaginary part on y-axis. The plane
on which complex numbers are represented is called a Complex Plane. For every point in a plane there is a complex number
& for every complex number there is a point in the plane. In a complex x-axis is called Real axis & y-axis Imaginary axis.
Properties
(1) Equality. Two complex numbers z1 = x1 + iy1 , z 2 = x2 + iy 2 are said to be equal if x1 = x2 , y1 = y 2
(3) Subtraction. z1 − z 2 = ( x1 − x2 ) + i ( y1 − y 2 )
= x1 x2 + ix2 y1 + ix1 y2 + i 2 y1 y2
= x1 x2 − y1 y 2 + i ( x1 y 2 + x2 y1 ) Q i 2 = −1
z1 ( x1 + iy1 )
(5) Division. z = ( x + iy )
2 2 2
multiply numerator & denominator by the conjugate of x2 + iy2 ie x2 – iy2, then
z1 ( x + iy1 )( x2 − iy 2 )
= 1
z 2 ( x2 + iy 2 )( x2 − iy 2 )
x1 x2 + y1 y 2 + i ( x2 y1 − x1 y 2 )
=
x22 + y 22
x1 x2 + y1 y 2 x2 y1 − x1 y 2
= +i
x22 + y 22 x22 + y 22
which is a complex number.
Note :- Product of a complex number with its conjugate is always a positive real number.
ie zz = ( x + iy)( x − iy) = x 2 + y 2 & z + z = 2 x, z − z = 2iy
z+z z−z y
∴ x= & y=
2 2i
OM x
cos θ = = ⇒ x = r cos θ
OP y
PM y
sinθ = = ⇒ y = r sinθ
OP r
∴ z = x + iy = r cos θ + ir sinθ = r (cosθ + i sinθ )
This form of the complex number is called 'Polar form'. Where r is called Modulus & θ is called argument which are
given by
y
r = x 2 + y 2 & θ = tan−1
x
r is always positive and argument θ varies from 0° to 360°. The value of argument satisfying –π < θ ≤ π is defined
as amplitude which is unique for a complex number.
Thus we have
Mod. z = z = r = x 2 + y 2
y
arg z = tan−1 and amp. z = θ where –π < θ ≤ π
x
x y
while finding the amplitude of a complex number, we have to find θ satisfying ie cos θ = and sinθ =
y x
Examples
2 + 3i
1. Express in the form x + iy
1+ i
2 + 3i ( 2 + 3i )(1 − i ) 2 + 3i − 2i − 3i 2
Solution : = =
1+ i (1 + i )(1 − i ) 1+1
2+i +3 5+i 5 1
= = = +i
2 2 2 2
(1 + i ) 2
2. Express in the form x + iy
3−i
(1 + i ) 2 1 + i 2 + 2i 2i (3 + i )
Solution : = = (using i 2 = −1)
3−i (3 − i ) (3 − i )(3 + i )
6i + 2i 2 − 2 + 6i − 2 6i 1 3i
= = = + =− +
9 +1 10 10 10 5 5
Solution : z = r = x 2 + y 2 = 1 + 1 = 2
x 1 y 1 π
cos θ = = and sinθ = = ∴ θ = 45° =
r 2 r 2 4
π
∴ Modulus is 2 and amplitude is
4
16 KSOU Complex Trigonometry
Solution : z = 3 + 1 = 4 = 2
− 3
cosθ =
2 5π
→θ =
1 6
sinθ =
2
5π
∴ Modulus is 2 & amplitude is
6
Solution : z = 1 + 3 = 4 = 2
1
cosθ = −
2 2π
→θ = −
3 3
sinθ = −
2
2π
∴ Modulus is 2 & amplitude is −
3
Solution : z = 1 + 1 = 2
x 1
cos θ = =
r 2 π
→θ = −
y 1 4
sinθ = = −
r 2
π
∴ Modulus is 2 & amplitude is −
4
1
7. If α + iβ = , then prove that (α 2 + β 2 )(a 2 + b 2 ) = 1
a + ib
1 ( a − ib)
Solution : α + iβ = =
a + ib (a + ib)(a − ib)
a − ib a b
= = −i
a +b
2 2
a +b
2 2
a + b2
2
a2 b2 (a 2 + b 2 ) 1
α2 + β2 = + = =
(a 2
+b ) (a
2 2 2
+b 2 2
) (a 2
+b )
2 2 a + b2
2
∴ cross - multiplying
(α 2 + β 2 )(a 2 + b 2 ) = 1
DeMoivre's Theorem
Statement : If n is a +ve or –ve integer, then (cosθ + i sinθ ) n = cos nθ + i sin nθ
If n is a +ve or –ve fraction, one of the values of (cosθ + i sinθ ) n is cos nθ + i sin nθ .
p
Case (ii) When n = where p & q are +ve integers.
q
q
p p
Consider cos θ + i sin θ
q q
p p
= cos ⋅ qθ + i sin ⋅ qθ = cos qθ + i sin pθ = (cosθ + i sinθ ) p
q q
q
p p
ie (cosθ + i sinθ ) p = cos θ + i sin θ
q q
p
p p
taking qth roots on both sides. One of the values of (cosθ + i sinθ ) q
= cos θ + i sin θ
q q
Case (iii) when n is –ve integer or –ve fraction. Let n = –m where m is a +ve integer or +ve fraction
(cosθ + i sinθ ) n = (cosθ + i sinθ ) − m
18 KSOU Complex Trigonometry
1 1
= =
(cos θ + i sinθ ) m
cos mθ + i sin mθ
cos mθ − i sin mθ
= = cos( −m)θ + i sin(− m)θ = cos nθ + i sin nθ .
1
Important Results
1
(i) = cosθ − i sinθ
If x = cos θ + i sin θ then
x
1 1
∴ x + = 2 cos θ and x − = 2i sinθ .
x x
1 1
= cosθ − i sinθ & n = (cos θ − i sinθ ) n
x x
= cos nθ − i sin nθ
1 1
∴ xn + = 2 cos nθ & x n − = 2i sin nθ
xn xn
Note :- For convenience cos θ + i sin θ can be written as cisθ .
z
1
n
=r
1
n
[cos(2kπ + θ ) + i sin(2kπ + θ )] 1
n
1 2 kπ + θ ( 2kπ + θ )
= r n cos + i sin
n n
where k = 0, 1, 2, ......... n – 1. Let us denote the nth roots of the complex number by z0, z1, z2, ..............zn – 1
Then,
1 θ θ 1 θ
for k = 0, z 0 = r n cos + i sin = r n cis
n n n
1 2π + θ 2π + θ 1 2π + θ
k = 1, z1 = r n cos + i sin = r n cis
n n n
1 4π + θ 4π + θ 1 4π + θ
k = 2, z 2 = r n cos + i sin = r n cis
n n n
n n n
the above n values gives nth roots of z = x + iy
MCA 11 - Mathematics SVT 19
1
Note :- If k = n, n + 1, n + 2 etc. The values will repeat. Hence these will be only n values of z n which are distinct.
Using the polar form of the complex number we can plot the nth roots of the complex in the following way.
1
z2
Draw a circle of radius r whose centre is O. Mark a point on the circle and take
n
θ z1
OA as intial line. Take a point B such that A OB = . Then B represent z0. Take a C
n
2π + θ z0
point C such that AOC = then C represent z1, like this all the nth roots can B
n
be represented. This diagram is called 'Argand Diagram'. O θ /n A
1
r n
Problems :
cos 5θ − i sin 5θ ) 2 (cos 7θ + i sin 7θ ) −3 zn
(1) Simplify
(cos 4θ − i sin 4θ ) (cosθ + i sinθ )
9 5
1 1
(3) If 2 cos θ = x + & 2 cos φ = y +
x y
Show that
1 xm yn
(i) x m y m + = 2 cos(mθ + nφ ) & (ii) + = 2 cos( mθ − nφ )
xm y m yn xm
1 x2 +1
Solution : 2 cos θ = x + =
x x
∴ x 2 + 1 = 2 cosθ ⋅ x ie x 2 − 2 cosθ x + 1 = 0
1
= 2 cos(mθ + nφ ) − i sin(mθ + nφ ) (2)
x yn m
xm yn
∴ = cos( mθ − nφ ) + i sin(mθ − nφ ) & = cos(mθ − nφ ) − i sin(mθ − nφ )
yn xm
adding
xm yn
n
+ = 2 cos(mθ − nφ )
y xm
n a
b
Solution : Let a + ib = r (cos θ + i sinθ ) where r = a 2 + b 2 & θ = tan−1
a
m m
= r n (cos θ + i sinθ )
m m m m
∴ ( a + ib) =r cos n θ + i sin n θ (1)
n n n
a − ib = r (cosθ − i sinθ )
m m
cos n θ − i sin n θ
m m
( a − ib) n
=r n
(2)
adding (1) & (2)
m m
= r n 2 cos θ
m m
( a + ib) n
+ ( a − ib) n
n
m
= a 2 + b 2
2n m b
tan−1
m m
∴ ( a + ib) n
+ ( a − ib) n
2 cos
n a
b b
Q r = a 2 + b 2 & tanθ = ⇒ θ = tan−1
a a
∴ ( a + ib)
m
n
+ ( a − ib)
m
n
(
= 2 a2 + b2 )m
2n m b
2 cos tan−1
n a
MCA 11 - Mathematics SVT 21
(5) Find the cube roots if 1 + i and represent them on Argand diagram.
r2 = 2 ⇒ r = 2
1
cos θ =
2 π
→θ =
1 4
sinθ =
2
π π
∴ z = 2 cos + i sin
4 4
π π
= 2 cos 2kπ + + i sin 2kπ + for k ∈ I
4 4
π
∴ z = 1 + i = 2 cis 2kπ +
4
( 2)
1
1 1 1 8kπ + π 3
8kπ + π
= (1 + i ) = cis
3 1
z 3 3
= 2 6 cis for k = 0, 1, 2
4 12
1 π 1
when k = 0, z 0 = 2 6 cis = 2 6 cis15°
12
1 9π 1 3π 1
k = 1, z1 = 2 6 cis = 2 6 cis = 2 6 cis135°
12 4
1 17π 1 17π 1
k = 2, z 2 = 2 6 cis k = 2, z 2 = 2 6 cis = 2 6 cis255°
12 12
To represent them on Argand diagram.
1
Draw a circle of radius 2 6
with centre O. Let OA be the initial line. C
Take a point B on the circle such that A OB = 15°, take a point C on the circle
B
such that A OC = 135° & take a point D on the circle such that A OD = 255° then 135°
the points B, C, D represent z0, z1, z2. 15° A
255° O 1
2 6
1
1 3
4
(6) Find all the values of − i & find their continued product.
2 2
1 3
Solution : Let z = −i = r (cosθ + i sinθ ) D
2 2
1 3
∴ r cosθ = , r sinθ = −
2 2
Squaring & adding
1 3 4
r2 = + = =1
4 4 4
22 KSOU Complex Trigonometry
1
cosθ =
2 π
→θ = −
3 3
sinθ = −
2
π π
∴ z = 1cos − + i sin −
3 3
π
z = cis 2kπ − for k ∈ I
3
1
1 3
1 1
π 6kπ − π
4 4 4
6kπ − π
= cis for k = 0, 1, 2, 3
12
π 5π
for k = 0, z 0 = cis − k = 1, z1 = cis
12 12
11π 17π
k = 2, z 2 = cis k = 3, z 3 = cis
12 12
Their continued product
π 5π 11π 17π
= ci s − cis cis cis
12 12 12 12
π 5π 11π 17π 32π 8π
= cis − + + + = cis = cis
12 12 12 12 12 3
2π 2π 2π 2π 1 3
= cis 2π + = cis = cos + i sin = − +i
3 3 3 3 2 2
Expansion of sin (nθ ) and cos (nθ ) in powers of sinθ & cosθ
= cos5 θ + 5C1 cos 4 θ ⋅ i sinθ + 5C2 cos3 θ (i sinθ ) 2 + 5C3 cos 2 θ (i sinθ ) 3 + 5C4 cosθ (i sinθ ) 4 + 5C5 (i sinθ ) 5
= cos 5 θ + 5i cos 4 θ sinθ − 10 cos 3 θ sin 2 θ − 10i cos 2 θ sin 3 θ + 5 cos θ sin 4 θ + i sin 5 θ
= cos5 θ − 10 cos3 θ sin 2 θ + 5 cosθ sin4 θ + i[5 cos 4 θ sinθ − 10 cos 2 θ sin3 θ + sin5 θ ]
MCA 11 - Mathematics SVT 23
= cos 6 θ + 6C1 cos5 θ (i sinθ ) + 6C2 cos 4 θ (i sinθ ) 2 + 6C3 cos3 θ (i sinθ ) 3 + 6C4 cos 2 θ (i sinθ ) 4
+ 6C5 cosθ (i sinθ ) 5 + 6C6 (i sinθ ) 6
= cos 6 θ + 6i cos 5 θ sinθ − 15 cos 4 θ sin 2 θ − 20i cos 3 θ sin 3 θ + 15 cos 2 θ sin 4 θ + 6i cos θ sin 5 θ − sin 6 θ
= cos6 θ − 15 cos 4 θ sin 2 θ + 15 cos 2 θ sin4 θ − sin6 θ + i[6 cos5 θ sinθ − 20 cos3 θ sin3 θ + 6 cosθ sin5 θ ]
Equating real and imaginary parts separately
ie cis(θ1 + θ 2 + θ 3 + ............ + θ n )
where S1 = ∑ tanθ 1
S 2 = ∑ tanθ1 tanθ 2
S1 − S 3 .................
∴ tan(è1 + è 2 + è3 + ............. + è n ) =
1 − S 2 + S 4 .................
24 KSOU Complex Trigonometry
Exercise
1. Find the continued product of fourth roots of unity.
2. If 'ω ' is the cube root of unity, then find the value of (4 − 3ω − 3ω 2 )3.
8
sin(π / 8) + i cos(π / 8)
3. Find the value of
sin(π / 8) − i cos(π / 8)
4. Write the conjugate of the multiplicative inverse of the complex number (sinθ + i cos θ ).
5. If z = x + iy then what does z + 1 = 1 represents?
11. If 'ω ' in the cube root of '1', find the value of (1 + ω ) 3 − (1 + ω 2 ) 3 .
2 4 2 4
12. Simplify + + + .
i i 2 i3 i 4
3π 3π
13. Express 2 cos + i sin in the Cartesian form.
4 4
−1
π π
14. Find the real part of 1 + cos + i sin .
5 5
a + ib a 2 + b2
15. If x + iy = , prove that ( x 2 + y 2 ) 2 = 2 .
c + id c + d2
n
1+ i
16. Find the least positive integer ' n' for which = 1.
1− i
z −1 π
19. If z = x + iy be such that amp = , show that x + y − 2 y = 1.
2 2
z + 1 4
20. Express 1 − i in the polar form.
3 −i
21. Express in the polar form.
2
2
2+i
22. Find the modulus and amplitude of .
3−i
(sinθ + i cosθ )3
23. Simplify
(cosα + i sinα ) 2
MCA 11 - Mathematics SVT 25
25. If ω is a cube root of unity, show that (1 − ω + ω 2 )(1 − ω 2 + ω 4 )(1 − ω 4 + ω 8 )(1 − ω 8 + ω 16 ) = 16.
1
26. If x = cis θ , then find x 5 + .
x5
27. If i 2 = −1, then find i 2 + i 4 + i 6 + L + (2n) terms.
x− y α − β
28. If x = cis α and y = cis β , prove that = i tan
x+ y 2
nπ
29. If α & β are the roots of x 2 − 2 x + 4 = 0, then prove that α n + β n = 2 n +1 cos .
3
30. Prove the following
i) (1 + i )3 + (1 − i )3 = −4 ( ) (
6
ii) 1 − i 2 + 1 + 2 ) = 46.
6
1 1
31. If Z = cos θ + i sinθ , show that Z n + n
= 2 cos nθ and Z n − = 2i sin nθ .
Z Zn
32. If x = cis α , y = cis β , prove that
1 1
i) xy + = 2 cos(α + β ) ii) xy − = 2i sin(α + β ).
xy xy
x 2n − 1
34. If x = cis θ , prove that = i tannθ .
x 2n + 1
35. Prove that (1 + cosθ + i sinθ ) 8 + (1 + cosθ − i sin θ )8 = 29 ⋅ cos(4θ ) ⋅ cos8 (θ / 2).
36. If cos α + 2 cos β + 3 cos γ = 0 = sinα + 2 sin β + 3 sin γ , prove that
i) cos 3α + 8 cos 3β + 27 cos 3γ = 18 cos(α + β + γ ) ii) sin 3α + 8 sin 3β + 27 sin 3γ = 18 sin(α + β + γ ).
1 1 1
37. If 2 cos α = x + , 2 cos β = y + and 2 cos γ = z + , prove that
x y z
1 1
i) xyz + = 2 cos(α + β + γ ) ii) xyz − = 2i sin(α + β + γ ).
xyz xyz
38. [
Show that (a + ib) 2n − (a − ib) 2n = 2i(a 2 + b 2 ) n ⋅ sin 2n ⋅ tan−1 (b / a) ]
n
1 + sinθ + i cos θ nπ nπ
39. Prove that = cos − nθ + i sin − nθ
1 + sinθ − i cos θ 2 2
7
1 − cos θ − i sinθ θ
40. Prove that = i tan7 ⋅ cis 7θ .
1 + cos θ − i sinθ 2
n
1 + i tanα 1 + i tanα
41. Prove that = .
1 − i tanα 1 − i tanα
26 KSOU Complex Trigonometry
46. Find the fourth roots of 1 + i 3 and represent them in the Argand diagram.
47. Solve x 7 − x = 0.
48. Solve x 5 + 1 = 0.
49. Express sin 7θ & cos 7θ in powers of sinθ & cos θ .
50. Express cos 8θ & sin 8θ in powers of cos θ & sinθ .
–––––––––––––
MATRIX THEORY
Examples
a1 b1
Matrix of order 3 × 2 is a2 b2
a b
3 3
a1 a 2 a3
b1 b2 b3
Matrix of order 4 × 3 is
c1 c2 c3
d d 2 d 3
1
a1 b1 c1
Matrix of order 3 × 3 is a 2 b2 c2
a3 b3 c3
Note :- Elements of Matrices are written in rows and columns with in the bracket ( ) or [ ].
Types of Matrices
(1) Equivalent Matrices : Two matrices are said to be equivalent if the order is the same.
(2) Equal Matrices : Two matrices are said to be equal if the corresponding elements are equal.
(3) Rectangular & Square Matrices : A matrix of order m × n is said to be rectangular if m ≠ n, square if m = n.
(4) Row Matrix : A matrix having only one row is called Row Matrix.
(5) Column Matrix : A matrix having only one column is called Column Matrix.
(6) Null Matrix or Zero Matrix : A matrix in which all the elements are zeros is called Null Matrix or Zero Matrix
denoted as O. [English alphabet O not zero where as elements are zeros]
(7) Diagonal Matrix : A diagonal matrix is a square matrix in which all elements except the elements in the principal
diagonal are zeros.
2 0 0
4 0
Example 0 1 0
0 6 0 0 4
are diagonal matrices of order 2 & 3.
(8) Scalar Matrix : A diagonal matrix in which all the elements in the principal diagonal are same.
28 KSOU Matrix Theory
8 0 0
4 0
Example 0 8 0
0 4 0 0 8
are Scalar Matrices of order 2 & 3.
(9) Unit Matrix or Identity Matrix : A diagonal matrix in which all the elements in the principal diagonal is 1 is
called Unit Matrix or Identity Matrix denoted by I.
1 0 0 0
1 0 0 1 0 0
Example : ,
0 1 0 0 1 0
0 0 0 1
are unit matrices of order 2 & 4.
(10) Transpose of a Matrix : If A is any matrix then the matrix obtained by interchanging the rows & columns of A is
called 'Transpose of A and it is written as A' or AT.
a b
a e
Example : If A = c d then A′ is
c
b f
e f d
Matrix addition
Two matrices can be added or subtracted if their orders are same.
a1 b1 c1 c d1 e1
Example : If A = & B = 1
a2 b2 c2 c2 d2 e2
a +c b1 + d1 c1 + e1
A + B = 1 1
a2 + c2 b2 + d 2 c2 + e2
a −c b1 − d1 c1 − e1
A − B = 1 1
a2 − c2 b2 − d 2 c2 − e2
Matrix Multiplication
If A is a matrix of order m × p and B is matrix of order p × n, then the product AB is defined and its order is m × n. (ie. for AB
to be defined number of columns of A must be same as number of rows of B)
α 1 β1
a b1 c1
Example : Let A = 1 & B = α 2 β2
a2 b2 c2 α β 3
3
which is of order 2 × 2.
Note :- In a skew symmetric matrix the elements in principal diagonal are all zeros.
2 3 5
Example : A = 3 7 6 is symmetric where A = A′
5 6 8
0 − 2 7
B= 2 0 6 is skew symmetric where B = − B′
− 7 − 6 0
Determinant
A determinant is defined as a mapping (function) from the set of square matrices to the set of real numbers.
If A is a square matrix its determinant is denoted as A .
a1 b1 c1 a1 b1 c1
Example : Let A = a2 b2
c2 then det. A or A = a2 b2 c2
a3 b3 c3 a3 b3 c3
a 22 a 23
Consider
a32 a33 which is a determinant formed by leaning all the elements of row and column in which all lies. This
determinant is called Minor of a11. Thus we can form nine minors. In general if A is matrix of order n × n then minor of aij is
30 KSOU Matrix Theory
obtained by leaning all the elements in the row and column in which aij lies in A . The order of this minor is n – 1where as the
order of given determinant is n if this minor is multiplied by (–1)i + j then it is called Co-factors of aij.
a 22 a 23
Minor of a11 =
a32 a33
a22 a23 a 22 a 23
Co - factor of a11 = ( −1)1+1 =
a32 a33 a32 a33
a12 a13
Minor of a 21 is
a32 a33
Value of a determinant
Consider a matrix A of order n × n. Consider all the elements of any row or column and multiply each element by its corresponding
co-factor. Then the algebraic sum of the product is the value of the determinant.
a1 b1
Example : Let A =
a2 b2
Co-factor of a1 is b2
Co-factor of b1 is –a2
∴ A = a1b2 − b1a2
a1 b1 c1
Let A = a2 b2 c2
a3 b3 c3
b2 c2 b2 c2
Co - factor of a1 is ( −1)1+1 =
b3 c3 b3 c3
a2 c2 a2 c2
Co - factor of b1 is ( −1)1+ 2 =−
a3 c3 a3 c3
a2 b2 a2 b2
Co - factor of c1 is ( −1)1+ 3 =
a3 b3 a3 b3
b2 c2 a2 c2 a2 b2
∴ A = a1 − b1 + c1
b3 c3 a3 c3 a3 b3
Properties of determinants
(1) If the elements of any two rows or columns are interchanged then value of the determinant changes only in sign.
(2) If the elements of two rows or columns are identical then the value of the determinant is zero.
(3) If all the elements of any row or column is multipled by a constant K, then the value of the determinant is multipled
by K.
(4) If all the elements of any row or column are written as sum of two elements then the determinant can be written as
sum of two determinants.
(5) If all the elements of any row or column are multiplied by a constant and added to the corresponding elements of
any other row or column then the value of the determinant donot alter.
Adjoint of a Matrix
a1 b1 c1
Let A = a2 b2 c2
a c3
3 b3
Let us denoted the co-factors of a1, b1, c1, a2, b2, c2, a3, b3, c3 as A1, B1, C1, A2, B2, C2, A3, B3, C3 transpose of matrix of
co-factors is called Adjoint of the Matrix.
A1 B1 C1
Matrix of Co - factors = A2 B2 C2
A C3
3 B3
A1 A2 A3
Adjoint of A = B1 B2 B3
C C C3
1 2
a1 b1 c1 A1 A2 A3
A. adj. A = a2 b2 c2 B1 B2 B3
a c3 C1 C 2 C3
3 b3
b2 c2 a2 c2 a2 b2
Now a1 A1 + b1 B1 + c1C1 = a1 − b1 + c1 =∆ The value of the det. A.
b3 c3 a3 c3 a3 b3
Similarly a 2 A2 + b2 B2 + c 2C 2 = ∆
a3 A3 + b3 B3 + c3C3 = ∆
b1 c1 a1 c1 a1 b1
a1 A2 + b1 B2 + c1C 2 = −a1 + b1 − c1
b3 c3 a3 c3 a3 b3
∆ 0 0
∴ A. adj. A = 0 ∆ 0 where ∆ = A
0 0 ∆
1 0 0
= ∆ 0 1 0 = ∆.I
0 0 1
∴ A. adj. A = A ⋅ I = adj. A
Inverse of a Matrix
Two non-singular matrices A & B of the same order is said to be inverse of each other if AB = I = BA. Inverse of A is denoted
as A–1. Inverse of B is denoted as B–1 and further (AB)–1 = B–1A–1.
A.adj. A = A ⋅ I
multiply by A −1 , AA −1.adj. A = A A −1
adj. A
ie adj. A = A A −1 ⇒ A −1 =
A
1 4 − 2
Example : Find the inverse of − 2 − 5 4
1 −2 1
1 4 − 2
Let A = − 2 − 5 4
1 −2 1
−5 4 −2 4 −2 −5
−
−2 1 1 1 1 −2
4 −2 1 −2 1 4
Matrix of Co - factors = − −
−2 1 1 1 1 −2
4 −2 1 −2 1 4
−
− 5 4 −2 4 − 2 − 5
(−5 + 8) − (−2 − 4) (4 + 5)
= − (4 − 4) (1 + 2) − (−2 − 4)
(16 − 10) − (4 − 4) (−5 + 8)
3 6 9
= 0 3 6
6 0 3
MCA 11 - Mathematics SVT 33
3 0 6
∴ adj.A = 6 3 0
9 6 3
1 4 −2
A = − 2 − 5 4 = 1( −5 + 8) − 4( −2 − 4) − 2( 4 + 5) = 3 + 24 − 18 = 9
1 −2 1
3 0 6 39 0 6
9
−1 1 1
A = adj.A = 6 3 0 = 6 9 3
9 0
A 9 9 6 3
9 6 3 9 9 9
13 0 2
3
−1
A = 23 1
3 0
2 1
1 3 3
Cramer's Rule
To solve the equations
a1 x + b1 y + c1 z = d1
a 2 x + b2 y + c2 z = d 2
a3 x + b3 y + c3 z = d 3
a1 b1 c1
Consider ∆ = a2 b2 c2
(1)
a3 b3 c3
first evaluate & if it is not zero then multiply both sides of (1) by x.
a1 b1 c1 a1 x b1 c1
∆x = x a2 b2 c 2 = a 2 x b2 c2
a3 b3 c3 a3 x b3 c3
multiply the elements of columns 2 & 3 by y & z and add to elements of column 1.
a1 x + b1 y + c1 z b1 c1
then ∆x = a2 x + b2 y + c2 z b2 c2
a3 x + b3 y + c3 z b3 c3
d1 b1 c1
= d2 b2 c2 = ∆1 (say) (2)
d3 b3 c3
a1 b1 c1 a1 b1 y c1
∆y = y a2 b2 c2 = a2 b2 y c2
a3 b3 c3 a3 b3 y c3
34 KSOU Matrix Theory
multiply the elements of columns 1 & 3 by x & z and add to the elements of column 2.
a1 a1 x + b1 y + c1 z c1 a1 d1 c1
= a2 a2 x + b2 y + c2 z c2 = a2 d2 c2 = ∆ 2 (say) (3)
a3 a3 x + b3 y + c3 z c3 a3 d3 c3
multiply both sides of (1) by z
a1 b1 c1 a1 b1 c1 z
∆z = z a2 b2 c2 = a2 b2 c2 z
a3 b3 c 3 a3 b3 c3 z
multiply the elements of columns 1 & 2 by x & y and add to the elements of column 3.
a1 b1 a1 x + b1 y + c1 z
= a2 b2 a 2 x + b2 y + c2 z
a3 b3 a3 x + b3 y + c3 z
a1 b1 d1
= a2 b2 d 2 = ∆ 3 (say) (4)
a3 b3 d3
∆1
then x = from (2)
∆
∆2
y= from (3)
∆
∆3
z= from (4)
∆
Note :- Verification of values of x, y, z can be done by substituting in the given equations.
Example - 1
Solve 2 x + y − z = 3
x + y + z =1
x − 2 y − 3y = 4
2 1 -1
Let ∆ = 1 1 1
(1)
1 −2 −3
= 2( −3 + 2) − 1( −3 − 1) − 1( −2 − 1) = −2 + 4 + 3 = 5
3 1 -1
=1 1 1 = 3( −3 + 2) − 1( −3 − 4) − 1( −2 − 4) = −3 + 7 + 6 = 10
4 −2 −3
10 10
∴ x= = =2
∆ 5
multiply both sides of (1) by y
2 3 -1 2 y -1
∆y = y 1 1 1 == 1 y 1
1 4 −3 1 − 2y − 3
multiply the elements of column 1 by x & 3 by z and to the corresponding elements of column 2.
2 2 x + y − z -1
then ∆y = 1 x + y + z 1
1 x − 2 y − 3z − 3
2 3 -1
= 1 1 1 = 2( −3 − 4) − 3( −3 − 1) − 1( 4 − 1) = −14 + 12 − 3 = −5
1 4 −3
−5 −5
∴ y= = = −1
∆ 5
multiply both sides of (1) by z
2 3 -1 2 3 -z
∆z = z 1 1 1 = 1 1 z
1 4 − 3 1 4 − 3z
multiply the elements of column 1 by x & column 2 by y and to the corresponding elements of column 3.
2 1 2x + y − z
then ∆z = 1 1 x+ y+z
1 − 2 x − 2 y − 3z
2 1 3
= 1 1 1 = 2( 4 + 2) − 1( 4 − 1) + 3( −2 − 1) = 12 − 3 − 9 = 0
1 −2 4
0 0
∴ z= = =0
∆ 5
Thus solution is x = 2, y = –1 & z = 0 which can be verified by substituting in the given equations.
Example - 2
Solve 4 x + y = 7
3 y + 4z = 5
5x + 3z = 2
4 1 0
∆ = 5 3 4 = 4(9 − 0) − 1(0 − 20) + 0(0 − 15) = 36 + 20 = 56
2 0 3
36 KSOU Matrix Theory
7 1 0
∆1 = 5 3 4 = 7(9 − 0) − 1(15 − 8) + 0(0 − 6) = 63 − 7 = 56
2 0 3
4 7 0
∆ 2 = 0 5 4 = 4(15 − 8) − 7(0 − 20) + 0(0 − 25) = 28 + 140 = 168
5 2 3
4 1 7
∆ 3 = 0 3 5 = 4(6 − 0) − 1(0 − 25) + 7(0 − 15) = 24 + 25 − 105 = −56
5 0 2
∆1 56
∴ x= = =1
∆ 56
∆ 2 168
y= = =3
∆ 56
∆ 3 − 56
z= = = −1
∆ 56
a1 b1 c1 x d1
Consider A = a 2 b2 c2 X = y & B = d2
a c3 z d
3 b3 3
then given equations can be written in Matrix form as AX = B. If A ≠ 0 solution exists multiply both sides by A–1
A−1 ( AX ) = A−1 B
A −1 AX = A −1 B
ie IX = A −1 B
∴ X = A−1B
Example
Solve 3 x − y + 2 z = 13
2x + y − z = 3
x + 3 y − 5z = −8
3 −1 2 x 13
Let A = 2 1 − 1 X = y , B = 3
1 3 − 5 z − 8
then given equations can be written as AX = B
MCA 11 - Mathematics SVT 37
∴ X = A−1B (1)
To find A–1
3 −1 2
A = 2 1 − 1 = 3( −5 + 3) + 1( −10 + 1) + 2(6 − 1) = −6 − 9 + 10 = −5
1 3 −5
1 −1 2 −1 2 1
−
3 −5 1 −5 1 3
−1 2 3 2 3 −1
Matrix of Co - factors = − −
3 −5 1 −5 1 3
−1 2 3 2 3 −1
−
1 − 1 2 −1 2 1
(−5 + 3) − (−10 + 1) (6 − 1) − 2 9 5
= − (5 − 6) (−15 − 2) − (9 + 1) = 1 − 17 − 10
1 − 2 − (−3 − 4) (3 + 2) − 1 − 7 5
− 2 1 − 1 − 2 1 − 1
1 1
adj.A = 9 − 17 7 ∴ A −1 = adj. A = − 9 − 17 7
5 − 10 5 A 5
5 − 10 5
Using this in (1)
− 2 1 − 1 13 − 26 3 8 − 15 3
1 1 1
X = − 9 − 17 7 3 = − 117 − 51 − 56 = − 10 = − 2
5 5 5
5 − 10 5 − 8 65 − 30 − 40 −5 1
∴ x = 3, y = –2, z = 1 is the solution
Example - 1
1 4
Find the eigen roots and eigen vectors of the matrix
2 3
1 4 1 4 1 0
Let A = Characteristic equation is 2 3 − λ 0 1 = 0
2 3
1− λ 4
ie =0 ⇒ (1 − λ )(3 − λ ) − 8 = 0
4 3− λ
⇒ 3 − 3λ − λ + λ2 − 8 = 0 ⇒ λ2 − 4λ − 5 = 0
38 KSOU Matrix Theory
⇒ (λ − 5)(λ + 1) = 0 ⇒ λ = −1, 5
x + 4y − x
⇒ =
2x + 3 y − y
x + 4 y = −x ix = −4 y x y
⇒ ⇒ ⇒∴ =
2x + 3y = − y ie x = −2 y −2 1
⇒ x1 + 4 x2 = 5 x1 x1 x2
ie x1 = x2 ie =
2 x1 + 3 x2 = 5 x2 1 1
Example - 2
6 −2 2
Find the eigen roots and eigen vectors of the matrix − 2 3 − 1
2 −1 3
6 −2 2 6−λ −2 2
Let A = − 2 3 − 1 Characteristic equation is − 2 3 − λ − 1 = 0
2 −1 3 −1 3 − λ
2
[ ]
ie (6 − λ ) (3 − λ ) 2 − 1 + 2[− 2(3 − λ ) + 2]+ 2[2 − 2(3 − λ )] = 0
ie (6 − λ )[9 + λ2
]
− 6λ − 1 + 2[− 6 + 2λ + 2]+ 2[2 − 6 + 2λ )] = 0
6 − 2 2 x1 x1 x1
ie − 2 3 − 1 x2 = 2 x2 where X = x2
2 − 1 3 x x x
3 3 3
⇒ 6 x1 − 2 x2 + 2 x3 = 2 x1 ⇒ 4 x1 − 2 x2 + 2 x3 = 0
−2 x1 + 3 x2 − x3 = 2 x2 −2 x1 + x2 − x3 = 0
2 x1 − x2 + 3 x3 = 2 x3 2 x1 − x2 + x3 = 0
a1 − λ b1 c1
then characteristic equation is a 2 b2 − λ c 2 =0
a3 b3 c3 − λ
Then as per Cayley Hamilton Theorem − A3 + a1 A 2 + a2 A + a3 I = 0 where I is a unit matrix of order 3 & 0 is a null
matrix of order 3.
In general if A is a square matrix of order n then characteristic equation will be of the form
(−1) n A n + aA n −1 + a2 A n − z + ........ + an I = 0
where I is a unit matrix of order n & 0 is a null matrix of order n.
∴ If an = 0, matrix A is singular & a n ≠ 0 the matrix A is non-singular & hence inverse exists and we can find the
inverse of A using Cayley Hamilton Theorem.
Example - 1
a b
Let A =
c d
a −λ b
Characteristic equation is =0
c d −λ
ie λ2 + a1λ + a2 = 0 where a1, a2 are constants.
A2 + a1 A + a2 I = 0
multiply both sides by A–1
ie A + a1 I + a2 A−1 = 0
∴ a2 A−1 = −( A + a1I )
1
∴ A −1 = − ( A + a1 I )
a2
MCA 11 - Mathematics SVT 41
Example - 2
The characteristic equation of a matrix A of order 2 is λ2 − 5λ + 10 = 0 find A .
Example - 3
2 − 1
Find the inverse of − 3 4 using Cayley Hamilton Theorem.
2 − 1 2−λ −1
Solution : Let A = C.E. is =0
−3 4 −3 4−λ
ie ( 2 − λ )(4 − λ ) − 3 = 0 ie 8 − 4λ − 2λ + λ2 − 3 = 0
ie λ2 − 6λ + 5 = 0
A 2 − 6 A + 5I = 0
multiply both sides by A–1
A − 6 I + 5 A −1 = 0
∴ 5 A−1 = − A + 6I
2 − 1 1 0 − 2 + 6 1 + 0 4 1
= − + 6 = =
− 3 4 0 1 3 + 0 − 4 + 6 3 2
1 4 1
∴ A −1 =
5 3 2
Diagonalisation of Matrices
If A is a square matrix of order n where all the eigen values are linearly independent then a matrix P can be found such that
P–1AP is a Diagonal Matrix.
Let A be a square matrix of order 3 and let λ1, λ2, λ3 be the eigen values, corresponding to these. Let X1, X2, X3 be three
vectors where
x1 y1 z1
X 1 = x2 , X 2 = y 2 , X 3 = z 2
x3 y3 z3
x1 y1 z1 λ1 0 0
Let P = x2 y2 z 2 Then P AP = 0 λ2
−1
0
x3 y3 z3 0 0 λ3
Example - 1
1 − 2
Let A =
− 5 4
1− λ −2
C.E. is = 0 ⇒ (1 − λ )( 4 − λ ) − 10 = 0
−5 4−λ
42 KSOU Matrix Theory
1 − 2 x1 6 x1
ie =
− 5 4 x2 6 x2
x1 − 2 x2 = − x1
⇒ 5 x1 = −2 x2
− 5 x1 + 4 x2 = − x2
x1 x − 2
ie = 2 ∴ eigen vector is
−2 5 5
1 − 2 1 5 2
Let P = Then P −1 =
1 5 7 − 1 1
1 5 2 1 − 2 1 − 2
∴ P −1 AP =
7 − 1 1 − 5 4 1 5
1 5 − 10 − 10 + 8 1 − 2 1 − 5 − 2 1 − 2
= − 1 − 5 2 + 4 1 5 = − 6 6 1 5
7 7
1 − 5 − 2 10 − 10 1 − 7 0 − 1 0
= = =
7 − 6 + 6 12 + 30 7 0 42 0 6
1 − 2 1 − 2
Thus P = diagonalize the matrix
1 5 − 5 4
Example - 2
1 1 3
Let A = 1 5 1
3 1 1
1− λ 1 3
Characteristic equation is 1 5−λ 1 =0
3 1 1− λ
x1
where X 1 = x2
x3
1 1 3 x1 − 2 x1
ie 1 5 1 x2 = − 2 x2
3 1 1 x − 2 x
3 3
ie x1 + x2 + 3 x3 = −2 x1 ie 3 x1 + x2 + 3 x3 = 0 (1)
x1 + 5 x2 + x3 = −2 x2 x1 + 7 x2 + x3 = 0 (2)
3 x1 + x2 + x3 = −2 x3 3 x1 + x2 + 3 x3 = 0 (3)
(1) & (3) are same.
Put x2 = 0 in (1) or (2), then x1 + x3 = 0
− 1
∴ eigen vector X 1 = 0
x x
ie x1 = − x3 ⇒ 1 = 3
−1 1
1
Let X2 be the eigen vector for λ = 3.
ie AX 2 = 3 X 2
1 1 3 y1 3 y1 y1
ie 1 5 1 y 2 = 3 y 2 where X 2 = y 2
3 1 1 y 3 y y3
3 3
ie y1 + y 2 + 3 y3 = 3 y1 ie −2 y1 + y 2 + 3 y3 = 0 (1)
y1 + 5 y 2 + 3 y3 = 3 y 2 y1 + 2 y 2 + y3 = 0 (2)
3 y1 + y 2 + 3 y3 = 3 y3 3 y1 + y 2 − 2 y3 = 0 (3)
y1 y3
⇒ 5 y1 = 5 y3 ⇒ y1 = y3 ∴ = (5)
1 1
44 KSOU Matrix Theory
1
∴ X 2 = − 1
y y y
From (4) & (5) 1 = 2 = 3
1 −1 1
1
Next, let X3 be the eigen vector for λ = 6
ie AX 3 = 6 X 3
1 1 3 z1 6 z1 z1
ie 1 5 1 z 2 = 6 z 2 where X 3 = z 2
3 1 1 z 6 z z3
3 3
ie z1 + z 2 + 3 z 3 = 6 z1 ie −5 z1 + z 2 + 3 z 3 = 0 (1)
z1 + 5 z 2 + z 3 = 6 z 2 z1 − z 2 + z 3 = 0 (2)
3 z1 + z 2 + z 3 = 6 z 3 3 z1 + z 2 − 5 z 3 = 0 (3)
z1 z 2
ie 2 z1 = z 2 ⇒ = (5)
1 2
1
∴ X 3 = 2
z z z
From (4) & (5), 1 = 2 = 3
1 2 1
1
− 1 1 1 − 2 0 0
Let P = 0 − 1 2 Then P AP = 0 3 0
−1
1 1 1 0 0 6
− 1 1 1 1 1 3
∴ P = 0 − 1 2 diagonalize 1 5 1
1 1 1 3 1 1
Exercise
1 2 3
1. Evaluate − 1 2 3 .
2 3 1
a −b b−c c−a
2. Evaluate b − c c − a a − b .
c −a a−b b−c
MCA 11 - Mathematics SVT 45
1 2 3
3. Evaluate 2 3 4 .
3 4 4
2 0 4
4. If 6 x 5 = 0, then find x.
−1 − 3 1
1 p −q
5. Evaluate − p 1 r .
q −r 1
2 0
6. Find the adjoint of .
2 3
1 2 3
.
7 If A = 2 3 4 find the co - factor of 1.
3 4 2
2 1 5
8. If A = find AA′ & A′A.
0 3 7
secθ tanθ
9. Find the inverse of .
tanθ secθ
2 3 4 1 − 2 3
10. If A = and B = find 5 A − 3B and 6 B − 7 A.
−1 0 5 0 4 2
4 − 2 5 2 0 4
11. If 3 A + B = and 2 B + A = find A and B.
3 7 6 −1 2 3
1 + x x x y − 4 5
12. If + = find x and y.
5 x 7 − y 12 4 x
5 6 7
13. If A = verify that (A ′)′ = A.
8 9 10
2 4
− 3 4 5
14. If A = and B = − 4 3 find A + B′ and A′ − B.
6 − 2 1 3 − 2
3 4
15. If A = prove that A 2 − 10 A + I = 0.
5 7
6 5
16. Find the inverse of .
3 2
1 2
17. Find the characteristic equation of .
0 4
46 KSOU Matrix Theory
2 3
18. Find the eigen values of .
0 4
19. Solve 2 x + z = −1, 2 y + x = 5, z − y = −2 by Cramer' s Rule.
20. Solve 5 x − y + 4 z = 5
2x + 3 y + 5z = 2
7 x − 2 y + 6z = 5
by matrix method.
2 − 1
21. Find the characteristic roots of .
0 1
1 0 − 1
22. Find the characteristic roots of 1 2 1 .
2 2 3
1 2
23. Verify Cayley - Hamilton Theorem for the matrix .
3 4
2 0
24. Verify Cayley - Hamilton Theorem for the matrix .
1 − 1
1 2
25. Find the eigen vectors for the matrix .
2 1
6 −2 2
26. Find the eigen values and eigen vectors for the matrix − 2 3 − 1.
2 −1 3
BCA 21 / IMCA 21 / Mathematics SVT 21
ALGEBRAIC STRUCTURES
Abreviations used
N : represent set of natural numbers.
Z or I : represent set of +ve and –ve integers including zero.
Z+ : represent non-negative integers ie. +ve integers including zero.
Q : represent set of rational numbers.
R : represent set of real numbers.
C : represent set of complex numbers.
Zn = {0, 1, 2, 3, .............. n – 1} ie. Zn represent set of integers modulo n.
Q+ : represent set of +ve rational numbers.
z - {o} : represent set of integers except 0.
Q - {o} : represent set of rational numbers except zero.
R - {o} : set of real numbers except zero.
A set in general is denoted by S.
∀ : for all
∈ : belongs to
Binary Operation
If S is a non-empty set then a mapping (function) from S × S to S is defined as Binary Operation (in short B.O.) and denoted
by ∗ (read as star). ie. : S × S → S (Star maps S cross S to S)
Another Definition
If S is non-empty set then ∗ (star) is said to be a Binary operation if ∀ a, b ∈ S, a ∗ b ∈ S.
Examples
(1) on N + and × (ie addition & multiplication) are B.O.
2 + 3 = 5∈ N 2×3 = 6∈ N
4 + 5 ∈ Z , 3 − 4 = 1∈ Z , 4 − 3 = 1 ∈ Z , 5 × 6 = 30 ∈ Z
a
(3) On Q & R +, – & × are B.O. but ÷ is not a B.O. on Q & R Q for 0, ∉ Q & R but on Q - {o} & R - {o} ÷ is a B.O.
0
(4) on C, + and × are B.O.
( x1 + iy1 ) + ( x2 + iy 2 ) = ( x1 + x2 ) + i ( y1 + y 2 ) ∈ C
( x1 + iy1 ) + ( x2 + iy 2 ) = ( x1 x2 − y1 y 2 ) + i ( x1 y 2 + x2 y1 ) ∈ C
48 KSOU Algebraic Structures
Definitions
(1) A non-empty set S with one or more binary operations is called an 'Algebraic Structure'.
(N, +), (Z, +, ×), (Q, +, ×) are all algebraic structures.
Examples
(i) + and × (addition and multiplication) are associative and commutative on N, Z, Q & R.
(iii) 1 is an identity for B.O. × on N but + has no identity on N. Where as O is an identity on Z, Q and R for the B.O. +.
1 0
(iv) If S is a set of 2 × 2 matrices and B.O. is matrix multiplication then I = is an identity element.
0 1
Group
A non-empty set G together with a B.O. ∗ ie (G, ∗) is said to form a group if the following axioms are satisfied.
G4. Inverse : ∀ a ∈ G, there exists an element b such that a ∗ b = e = b ∗ a. This b is called inverse of a and usually denoted
as a–1
ie. a ∗ a–1 = e = a–1 ∗ a.
In addition to the above four axions if ∀ a, b ∈ G, a ∗ b = b ∗ a. Then (G, ∗) is called an 'abelian group' or 'commutative
group'.
Examples
(i) (N, +) is a groupoid and semigroup.
Note :- Every group is a monoid but the converse is not true, (Z, +) is a group and also a monoid but (N, ×) is a monoid but
not a group.
MCA 11 - Mathematics SVT 49
Properties of Groups
1. Cancellation laws are valid in a group
ie if (G , ∗) is a group then ∀ a, b, c ∈ G ,
(i) a ∗ b = a ∗ c ⇒ b = c (left cancellation law)
(ii ) b ∗ a = c ∗ a ⇒ b = c ( right cancellation law)
Proof :- a ∗ b = a ∗ c, as a ∈ G, a −1 ∈ G
∴ a −1 ∗ (a ∗ b) = a −1 ∗ (a ∗ c)
ie (a −1 ∗ a) ∗ b = (a −1 ∗ a) ∗ c
ie e ∗ b = e ∗ c where e is the identity.
⇒b=c
Similarly by considering
(b ∗ a) ∗ a −1 = (c ∗ a) ∗ a −1
we get b = c
a −1 ∗ (a ∗ x) = a −1 ∗ b
ie (a −1 ∗ a) ∗ x = a −1 ∗ b
ie e ∗ x = a −1 ∗ b
∴ x = a −1 ∗ b
ie (a −1 ∗ a ) ∗ x1 = (a −1 ∗ a) ∗ x2
ie e ∗ x1 = e ∗ x2
⇒ x1 = x2
∴ solution is unique.
Proof :- To prove identity is unique. If possible let e1 & e2 are two identities then
∀a ∈ G, a ∗ e1 = a = e1 ∗ a (1)
& a ∗ e2 = a = e2 ∗ a (2)
From LHS of (1), a ∗ e1 = a = e2 ∗ a (using (2))
ie a ∗ e1 = e2 ∗ a = a ∗ e2 (using LHS of (2))
50 KSOU Algebraic Structures
4. ( )
In a group G, a −1
−1
= a ∀a ∈ G
We have a ∗ a −1 = e = a −1 ∗ a
it can be easily seen from above relation that inverse of a–1 is a ie a −1( ) −1
=a.
Note :- If b & c are elements of G, such that b ∗ c = e = c ∗ b then each is the inverse of the other.
5. In a group (G , ∗)
∀a, b ∈ G, (a ∗ b) −1 = b −1 ∗ a −1
Proof :- Consider (a ∗ b) ∗ (b −1 ∗ a −1 )
= a ∗ (b ∗ b −1 ) ∗ a −1 = a ∗ e ∗ a −1 = a ∗ a −1 = e
∴ (a ∗ b) −1 = b −1 ∗ a −1
Proof :- Order of group means the number of elements in a group. If a group G has n elements. The order of G is n, which
is denoted as O(G) = n.
If n is finite it is called finite group and n is Infinite then it is called Infinite group.
Subgroups
A non-empty subset H of a group G is said to form a subgroup with respect to the same binary operation ∗ if ( H , ∗) is a
group.
Eg. (1) (z, +) is a subgroup of (Q, +)
(2) H = { 0, 2, 4} is a subgroup of G = { 0, 1, 2, 3, 4, 5} with B.O. + mod 6 ie ⊕ 6
(3) H = { −1, 1} is a subgroup of G = {1, − 1, i, − i} with respect to the B.O. multiplication.
Theorem
A non-empty subset H of a group G is a subgroup of G if and only if ∀a, b ∈ H , ab −1 ∈ H .
Permutation group
Let S = { a1 , a 2 , a3 , .......... a n }
Then a one-one and onto mapping or function from S onto itself is called a Permutation.
a1 a2 a3 .......... a n
Permutation is denoted as
f ( a1 ) f (a 2 ) f ( a3 ) .......... f ( a n )
There will be n ! ie ∠n permutations the set of permutations is denoted by Sn.
Let f , g ∈ S n . There is a composite mapping for f & g denoted as f o g , this can be taken as binary operation. Then the
set Sn with binary operation 'O' (ie composite mapping) will form a group. For convenience f o g is denoted as gf.
1 2 3 4 1 2 3 4
Let f = & g = ∈ S 4 the B.O. composite function is given by
3 1 4 2 2 3 4 1
1 2 3 4 1 2 3 4 1 2 3 4
f o g = o =
3 1 4 2 2 3 4 1 ? ? ? ?
to fill up the second row, following is the procedure.
in g : g (1) = 2, g ( 2) = 3, g (3) = 4, g ( 4) = 1
& in f : f (1) = 3, f ( 2) = 1, f (3) = 4, f ( 4) = 2
Now f o g (1) = f [ g (1)] = f ( 2) = 1
f o g ( 2) = f [ g ( 2)] = f (3) = 4
f o g (3) = f [ g (3)] = f ( 4) = 2
f o g ( 4) = f [ g (4)] = f (1) = 3
52 KSOU Algebraic Structures
1 2 3 4
∴ f o g =
1 4 2 3
for convenience f o g is written as gf
1 2 3 4 1 2 3 4 1 2 3 4
ie f o g = o =
3 1 4 2 2 3 4 1 1 4 2 3
1 2 3 4 1 2 3 4 1 2 3 4
& gf = =
2 3 4 1 3 1 4 2 1 4 2 3
the composite function is also called product function.
1 2 1 2 1 2
then =
1 2 2 1 2 1
1 2 1 2 1 2
=
1 2 1 2 1 2
1 2 1 2 1 2
=
2 1 2 1 1 2
1 2
closure law is satisfied, associative law can be easily verified. inverse e =
1 2
−1
1 2 1 2
=
1 2 1 2
−1
1 2 1 2
=
2 1 2 1
∴ S2 forms a group.
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
S 3 =
1 2 3 1 3 2 3 2 1 2 3 1 3 1 2 3 2 1
Let us denote the elements as f1 , f 2 , f 3 , f 4 , f 5 & f 6 respectively.
ie S 3 = {f1 , f 2 , f 3 , f 4 , f 5 , f 6 }
identity e = f1
Q f 3 f 4 = f 6 but f 4 f 3 = f 2 ∴ f 3 f 4 ≠ f 4 f 3 .
Examples
(1) Show that the set R - {o} with B.O. × forms a group.
Solution : For any elements a, b ∈ R - {o}. a ∗ b ∈ R - {o}
2, 3 ∈ R - {o}, 2 × 3 = 6 ∈ R - {o}
∴ closure law is satisfied.
For any three elements a, b, c ∈ R - {o}
(a × b)× c = a × (b × c)
(–3 × 4) × 5 = –12 × 5 = –60.
–3 × (4 × 5) = –3 × 20 = –60.
∴ associative law is satisfied.
Identity element is 1,
ie. ∀ a ∈ R - {o}, a × 1 = a = 1 × a.
Let a ∈ R - {o} then there exists
1 1 1
∈ R - { o} such that a × = 1 = × a
a a a
∴ inverse exists for all elements R - {o}.
∴ (R - {o}, ×) forms an abelian group.
⊕5 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
54 KSOU Algebraic Structures
From the above table it can be easily seen that closure law is satisfied.
2 ⊕ 5 (3 ⊕ 5 4) = 2 ⊕ 5 2 = 4
( 2 ⊕ 5 3) ⊕ 5 4 = 0 ⊕ 5 4 = 4
ie 2 ⊕ 5 (3 ⊕ 5 4) = ( 2 ⊕ 5 3) ⊕ 5 4
a ⊕5 b = b ⊕5 a
∴ ( Z 5 , ⊕ 5 ) is an abelian group.
(3) Show that G = {1, 2, 3, 4} with B.O. multiplication mod 5 ie ⊗ 5 is an abelian group.
⊗5 1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1
From the above table it can be seen that closure law is satisfied.
( 2 ⊗5 3) ⊗5 4 = 1 ⊗5 4 = 4
2 ⊗5 (3 ⊗5 4) = 2 ⊗5 2 = 4
∴ associate law is satisfied.
identity is 1.
inverse of 1 is 1
inverse of 2 is 3
inverse of 3 is 2
inverse of 4 is 4
∴ (G , ⊗5 ) forms a group and it can be seen from the table that it forms an abelian group.
y ∗ x = 1 + yx ∴ x∗ y = y∗ x
Associative property ( x ∗ y ) ∗ z = x ∗ ( y ∗ z )
LHS = (1 + xy ) ∗ z = P ∗ z = 1 + pz = 1 + (1 + xy ) z = 1 + z + xyz
RHS = x ∗ (1 + yz ) = x ∗ Q = 1 + xQ = 1 + x(1 + yz ) = 1 + x + xyz
∴ LHS ≠ RHS ∴ ∗ is not a associative
ab
6. In a group (G , ∗), a ∗ b = . Find the identity element, inverse of 4 and solve 4 ∗ x = 5
2
To find e : a ∗ e = a = e ∗ a
ae
a∗e = = a ∴ e = 2 i.e. identity element is 2.
2
a ∗ a −1 = e = a −1 ∗ a
aa −1 4
a ∗ a −1 = 2 ⇒ = 2 ∴ a −1 =
2 a
4
4 −1 = = 1. ∴ inverse of 4 is 1.
4
4x 10 5
4∗ x = 5 ⇒ = 5 ∴x = =
2 4 2
= cos{(α + β ) + γ } + i sin{(α + β ) + γ }
= cos{α + ( β + γ )} + i sin{α + ( β + γ )}
= x (yz) Q multiplication is associative on R ∴ Associative axiom is satisfied.
iii) 1 = cos 0 + i sin 0 ∈ G is the identity element
56 KSOU Algebraic Structures
iv) (cos θ + i sinθ )(cosθ − i sinθ ) = 1 ⇒ cos θ − i sinθ is the multiplicative inverse of cos θ + i sinθ .
8. Show that the cube roots of unity form an abelian group under multiplication
We know that the cube roots of unity are 1, ω and ω 2 . Let G = {1, ω , ω 2 }
. 1 ω ω2
1 1 ω ω2 here ω 3 = 1
ω ω ω2 1 & ω 4 = ω 3 ⋅ ω = ω.
ω2 ω2 1 ω
i) All the entries in the table are the same as the elements of the set. This means the closure law is satisfied.
(1⋅ ω ) ⋅ ω 2 = ω ⋅ ω 2 = 1
1 0 −1 0 1 0 −1 0
9. Show that the four matrices , , and form an abelian group under matrix
0 1 0 1 0 − 1 0 − 1
multiplication.
1 0 −1 0 1 0 −1 0
Take = I , = A, = B, = C ; G = {1, A, B, C}
0 1 0 1 0 − 1 0 − 1
IA = AI = A, IB = BI = B, IC = CI = C
−1 0 1 0 −1 + 0 0 + 0 −1 0
AB = = = = C
0 1 0 − 1 0 + 0 0 − 1 0 − 1
Similarly, it can be shown that BA = C , AC = CA = B, BC = CB = A, A ⋅ A = I etc.
MCA 11 - Mathematics SVT 57
i) The entries in the table are the same as the elements of the set G. ∴ Closure law is satisfied.
ii) A( BC ) = A( A) = I ; ( AB)C = (C )C = I ∴ Associative law is satisfied.
iii) I is the identity element.
iv) Inverses of I , A, B are respectively I , A, B, C. ∴ G is a group under matrix multiplication
∴ (G , ⋅) is a group.
v) Since the entries on either side of the leading diagonal are symmetric, (G, ⋅) is an abelian group.
10. If every element of a group G has its own inverse, show that G is abelian
Given a −1 = a, ∀ a, b ∈ G (1)
11. In a group (G, ⋅) if (ab) 2 = a 2b 2 , ∀ a, b ∈ G. Prove that (G, ⋅) is abelian and conversely.
12. { }
Given Q0, the set of non zero rational numbers is a multiplicative group and H = 2 n n ∈ Z , show that H is a subgroup
of Q0 under multiplication.
{ } {
H = 2 n n ∈ Z = ...2 −2 , 2 −1 , 2 0, 21..... }
i) 2 m , 2 n ∈ H , 2 m ⋅ 2 n = 2 m + n ∈ H ∴ closure law is satisfied
ii) (2 m ⋅ 2 n )2 r = 2 m ⋅ (2 n ⋅ 2 r ), ∀ m, n, r ∈ z ie (2 m+ n )2 r = 2 m (2 n + r )
Exercise
1. If N = {1, 2, 3, ....}, which of the following are binary operation of N.
a
(1) a ∗ b = a + 2b ( 2) a ∗ b = 3a − 4b (5) a ∗ b =
b
2. Which of the following operations on the given set are binary
(1) on I, the set of integers, a ∗ b = 3a − 4b
(2) on R, a * b = a 2 − b 2
(3) on R, a ∗ b = ab
6. (
In a group G, ∀ a, b ∈ G, find a −1b −1 )−1
.
7. If the binary operation ∗ on the set Z is defined by a ∗ b = a + b + 5, find the identity element.
8. In the group of non zero integers mod 5. Find the multiplicative inverse of 4.
1 2 3 1 2 3
9. If f = and g = are permutations in S 3 , find f o g .
1 2 3 2 3 1
10. If S = {1, 2, 3, 4, 5, 6} w.r.t. multiplication (mod 7), solve the equation 3x = 5 in S.
11. Show that S = {1, 2, 3} under multiplication (mod 4) is not a group.
1 2 3 4 1 2 3 4
12. If f = and g = find gf .
3 4 1 2 2 3 1 4
ab
13. The binary operation ∗ is defined by a ∗ b = , on set of rational numbers, show that ∗ is associative.
7
ab
14. If ∗ is defined by a ∗ b = , on the set of real numbers, show that ∗ is both commutative and associative.
2
17. On the set of real numbers, R, ∗ is defined by a ∗ b = 2a − 3b + ab, examine whether ∗ is commutative and associative.
18. In the set of rationals except 1, binary operation ∗ is defined by a ∗ b = a + b − ab. Find the identity and inverse of 2.
ab
19. On the set of positive rational numbers Q + , a ∗ b = , ∀ a, b ∈ Q + . Find the identity element and the inverse of 8.
4
20. In a group of integers, an operation ∗ is defined by a ∗ b = a + b − 1. Find the identity element.
MCA 11 - Mathematics SVT 59
Scalar : A physical quantity which has only magnitude and no direction is called a 'Scalar'. A
Eg. speed, volume, mass, density, temperature etc.
Vectors are represented by directed line segments. Let AB be the line segment. Vector from A to B is denoted by AB and
vector from B to A is denoted by BA. AB can also be represented by a. The length of AB is magnitude of the vector denoted
as AB or a or simply a. For AB, A is the initial point and B the terminal point.
Co-initial vectors : Vectors having the same initial point are called co-initial vectors.
Coplanar vectors : Vectors in the same plane are called 'coplanar vectors'.
Let OA = a & OB = b
Complete the parallelogram OACB.
then OA + AC = OC by triangle law
B C
but AC = OB (parallel vectors)
∴ OC = OA + OB = a + b
b
Note : - AB = OB − OA
& BA = OA − OB
O A
a
60 KSOU Algebraic Structures
Properties
(ii) ( )( )
Vector addition is associative ie a + b + c = a + b + c.
(iii) Set of vectors V, with binary operation vector addition will form a 'Group'. The identity being 0 (null vector) and
inverse of a is − a.
Position vectors
Y
(i) Let P be a point in a plane where O is the origin and OX & OY are co-
ordinate axes. OP is called position vectors of P.
OP = x 2 + y 2
Note :- A plane vector is an ordered pair of real numbers and the distance between O & P is the magnitude of OP.
(ii) Let P be a point in three dimensional space where OX, OY & OZ are co-ordinate axes. Let (x, y, z) be the co-ordinates of
P. Draw PQ ⊥ r to the plane XOY & QA & QB parallel to OY & OX respectives to meet OX at A & OY at B.
j B
OP = OQ + QP (by triang le law) O Y
i
= xi + yj + zk
A
OP = x 2 + y 2 + z 2 Q
= ( x2 − x1 )i + ( y 2 − y1 ) j + ( z 2 − z1 ) k
O Y
and unit vector in the direction of
( x2 − x1 )i + ( y 2 − y1 ) j + ( z 2 − z1 ) k
PQ is
( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 + ( z 2 − z1 ) 2
If a = a1i + a 2 j + a3 k and b = b1i + b2 j + b3 k are two non-zero vectors, then a1b1 + a 2 b2 + a3b3 is defined as 'scalar product'
of two vectors a & b, denoted as a ⋅ b also known as 'dot product'.
i j k
ie a1 a2 a3 is defined as ' vector product' of two vector a & b, denoted as a × b also known as cross product.
b1 b2 b3
i j k
ie a × b = a1 a2 a3 = ∑ (a 2b3 − a3b2 ) i
b1 b2 b3
∴ a × b or − b × a is a null vector, hence two vectors are said to be parallel or coincident if a × b = 0 (null vector)
62 KSOU Algebraic Structures
Note :- N
B
n
b a × b = ab sinθn represent anti - clockwise rotation.
θ
O A
a
B
O θ A
a
− n
Projection of b upon a
B
Let OA = a , OB = b & A OB = θ . Draw BD ⊥ r to OA then OD is the projection of b upon a.
b
a ⋅b
Since a ⋅ b = ab cos θ , cos θ =
ab θ
O A
a D
OD OD
from the ∆le OBD cos θ = =
OB b
a ⋅b
∴ OD = b cos θ = b ⋅
ab
a
= ⋅ b = a ⋅ b
a
further area of parallelogram whose adjacent sides are OA & OB is given by
BD
OA ⋅ BD = ab sinθ Q = sinθ but a × b = ab sinθ .
OB
1
and therefore area of ∆le OAB = a×b .
2
1
Also area of the parallelogram whose diagonals are d1 & d 2 is d1 × d 2
2
MCA 11 - Mathematics SVT 63
a1 a2 a3
which is b1 b2 b3
c1 c2 c3
Thus a ⋅ (b × c ) or (a × b) ⋅ c
which is 'Scalar Triple Product' which is usually denoted as [ a b c ] also called 'Box-Product'.
∴ OP =
b×c
⋅a =
( )
a ⋅ b×c O
b B
b×c b×c
= b×c
( )
a⋅ b×c
b×c
( )
= a ⋅ b×c = [ abc ]
∴ Volume of parallelopiped whose coterminus edges are given by a, b, c is
a1 a2 a3
[ a b c ] = b1 b2 b3
c1 c2 c3
( ) ( ) ( )
a × b× c = a ⋅c b − a ⋅b c
Examples
Solution : a ⋅b = −4 + 15 = 11
2. ( )(
If α = i + 2 j − 3k, β = 3i − j − 2k, find 2α + β ⋅ α + 2 β )
( )( )
Solution : 2α + β ⋅ α + 2 β = (5i + 3 j − 8k ) ⋅ (7i − 7 k ) = 35 + 0 + 56 = 91
3. Prove that the vectors a = 3i − 2 j − k, b = 2i + j − 4k are perpendicular to each other.
4. Find m such that 3i + mj + k and 2i − j − 8k are orthogonal .
Solution : Given a + b = a − b
( )( ) ( )( )
Squaring both the sides, a + b ⋅ a + b = a − b ⋅ a − b
a ⋅ a + a ⋅b + b ⋅a + b ⋅b = a ⋅ a − a ⋅b − b ⋅ a + b⋅b ( )
i.e. 2 a ⋅ b = 0
∴ a ⋅ b = 0 ∴ a is perpendicular to b.
a⋅b −3+3+ 5 5
Solution : Projection of a on b = = = .
b 9 +1+1 11
( )
Solution : Projection of a + c on b =
(a + c)⋅ b = (5i − 3 j + 3k)⋅ (i − 2 j + 2k) = 5 + 6 + 6 = 17
b 1+ 4 + 4 3 3
8. Find the cosine of the angle between vectors a = 4i − 3 j + 3k and b = 2i + j − k
a ⋅b 8−3−3 2
Solution : cosθ = = =
a b 16 + 9 + 9 4 + 1 + 1 34 6
i j k
Solution : a × b = 9 − 1 4 = i(−1 + 4) − j (9 − 32) + k(−9 + 8) = 3i + 23 j − k
8 −1 1
iˆ ˆj kˆ
Solution : a × b = 1 2 2 = iˆ(−4 − 2) − ˆj (−2 − 4) + kˆ(1 − 4) = −6iˆ + 6 ˆj − 3kˆ
2 1 −2
a × b = 36 + 36 + 9 = 81 = 9
11. Find the unit vecto r perpendicular to the pair of vectors a = 6i − 2 j + k and b = 3i + j − 2k
i j k
Solution : Vector perpendicular to a & b is a × b = 6 − 2 1 = i(4 − 1) − j (−12 − 3) + k(6 + 6) = 3i + 15 j + 12k
3 1 −2
12. ( )( ) ( )
Prove that 2a + b × a + 2b = 3 a × b
Solution : a × b + a × c + b × c + b × a + c × a + c × b = a × b + a × c + b × c − a × b − a × c − b × c = 0.
14. Find the sine of the angle between th e vectors a = i + j + k and b = 2i − 3 j + 2k
i j k
Solution : a × b = 1 1 1 = i(2 + 3) − j (2 − 2) + k(−3 − 2) = 5i − 5k , a × b = 5 2 + 5 2 ,
2 −3 2
52 + 52 5 2
∴ sinθ = =
1+1+1 4 + 9 + 4 3 17
Solution : Given a + b + c = 0
( ) ( )
a × a + b + c = a × a + a × b + a × c = 0 ∴ a × b = − a × c [Q a × a = 0 ] ∴ a × b = c × a
Similarly b × a = c × b or c × b = − a × b or a × b = b × c
Similarly c × a = b × c ∴ a × b = b × c = c × a
16. Find the area of the triangle whose sides are a = 3i − 2 j + k and b = i − 3 j + 5k
1
Solution : A = a×b
2
i j k
a × b = 3 − 2 1 = i(−10 + 3) − j (15 − 1) + k(−9 + 2) = −7i − 14 j − 7k
1 −3 5
66 KSOU Algebraic Structures
49 + 196 + 49 294
area = = sq.units
2 2
17. Position v ectors of the points A, B and C are respectively i − j + k, 2i + j − k and 3i − 2 j + k. Find the area of triangle
ABC.
Solution : AB = OB − OA = ( 2i + j − k) − (i − j + k) = i + 2 j − 2k.
i j k
AB × AC = 1 2 − 2 = i(0 − 12 − j (4) + k(−1 − 4) = −2i − 4 j − 5k.
2 −1 0
1 45
area = 4 + 16 + 25 = sq.units
2 2
18. Find the area of a parallelogram whose adjacent sides are a = 3i + 2 j − k and b = i + 2 j + 3k.
i j k
Solution : a × b = 3 2 − 1 = i(6 + 2) − j (9 + 1) + k(6 − 2) = 8i − 10 j + 4k
1 2 3
19. Find the area of a parallelogram whose diagonals are d1 = 3i + j + 2k and d 2 = i − 3 j + 4k.
1
Solution : A = d1 × d 2
2
i j k
d1 × d 2 = 3 1 2 = i(4 + 6) − j (12 − 2) + k(−9 − 1) = 10i − 10 j − 10k
1 −3 4
1 1
d1 × d 2 = 100 + 100 + 100 ; d1 × d 2 = 300 sq.units = 5 3 sq.units.
2 2
20. Find the scalar triple product of vectors a = 2i − j + 3k, b = i + 2 j + 3k and c = 3i + j − k.
2 −1 3
( )
Solution : a ⋅ b × c = 1 2 3 = 2(−2 − 3) + 1(−1 − 9) + 3(1 − 6) = −10 − 10 − 15 = −35.
3 1 −1
1 −1 1
( )
Solution : a ⋅ b × c = 2 − 4 5 = 1(−4 + 25) + 1(2 − 15) + 1(−10 + 12) = 21 − 13 + 2 = 10 cubic units
3 −5 1
2 −3 4
[ ]
Solution : a b c = 1 2 − 1 = 0
λ −1 2
i.e., 2(4 − 1) + 3( 2 + λ ) + 4( −1 − 2λ ) = 0
8
6 + 6 + 3λ − 4 − 8λ = 0 ⇒ 5λ = 8 ; λ = .
5
24. Show that the points A( 4, 5, 1), B(0, − 1, − 1), C (3, 9, 4) and D (−4, 4, 4) are coplanar.
Exercise
1. The position vectors of the points A, B and C are respectively a , b and 2a − 3b. Express vectors BC , AC , AB in terms of
a and b.
2. Position v ectors are A, B, C and D are 2a + 4c, 5a + 3 3 b + 4c, − 2 3 b + c and 2a + c respectively.
3
Show that AB || CD and AB = CD
2
4. Find the unit vecto r in the direction of a + b + c where a = i + 4 j + 2k , b = 3i − 3 j − 2k and c = −2i + 2 j + 6k.
6. Show that the vectors (–1, 2, 3) and (2, –5, 4) are orthogonal.
7. Find the values of λ such that λi − 3 j + k and λi + λj + 2k may be orthogonal .
8. If a and b are unit vecto r show that the vectors a + b and a − b are orthogonal .
13. If a = ( 2, − 1, 3), b = ( −2, 1, 4) and c ≡ ( 2, 1, − 7), find the unit vecto r in the direction of a + b + c.
19. Find the sine of the angle between the vectors 2i − 3 j + k and 3i + j − 2k.
20. Find the unit vecto r perpendicu lar to the vectors 3i + j + 2k and 2i − 2 j + 4k.
21. Find the volume of the parallelopiped whose co-terminal edges are represented by a = 2i − 3 j + 4k ; b = i + 2 j − k and
c = 3i + j + 2k.
22. Show that vectors 2i − j + k , i + 2 j − 3k and 3i − 4 j − 5k are coplanar.
23. If the vectors i + 2 j + 5k , 2 i + xj − 10k and 3i + 9 j − 2k are coplanar. find x.
24. Show that points A (2, 3, –1), B (1, –2, 3), C (3, 4, –2) and D (1, –6, 6) are coplanar.
25. Find the scalar tri ple product of, a = i − 2 j + k , b = 2i + j + k , c = i + 2 j + k.
26. ( )( ) ( )
Prove that 2a + 3b × a + 4b = 5 a × b
27. Find the area of the triangle whose 2 sides represented by a = i − 3 j − 2k, b = i + 2 j + k.
28. Find the area of the triangle whose vertices are i − j + 2k, 2 j + k and j + 3k.
29. Find the area of parallelogram whose diagonals are a = 3i + j − 2k and b = i − 3 j + 4k.
30. ( ) ( )
If a = (1, − 1, 2), b = (1, 2, 3) & c = (3, − 2, 4). Find a × b × c and a × b × c.
BCA 21 / IMCA 21 / Mathematics SVT 65
DIFFERENTIAL CALCULUS
Limits of functions
x 2 −1
Consider y = f ( x) = the function is defined for all values of x except for x = 1.
x −1
1 −1 0
∴ for x = 1, f ( x) = = which is indeterminate.
1 −1 0
Let us consider the values of f (x) as x approaches 1
x 2 −1
x f ( x) =
x −1
.9 1.9
.99 1.99
.999 1.999
1.01 2.01
1.001 2.001
1.0001 2.0001
x −1
x 2 −1 x 2 −1 x2 −1
This value 2 is called " Limit of as x approaches 1" which can be written as lim or Lt = 2.
x −1 x →1 x − 1 x →1 x − 1
In general the limit of a function f (x) as x approaches a is denoted as l and which is written as
lim f ( x) = l or Lt f ( x) = l
x→a x →a
Properties
(1) lim[ f ( x) ± g ( x)] = lim f ( x) ± lim g ( x)
x →a x →a x→a
(2) lim[ f ( x) g ( x)] = lim f ( x) lim g ( x) in particular lim kf ( x) = k ⋅ lim f ( x) where k is a contant
x →a x →a x →a x →a x →a
f ( x) xlim f ( x)
(3) lim = →a provided lim g ( x) ≠ 0.
x→ a g ( x) lim g ( x) x →a
x→a
Standard Limits
xn − an
(1) lim = na n −1
x→a x−a
70 KSOU Differential Calculus
sinθ tanθ
(2) lim = 1 (θ in radians) also lim =1
θ →0 θ θ→ 0 θ
n
1 1
(3) lim 1 + = e 2 < e < 3 or lim(1 + x) x = e
n →∞ n x →0
a x −1 e x −1
(4) lim = log e a ( a > 0) in particular lim = 1.
x →0 x x →0 x
Examples
x2 + 4x + 3 0+0+3 3
(1) lim = =
x →0 x − 5x + 4
2
0−0+4 4
3 4
+ 2−
(2) lim 2
2 x 2 − 3x + 4
(dividing Nr & Dr by x 2 ) = lim
x x2 = 2 − 0 + 0 = 2
x →∞ 3 x + 2 x + 1 x →∞ 2
3+ +
1 3+0+0 3
x x2
x 4 − 81 x 4 − 34
(3) lim = lim = 4(3) 3 = 108
x →3 x − 3 x →3 x − 3
x7 + a7 x 7 − (−a) 7
x +a
7 7
x − (−a) 7( − a ) 6
= lim 5x + a 5 = lim
7
(4) lim 5
= = a2
x →− a x 5 + a 5 x →− a x + a x →− a x − (−a) 5
5 − (−a) 4
5
x+a x − (−a)
sin 7 x sin 7 x
(5) lim = lim × 7 = 1× 7 = 7
x →0 x x →0 x
1 − cos 2θ 2 sin 2 θ
(6) lim = lim =2
θ →0 θ2 θ →0 θ2
tan3 x
−1
tan3 x − x x 3 −1
(7) lim (dividing Nr & Dr by x) = lim = =1
x →0 3 x − sin x x →0 sin x 3 −1
3−
x
ab
b 1
(8) lim(1 + ax) x = lim (1 + ax)
ax = e ab
x →0 x →0
3
n n
3 3 3
(9) lim 1 + = lim 1 + = e3
n →∞ n n →∞ n
−2
1 −
1
(10) lim(1 − 2 x) x = lim (1 − 2 x) 2 x = e −2
x →0 x →0
ax − bx ( a x − x ) − (b x − x ) ax − x bx − x a
(11) lim = lim = lim − lim = log e a − log e b = log e
x →0 x x →0 x x →0 x x →0 x b
MCA 11 - Mathematics SVT 71
2x −1
2 −1
x
log e 2
(12) lim = lim x = = log e 2
x →0 sin x x →0 sin x 1
x
Continuity of a function
A function f ( x) is said to be continuous at x = a if lim f ( x) = f (a)
x →a
∴ A function f ( x) is said to be continuous if lim f ( x) exists, f (a) exists and they are equal.
x→0
If these donot happen then the function is said to be not continuous or discontinuous.
A function f (x) is said to be continuous in an interval if it is continuous at all points in the interval.
Examples
x 2 −1 x 2 −1 0
(1) f ( x) = is not continuous at x = 1 Q lim = 2 exists but f (0) = donot exists.
x −1 x →1 x −1 0
where as it is continuous at all other values of x.
4 x + 3 for x ≥ 4
(2) Discuss the continuity of function f ( x) = at x = 4.
3x + 7 for < 4
Solution : While finding the limit of a function f (x) as x approaches a, if we consider the limit of the function as x approaches
a from left hand side, the limit is called 'Left Hand Limit' (LHL) and if x approaches a from right hand side the
limit is called 'Right Hand Limit' (RHL) and the limit of the function is said to exists if both LHL & RHL exists
and are equal, for convenience LHL & RHL are denoted as lim− f ( x) and lim+ f ( x) and further
x →a x →a
and f ( 4) = 19
∴ The function is continuous at x = 4
sin x
for x ≠ 0
(3) Examine the continuity of f ( x) = x at x = 0.
2 for x = 0
sin x
Solution : lim = 1, but f (0) = 2
x →0 x
∴ The function is discontinuous at x = 0.
5 x − 4 for 0 < x ≤ 1
(4) Examine the continuity of the function f ( x) = 4 x 2 − 2 x for 1 < x < 2
4 x + 4 for x ≥ 2
at x = 1 and x = 2.
72 KSOU Differential Calculus
LHL ≠ RHL at x = 1
and f (2) = 4 × 2 + 4 = 12
∴ lim f ( x) = f (2)
x →2
Differentiability of a function
f ( a + h) − f ( a )
A function f ( x) is said to be differentiable at a point a if lim exists and the derivative is denoted as f '( a ).
h →0 h
f ( x + δx ) − f ( x )
A function f ( x) is said to be differentiable at x if lim exists and the derivative is denoted as f '( x).
δx
δx →0
To find the derivates of xn, loge x, ax, sin x, cos x and a constant C with respect to x.
dy ( x + δx ) n − x n
(1) Let y = x n , = lim = nx n −1
dx δx →0 x + δx − x
Eg. (i)
d 7
dx
( )
x = 7 x 6 , (ii )
d −4
dx
( )
x = −4 x − 5 (iii)
d 94
dx
( )
9 9 1 9 5
x = x 4− = x 4
4 4
(iv)
dx
x ( )
d − 53 5 5 1
3
5 8
= − x− 3− = − x− 3
3
( v)
d
dx
( x )= 2 1 x
dy log e ( x + δx ) − log e x
(2) Let y = log e x then = lim
dx δx →0 δx
x
1 x x + δx 1 δx δx 1 1
= lim ⋅ log = lim ⋅ log1 + = log e e = .
δx →0 x δx x δx →0 x x x x
(3) Let y = a x
dy a x + δx − a x ( a δx − 1) d x
= lim = lim a x = a x log e a ( a > 0) in particular (e ) = e x
dx δx →0 δx δx →0 δx dx
MCA 11 - Mathematics SVT 73
x + δx + x x + δx − x δx
2 cos sin sin
2 2 δx 2 = cos( x + 0) ⋅1 = sin x.
= lim = lim cos x + ⋅
δx →0 δx δx →0 2 δx
2
(5) Let y = cos x
dy cos( x + δx ) − cos x
= lim
dx δx →0 δx
δx
sin
δx 2 = − sin(x + 0) ⋅1 = − sin x.
= lim − sin x + ⋅
δx →0 2 δx
2
dy c−c
(6) Let y = c (a constant) = lim =0
dx δ x →0 δx
Thus derivative of a constant is zero.
ax a x log e a
sin x cos x
cos x − sin x
constant zero
dy du dv
ie = +
dx dx dx
dy du dv dw
Note : - If y = u + v − w then = + −
dx dx dx dx
dy d 6 d d
Eg. (1) If y = x 6 + sin x − cos x then = ( x ) + (sin x) − (cos x)
dx dx dx dx
= 6 x 5 + cos x − (− sin x) = 6 x 5 + cos x + sin x.
dy 1
( 2) If y = 3 x − log e x + c then = 3 x log e 3 − + 0.
dx x
II Product Rule
dy
If y = uv where u & v are functions of x then to find .
dx
Give a small increment δx to x, let the corresponding increments in u , v & y be δu , δv & δy respectively.
∴ δy = y + δy − y = uδv + vδu + δu ⋅ δv
divide through out by δx,
δy δv δu δv
=u +v + δu ⋅
δx δx δx δx
take limit on both sides as δx → 0,
δy δv δu δv
then lim = u lim + v lim + lim δu ⋅
δx →0 δx δx →0 δx δx →0 δx δx →0 δx
dy dv du dv
ie =u +v +o
dx dx dx dx
dy dv du
ie =u +v
dx dx dx
dy dv
Note (1) If y = kv where k is a constant then =k
dx dx
dy dw dv du
( 2) If y = uvw, then = uv + uw + vw
dx dx dx dx
3 dy 3 d d 3
Eg. (1) If y = x 2 sin x then =x 2 (sin x) + sin x x 2
dx dx dx
3 3 12
=x 2 cos x + sin x ⋅ x
2
dy
( 2) If y = 8 cos x then = −8 sin x
dx
dy d d d
(3) If y = e x sin x ⋅ log x then = e x sin x (log x) + e x log x ⋅ (sin x) + sin x ⋅ log x (e x )
dx dx dx dx
1
= e x sin x ⋅ + e x log x ⋅ cos x + sin x ⋅ log x ⋅ e x .
x
MCA 11 - Mathematics SVT 75
u + δu
then y + δy =
v + δv
u + δu u v(u + δu ) − u (v + δv) vu + vδu − uv − uδv
∴ δy = − = =
v + δv v v ( v + δv ) v ( v + δv )
divide through out by δx
δu δv
v −u
δy
= δx δx
δx v ( v + δv )
du dv
v −u
dy dx dx
ie =
dx v2
This rule can be easily remembered in the following manner
u Nr
If y = = (say)
v Dr
dy Dr (derivative of Nr ) − Nr (derivative of Dr )
Then =
dx ( Dr ) 2
k dy k dv
Note : - If y = where k is a constant then =− 2
v dx v dx
sin x
Eg. (1) If y = tan x =
cos x
d d
cos x (sin x) − sin x (cos x)
dy dx dx
then =
dx cos 2 x
cos x ⋅ cos x − sin x( − sin x) cos 2 x + sin 2 x 1
= = = = sec 2 x.
cos x
2
cos x
2
cos 2 x
dy
∴ If y = tan x, then = sec 2 x.
dx
cos x
( 2) If y = cot x =
sin x
dy sin x( − sin x) − cos x(cos x)
then =
dx sin 2 x
− sin 2 x − cos 2 x −1
= = = −cosec 2 x.
sin x
2
sin 2 x
76 KSOU Differential Calculus
dy
∴ If y = cot x, then − cosec 2 x.
dx
1
(3) If y = sec x =
cos x
dy 1 d
then =− ⋅ (cos x)
dx cos x
2
dx
1 1 1 sin x
=− ( − sin x) = (sin x ) = ⋅ = sec x tan x.
cos x
2
cos x cos x cos x
dy
∴ If y = sec x, then = sec x tan x.
dx
1
( 4) If y = cosec x =
sin x
dy 1
then = − 2 ⋅ cos x
dx sin x
1 cos x
=− ⋅ = −cosec x ⋅ cot x.
sin x sin x
dy
∴ If y = cosec x, then = −cosec x ⋅ cot x.
dx
x 2 −1
(4) If y = tan 2
x +1
dy x2 −1 d x2 −1
then = sec 2 2 2
dx x + 1 dx x + 1
x 2 − 1 ( x 2 + 1)(2 x ) − ( x 2 − 1)2 x 2
x − 1 × 2 x( x + 1 − x + 1)
2 2 2
= sec 2 2 × = sec
x2 +1
x +1 ( x 2 + 1) 2 ( x 2 + 1) 2
x 2 −1 4x
= sec 2 2
x + 1 × ( x 2 + 1) 2
dy d d e x + e−x e x − e−x
then = (sinh x) = = = cosh x.
dx dx dx 2
2
( 2) If y = cosh x,
dy d d e x − e−x e x + e−x
then = (cosh x) = = = sinh x.
dx dx dx 2
2
sinh x
(3) If y = tanhx = ,
cosh x
dy d sinh x
then =
dx dx cosh x
cosh x
( 4) If y = coth x = ,
sinh x
dy d cosh x
then =
dx dx sinh x
dy d 1 1 d
then = =− (cosh x)
dx dx cosh x cosh 2 x dx
−1 −1 sinh x
= ⋅ sinh x = ⋅ = −sech x tanhx.
cosh 2 x sech x cosh x
78 KSOU Differential Calculus
1
(6) If y = cosech x = ,
sinh x
dy −1 d
then =− (sinh x)
dx sinh2 x dx
−1 − 1 cosh x
= ⋅ cosh x = ⋅ = −cosech x ⋅ coth x.
sinh x
2
sinh x sinh x
Implicit Functions
Function of the type f ( x, y ) = 0 is called Implicit function.
dy
To find treat y as a function of x & use chain rule.
dx
Eg. (1) If ax 2 + 2hxy + by 2 = 0
dy dy
then 2ax + 2h x + y + 2by =0
dx dx
dy
ie (2hx + 2by ) = −2ax − 2hy
dx
dy − 2( ax + hy ) − ( ax + hy)
∴ = =
dx 2(hx + by ) hx + by
Eg. (2) If x sin y + y sin x = 10
differentiating w.r.t. x
dy dy
x cos y + sin y + y cos x + sin x =0
dx dx
dy
ie ( x cos y + sin x) = − sin y − y cos x
dx
dy − (sin y + y cos x)
∴ =
dx ( x cos y + sin x )
Parametric functions
Functions of the type x = f (t ), y = g (t ) taken together is called Parametric function, where t is the paramter. Parametric
functions are also denoted as x = f (θ ), y = f (θ ) where θ is the parameter.
dy dx dy dx dy
To find , consider & or &
dx dt dt dθ dθ
dy dy dt dy dy dθ
then = or =
dx dx dt dx dx dθ
dy
Eg. (1) If x = a cos 3 t , y = a sin 3 t , find
dx
dx dy
Solution : x = a cos 3 t , = −3a cos 2 t sin t , y = a sin 3 t , = 3a sin 2 t cos t
dt dt
dy dy dt 3a sin 2 t cos t
∴ = = = − tant
dx dx dt − 3a cos 2 t sin t
MCA 11 - Mathematics SVT 79
dy
( 2) If x = a (3 cos θ − 4 sin 3 θ ) & y = a (3 sinθ − 4 cos 3 θ ), find
dx
dy dy dθ a (3 cosθ + 12 cos 2 θ sin θ )
Solution : = =
dx dx dθ a (−3 sin θ − 12 sin 2 cos θ )
3 cosθ (1 + 4 sin θ cos θ )
= = − cot θ
− 3 sin θ (1 + 4 sin θ cosθ )
If y = f ( x) g ( x )
use logarithms on both sides
dy g ( x) f '( x)
∴ = f ( x) g ( x ) + g '( x ) log f ( x )
dx f ( x)
dy
Eg. (1) Find if y = x x
dx
Solution : log y = x log x
1 dy 1
∴ = x ⋅ + log x ⋅1
y dx x
dy
ie = x x [1 + log x]
dx
dy
( 2) If y = x sin x + (cos x) tanx , find .
dx
Solution : Let y = u + v where u = x sin x & v = (cos x) tanx
log u = sin x log x
1 du sin x
∴ = + cos x log x
u dx x
du sin x
ie = x sin x + cos x log x
dx x
v = (cos x ) tanx
log v = tan x log cos x
1 dv 1( − sin x)
∴ = tan x + sec 2 x log cos x
v dx cos x
dv
∴ = (cos x) tan x [ − tan2 x + sec 2 x log cos x]
dx
dy du dv sin x
∴ = + = x sin x + cos x log x + (cos x) tan x [ − tan2 x + sec 2 x log cos x]
dx dx dx x
80 KSOU Differential Calculus
d 1
(3) (tanh−1 x) = for x < 1
dx 1− x2
d −1
( 4) (coth−1 x) = 2 for x > 1
dx x −1
d −1
(5) (sech −1 x) = for x < 1
dx x 1− x2
d −1
( 6) (cosech −1 x) =
dx x 1+ x2
Exercise
dy
Find of the following
dx
1. y = x 2 + a 2 2. y = loge (3x + 2) 3. y = log e cos x
x +1
4. y = 5. y = x 2 e x 6. y = ( 2 x + 3) 2
x −1
7. y = ax 2 + bx + c 8. y = e x
9. y = (3 x + 5)1 / 3
x2 + x +1
16. y = (1 + x 2 ) tan−1 x 17. y = (1 + x 2 ) sin x 18. y =
x
ex e x + e− x
22. y = 2
23. y = 24. y = sin x ⋅ sin 2 x
x 2
2x x+a
25. y = ( x + a)( x + b)( x + c) 26. y = sin −1 27. y = tan−1
1+ x 2
1 − ax
1 − cos x
28. y = 29. y = e sinh x 30. y = log e sin x
1 + cos x
1+ x
31. y = log e ( x 2 tan x) 32. y = (1 − x 2 ) cos −1 x 33. y = tan−1
1− x
1− x
34. y = tan−1 35. xy = c 2 36. sin −1 x + sin −1 y = 0
1+ x
1
40. x = t , y = 41. x = 4 cosh t , y = 4 sinh t 42. x = a cos t , y = b sin t
t
( x 2 + 1)e x e x − e−x
46. y = 47. y = 48. y = (1 + x 2 ) tan x
log e x ⋅ cosec x e x + e−x
xe x x2 + ex cos −1 x
49. y = 50. y = 51. y =
1 + cos x 1 − x log e x log e x
x 3e x x2 +1 3x − x 3
52. y = 53. y = sec −1 2 54. y = tan−1 2
cos −1 x x −1 1 − 3x
Successive Differentiation
dy
If y = f ( x) then = f ′( x ) is also a function of x, hence further derivatives can be obtained.
dx
d2y
The second derivative is denoted by or f ′′( x) or y 2 or D 2 y.
dx 2
d3y
The third derivative is denoted by or f ′′′( x) or y3 or D 3 y.
dx 3
dny
In general n th derivative is denoted as or f ( n ) ( x) or y n or D n y.
dx n
Examples
d2y
1. If y = (1 + x 2 ) tan−1 x find at x = 1
dx 2
Solution : y = (1 + x 2 ) tan−1 x
dy 1
= (1 + x 2 ) × + tan−1 x ⋅ ( 2 x) = 1 + 2 x tan−1 x.
dx 1+ x 2
d2y 1
= 0 + 2x × + 2 tan−1 x
dx 2 1+ x2
d2y 2 π π
at x = 1, 2 = + 2× = 1+ .
dx x =1 1 + 1 4 2
d2y − 4a
2. If y 2 = 4ax, show that 2
=
dx y3
Solution : y 2 = 4ax
differentiating w.r.t. x
dy dy 2a
2y = 4a ⇒ =
dx dx y
again differentiating w.r.t. x
d2y 2a dy 2a 2a − 4a 2
=− = − =
dx 2 y 2 dx y 2 y y3
t d2y sin t
3. If x = a cos t + log tan & y = a sin t , then show that =
2 dx 2
a cos 4 t
t
Solution : x = cos t + log tan
2
dx 1 1 t
∴ = a − sin t + sec 2
dt 1
tan 2 2
2
MCA 11 - Mathematics SVT 83
1 1 1 − sin 2 t + 1 a cos 2 t
= a − sin t + = a − sin t + = a − sin t + = a =
1 t t t sin t sin t sin t
2 tan cos 2 2 sin cos
2 2 2 2
dy
y = a sin t , ∴ = a cos t
dt
dy dy dt a cos t
∴ = = × sin t = tant
dx dx dt a cos 2 t
d2y d dy d dy dt 1 sec 2 t
∴ = = = sec 2 t × =
dx 2 dx dx dt dx dx dx dt a cos 2 t
sin t
d2y sin t
∴ =
dx 2
a cos 4 t
d2y
4. If y = sin 2 x, find
dx 2
dy
Solution : = 2 sin x cos x = sin 2 x
dx
d2y
= cos 2 x ⋅ 2 = 2 cos 2 x
dx 2
d2y
5. If y = x 2 log e x find
dx 2
dy 1
Solution : = x 2 ⋅ + log e x ⋅ 2 x = x( 2 log e x + 1)
dx x
d2y 2
= x + (2 log e x + 1) = 3 + 2 log e x
dx 2 x
d2y
6. If y = e ax ⋅ sin(bx + c) find
dx 2
dy
Solution : = e ax ⋅ cos(bx + c) ⋅ b + sin(bx + c) ⋅ ae ax
dx
d2y
dx 2
{ }
= e ax − b 2 sin(bx + c) + ab cos(bx + c) + ae ax {b cos(bx + c) + a sin(bx + c)}
{ }
= e ax 2ab cos(bx + c) + (a 2 − b 2 ) sin(bx + c)
d2y
7. If x = a (θ − sinθ ), y = b(1 − cos θ ) find
dx 2
dx dy
Solution : = a (1 − cos θ ), = b sinθ
dθ dθ
dy b sinθ b ⋅ 2 sin(θ 2) cos(θ 2) b
= = = cot(θ 2)
dx a (1 − cos θ ) a ⋅ 2 sin 2 (θ 2) a
84 KSOU Differential Calculus
d2y b θ 1 dθ b θ 1 b
= − cosec 2 ⋅ ⋅ = − cosec 2 ⋅ = − 2 cosec 4 (θ 2).
dx 2
a 2 2 dx 2 a 2 a (1 − cos θ ) 4a
−1
8. If y = e m sin x
, then prove that (1 − x 2 ) y 2 − xy1 − m 2 y = 0
dy −1 m
Solution : = e m sin x ⋅
dx 1− x2
cross multiplying & squaring, we have
(1 − x 2 ) y12 = m 2 y 2
differentiating w.r.t. x
(1 − x 2 ) ⋅ 2 y1 y 2 + y12 (−2 x) = m 2 ⋅ 2 yy1
m
Solution : y1 = cos(m sin −1 x) ⋅
1− x2
m 1− y2
y1 = ⇒ (1 − x 2 ) y12 = m 2 (1 − y 2 )
1− x 2
differentiating w.r.t. x
(1 − x 2 )2 y1 y 2 + y12 (−2 x) = m 2 (−2 yy1 )
dy 1
Solution : = ex ⋅ + e x tan−1 x
dx 1+ x 2
(1 + x 2 )[ y1 − y] = e x
differentiating w.r.t. x
(1 + x 2 )( y2 − y1 ) + 2 x( y1 − y) = e x
(1 + x 2 ) y 2 − (1 + x 2 − 2 x) y1 − 2 xy = (1 + x 2 ) y1 − (1 + x 2 ) y
⇒ (1 + x 2 ) y 2 − 2(1 + x 2 − x) y1 + (1 − x) 2 y = 0
Exercise
d2y 16 d2y −2
(1) If 4 x 2 + 9 y 2 = 36 show that =− ( 2) If x 2 + 2 xy + 3 y 2 = 1 prove that =
dx 2
9y 2
dx 2
( x + 3 y) 3
d2y 1 d2y
(3) If x = a cos 3 θ , y = a sin 3 θ find ( 4) If x = a tanθ , y = a sin 2θ find
dx 2 2 dx 2
MCA 11 - Mathematics SVT 85
d2y sec 3 θ
(5) If x = a (cos θ + θ sinθ ), y = a (sinθ − θ cos θ ) show that =
dx 2
aθ
d2y d2y
(6) If y = sin −1 2 x, then find 2
(7) If y = x log e x, then find
dx dx 2
d2y d2y
(8) If y = e 4 x ⋅ sec 3 x, then find 2
(9) If y = a x , then find
dx dx 2
d2y d2y
(10) If x = at 2 , y = 2at , then find 2
(11) If x 3 y 3 = a x , then find
dx dx 2
d2y
(12) If y = a cos mx + b sin mx, prove that 2
+ m2 y = 0
dx
m
(13) If y = x + x 2 + 1 , prove that ( x 2 + 1) y 2 + xy1 − m 2 y = 0
b
(18) If y = ax + 2
then show that x 2 y 2 + 2( xy1 − y ) = 0
x
b
(19) If y = ax n +1 + then show that x 2 y 2 = n( n + 1) y
xn
d 2x
( 20) If x = a cos nt + b sin nt then show that + n2 x = 0
dt 2
1
in particular if m = −1, ie y =
ax + b
differentiating ( n − 1) times
( −1) n −1 a n −1 ( −1) n −1 a n
yn = ⋅a =
( ax + b) n ( ax + b) n
3. If y = a mx to find yn
∴ In general, yn = m n (log a) n a mx
4. If y = sin(ax + b) to find y n
π
Solution : y1 = a cos(ax + b) = a sin ax + b +
2
again differentiating w.r.t. x
π π
y 2 = a 2 cos ax + b + = a 2 sin ax + b + 2 ⋅
2 2
π π
y3 = a 3 cos ax + b + 2 ⋅ = a 3 sin ax + b + 3 ⋅
2 2
π
∴ In general, y n = a n sin ax + b + n ⋅
2
5. If y = cos(ax + b) to find y n
π
Solution : y1 = −a sin(ax + b) = a cos ax + b +
2
again differentiating w.r.t. x
π π
y 2 = −a 2 sin ax + b + = a 2 cos ax + b + 2 ⋅
2 2
π
∴ In general, y n = a n cos ax + b + n ⋅
2
6. If y = e ax cos(bx + c) to find y n
y1 = ae ax cos(bx + c) − be ax sin(bx + c)
y2 = r 2 e ax cos(bx + c + 2α )
b
In general, y n = r n e ax cos(bx + c + nα ) where r = a 2 + b 2 & α = tan−1
a
MCA 11 - Mathematics SVT 87
7. If y = e ax sin(bx + c) to find yn
Solution : differentiating w.r.t. x
y1 = ae ax sin(bx + c) + be ax cos(bx + c)
put a = r cos α , b = r sin α
y 2 = r 2 e ax sin(bx + c + 2α )
b
In general, y n = r n e ax sin(bx + c + nα ) where r = a 2 + b 2 & α = tan−1
a
8. Statement of Leibmitz's Theorem on nth derivative of a product
If u & v are functions of x, the n th derivative of the product uv is given by
(uv) n = u n v + nc1u n −1v 2 + nc2 u n − 2 v2 + LL + ncn uvn
where suffixes of u & v represent order of the derivatives of u & v.
Examples
1
1. Find the n th derivative of
x − 6x + 8
2
1 1 A B
Solution : Let y = = = + (Say)
x − 6x + 8
2
( x − 2)( x − 4) ( x − 2) ( x − 4)
multiplying throughout by ( x − 2)( x − 4)
1 = A( x − 4) + B ( x − 2)
1
put x = 2, 1 = A( −2) ⇒ A = −
2
1
put x = 4, 1 = 2 B ⇒ B =
2
1 1
−
∴ y= 2 + 2
( x − 2) ( x − 4)
1 1
− ( −1) n n ! ( −1) n n !
differentiating n times, we have y n = 2 + 2
( x − 2) n +1 ( x − 4) n +1
2. Find the n th derivative of sin 2 x cos 3 x
(1 − cos 2 x) cos 3 x + 3 cos x
Solution : Let y = sin 2 x cos 3 x = ×
2 4
1
ie y = [cos 3 x + 3 cos x − cos 3 x cos 2 x − 3 cos 2 x cos x]
8
1 1 3
= cos 3 x + 3 cos x − (cos 5 x + cos x) − (cos 3 x + cos x)
8 2 2
1 1 1 3 3
= cos 3 x + 3 cos x − cos 5 x − cos x − cos 3 x − cos x)
8 2 2 2 2
88 KSOU Differential Calculus
1 1 1
∴ y= cos x − cos 3 x − cos 5 x
8 2 2
1 π 1 n π 1 n π
differentiating n times, we have y n = cos x + − ⋅ 3 cos 3x + n − ⋅ 5 cos 5 x + n
8 2 2 2 2 2
3. If y = e 2 x sin2 x to find yn
1 2x 1 2x 1 2x
Solution : y = e (1 − cos 2 x) y= e − e cos 2 x
2 2 2
1 n 2x 1 n 2x
differentiating n times, we have y n = 2 e − ⋅ r e cos( 2 x + n − α )
2 2
2 π
where r = 4 + 4 = 8 , α = tan−1 = tan−1 1 =
2 4
∴ y n = 2 n −1 e 2 x −
1
2
( 8) n π
cos 2 x + n ⋅
2
ie y n =
1
2
( 80 ) e n 4x
sin(8 x + n tan−1 2) +
1
2
( 20 ) en 4x 1
sin 2 x + n tan−1
2
5. If y = x 2 log 3x find yn
Solution : y = x 2 log 3x
Let u = log 3 x = log 3 + log x
( −1) n −1
∴ un = , v = x2
xn
differentiating n times, using Leibnitz' s Theorem,
=
( −1) n −3
x n−2
[(−1) 2
+ ( −1)1 2n + n( n − 1) = ] ( −1) n −3
x n−2
[1 − 2n + n 2
−n ]
ie y n =
( −1) n −3
x n−2
[n 2
− 3n + 1 ]
6. If y = a cos(log x) + b sin(log x) show that x 2 y n + 2 + (2n + 1) x y n +1 + (n 2 + 1) y n = 0
Solution : Let y = a cos(log x) + b sin(log x)
differentiating w.r.t. x
1 1
y1 = −a sin(log x) + b cos(log x)
x x
MCA 11 - Mathematics SVT 89
1 1
xy 2 + y1 = −a cos(log x) − b sin(log x )
x x
ie x 2 y2 + xy1 + y = 0
differentiating n times, using Leibnitz' s Theorem, we have
x 2 y n + 2 + nC1 2 xyn +1 + nC 2 ⋅ 2 y n
+ xy n +1 + nC1 ⋅1 y n
+ yn = 0
adding, x 2 y n + 2 + (2n + 1) xy n +1 + [ n(n − 1) + n + 1] y n = 0
ie x 2 y n+ 2 + (2n + 1) xyn +1 + (n 2 + 1) y n = 0
−1
7. If y = e m cos x
, prove that (1 − x 2 ) y n + 2 − ( 2n + 1) xy n +1 − ( n 2 + m 2 ) y n = 0
−1
Solution : y = e m cos x
differentiating w.r.t. x
−m
y1 = e m cos x ⋅ ie 1 − x 2 y1 = −my
1− x2
dividing by 2 y1 , we have
(1 − x 2 ) y2 − xy1 − m 2 y = 0
(1 − x 2 ) y n + 2 + nC1 y n +1 ( −2 x) + nC 2 y n ( −2)
− xy n +1 + nC1 y n ( −1)
− m 2 yn = 0
adding, (1 − x 2 ) y n + 2 − ( 2n + 1) xy n +1 − ( n 2 − n + n + m 2 ) y n = 0
ie (1 − x 2 ) yn + 2 − (2n + 1) xyn +1 − (n 2 + m 2 ) yn = 0
Solution : y1 m + y −1 m = 2 x
( )
ie y1 m
2
+ 1 = 2 xy1 m
∴ (y ) 1m 2
− 2 xy1 m + 1 = 0
90 KSOU Differential Calculus
2x ± 4x2 − 4
which is a quadratic equation in y1 m ∴ y1 m = = x ± x 2 −1
2
m
Consider, y1 m = x + x 2 − 1 ∴ y = x + x 2 − 1
differentiating w.r.t. x
2
m −1 m −1 x − 1 + x
y1 = m x + x − 1 2 1 + = m x + x − 1
2x 2
2 x −1 x −1
2 2
ie x 2 − 1 y1 = my
( x 2 − 1) y2 + xy1 − m 2 y = 0
differentiating n times using Leibnitz' s Theorem
( x 2 − 1) y n + 2 + nC1 y n +1 2 x + nC 2 y n 2
+ xy n +1 + nC1 y n 1
− m 2 yn = 0
adding, ( x 2 − 1) y n + 2 + ( 2n + 1) xy n +1 + ( n 2 − n + n − m 2 ) y n = 0
ie ( x 2 − 1) yn + 2 + (2n + 1) xyn +1 + (n 2 − m 2 ) y n = 0
m
We obtain the same result if y1 m = x − x 2 − 1 ie y = x − x 2 − 1
Exercise
1
1. Find the n th derivative of
x − 5x + 6
2
2. Find the n derivative of (i) sin3 x (ii) cos3 x (iii) sin 4 x cos 3x (iv) sin 8 x sin 4 x (v) cos 5x cos x.
th
3. Find the n th derivative of (i) e3 x sin2 x (ii) e 2 x cos 2 x (iii) e x sin 5x cos 2 x
Polar Co-ordinates r = f ( q)
Let O be a fixed point and OA the initial line. Let P be any point on the curve.
rdθ
To prove that tanφ = (important result)
dr
dy
If the Cartesian co - ordinates of P are ( x, y ) then x = r cos θ y = r sinθ and tanψ =
dx
dr
r cos θ + sinθ
dy dθ dθ
∴ tanψ = =
dx dθ cos θ dr − r sinθ
dθ
dr
using (1) & dividing Nr & Dr of RHS by cos θ , we have
dθ
dθ dθ
r + tanθ tanθ + r
tanθ + tanφ
tan(θ + ϕ ) = dr ie = dr
dθ 1 − tanθ tanφ 1 − tanθ ⋅ r dθ
1− r tanθ
dr dr
dθ
Comparing LHS & RHS, we have tanφ = r (3)
dr
[ ]
2
1 1 1 1 1 dr
from (2) p = r sin φ ∴ = 2 cosec 2φ = 2 1 + cot 2 φ = 2 + 2 using (3)
p2 r r r r rdθ
2
1 1 1 dr
= + (4)
p2 r 2 r 4 dθ
using r = f (θ ) & ( 4) we can eliminate θ and obtain a relation between p & r for the curve, which is called Pedal
Equation or (p, r) equation of the curve.
92 KSOU Differential Calculus
Let PT1 & PT2 be the tangents to the two curves and let φ1 & φ 2 bet the angles
made by the tangents with the radius vector OP. P
f1
∴ Angle between tw o curves is given by φ1 − φ 2 f2
f1 -f2
tanφ1 − tanφ 2
Now tan(φ1 − φ 2 ) = (5)
1 + tanφ1 tanφ 2
using the result in (3) we can find tanφ1 & tanφ 2 and hence angle between two T1
T2
curves at the point of intersection can be found out.
If tanφ1 = tanφ 2 then tan(φ1 − φ 2 ) = 0
ie φ1 = φ 2 ∴ the two curves touch each other at P.
π
If tanφ1 tanφ 2 = −1 then φ1 − φ 2 =
2
∴ the two curves are said to intersect orthogonally.
Examples
(1) Show that in the equiangular spiral r = aeθ cotα the tangent is inclined at a constant angle to the radius vector.
Solution : r = aeθ cotα
dr
differentiating w.r.t. θ , = aeθ cot α cot α = r cot α
dθ
dθ r r
tanφ = r = = = tanα ⇒ φ = α hence the result.
dr dr r cot α
dθ
π
( 2) Show that the tangent to the Cardiod r = a (1 + cos θ ) at the point θ = is parallel to the initial line.
3
Solution : r = a (1 + cos θ )
dr
differentiating w.r.t. θ , = a (0 − sin θ )
dθ
θ
2 cos 2
dθ r a (1 + cos θ ) 2 = − cot θ = tan π + θ
tanφ = r = = =
dr dr − a sinθ θ θ 2 2 2
− sin cos
dθ 2 2
π θ π π π π π π
∴φ= + when θ = , φ= + & ψ =θ +φ = + + = π
2 2 3 2 6 3 2 6
∴ the tangent is parallel to the initial line.
2a θ θ
(3) For the curve = 1 − cos θ show that (i ) φ = π − and (ii ) p = a cosec
r 2 2
2a
Solution : = 1 − cos θ
r
MCA 11 - Mathematics SVT 93
2a dr dr − r 2 sin θ
differentiating w.r.t. θ , − = sin θ ∴ =
r dθ
2
dθ 2a
θ
2 sin 2
dθ r r 2a 1 − cosθ 2 = − tanθ
Now tanφ = r = = =− = =
dr dr − r 2 sinθ r sinθ − sinθ θ θ 2
− sin cos
dθ 2a 2 2
θ θ θ θ θ
ie tanφ = − tan = tan π − ∴ φ = π − & p = r sin φ = r sin π − = r sin
2 2 2 2 2
θ 2a θ 2a a θ
ie p = sin ⋅ = sin ⋅ = ∴ p = a cosec
2 1 − cos θ 2 2 sin 2 θ sin θ 2
2 2
(4 ) Find the angle between the curves r = 2 sinθ & r = sinθ + cos θ .
dr
Solution : r = 2 sinθ , = 2 cos θ
dθ
Let φ1 be the angle made by the tangent at the point of intersection with radius vector
dθ r 2 sinθ
∴ tanφ1 = r = = = tanθ
dr dr 2 cos θ
dθ
∴ φ1 = θ (1)
for the curve r = sin θ + cos θ
dr
= cos θ − sinθ
dθ
Let φ 2 be the angle w.r.t. this curve
(5) Prove that the curves r = a (1 + cos θ ) & r = b(1 − cos θ ) intersect at right angles.
dθ r a (1 + cos θ ) (1 + cos θ )
∴ tanφ1 = r = = =
dr dr − a sinθ − sinθ (1)
d θ
94 KSOU Differential Calculus
dr
Consider r = b(1 − cos θ ), = b sinθ
dθ
2a
(6) Find the pedal equation of = 1 + cos θ
r
2a
Solution : = 1 + cos θ
r
differentiating w.r.t. θ
2a dr
− = − sinθ
r 2 dθ
1 dr sinθ
ie = (1)
r 2 dθ 2a
2 2
1 1 1 dr 1 1 dr 1 sin 2 θ
Now = + = 2 + 2 = 2+ using (1)
p2 r 2 r 4 dθ r r dθ r 4a 2
2a 2 1
=
1
+
1
[1 − cos θ ]= r1 + 4a1
2
1 −
1 4a 2 4a
− 1 = 2 + 2 1 − 2 − 1 +
r2 4a 2 r r
2 2
4a r r
1 1 1 − 4a 2 4a 1 1 1 1
ie = + 2
+ = 2 − 2 + =
p2 r2 4a r 2 r r r ar ar
Exercise
3θ
1. Find the length of the perpendicular from the pole on the tangent to the curve. r = a (1 − cosθ ) Ans : 2a sin
2
π
3. Find the angle of intersection of the curves r = 2 sinθ , r = 2 cos θ Answer :
2
4. Show that the curves r = a (1 + sinθ ) & r = a (1 − sinθ ) intersect orthogonally.
Indeterminate Forms
f ( x) 0 ∞
While evaluating lim or lim[ f ( x) − g ( x) ] or lim f ( x) g ( x ) when it takes the forms , , ∞ − ∞, 1∞ , ∞°, 0° they
x→a g ( x) x→a x→a 0 ∞
are called Indeterminate forms and to evaluate such forms the following rule known as L' Hospital's Rule is used.
f ( x) 0 ∞ f ( x) f ′( x) 0 ∞
If lim is of the form or then lim = lim again if this is of the form or then
x→a g ( x) 0 ∞ x → a g ( x) x → a g ′( x ) 0 ∞
f ( x) f ′′( x) 0 ∞
lim = lim whenever it is of the or This rule can be applied.
x→a g ( x) x → a g ′′( x) 0 ∞
1 1
−
g ( x) f ( x)
Consider lim [ f ( x) − g ( x)] = lim
0
which is of the form & hence L' Hospital' s Rule can be applied.
x→a x→a 1 0
g ( x) f ( x)
Consider lim f ( x ) g ( x ) = y (say)
x →a
log f ( x)
then log y = lim log f ( x) g ( x ) = lim g ( x) log f ( x) = lim
x →a x →a x →a 1
g ( x)
0 ∞
which is of the form or and hence L' Hospital' s rule can be applied.
0 ∞
Examples
log x
(1) Evaluate lim
x →1 x 2 − 3x + 2
log x 0
Solution : lim this is of the form ∴ using L' Hospital' s rule
x →1 x − 3x + 2
2
0
1
1
= lim x = = −1
x →1 2 x − 3 2−3
2 sin x − sin 2 x
(2) Evaluate lim
x →0 x3
2 sin x − sin 2 x 0
Solution : lim This is of the form ∴ applying L' Hospital' s rule
x →0 x3 0
2 cos x − 2 cos 2 x 0
= lim again it is of the form ∴ applying L' Hospital' s rule again
x →0 3x 2 0
1 4
= − + =1
3 3
96 KSOU Differential Calculus
tan x − x
(3 ) Evaluate lim
x→0 x 2 tan x
tan x − x tan x − x x tan x − x x
Solution : lim = lim × = lim Q lim =1
x →0 x 2 tan x x →0 x3 tan x x → 0 x 3 x → 0 tan x
0
This is of the form ∴ using L' Hospital' s rule
0
1 1
(4) Evaluate lim − x
x→0 x e − 1
Solution : This is of the form ∞ − ∞
1 1 (e x − 1) − x 0
∴ lim − x = lim which is of the form ∴ using L' Hospital' s rule
x →0 x e − 1 x → 0 (e − 1) x
x
0
ex −1
= lim
x →0 e x
− 1 + xe x
0
which is again of the form ∴ using the rule again
0
ex 1 1
= lim = =
x→0 e x + xe + e x x
1+ 0 +1 2
x 1
(5) Evaluate lim −
x →1 x − 1 log x
Solution : This of the form ∞ − ∞
x −1
x log x −
1 x
∴ lim − = lim
x →1 x − 1 log x
x →1 x − 1
log x
x
0
This is of the form ∴ using L' Hospital' s rule
0
1 1
−
x x2 x −1 0
= lim = lim is form again applying the L' Hospital' s rule
x →1 1 x − 1 1 x →1 log x + x − 1 0
log x × 2 + ×
x x x
1 1
= lim =
x →1 1
+1 2
x
log tan x
log y = lim log(tan x) cot x = lim cot x log tan x = lim using L' Hospital' s rule
x→
π
x→
π
x→
π tan x
2 2 2
1
⋅ sec 2 x
= lim tan x = lim
1
= lim cot x = 0
x → tan x
π sec x2 π π
x→ x→
2 2 2
∴ y = e0 = 1
1
sin x x
( 7) Evaluate lim
x →0 x
sin x
log
1 sin x x
log y = lim log = lim
x →0 x x x →0 x
0
This is of the form ∴ applying the L' Hospital' s rule
0
x x cos x − sin x
×
sin x x2 x x cos x − sin x x cos x − sin x
= lim = lim × = 1 × lim
x →0 1 x →0 sin x x 2 x → 0 x2
0
This is of the form ∴ again applying the L' Hospital' s rule
0
cos x − x sin x − cos x − x sin x − sin x
= lim = lim = lim =0
x →0 2x x → 0 2x x →0 2
ie log y = 0 ∴ y = 1
a sin x − sin 2 x
(8) If lim is finite, find the value of a and the limit.
x →0 tan3 x
0
Solution : The given limit is of the form ∴ applying the L' Hospital' s rule
0
a cos x − 2 cos 2 x
= lim
x →0 3 tan2 x ⋅ sec 2 x
0
limit exists if this is of the form
0
∴ a cos x − 2 cos 2 x = 0 for x = 0 ∴ a = 2
Exercise
Evaluate the following
cosh x − cos x e x − e sin x
(1) lim ( 2) lim
x →0 x sin x x →0 x − sin x
e x − sin x − 1 xe x − log(1 + x)
(3) lim ( 4) lim
x →0 log(1 + x) x →0 x2
1 1 1 1
(5) lim 2 − ( 6) lim −
x →0 x sin 2 x x →0 sin x
x
1 1
(7) lim(cos x) x2 (8) lim( x) 1− x
x →0 x →1
1
1
ax + bx + cx x
(9) lim(cot x) log x
(10) lim
x →0 x →0 3
3 1 1 1 1 1
Answers : (1) 1 (2) 1 (3) 2 (4) (5) − (6) 0 (7) (8) (9) (10) ( abc) 3
2 3 e e e
MCA 11 - Mathematics SVT 99
Partial Derivatives
A function of two independent variables and a dependent variable is denoted as z = f ( x, y ) which is explicit function where
x & y are independent variables and z a dependent variable. Implicit function is denoted by φ ( x, y, z ) = C
f ( x + δx, y ) − f ( x, y ) ∂z ∂f
If lim exists then it is called Partial derivative of z or f w.r.t. x and denoted by or
δ x →0 δx ∂x ∂x
f ( x, y + δy ) − f ( x, y ) ∂z ∂f
If lim exists then it is called Partial derivative of z or f w.r.t. y and denoted by or
δy →0 δy ∂y ∂y
∂z ∂z
while obtaining the derivative differentiate the given function w .r.t. x treating y as a constant and while finding
∂x ∂y
differentiate the given function with respect to y, treating x as a constant.
∂z ∂z
Eg. (1) If z = x 2 + xy − y 2 then = 2x + y + 0 = 2x + y & = x − 2y
∂x ∂y
∂z ∂z
( 2) If z = x 2 y − x sin xy then = 2 xy − x cos xy ⋅ y − sin xy & = x 2 − x cos xy ⋅ x = x 2 (1 − cos xy)
∂x ∂y
2 xy ∂ ( x 2 − y 2 ) 2 y − 2 xy ⋅ 2 x
(3) If z = tan−1 2 then z = 1
×
x −y
2
∂x
1+
4x2 y 2 (x 2
− y2 ) 2
(x 2
− y2 ) 2
=
(x 2
− y2 ) 2
×
2x2 y − 2 y3 − 4x2 y
=
− 2 y3 − 2x2 y
=
− 2 y( y 2 + x 2 )
=
− 2y
(x 2
−y 2 2
) + 4x y 2 2
(x 2
−y 2 2
) (x 2
+y )
2 2
(x 2
+y )
2 2 x + y2
2
∂z 1 ( x 2 − y 2 )(2 x) − 2 xy (−2 y )
& = ×
∂y
1+
4x y 2 2
(x 2
− y2 ) 2
(x 2
− y2 ) 2
=
(x 2
− y2 ) 2
×
2 x 3 y − 2 xy 2 + 4 xy 2
=
2 x 3 + 2 xy 2
=
2 x( x 2 + y 2 )
=
2x
(x 2
−y 2 2
) + 4x y 2 2
(x 2
−y 2 2
) (x 2
+y )
2 2
(x 2
+y )
2 2 x + y2
2
Successive derivatives
∂z ∂z
For the function z = f ( x, y ) & are first order partial derivatives, the second order partial derivatives are
∂x ∂y
∂2z ∂2z
In example (2) = 2 x + x 2 y sin xy − x cos xy − x cos xy & = 2 x + x 2 y sin xy − 2 x cos xy
∂y∂x ∂x∂y
100 KSOU Differential Calculus
∂2z ∂2z
∴ =
∂x∂y ∂y∂x
∂2z ∂ − 2y ( x 2 + y 2 )(−2) + 2 y ⋅ 2 y − 2 x 2 − 2 y 2 + 4 y 2 − 2( x 2 − y 2 )
= = = =
( ) ( ) ( )
In example (3)
∂y∂x ∂y x 2 + y 2
x2 + y2
2
x2 + y2
2
x2 + y2
2
∂2z ∂ 2x ( x 2 + y 2 ) 2 − 2 x ⋅ 2 x 2 y 2 − 2 x 2 − 2( x 2 − y 2 )
= 2 = = =
( ) ( ) ( )
and
∂x∂y ∂x x + y 2 x2 + y2
2
x2 + y2
2
x2 + y 2
2
∂2z ∂2z
Thus in general, always =
∂x∂y ∂y∂x
Exercise
∂z ∂z ∂2z ∂2z
(1) Find , for z = log( x 2 + y 2 ) and show that =
∂x ∂y ∂x∂y ∂y∂x
∂2z ∂2z
( 2) If x = f ( x + ct ) + φ ( x − ct ) show that = c2 where c is a constant.
∂t 2 ∂x 2
∂z ∂z
(3) If z = e ax +by f (ax − by ) then show that b +a = 2abz
∂x ∂y
2
x2 + y2 ∂u ∂u ∂u ∂u
( 4) If u = then show that − = 41 − −
x+ y ∂x ∂y ∂x ∂y
y ∂ 2u ∂ 2u
(5) If u = sin −1 then show that = .
x ∂x∂y ∂y∂x
105 KSOU Matrix Theory
INTEGRAL CALCULUS
dy
Given = f ( x), the process of finding y is called 'Integration' and the resulting function is called 'Integral'. If g (x) is
dx
the integral then ∫ f ( x)dx = g ( x) is the notation used to represent the process.
In the above notation f (x) is called 'Integrand' and further
d
[g ( x)] = f ( x).
dx
But
d
[g ( x) + c] = g ′( x) when c is a constant ∴ ∫ f ( x)dx = g ( x) + c
dx
Thus integral of a function is not unique and two integrals always differ by a constant.
Properties
x n+1 d x n +1
1. ∫ x n dx =
n +1
+ c (n ≠ − 1) Q
+ c = xn
dx n + 1
∫ x dx = log
1 d 1
2. e x+c Q (log x + c) =
dx x
ax d ax
3. ∫ a x dx = +c Q + c = a x in particular ∫ e dx = e +c
x x
log a
dx log a
4. ∫ sin x dx = − cos x + c
5. ∫ cos x dx = sin x + c
6. ∫ sec x tanx dx = sec x + 1
7. ∫ cosec x cot x dx = −cosec x + c
8. ∫ sinh x dx = cosh x + c
9. ∫ cosh x dx = sinh x + c
10. ∫ sech x tanhx dx = −sech x + c
102 KSOU Integral Calculus
∫
1
13. dx = sin −1 x + c or − cos −1 x + c
1− x 2
∫
1
14. dx = sinh−1 x + c
1+ x 2
∫
1
15. dx = cosh −1 x + c
x −12
∫x
1
16. dx = sec −1 x + c or − cosec −1 x + c
x −12
∫x
1
17. dx = −sech −1 x + c
1− x 2
∫x
1
18. dx = −cosech −1 x + c
1+ x 2
Methods of Integration
There are two methods (1) Integration by substitution & (2) Integration by parts.
1. Integration by substitution
∫ f ( x) dx
dx
Consider put x = φ (t ) then = φ ′(t ) ie dx = φ ′(t ) dt
dt
∴ ∫ f ( x) dx = ∫ f [φ (t )]φ ′(t ) dt
now for the new integrand, we can use the standard forms, ie. we have to make a proper substitution so that the given
integrand reduced to a standard one.
Examples
sin x
1. ∫ tan x dx = ∫ cos x dx
dt
put cos x = t , then − sin x = ie sin x dx = −dt
dx
∫ tan x dx = ∫ − t
dt
∴ = − log t = − log cos x = log sec x + c
tan x sec x
or ∫ tan x dx = ∫ sec x
dx
dt
sec x tan x = ∴ sec x tan x dx = dt
dx
MCA 11 - Mathematics SVT 103
∫ tan x dx = ∫ t
dt
∴ = log t = log sec x + c
cos x
2. ∫ cot x dx = ∫ sin x dx
dt
put sin x = t , differentiating w.r.t. x cos x = ie cos x dx = dt
dx
∫ cot x dx = ∫ t
dt
∴ = log t = log sin x + c
2. Integration by parts
d dv du
If u & v are functions of x, we know that, (uv) = u +v
dx dx dx
∴ By definition of Integration
dv du
∫ ∫ ∫
dv du
uv = u + v dx = u dx + v dx using property (1)
dx dx dx dx
∫ u dx dx = uv − ∫ v dx dx
dv du
∴
The result can be used as the standard result. Out of the two functions of the product, one has to be taken as u & another
dv
then the RHS after evaluation gives the integral or if both functions have taken as u & v then the result is as follows
dx
du
∫ ∫ ∫
uv dx = u v dx − ⋅ v dx dx
dx ∫
any one form can be used depending on convenience. The first one can also be written as ∫ uv′ dx = uv − ∫ u′v dx
Examples
1. ∫ xe dx put u = v, v′ = e x , u ′ = 1, v = e x
x
∴ ∫ xe ∫ ∫
dx = uv − u ′v dx = xe x − 1 ⋅ e x dx = xe x − e x + c
x
∴ ∫ uv′ dx = uv − ∫ u′v dx
∫ log x dx = x log x − ∫ x ⋅ x dx = x log x − ∫1 ⋅ dx = x log x − x + c
1
ie
∫ sin
−1 1
4. x dx put u = sin −1 x, v′ = 1, u ′ = , v=x
1 − x2
∫ sin ∫
−1 1
∴ x dx = x sin −1 x − x dx
1 − x2
−x
to evaluate ∫ 1− x 2
dx put 1 − x 2 = t 2 differentiating w.r.t. x
− 2 x dx = 2t dt ⇒ − x dx = t dt
− x dx
∫ ∫ ∫
t dt
∴ = = 1 ⋅ dt = t = 1 − x 2
1− x 2
t 2
∫ sin
−1
∴ x dx = x sin −1 x − 1 − x 2 + c
∫a ∫x ∫a ∫ Ax
dx dx dx dx
(1) , ( 2) , (3) & ( 4)
2
+ x2 2
− a2 2
− x2 2
+ Bx + C
to evaluate (1) put x = at , dx = a dt
∫a ∫a ∫
dx a dt 1 dt 1 1 x
∴ = = = tan−1 t = tan−1 + c
2
+ x2 2
+a t2 2
a 1+ t 2
a a a
to evaluate (2) & (3) use partial fractions
1 1 A B
= = + (Say)
x −a
2 2
( x + a )( x − a ) x + a x − a
multiply throughout by x 2 − a 2
1 = A( x − a) + B( x + a)
1
put x = a, 1 = 0 + B ⋅ 2a ∴ B =
2a
−1
put x = −a, 1 = A(−2a) + 0 ∴ A =
2a
−1 1
1 2 a
∴ 2 = + a
2
x − a2 x + a x − a
x−a
∫x
1 1
∫ x + a + 2a ∫ x − a = − 2a log( x + a) + 2a log( x − a) = 2a log x + a
dx 1 dx 1 1 1
∴ dx = −
2
−a 2
2a
x−a
∫x
dx 1
∴ = log +c
2
− a2 2a x+a
MCA 11 - Mathematics SVT 105
1 1 A B
next, = = + (Say)
x −a
2 2
( a + x)(a − x) a + x a − x
a+x
∫a
1 1
∫ a + x dx + 2a ∫ a − x dx = 2a log(a + x) − 2a log(a − x) = 2a log a − x
1 1 1 1 1 1
∴ dx =
2
−x 2
2a
a+x
∫a
1 1
∴ dx = log +c
−x a−x
2 2
2a
∫ Ax
dx
to evaluate (4) 2
+ Bx + c
∫x ∫ ∫
1 dx 1 dx 1 dx
G.I. = = 2
= 2
A B C A B 2 A B B 2 − 4 AC
2
+ x+ x + −
B
+
C
x + −
A A 2A 4 A2 A 2A 4 A2
This integral will take any one of (1), (2) or (3) and hence can be evaluated.
Examples
∫ 3x
dx
(1) Evaluate 2
− 2x + 4
dx 1 dx 1 dx 1 dx
Solution : ∫ 3x 2
− 2x + 4
=
3 ∫x 2 2
− x+
=
4 3 1
2∫ =
4 4 3 1
2
− 4 + 12
∫
3 3 x− − + x − +
3 9 3 3 9
1
x−
3 = 1 tan−1 3 x − 1 + c
∫ ∫
1 dx 1 dx 1 1
= 2
= 2
= × tan−1
3 8 2 3 2 2 2 3 2 2 2 2 2 2 2 2
+ x − 1 + x − 1
3 3 3 3 3 3
∫x
dx
( 2) Evaluate 2
− 10 x + 21
x −5−2 1 x−7
∫x ∫ ( x − 5) ∫ ( x − 5)
dx dx dx 1
Solution : = = = log = log +c
2
− 10 x + 21 2
− 25 + 21 2
−2 2
2× 2 x −5+ 2 4 x −3
dx
(3) Evaluate ∫ 6 − 4x − 2x 2
106 KSOU Integral Calculus
∫ 6 − 4x − 2x ∫ ∫ ∫
dx 1 dx 1 dx 1 dx
Solution : = = =
2
2 3 − ( x 2 + 2 x) 2 3 − ( x + 1) 2 + 1 2 2 2 − ( x + 1) 2
1 1 2 + ( x + 1) 1 3 + x
= × log = log +c
2 2× 2 2 − ( x + 1) 8 1 − x
Type II
∫ ∫ ∫ ∫
dx dx dx dx
(1) , ( 2) , (3) , (4)
a2 − x2 a2 + x2 x2 − a2 Ax 2 + Bx + C
∫
dx
to evaluate put x = a sinθ , dx = a cosθ dθ
a − x2
2
dx a cosθ dθ a cosθ dθ x
∴ ∫ a2 − x2
= ∫ a 2 − a 2 sin 2 θ
= ∫ a cosθ ∫
= 1 ⋅ dθ = θ = sin−1
a
∫
dx x
∴ = sin −1 +c
a2 − x2 a
∫
dx
to evaluate put x = a sinhθ , dx = a coshθ dθ
a + x2
2
dx a coshθ dθ a coshθ dθ x
∴ ∫ a +x2 2
= ∫ a + a sinh θ
2 2 2
= ∫ a coshθ ∫
= 1 ⋅ dθ = θ = sinh−1
a
∫
dx x
∴ = sinh−1 +c
a2 + x2 a
∫
dx
to evaluate put x = a coshθ , dx = a sinhθ dθ
x − a2
2
∫
dx x
∴ = cosh−1 +c
x −a2 2 a
dx a sinhθ dθ a sinhθ dθ x
∴ ∫ x2 − a2
= ∫ a 2 cosh2 θ − a 2
= ∫ a sinhθ ∫
= 1 ⋅ dθ = θ = cosh−1
a
∫
dx
to evaluate
Ax + Bx + C
2
A∫ A∫ A∫
1 dx 1 dx 1 dx
G.I. = = =
B C 2 2
x2 + x+ B B2 C B B 2 − 4 AC
A A x+ − + x + −
2 A 4A 2
A 2A 4 A2
This will reduce to any one of (1), (2) & (3) and hence can be evaluated.
Examples
∫
dx
(1) Evaluate
2x − 5x2
MCA 11 - Mathematics SVT 107
1
x−
∫ ∫ ∫
dx 1 dx 1 dx 1 5 = 1 × sin −1 (5 x − 1) + c
Solution : = = = sin −1
1 2 5 2 5 2 2 5 1 5
x − x2 − x2 − x 1 1
−x− 5
5 5 5
5 5
∫
dx
(2) Evaluate
x − 2x + 5
2
x −1
∫
dx
∫
dx
∫
dx
Solution : = = = sinh−1 +c
x2 − 2x + 5 ( x − 1) 2 + 2 2 2 2 + ( x − 1) 2 2
∫
dx
(3) Evaluate
4 x 2 − 12 x + 8
1 dx 1 dx 1 dx
Solution : G.I. =
4 ∫ x − 3x + 2
2
=
2 ∫ 3 9
2
=
2 ∫
2
3 9−8
x − − +2 x− −
2 4 2 4
3
x−
∫
1 dx 1 2 = 1 cosh −1 ( 2 x − 3) + c
= = cosh −1
2 2 2 2 1 2
3 1
x − − 2
2 2
Type III
px + q px + q
∫ Ax 2
+ Bx + C
dx and ∫ Ax 2 + Bx + C
to evaluate put px + q = l (derivative of Ax 2 + Bx + C ) + m = l (2 Ax + B) + m
where l & m are the constants to be found out by equating the co-efficients of corresponding terms on both sides. ie. to
solve for m & n from the equations
2 Al = p and lB + m = q
px + q 2 Ax + B
∫ Ax ∫ Ax ∫ ∫
dx dx
then dx = l dx + m = l ⋅ log( Ax 2 + Bx + C ) + m
2
+ Bx + c 2
+ Bx + C Ax 2 + Bx + C Ax 2 + Bx + C
the second integral in RHS is Type I and hence can be evaluated.
px + q px + q
∫ ∫ ∫ ∫
dx dl
dx = l dx + m = 2l Ax 2 + Bx + C + m
Ax + Bx + C
2
Ax + Bx + C
2
Ax + Bx + C
2
Ax + Bx + C
2
Examples
2x + 3
(1) Evaluate ∫ 3x 2
− 4x + 5
dx
2x + 3 1 6x − 4 13 dx 1 13 dx
∴ ∫ 3x 2
− 4x + 5
dx = ∫
3 3x 2 − 4 x + 5
dx + ∫ = log(3x 2 − 4 x + 5) +
3 3x 2 − 4 x + 5 3 9 ∫x 2 4
− x+
5
3 3
∫ ∫
1 13 dx 1 13 dx
= log(3 x 2 − 4 x + 5) + 2
= log(3 x 2 − 4 x + 5) + 2
3 9 2 4 5 3 9 2 11
2
x− − +
x− +
3 9 3 3 3
x− 2
1 13
= log(3 x 2 − 4 x + 5) + ×
3
tan−1 3 = 1 log(3 x 2 − 4 x + 5) + 13 tan−1 3 x − 2 + c
3 9 11 11 3 3 11 11
3
5x − 7
( 2) Evaluate ∫ 3x − x 2 − 2
dx
5x − 7 3 − 2x
∫ ∫ ∫
5 1 dx
∴ dx = − dx +
3x − x − 2 2 2 3x − x − 2
2 2 − 2 − ( x 2 − 3x)
1
∫ ∫
5 1 dx dx
= − ⋅ 2 3x − x 2 − 2 + = −5 3 x − x 2 − 2 +
2 2 2 2 2 2
3 9 1 3
−2−x− + −x−
2 4 2 2
3
1 x−
= −5 3 x − x 2 − 2 + × sin−1 2 = −5 3x − x 2 − 2 + 1 sin −1 (2 x − 3) + c
2 1 2
2
Type IV
∫ a cos x + b sin x + c
dx
x 1 x dt
to evalute put tan = t then differentiating w.r.t. x sec 2 =
2 2 2 dx
x x t 1 2t
& sin x = 2 sin cos = 2 × =
2 2 1+ t 2
1+ t 2 +
1 t2
x 2dt 1− t2 2t
∴ when tan = t , dx = , cos x = & sin x =
2 1+ t2 1+ t2 1+ t2
MCA 11 - Mathematics SVT 109
2dt
(1 + t 2 )
∫ a cos x + b sin x + c = ∫ ∫ a(1 − t ∫
dx 2dt 2dt
∴ = =
(1 − t 2 ) 2t 2
) + 2bt + c(1 + t 2 ) (c − a )t 2 + 2bt + a + c
a +b +c
(1 + t 2 ) (1 + t 2 )
which is Type I and hence can be evaluated.
Examples
∫ 2 cos x − 3 sin x + 5
dx
(1) Evaluate
x 2dt 1− t2 2t
Solution : Put tan = t , then dx = , cos x = & sin x =
2 1+ t 2
1+ t 2
1+ t2
2dt
1+ t2 2
∫ ∫ 2(1 − t ∫ 3t ∫
2dt 2dt dt
G.I. = = = =
2(1 − t 2 ) 3 × 2t 2
) − 6t + 5(1 + t 2 ) 2
− 6t + 7 3 t − 2t +
2 7
− +5
1+ t2 1+ t2 3
−1 − = 1 tan−1 3 (t − 1)
∫ ∫
2 dt 2 dt 2 1 t 1
= = = × tan
3 (t − 1) 2 − 1 + 7 3 2
2
3 2 2 3 2
(t − 1) 2 +
3
3
3 3
3 x
∫ 2 cos x − 3 sin x + 5 =
dx 1
∴ tan−1 tan − 1 + c
3 2 2
∫ 3 − 5 cos x
dx
(2) Evaluate
x 2dt 1− t2
Solution : Put tan = t , then dx = & cos x =
2 1+ t 2
1+ t2
2dt
(1 + t 2 )
∫ 3 − 5 cos x = ∫ ∫ 3(1 + t ∫ ∫ ∫
dx 2dt 2dt 2 dt 1 dt
∴ = = = =
5(1 − t 2 ) 2
) − 5(1 − t 2 ) 8t 2
− 2 8 t −
2 2 4 2 1
2
3− t −
1+ t2 8 2
1 x 1
t− tan −
1 1 1
2 = log 2 2
= × log
4 2× 1 t+
1 4 x 1
tan +
2 2 2 2
x
2 tan − 1
∫
dx 1 2
∴ = log +c
3 − 5 cos x 4 x
2 tan + 1
2
∫ 3 + 2 sin x
dx
(3) Evaluate
x 2dt 2t
Solution : Put tan = t , then dx = and sin x =
2 1+ t 2
1+ t2
110 KSOU Integral Calculus
2dt
1+ t2 =
∫ ∫ ∫ ∫ ∫
dx 2dt 2 dt 2 dt
∴ = = =
3 + 2 sin x 3+
4t 3(1 + t ) + 4t
2
3 4
1+ t2 + t 3 2
2
4
t + − +1
1+ t 2
3 3 9
2
t+
3 = 2 tan−1 3t + 2
∫
2 dt 2 1
= 2
= × tan−1
3 2 5 3
2 5 5 5 5
t + +
3 3 3 3
x
3 tan + 2
∫
dx 2 −1 2 +c
∴ = tan
3 + 2 sin x 5 5
Type V
a cos x + b sin x
∫ c sin x + e cos x dx
Solution : to evaluate put a cos x + b sin x = l (Denominator) + m(derivative of denominator)
where l & m are constants to be foundout by equating the co - efficients of sin x & cos x separately.
ie from the equations lc − me = b & le + mc = a
a cos x + b sin x c sin x + e cos x c cos x − e sin x
then ∫ c sin x + e cos x dx = l ∫ c sin x + e cos x dx + m∫ c sin x + e cos x dx = lx + m log(c sin x + e cos x) + c
Examples
3 cos x − 2 sin x
Evaluate ∫ 4 sin x + cos x dx
Solution : Put 3 cos x − 2 sin x = l (4 sin x + cos x) + m(4 cos x − sin x)
∴ 4l − m = −2 (1)
l + 4m = 3 (2)
(1) × 4 16l − 4m = −8
(2) × 1 l + 4m = 3
5
adding 17l = −5 ⇒ l = −
17
20 − 20 + 34 14
from (1), m = 4l + 2 = − +2= =
17 17 17
3 cos x − 2 sin x 5 4 sin x + cos x 14 4 cos x − sin x 5 14
∴ ∫ 4 sin x + cos x dx = − 17 ∫ 4 sin x + cos x dx + 17 ∫ 4 sin x + cos x = − 17 ∫1⋅ dx + 17 log(4 sin x + cos x)
5 14
=− x + log(4 sin x + cos x) + c
17 17
MCA 11 - Mathematics SVT 111
Type VI
∫ f ( x) e dx where f ( x) = φ ( x) + φ ′( x)
x
∫ f ( x) e ∫
dx = φ ( x) e x dx + φ ′( x) e x dx ∫
x
Solution : (1)
∫ φ ( x) e
x
Consider dx
∫
∴ φ ( x) e x dx = φ ( x) e x − φ ′( x) e x dx ∫
substituting this in (1), we have
∫ f ( x) e ∫ ∫
dx = φ ( x) e x − φ ′( x) e x dx + φ ′( x) e x dx = φ ( x) e x + c
x
Examples
xe x
(1) Evaluate ∫ (1 + x) 2
dx
xe x (1 + x − 1) e x (1 + x) e x ex ex ex
Solution : ∫ (1 + x) 2
dx = ∫ (1 + x) 2
dx = ∫ (1 + x) 2
dx − ∫ (1 + x) 2
dx = ∫ 1+ x
dx −∫(1 + x) 2
dx (1)
ex
Consider ∫ (1 + x)
dx
1 −1
put u = , v′ = e x , u ′ = , v = ex
1+ x (1 + x) 2
ex ex − ex ex ex
∴ ∫ (1 + x)
dx =
(1 + x)
−
(1 + x) 2
dx =
(1 + x)∫+
(1 + x) 2
dx ∫
substituting in (1)
xe 2 ex ex ex ex
∫ (1 + x) 2
dx =
(1 + x)
+
(1 + x) 2 ∫
dx −
(1 + x) 2
dx = ∫
1+ x
+c
x − sin x
( 2) Evaluate ∫ 1 − cos x dx
x x
2 sin cos
x − sin x sin x
∫ ∫ ∫ ∫ ∫
x x 2 2 dx
Solution : dx = dx − dx = dx −
1 − cos x 1 − cos x 1 − cos x 2 sin 2 x
2 sin 2 x
2 2
∫ 2 x cosec ∫
1 x x
= 2
dx − cot dx (1)
2 2
∫ 2 x cosec x dx
1 2
Consider
1 x x
put u = x, v′ = cosec 2 , u ′ = 1, v = − cot
2 2 2
112 KSOU Integral Calculus
∫ 2 x cosec ∫
1 x x x
2
dx = − x cot + cot dx
2 2 2
substituting in (1)
x − sin x x x x x
∫ 1 − cos x dx = − x cot 2 + ∫ cot 2 dx − ∫ cot 2 dx = − x cot 2 + c
Other examples
sin x cos x
(1) Evaluate ∫ 1 + sin 4
x
dx
1
dt
sin x cos x
∫ ∫
1 1
∴ dx = 2 = tan−1 t = tan−1 (sin 2 x) + c
1 + sin 4 x 1+ t2 2 2
x2 + 1
(2) Evaluate ∫ ( x + 1)( x 2 + 2)
dx
x2 + 1 A Bx + C
Solution : Let = + 2
( x + 1)( x + 2) x + 1 x + 2
2
then x 2 + 1 = A( x 2 + 2) + ( Bx + C )( x + 1)
2
put x = −1, 2 = A(1 + 2) + 0 ⇒ A =
3
4 1
put x = 0, 1 = 2 A + C ⇒ C = 1 − =−
3 3
Equating co - efficient of x 2 on both sides
2 1
A + B = 1⇒ B =1− A = 1− =
3 3
2 1 1
x−
x2 + 1
∴ = 3 +3 3
( x + 1)( x 2 + 2) x + 1 x 2 + 2
x2 + 1 x −1
∫ ∫ x +1 + 3 ∫ x ∫ x +1 + 6 ∫ x ∫
2 dx 1 2 dx 1 2x 1 dx
∴ dx = dx = dx −
( x + 1)( x 2 + 2) 3 2
+2 3 2
+2 3 x2 + 2
2 1 1 x
= log( x + 1) + log( x 2 + 2) − tan−1 +c
3 6 3 2 2
Type VII
(1) ∫ a 2 − x 2 dx ( 2) ∫ a 2 + x 2 dx (3) ∫ x 2 − a 2 dx ( 4) ∫ Ax 2 + Bx + C dx
(5) ∫ ( px + q) Ax 2 + Bx + C dx
MCA 11 - Mathematics SVT 113
a2
∫ a 2 − x 2 dx = ∫ ∫ ∫
a 2 − a 2 sin 2 θ ⋅ a cosθ dθ = a cosθ ⋅ a cosθ dθ = a 2 cos 2 θ dθ =
2 ∫
(1 + cos 2θ ) dθ
a2 a 2 sin 2θ a 2 a2 a2 a2
= θ+ × = θ+ sinθ cosθ = θ + ⋅ sinθ ⋅ 1 − sin 2 θ
2 2 2 2 2 2 2
a2 x a2 x x2 a2 x a2 x
= sin −1 + ⋅ 1− 2 = sin −1 + ⋅ a2 − x2
2 a 2 a a 2 a 2 a2
a2
∫
x 2 x
∴ a 2 − x 2 dx = a − x2 + sin −1 + c
2 2 a
a2
∫ a 2 + x 2 dx = ∫ ∫
a 2 + a 2 sinh2 θ ⋅ a coshθ dθ = a 2 cosh2 θ dθ =
2 ∫ (1 + cosh 2θ ) dθ
a2 a2 a2 a 2 sinh 2θ a 2 a2
=
2 ∫
1 ⋅ dθ +
2 ∫ cosh 2θ dθ =
2
θ+
2
⋅
2
= θ+
2 2
× sinhθ coshθ
a2 a2 a2 x a2 x x2 a2 x x
= θ+ sinhθ 1 + sinh2 θ = sinh−1 + ⋅ 1+ 2 = sinh−1 + a2 + x2
2 2 2 a 2 a a 2 a 2
a2
∫
x 2 x
∴ a 2 + x 2 dx = a + x2 + sinh−1 + c
2 2 a
a2
∫ x 2 − a 2 dx = ∫ ∫ ∫
a 2 cosh2 θ − a 2 ⋅ a sinhθ dθ = a sinhθ ⋅ a sinhθ dθ = a 2 sinh2 θ dθ =
2 ∫ (cosh 2θ − 1) dθ
a2 a2 a 2 sinh 2θ a 2 a2 a2
∫ ∫
x
= cosh 2θ dθ − 1 ⋅ dθ = ⋅ − θ= sinhθ coshθ − cosh−1
2 2 2 2 2 2 2 a
a2 a2 x a2 x2 x a2 −1 x x a2 x
= cosh 2 θ − 1 coshθ − cosh −1 = 2
− 1 ⋅ − cosh = x 2
− a 2
− cosh −1
2 2 a 2 a a 2 2 2 2 a
a2
∫
x 2 x
∴ x 2 − a 2 dx = x − a2 − cosh−1 + c
2 2 a
∫ ∫
B C
( 4) To evaluate Ax 2 + Bx + C dx = A x2 + x + dx. This will take the form (1), (2) or (3) and hence can be
A A
evaluated.
(5) To evaluate ∫ ( px + q) Ax 2 + Bx + C dx
put px + q = l (derivativeof Ax 2 + Bx + C ) + m = l (2 Ax + B) + m
where l & m are constants to be found out,
then, ∫ ( px + q) ∫
Ax 2 + Bx + C dx = l ( 2 Ax + B) Ax 2 + Bx + C dx + m ∫ Ax 2 + Bx + C
114 KSOU Integral Calculus
Β
∫
2l 3 C
= ( Ax 2 + Bx + C ) 2 + m A x2 + x + dx
3 A A
the second integral reduces to (1), (2) or (3) and hence can be evaluated.
Type VIII
∫ ( px + q)
dx
Ax 2 + Bx + C
1 −1 1 1
put px + q = then p dx = 2 dt & x = − q
t t pt
dt
−
− dt
∫ ( px + q) ∫1 t2
∫
dx
∴ = =
Ax 2 + Bx + C A 1
2
Β 1 A B
(1 − tq ) 2 + (t − qt 2 ) + Ct 2
2
− q + − q + C p 2
p
t p t pt
This integral reduces to any one of Type II and hence can be solved.
Type IX
∫
Let C = e ax cos(bx + c) dx & S = e ax sin(bx + c) dx∫
∫
Consider C = e ax cos(bx + c) dx
sin(bx + c )
put u = e ax , u ′ = ae ax , v′ = cos(bx + c), v =
b
e ax sin(bx + c) a ax
C=
b
−
b ∫
e sin(bx + c) dx
bC = e ax sin(bx + c) − aS
∴ aS + bC = e ax sin(bx + c) (1)
∫
Consider S = e ax sin(bx + c) dx
− cos(bx + c)
put u = e ax , u ′ = ae ax , v′ = sin(bx + c), v =
b
− e ax cos(bx + c) a ax
∴S=
b
+
b ∫
e cos(bx + c) dx
bS = −ea ax cos(bx + c) + aC
ie bS − aC = −e ax cos(bx + c) (2)
Examples
∫e ∫ ∫ ∫
1 2x 1 2x 1 2x
(1) 2x
sin 3x cos 2 x dx = e [sin 5 x + sin x] dx = e sin 5 x dx + e sin dx
2 2 2
1 2 x (2 sin 5 x − 5 cos 5 x) 1 2 x (2 sin x − cos x)
= e + e +c
2 29 2 5
1 e 3x 1 3 x (3 cos 2 x + 2 sin 2 x)
∫ ∫ ∫ ∫
1 3x 1 3x 1 3x
(2) e 3x cos 2 x dx = e (1 + cos 2 x) dx = e dx + e cos 2 x dx = × + e
2 2 2 2 3 2 9+4
e3x e3x
= + (3 cos 2 x + 2 sin 2 x) + c
6 26
Exercise
Integrate the following w.r.t. x
−1
sin −1 x 1 esin x
(1) ( 2) (3)
1 − x2 x cos (log x)
2
1 − x2
sec 2 x 3x − 2 x
( 4) (5) (6)
tan x(2 + tan x) ( x + 1) 2 ( x + 3) ( x − 1)( x 2 + 4)
x3 − x − 2 4x + 5 4x + 1
(7) (8) (9)
x2 −1 x + 22 x + 2
2
x 2 − 6 x + 18
(1 + x) x 1 1
(10) e (11) (12)
( 2 + x) 2 2 + cos x − sin x 3 + 4 cos x
1 1
(19) (20) (21) e 2 x sin 4 x sin 2 x
( x + 1) 2 x + 3x + 4
2
( x + 1) x − 1 2
Definite Integrals
b
ie If ∫ f ( x) dx = g ( x) then ∫ f ( x) dx = g (b) − g (a)
a
Examples
2
∫ (x − 2 x 2 + 3) dx
3
(1) Evaluate
1
2
2 x4 x3 24 23 1 2
1
∫
16 1 2
Solution : ( x − 2 x + 3) dx = − 2 ×
3 2
+ 3 x = − 2× + 3 ⋅ 2 − − + 3 = 4 = +6− + −3
1
4 3 1 4 3 4 3 3 4 3
16 1 2 84 − 64 − 3 + 8 25
=7− − + = =
3 4 3 12 12
1 (sin −1 x) 2
( 2) Evaluate ∫0
1 − x2
dx
1
Solution : Put sin −1 x = t then dx = dt
1 − x2
π
when x = 0, t = sin −1 0 = 0 when x = 1, t = sin −1 1 =
2
π
1 (sin −1 π
t3 2 1 π
3
x) 2 π3
∴ ∫ dx = ∫
2 2
t dt = = =
0
1 − x2 0 3 0 3 2 24
1
∫
dx
(3) Evaluate dx
−1 x 2 + 2x + 5
1
1 1 x + 1 −1 1 + 1 1 −1 − 1 + 1
∫ ∫
dx dx 1 1
Solution : dx = = tan−1 = tan − tan
− x + 2x + 5
1 2 − ( x + 1) + 2
1 2 2
2 2 −1 2 2 2 2
1 1 1π π
= tan−1 1 − tan−1 0 = −0=
2 2 24 8
π
∫
dx
( 4) Evaluate
0 4 + 3 cos x
x 2dt 1− t2
Solution : Put tan = t then dx = , cos x =
2 1+ t2 1+ t2
π
when x = 0, t = tan0 = 0 when x = π , t = tan = ∞
2
MCA 11 - Mathematics SVT 117
2dt
∞
π ∞
1+ t2
∞ ∞ t
∫ ∫ ∫ ∫
dx 2dt 2dt 1 −1
= = = = tan
0 +
4 3 cos x 0
4+
3(1 − t 2 ) 0 4(1 + t ) + 3(1 − t 2 )
2 0
t2 + ( 7) 2
7
7 0
(1 + t 2 )
1 1 1 π π
= tan−1 ∞ − tan−1 0 = × =
7 7 7 2 2 7
b a
2. ∫a
f ( x) dx = − ∫ b
f ( x) dx
b c b
3. ∫a
f ( x) dx = ∫ a
f ( x) dx + ∫ f ( x) dx
c
where a < c < b
a a b b
4. ∫0
f ( x) dx = ∫ 0
f (a − x) dx also ∫a
f ( x) dx = ∫a
f ( a + b − x ) dx
2 ∫ af ( x ) dx
a 0
5. ∫−a
f ( x) dx =
0
if f ( x) is an even function
if f ( x) is an odd function
2 ∫ af ( x ) dx
2a 0 if f ( 2a − x) = f ( x )
6. ∫0
f ( x) dx =
0
if f (2a − x) = − f ( x)
Examples
π
sin n x
∫
2
(1) Evaluate dx
0 cos n x + sin n x
π
π π sin n − x
sin x n
2
∫ ∫ ∫
a
∫
a
Solution : Let I = dx = f ( x) dx = f ( a − x) dx
2 2
dx using
cos x + sin n x n π nπ
0 n 0 0 0
cos − x + sin − x
2 2
π
cos n x
= ∫
2
dx
0 sin n x + cos n x
π π π π π
sin n x cos n x (sin n x + cos n x) π
∴ 2I = ∫ 0
2
cos n x + sin n x
dx + ∫ 0
2
sin n x + cos n x
dx = ∫ 0
2
(cos n x + sin n x)
dx = ∫ 0
2
1 ⋅ dx = x ] 0
2
=
2
π
∴ I=
4
π x sin x
( 2) Evaluate ∫ 0 1 + sin x
dx
π x sin x π (π − x) sin(π − x) a a
Solution : Let I = ∫ 0 1 + sin x
dx = ∫ 0 1 + sin(π − x)
dx using ∫ 0
f ( x) dx = ∫0
f (a − x) dx
118 KSOU Integral Calculus
π (π − x) sin x
ie I = ∫0 1 + sin x
dx
π π π
∫ ∫ ∫ 1 dx = π [sec x − tan x + x] = π [sec π − tanπ + π ] − π [sec 0 − tan0 + 0]
π
= π sec x tan x − sec 2 x dx + 0
0 0 0
= π [− 1 + π ] − π [1] = π [− 2 + π ]
π
∴ I= (π − 2)
2
π
∫
3 dx
(3) Evaluate
π
6 1 + tan x
π
π π
cos − x dx
cos x dx 2 b b
Solution : Let I = ∫ = ∫ ∫ f ( x) dx = ∫ f (a − x) dx
3 3
using
π
cos x + sin x π π π a a
6 6 cos − x + sin − x
2 2
π
sin x dx
∴ I= ∫
3
dx
π
6 sin x + cos x
π
π cos x dx sin x dx π cos x + sin x π 3
∴ 2I = ∫ + dx = ∫ dx =
∫ 1 ⋅ dx = x
3 3 3
π
π cos x + sin x π
6 cos x + sin x sin x + cos x 6 6 π
6
π π π
= − =
3 6 6
π
∴ I=
12
Exercise
Evaluate the following
π
sin x e π
e tan x
∫ ∫ ∫
2 4
(1) dx ( 2) log x dx (3) dx
0 1 + cos 2 x 1 0 cos 2 x
1 1 1
∫ ∫ ∫
dx
x tan−1 x dx xe − x dx
2
( 4) (5) (6)
− 12 9 − x2 0 −1
π ∞ π
∫ ∫ ∫
dx x dx
log(1 + tanθ ) dθ
2 4
( 7) (8) (9)
0 a cos x + b 2 sin 2 x
2 2 0 ( x + 1)( x 2 + 1) 0
2a f ( x) π 1 log(1 + x)
∫ ∫ ∫
2 x dx
(10) dx (11) (12) dx
0 f ( x ) + f ( 2a − x ) 0 sin x + cos x 0 (1 + x 2 )
π 1 7 π 1 2 π π
Answers : (1) , (2) 1, (3) e − 1, (4) log , (5) − , (6) − , (7) , (8) ,
4 3 5 4 2 e 2ab 4
(9)
π
8
log 2 (10) a (11)
π
2 2
log 2 + 1 ( ) (12)
π
8
log 2
MCA 11 - Mathematics SVT 119
Reduction formulae
π
∫ ∫
2
I. To obtain the reduction formula for I n = sin n x dx and hence to evaluate sin n x dx
0
∫
Solution : I n = sin n−1 x ⋅ sin x dx
put u = sin n−1 x & v′ = sin x, u ′ = (n − 1) sin n−2 x cos x & v = − cos x
∫
∴ I n = − sin n−1 x cos x + (n − 1) sin n−2 x cos 2 x dx = − sin n−1 cos x + (n − 1) sin n−2 x(1 − sin 2 x) dx ∫
∫ ∫
= − sin n−1 x cos x + (n − 1) sin n−2 x dx − (n − 1) sin n x dx = − sin n−1 cos x + (n − 1) I n−2 − (n − 1) I n
If n is even I 0 = 1dx = x ∫ ∫
If n is odd I1 = sin x dx = − cos x
∫
2
If I n = sin n x dx then
0
π
− sin n −1 x cos x 2
n −1 n −1
In = + I n−2 = 0 + I n−2
n 0
n n
n −1 n −1 n − 3 n −1 n − 3 n − 5
∴ In = In−2 = × In−4 = × × I n−6
n n n−2 n n−2 n−4
in general
n −1 n − 3 1 π
n × n − 2 × LL × 2 × 2 if n is even.
In =
n −1 n − 3 2
× × LL × × 1 if n is odd.
n n−2 3
π
5 3 1 π 5π
∫
2
Eg. (1) I 6 = sin 6 x dx = × × × =
0 6 4 2 2 32
π
∫
2 4 2 8
( 2) I 5 = sin5 x dx = × ×1 =
0 5 3 15
∫ ∫
2
II. To obtain the reduction formula for I n = cos n dx and to evaluate cos n x dx
0
cos n −1 x sin x n − 1
In = + In
n n
π π
π a a
∫ ∫ ∫ ∫
2 2
and further if I n = cos n x dx = cos n − x dx using f ( x) dx = f (a − x) dx
0 0 2 0 0
120 KSOU Integral Calculus
∫
2
= sin n x dx which is I
0
∫
2 6 4 2 16
∴ Eg. (1) cos 7 x dx = × × ×1 =
0 7 5 3 35
π
7 5 3 1 π 35π
∫
2
( 2) cos 8 x dx = × × × × =
0 8 6 4 2 2 256
tann −1 x
In = − I n − 2 which is the required formula.
n −1
− cotn −1 x
In = − I n−2
n −1
∫ ∫
∴ I n = sec n− 2 tan x − (n − 2) sec n − 2 x ⋅ tan 2 x dx = sec n − 2 x tan x − (n − 2) sec n − 2 (sec 2 x − 1) dx
∫
= sec n − 2 x tan x − (n − 2) sec n x dx + (n − 2) sec n − 2 x dx ∫
∴ I n + (n − 2) I n = sec n − 2 x tan x + (n − 2) I n − 2
(1 + n − 2) I n = secn − 2 x tan x + (n − 2) I n − 2
sec n − 2 x tan x n − 2
∴ In = + I n − 2 which is the required reduction formula.
n −1 n −1
put u = cosecn − 2 x & v′ = cosec2 x, u′ = −(n − 2)cosecn − 3 x ⋅ cosec x cot x & v = − cot x
∫ ∫
∴ I n = −cosec n − 2 x cot x − (n − 2)cosec n − 2 x ⋅ cot2 x dx = −cosec n − 2 x cot x − (n − 2) cosec n − 2 x(cosec 2 x − 1) dx
∫ ∫
= −cosecn − 2 x cot x − (n − 2) cosecn x dx + (n − 2) cosec n − 2 x dx = −cosec n − 2 x cot x − (n − 2) I n + (n − 2) I n − 2
MCA 11 - Mathematics SVT 121
ie I n + (n − 2) I n = −cosecn − 2 x cot x + (n − 2) I n − 2
ie (n − 1) I n = −cosecn − 2 x cot x + (n − 2) I n − 2
− cosec n − 2 x cot x (n − 2)
ie I n = + I n − 2 which is the required reduction formula.
(n − 1) (n − 1)
∫ ∫
2
VII. To obtain the reduction formula of I m, n = sin m x cos n x dx and hence to evaluate sin m x cos n x dx
0
∫
Solution : I m, n = sin m x cos n −1 x ⋅ cos x dx
put u = sinm x cos n −1 x & v′ = cos x, u′ = m sinm −1 x ⋅ cos n x − (n − 1) sinm +1 x ⋅ cos n − 2 x & v = sin x
∫ ∫
= sin m +1 x cos n −1 x − m sin m x cos n x dx + (n − 1) sin m + 2 x ⋅ cos n − 2 x dx
∫
= sin m +1 x cos n −1 x − mI m, n + (n − 1) sin m x(1 − cos 2 x) cos n − 2 x dx
∫
= sin m +1 x cos n −1 x − mI m, n + (n − 1) (sinm x cos n − 2 x − sin m x cos n x) dx
= sin m +1 x cos n −1 x − mI m , n + (n − 1) I m, m − 2 − (n − 1) I m , n
ie I m , n + mI m, n + ( n − 1) I m , n = sin m +1 x cos x n −1 x + (n − 1) I m , n − 2
sin m +1 x cos n −1 x (n − 1)
ie I m, n = + I m, n − 2
( m + n) m+n
π
∫
2
which is the required reduction formula, if I m, n = sin m x cos n x dx
0
π
sin m +1 x cos n −1 x 2
n −1 n −1 n −1
then I m, n = + I m, n − 2 = 0 + I m , n − 2 ∴ I m, n = I m, n − 2
(m + n) m+n m+n m+n
0
n −1 n −3 2 1
m + n × m + n − 2 × LL × m + 3 × m + 1 if n is odd & m odd or even
n −1 n −3 1 m −1 m − 3 2
I m, n = × × LL × × × × LL × × 1 if n is even & m is odd
m + n m + n − 2 m+2 m m−2 3
n −1 n−3 1 m −1 m − 3 1 π
m + n × m + n − 2 × LL × m + 2 × m × m − 2 × LL × 2 × 2 if n is even & m is even
Examples
π
∫
2 4 2 1 1
(1) I 5, 5 = sin 5 x cos 5 x dx = × × =
0 10 8 6 60
π
∫
2 4 2 1 8
( 2) I 6, 5 = sin 6 x cos 5 x dx = × × =
0 11 9 7 693
122 KSOU Integral Calculus
∫
2 3 1 6 4 2 48
(3) I 7, 4 = sin 7 x cos 4 x dx = × × × × ×1 =
0 11 9 7 5 3 3465
π
5 3 1 5 3 1 π 5π
∫
2
( 4) I 6, 6 = sin 6 x cos 6 x dx = × × × × × × =
0 12 10 8 6 4 2 2 2048
1 x9
(5) Evaluate ∫
0
1 − x2
dx
π
Solution : put x = sinθ , dx = cosθ dθ when x = 0, θ = 0 when x = 1, θ =
2
1 x9 π
sin 9 θ ⋅ cosθ dθ π
sin 9 θ ⋅ cosθ dθ π
∫ ∫ ∫ ∫
8 6 4 2 128
dx = = = sin 9 θ dθ = × × × ×1 =
2 2 2
0
1− x 2 0
1 − sin θ 2 0 cosθ 0 9 7 5 3 315
2a x 3 dx
(6) Evaluate ∫
0
2ax − x 2
2a x 3 dx 2a x 3 dx
Solution : ∫0
2ax − x 2
= ∫
0
a 2 − ( x − a) 2
put x − a = a sinθ , dx = a cosθ dθ
π π
when x = 0, sinθ = −1 ⇒ θ = − when x = 2a, sinθ = 1 ⇒ θ =
2 2
π
( a + a sinθ ) 3 ⋅ a cosθ dθ π
a 3 (1 + sinθ ) 3 a cosθ dθ π
−π 2
a − a sin θ
2 2 2 −π 2 a cosθ −π 2
π π π π
π π
= a3 ∫ 1 dθ + ∫ sin 3 θ dθ + 3 ∫ sinθ dθ + 3 ∫ sin 2 θ dθ = a 3 θ ∫ +0+0+6 ∫ sin 2 θ dθ
2 2 2 2 2 2
−π 2 −π 2 −π 2 −π 2 −π 2 0
π π 1 π 3π 5π 3
= a 3 − + 6 × × = a 3 π + = a
2
2 2 2 2 2
Exercise
Evaluate the following
∫ ( )
π π 1
∫ ∫
3
sin5 3θ dθ
6
(1) ( 2) x sin 7 x dx (3) x4 1 − x2
2
dx
0 0 0
∞ 1 2
∫ (1 + x ) ∫ ∫
dx 5
( 4) 7
(5) x 6 1 − x 2 dx ( 6) x 2
2 − x dx
0 2 2 0 0
π π π 1 − cosθ
∫ ∫ ∫ sin 2 θ dθ
2 2
(7) cot 4 x dx (8) cosec 5 x dx (9)
π
4
π
6 0 1 + cosθ
1 x7 1 x3 a x 7 dx
(10) ∫
0
1 − x4
dx (11) ∫ (1 + x )
0 2 4
dx (12) ∫0
a2 − x2
Answers : (1)
8
45
, ( 2)
16π
35
, (3)
3π
256
, ( 4)
8
15
, (5)
5π
256
, (6)
5π
8
, ( 7)
3π − 8
12
, (8)
11 3 3
4
+ log 2 + 3 , (9)
8
8 2
3
, ( )
1 1 16a 7
(10) , (11) , (12)
3 24 35
127 KSOU Matrix Theory
DIFFERENTIAL EQUATIONS
An equation which consists of one dependent variable and its derivatives with respect to one or more independent variables is
called a 'Differential Equation'. A differential equation of one dependent and one independent variable is called 'Ordinary
Differential Equation'. A differential equation having one dependent and more than one independent variable is called
'Partial Differential Equation'.
d2y
3. ax dx + by dy = 0 4. =0
dx 2
dy 2 x + 3 y − 7 dy 2 x − 3 y + 4
5. = 6. =
dx 3 x − y + 4 dx 4 x − 6 y + 1
3
d 2 y dy
2 2
7. a 2 = 1 + 8. y dx + x dy = 0
dx dx
2
d2y dy d2y dy
9. − 4 + 3 y = 0 10. x2 − 2x + 3y = 0
dx
2 2
dx dx dx
Degree
The highest degree of the highest order derivative occurring in a differential equation (after removing the radicals if any) is
called 'Degree' of the differential equation.
In the examples given above (1), (2), (3), (5), (6) & (8) are of order one and degree one, (4), (9) & (10) are of order two
and degree one where as (7) is of order two and degree two after removing the radicals.
Examples
Solution : x 2 + y 2 = a 2
differentiating w.r.t. x, we have
dy
2x + 2 y = 0 ⇒ x dx + y dy = 0 which is the differential equation.
dx
124 KSOU Differential Equations
(2) Form the differential equation by eliminating ' m ' & 'c ' from y = mx + c
Solution : y = mx + c
dy
differentiating w.r.t. x, we have =m
dx
again differentiating w.r.t. x, we have
d2y
= 0 is the required differential equation.
dx 2
(3) Obtain the differential equation by eliminating ' a ' & 'b ' from y = a cos 3 x + b sin 3 x
Solution : y = a cos 3 x + b sin 3 x
differentiating w.r.t. x, we have
dy
= −3a sin 3x + 3b cos 3x
dx
again differentiating w.r.t. x, we have
d2y
= −9a cos 3 x − 9b sin 3 x = −9 y
dx 2
d2y
ie + 9y = 0 which is the required differential equation.
dx 2
Note :- It can be seen from the above examples that the order of the differential equation depends on the number of
arbitrary constants in the equation. ie. if arbitrary constant is one then order is one and if the arbitrary constants are
two then the order is two.
ie ∫ f ( x) dx + ∫ g ( y) dy = Constant
Eg. 1. Solve y 1 − x 2 dy + x 1 − y 2 dx = 0
y dy x dx
equation becomes + =0
1− y 2
1 − x2
∫ ∫
y dy x dx
integrating + = Constant
1− y 2
1 − x2
ie − 1 − y 2 − 1 − x 2 = −C
ie 1 − y 2 + 1 − x 2 = C is the solution.
3e x sec 2 y
dx + dy = 0
1− ex tan y
3e x sec2 y
integrating, ∫ 1− e x
dx + ∫ tan y
dy = Constant
1 + y −1 1+ x
ie dy = dx
1+ y x
1 1
ie 1 − dy = + 1 dx
1+ y x
integrating, y − log(1 + y ) = log x + x + c is the soluton
dy
Eg. 4. Solve ( x − y ) 2 = a2
dx
Solution : put x − y = u
dy du du dy
then 1 − = ie 1 − =
dx dx dx dx
du
given equation becomes u 2 1 − = a2
dx
du
ie − u 2 = a2 − u2
dx
u 2 du
ie = dx
u2 − a2
u2 − a2 + a2
ie du = dx
u2 − a2
a 2
ie 1 + 2 du = dx
u − a2
a2 u−a
integrating u + log = x+c
2a u+a
126 KSOU Differential Equations
a x− y−a
ie x − y + log = x+c
2 x− y+a
a x− y−a
ie log = y + c is the solution.
2 x− y+a
dy
= xe y − x given y = 0 when x = 0
2
Eg. 5. Solve
dx
dy
= xe y ⋅ e − x
2
Solution : given equation is
dx
ie e − y dy = xe − x dx
2
1
integrating, − e − y = − e − x + c
2
2
1 1
when x = 0, y = 0 ∴ −1 = − +c ⇒ c = −
2 2
1 1
∴ solution is − e − y = − e − x − ie 2e − y = e − x + 1
2 2
2 2
dy f ( x, y )
Equation of the type = or f ( x, y ) dx + g ( x, y ) dy = 0 where f ( x, y ) & g ( x, y ) are homogeneous expressions
dx f ( x, y )
in x & y of same degree is called a ' Homogeneous Equation'.
To solve put y = vx
dy dv
then = v+ x or dy = v dx + x dv
dx dx
by substituti ng this, the given equation reduces to variable separable form and hence can be solved.
dy
Eg. (1) Solve 2 xy = 3y 2 + x2
dx
dy dv
Solution : put y = vx then =v+x
dx dx
dv
given equation becomes 2 x 2 v v + x = 3 x 2 v 2 + x 2
dx
divide throughout by x 2
dv
then 2v v + x = 3v 2 + 1
dx
dv
ie 2v 2 + 2vx = 3v 2 + 1
dx
dv 2v dv dx
ie 2vx = v2 +1 ie =
dx v +1
2
x
integrating, we have log(v 2 +1) = log x + log c
MCA 11 - Mathematics SVT 127
y2
ie log 2 +1 = log x + log c
x
( y2 + x2 )
ie log = log x + log c
x2
ie log ( y 2 + x 2 ) − log x 2 = log x + log c
∴ solution is y 2 + x 2 = cx3
x x
Eg. (2) Solve 1 + e y dx + e y 1 − dy = 0
x
y
ie (v + vev + ev − ev ) dy + y (1 + ev ) dv = 0
ie (v + ev ) dy + y(1 + ev ) dv = 0
dy (1 + e v ) dv
∴ + =0
y v + ev
integrating, log y + log(v + ev ) = log c
ie log y (v + ev ) = log c
ie y (v + e v ) = c
x x
y + e y = c
y
x
ie x + ye y
= c is the solution
dy a1 x + b1 y + c1
Give = or ( a1 x + b1 y + c1 ) dx + ( a2 x + b2 y + c2 ) dy = 0
dx a2 x + b2 y + c2
a1 b1
Case (i) If = put a1 x + b1 y = t then it reduces to homogeneous equation and hence can be solved.
a2 b2
a1 b1
Case (ii) If ≠ put x = X + h & y = Y + k
a2 b2
where h & k are constants to be found out such that a1h + b1k + c1 = 0 & a2 h + b2 k + c2 = 0
then given equation reduces to
dY a X + b1Y
= 1 or ( a1 X + b1Y ) dX + ( a 2 X + b2Y ) dY = 0 which is homogeneous and hence can be solved.
dX a 2 X + b2Y
128 KSOU Differential Equations
dy x + y −1
Eg. (1) Solve =
dx 2 x + 2 y + 3
dy dt
Solution : put x + y = t then 1 + =
dx dx
dt t −1
∴ given equation becomes −1 =
dx 2t + 3
dt t −1 t − 1 + 2t + 3 3t + 2
= +1 = =
dx 2t + 3 2t + 3 2t + 3
2t + 3
ie dt = dx
3t + 2
2t + 3
integrating ∫ 3t + 2 dt = x + c
put 2t + 3 = l (3t + 2) + m = 3lt + 3l + m
2 4 5
∴ 3l = 2 ⇒ l = 2l + m = 3 ⇒ m = 3 − 2l = 3 − =
3 3 3
2 3t + 2 5 2
∫ ∫ ∫ ∫
5 dt 2 5 dt 2
∴ x+c = dt + = 1 dt + = t + log t +
3 3t + 2 3 3t + 2 3 9 t + 2 3 9 3
3
2 5 2
∴ x+c = ( x + y ) + log x + y +
3 9 3
1 2 5 2
ie x + c = y + log x + y + is the solution
3 3 9 3
put Y = vX then dY = v dX + X dv
∴ (2 X + vX )(v dX + X dv) = ( X + 2 vX ) dX
divide throughout by X
then (2 + v)v dX + 2(2 + v) X dv = (1 + 2v) dX
∴ (2v + v 2 − 1 − 2v) dX + (2 + v) X dv = 0
ie (v 2 − 1) dX + (v + 2) X dv = 0
dX (v + 2) dv
ie + =0
X v2 −1
MCA 11 - Mathematics SVT 129
(v + 2) dv
integrating, log X + ∫ v2 −1
= Constant
∫ ∫
1 2v dv dv
ie log X + +2 2 = Constant
2 v2 −1 v −1
1 1 v −1
ie log X + log(v 2 − 1) + 2 × log = log c
2 2 v +1
Y
Y2 −1
ie 2 log X + log 2 − 1 + 2 log X = 2 log c
X Y
+1
X
ie 2 log X + log(Y 2 − X 2 ) − 2 log X + 2 log(Y − X ) − 2 log(Y + X ) = 2 log c
(Y − X ) 3
ie log = log c 2
Y+X
∴ (Y − X )3 = c 2 (Y + X ) but X = x − 1, Y = y − 1
then + Py e ∫ = Qe ∫
dy P dx P dx
dx
dy ∫ P dx
+ yPe ∫ = Qe ∫
P dx P dx
ie e
dx
d ∫ P dx ∫ P dx
ie ye = Qe
dx
∴ ye ∫ = Qe ∫
∫
P dx P dx
dx + c is the required solution.
+ Px = Q where P & Q are function of y is also a linear equation and its solution is xe ∫ = Qe ∫
∫
dx P dy P dy
Note : - dy + c
dy
dy
Eg. (1) Solve the equation + y tan x = cos x
dx
Solution : Comparing with the standard equation
P = tanx & Q = cos x
∴ Solution is ye ∫ = Qe ∫
∫
P dx P dx
dx + c
∫
ie y sec x = cos x ⋅ sec x dx + c = 1 dx + c = x + c ∫
−1
Eg. (2) (1 + y 2 ) dx + ( x − e − tan y
) dy = 0
∫ P dy = tan y ∴ Solution is xe ∫ = Qe ∫
∫
−1 P dy P dy
dy + c
−1
e − tan y
∫ (y
tan −1 y −1
ie xe = e tan y
dy + c
2
+ 1)
∫y
−1 dy
ie xe tan y
= + c = tan −1 y + c
2
+1
−1
∴ Solution is xe tan y
= tan −1 y + c
Bernoulli's Equation
dy
The equation + Py = Qy n where P & Q are functions of x is called Bernoulli' s Equation.
dx
1 dy P
To find the solution divide by y n then n
+ n −1 = Q
y dx y
1
put n −1
= z ie y − n +1 = z
y
differentiating w.r.t. x
dy dz 1 dy 1 dz
( −n + 1) y − n = ie = ( n ≠ −1)
dx dx y dx ( −n + 1) dx
n
1 dz
∴ equation becomes + Pz = Q
(−n + 1) dx
dz
ie + ( −n + 1) Pz = ( −n + 1)Q which is a linear equation and hence can be solved.
dx
dx dz
Note : - + Px = Qx n is also Bernoulli' s equation, whose solution is given by + ( −n + 1) Pz = ( −n + 1)Q
dy dy
where P & Q are functions of y
MCA 11 - Mathematics SVT 131
dy
Eg. (1) Solve + y tan x = y 2 sec x
dx
1 dy 1
then + tan x = sec x
y 2 dx y
1 1 dy dz
put = z then − 2 =
y y dx dx
dz
∴ equation becomes − + tan x ⋅ z = sec x
dx
dz
ie − tan x ⋅ z = − sec x
dx
which is linear where P = − tanx, Q = − sec x
∫ ∫
∴ Solution is z cos x = − sec x ⋅ cos x dx + c = − 1 dx + c = − x + c
cos x
∴ Solution is = −x + c
y
dy tan y
Eg. (2) Solve − = (1 + x) e x sec y
dx (1 + x)
dy sin y
Solution : the given equation is cos y − = (1 + x) e x
dx (1 + x)
dy dz
put sin y = z , then cos y =
dx dx
dz z
∴ equation becomes − = (1 + x) e x which is a linear equation where
dx 1 + x
1
P=− , Q = (1 + x) e x
1+ x
∫ P dx = ∫ − (1 + x) dx = − log(1 + x)
1
1
log 1
∴ e ∫ P dx = e − log(1+ x ) = e 1+ x =
1+ x
∫ ∫
1 1
∴ Solution is z ⋅ = (1 + x) e x dx = e x dx = e x + c
1+ x (1 + x)
sin y
∴ Solution is = ex + c
1+ x
132 KSOU Differential Equations
Consider ∫ M dx (1)
where the integration is done w.r.t. x treating y as a constant and take ∫ N dy (2)
M = 2 x 3 − xy 2 − 2 y + 3, N = − x 2 y − 2 x
∂M ∂N ∂M ∂N
= −2 xy − 2, = −2 xy − 2 ∴ = hence equation is exact
∂y ∂x ∂y ∂x
x4 x2 x4 x2 y 2
∫ ∫
M dx = (2 x 3 − xy 2 − 2 y + 3) dx = 2 ⋅
4
− y2 ⋅
2
− 2 xy + 3x =
2
−
2
− 2 xy + 3x (1)
∫ N dy = ∫ (− x ∫
y − 2 x) dy = 0 ⋅ dy (omitting the terms which contain x)
2
= Constant
x4 x2 y2 c
∴ solution is − − 2 xy + 3x =
2 2 2
ie x 4 − x 2 y 2 − 4 xy + 6 x = c
M = x 3 + xy 2 − a 2 x, N = x 2 y − y 3 − b 2 y
∂M ∂N
= 2 xy, = 2 xy
∂y ∂x
∂M ∂N
= ∴ equation is exact
∂y ∂x
x4 x2 y 2 a2 x2
for solution, consider ∫ ∫
M dx = ( x 3 + xy 2 − a 2 x) dx =
4
+
2
−
2
(treating y as a constant)
∫ N dy =∫ ( x y − y ∫
− b 2 y ) dy = (− y 3 − b 2 y ) dy omitting the terms which contain x
2 3
y 4 b2 y 2
=− −
4 2
x4 x2 − y 2 a 2 x 2 y 4 b2 y 2 c
∴ solution is + − − − =
4 2 2 4 2 4
ie x 4 + 2 x 2 y 2 − 2a 2 x 2 − y 4 − 2b 2 y 2 = c
MCA 11 - Mathematics SVT 133
Exercise
Solve the following
dy dy
(1) x 2 (1 − y ) + y 2 (1 + x) = 0 ( 2) = e 2 x − 3 y + 4 x 2e − 3 y
dx dx
dy x( 2 log x + 1)
(3) = ( 4) cos( x + y ) dy = dx
dx (sin y + y cos y )
dy dy dy
(5) y−x = a y 2 + (6) ( x + 1) + 1 = e− y
dx dx dx
dy y y
(7) x dy − y dx = x 2 + y 2 dx (8) = + sin
dx x x
(9) x 2 y dx − ( x 3 + y 3 ) dy = 0 (10) (2 x + 5 y + 1) dx − (5x + 2 y − 1) dy = 0
dy 2 y − x − 4
(11) ( 4 x − 6 y − 1) dx + (3 y − 2 x − 2) dy = 0 (12) =
dx y − 3 x + 3
dy dy
(13) x log x + y = (log x ) 2 (14) = x 3 − 2 xy if y = 2 when x = 1
dx dx
dy dy
(15) + y cos x = y 3 sin 2 x (16) x + y = x3 y6
dx dx
(17) (1 + y 2 ) dx = (tan−1 y − x) dy (18) (2 xy + y − tan y ) dx + ( x 2 − x tan2 y + sec2 y) dy = 0
2x y 2 − 3x 2 1
(19) dx + dy = 0 ( 20) y1 + + cos y dy + ( x + log x − x sin y ) dy = 0
x
3 4
y y
Answers
x 1 1 x+ y
(1) log − − = c ( 2) 3e 2 x − 2e 3 y + 8 x 3 = c (3) y sin y = x 2 log x + c ( 4) y = tan + c (5) ( x + 1)(2 − e y ) = c
y x y 2
3 3
−1 x 2
(6) ( x + 1)(1 − e ) = c (7) y + x + y = cx
y 2 2 2
(8) y = 2 x tan (cx) (9) = 3 log cy (10) ( x + y ) 7 = c x − y −
y 3
( )
1
5 2Y + 5 + 21 X 21
(11) ( 2 x − y ) + log(7 − 8 x + 12 y ) = c (12) ( X 2 − XY + Y 2 ) = c
4
2Y − 5 − 21 X ( ) where X = x − 2, Y = y − 3
1 5
(13) y log x = (log x) 3 + c (14) 2 y − x 2 + 1 = 4e1− x (15) y − 2 = ce 2 sin x + 2 sin x + 1 (16) x 3 y 5 + cx 2 = 1
2
3 2
−1
(17) x = tan−1 y − 1 + ce − tan y
(18) x 2 y + xy − x tan y + tan y = c (19) x 2 − y 2 = cy 3 ( 20) y ( x + log x ) + x cos y = c