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MATHEMATICS

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SIX MARKS

MARCH 2006

Page 2

JUNE 2006

OCTOBER 2006

P represents the variable complex number z. Solve x  4  0 , if 1+I is one of the roots. It is Solve ( D  2D  2) ySin2 x5 2

4

Find the locus of P if 2 z  1 z  2 .

given that 1+I is a root

 1 – i is also a root

Let zx  iy

Sum of the roots

2( x  iy)  1 x iy 2

Product of the roots

(2 x  1)i(2 y)  ( x  2)iy 2

2

2

The corresponding factor is

(2 x 1)2  (2 y)2 ( x  2)2  y 2 4 x  4 x  1  4 y x  4 x  4  y 2

2

2

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

3x2  3 y 2  3

x2  2 x  2

2

2

2

Equating x term we get

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

O2 p 4

2p = 4 p = 2

Other factor is x  2 x  20 2

To find the roots of x2  2 x  2  0 a =1, b=2, c = 2 2 (2) 2  4(1)(2) b b  4ac ⇒ x  x 2(1) 2a 2

x

2 4(1) 2 1 2i x x 2 2 2

x 1 i

a =1, b = -2, c = 2 (2) (2) 2  4(1)(2) p 2(1)

 4  4 1  p  p1 i 2 2   1,  1  p

Complementary function

x 4( x  2 x  2)( x  px  2) 2

x  y  1 2

= (1+i) (1+i) =

12 (i)2 1 (1)112

(2 x  1)  (2 y)  ( x  2)  y  2

= 1 + i+1-i = 2

Characteristic equation is p 2  2 p 2  0

Other roots are -1 + i and -1 –i

ye x [ ACos xBSin x] ye x [ ACosxBSinx] 1 1        Sin2 x PI1 2 Sin 2 x 2    D  2D  2   2  2 D  2  1    1   Sin2 x Sin2 x   4  2 D  2   2 D  2   1    D 1    Sin2 x   2 Sin2 x    D  1   D  1      5  [2Cos2xsin2 x]  2     D  2D  2 

General solution y = C.F + P.I1 + P.I2   ye x [ ACosxBSinx] [2Cos2 xSin2 x]  

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MARCH 2007

JUNE 2007

Find k , , the normal distribution

Solve the equation x  4 x  6 x  4 0if 1  iisaroot

whose probability distribution function

The given equation has three roots.

3

OCTOBER 2007 Page 3

2

( x  y)2 

dy  1 dx

is given by f ( x)ke2 x 4 x2 2

f ( x)ke2 x

2

ke2( x

2

ke

Sum of these roots

2

( x 1) 2 1  4 1 4

2

 ( x 1)   2 1  4 1   4 

Product of these roots 2

...............(1)

We have 1  x   

  1 f ( x) e 2    2

............(2)

1-i + 1+i

=

2

=

(1-i)(1+i)

=

1-i2

=

1-(-1)

=

1+1 = 2

– x (sum) + product ⇒i.e,

i.e,

 1,  1 2

=

The factor corresponding to these roots is x2

From (1) and (2)

1 1 2      2 1 2  2

x 2  2 x2

 The other factor is x-2 k 



x  x(2)  2

 x3  4 4 x 2  6 x  4 x 2  2 x2)( x  2) 2  x  2 2  x 

Ans :  1  1

2

 The corresponding factor is x2  x(2)  2

k 

2

( x  y)2 

 Its Conjugate 1+i also is a root

 4 x  2 x 1)

ke2( x1) ke

It is given that 1-i is a root

 4 x2

 The root is x = 2 2

 The roots are 1+i, 1-i and 2.

dy  1(1) dx

Let Z = x+ y

(2) Differentiate w.r, t x 

dz dy dy dz 1 ⇒   1(3) dx dx dx dx

 dz   1 1  dx 

Put (2), (3) in (1) (1)z 2 

dz 1 dz 1 dz 1  z 2  1 2   2  1  2 dx z dx z dx z

z2 dzdx 1 z2

⇒

(1  z 2 )  1 dz  dx 1 z2



1

 1 1  z  dz  z

2

 dz  dx 

dz  dx  C 1 

2

ztan 1 zxc

x ytan 1 ( x  y)x c x ytan 1 ( x  y)x c

MARCH 2008 P represents the variable Z  z 1  0  z 1 

Find the locus of P if Re  Let z = x+iy

 x  1)  iy x  i y  1  z 1      x  i( y  i) x  i y  i  z 1 

 x  1)  i yxx  i( y  1)   x  i( y  1) x  i( y  1) =

x( x  1  y( y  1)

i  xy  ( x  1)( y  1)  2

x  ( y  1)

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JUNE 2008 Find the order of all the elements 0t the group (Z6,+6)

OCTOBER 2008

      LHS a  b,b  c,c  a 

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

 z 1 

It is given that Re  0  z 1  x( x  1)  y( y  1) 0  x 2  1( y  1)2 1

ie x( x  1)  y( y  1)0 ie x2  x  y 2  y0 Locusof x2  y 2  x  y0

 

       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



2

x( x  1)  y ( y  1)  z 1  Re  x 2  ( y  1) 2  z 1 

Page 4

      Show that a  b,b  c,c  a 0





              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



















   abc 0  0  0  0 bca     abc  cba     abc   abc 0RHS







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MARCH 2009

JUNE 2009

OCTOBER 2009

P presents the variable complex number z.

Solve the following system of linear

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

equations by determinant method

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

Find the locus of p, if z  3i  z  3i .

2x-3y=7; 4x-6y=14. 2 3  12  120 4 6

Let z  x  iy z  3i  z  3i x  iy  3i  x  iy  3i

x

x  i( y  3)  x  i( y  3)

x  ( y  3)  x  ( y  3) 2

2

2

y

2

Squaring both sides

2 7   280 4 14

Since   0 and x y0 . But atleast one

y 2  6 y  9 y 2  6 y  9 12 y0

7 3 2  420 14 6

aij is not equal to zero.

 The system is consistent and has infinitely many solutions.

y0

 The locus of P is x – axis.

Put y = k (1)2 x  3k 7

3x  7  3k x

7  3k 7  3k Thesolutionisx , yk 2 2

wherekR

    OAi  2 j3k     OB 3i  jk     OC 2i  3 jk     OD 6i  4 j 2k       AB OB OA2i3 jk       AC OC OA 3i jk       AD OD OA 5i  jk

2 3 1     AB, AC , AD  3 1 2   5

6 1

= 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

Verify that  A

  A 

1 T

T

1

for the matrix

A1

adj A

1

A 121527

 6 3  Adj( A)   5 2 1  6 3  A   27  5 2 1

A 

1 T

With usual symbol, prove that (Z5 {0},5 )

 2 1  4  Find the rank  0 1 2   1 3  7 

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]

i.

 The closure axiom is true. ii.

iii.

The identity elements [1] G and it satisfies the identity axiom.

iv.

 6 5 adj AT    3 2

A 

Multiplication modulo 5 is always associative.

Adj( AT )  AT

AT 121527

T 1

All the elements of the composition table are the elements of G.

 2 5  AT    3 6

A 

OCTOBER 2010

From the table

1  6 3    ...(1) 27  5 2

1 T

JUNE 2010 is a group Let G = { [1], [2]. [3],[4]}

 2 3 A .  5 6

From (1) and (2)   AT   AT  1

Inverse of [1] is [1]

Inverse of [2] is [3] ; Inverse of [3] is [2] Inverse of [4] is [4] and it satisfies the

1  6 5   ...(2) 27  3 2

inverse axiom.  The given set forms a 1

Page 6

group under multiplication modulo 5.

 1 3  7  1 3  7   ⇒ ~  0 1 2 R1 R3 ~  0 1 2R3  R3  R2    2 1  4   2 1  4 

 1 3  7 ~  0 1 2 R1 R1  R3  2 1  4 

 1 3  7 ~  0 1 2 R3  R3  2  2 1  4   1 1  ~0 1  1 0  

 5  3   2 R1 R1  3  1  

 1 1  ~0 1   1 0 

 5  3   2 R3  R1  R3 5 8  3 3

 1 1  ~0 1   1 0 

 5  3    2 Rank of thematrixis3  2    3

www.em.tnschools.co.in

MARCH 2011 Evaluate lim x cos x 0

x

JUNE 2011

OCTOBER 2011

Page 7

P represents the variable complex number z. Solve ( D  5D4) ysin5x 2

Find the locus of P it 3z  5 3 z  1 Given lim x cos x x 0

lim x cos x  cos x x 0

=0

Let z = x+iy

Characteristic equation is P2  5 p40

3z  53( x  iy)  5

P2  4P  P  4  0 ⇒P (P+4) + +1 (P+4)=0

x  5)  3 yi

(P+4) (P+1) = 0 ⇒P +4 = 0 or P+1 = 0

3z  5  x  5)2 (3 y) 2

P = -4 or P = -1

 9 x 2 30 x  25  9 y 2 ......................(1)

Complementary function is Ae4x Be x

1

=0

1 1     P.I  2 Sin 5 x    Sin5 x 2    5  5D  4   D  5D  4 

3 z  13 x  iy  1    x  1)  i y

 1   Sin5 x  5D  21

3  x  1)2  y 2   x2  2 x  1  y 2 ......................(2)

3z  5  3 z  1 9 x 2 30 x  25  9 y 2  x 2  y 2  2 x  1

by(1)and (2 Squaring both sides we get

  5D  21  Sin5 x  (5D  21)    (5D  21)

 5D  21   Sin5 x 2    25 ( 5 ) 441   

9 x2  9 y 2  30 x  25 9( x2  y 2  2 x  1) 9 x2  9 y 2  30 x  25  9 x2  9 y 2  18x  90

48x  140 24 x  7  0

 Locus of P is a straight line

(5D  21)sin5 x 625  441

1  [5.5cos x 21sin  5 x] 1066 

(25cos5 x 21sin  5 x) 1066

General Solution is y Ae4 x Be x 

(25cos5 x 21sin  5 x)  1066

PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2012 Find the order of each element of the group, (Z7 {[0]},7 ) G = {[1], [2], [3], [4], [5], [6]} i).

The identify element is [1] So its order is 1

 0 ([1]) = 1 [2]3 = [2] 7 [2] 7 [2] = [8] = [1]

ii).

 0 ([2]) = 3 iii). [3]6 = [3] 7 [3] 7 [3] 7 [3] 7 [3] 7 [3] 7 = [729] = [1]

 0 ([3] = 6 iv).

[4]3 = [4] 7 [4] 7 [4] 7

= [64] = [1]

 0 ([4] = 3 v). [5]6 = [5] 7 [5] 7 [5] 7 [5] 7 [5] 7 [5] 7 = [15625] = [1]

 0 ([5]) = 6 vi).

[6]2 = [6] 7 [6]

= [36] = [1]

 0 ([6]) = 2 www.TrbTnpsc.com

JUNE 2012     4  1  Find the rank of matrix     4  4      0 

    4      1  1    Let A     4    1  4        0      0      1    ~    1 R3  R1  R3     0       1    ~    1 R3  2 R2  R3  ( A)3     0  

( x  y)2 

www.Padasalai.Net Page 8 OCTOBER 2012 dy ( x  y )2   1 dx

dy  1(1) dx

Let Z = x+ y

Differentiate w.r, t x 

dz dy dy dz 1 ⇒   1(3) dx dx dx dx

 dz   1 1  dx 

Put (2), (3) in (1) (1)z 2 

dz 1 dz 1 dz 1  z 2  1 2   2  1  2 dx z dx z dx z

z2 dzdx 1 z2

⇒

(1  z 2 )  1 dz  dx 1 z2



1

 1 1  z  dz  z

2

 dz  dx 

dz  dx  C 1 

2

ztan 1 zxc

x ytan 1 ( x  y)x c x ytan 1 ( x  y)x c

(2)

PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2013 Solve the following system of linear equations by determinant method x+y+3z=4; 2x+2y+6z =7; 2x+y+z=10. 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 x 7

1 3 2 6

10 1 1

JUNE 2013

         If ai j,b  j  k ,c  k  j , then find        a  b,b  c,c,a          a  b(i  j ) j  k )       a  bi  j  j  k      a  bi  2 j  k      b  ci j  k      c  a 2i j  k                a  b,b  c,c  a [i  2 j  k ,i  j 2k , i  j  k ]  

2 1 1 2

1  1 2

1

1

= 1 (1+2) – (-2) (1-4) +1 (1+2)

= 4 (2-6) – 1 (7-60) + 3 (7-20)

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

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

= 3-6+3

= - 16 +53 – 39

= 6-6

= -55 + 53 = -2

=0

x = - 2  0

       a  b,b  c,c  a 0

   and x   and so The given equations are in consistent and hence no solution. www.TrbTnpsc.com

www.Padasalai.Net Page 9

PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

www.Padasalai.Net Page 10

TEN MARKS

MARCH 2006 Find

the

volume

of

the

solid

JUNE 2006

OCTOBER 2006

Find the vector and Cartesian equation of the

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

plane passing through the point

margins at the bottom and sides and a 2cm

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

margin on the top. What dimension will give the

generated by the revolution of the loop of the curve x  t 2 and yt 

t3 3

x+2y+4z+7=0 and 2x-y+3z+3=0. The vector normal to the planes x+2y+4z+7 = 0 and 2x-

 t3  t3 x  t yt  t1   3 3 

      y+3z+3=0 are i  2 j 4k and 2i  j  3k .

2

 x ie y 2 t 2 1    3

largest printed area ?(Diagram also needed)

The required plane is

2

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 180cm2

 r to the planes

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

 The required plane parallel to the above two vectors

 (x+2) (y+3) = 180⇒ y3 180 y 180  3 (1) x2

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

      i  2 j  4k and 2i  j  3k and passes through the point (-1, -2, -1).

x Let y 0x0or1  0 3











180  180   3x Ax  3 A x2 x2 



Vector equation is r  a  sutv 









 



Iet r i  2 j  k s i  2 j  4k  t 2i  j  3k 

dA 360  3 360( x  2) 2 3 dx ( x  2)2

3

x0orx3 Volume   y 2 dx 0

x  x1 11

2

 2 x2   x   x1   dx   x1  x   dx 3 9  3 0  0 3

3

2

2

12 3

3   x 2 2 x3 1 x3  2 x3     x  x 2   dx   .  .  3 9  2 3 3 9 4 0 0 

Its Cartesian equation is ( x1 , y1 , z1 )(1, 2,1)  y  y1 z  z1 (11 ,m1 ,n1 )  ) m1 n1 0 (12 ,m2 ,n2 )(2, m2 n2

 x2 2 3 1 4   9 9     x  x     6 (0)   36 0 4   2  2 9 9 9  18  24  9  3   6     4 4  2  

x  1 y  2 z 1 4 0

1

2

2

1 (11 ,m 31 ,n1 )  )

dA 0  360 2  30 3( x  2)2 360by3,( x  2)2 120 dx ( x  2)

x2 120x 2  d A  2  x2  dx  2

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

(12 ,m2 ,n2 )(2,

30  2



720 720  Negative  Printed 3 (2 30  2  2) (2 30)3

area is maximum when

(x+1) [6+4] – (y+2) [3-8] + (z-1) [-1-4] = 0 10x+5y-5z+25 = 0⇒  2x+y-z+5 = 0

www.TrbTnpsc.com

d2A 720 360(2)( x  2) 3  0 2 dx ( x  2)3

Thus the Cartesian equation is

3

x2

 y

x2 302 (1)

180 3 2 3022

90 90 30 3 3 = 3 30 3 The he printed area 30 30 are

2 302 cm and 3 30 3

cm.

PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2007

JUNE 2007

www.Padasalai.Net Page 11 OCTOBER 2007

Find the common area between y  x Find the axis focus, latusreetum, equation of Find the eccentricity, centre, foci and vertices of the 2

2 2 the latus rectum, vertex and directrix for the hyperbola 12 x  4 y  24 x  32 y 127 0 and draw the

and x 2  y .

parabola y 2  8x  2 y  17  0

diagram.

y 2  8x  2 y  17  0 ⇒ y 2  2 y  8x  17

12 x2  4 y 2  24 x  32 y 127 0

y 2  2 y  12  12 8x  17 ⇒ ( y  1)2  1  8x  17

12( x2  2 x  12 12 )  4( y 2  8 y  42  42 )127

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

( y  1)2  8x  17  1 ⇒ ( y  1)2  8( x  2)

12[(  x  1)2  1]4[(  y  4)2  16]127

intersection Put (2) in (1)

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

12( x  1)2  4( y  4)2  127  12  64

y2  x

x2  y

(1) (2)

( x 2 )2  x ⇒ x4  x 0 ⇒ x( x3  1)0

About X,Y

Referred to x.y

x 0x3  1  0

Axis

Y=0

Y=0  y-1=0  y=1

x =0

Focus

(a,0) ie (2,0)

X=2  x-2=2  x=2+2 x=4

x= 1

Y=0  y-1=0⇒y=1⇒F(4,1)

x = 0  in (2), y =0 x = 1  in (2), y = 1

Latus

X=a

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

rectum

ie X=2

Vertex

(0,0)

b

Required area  [ f ( x)g( x)]dx a

f ( x)  y 2  x, g ( x):x 2  y 

 





x  x 2 dx

 x 3 2 x3  2 3  1       x 2 x3  3 3  3 3    2 

1  2 1 2 1 1 2  .1 .1 0  0)  .   3  3 3 3 3 3

sq. units www.TrbTnpsc.com

X=2  x-2=2  x=2+2

X2 Y2  Where X x  1 Y  y  4  75   75       12   4 

5 5 3 2 75 4 a  b  e  1  2 2 4 25 e2  1 3e2 4e 4e2

x=4

Referred to X, Y

X=0  x-2=0  x=2

Referred to x,y

Y=0  y-1=0⇒ y=1V (2,1)

Centre

C (0,0 )

C (1,4)

Foci

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

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

Directri

X = -a

X  2x2 2

x

X = -2

x 22 ⇒ x0

Graph referred to x,y

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

5  A , 0  2 

 5  A  , 0   2 

Graph Referred to x, y

7 ,4 2



A  3 , 4 2



PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2008

JUNE 2008

Find the vector and Cartesian equation of the plane passing through the point

www.Padasalai.Net Page 12 OCTOBER 2008

Find the common area between y 2  x and Find the local minimum and maximum values of f ( x)2 x3  3x2  36 x10

x2  y .

(-1, -2, 1) and perpendicular to two planes x+2y+4z+7=0 and 2x-y+3z+3=0. The vector normal to the planes x+2y+4z+7 = 0







 



and 2x-y+3z+3=0 are i  2 j 4k and 2i  j  3k . 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 i  2 j  4k and 2i  j  3k and passes through the point (-1, -2, -1).      Vector equation is r  a  sutv           Iet r i  2 j  k s i  2 j  4k t 2i  j  3k



 



Its Cartesian equation is x  x1 11 12

y  y1 m1

z  z1 n1 0

m2

n2

( x1 , y1 , z1 )(1, 2,1) 

(11 ,m1 ,n1 )  ) (12 ,m2 ,n2 )(2,

Thus the Cartesian equation is x  1 y  2 z 1 1

2

2

1

4 0

(11 ,3m1 ,n1 )  )

(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

www.TrbTnpsc.com (12 ,m2 ,n2 )(2,

y2  x

x2  y

(1)

(2)

f ( x)2 x3  3x2  36 x10

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

f '( x)6 x2  6 x  36

intersection Put (2) in (1)

f "( x)12 x6 Let f '( x)0

( x 2 )2  x ⇒ x4  x 0 ⇒ x( x3  1)0

2 6 x2  6 x  360 ⇒ i.e,x  x  60

x 0x3  1  0

x2  3x  2 x  6  0 ⇒ x( x  3)  2( x  3)0

x =0

( x  3)( x  2)0 ⇒ x  30,x2  0

x= 1

x = 0  in (2), y =0

x  3,x 2

x = 1  in (2), y = 1

[ f "( x)]x3 12(3)6 366 30

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

negative

b

Required area  [ f ( x)g( x)]dx a

f ( x)  y  x, g ( x):x  y 2



 



2

 x 3 2 x3   2 3  1 3  2 x  x dx     x 2 x 3  3 3  3  2 



1  2 1 2 1 2  .1 .1 0  0)  .  3  3 3 3 3 

1 sq. units 3

f(x) attains maximum when x= 3 maximum value is 2(3)3 3(3)2  36(3)10 = 2(27)(9)10810 54271081091 [ f "( x)]x2 12(2)624 630 positive  f ( x) attains minimum when x = 2

Minimum value is 2(2)3 3(2)2  36(2) 10ie2(8) 3(4)7210

ie16127210ie34 Minimum value is -34.

PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2009 P represents the variable Z  z 1  0  z 1 

Find the locus of P if Re 

JUNE 2009

www.Padasalai.Net Page 13 OCTOBER 2009

Find the eccentricity, centre, foci and Solve : ( D2  5D 6) ySinx2e3 x vertices of the ellipse 16x2 9 y 2  32x  36 y 92  and draw the diagram .

Let z = x+iy

16 x2 9 y 2  32 x  36 y 92 

Characteristic equation is P2  5 p0

 x  1)  iy x  i y  1  z 1      x  i( y  i) x  i y  i  z 1 

16( x2  2 x  1  1)9( y 2  4 y  4  4)92

(p-3) (p-2) = 0⇒P = 3 or p = 2

16( x  1)2 9( y  2)2 144

Complementary function is Ae3 x Be2 x

( x  1)2 ( y  2) 2  1  ing by 144 9 16

1 1     PI1 2 Sinx  2 Sinx   D  5D  6   1  5D  6 

 x  1)  i yxx  i( y  1)  x  i( y  1) x  i( y  1)



=

x( x  1  y( y  1)

i  xy  ( x  1)( y  1) 

x 2  ( y  1) 2 x( x  1)  y ( y  1)  z 1  Re  x 2  ( y  1) 2  z 1   z 1 

It is given that Re  0  z 1  x( x  1)  y( y  1) 0  x 2  1( y  1)2 1

ie x( x  1)  y( y  1)0 ie x2  x  y 2  y0 Locusof x2  y 2  x  y0

x2 y 2 a 2 b 2 7  1 a 2 16,b2 9 e  9 16 a 4

w.r. to X, Y

w.r to x,y xx1, y y2

Centre

(1, -2)

(0, 0) Foci (0,ae)

(1, 2 7)and (1, 2 7)

(0, 7)

Vertices (0, ±a)(0, ±4)

(1,2)and (1  6)

1    1    Sin  x   5D  5 Sinx  1 5D  6 

 1 D 1 1  D 1    Sinx  2 Sinx 5  ( D  1)( D  1)  5  D  1 1  D 1    Sinx  1 D  1Sinx 1  D  1)Sinx  5  1  1 5  2  10 1 1   DSinxSinx] CosxSinx] 10 10 1 1   3x   3 x  1 e3 x PI 2  2 2e  2 e   0    D  5D  6   3  5.3  6 

  3x   3x 1 1 PI 2 2 e  e   ( D  3)( D  2)   ( D  3  3)(3  2)   1  3x  1  3x 3x 2 e  e  xe   D.1  D

General Solution is y = C.F + P.I1 + P.I 2 1 y Ae3 x Be2 x  [Cosx Sinx]2 xe3 x  10

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PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2010

www.Padasalai.Net Page 14 OCTOBER 2010

JUNE 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 3x  4 y  0

(1  2 x3 )

dy  6 x 2 yCo sec2 x dx

Find the vector and Cartesian equations of the plane, through the point (1,2-2) and parallel to the line

x  2 y 1 z  4 and perpendicular to   3 2 4

the plane 2x+3y-3z=8. (1  2 x3 )

dy  6 x 2 yCo sec2 x dx

The normal vector to the plane 2x+3y+3z =8 is    2i  3 j  3k .

dy 6 x2 Co sec2 x   dx 1  2 x3 1  2 x3

This vector is parallel to the required plan. The

1  2 x3 ,

6x2 Co sec2 x P Q  1  2 x3 1  2 x3

 Pdx

about 3x  4 y  0 ⇒y = 0  3 x = 0  x = 0

x = 0 to 4 4y = 3x

3x – 4y = 0

3 9 ⇒ y x ⇒ y 2  x 2  4 16



Pdx

  y 2 dx

 elog(1 2 x )   x3 3

Solution is y( IF ) Q( IF )dxC 2

Co sec x (1  2 x3 )dxC 2 1  2x

y(1  2 x ) Co sec xdxC 3

a

y(1  2 x )CotxC 3

4

 

4 9 9  x3   x 2 dx   16 0 16  3  0

3 3  4  012 cu.units 16 

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







plane is ra  su  tv i.e.

          ri  2 j  2k s(3i  2 j.4k )t (i  3 j  3k )

Cartesian from x  x1 11

y  y1 m1

12

m2

( x1 , y1 , z1 )(1, 2, 2) z  z1 n1 0 (11 ,m1 ,n1 )  ) (11 ,m2 ,n2 )(2, n 2

2

4

9    x 2 dx 16 0

The vector equation of the required

6x2 log(1  2x3 ) 3 1  2x

y(1  2 x3 )

b

Volume

e

required plane passes through (1,2-2) and parallel         to u 3i  2 j  4k and v2i  3 j  3k .

y(1  2 x3 )CotxC

x 1 y  2 z  2 3 2 4 0 3

3

3

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

PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2011

JUNE 2011

Find the vertex, axis, focus, equation of latus

Find the vertex, axis, focus, equation of latus rectum, equation of directrix and length of latus rectum of the parabola x2  4 x  4 y

rectum, equation of directrix and length of latus rectum of the parabola y 2  4 y  4 x  8  0 and

www.Padasalai.Net Page 15 OCTOBER 2011

Solve the system to equations. X+2y+z=2, 2x+4y+2z=4, x-2y-z=0

hence. Draw the diagram. y2  4 y  4x  8  0

x 2  4 x  4 y ⇒ x2  4 x  22  22  4 y

y 2  4 x  4  4 4x ⇒  y  2)2  4x4

( x  2)2  4( y  1)

 y  2)  4 x Y  4 X 

X 2 4Y WhereY  y1

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

4a4 X x 2 a1

The type is open left ward.

The type is open downward.

2

2

Referred

Referred to x,y

X, Y

X = x +1 Y = y+2

toX, Y

X = x +1 Y = y+2

Vertex

(0, 0)

(-1,-2)

axis

Y=0

Y = -2

axis

X=0

 x- 2 = 0

Focus

(-a,0)i.e (-1,0)

Focus = = (-2, -2)

Vertex

(0,0)

(2, 1)

Equation of

X = -a

X = -1  x +1 = -1

Focus

(0, -1)

Focus = (2, 0)

latus rectum i.e X = -1

 x = -2

Equ.ofL.R

Y=1

y=2

Equation of X =a

X = 1  x+1 = 1

Equ.of dir.

Y = -1

y=0

directrix

 x=0

Length of L.R 4a = 4

4a =4

3 2 1(  4  4)2( 2  2)3( 4  4) 2(4)3(8) 16

Referred to x,y

Len. L.R 4a=4

2 4

1 2 1

Referred to

i.e X = 1

1  2

2 x 4

2 4

3 2 (4  4)2( 4  0)3( 8  0) = 8

0 2 1

– 24 = -16 1 y 2

2 4

3 2 (4  0)2(2  2)3( 0  4) = -4

1 2 1

+ 8 – 12 = -8 1 z 2

2 4

2 4 (0  8)2(0  4)( 4  4)

1 2 0

4a = 4

= 8 + 8 – 16 = 0 x

x 16 y 8 1 ⇒ y  x 1 y   16  16 2 z

z 0  0  16

x 1, y 1 ,z 0 2

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PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2012 Find

the

area

between

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JUNE 2012 the

curves Find the eccentricity, centre, foci and

Find the vector and Cartesian equations of the plane which contains the line

yx 2  2 x  3 x – axis and the lines x = -3 vertices of the hyperbola

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

(9 x2  36 x)(7 y 2  14 y) 92 and draw the

and x =5.

x 1  y z  1 and   2 3 1

diagram. (9 x2  36 x)(7 y 2  14 y) 92

The required plane contains the line

9( x2  4 x)7( y 2  2 y) 92

the line x  1  y  0  z  (1)  The plane passes through 2

9[(  x  2)2  4]7[(  y 1)2 1] 92

 A1  A2  A3

Required area 





  ydx   ( y )dx  ydx 











  ( x 2  2 x  3)dx   (3  2 x  x 2 dx  ( x 2  2 x  3)dx 



2



Referred

to X, Y 5 3 3  x3      x x 2 2 2   x  3x   3x  x     x  3x  Centre C (0,0 ) 3 1  3 3 3  3 Foci F (0, 6 )  1    1  3             3  Vertices V(0, 3) 1

3

1   125    3  1    25  15       3  3         2  9  9  2  40 9  = 32    

sq.unit www.TrbTnpsc.com

2

2

Where X x  2;Y  y  1

a2  9 a 3 b2 7 e 1 

1



7( y  1) 9( x  2) ( y  1) ( x  2)  1 i.e  1 63 63 9 7

Y2 X2  1 9 7

3







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

7( y  1)2  9( x  2)2 63 2

x 1  y z  1 i.e,   2 3 1

b2 4  12 16   42 2  a 4 4

Referred to x,y X x  2;Y  y  1

(-2, 1)

This plane is perpendicular to x-2y+3z-2 = 0.    That is the plane is parallel to i  2 j  3k . The required plane passes through (1,0, -1) whose    postion vector ai  j and parallel to two vector        u2i  3 j  k and L2 j  3k . 





  



 







i.e, r  i  j  s(2i  3 j  k )t(i  2 j  3k ) Its Cartesian form is

(-2, 7) & (-2, - 5) (-2, 4) and (-2, -2)



Vector equation r  a  su  tv

Ie

x 1 y  0

x  x1 11

y  y1 m1

12

m2

z  z1 n1 0 n2

z 1

2

3

1 0

1

2

3

(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

PREPARED BY :S.MANIKANDAN ., VICE PRINCIPAL ., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM ., 9047039080

MARCH 2013 Verify

u u for the function n  xy yx 2

2

x usin   y

x usin   y

JUNE 2013  x y  u u 1  x  y   y   Cos if    x  y  x y 2  x  y     x y  usin Pr ove.  x  y    x

 x y  x y 1 usin  x  y  i.esin (u ) x  y  

x1 u cos   x y y

f Sin1(u )

x y x y

x  1  1 x  x   2u cos .  2   Sin   2  xy y  y  y y  y 

Put x = tx, y = ty⇒ f 

1  x x  x  2  cos  3  Sin ....(1) y  y y   y

f 

 x   x  u cos  2  y  y  y 

x 1   x   x  1   2u cos   2   2  sin   xy y y   y   y  y  

1  x x  x  cos  3  Sin  2  y  y y   y

1 t ( x  y) ( x  y) ⇒ f t 2  t ( x  y ) x y

f is a homogeneous function in x and y of degree 1 &By Euler’s theorem

x

x

2

f f 1  y  . f x y 2

  1 (sin 1 ) y ( Sin 1u ) Sin 1u x y 2

x.

1 1 u

x.

 2u  2u    yx xy

2

.

u 1 u 1  y   Sin1u 2 x 1  u y 2

u u 1  y   u 2 . Sin 1 u x y 2

 1 Sin2 

 Cos 2 

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tx  ty tx  ty

x  y 1 1  x y  . Sin  Sin   x y 2 x  y  

x y 1  x y .  x  y 2  x  y

 1  x y   x y     Cos    2 x y  x y

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