+2 · 2014. 1. 22. · page 2six marks march 2006 june 2006 october 2006 p represents the variable...

+2 MATHEMATICS

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SIX MARKS

MARCH 2006 JUNE 2006 OCTOBER 2006 P represents the variable complex number z.

Find the locus of P if 2 1 2z z .

Solve 4 4 0x , if 1+I is one of the roots. It is

given that 1+I is a root

Solve 2( 2 2) 2 5D D y Sin x

Let z x iy

2 ( ) 1 2x iy x iy

(2 1) (2 ) ( 2)x i y x iy

2 2 2 2(2 1) (2 ) ( 2)x y x y

2 2 2 2(2 1) (2 ) ( 2)x y x y

2 2 2 24 4 1 4 4 4x x y x x y

2 23 3 3x y

2 2 1x y

Which is a circle whose centre is origin and radius is 1 unit.

1 – i is also a root

Sum of the roots = 1 + i+1-i = 2

Product of the roots = (1+i) (1+i) =

2 21 ( ) 1 ( 1) 1 1 2i

The corresponding factor is

2x x (sum of roots) + product of roots

2 2 2x x

2 2 24 ( 2 2) ( 2)x x x x px

Equating x term we get

2 4O p 2p = 4 p = 2

Other factor is 2 2 2 0x x

To find the roots of 2 2 2 0x x

a =1, b=2, c = 2

2 4

2

b b acx

a

⇒

22 (2) 4 (1) (2)

2 (1)x

2 4 ( 1) 2 1 2

2 2 2

ix x x

1x i Other roots are -1 + i and -1 –i

Characteristic equation is 2 2 2 0p p

a =1, b = -2, c = 2

2( 2) ( 2) 4(1) (2)

2 (1)p

4 4 1

2 2p p

1p i

1, 1

Complementary function

[ ]xy e ACos x B Sin x

[ ]xy e ACos x B Sin x

1 2 2

1 12 2

2 2 2 2 2PI Sin x Sin x

D D D

1 12 2

4 2 2 2 2Sin x Sin x

D D

12

1Sin x

D

2

12

1

DSin x

D

[ 2 2 sin 2 ]Cos x x

25

2 2D D

General solution y = C.F + P.I1 + P.I2

[ ] [2 2 2 ]xy e ACos x B Sin x Cos x Sin x


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MARCH 2007 JUNE 2007 OCTOBER 2007 Find , ,k the normal distribution

whose probability distribution function

is given by 22 4 2( ) x xf x k e

Solve the equation 3 24 6 4 0 1x x x if i is a root

The given equation has three roots.

2( ) 1dy

x ydx

22 4 2( ) x xf x k e

22( 4 2 1)x x xk e

22( 1)xk e

2( 1)124 1

4

x

ke

2

( 1)124 1

4 ............... (1)

x

ke

We have

1

21( ) ............ (2)2

x

f x e

From (1) and (2)

11,2

1 1 2

12 22

k

Ans :

2 112

k

It is given that 1-i is a root

Its Conjugate 1+i also is a root

Sum of these roots = 1-i + 1+i

= 2

Product of these roots = (1-i)(1+i)

= 1-i2

= 1-(-1)

= 1+1 = 2

The factor corresponding to these roots is

x2 – x (sum) + product ⇒i.e, 2 (2) 2x x

i.e, 2 2 2x x

The corresponding factor is 2 (2) 2x x

32 2

2

44 6 4 2 2) ( 2) 2

2

xx x x x x x

x

The other factor is x-2

The root is x = 2

The roots are 1+i, 1-i and 2.

2( ) 1 (1)dy

x ydx

Let Z = x+ y

(2)

Differentiate w.r, t x

1dz dy

dx dx ⇒ 1 (3)

dy dz

dx dx

Put (2), (3) in (1) 2(1) 1 1dz

zdx

2

2 2 2

1 1 11 1

dz dz dz z

dx z dx z dx z

2

21

zdz dx

z

⇒2

2

(1 ) 1

1

zdz dx

z

2

11

1dz dx

z

2 1

dzdz dx C

z

1tanz z x c

1tan ( )x y x y x c

1tan ( )x y x y x c


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MARCH 2008 JUNE 2008 OCTOBER 2008 P represents the variable Z

Find the locus of P if Re1

01

z

z

Find the order of all the elements 0t the group (Z6,+6)

Show that , , 0a b b c c a

Let z = x+iy

1 1

1

z x i y

z x i y i

1)

( )

x iy

x i y i

1) ( 1)x

( 1) ( 1)

x i y x i y

x i y x i y

=

2 2

2 2

( 1 ( 1) ( 1)( 1)

( 1)

1 ( 1) ( 1)Re

1 ( 1)

x x y y i xy x y

x y

z x x y y

z x y

It is given that Re 1

01

z

z

2 2

( 1) ( 1) 0

1( 1) 1

x x y y

x y

ie ( 1) ( 1) 0x x y y

ie 2 2 0x x y y

2 2 0Locus of x y x y

Z6={[0],[1],[2],[3],[4],[5]} O[0]=1 O[1]=6 ∴[1]+[1]+[1]+[1]+[1]+[1]=0 O[2]=3 ∴[2]+[2]+[2]=0 O[3]=2 ∴[3]+[3]=0 O[4]=3 ∴[4]+[4]+[4] O[5]=6 ∴[5]+[5]+[5]+[5]+[5]+[5]=0

, ,LHS a b b c c a

.a b b c c a

.a b b c a c c a

.a b b c b a c c c a

.a b b c a b c a

. .a b c a b c a b b c a b c a

. . . . . .a b c a a b a c a b b c b a b b c a

0 0 0 0a b c b c a

a b c c b a

0a b c a b c RHS


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MARCH 2009 JUNE 2009 OCTOBER 2009 P presents the variable complex number z.

Find the locus of p, if 3 3z i z i .

Solve the following system of linear

equations by determinant method

2x-3y=7; 4x-6y=14.

Show that the points A (1,2,3), B (3, -1, 2), C (-2, 3,

1) and D (-6, -4, 2) are laying on the same plane.

Let z x iy

3 3z i z i

3 3x iy i x iy i

( 3) ( 3)x i y x i y

2 2 2 2( 3) ( 3)x y x y

Squaring both sides

2 26 9 6 9y y y y

12 0y

0y

The locus of P is x – axis.

2 312 12 0

4 6

7 32 42 0

14 6x

2 728 0

4 14y

Since 0 and 0x y . But atleast one

aij is not equal to zero.

The system is consistent and has

infinitely many solutions.

Put y = k

(1) 2 3 7x k

3 7 3x k

7 3 7 3,

2 2

k kx The solution is x y k

where k R

2 3OA i j k

3OB i j k

2 3OC i j k

6 4 2OD i j k

2 3AB OB OA i j k

3AC OC OA i j k

5AD OD OA i j k

2 3 1

, , 3 1 2

5 6 1

AB AC AD

= 2(-1-12) – (-3) (3-10) – 1 (18-5)

= 2 (-13) + 3 (13) – 1 (13)

= -26 + 39 – 13 = -39 + 39 = 0

, ,AB AC AD

are Co – planar Hence the points

A,B,C,D are coplanar.


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MARCH 2010 JUNE 2010 OCTOBER 2010

Verify that 1

1T

TA A

for the matrix

2 3

5 6A

.

With usual symbol, prove that 5 5( {0}, )Z

is a group Let G = { [1], [2]. [3],[4]}

Find the rank

2 1 4

0 1 2

1 3 7

1 adjAA

1 12 15 27A

6 3( )

5 2Adj A

16 31

5 227A

16 31

... (1)5 227

T

A

2 5

3 6

TA

1( )TTT

Adj AA

A

12 15 27TA

6 5

3 2

Tadj A

1 6 51

... (2)3 227

TA

From (1) and (2) 1 1

T TA A

5 [1] [2] [3] [4]

[1] [1] [2] [3] [4]

[2] [2] [4] [1] [3]

[3] [3] [1] [4] [2]

[4] [4] [3] [2] [1]

From the table

i. All the elements of the composition

table are the elements of G.

The closure axiom is true.

ii. Multiplication modulo 5 is always

associative.

iii. The identity elements [1] G and it

satisfies the identity axiom.

iv. Inverse of [1] is [1]

Inverse of [2] is [3] ; Inverse of [3] is [2]

Inverse of [4] is [4] and it satisfies the

inverse axiom. The given set forms a

group under multiplication modulo 5.

1 3

1 3 7

~ 0 1 2

2 1 4

R R

⇒3 3 2

1 3 7

~ 0 1 2

2 1 4

R R R

1 1 3

1 3 7

~ 0 1 2

2 1 4

R R R

3 3

1 3 7

~ 0 1 2 2

2 1 4

R R

1 1

51 1

3

~ 0 1 2 3

1 0 1

R R

3 1 3

51 1

3

~ 0 1 2

5 81 0

3 3

R R R

51 1

3

~ 0 1 2 3

21 0

3

Rank of the matrix is


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MARCH 2011 JUNE 2011 OCTOBER 2011 Evaluate

0lim cosxx

x

P represents the variable complex number z.

Find the locus of P it 3 5 3 1z z

Solve 2( 5 4) sin 5D D y x

Given 0

lim cosxx

x

0lim cos cosx xx

x

= 10

= 0

Let z = x+iy

3 5 3( ) 5z x iy

5) 3x yi

2 23 5 5) (3 )z x y

2 29 30 25 9 ......................(1)x x y

3 1 3 1z x iy

1)x i y

2 23 1)x y

2 22 1 ......................(2)x x y

2 2 2 23 5 3 1 9 30 25 9 2 1z z x x y x y x

(1) (2by and

Squaring both sides we get

2 2 2 29 9 30 25 9( 2 1)x y x x y x

29 x 29 y 230 25 9x x 29 y 18 9 0x

48 14 0x

24 7 0x

Locus of P is a straight line

Characteristic equation is 2 5 4 0P p

2 4 4 0P P P ⇒P (P+4) + +1 (P+4)=0

(P+4) (P+1) = 0 ⇒P +4 = 0 or P+1 = 0

P = -4 or P = -1

Complementary function is 4x xAe Be

2

1. 5

5 4P I Sin x

D D

2

15

5 5 4Sin x

D

15

5 21Sin x

D

5 215

(5 21) (5 21)

DSin x

D D

2

5 215

25 ( 5 ) 441

DSin x

(5 21) sin 5

625 441

D x

1[ 5.5 cos 21sin 5 ]

1066x x

(25 cos 5 21sin 5 )

1066

x x

General Solution is

4 (25 cos 5 21sin 5 )

1066

x x x xy Ae Be


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MARCH 2012 JUNE 2012 OCTOBER 2012 Find the order of each element of the

group, 7 7( {[0]}, )Z

G = {[1], [2], [3], [4], [5], [6]}

Find the rank of matrix

41

44

0

2( ) 1dy

x ydx

i). The identify element is [1]

So its order is 1

0 ([1]) = 1

ii). 3[2] = [2] 7 [2] 7 [2] = [8] = [1]

0 ([2]) = 3

iii). 6[3] = [3] 7 [3] 7 [3] 7 [3] 7 [3] 7 [3] 7

= [729] = [1]

0 ([3] = 6

iv). 3[4] = [4] 7 [4] 7 [4] 7

= [64] = [1]

0 ([4] = 3

v). 6[5] = [5] 7 [5] 7 [5] 7 [5] 7 [5] 7 [5] 7

= [15625] = [1]

0 ([5]) = 6

vi). 2[6] = [6] 7 [6]

= [36]

= [1]

0 ([6]) = 2

4 11

4 14

0 0

Let A

3 1 3

1

~ 1

0

R R R

3 2 3

1

~ 1 2 ( ) 3

0

R R R A

2( ) 1 (1)dy

x ydx

Let Z = x+ y (2)

Differentiate w.r, t x

1dz dy

dx dx ⇒ 1 (3)

dy dz

dx dx

Put (2), (3) in (1) 2(1) 1 1dz

zdx

2

2 2 2

1 1 11 1

dz dz dz z

dx z dx z dx z

2

21

zdz dx

z

⇒2

2

(1 ) 1

1

zdz dx

z

2

11

1dz dx

z

2 1

dzdz dx C

z

1tanz z x c

1tan ( )x y x y x c

1tan ( )x y x y x c


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MARCH 2013 JUNE 2013 Solve the following system of linear

equations by determinant method

x+y+3z=4; 2x+2y+6z =7; 2x+y+z=10.

If , ,a i j b j k c k j

, then find

, , ,a b b c c a

1 1 3

2 2 6

2 1 1

= 1 (2-6) – 1 (2-12) + 3(2-4)

= 1 (-4) – 1 (-10) + 3 (-2)

= - 4 + 10- 6

= - 10 + 10 = 0

=0

4 1 3

7 2 6

10 1 1

x

= 4 (2-6) – 1 (7-60) + 3 (7-20)

= 4 (-4) -1 (-53) + 3 (-13)

= - 16 +53 – 39

= -55 + 53 = -2

x = - 2 0

and x and so

The given equations are in consistent and

hence no solution.

( ) )a b i j j k

a b i j j k

2a b i j k

b c i j k

2c a i j k

, , [ 2 , 2 , ]a b b c c a i j k i j k i j k

1 2 1

1 1 2

2 1 1

= 1 (1+2) – (-2) (1-4) +1 (1+2)

= 1 (3) + 2 (-3) + 1 (3)

= 3-6+3

= 6-6

= 0

, , 0a b b c c a


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TEN MARKS

MARCH 2006 JUNE 2006 OCTOBER 2006 Find the volume of the solid

generated by the revolution of the

loop of the curve 2x t and 3

3

ty t

Find the vector and Cartesian equation of the

plane passing through the point

(-1, -2, 1) and perpendicular to two planes

x+2y+4z+7=0 and 2x-y+3z+3=0.

A poster is to have an area of 180 cm2 with 1 cm

margins at the bottom and sides and a 2cm

margin on the top. What dimension will give the

largest printed area ?(Diagram also needed)

2x t

3 3

13 3

t ty t t

2

2 2 13

xie y t

0 0 1 03

xLet y x or

0 3x or x Volume 3

2

0

y dx

23

0

13

xx dx

23 2

0

21

3 9

xx x dx

2 33 3 2 3 32

0 0

2 2 1. .

3 9 2 3 3 9 4

x x x xx x dx

32

3 4

0

2 1 9 96 (0)

2 9 36 2 4

xx x

9 9 18 24 9 3642 4

The vector normal to the planes x+2y+4z+7 = 0 and 2x-

y+3z+3=0 are 2 4i j k

and 2 3i j k

.

The required plane is r to the planes

x+2y+4z+7 = 0 and 2x-y+3z+3= 0

The required plane parallel to the above two vectors

2 4i j k

and 2 3i j k

and passes through the point

(-1, -2, -1). Vector equation is r a su tv

Iet 2 2 4 2 3r i j k s i j k t i j k

Its Cartesian equation is

1 1 1

1 1 1

2 2 2

1 0

1

x x y y z z

m n

m n

Thus the Cartesian equation is

1 2 1

1 2 4 0

2 1 3

x y z

(x+1) [6+4] – (y+2) [3-8] + (z-1) [-1-4] = 0

10x+10+5y+10-5z+5 = 0

10x+5y-5z+25 = 0⇒ 2x+y-z+5 = 0

Let the length of printed portion ABCD be x can and its

breadth be y cm (x+2) cm and its breadth is (y+3) cm.

Poster area is given as 2180 cm

(x+2) (y+3) = 180⇒180

32

yx

1803

2y

x

(1)

Let the area of the printed portion be A A = xy

1803

2A x

x

1803

2A x

x

2

2

3603 360 ( 2) 3

( 2)

dAx

dx x

23

2 3

720360 ( 2) ( 2) 0

( 2)

d Ax

dx x

0dA

dx 2 2

2

3603 0 3( 2) 360 3, ( 2) 120

( 2)x by x

x

2 120 2x x 2

2 2 30 2 3 3

720 720

(2 30 2 2) (2 30)x

d ANegative

dx

Printed

area is maximum when 2 30 2x 180(1) 32 30 2 2

90 90 303 3

3030y = 3 30 3 The he printed area

are 2 30 2 cm and 3 30 3 cm.

1 1 1( , , ) ( 1, 2,1)x y z

1 1 1(1 , , ) )m n

2 2 2(1 , , ) (2,m n

1 1 1(1 , , ) )m n

2 2 2(1 , , ) (2,m n


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MARCH 2007 JUNE 2007 OCTOBER 2007

Find the common area between 2y x

and 2x y .

Find the axis focus, latusreetum, equation of

the latus rectum, vertex and directrix for the

parabola 2 8 2 17 0y x y

Find the eccentricity, centre, foci and vertices of the

hyperbola 2 212 4 24 32 127 0x y x y and draw the

diagram.

2y x (1) 2x y

(2)

Solve (1) and (2) to get the point of

intersection Put (2) in (1)

2 2( )x x ⇒ 4 0x x ⇒ 3( 1) 0x x

30 1 0x x

x =0 x= 1

x = 0 in (2), y =0

x = 1 in (2), y = 1

The point of intersection are (0,0) and (1,1)

Required area [ ( ) ( )]b

a

f x g x dx

2( ) ,f x y x 2( ) :g x x y

2x x dx

332 3

322 1

3 3 3 32

x xx x

2 1.1 .1 0 0)

3 3

2 1 2 1.

3 3 3

1

3

sq. units

2 8 2 17 0y x y ⇒ 2 2 8 17y y x

2 2 22 1 1 8 17y y x ⇒ 2( 1) 1 8 17y x

2( 1) 8 17 1y x ⇒ 2( 1) 8( 2)y x

2 8 Y=y-1X=x-2Y X where 4a 8 2a

About X,Y Referred to x.y

Axis Y=0 Y=0y-1=0y=1

Focus (a,0) ie (2,0) X=2x-2=2x=2+2 x=4

Y=0y-1=0⇒y=1⇒F(4,1)

Latus

rectum

X=a

ie X=2

X=2x-2=2x=2+2

x=4

Vertex (0,0) X=0x-2=0 x=2

Y=0y-1=0⇒ y=1V (2,1)

Directri

x

X = -a

X = -2

2 2 2X x

2 2x ⇒ 0x

Graph referred to x,y

2 212 4 24 32 127 0x y x y

2 2 2 2 2 212( 2 1 1 ) 4 ( 8 4 4 ) 127x x y y

2 212[( 1) 1] 4[( 4) 16] 127x y

2 212 ( 1) 4( 4) 127 12 64x y

2 2

175 75

12 4

X YWhere X x

4Y y

5 5 3

2 2a b

2

75 41

4 25e

2 21 3 4 4 2e e e e

Referred to X, Y Referred to x,y

Centre C (0,0 ) C (1,4)

Foci F (ae, 0) = (5,0)

1F (-a e, 0) = (-5,0)

F (6,4)(-4, 4)

Vertics 5,0

2A

5,0

2A

7

2, 4 3 ,42A

Graph Referred to x, y


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MARCH 2008 JUNE 2008 OCTOBER 2008 Find the vector and Cartesian equation of

the plane passing through the point

(-1, -2, 1) and perpendicular to two planes

x+2y+4z+7=0 and 2x-y+3z+3=0.

Find the common area between 2y x and

2x y .

Find the local minimum and maximum values of

3 2( ) 2 3 36 10f x x x x

The vector normal to the planes x+2y+4z+7 = 0

and 2x-y+3z+3=0 are 2 4i j k

and 2 3i j k

.

The required plane is r to the planes

x+2y+4z+7 = 0 and 2x-y+3z+3= 0

The required plane parallel to the above two

vectors 2 4i j k

and 2 3i j k

and passes

through the point (-1, -2, -1).

Vector equation is r a su tv

Iet 2 2 4 2 3r i j k s i j k t i j k

Its Cartesian equation is

1 1 1

1 1 1

2 2 2

1 0

1

x x y y z z

m n

m n

Thus the Cartesian equation is

1 2 1

1 2 4 0

2 1 3

x y z

(x+1) [6+4] – (y+2) [3-8] + (z-1) [-1-4] = 0

10x+10+5y+10-5z+5 = 0

10x+5y-5z+25 = 0⇒ 2x+y-z+5 = 0

2y x (1) 2x y (2)

Solve (1) and (2) to get the point of

intersection Put (2) in (1)

2 2( )x x ⇒ 4 0x x ⇒ 3( 1) 0x x

30 1 0x x

x =0 x= 1

x = 0 in (2), y =0

x = 1 in (2), y = 1

The point of intersection are (0,0) and (1,1)

Required area [ ( ) ( )]b

a

f x g x dx

2( ) ,f x y x 2( ) :g x x y

2x x dx

3

32 332

2 1

3 3 3 32

x xx x

2 1.1 .1 0 0)

3 3

2 1 2 1.

3 3 3

1

3 sq. units

3 2( ) 2 3 36 10f x x x x

2'( ) 6 6 36f x x x

"( ) 12 6f x x '( ) 0Let f x

26 6 36 0x x ⇒ 2. , 6 0i e x x

2 3 2 6 0x x x ⇒ ( 3) 2 ( 3) 0x x x

( 3) ( 2) 0x x ⇒ 3 0 , 2 0x x

3, 2x x

3[ "( )] 12 ( 3) 6xf x 36 6 30

negative

f(x) attains maximum when x= 3

maximum value is 3 22( 3) 3( 3) 36 ( 3) 10

= 2 ( 27) (9) 108 10 54 27 108 10 91

2[ "( )] 12 (2) 6 24 6 30xf x positive

( )f x attains minimum when x = 2

Minimum value is

3 22(2) 3(2) 36(2) 10 2 (8) 3(4) 72 10ie

16 12 72 10 34ie ie

Minimum value is -34.

1 1 1( , , ) ( 1, 2,1)x y z

1 1 1(1 , , ) )m n

2 2 2(1 , , ) (2,m n

1 1 1(1 , , ) )m n

2 2 2(1 , , ) (2,m n


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MARCH 2009 JUNE 2009 OCTOBER 2009 P represents the variable Z

Find the locus of P if Re1

01

z

z

Find the eccentricity, centre, foci and

vertices of the ellipse 2 216 9 32 36 92x y x y

and draw the diagram .

Solve : 2 3( 5 6) 2 xD D y Sin x e

Let z = x+iy

1 1

1

z x i y

z x i y i

1)

( )

x iy

x i y i

1) ( 1)x

( 1) ( 1)

x i y x i y

x i y x i y

=

2 2

2 2

( 1 ( 1) ( 1)( 1)

( 1)

1 ( 1) ( 1)Re

1 ( 1)

x x y y i xy x y

x y

z x x y y

z x y

It is given that Re 1

01

z

z

2 2

( 1) ( 1) 0

1( 1) 1

x x y y

x y

ie ( 1) ( 1) 0x x y y

ie 2 2 0x x y y

2 2 0Locus of x y x y

2 216 9 32 36 92x y x y

2 216( 2 1 1) 9 ( 4 4 4) 92x x y y

2 216 ( 1) 9 ( 2) 144x y

ing by 144 2 2( 1) ( 2)

19 16

x y

2 2

19 16

x y 2 216, 9a b

2 2 7

4

a be

a

w.r. to X, Y w.r to x,y

1, 2x x y y

Centre

(0, 0)

(1, -2)

Foci (0, )ae

(0, 7)

(1, 2 7) (1, 2 7)and

Vertices

(0, ±a)(0, ±4)

(1, 2) (1 6)and

Characteristic equation is 2 5 0P p

(p-3) (p-2) = 0⇒P = 3 or p = 2

Complementary function is 3 2x xAe Be

1 2

1

5 6PI Sin x

D D

2

1

1 5 6Sin x

D

1

1 5 6Sin x

D

1

5 5Sin x

D

2

1 1 1 1

5 ( 1) ( 1) 5 1

D DSin x Sin x

D D D

1 1

5 1 1

DSinx

1 1 11)

5 2 10

DSin x D Sin x

1 1] ]

10 10D Sin x Sin x Cos x Sin x

3 3

2 2 2

1 12

5 6 3 5.3 6

x xPI e eD D

31

0

xe

3 3

2

1 12

( 3) ( 2) ( 3 3) (3 2)

x xPI e eD D D

3 3 31 12.1

x x xe e xeD D

General Solution is y = C.F + P.I1 + P.I 2

3 2 31 [ ] 210

x x xy Ae Be Cosx Sinx xe


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MARCH 2010 JUNE 2010 OCTOBER 2010 Find the volume of the solid obtained by

revolving the area of the triangle whose sides are having the equation y = 0, x=4 and 3x-4y=0,

about 3 4 0x y

3 2 2(1 2 ) 6 secdy

x x y Co xdx

Find the vector and Cartesian equations of the

plane, through the point (1,2-2) and parallel to

the line 2 1 4

3 2 4

x y z

and perpendicular to

the plane 2x+3y-3z=8.

about

3 4 0x y ⇒y = 0 3 x = 0 x = 0

x = 0 to 4 3x – 4y = 0

4y = 3x ⇒3

4y x ⇒ 2 2

9

16y x

Volume 2b

a

y dx

4

2

0

9

16x dx

44 32

0 0

9 9

16 16 3

xx dx

33

4 0 12 .16

cu units

3 2 2(1 2 ) 6 secdy

x x y Co xdx

2 23

3 3

6 sec1 2 ,

1 2 1 2

dy x Co xx

dx x x

2 2

3 3

6 sec

1 2 1 2

x Co xP Q

x x

23

3

6log (1 2 )

1 2

xPdx x

x

3log (1 2 ) 3Pdx x

e e x

Solution is ( ) ( )y IF Q IF dx C

23 3

2

sec(1 2 ) (1 2 )

1 2

Co xy x x dx C

x

3 2(1 2 ) secy x Co x dx C

3(1 2 )y x Cot x C

3(1 2 )y x Cot x C

The normal vector to the plane 2x+3y+3z =8 is

2 3 3i j k

.

This vector is parallel to the required plan. The

required plane passes through (1,2-2) and parallel

to 3 2 4u i j k

and 2 3 3v i j k

.

The vector equation of the required

plane is r a su tv

i.e. 2 2 (3 2 .4 ) ( 3 3 )r i j k s i j k t i j k

Cartesian from

1 1 1

1 1 1

2 2 2

1 0

1

x x y y z z

m n

m n

1 2 2

3 2 4 0

3 3 3

x y z

i.e. (x-1) (-6+12) – (y-2) (9+8) + (z+2) (9+4) = 0

6 (x-1) – 17 (y-2) + 13 (z+2) = 0

6x-6 -17y+34+13z+26 = 0⇒6x-17y+ 13z+ 54 = 0

1 1 1( , , ) (1,2, 2)x y z

1 1 1(1 , , ) )m n

1 2 2(1 , , ) (2,m n


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MARCH 2011 JUNE 2011 OCTOBER 2011 Find the vertex, axis, focus, equation of latus

rectum, equation of directrix and length of latus

rectum of the parabola 2 4 4 8 0y y x and

hence. Draw the diagram.

Find the vertex, axis, focus, equation of latus rectum, equation of directrix

and length of latus rectum of the

parabola 2 4 4x x y

Solve the system to equations.

X+2y+z=2, 2x+4y+2z=4, x-2y-z=0

2 4 4 8 0y y x

2 4 4 4 4y x x ⇒ 22) 4 4y x

22) 4y x 2 4Y X

Where Y = y +2 ; X = x +1 ; 4a = 4 ; a = 1

The type is open left ward.

Referred to

X, Y

Referred to x,y

X = x +1 Y = y+2

Vertex (0, 0) (-1,-2)

axis Y = 0 Y = -2

Focus (-a,0)i.e (-1,0) Focus = = (-2, -2)

Equation of

latus rectum

X = -a

i.e X = -1

X = -1 x +1 = -1

x = -2

Equation of

directrix

X =a

i.e X = 1

X = 1 x+1 = 1

x = 0

Length of L.R 4a = 4 4a =4

2 4 4x x y ⇒2 2 24 2 2 4x x y

2( 2) 4( 1)x y

2 4 1X Y Where Y y

4 4 2a X x 1a

The type is open downward.

Referred

toX, Y

Referred to x,y

X = x +1

Y = y+2

axis X = 0 x- 2 = 0

Vertex (0,0) (2, 1)

Focus (0, -1) Focus = (2, 0)

Equ.ofL.R Y = 1 y = 2

Equ.of dir. Y = -1 y = 0

Len. L.R 4a=4 4a = 4

1 2 3

2 4 2 1( 4 4) 2 ( 2 2) 3 ( 4 4)

1 2 1

2( 4) 3( 8) 16

2 2 3

4 4 2 ( 4 4) 2 ( 4 0) 3 ( 8 0)

0 2 1

x

= 8

– 24 = -16

1 2 3

2 4 2 ( 4 0) 2 ( 2 2) 3 ( 0 4)

1 2 1

y

= -4

+ 8 – 12 = -8

1 2 2

2 4 4 (0 8) 2 (0 4) ( 4 4)

1 2 0

z

= 8 + 8 – 16 = 0

161

16

xx x

8

16

yy

⇒

1

2y

00

16

zz

11, , 02

x y z


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MARCH 2012 JUNE 2012 OCTOBER 2012 Find the area between the curves

2 2 3y x x x – axis and the lines x = -3

and x =5.

Find the eccentricity, centre, foci and

vertices of the hyperbola

2 2(9 36 ) (7 14 ) 92x x y y and draw the

diagram.

Find the vector and Cartesian equations of the

plane which contains the line 1 1

2 3 1

x y z and

perpendicular to the plane x-2y+3z-2=0.

Required area 1 2 3A A A

( )y dx y dx y dx

2 2 2( 2 3 ) (3 2 ( 2 3)x x dx x x dx x x dx

1 3 5

3 3 32 2 2

3 1 3

3 3 33 3 3

x x xx x x x x x

11 3

3

1 1253 1 25 15

3 3

2 9 9 2 40 9

= 32

sq.unit

2 2(9 36 ) (7 14 ) 92x x y y

2 29 ( 4 ) 7 ( 2 ) 92x x y y

2 29[( 2) 4] 7[( 1) 1] 92x y

2 27 ( 1) 9( 2) 63y x

2 27( 1) 9( 2)1

63 63

y x

2 2( 1) ( 2). 1

9 7

y xi e

2 2

19 7

Y X Where 2; 1X x Y y

2 9 3a a 2 7b 2

21

be

a

4 12 164 2

4 4

Referred

to X, Y

Referred to x,y

2; 1X x Y y

Centre C (0,0 ) (-2, 1)

Foci F (0, 6 ) (-2, 7) & (-2, - 5)

Vertices (0, 3)V (-2, 4) and (-2, -2)

The required plane contains the line 1 1

2 3 1

x y z i.e,

the line 1 0 ( 1)

2 3 1

x y z

The plane passes through

the point (1, 0, -1) and parallel to the vector 2 3u i j k

This plane is perpendicular to x-2y+3z-2 = 0.

That is the plane is parallel to 2 3i j k

.

The required plane passes through (1,0, -1) whose

postion vector a i j

and parallel to two vector

2 3u i j k

and 2 3L j k

.

Vector equation r a su tv

i.e, (2 3 ) ( 2 3 )r i j s i j k t i j k

Its Cartesian form is 1 1 1

1 1 1

2 2 2

1 0

1

x x y y z z

m n

m n

Ie1 0 1

2 3 1 0

1 2 3

x y z

(x-1) (-9+2) – y (6-1) + (z+1) (-4+3) = 0

(x-7) (-7) – y(5) + (z+1) (-1) = 0

-7x – 5y –z +6 = 0⟹7x+5y+z-6 = 0


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MARCH 2013 JUNE 2013

Verify 2 2u u

x y y x

for the function n

sinx

uy

1

2

sin Pr .

u u x y x yx y Cos if

x y x y x y

x yu ove

x y

sinx

uy

1cos

xu

x y y

2

2 2

1 1cos .

x x xuSin

x y y y y y y

2 3

1cos .... (1)

x x xSin

y y y y

2cos

x xu

y y y

2

2 2

1 1cos sin

x x xu

x y y y y y y

2 3

1cos

x x xSin

y y y y

2 2u u

y x x y

sinx y

ux y

1. sin ( )x y

i e ux y

1 ( )x y

f Sin ux y

Put x = tx, y = ty⇒tx ty

ftx ty

( )

( )

t x yf

t x y

⇒

12

( )x yf t

x y

f is a homogeneous function in x and y of degree 12

&By Euler’s theorem 1

.2

f fx y f

x y

1 1 11(sin ) ( )2

x y Sin u Sin ux y

1

2 2

1 1 1. .

21 1

u ux y Sin u

x yu u

2 11. .2

u ux y u Sin u

x y

2 111 .2

x y x ySin Sin Sin

x y x y

2 1.2

x y x yCos

x y x y

1

2

x y x yCos

x y x y


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+2 · 2014. 1. 22. · page 2six marks march 2006 june 2006 october 2006 p represents the variable...

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