math -1516 em
TRANSCRIPT
-
7/25/2019 Math -1516 EM
1/81
1
SPEED AND ACCURACY BRING SUCCESS IN MATHEMATICS
Easy to score good marks for average students;
Higher Secondary Second Year Mathematics
BLUE PRINT
Chapter
No.
Chapters No. of Questions Total
Marks1 Mark 6 Marks 10 Marks
1 Application of Matrices andDeterminants
4 2 1 26
2 Vector Algebra 6 2 2 38
3 Complex Numbers 4 2 1 26
4 Analytical Geometry 4 1 3 40
5 Differential Calculus Application - 1 4 2 2 36
6 Differential Calculus Application - 2 2 1 1 18
7 Integral calculus and its applications 4 1 2 30
8 Differential Equations 4 1 2 30
9 Discrete Mathematics 4 2 1 26
10 Probability Distributions 4 2 1 26
Total Number of Questions 40 16 16 296
TABLE - I
Chapter
No.Chapters
No. of
6 & 3 Mark
Questions
No. of
10 Mark
Questions
Total
Marks
2 Vector Algebra --- 20(2) 20
4 Analytical Geometry --- 28(3) 30
50 marks can be scored if we practice 20 Ten mark questions in Lesson-2 and 28 Tenmark questions in Lesson4
TABLE - II
Chapter
No.Chapters
No. of
6 & 3 Mark
Questions
No. of
10 Mark
Questions
Total
Marks
9 Discrete Mathematics 33+12(2) 15(1) 22
6 Differentials calculus Application - II --- 11(1) 10
3 Complex Number --- 16(1) 10
-
7/25/2019 Math -1516 EM
2/81
2
30 marks can be scored if we study all the 271 one mark questions in the text book. Ifwe practice completely from table I to VI we can score 134 marks.
42 Marks can be scored if we practice 15 Ten mark questions 31 Six mark
questions, 12 Three mark questions in Lesson-9and 11 Ten mark questions in Lesson6
and 16 Ten mark question in Lesson 3
TABLE - III
Chapter
No.Chapters
No. of
6 & 3 Mark
Questions
No. of
10 Mark
Questions
Total
Marks
1 Application of Matrices andDeterminants
35+13(2) --- 12
10 Probability Distributions --- --- ---
12 Marks can be scored if we practice 35 Six mark questions and 13 Three mark questionsin Lesson1.
TABLE - IV
Chapter
No.Chapters
No. of
6 & 3 Mark
Questions
No. of
10 Mark
Questions
Total
Marks
5 Differentials calculus Application - I --- --- ---
7 Integral calculus and its applications --- --- ---8 Differential Equations --- --- ---
-
7/25/2019 Math -1516 EM
3/81
3
Slip tests should be conducted repeatedly on the 271 one words questions in the textbook and 380 questions in the COMEbook
We can easily answer 9 out of the 10 six mark questions in the question paper if wepractice all the question in Lesson, 1,2,3,9,10
For the remaining questions you have to concentrate on all the lessons
We can easily answer 9 out of the 10 ten mark questions in the question paper if wepractice all the 10 mark questions in lesson 1,2,3,4,6,9,10
For the remaining 1 question you have to concentrate on all the lessons
Created questions can be answered easily if we have practice on all the lessons
At least 5 full portion tests should be written before the public exam. It will bring thefollowing results
oWe can assess if we could answer all the questions with in the stipulated time(3hours)
oWe can analyze whether we could answer all the questions to the extent ofscoring full marks
oWe can identify the hurdles to score full marks accordingly.
Avoid writing without reading the questions thoroughly
o (E.g) Without reading the questions 4.35 and 5.7, if we just read ladderwemay give a completely wrong answer.
Avoid answering in a hurry without reading the questions completely and observingthe pictures promptly
o (E.g) It is possible to answer using the formula of parabolainstead ofellipse
Use pen for writing the answers and pencil for drawing the diagrams
Dont waste your precious time on colouring the pages
Those who aim at centum marks should give extra attention to one mark questions.
GUIDELINES TO GET;100% MARKS
-
7/25/2019 Math -1516 EM
4/81
4
MATHEMATICS2. VECTOR ALGEBRA (10 MARK)
Two questions for full test Total number of questions : 20
1)Prove that
Cos(A-B) = CosACosB + SinAsinB
Solution:
Let P(CosA, SinA) and
Q(CosB, SinB) be any two
points on the unit circle with
centre at the origin O.
Let and be the unit
vectors along the
co-ordinate axes.
=CosA +SinA
=CosB +SinB
. = Cos(AB)= Cos(AB)...(1)
. =CosACosB+SinASinB (2)
(1),(2) Cos (AB)=CosACosB+SinASinB
3) Prove that
Cos(A+B) = CosACosB SinASinBSolution:
Let P(CosA, SinA) and
Q(CosB, -SinB) be any two
points on the unit circle with
centre at the origin O. Let
and be the unit vectors
along the co-ordinate axes.
=CosA +SinA
=CosB SinB
. = Cos (A+B)
=Cos(A+B).(1)
. =CosACosB SinASinB .(2)
(1)=(2) Cos (A+B)=CosACosB SinASinB
4) Prove thatSin (A+B) = SinACosB +CosASinB
Solutons:
Let P(CosA, SinA) and Q (CosB, -SinB) be any two
points on the unit circle
with centre at the origin
O.
Let and be the unit
vectors along the
co-ordinate axes.
=CosA +SinA
=CosB SinB
= Sin (A+B)
= Sin (A+B) ..(1)
=
=(SinACosB+CosASinB) .(2)
(1),(2) Sin (A+B)=SinACosB+CosASinB
2) Prove that Sin(A-B) = SinACosB CosASinBSolution:
Let P(CosA, SinA) and
Q(CosB,SinB) be any two
points on the unit circle
with centre at the origin O.
Let and be the unit
vectors along the
co-ordinate axes
=CosA +SinA
=CosB +SinB
x = Sin (AB)
= Sin (AB) .(1)
=
= (SinACosB CosASinB)(2)
(1),(2) Sin (AB)=SinACosB CosASinB
-
7/25/2019 Math -1516 EM
5/81
5
5)Altitides of a triangle are concurrent prove by
vector method.
Solution:
Let ABC be the given triangle.
Let the altitudes AD and BE intersecting at O and take
it as the origin.
To prove that CO is
perpendicular to AB.
= , = , =
AD BC
/ . =0 . =0. (1)
BE CA
/ =0 . )=0..........(2)
(1) +(2)
=0
( ) =0 . =0
OC ABHence the altitudes of a triangle are concurrent.
7) If = + , = + , = + +
and = + + 2 , then verify that
) ( ) = ] -[ ]
Solution:
=
= (10) (12)+ (02)
=
=
= (21) (41)+ (21)
=
) ( )=
= (16) (1+2)+ (31)
= .(1)
[ ] =
=1(01)1(22)+1(20)=1
] =
=1(01)1(41)+1(20)=2
] [ ]
= 2( + + ) 1 ( + +2 )
= 5 3 4 ..(2)
(1)=(2)
)( )= ] [ ]
6) If = + - , = + ,
= - . Verify that
) =
Solution:
=
= (0 5) (6 0)+ (2 0)
=
( )=
= (6 6) (4 5)+ (12 + 15)
= (1)
. = =6
. = )=9
=6( )+9(
= (2)
(1) =(2)
)=
-
7/25/2019 Math -1516 EM
6/81
6
8) Show that the lines = = and
= = intersect and hence find the point
of intersection.
Solution:
Let = +
= 4
= 3 ,
= 2 + 3
= 3
[ , , ] = = 0
The lines are intersecting.
Let = = = Any point on theline is
( .(1)
Let = = = Any point on theline is
(2 ..(2)
From (1) and (2) + 1 = 0 (or) 3 1 = 1
, = 0
Point of intersection is (4, 0, 1)
10) Find the vector and cartesian equations of
the plane through the point (2,1,3) and
parallel to the lines = = and
= =
Solution:
Let = 3
= 4 ;
= 2 3 + 2
Vector equation is = + + t
= 2 3 + s( 4 )+ t (2 3 + 2 )
Cartesian equation is
=0
=0
(x2)(8) (y+1)(14)+(z+3)(13)=0
8x+16 14y1413z39=08x+14y+13z+37 =0
11) Find the vector and cartesian equations of
the plane through the point (1,3,2) and
parallel to the lines
= = and = =
Solution:
Let = + + 2= + 3 ;
= + 2 + 2
Vector equation is
= + + t
= + + 2 + s( + 3 )+ t( + 2 + 2 )
Cartesian equation is
= 0
= 0
(x1)(8)(y3)(1)+(z2)(5)=08x+y5z1=0
9) Show that the lines = = and
= = intersect and find their point of
intersection.
Solution:
Let = ;
= 2 +
= + 3 ;
= + 2
= + 2
[ , , ] = = 0
The lines are intersecting
Let = = = Any point on the line is
( .(1)
Let = = = Anypoint on the line is
( ..(2)
From (1) and (2)
= 1 .(3)
1=2 +1 2 =2 .(4)(3)+(4) =1, =0Point of intersection is (1, 1, 0)
-
7/25/2019 Math -1516 EM
7/81
7
12) Find the vector and cartesian equations of
the plane containing the lines = =
and parallel to the line = =
Solution: = +2 +
= +3 , = +2 +
Vectors equation is = + +t
= 2 + 2 + + s( +3 ) + t( +2 + )
Cartesian equation is =0
=0
(x 2) ( 3) (y 2) ( 7) + (z 1) ( 5) = 03x+ 6 + 7y 14 5z+ 5 = 0
3x 7y+5z+ 3 = 0
15) Find the vector and cartesian equations of
the plane through the points (1,1,1) and(1,1,1) and perpendicular to the plane
x+2y+2z= 5
Solution: = + +
= + ; = +2 +2
Vector equation is =(1s) + +t
= (1 s)( + + + s( + ) + t ( + + )
Cartesian equation is =0
=0
(x+ 1) ( 4) (y 1) (4) + (z 1) (6) = 0 4x 4 4y+ 4 + 6z 6 = 0
2x+ 2y 3z+ 3 = 0
13) Find the vector and cartesian equations to theplane through the point (1,3,2) and perpendicularto the planesx+ 2y+ 2z= 5 and
3x+ y+ 2z= 8
Solution: Let = + + 2 ; + 2 ;
= 3 + + 2
Vector equation is = + + t
= + + 2 + s( + 2 )+ t ( + + )
Cartesian equation is = 0
= 0
(x+1)(2) (y3) (4) + (z2) (5) = 0
2x+4y 5z= 0
16) Find the vector and cartesian equations of
the plane through the points (1,2,3) and
(2,3,1) perpendicular to the plane
3x 2y+4z 5 = 0
Solution : Let = + + ;
= + ; = 2 + 4
Vector equation is = (1s) + + t
= (1s)( + + 3 )+ s(2 + )+ t(3 + )
Cartesian equation is = 0
= 0
(x1) (0) (y2) (10) + (z3) (5) = 0
10y 5z+ 35 = 0 by 5 2y+z 7 = 0
-
7/25/2019 Math -1516 EM
8/81
8
14) Find the vector and cartesian equations of
the plane passing through the points
A (1,2,3) and B(1,2,1) and parallel to
the line = =
Solution:Let = 2 + 3
= ; = 2 + 3 + 4
Vector equation is = (1s) + + t
= (1s)( 2 + 3 )+ s( )+ t( + + )
Cartesian equation is = 0
= 0
(x1) (28) (y+2) (0) + (z3) (14) = 028x 14z+ 14 = 0
2x z + 1 = 0
17) Find the vector and cartesian equations of
the plane containing the line = =
and passing through the point (1, 1, 1)
Solution: = + ;
= + ; = +3 2
Vector equation is = (1 s) + +t
= (1s)( + ) +s(2 + )+t(2 + )
Cartesian equation =0
=0
(x+ 1) ( 8) (y 1) ( 10) + (z+ 1) (7) = 08x+ 10y+ 7z 11 = 0 8x-10y-7z+ 11= 0
18) Find the vector and cartesian equations of
the plane passing through the points
(2, 2,1),(3, 4, 2) and (7, 0, 6)Solution :
Let = +
= +
= + 6
Vector equation is
= (1st) + + t
= (1st) (2 + + s(3 + )
+ t (7 + )
Cartesian equation is
= 0
= 0
(x2) (20) (y2) (8) + (z+1) (12) = 020x 40 + 8y 16 12z 12 = 020x+8y 12z 68 = 05x+ 2y 3z 17 = 019) Find the vector and Cartesian equations of
the plane passing through the points with
position vectors + 4 + 2 , 2 ,
7
Solution :
Let = + 4 + 2
= 2
= 7
20) Derive the equation of the plane in the
intercept form
Solution:
Let a, b, c be the x, y and z intercepts of the plane
=
=
=
(i) Vector equation is= (1st) + + t
= (1st) a + + tc
(ii) Cartesian equation is
= 0
= 0
(x a) (bc0)y(ac0)+z(0+ab) = 0xbc abc+ yac+ zab= 0xbc+yac+ zab= abc
Dividing by abc
-
7/25/2019 Math -1516 EM
9/81
9
Vector equation is = (1 s t) + + t
= (1st) + 4 + 2 )+ s(2 )+ t(7 + )
Cartesian equation is
= 0
= 0
(x 3) ( 6) (y 4) (13) + (z 2) (28) = 06x 13y + 28z + 14 = 0
6x + 13y 28z 14 = 0
-
7/25/2019 Math -1516 EM
10/81
10
4. ANALYTICAL GEOMETRY (10 MARK)
Three questions for full test Total number of questions 28
1) Find the axis, vertex, focus, directrix, equation
of the latus rectum and length of the latus
rectum of the parabola
and hence draw their graph,
Solution:
(y+ 3) 2= 8x, [ Y2 = 4aX]
X= x x= X Y= y+ 3 y= Y 3; a= 2The type is open rightward
Referred to
X, Y
Referred to
x,y
x= X, y= Y 3
Axis Y= 0 y= 3
Vertex (0, 0) V (0, 3)
Focus (a, 0) = (2, 0) F(2, 3)
Directrix X = a, X = 2 x= 2
LatusRectum X = a, X = 2 x= 2
Length L.R 4a 8
2) Find the axis, vertex, focus, directrix,
equation of the latus rectum and length latus
rectum of the parabola
and hence draw their graph,
Solution:
= 12( y +1) ); [ ]
X =x 3 x= X + 3; Y = y+ 1 y= Y 1; a = 3The type is open upward
Referred to
X, Y
Referred to x, y
x= X+3, y= Y 1
Axis X = 0 x= 3
Vertex (0, 0) V (3, 1)
Focus (0, a) = (0, 3) F(3, 2)
Directrix Y = a, Y = 3 y= 4
Latus
Rectum
Y = a, Y = 3 y= 2
Length L.R 4a 12
3) Find the axis, vertex, focus, directrix,
equation of the latus rectum and length
latus rectum of the parabola
and hence draw their
graph,
Solution:
, [
X =x 1 x= X + 1; Y = y 3 y= Y + 3; a = 2The type is open leftward
Referred to
X, Y
Referred to x, y
x = X + 1,
y = Y + 3
Axis Y = 0 y = 3
Vertex (0, 0) V (1, 3)
Focus (a, 0)=(2, 0) F(1, 3)
Directrix X = a, X = 2 x= 3Latus Rectum X = a, X= 2 x = 1
Length L.R 4a 8
(4) Find the axis, vertex, focus, directrix,
equation of the latus rectum and length latus
rectum of the parabola
and hence draw their graph,
Solution:
= 8( y +2); [ ]
X =x 1 x= X + 1;Y = y+ 2 y= Y 2; a = 2The type is open downward
Referred to
X, Y
Referred to x, y
x= X+1, y= Y 2
Axis X = 0 x= 1
Vertex (0, 0) V (1, 2)
Focus (0, a)= (0, 2)
F (1, 4)
Directrix Y = a, Y = 2 y= 0
Latus Rectum Y = a, Y =2 y= 4Length L.R 4a 8
-
7/25/2019 Math -1516 EM
11/81
11
5) Find the eccentricity, centre, foci, vertices of
the ellipse and draw the diagram
Solution:
( + 4(
X =x 4 x= X + 4; Y = y 2 y= Y + 2= 100 a = 10; = 25 b = 5
Major axis is parallel to xaxis
Eccentricity = = = , ae = 5
Referred to
X, Y
Referred tox,
y
x= X + 4,
y= Y + 2
Centre (0,0) C (4,2)
Foci (ae,0)=(5 ,0)
(ae,0)=(5 ,0),
(4+5 ,2)
(45 ,2)
Vertices A (a,0)=A (10,0)
(a,0), (10,0)
A (14,2),
(6,2)
Referred to
X, Y
Referred to x,
y
x = X+1,
y = Y4
Centre C (0,0) C(1,4)
Foci (0,ae)=(0, )
(0,ae)=(0, )
(1,4+ )
(1,4 )
Vertic
es
A (0,a)=A (0,6)
(0,a)= (0,6)
A (1,2),
(1,10)
7) Find the eccentricity, centre, foci, vertices of
the ellipse and draw the diagram
Solution:
16(
16
X =x+ 1 x= X 1; Y =y 2 y= Y + 2
; = 9 b = 3
The major axis is parallel to yaxis
eccentricity = = =
ae = 4 =
Referred to X, Y
Referred to
x, y
x =X1,
y =Y +2
Centre C (0,0) C (1,2)
Foci (0,ae) =(0, )
(0,ae)=(0, )
(1,2+ )
(1,2 )
Vertices A (0,a)=A (0,4)
(0,a)= (0,4)
A (1,6)
(1,2)
6) Find the eccentricity, centre, foci, vertices ofThe ellipse and draw the diagram
Solution:
36(
36( + 4 = 144
;
X =x 1 x= X + 1; Y = y+ 4 y = Y 4
= 36 a= 6; = 4 b = 2
The major axis is parallel toyaxis
eccentricity =
=
=
ae = 4
-
7/25/2019 Math -1516 EM
12/81
12
8) Find the eccentricity centre, foci and vertices
of the hyperbola
= 0 and also
trace the curve
Solution:
9( 16( = 199
9 16 = 144
X =x 1 x= X + 1 ; Y = y+ 2 y= Y 2
= 16 a= 4 ; = 9 b =3
Transverse axis is parallel toxaxis
eccentricity = = =
ae = 5
Referred to
X, Y
Referred to x, y
x = X+ 1,
y = Y2
Centre C (0, 0) C (1, 2)Foci (ae, 0) = (5, 0)
(ae, 0) = (5, 0)
(6, 2)
(4, 2)
Vertices A (a, 0) = A (4, 0)
(a, 0)= (4, 0)
A (5, 2),
(3, 2)
9) Find the eccentricity, centre, foci and
vertices of the hyperbola
and draw the
diagram
Solution:
4 = 4
;
X =x+ 3 x= X 3; Y = y 2 y= Y + 2
= 4 a= 2 , = 1 b=1
Transverse axis is parallel toxaxis
eccentricity = = =
ae =
Referred to
X, Y
Referred to x, y
x = X3,y = Y + 2
Centre C (0, 0) c (3,2)
Foci F1(ae, 0) =( 0)
F2(ae,0)=( 0)
T e e uation
( 2)
( , 2)
Vertices A (a, 0)=(2, 0)
(a, 0)=(2, 0)
A (1,2),
(5, 2)
10) Find the eccentricity, centre, foci and vertices
of the hyperbola
and draw
their diagram.
Solution :
9( 16( = 164
9 16 = 144
= 1 ,
X =x+ 2 x= X 2, Y = y 1 y= Y + 1
= 9 a= 3, = 16 b =4
11) Find the eccentricity, centre, foci and
vertices of the hyperbola
and draw their
diagram.
Solution:
Given :
( 3( = 183 = 12
= 1
X =x+ 3 x= X 3 Y = y 1 y= Y + 1
= 4 a= 2, = 12 b=2
Transverse axis is parallel toyaxis
eccentricity = = = 2
-
7/25/2019 Math -1516 EM
13/81
13
Transverse axis is parallel toyaxis
eccentricity = = =
ae = 3 = 5
Referred to
X, Y
Referred
to
x, y
x = X2,y = Y +1
Centre C (0, 0) c (2, 1)
Foci (0, ae) = (0,5)
(0, ae) = (0, )
(2, 6)
(2, 4)
Vertices A (0, a) = A (0, 3)
(0, a)= (0, 3)
A (2, 4),
(2, 2)
ae = 4
Referred to
X, Y
Referred
to
x, y
x = X 3,y = Y + 1
Centre C (0, 0) c (3 1)
Foci (0, ae) = (0,4)
(0, ae)= (0 )
(3, 5)
(3, 3)
Vertices A (0, a) = A (0, 2)
(0, a)= (0, 2)
A (3, 3),
(3, 1)
12) A comet is moving in a parabolic orbit
around the sun which is at the focus of a
parabola. When the comet is 80 million Kmsfrom the sun, the line segment from the sun
to the comet makes an angle of radians
with the axis of the orbit. Find (i) the
equation of the comets orbit (ii) how close
does the comet come nearer to the sun?
Solution:
Equation of the parabola .(1)
From the right FQP
cos ( =
= FQ 40= 1 FP PM
80 = 2a + 40
2a = 40 a 20(i) The equation of the comets orbit is
(ii) The shortest distance between the sun and
The comet = 20 million kms.
14) On lighting a rocker cracker it gets projected
in a parabolic path and reaches the ground
12mts away from the starting point. Findthe angle of projection.
Solution:The equation is .(1)
The point (6, 4) lies on the parabola= 4a( 4)
a = 9
(1) = 9y
= 9
=
= =
=
=
Angle of projection is
-
7/25/2019 Math -1516 EM
14/81
14
15) Assume that water issuing from the end of a
horizontal pipe, 7.5m above the ground,
describes a parabolic path. The vertex of the
parabolic path is at the end of the pipe. At a
position 2.5m below the line of the pipe, the
flow of water has curved outward 3m
beyond the vertical line through the end of
the pipe, How far beyond this vertical line
will the water strike the ground?
Solution:
The equation = 4ay ..(1)
The point (3, 2.5) lies on the parabola= 4a( 2.5)
a =
= y ..(2)
The point ( , 7.5) lies on the parabola
= (7.5)
= 27
= 3 m
The water strikes the ground 3 m beyond
the vertical line.
13) The girder of a railway bridge is in the
parabolic form with span 100ft. and the
highest point on the arch is 10ft. above the
bridge. Find the height of the bridge at 10ft.
to the left or right from the midpoint of the
bridge.
Solution:
The equation is .(1)
The point (50, 10) lies on the parabola= 4a (10),
4a = 250
(1)
..(2)
B (10, ) lies on the parabola
100
ft
Height of the bridge at the required place
= 10
= 9 feet
-
7/25/2019 Math -1516 EM
15/81
15
16) A cable of a suspension bridge hangs in the
form a parabola when the load is uniformly
distributed horizontally. The distance
between two towers is 1500 ft, the points of
support of the cable on the towers are 200ft
above the road way and the lowest point on
the cable is 70 ft above the roadway. Find the
vertical distance to the cable from a pole
whose height is 122ft.
Solution:
The equation is
= 4ay ..(1)
The point (750, 130) lies
on the parabola
= 4a(130)
4a =
(1) y .(2)
The point P( , 52) lies on the parabola
(2) = 52
= 150 ft
Vertical distance to the cable from a pole =
= 300 ft
17) A cable of a suspension bridge is in the form
of a parabola whose span is 40 mts. The road
way is 5 mts below the lowest point of the
cable. If an extra support is provided across
the cable 30 mts above the ground level, find
the length of the support if the height of the
pillars are 55 mts.
Solution:
The equation is
= 4ay ..(1)
The point (20, 50) lies on the parabola
= 4a(50)
a = 2
(1) = 8y (2)
The point ( , 25) lies on the parabola
(2) = 8 25
= 200
=
= 10 ft
Length of the support =
= 20 ft
18) A Khokho player in a practice session whilerunning realizes that the sum of the distances
from the two khokho poles from him is
always 8 m. Find the equation of the path
traced by him if the distance between thepoles is 6m.
Solution:
The equation is
..(1)
2a = 8 a = 4
4e = 6
e =
= 7
(1)
Which is the equation of the path traced by the
khokho player.
19)The orbit of the planet mercury around the
sun is in elliptical shape with sun at a focus.
The semimajor axis is of length 36 million
miles and the eccentricity of the orbit is
0.206. Find (i) How close the mercury getsto sun? (ii) The greatest possible distance
between mercury and the sun.
Solution:
The equation is
..(1)
Semi major axis = a
= 36 million miles
e = 0.206 ae = 7.416
A = a ae = 36 7.416
= 28.584 milion miles
= a+ae = 36 + 7.416
= 43.416 million miles
(i) The closest distances of the mercury from
the sun = 28.584 million miles
(ii) The greatest distance of the mercury from
thesun = 43.416 million miles.
-
7/25/2019 Math -1516 EM
16/81
16
20) A satellite is travelling around the earth in an
elliptical orbit having the earth at a focus and
of eccentricity . The shortest distance that
the satellite gets to the earth is 400 kms. Find
the longest distance that the satellite gets
from the earth.
Solution:
CA C =400a ae=400
ae=800 =400
Longest distance between the satellite from
the earth = a+ ae = 800 + 400 = 1200 km
22) The arch of a bridge is in the shape of a
semiellipse having horizontal span of 40 ftand 16 ft high at the centre. How high is the
arch, 9 ft from the right or left of the centre.
Solution:
The equation is
..(1)
2a = 40 a= 20 b= 16
..(2)
Point (9, ) lies on the ellipse
= 1 =
= ft
The required height = ft
21) An arch is in the form of a semiellipe whosespan is 48 feet wide. The height of the arch is
20 feet. How wide is the arch at the height of
10 feet above the base?
Solution:
The equation is
..(1)
2a = 48 a = 24, b = 20
..(2)
Point ( lies on the ellipse
= 576
= 24
= 12
The required width = 24 ft.
23) The ceiling in a hallway 20 ft wide is in the
shape of a semi ellipse and 18 ft high at the
centre. Find the height of the ceiling 4 feet
from either wall if the height of the side
walls is 12
ft.Solution:
The equation is
..(1)
2a= 20 a= 10,b = 6
(1) ..(2)
Point (6, lies on the ellipse
= 1
=
=
= 4.8
Required height of the ceiling = 12 + 4.8 = 16.8 ft
-
7/25/2019 Math -1516 EM
17/81
17
24) A ladder of length 15m moves with its ends
always touching the vertical wall and the
horizontal floor. Determine the equation of
the locus of a point P on the ladder, which is
6m from the end of the ladder in contact with
the floor.
Solution:
Let P( be any point on the line AB such that AP=6
and BP=9
Assum:
___PAO = ___BPQ =
From the right PQB
Cos =
From the right ARP;
Sin =
Cos 2 + Sin2 = 1
+ = 1
Locus P( is
26) Find the equation of the rectangular
hyperbola which has for one of its
asymptotes the linex+ 2y 5 = 0 and passesthrough the points (6, 0)and (3, 0)
Solution:
Equation of the asymptote is x+ 2y 5 = 0The other asymptote is of the form 2x y+ l = 0Combined equation of the asymptote is
(x+ 2y 5) (2x y+ l) = 0Equation of the rectangular hyperbola is of the form
(x+ 2y 5) (2x y+ l) + k= 0
It passes through (6, 0)
(1) (6 + 0 5) (12 0 + l ) + k= 0l + k= 12 (2)
Again it passes through (3, 0)(1) (3 + 0 5) (6 0 + l) + K = 0
(8) (6 + l) + k= 0
8 + k= 48 .(3)
Solving (2)&(3) l = 4, k= 16
(1) (x+ 2y 5 ) (2x y+ 4) 16 = 0This is the required equation
27) Prove that the line 5x + 12y = 9 touches the
hyperbola and find its point of contact.
Solution:
5x+ 12y= 9 , m = , c=
= 9 1 = =
The line touches the hyperbola
Point of contact= =
Point of contact = (5, )
25) Find the equation of the hyperbola if the
asymptotes are parallel tox+ 2y 12 = 0,x 2y+ 8 = 0, (2, 4) is the centre of thehyperbola and it passes through (2, 0)
Solution:
The asymptotes are parallel to
x+ 2y 12 = 0, x 2y+ 8 = 0The asymptotes are of the form
x+ 2y+ = 0 ....(1)
x 2y+ m = 0 (2)It passes through the centre(2, 4)
(1) = 10
(2) m = 6
Equations of the asymptotes are
(1) x+ 2y 10 = 0(2) x 2y+ 6 = 0Combined equation of the asymptote is
(x+ 2y 10) (x 2y+ 6) = 0Equation of the hyperbola is of the form
(x+ 2y 10) (x 2y+ 6) + k= 0 (3)It passes through (2, 0)
(8) (8) + k= 064 + k= 0k= 64(3) (x+ 2y 10)(x 2y+6) +64 = 0This is the required equation of the hyperbola.
-
7/25/2019 Math -1516 EM
18/81
18
28) Show that the linex y+ 4 =0 is a tangent to
the ellipse . Find the
coordinates of the point of contact
Solution:
x y+4 = 0 y= x+ 4 m= 1, c= 4
,
= 12 1 + 4, = 16 =
The line touches the ellipse,
Point of contact = =
Point of contact = (3, 1)
6. DIFFERENTIAL CALCULUS APPLICATIONS II (10 Mark Questions)One Question for full Test Total number of questions: 11
1) Trace the curve y = 2) Trace the curvey = 3) Trace the curve
Domain:(-
Extent :Horizontal extent : (-
Vertical extent : (-
Intercepts :
xintercept = -1
yintercept = 1Origin :
Does not pass
through the origin
Domain :(-
ExtentHorizontal extent : (-
Vertical extent : (-
Intercepts :
xintercept = 0
yintercept = 0Origin :Passes through the
origin
Domain : [0,
ExtentThe curve exists in
first and fourth quadrantIntercepts :
xintercept = 0
yintercept = 0
Origin :Passes through the origin
SymmetryNot symmetrical
about any axis
Symmetry Symmetrical about
the origin
Symmetry Symmetrical about
thexaxis
AsymptotesNo asymptote
AsymptotesNo asymptote
AsymptotesNo asymptote
Monotonicity
The curve is increasing
in (
Monotonicity
The curve is increasing
in(
Monotonicity
y= the curve is increasing
y= - , the Curve is decreasing
Special pointsConcave downward in (-
Concave upward in (0,
Point of inflection (0, 1)
Special pointsConcave downward in (-
Concave upward in (0,
Point of inflection (0, 0)
Special points(0, 0) is not a point of inflection.
-
7/25/2019 Math -1516 EM
19/81
19
4) Using Eulers theorem, prove that
tanu, if
Solution;
= = = f(x,y)
fis a homogeneous function in xand yofdegree
By Eulers theorem,
f
x sin u
xcosu + ycos u = sin u
x tanu
5) Using Eulers theorem, prove that
x , if u=
Solution:
u= tan u= = f(x,y)
fis a homogeneous function in xand yofdegree 2
By Eulers theorem,
= 2f
= 2 tan u
+ = 2 tan u
6) Verify Eulers theorem forf(x,y) =
Solution;
f(tx, ty) = = = f(x,y)
f is a homogeneous function inxand yof degree 1 To verify Eulers theorem, we have to prove that
= f
= =
=
= = = f
Eulers theorem is verified
9) Verify for the functionu=
Solution:
= ;
=
= = (1)
= = (2)
(1)=(2)
10) If w= and v= ylog x, find and
Solution:
w=
u=
v= ylog x
=
=
=
=
7. Verify for the function
Solution:
= . = . =
= =
=
= = (1)
=
= = . (2)
From (1) & (2)
-
7/25/2019 Math -1516 EM
20/81
20
8. Verify for the function
u = sin 3x cos 4y
Solution:
u= sin 3xcos 4y
= 3 cos 3xcos 4y
= 4sin3xsin 4y
= = 4 cos 3x3 sin 4y
= 12 cos 3xsin 4y .(1)
= = 3 cos 3x( sin 4y) 4
= 12 cos 3xsin 4y .(2)
(1)=(2)
11. Use differentials to find an approximate
value for the given numbery= +
Solution: Consider
= 1,
1+ 0.0066 = 1.0066
Consider
x= 1,
y=
dy=
1+ 0.005 = 1.005
(1) + (2) + 1.0066 + 1.005
= 2.0166
-
7/25/2019 Math -1516 EM
21/81
21
9. Discrete Mathematics (10 Mark)
One question for full test Total number of questions : 15
1) Prove that the set of four functions
on the set of non zero complex numbers C
defined by and
forms an abelian group
with respect to the composition of functions.
Solution: Let G =o Composition of functions
o
1)Closure axiom:
From the table closure axiom is true.
2)Associative axiom
Composition of functions is always associative3) Existence of identity
G is the identity element.
4) Existence of inverse
Inverses of are
respectively.
5) Commutative axiom
From the table commutative axiom is true.
(G, o is an abelian group.
2) Show that
where w 1 form a group with respect
to matrix multiplication.
Solution:I = , A = , B = ,
C = , D = E =
Let G =
. I A B C D E
I I A B C D E
A A B I E C D
B B I A D E C
C C D E I A B
D D E C B I A
E E C D A B I
1) Closure axiom
Form the table closure axiom is true
2) Associative axiom
Matrix multiplication is always associative
3) Existence of identity
I= G is the identity element.
4) Existence of inverse:Inverses of
I, A, B, C, D, E are I, B, A, C, D, E respectively
G is a group under multiplication of matrices.
(4) Show that the set forms an
abelian group under multiplication modulo 11.Solution: Let G =
11 Multiplication modulo 1111 [1] [3] [4] [5] [9]
[1] [1] [3] [4] [5] [9]
[3] [3] [9] [1] [4] [5]
[4] [4] [1] [5] [9] [3]
[5] [5] [4] [9] [3] [1]
[9] [9] [5] [3] [1] [4]
1) Closure axiom:
From the table closure axiom is true.
2) Associative axiomMultiplication modulo 11 is always associative
3) Existence of identity
[1] G is the identity element.
4) Existence of inverse
Inverses of
respectively.
5) Commutative axiom
From the table commutative axiom is true.
(G,11 is an abelian group.
3)Show that ( forms a group
Solution:Let G =
7Multiplication modulo 7
7 [1] [2] [3] [4] [5] [6]
[1] [1] [2] [3] [4] [5] [6]
[2] [2] [4] [6] [1] [3] [5]
[3] [3] [6] [2] [5] [1] [4]
[4] [4] [1] [5] [2] [6] [3]
[5] [5] [3] [1] [6] [4] [2]
[6] [6] [5] [4] [3] [2] [1]
1) Closure axiom:From the table closure axiom is true.
2) Associative axiom
Multiplication modulo 7 is always associative
3) Existence of identity
[1] G is the identity element.
4) Existence of inverse
Inverses of are
respectively (G,7) is a group.
-
7/25/2019 Math -1516 EM
22/81
22
5) Show that the set G of all matrices of the form
where x , is a group under matrix
multiplication.
Solution:
Let G =
1) Closure axiom :
X =
Y = G, x 0, y 0
XY = G, [2xy 0
Closure axiom is true.2) Associative axiom:
Matrix multiplication is always associative
3) Identity axiom:
Let E = be the identity element
XE X 2xe= x, e =
E= G is the identity element.
4) Inverse axiom :
Let = be the inverse of X
= , =
G is the inverse of
G is a group under matrix multiplication.
6) Show that the set of all matrices of the form
, a R , forms an abelian group under
matrix multiplication
Solution:
Let G =
1) Closure axiom:
A = , B = G,where a, b R .
AB = G[ ab 0
Closure axiom is true.2) Associative axiom:
Matrix multiplication is always associative
3) Identity axiom:
Let E = be the identity element
AE = ae= a e= 1
G is the identity element.
4) Inverse axiom :
Let = be the inverse of
= E = 1
G is the inverse of
5) Commutative axiom:
AB = = = BA
Commutative axiom is true
G is an abelian group under matrix multiplication.
-
7/25/2019 Math -1516 EM
23/81
23
7) Show that (Z, ia an infinite abelian group where
is defined as a b= a + b + 2
Solution:
Z = The set of all integers
a b= a+ b+ 2
1) Closure axiom
a, b Z, a b= a+ b+ 2 Z
Closure axiom is true.
2) Associative axiom: a, b, c Z
a (b c) = a ( b+ c+ 2)
= a+ (b + c + 2) + 2
= a+ b+ c+ 4
(a b c= ( a+ b+ 2) c
= a+ b+ c+ 4
a (b c) = (a b c
Associative axiom is true.3) Existence of Identity
Let e be the identity element
= a+ e+ 2 = a e= 2
2 Z is the identity element.
4) Existence of Inverse
Let be the inverse of a
= 2
Z is the inverse of a
5) Commutative axiom:
a, b Z
a b= a+ b+ 2 = b+ a+ 2 =
Commutative axiom is true.
Z is an infinite set. Z, is an infinite abelian group
8) Show that the set G of the positive rationals
forms a group under the composition defined
by a = for all a, b G
Solution:
G = The set of all positive rationals
a =
1) Closure axiom: a, b G , a = G
Closure axiom is true.2) Associative axiom: a, b, c G
= = =
=
a (b c) = (a b c
Associative axiom is true3) Existence of Identity
Let e be the identity element
= a, e= 3
G is the identity element.
4) Existence of Inverse
Let be the inverse of a
= 3 = 3 =
G is the inverse of a
(G, is a group
-
7/25/2019 Math -1516 EM
24/81
24
9) Let G be the set of all rational numbers except 1
and be defined on G by
for all a, b G. Show that (G, is an infinite
abelian group.
Solution:
G = The set of all rational numbers except 1
1) Closure axiom:
a, b, G, a 1, b 1Suppose
a= 1 (or) b= 1 to a, b, G,
G
Closure axiom is true
2) Associative axiom
a, b, c G
=
=
== (
=
=
=
Associative axiom is true.3) Existence of Identity
Let ebe the identity element
e(1 a) = 0 e= 0, since a 1
0 is the identity element
4) Existence of Inverse
Let be the inverse of a
a 1,
is the inverse of a
Inverse axiom is true
5) Commutative axiom
a, b, G
Commutative axiom is true.
(G, ) is an abelian group.
G contains infinite number of elements.
G, ) is an infinite abelian group
10) Show that the set G of all rational numbers
except 1 forms an abelian group with respect
to the operation given by =
for all a, b G.
Solution:
G = The set of all rational numbers except 1
=
1) Closure axiom:a, b, G, a 1 and b 1
Suppose
a = 1 (or) b = 1 to a, b, G,
G
Closure axiom is true
2) Associative axiom
a, b, c G
=
=
=
= (
= (
=
=
=
Associative axiom it true3) Existence of Identity
Let e be the identity element
=
e(1+ a) e= 0, [ a 1]
0 G is the identity element
4) Existence of Inverse
Let be the inverse of a
G is the inverse of a
Inverse axiom is true
5) Commutative axiom
a, b, G
a*b= a+ b+ ab= b+ a+ ba= b*a Commutative axiom is true.
(G, ) is an abelian group.
G contains infinite number of elements.
G, ) is an infinite abelian group
-
7/25/2019 Math -1516 EM
25/81
25
11) Show that the set G = is an
infiniteabelion group with respect to addition.
Solution:
1) Closure axiom
a+ , c+ d G Where a, b, c, d Q
Sincea
+c,b
+d
QClosure axiom is true.2) Associative axiom
Addition is always associative
3) Identity axiom
a+ G, there exist an element
0 = 0 + 0 G such that
0 G is the identity element
4) Inverse axiom
a + G, there exist an element
G such that
= +
is the inverse of
5) Commutative axiom
a+ , c+ G
(a+ ) + (c+ )= ( a+ c) + (b+ d)
= (c+ a) + (d+ b)
= (c+ ) + ( a+ )
Commutative axiom is true.
G Contains infinite number of elements.
G, is an in inite abelian group
12) Show that the set G = is an abelian
group under multiplication.
Solution:
Given G =
1) Closure axiom
G, Where a, b z
= G, since a+ b z
Closure axiom is true.2) Associative axiom
G
)= . =
= . =
Associative axiom is true.3) Identity axiom
G, there exists an element
= 1 G such that .1 =
1 G is the identity element.4) Inverse axiom
G, there existy an element
G such that = = = 1
is the inverse of
Inverse axiom is true
5) Commutative axiom
G
= = = .
Commutative axiom is true
(G, .) is an abelian group
-
7/25/2019 Math -1516 EM
26/81
26
13) Show that the set M of complex numbers z with
the condition = 1 forms a group with respect
to the operation of multiplication of complex
numbers.
Solution:
M = Set of all complex numbers having
modulus value 1.
1) Closure axiom
M M
since = = 1
Closure axiom is true.2) Associative axiom
Multiplication of complex numbers is always
associative
3) Identity axiom
M there exists an element 1 M such that
z . 1 = 1 . z = z
1 G is the identity element.4) Inverse axiom
z M, there exists an element M such that
z . = . z = 1 [
is the inverse of z
(M, .) is a group
15) Show that forms group
Solution:
Let =
1) Closure axiom
[ ] =
[ ], [m] , 0 , m < n
[ ]
Closure axiom is true.2) Associative axiom
Addition modulo n is always associative
3) Identity axiom
[0] is the identity element.
4) Inverse axiom :[ ] , there exist an element
[n ] such that
[ ] = = [0]
is the inverse of [ ]
Inverse axiom is true.
, ) is a group.
14) Show that the roots of unity form an abelian
group of finite order with usual multiplication.
Solution:
Let G = ,
1) Closure axiom
Let , G, 0 , mTo prove , = G
Case (i) if + m n then G
Case (ii) if + m n
By division algoritham
+ m = (q.n) + r where 0 r < n
= = . = . G
Closure axiom is true.2) Associative axiom
Multiplication is always associative for the
set of complex numbers.
3) Identity axiom
G, there exists an element l G such
that . 1=1. =
1 G is the identity element.
4) Inverse axiom
G, there exists an element G
such that . = = = l
is the invese of
Inverse axiom is true
5) Commuative axiom G
. = = =
Commutative axiom is true.
G. contains finite number of elements.
G, . is a inite abelian group.
-
7/25/2019 Math -1516 EM
27/81
27
3. COMPLEX NUMBERS. (10MARK)
One question for full testTotal number of questions : 16
1) P represents the variable complex numberz.
Find the locus of P, if Im = 2
Solution:
Letz= x+ iy
= =
=
Im
= 2
x(2x+ 1) +2y(1 y) = 2[(1 y)2+ x2]
2x2 x+ 2y 2y2= 2(1 +y2 2y+ x2)x+ 2y= 2 + 4y
x+ 2y 2 = 0
Locus of P is a straight line
4) P represents the variable complex numberz.
Find the locus of P if arg =
Solution : Let z= x+ iy
arg = arg(z 1) arg(z+ 1) =
arg(x+ iy 1) arg(x+ iy+ 1) = arg[(x 1) + iy] arg[(x+ 1)+ iy] =
=
=
= tan
=
2y= ( )
= 0
Locus of P is a circle
2) P represents the variable complex numberz.
Find the locus of P, if Re = 1
Solution :
Letz= x+iy
= =
=
Re = 1
(x 1)x +y(y + 1) =x2 + (y + 1)2
x2 x+ y2+ y= x2 +y2 + 1 + 2y
x+ y + 1 = 0
Locus of P is a straight line
5) P represents the variable complex numberz.
Find the locus of P if arg =
Solution :
Let z = x + iy
arg =
arg(z 1) arg(z+ 3) =
arg (x+ iy 1) arg(x+ iy+ 3) =
arg[(x 1 + iy] arg[(x+ 3) + iy] =
=
=
= tan
=
0 =
= 0
Locus of P is a circle
3) P represents the variable complex numberz.
Find the locus of P if Re = 1
Solution :
Let z = x + iy= =
=
Re = 1
(x+ 1)x+ y(y+ 1) = x2+ (y+ 1) 2
x2 +x+ y2+ y= x2 +y2 + 1 + 2y
x y 1 = 0 Locus of P is a straight line
tan = =
-
7/25/2019 Math -1516 EM
28/81
28
6) If and are the roots of x22x + 2 = 0 and
cot = y+ 1, show that =
Solution :x2 2x+ 2 = 0x = 1 i, Let = 1 + i and = l i = 2i
Given cot = y + 1 y= cot 1 = 1
(y + )n =
=
(y + )n =
(y + )n =
= =
9) Find all the values of
Solution :
consider
2 ( cos + i sin )
=
=
=
= [ cos( ) + i sin( )], k= 0, 1, 2
7) If and are the roots of the equation
x22px+ (p2+ q2) = 0 and tan =
show that = qn1
Solution :x2 2px+ (p2+ q2) = 0 x= p qi Let = p + qiand = p qi = 2 iq
tan = y + p = y = p
(y + )n = = q n
(y + )n = (
(y + )n = (
=
= qn1
10) Find all the values of
Solution : consider
2 ( cos + i sin )
=
=
=
=
=
= ;k= 0, 1, 2
8) If and are the roots of the equationx22x+ 4 = 0. Prove that
n n = i2n+1sin and deduct 9 9
Solution :
x2 2x+ 4 =0
x = 1 i
= 1+ i = 2 ( cos + i sin )
b = 1 i = 2 ( cos i sin )
n= 2 n( cos + isin )
bn = 2n(cos i sin )
n n = 2n 2 i sin
= i 2n+1sin
n = 9
9 9 = i2 9+1 sin 9 9 = 0
11) Find all the values of and hence
prove that the product of the values is 1
Solution: i = cos ( ) + i sin ( )
=
=
=
= , k= 0, 1, 2, 3
When k= 0,
When k= 1,
When k= 2,
When k= 3,
Product =
= = 1
-
7/25/2019 Math -1516 EM
29/81
29
12) Solve the equationx9 +x5x41 =0
Solution :
x9 +x5 x4 1 = 0x5(x4 + 1) 1(x4 + 1) = 0
(x5 1)(x4 + 1) = 0
x5 1 = 0 x= =
=
, k= 0, 1, 2, 3, 4
x4 + 1 = 0 x=
=
=
, k= 0, 1, 2, 3
13) Solve the equationx7 +x4 +x3 + 1 = 0
Solution :
Givenx7 +x4 +x3 + 1 = 0
x4(x3 + 1)+1(x3 + 1) = 0
(x4 + 1)(x3 + 1) = 0
x4 + 1= 0 x=
=
x=
, k= 0, 1, 2, 3
x3 + 1= 0 x=
=
x= [
, k= 0, 1, 2
14) Solve the equationx4x3 +x2 x+ 1 = 0Solution:
x4 x3 +x2 x+1 = 0
= 0
= 1,x 1
x=
x=
=
x= cos + isin , k= 0, 1, 3, 4 as x 1
When k= 0, x= + i When k= 1, x= +
When k= 3, x= + i
When k= 4, x= + i (The root formed by k= 2
excluded, sincex 1)
15) If =2 cos and = 2 cos , show that
(i) n )
(ii) n )
Solution:
Letx= cos ; y= cos+ isin
=
= cos(m n) + i sin(m n ).(1)
= cos(m n) i sin(m n )..(2)
(1) + (2) = n )
(1) (2) n )
16) If a= cos2 + i sin2 , b = cos2 + isin2
and c = cos2 + i sin2 prove that
(i) + = 2 cos (
(ii) = 2 cos (
Solution:
i)abc=(cos2 +isin2 (cos2 +isin2
(cos2 +isin2
=cos2( +i sin 2(
=cos( +i sin ( .(1)
=cos( i sin ( (2)
(1)+(2)
+ =2cos(
(ii) = +
=
=cos2( +i sin 2( .(3)
=cos2( i sin2( (4)
(3)+(4)
+ =2cos2(
=2cos2(
-
7/25/2019 Math -1516 EM
30/81
30
9.Discrete Mathematics- 6 Marks questions and answers
1) Example 9/4 (iii) construct the truth table for (pq)(~q) .
p q (pq) ~q (pq)(~q)
T T T F FT F T T T
F T T F F
F F F T F
2) Example 9/4 (iv)construct the truth table for~ [(~ p)(~ )]/p q ~p ~q
T T F F F T
T F F T F T
F T T F F T
F F T T T F
3) Example9.5: construct the truth table for(pq)(~ r )/p q r (pq) (pq)(~ r )T T T T F T
T T F T T T
T F T F F F
T F F F T T
F T T F F F
F T F F T T
F F T F F F
F F F F T T
4) Example9.6: construct the truth table for(pq) r / p q r (pq) (pq) r
T T T T TT T F T F
T F T T T
T F F T F
F T T T T
F T F T F
F F T F F
F F F F F
5) Exercise 9.27construct the truth table for (p q ) [~ (pq )]/
p q (p q ~ (pq )] (p q ) [~ (pq )]
T T T F T
T F F T T
F T F T T
F F F T T
6) Exercise9.2 - 9 construct the truth table for(p q ) r /p q r (p q ) (p q ) rT T T T T
T T F T T
T F T T T
1. Matrices and Determinants- 6 and 3 mark Questions
-
7/25/2019 Math -1516 EM
31/81
31
T F F T T
F T T T T
F T F T T
F F T F T
F F F F F
7) Exercise 9.2-10 construct the truth table for(pq ) r /p q r (pq ) (pq ) rT T T T T
T T F T T
T F T F TT F F F F
F T T F T
F T F F F
F F T F T
F F F F F
8) Example 9.7: Show that ~ (p q ) (~ p)( ~q )p q p q ~ (p q )T T T F
T F T F
F T T F
F F F T
p q
T T F F F
T F F T F
F T T F F
F F T T T
The last columns are identical/ /
9) Example 9.10: (i)Show that the statement [ (~ p ) (~ q ) ] p is a tautology/
p q
T T F F F T
T F F T T T
F T T F T T
F F T T T T
The last column contains onlyT / is a tautology.
10) Example 9.10:(ii) show that [(~ q)p) ]q is a contradiction/p q
T T F F F
T F T T F
F T F F F
F F T F F
The last column contains onlyF / is a contradiction.
-
7/25/2019 Math -1516 EM
32/81
32
11) Example 9.11: use the truth table to determine whether the statement [(~ p)q] [ p(~q)]is a tautology/
p q
T T F F F
T F F T F
F T T F F
F F T T F
The last column contains onlyT / the given statement is a tautology.12) EXERCISE 9.3 (i) Use the truth table to establish the following statement is a tautology or a
contradiction [(~ p) q ) ] p.p q (~ p) q ) (~ p) q ) ] p
T T F F F
T F F F F
F T T T F
F F T F F
The last column contains onlyF / is a contradiction.
13) EXERCISE 9.3 (ii)Use the truth table to establish the following statement is a tautology or acontradiction (p q ) [~ (p q ) ]p q
T T T F T
T F T F T
F T T F T
F F F T T
The last column contains onlyT / is a tautology.
14) EXERCISE 9.3 (iii) Use the truth table to establish the following statement is a tautology or a
contradiction [ p (~q ) ] [ ( ~ p ) q ]/
p q
T T F F F T T
T F F T T F T
F T T F F T T
F F T T F T T
The last column contains onlyT / is a tautology.
15) EXERCISE 9.3 ( iv) Use the truth table to establish the following statement is a tautology or a
contradiction q [p (~q ) ] /p q
T T F T T
T F T T T
F T F F T
F F T T T
The last column contains onlyT / is a tautology.
-
7/25/2019 Math -1516 EM
33/81
33
16) EXERCISE 9.3 (v)Use the truth table to establish the following statement is a tautology or a
contradiction [ p ( ~p ) ] [ (~ q) p ] /p q (~ q) p [ p ( ~p ) ] [ (~ q) p ]T T F F F F F
T F F T F T F
F T T F F F F
F F T T F T F
The last column contains onlyF/ is a contradiction.
17) EXERCISE 9.3 - 2. Show that p q (~p ) q.p q
T T T
T F F
F T T
F F T
p q
T T F T
T F F F
F T T T
F F T T
The last columns are identical / p q (~p ) q .
18) EXERCISE 9.3 - 3. Show that p q ( p q ) ( q p ).p q p qT T T
T F F
F T F
F F T
p q p q q p ( p q ) ( q p )
T T T T TT F F T F
F T T F F
F F T T T
The last columns are identical /p q ( p q ) ( q p ) .
19) EXERCISE 9.3 - 4.Show thatpq[ (~p ) q ) ][ (~q ) p ) ].p q pqT T T
T F F
F T F
F F T
p q
T T F F T T T
T F F T F T F
F T T F T F F
F F T T T T T
The last columns are identical /pq [(~p ) q ) ][ (~q ) p ) ];.
-
7/25/2019 Math -1516 EM
34/81
34
20) EXERCISE 9.3 - 5. Show that ~ (pq ) (~p)(~q ).
p q (pq ~ (pq )
T T T F
T F F T
F T F T
F F F T
p q
T T F F F
T F F T T
F T T F T
F F T T T
The last columns are identical / ~ (pq ) (~p)(~q ) .
21) EXERCISE 9.3 - 6.Show that and are not equivalent.
p q
T T T TT F F T
F T T F
F F T T
/ p q and q p are not equivalent.
22). EXERCISE 9.3 - 7. Show that (pq) ( p q) is a tautology.
p q pq p q (pq) ( p q)
T T T T T
T F F T T
F T F T T
F F F F T
The last column contains onlyT / (pq) ( p q) is a tautology .
-
7/25/2019 Math -1516 EM
35/81
35
23).Group:Definition :
A non-empty set G, together with an operation * i.e., (G, *) is said to be agroup if it satisfies the following axioms
(1) Closure axiom : a, b G a * b G
(2) Associative axiom : a, b, c G, (a * b) * c = a * (b * c)
(3) Identity axiom : There exists an element e G such that a * e = e * a = a, a G.
( 4) Inverse axiom : a G there exists an element a1G such that a1* a = a * a1= e.
e is called the identity element of G and a1
is called the inverse of a in G.
24) State and prove cancellation laws of a group:
Proof
Let G be a group. Then for all a, b, c G.
(i) (Left Cancellation Law)
(ii) (Right Cancellation Law)
Proof:
(i) (ii)
(
b
25) State and prove reversal law of inverses of a group.
Proof:
Let G be a group a, b G. Then =
It is enough to prove is the inverse of
To prove (i)
(ii) = e=
=
=
=
=
is the inverse of
i.e., =
26)Prove that (Z,+) is an infinite abelian group.
Solution :Z = set of all integers
Closure axiom Sum of 2 integers is also an integer
Associative axiom Usual addition is always associative
Identity axiom 0 Z is the identity element
Inverse axiom is the inverse of
Commutative axiom Addition is always commutative
Z contain infinite number of elements.
(Z,+) is an infinite abelian group.
-
7/25/2019 Math -1516 EM
36/81
36
27)Example 9.13: (R{0}, . ) is an infinite abelian group.
Solution:
Closure axiom Product of two non zero real numbers is also a non zero real
number.
Associative axiom Usual multiplication is always associative
Identity axiom 1 R{0} is the identity element
Inverse axiom 1/a R{0} is the inverse of a R{0}
Commutative axiom Multiplication is always commutative
R
{0} contain infinite number of elements.(R{0},.) is an infinite abelian group.
28)Show that the cube roots of unity forms a finite abelian group under
multiplication.
Solution :
1
1 1
1
1
Closure axiom from the table closure axiom is trueAssociative axiom Usual multiplication is always associative
Identity axiom 1 is the identity element
Inverse axiom The inverse of 1 is 1
The inverse of is
The inverse of is
Commutative axiom Multiplication is always commutative
G is finite group. (G,.) is a finite abelian group
29)Show that the fourth roots of unity forms a finite abelian group under multiplication.
Solution :
1 -1 i -i1 1 -1 i -i
-1 -1 1 -i i
i i -i -1 1
-i -i i 1 -1
Closure axiom from the table closure axiom is true
Associative axiom Usual multiplication is always associative Identity axiom 1 is the identity element
Inverse axiom The inverse of 1 is 1
The inverse of -1is -1
The inverse of i is iThe inverse of -i is iCommutative axiom Multiplication is always commutative
G is finite group. (G,.) is a finite abelian group
-
7/25/2019 Math -1516 EM
37/81
37
30)Example 9.16 : Prove that (C, +) is an infinite abelian group.Solution:(i) Closure axiom : Sum of two complex numbers is always a complex number.
Closure axiom is true.(ii) Associative axiom : Addition is always associative in C
(iii) Identity axiom :o C is the identity element .
(iv) Inverse axiom : z C is the inverse of z C
(v) Commutative property : Addition of complex numbers is always commutativeSince C is an infinite set (C, +) is an infinite abelian group.
31) Example 9.17 : Show that the set of all non-zero complex numbers is an abeliangroup under the usual multiplication of complex numbers.Solution:
(i) Closure axiom : Let G = C {0} Product of two non-zero complex
numbers is again a non-zero complex number.(ii) Associative axiom :Multiplication is always associative.
(iii) Identity axiom : 1 Gis the identity element .
(iv) Inverse axiom :
1/z Gis the inverse of z G.
(v) Commutative property :Multiplication of complex numbers is always commutative.
(C-{0} , .) is an abelian.
32) Example 9.19 : Show that the set of all 2 2 non-singular matrices forms a non-abelian infinite group
under matrixmultiplication, (where the entries belong toR).
Solution:Let Gbe the set of all 2 2 non-singular matrices, where the entries belongtoR.
(i) Closure axiom : Since product of two non-singular matrices is again non-singular and the
order is 2 2,
the closure axiom is satisfied.(ii) Associative axiom : Matrix multiplication is always associative
and hence associative axiom is true.
(iii) Identity axiom : is the identity element .
(iv) Inverse axiom : the inverse ofA G, exists i.e.A1 exists and is of order 2 2 .
Thus the inverse axiom is satisfied.
Hence the set of all 2 2 non-singular matrices forms a group under matrix
multiplication. Further, matrix multiplication is non-commutative and the setcontain infinitely many elements.The group is an infinite non-abelian group.
33) Example 9.21 : Show that the set G=
-
-
-
-
10
01,
10
01,
10
01,
10
01
form an abeliangroup, under multiplication of matrices.
Solution :
Let .
I A B C
I I A B C
A A I C B
B B C I A
C C B A I
(i) Closure axiom : All the entries in the multiplication table are members of G.
-
7/25/2019 Math -1516 EM
38/81
38
Closure axiom is true.
(ii) associative axiom : Matrix multiplication is always associative(iii) identity axiom :Iis the identity element in G.(iv) inverse axiom :I is the inverse ofI
A is the inverse ofAB is the inverse ofBC is the inverse of C
From the table it is clear that .is commutative. G is an abelian group
under matrix multiplication.3 Mark Questions:-
34. Theorem : In a group G, (a1)1= a for every a G.
Proof :
We know that a1G and hence (a1)1G. Clearly a * a1= a1* a = e
a1*(a1)1= (a1)1* a1= e
a * a1= (a1)1* a1
a = (a1)1(by Right Cancellation Law)
35. Theorem :
The identity element of a group is unique.
Proof : Let Gbe a group. If possible let e1 and e2 be identity elements in G.
Treating e1 as an identity element we have e1* e2 = e2 (1)
Treating e2as an identity element, we have e1* e2= e1 (2)
From (1) and (2), e1 = e2
Identity element of a group is unique.
36.Theorem :
The inverse of each element of a group is unique.
Proof :
Let Gbe a group and let aG.
If possible, let a1and a2be two inverses of a.
Treating a1 as an inverse of awe have a * a1 = a1* a = e.
Treating a2as an inverse of a, we have a * a2= a2* a = e
Now a1 = a1 * e = a1 * (a * a2) = (a1* a) * a2 = e * a2= a2
Inverse of an element is unique.
Do it your self:- Example 9.4: Construct the truth table for the following statement:
(i) ( ) ( )( )qp ~~ (ii) ( )( )qp ~~
EXERCISE 9.2 Construct the truth table for the following statement:
(1) ( )qp ~ (2) ( ) ( )qp ~~ (3) ( )qp ~
(4) ( ) ( )pqp ~ (5) ( ) ( )qqp ~
(6) ( )( )qp ~~
(7) ( ) ( )qqp ~
-
7/25/2019 Math -1516 EM
39/81
39
XII Maths (Book Back 1 Mark Questions)
Unit 1 Matrix and Determinents
1. The rank of the matrix is
1) 1 2) 2 3) 3 4) 4
2. The rank of the diagonal matrix is
1) 0 2) 2 3) 3 4) 5
3. IfA= [2 0 1], then the rank of the A AT
1) 1 2) 2 3) 3 4) 0
4. IfA= then the rank of the A AT
1) 3 2) 0 3) 1 4) 2
5. If the rank of the matrix is 2, then is
1) 1 2) 2 3) 3 4) any real number
6. IfAis a scalar matrix with scalar k 0, of order 3, thenA1is
1) I 2) I 3) I 4) kI
7. If the matrix has an inverse then the values of k
1) kis any real number 2) k= 4 3) k 4 4) k 4
8. IfA= , then (adjA)A=
1) 2) 3) 4)
9. IfAis a square matrix of order n, then | adjA| is
1) |A| 2 2) | A| n 3) |A| n1 4) | A|
10. The inverse of the matrix is
1) 2) 3) 4)
11. IfAis a matrix of order 3, then det ( k A)
1) k3(detA) 2) k2(detA) 3) k(detA) 4) (detA)
12. If I is the unit matrix of order n, where k 0 is a constant, then adj(kI) =1) kn(adjI) 2) k(adjI) 3) k2 (adjI) 4) kn - 1(adjI)
13. IfAand Bare any two matrices such that AB= Oand Ais non-singular, then1) B= O 2) B is singular 3) B is non-singular 4) B= A
14. IfA= then A12is
1) 2) 3) 4)
15. The inverse of is
1) 2) 3) 4)
-
7/25/2019 Math -1516 EM
40/81
40
16. In a system of 3 linear non-homogenous equation with three unknowns, if = 0 and x= 0, y0 andz= 0
then the system has
1) unique solution 2) two solutions 3) infinitely many solutions 4) no
solution
17. The system of equations ax+ y+ z= 0, x+ by+ z= 0, x+ y+ cz= 0 has a non-trivial solution, then
=
1) 1 2) 2 3) 1 4) 0
18. If aex+ bey= c,pex+ qey= d and 1= , 2= , 3= , then the value of (x,y) is
1) 2) 3) 4)
19. If the equation 2x+ y+ z= l, x 2y+ z= m, x +y 2z= nsuch that l+ m+ n= 0, then the system has1) a non-zero unique solution 2) trivial solution 3) infinitely many solution 4)
No solution
Unit 2 Vector Algebra20. If is a non-zero vector and mis a non-zero scalar then m is a unit vector, if
1) m = 1 2) a= | m| 3) a= 4) a =1
21. If are two unit vectors and is the angle between them, then ( ) is a unit vector if
1) = 2) = 3) = 4) =
22. If and include an angle 120 and their magnitude are 2 and then . is equal to
1) 2) 3) 2 4)
23. If = ( ) + ( ) + ( ) then
1) uis an unit vector 2) = + + 3) = 4)
24. If + + = 0, | | = 3, | | = 4, | | = 5 then the angle between and is
1) 2) 3) 4)
25. The vectors 2 + 3 + 4 and a + b + c are perpendicular when
1) a= 2, b= 3, c= 4 2) a= 4, b= 4, c= 5 3) a= 4, b= 4, c= 5 4) a= 2, b= 3, c= 426. The area of the parallelogram having a diagonal 3 + and a side 3 + 4 is
1) 10 2) 6 3) 4) 3
27. If then
1) is parallel to 2) is perpendicular to 3) | | = | | 4) and are unit vectors
28. If , and + are vectors of magnitude then the magnitude of is
1) 2 2) 3) 4) 1
29. If ( ) + ( ) + ( ) = then
1) = 2) = 3) and are parallel 4) = or = or and are
parallel
30. If = 2 + + , = + 3 + 2 then the area of the quadrilateral PQRS is
1) 5 2) 10 3) 4)
31. The projection of on a unit vector equals thrice the area of parallelogram OPRQ. Then
is
1) tan1 2) cos1 3) sin1 4) sin1
32. If the projection of on and the projection of on are equal then the angle between + and
is
-
7/25/2019 Math -1516 EM
41/81
41
1) 2) 3) 4)
33. If ( ) = ( ) for non-coplanar vectors , , then
1) parallel to 2) parallel to 3) parallel to 4) + + =
34. If a line makes 45, 60 with positive direction of axesxand ythen the angle it makes with the zaxis
is
1) 30 2) 90 3) 45 4) 60
35. I f [ , , ] = 64 then [ , , ] is
1) 32 2) 8 3) 128 4) 0
36. If [ + , + , + ] = 8 then [ , , ] is1) 4 2) 16 3) 32 4) 4
37. The value of [ + , + , + ] is equal to
1) 0 2) 1 3) 2 4) 4
38. The shortest distance of the point (2, 10, 1) from the plane . (3 + 4 ) = 2
1) 2 2) 3) 2 4)
39. The vector ( ) ( ) is
1) perpendicular to , , and
2) parallel to the vectors ( ) and ( )
3) parallel to the line of intersection of the plane containing and and the plane containing
and
4) perpendicular to the line of intersection of the plane containing and and the plane containing
and
40. If , , are a right handed triad of mutually perpendicular vectors of magnitude a, b, cthen the
value of
[ , , ] is
1) a2b2c2 2) 0 3) abc 4) abc
41. If , , are non-coplanar and [ , , ] = [ + , + , + ] then [ , , ] is
1) 2 2) 3 3) 1 4) 0
42. = s + t is the equation of1) a straight line joining the points and 2)xoyplane 3) yozplane 4) zoxplane
43. If the magnitude of the moment about the point + of a force + a ) acting through the point
+ is then the value of ais
1) 1 2) 2 3) 3 4) 4
44. The equation of the line parallel to and passing through the point (1, 3. 5) in
vector form is
1) = ( + 5 + 3 ) + t( + 3 + 5 ) 2) = ( + 3 + 5 ) + t( + 5 + 3 )
3) = ( + 5 + ) + t( + 3 + 5 ) 4) = ( + 3 + 5 ) + t( + 5 + )
45. The point of intersection of the line = ( ) + t (3 + 2 + 7 ) and the plane . ( + ) = 8 is
1) (8, 6, 22) 2) (8, 6, 22) 3) (4, 3, 11) 4) ( 4, 3, 11)46. The equation of the plane passing through the point (2, 1, 1) and the line of intersection of theplanes
. ( + 3 ) = 0 and .( + 2 ) = 0 is
1)x+ 4y z= 0 2)x+ 9y+ 11z= 0 3) 2x+ y z + 5 = 0 4) 2x y+ z= 0
47. The work done by the force = + + acting on a particle, if the particle is displaced from A(3, 3,
3) to the point B(4, 4, 4) is
1) 2 units 2) 3 units 3) 4 units 4) 7 units
-
7/25/2019 Math -1516 EM
42/81
42
48. If = 2 + 3 and = 3 + + 2 then a unit vector perpendicular to and is
1) 2) 3) 4)
49. The point of intersection of the lines and is
1) (0, 0, 4) 2) (1, 0, 0) 3) (0, 2, 0) 4) (1, 2, 0)50. The point of intersection of the lines
= ( + 2 + 3 ) + t( 2 + + ) and = (2 + 3 + 5 ) + 5( + 2 + 3 )
1) (2, 1, 1) 2) (1, 2, 1) 3) (1, 1, 2) 4) (1, 1, 1)
51. The shortest distance between the lines and is
1) 2) 3) 4)
52. The shortest distance between the parallel lines and is
1) 3 2) 2 3) 1 4) 0
53. The following two lines are and
1) parallel 2) intersecting 3) skew 4) perpendicular
54. The centre and radius of the sphere given byx2+ y2+ z2 6x+ 8y 10z+1 = 0 is1) (3, 4, 5), 49 2) ( 6, 8, 10) 1 3) (3, 4, 5), 7 4) (6, 8, 10), 7
Unit 3 Complex Numbers
55. The value of + is
1) 2 2) 0 3) 1 4) 156. The modulus and amplitude of the complex number [e3 - i/4 ]3 are respectively
1) e9, 2) e9, 3) e6, 4) e9,
57. If (m 5) + i(n+ 4) is the complex conjugate of (2 m+ 3) + i(3n2) then ( n, m) are
1) 2) 3) 4)
58. Ifx2+ y2= 1 then the value of is
1)x iy 2) 2x 3) 2 iy 4)x+ iy
59. The modulus of the complex number 2 + i is
1) 2) 3) 4) 760. IfA+ iB= ( a1+ ib1) (a2+ ib2) (a3+ ib3) thenA2+ B2is
1) a12+ b12+ a22+ b22+ a32+ b32 2) ( a1+ a2+ a3 )2 + (b1+ b2+ b3 )2
3) (a12+ b12) (a22+ b22) (a32+ b32) 4) ( a12+ a22+ a32) (b12+ b22+ b32)
61. If a= 3 + i and z= 2 3ithen the points on the Argand diagram representing az, 3azand azare1) Vertices of a right angled triangle 2) Vertices of an equilateral triangle
3) Vertices of an isosceles triangle 4) Collinear
62. The points z1, z2,z3,z4 in the complex plane are the vertices of a parallelogram taken in order if
and only if
1) z1+ z4 = z2+ z3 2) z1+ z3 = z2+ z4 3) z1+ z2 = z3+ z4 4) z1 z2 = z3 z463. Ifzrepresents a complex number then arg(z) + arg( ) is
1) /3 2) /2 3) 0 4) /4
64. If the amplitude of a complex number is /2 then the number is1) purely imaginary 2) purely real 3) 0 4) neither real nor
imaginary
65. If the point represented by the complex number iz is rotated about the origin through the angle /2in the counter clockwise direction, then the complex number representing the new position is
1) iz 2) iz 3) z 4) z
66. The polar form of the complex number ( i25)3 is ---------
1) cos + isin 2) cosp+ isinp 3) cosp isinp 4) cos isin
-
7/25/2019 Math -1516 EM
43/81
43
67. If Prepresents the variable complex number zand if | 2z 1| = 2|z| then the locus of Pis1) the straight linex= 1/4 2) the straight line y= 1/4
3) the straight linez= 1/2 3) the circle x2+ y2 4x 1 =0
68. =
1) cosq+isin q 2) cosq isin q 3) sin q icosq 4) sin + icosq
69. Ifzn= cos + isin then z1,z2 .z6is
1) 1 2) 1 3) i 4) i 70. If lies in the third quadrant, then z lies in the ---------
1) first quadrant 2) second quadrant 3) third quadrant 4) fourth quadrant
71. Ifx= cos+ isin, the value ofxn+
1) 2 cosn 2) 2 i sin n 3) 2 sin n 4) 2 icosn72. If a= cos i sin , b= cos isin , c= cos isin then ( a2 c2 b2) / abcis
1) cos 2( + ) + isin 2( + ) 2) 2 cos ( + ) 3) 2 isin ( + ) 4) 2 cos ( + )
73. z1= 4 + 5 i,z2 = 3 + 2i, then is
1) 2) 3) 4)
74. The value of i + i22+ i23+ i24+ i25is
1) i 2) i 3) 1 4) 1
75. The conjugate of i13+ i14+ i15+ i16is
1) 1 2) 1 3) 0 4) i
76. If i+ 2 is one root of the equation ax2 bx+ c= 0, then the other root is1) i 2 2) i 2 3) 2 + i 4) 2i+ i
77. The quadratic equation whose roots are i is
1)x2+ 7 = 0 2)x2 7 = 0 3) x2+ x+ 7 = 0 4) x2 x 7 = 0
78. The equation having 4 3 i and 4 + 3 i as roots is1)x2+ 8x+ 25 = 0 2) x2+ 8x 25 = 0 3)x2 8x+ 25 = 0 4) x2 8x 25 = 0
79. If is the root of the equation ax2+ bx+ 1 = 0, where a, bare real then ( a, b) is
1) (1, 1) 2) (1, 1) 3) (0, 1) 4) (1, 0)80. If i+ 3 is a root of x2 6x+ k= 0, then the value of kis
1) 5 2) 3) 4) 10
81. If is a cube root of unity then the value of (1 + 2)4+ (1+ 2)4is1) 0 2) 32 3) 16 4) 32
82. If is the nth root of unity then1) 1+ 2+ 4+ = + 3+ 5+ 2) n= 0 3) n= 1 4) = n 1
83. If is the cube root of unity then the value of (1 ) (1 2) (1 4) (1 8) is1) 9 2) 9 3) 16 4) 32
Unit 4 Analytical Geometry84. The axis of the parabolay2 2y+ 8x 23 = 0 is
1)y= 1 2)x= 3 3)x= 3 4)y= 1 85. 16x2 3y2 32x 12y 44 = 0 represents
1) an ellipse 2) a circle 3) a parabola 4) a hyperbola
86. The line 4x+ 2y= cis a tangent to the parabola y2= 16xthen cis
1) 1 2) 2 3) 4 4) 4
87. The point of intersection of the tangents at t1 = tand t2 = 3tto the parabolay2= 8xis
1) (6t2, 8t) 2) (8 t, 6t2) 3) (t2, 4t) 4) (4t,t2)
88. The length of the latus rectum of the parabolay2 4x+ 4y +8 = 01) 8 2) 6 3) 4 4) 2
-
7/25/2019 Math -1516 EM
44/81
44
89. The diretrix of the parabolay2= x+ 4 is
1)x= 2) x= 3)x= 4)x=
90. The length of the latus rectum of the parabola whose vertex is (2, 3) and the diretrix isx= 4 is1) 2 2) 4 3) 6 4) 8
91. The focus of the parabolax2= 16yis
1) (4, 0) 2) (0, 4) 3) ( 4, 0) 4) (0, 4)92. The vertex of the parabolax2= 8y 1 is
1) 2) 3) 4)
93. The line 2x+ 3y+ 9 = 0 touches the parabola y2
= 8xat the point1) (0, 3) 2) (2, 4) 3) 4)
94. The tangents at the end of any focal chord to the parabolay2= 12xis intersect on the line
1)x 3 = 0 2)x+ 3 = 0 3) y+ 3 = 0 4) y 3 = 095. The angle between the two tangents drawn from the point (4, 4) to y2= 16xis
1) 45 2) 30 3) 60 4) 90
96. The eccentricity of the conic 9x2+ 5y2 54x 40y+ 116 = 0 is
1) 2) 3) 4)
97. The length of the semi-major and the length of semi-minor axis of the ellipse
1) 26, 12 2) 13, 24 3) 12, 26 4) 13, 12
98. The distance between the foci of the ellipse 9x2+ 5y 2=180
1) 4 2) 6 3) 8 4) 2
99. If the length of major and semi-minor axes of an ellipse are 8, 2 and their corresponding equations
arey 6 = 0 andx+ 4 = 0 then the equations of the ellipse is
1) 2) 3) 4)
100. The straight line 2x y+ c= 0 is a tangent to the ellipse 4x2+ 8y2= 32, if cis
1) 2) 6 3)36 4) 4
101. The sum of the distance of any point on the ellipse 4x2+ 9y2= 36 from ( , 0) and ( , 0) is
1) 4 2) 8 3) 6 4) 18
102. The radius of the director circle of the conic 9x2+ 16y2= 144 is
1) 2) 4 3) 3 4) 5
103. The locus foot of the perpendicular from the focus to a tangent of the curve 16x2+ 25y2= 400 is
1)x2+ y2 = 4 2)x2+ y2 = 25 3) x2+ y2 = 16 4) x2+ y2 = 9
104. The eccentricity of the hyperbola 12y2 4x2 24x+ 48y 127 = 01) 4 2) 3 3) 2 4) 6
105. The eccentricity of the hyperbola whose latus rectum is equal to half of its conjugate axis is
1) 2) 3) 4)
106. The difference between the focal distance of any point on the hyperbola is 24
and the eccentricity is 2. Then the equation of the hyperbola is
1) 2) 3) 4)
107. The directrices of the hyperbolax2 4(y 3) 2= 16 are
1)y= 2)x= 3) y= 4) x=
108. The line 5x 2y + 4k= 0 is a tangent to 4x2y2= 36 then kis
1) 2) 3) 4)
109. The equation of the chord of contact of tangents from (2, 1) to the hyperbola is
1) 9x 8y 72 = 0 2) 9x+ 8y+ 72 = 0 3) 8x 9y 72 = 0 4) 8x + 9y+ 72 = 0
-
7/25/2019 Math -1516 EM
45/81
45
110. The angle between the asymptotes to the hyperbola is
1) 2) 3) 4)
111. The asymptotes to the hyperbola 36y2 25x2+ 900 = 0 are
1)y= 2)y= 3)y= 4) y=
112. The product of the perpendiculars drawn from the point (8, 0) on the hyperbola to its asymptotes is
is
1) 2) 3) 4)
113. The locus of the point of intersection of perpendicular tangents to the hyperbola is
1)x2+ y2 = 25 2) x2+ y2 = 4 3) x2+ y2 = 3 4)x2+ y2 = 7
114. The eccentricity of the hyperbola with asymptotes x+ 2y 5 = 0, 2x y+ 5 = 0
1) 3 2) 3) 4) 2
115. Length of the semi-trasverse axis of the rectangular hyperbola xy = 8 is
1) 2 2) 4 3) 16 4) 8
116. The asymptotes of the rectangular hyperbolaxy= c2are
1)x= c,y= c 2) x= 0, y= c 3) x= c,y= 0 4)x= 0, y= 0
117. The co-ordinate of the vertices of the rectangular hyperbolaxy= 16 are
1) (4, 4), (4, 4) 2) (2, 8), ( 2, 8) 3) (4, 0), (4, 0) 4) (8, 0), (8, 0)118. One of the foci of the rectangular hyperbola xy= 18 is
1) (6, 6) 2) (3, 3) 3) (4, 4) 4) (5, 5)
119. The length of the latus rectum of the rectangular hyperbola xy= 32 is
1) 1) 2) 32 3) 8 4) 16
120. The area of the triangle formed by the tangent at any point on the rectangular hyperbolaxy= 72 and
its asymptotes is
1) 36 2) 18 3) 72 4) 144
121. The normal to the rectangular hyperbolaxy= 9 at meets the curve again at
1) 2) 3) 4)
Unit 5 Differential Calculas and its Applications I1. The gradient of the curvey= 2x3+3x+ 5 at x= 2 is
1) 20 2) 27 3) 16 4) 21 2. The rate of change of area A of a circle of radiusris
1) 2r 2) 2r 3) r2 4)
3. The velocity vof a particle moving along a straight line when at a distance xfrom the origin is given
by a+ bv2= x2where aand bare constants. Then the acceleration is
1) 2) 3) 4)
4. A spherical snowball is melting in such a way that its volume is decreasing at a rate of 1 cm3/ min.
The rate at which the diameter is decreasing when the diameter is 10 cm is
1) cm / min 2) cm / min 3) cm / min 4) cm / min
5. The slope of the tangent to the curvey= 3x2+ 3sin xat x= 0
1) 3 2) 2 3) 1 4) 16. The slope of the normal to the curvey= 3x2at the point whose xcoordinate is 2 is
1) 2) 3) 4)
7. The point on the curvey= 2x2 6x 4 at which the tangent is parallel to the x-axis is
1) 2) 3) 4)
-
7/25/2019 Math -1516 EM
46/81
46
8. The equation of the tangent to the curvey= at the point ( 1, 1/5) is
1) 5y+ 3x= 2 2) 5y 3x= 2 3) 3x 5y= 2 4) 3x+3y= 2
9. The equation of the normal to the curve = at the point ( 3, 1/3) is
1) 3= 27 t 80 2) 5 = 27 t 80 3) 3= 27 t+ 80 4) =
10. The angle between the curves and is
1) 2) 3) 4)
11. The angle between the curvey= emx
and y= emx
for m> 1 is1) tan1 2) tan1 3) tan1 4) tan1
12. The parametric equations of the curvex2/3+ y2/3= a2/3 are
1)x= asin 3 ;y= acos 3 2)x= acos 3 ;y= asin 3
3)x= a3sin ;y= a3cos 4) x= a3cos ;y= a3sin 13. If the normal to the curvex2/3+ y2/3= a2/3makes an angle with the x- axis then the slope of thenormal is
1) cot 2) tan 3) tan 4) cot 14. If the length of the diagonal of a square is increasing at the rate of 0.1 cm /sec. What is the rate of
increase of its area when the side is cm?
1) 1.5 cm2/sec 2) 3 cm 2/sec 3) 3 cm2/sec 4) 0.15 cm2/sec
15. What is the surface area of a sphere when the volume is increasing at the same rate as its radius
1) 1 2) 3) 4 4)
16. For what values ofxis the rate of increase of x3 2x2+3x+8 is twice the rate of increase of x
1) 2) 3) 4)
17. The radius of a cylinder is increasing at the rate of 2 cm /sec and its altitude is decreasing at the rate
of 3 cm /sec. The rate of change of volume when the radius is 3 cm and the altitude is 5 cm is
1) 23 2) 33 3) 43 4) 53
18. Ify= 6x x3and xincreases at the rate of 5 units per second, the rate of change of slope when x= 3 is1) 90 units / sec 2) 90 units / sec 3) 180 units / sec 4) 180 units / sec
19. If the volume of an expanding cube is increasing at the rate of 4 cm3/sec then the rate of change of
surface area when the volume of the cube is 8 cubic cm is
1) 8 cm2/sec 2) 16 cm 2/sec 3) 2 cm2/sec 4) 4 cm2/sec
20. The gradient of the tangent to the curvey= 8 + 4x- 2x2at the point where the curve cuts the y-axis is
1) 8 2) 4 3) 0 4) 421. The angle between the parabolasy2= xand x2 =yat the origin is
1) 2tan1 2) tan1 3) 4)
22. For the curvex = etcost;y = et sin tthe tangent line is parallel to the x-axis when tis equal to
1) 2) 3) 0 4)
23. If the normal makes an angle with positive x-axis then the slope of the curve at the point where thenormal is drawn is
1) cot 2) tan 3) tan 4) cot
24. The value of a so that the curves y = 3ex andy = ex intersect orthogonally is
1) 1 2) 1 3) 4) 3
25. If s= t3 4 t2+ 7, the velocity when the acceleration is zero is
1) m/sec 2) m/sec 3) m/sec 4) m/sec
-
7/25/2019 Math -1516 EM
47/81
47
26. If the velocity of a particle moving along a straight line is directly proportional to the square of its
distance from a fixed point on the line. Then its acceleration is proportional to
1) s 2) s2 3)s3 4) s4
27. The Rolles constant for the functiony =x2on [ 2, 2] is
1) 2) 0 3) 2 4) 2
28. The c of Lagranges Mean Value Theorem for the function f(x) =x2+ 2x 1 ; a= 0, b= 1 is
1) 1 2) 1 3) 0 4)
29. The value of cinRolles Theorem for the functionf(x) = cos on [, 3] is
1) 0 2) 2 3) 4)
30. The value c of Lagranges Mean Value Theorem for the function f(x) = when a= 1and b= 4 is
1) 2) 3) 4)
31.
1) 2 2) 0 3) 4) 1
32.
1) 2) 0 3) log 4)
33. Iff(a) = 2; f(a) = 1;g(a) = 1;g(a) = 2 then the value of is
1) 5 2) 5 3) 3 4) 334. Which of the following function is increasing in (0, )
1) ex 2) 3) x2 4) x 2
35. The function off(x) =x2 5x+ 4 is increasing in1) ( , 1) 2) (1, 4) 3) (4, ) 4) everywhere
36. The function off(x) =x2is decreasing in
1) ( , ) 2) (, 0) 3) (0, ) 4) ( 2, )
37. The function y= tan x xis
1) an increasing function in 2) a decreasing function in
3) increasing in and decreasing in 4) decreasing in and increasing in
38. In a given semi circle of diameter 4 cm a rectangle is to be inscribed. The maximum area of the
rectangle is
1) 2 2) 4 3) 8 4) 16
39. The least possible perimeter of a rectangle of area 100 m2is
1) 10 2) 20 3) 40 4) 60
40. Iff(x) =x2 4x+ 5 on [0, 3] then the absolute maximum value is1) 2 2) 3 3) 4 4) 5
41. The curvey= exis1) concave upward forx>0 2) concave downward forx>0
3) everywhere concave upward 4) everywhere concave downward42. Which of the following curves is concave downward?
1)y= x2 2)y= x2 3) y = e x 4) y= x2+ 2x 343. The point of inflexion of the curvey= x4is at
1)x= 0 2) x= 3 3) x= 12 4) nowhere
44. The curvey= ax3+ bx2+ cx+ dhas a point of inflexion at x = 1 then
1) a+ b= 0 2) a+ 3 b= 0 3) 3a+ b= 0 4) 3 a+ b= 1
-
7/25/2019 Math -1516 EM
48/81
48
Unit 6 Differential Calculas and its Applications II
45. If u= xythen is equal to
1)yxy1 2) ulog x 3) ulog y 4)xyx1
46. If andf= sin uthen f is a homogenous function of degree
1) 0 2) 1 3) 2 4) 4
47. If then + is equal to
1) u 2) u 3) u 4) u
48. The curvey2(x 2) = x2(1 + x) has1) an asymptote parallel tox-axis 2) an asymptote parallel toy-axis
3) asymptotes parallel to both axes 4) no asymptote
49. Ifx = rcos;y = rsin , then is equal to
1) sec 2) sin 3) cos 4) cosec 50. Identify the true statements in the following
(i) If a curve is symmetrical about the origin, then it is symmetrical about both axes
(ii) If a curve is symmetrical about both the axes, then it is symmetrical about the origin
(iii) A curvef(x,y) = 0 is symmetrical about the liney= xif f(x,y) = f(y,x)
(iv) For the curvef(x,y) = 0, iff(x,y) = f(y, x), then it is symmetrical about the origin1) (ii), (iii) 2) (i), (iv) 3) (i), (iii) 4) (ii), (iv)
51. If then + is
1) 0 2) u 3) 2 u 4) u1
52. The percentage error in the 11th root of the number 28 is approximately ______ times the percentage
error in 28
1) 2) 3) 11 4) 28
53. The curve a2y2= x2( a2 x2) has1) only one loop betweenx= 0 and x= a 2) two loops between x= 0 and x= a
3) two loops betweenx= aand x= a 4) no loop54. An asymptote to the curvey2( a+ 2x) =x2(3 a x) is
1)x= 3 a 2)x= a/2 3) x= a/2 4)x= 055. In which region the curvey2( a+ x) =x2(3 a x) does not lie
1)x> 0 2) 0 3a 4) a
-
7/25/2019 Math -1516 EM
49/81
49
62. The value of is
1) 2) 3) 4)
63. The value of is
1) 0 2) 2 3) log 2 4) log 4
64. The value of is
1) 3/16 2) 3/16 3) 0 4) 3/8
65. The value of is
1) 2) 3) 0 4)
66. The value of is
1) 2) /2 3) /4 4) 0
67. The area bounded by the liney= x, thex-axis, the ordinatesx=1, x= 2 is
1) 2) 3) 4)
68. The area of the region bounded by the graph of y= sin xand y= cosxbetween x= 0 and x= is
1) + 1 2) 1 3) 2 2 4) 2 + 2
69. The area between the ellipse and its auxillary circle is
1) b(a b) 2) 2a(a b) 3) a(a b) 4) 2 b(a b)70. The area bounded by the parabolay2 = xand its latus rectum is
1) 2) 3) 4)
71. The volume of the solid obtained by revolving about the minor axis is
1) 48 2) 64 3) 32 4) 128
72. The volume, when the curvey= from x= 0 to x= 4 is rotated about x-axis
1) 100 2) 3) 4)
73. The volume generated when the region bounded byy= x, y= 1, x= 0 is rotated about y-axis
1) 2) 3) 4)
74. Volume of solid obtained by revolving the area of the ellipse about major and minor
axes are in the ratio
1) b2:a2 2) a2:b2 3) a:b 4) b: a
75. The volume generated by rotating the triangle with vertices at (0, 0), (3, 0) and (3, 3) aboutx-axis is
1) 18 2) 2 3) 36 4) 976. The length of the arc of the curvex2/3+ y2/3 = 4 is
1) 48 2) 24 3) 12 4) 96
77. The surface area of the solid of revolution of the region bounded byy= 2x,x= 0 and x= 2 about x-
axis is
1) 8 2) 2 3) 4) 4
-
7/25/2019 Math -1516 EM
50/81
50
78. The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance
2 and 4 from the centre is
1) 20 2) 40 3) 10 4) 30
Unit 8 Differential Equations
79. The integrating factor of + 2 = e4xis
1) logx 2)x2 3) ex 4) x
80. If cosxis an integrating factor of the differential equation + Py= Q, then P=
1) cot x 2) cot x 3) tan x 4) tan x81. The integrating factor of dx+ xdy= e y sec 2ydyis1) ex 2) e x 3) ey 4) e y
82. The integrating factor of is
1) ex 2) logx 3) 4) e x
83. Solution of + mx= 0, where m< 0 is
1)x= cemy 2)x= cemy 3) x= my+ c 4) x= c
84. y= cx c2is the general solution of the differential equation1) (y)2 xy+ y= 0 2) y= 0 3)y= c 4) (y)2+ xy+y= 0
85. The differential equation + 5y1/3= xis
1) of order 2 and degree 1 2) of order 1 and degree 2
3) of order 1 and degree 6 4) of order 1 and degree 3
86. The differential equation of all non-vertical lines in a plane is
1) = 0 2) = 0 3) = m 4) = m
87. The differential equation of all circles with the centre at the origin is
1)xdy+ ydx= 0 2) xdy ydx= 0 3)xdx+ ydy= 0 4) xdx ydy= 0
88. The integrating factor of the differential equation +py= Q
1) pdx 2) Q dx 3) e Qdx 4) epdx89. The complementary function of (D2+ 1)y= e2xis
1) (Ax+ B) ex 2)Acosx+ Bsin x 3) (Ax+ B) e2x 4) (Ax+ B) ex
90. A particular integral of (D2 4 D +4)y= e 2xis
1) e2x 2)xe2x 3) xe2x 4) e 2x
91. The differential equation of the family of linesy =mxis
1) = m 2)ydxxdy= 0 3) = 0 4)ydx+xdy= 0
92. The degree of the differential equation
1) 1 2) 2 3) 3 4) 6
93. The degree of the differential equation where cis a constant is
1) 1 2) 3 3) 2 4) 294. The amount present in a radio active element disintegrates at a rate proportional to its amount. The
differential equation corresponding to the above statement is (kis negative)
1) 2) = kt 3) = kp 4) = kt
95. The differential equation satisfied by all the straight lines inxyplane is
-
7/25/2019 Math -1516 EM
51/81
51
1) = a constant 2) = 0 3) y+ = 0 4) + y = 0
96. Ify= kexthen its differential equation is
1) = y 2) = ky 3) + ky= 0 4) = ex
97. The differential equation obtained by eliminating aand bfrom y= ae3x+ be3xis
1) + ay= 0 2) 9y= 0 3) 9 = 0 4) + 9x= 0
98. The differential equation formed by eliminatingAand Bfrom the relation y= ex(Acosx+ Bsin x) is
1)y2+ y1= 0 2) y2 y1= 0 3)y2 2y1+ 2y= 0 4) y2 2y1 2y= 0
99. If then
1) 2xy+ y2+ x2= c 2) x2+ y2 x+ y= c 3) x2+ y2 2xy= c 4)x2 y2 2xy= c
100. Iff (x) = andf (1) = 2 thenf (x) is
1) (x +2) 2) (x +2) 3) (x +2) 4) x( +2)
101. On puttingy =vx, the homogenous differential equationx2dy+ y(x+ y)dx= 0 becomes
1)xdv+ (2 v+ v2)dx= 0 2) vdx+ (2x+ x2)dv= 0 3) v2dx (x +x2)dv= 0 4) vdv+ (2x+ x2)dx=0
102. The integrating factor of the differential equation -ytan x= cosxis
1) secx 2) cosx 3) etanx 4) cot x
103. The P.I. of (3D2+ D 14)y= 13 e2xis
1) 26xe2x
2) 13xe2x
3)xe2x
4)x2
/ 2 e2x
104. The particular integral of the differential equation f(D)y= eaxwhere f(D) = (D a) g(D),g(a) 0is
1) meax 2) 3) g(a) eax 4)
Unit 9 Discrete Mathematics105. Which of the following are statements?
(i) May God bless you (ii) Rose is a flower (iii) Milk is white (iv) 1 is a prime
number
1) (i), (ii), (iii) 2) (i), (ii), (iv) 3) (i), (iii), (iv) 4) (ii), (iii), (iv)
106. If a compound statement is made up of three simple statements, then the number of rows in the
truth table is
1) 8 2) 6 3)