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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

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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 --- --- ---

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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

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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:


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:


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

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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)

)=

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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


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 )


= 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)

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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


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


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 + )


= 0

(x1) (0) (y2) (10) + (z3) (5) = 0

10y 5z+ 35 = 0 by 5 2y+z 7 = 0

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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( + + )


= 0

(x1) (28) (y+2) (0) + (z3) (14) = 028x 14z+ 14 = 0

2x z + 1 = 0


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


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 + )


= 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

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Vector equation is = (1 s t) + + t

= (1st) + 4 + 2 )+ s(2 )+ t(7 + )


= 0

= 0

(x 3) ( 6) (y 4) (13) + (z 2) (28) = 06x 13y + 28z + 14 = 0

6x + 13y 28z 14 = 0

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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



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



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

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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

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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

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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

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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

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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.

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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

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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.

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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



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.

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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)

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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

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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


[1] G is the identity element.


Inverses of

respectively.


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.


Multiplication modulo 7 is always associative


[1] G is the identity element.


Inverses of are

respectively (G,7) is a group.

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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:


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:


3) Identity axiom:

Let E = be the identity element

AE = ae= a e= 1


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.

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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


= a, e= 3




= 3 = 3 =

G is the inverse of a

(G, is a group

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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


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



a 1,

is the inverse of a

Inverse axiom is true


a, b, G


(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


a, b, c G

=

=

=

= (

= (

=

=

=

Associative axiom it true3) Existence of Identity


=

e(1+ a) e= 0, [ a 1]

0 G is the identity element



G is the inverse of a



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

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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


a+ , c+ G

(a+ ) + (c+ )= ( a+ c) + (b+ d)

= (c+ a) + (d+ b)

= (c+ ) + ( a+ )


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



G

= = = .

Commutative axiom is true

(G, .) is an abelian group

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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


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

[ ]


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


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


5) Commuative axiom G

. = = =


G. contains finite number of elements.

G, . is a inite abelian group.

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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


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


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


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


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 = =

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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

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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(

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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

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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.

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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


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


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


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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 ) ];.

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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 .

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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.

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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

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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.

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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 ~

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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)

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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

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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

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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

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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

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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

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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)

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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

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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

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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

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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

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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

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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)

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