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March 2015 – Higher Secondary Mathematics A

I Answer all the Questions. 40×1=40

1. Answer : c. Infinitely many solution

2. Answer : d. all the 3 above 3. Answer : c)–8 4. Answer : c. (1, 1, 2) 5. Answer : d. (p→q) ∧ (q→p) 6. Answer : c.

5125

7. Answer : d. both the axes 8. Answer : d. nowhere 9. Answer : a. – 2 10. Answer : d. b:a 11. Answer : a. 8 12. Answer : c.4 13. Answer : a. 48 14. Answer : a. 2cos nθ 15. Answer : c. 2

16. Answer : c.

0u

17. Answer : b. a is

perpendicular to b

18. Answer : c) consistent 19. Answer : b.Acosx+Bsinx 20. Answer : c. 0 21. Answer : d. 13, 12 22. Answer : c. xdx+ydy=0 23. Answer : b. (– ,0)

24. Answer : b. 0.3125 25. Answer : b.

cdyQFIFIx .).(.).(

26. Answer : d. x=0 27. Answer : a. B = 0

28. Answer : d. 25

29. Answer : b. 301

30. Answer : d.0<θ<1

31. Answer : d. 439

,e

32. Answer : c.a*b= ab

33. Answer : c. 12

1

34. Answer : d. –4 35. Answer : c. 4k

36. Answer : d.nb

nbsin

37. Answer : d. 6 38. Answer : c. 1n

39. Answer : b) 3 40. Answer : d.4c2lm=n2

www.Padasalai.Net www.TrbTnpsc.com



1. If the equation nzyxmzyx

lzyx

22

2 such that ,0 nml then the system has

a. a non zero unique solution b. trivial solution c. Infinitely many solution d. No Solution. 2. If X is a continuous random variable then P(a<X<b)= a.P(a≤X≤b) b.P(a<X≤b) c.P(a≤X<b) d. all the 3 above 3. If p+iq=(2–3i) (4+2i) then q is a)14 b)–14 c)–8 d)8

4. The point of intersection of the lines )()(

kjitkjir 232 and

)()(

kjiskjir 32532 is

a. (2, 1, 1) b. (1, 2, 1) c. (1, 1, 2) d. (1, 1, 1) 5. p↔q is equivalent to a.p→q b.q→p c.(p→q) ∨ (q→p) d. (p→q) ∧ (q→p) 6. If 2 cards are drawn from a well shuffled pack of 52 cards, the prob-ability that they are of

the same colours without replacement, is

a.21 b.

5126 c.

5125 d.

10225

7. The curve 9y2=x2(4–x2) is symmetrical about a. y-axis b. x-axis c. y=x d. both the axes 8. The point of inflexion of the curve y=x4 is at a. x=0 b. x=3 c. x=12 d. nowhere 9. Given E(X+c) = 8 & E(X – c) = 12 then value of c is a. – 2 b. 4 c. – 4 d. 2

10. Volume of solid obtained by revolving the area of the ellipse 12

2

2

2

b

y

a

x about major and

minor axes are in the ratio a. b2:a2 b. a2:b2 c. a:b d. b:a 11. If a compound statement is made up of three simple statement, then the number of rows in

the truth table is. a. 8 b. 6 c. 4 d. 2 12. In the group (Z5–{[0]}, ·5), 0([3]) is a.5 b.3 c.4 d.2 13. The length of the arc of the curve x2/3 +y2/3=4 is a. 48 b. 24 c.12 d.96

14. If the value of

is

a. 2cos nθ b. 2i sin nθ c. 2 sin nθ d. 2i cos nθ

15. The value of [

ikkjji ,, ] is equal to

a. 0 b. 1 c. 2 d. 4

16. If )()()(

bacacbcbau , then

a. u is a unit vector b.

cbau c.

0u d.

0u

17. If baba then

a. a is parallel to

b b.

a is perpendicular to

b

c. ba d.

a and

b are unit vectors

18. If ρ(A)=ρ*A B+ then the system is a) consistent and has infinitely many solution b) consistent and has a unique solution c) consistent d) inconsistent 19. The complementary function of (D2+1)y=e2x is a.(Ax+B)ex b.Acosx+Bsinx c.(Ax+B)e2x d.(Ax+B)e–x

20. If )(xf is an odd function then dxxfa

a

)(

is

a. dxxfa

)(0

2 b. dxxfa

)(0

c. 0 d. dxxafa

)( 0

21. The length of the semi-major and the length of semi minor axis of the ellipse 1

169144

22

yx are




a. 26, 12 b. 13, 24 c. 12, 26 d. 13, 12 22. The differential equation of all circles with centre at the origin is a. xdy+ydx=0 b.

xdy–ydx=0 c. xdx+ydy=0 d. xdx–ydy=0 23. The function f(x) = x2 is decreasing in a. (– , ) b. (– ,0) c. (0, ) d. (–2, )

24. For a Poisson distribution with parameter =0.25 the value of the 2nd moment about the origin is

a. 0.25 b. 0.3125 c. 0.0625 d. 0.025

25. The solution of a linear differential equation QPxdy

dx where P and Q are

functions of y, is a. cdxQFIFIy .).(.).( b. cdyQFIFIx .).(.).(

c. cdyQFIFIy .).(.).( d. cdxQFIFIx .).(.).(

26. The curve ay2=x2(3a-x) cuts the y-axis at a.x=–3a, x=0 b. x=0,x=3a c. x=0, x=a d. x=0 27. If A and B are any two matrices such that AB = 0 and A is non

singular, then a. B = 0 b. B is singular c. B is non singular d. B = A 28. The eccentricity of the hyperbola whose latus rectum is equal to half of its

conjugate axis is

a. 23 b.

35 c.

23 d.

25

29. The value of 1

0

41 dxxx )( is

a. 121 b.

301 c.

241 d.

201

30. In the law of mean, the value of ‘ ’ satisfies the condition a.θ>0 b.θ<0 c.θ<1 d.0<θ<1

31. The modulus and amplitude of the complex number 343 ie

are respectively

a. 2

9 ,e b.

29

,e c. 4

36 ,e d.

439

,e

32. Which of the following is not a binary operation on R

a.a*b=ab b.a*b=a–b c.a*b= ab d.a*b= 22 ba

33. The slope of the normal to the curve y=3x2 at the point whose x

coordinate is 2 is

a.131

b.141

c. 12

1 d. 121

34. The line cyx 24 is a tangent to the parabola xy 162 then c is a.–1 b.–2 c.4 d.

–4

35. If the matrix

54131

231k has an inverse then the values of k a. k is any real number

b. 4k c. 4k d. 4k

36. The angle between the line btar

and the plane qnr

is connected by the relation

a. q

na

cos b.nb

nb

cos c.n

ba

sin d.

nb

nbsin

37. Degree of the differential equation 2

231

1dx

yd

dx

dy

/

a. 1 b. 2 c. 3 d. 6 38. If is the nth root of unity then

a. 0n b. ...... 53421 c. 1n d. 1 n

39. The work done by the force kjiaF

in moving the point of application from (1,1,1) to

(2,2,2) along a straight line is given to be 5 units. The value of a is a) –3 b) 3 c)8 d) –8

40. The condition that the line lx+my+n=0 may be a tangent to the rectangular hyperbola xy=c2 is




a.a2l2+b2m2=n2 b.am2=ln c.a2l2-b2m2=n2 d.4c2lm=n2

II. Answer Any NINE AND 55TH 10×6=60

41. Find the rank of the matrix

101274363124

Let A =

101274363124

∴A~31

121076343421 CC

∴A~ 122 4121051050

3421RRR

∴A~233 50000

510503421

RRR

The last equivalent matrix is in echelon form.

And it has two non-zero rows. ∴ 𝛒 (A) = 2 42. Solve using det. method 2x–y=7, 3x+2y=11

The matrix eqn. is

117

2312

yx

Let 01

342312

31114

21117

x

1

212211372

y

By Crammer’s rule

131

113

yx yandx

The solution set is {x=3, y=–1}

43. Verify )(.)(

dcba , if ,,,

kjickibkjia 22

kjid 2

kji

kjikji

ba

2

211

102111

)()()(

kji

kjikji

dc

3

131

211112

)()()(

4231123111

))(())(())(()).(( dcba

44. (a) If A(–1,4,–3) is one end of a diameter AB of the sphere x2+y2+z2–3x–2y+2z–15=0, then find the coordinates of B.

(b) Find the angle between the line23

1

1

32

zyx and the plane 3x+4y+z+5=0

a) The eqn. of sphere is x2+y2+z2–3x–2y+2z–15=0 Here 2u = Co-eff.of x = – 3u = – 3/2 2v = Co-eff.of y = – 2v = –1 2w= Co-eff.of z = +2w = +1 Centre is (–u, –v, –w) = (3/2, –1, 1) If A(–1, 4, –3) is one end of a diameter AB Let the coordinates of B be (x2, y2, z2) The mid point of AB is the centre (

3/2, –1, 1)




1;2;2

23;24;31

1;1;

)1,1,(,,.,.

222

222

2

3

2

4

2

3

2

1

2

3

2

3

2

4

2

1

222

222

zyx

zyx

ei

zyx

zyx

The coordinates of B are (x2, y2, z2)=(–2,–2,1)

b) Let Ɵ be the angle between the line and the plane. nb

nb

sin

kjinkjib

4323 ,

nb = 9 – 4 – 2 = 3

91226141169419 nb

912

3

912

3 1sinsin angle

45. P represents the variable complex number Z. Find the locus of P, if iziz 55 .

Let z = x + iy

iiyxiiyxiziz 5555

)()( 55 yixyix

2222 55 )()( yxyx

2222 55 )()( yxyx

25102510 22 yyyy 0252520 y

0 y Which is the locus of P.

46. Find the square root of (– 7+24 i )

Let iyxi 247

On squaring both sides we get, –7+24i=(x2–y2)+(2xy)i Equating the real and imaginary parts x2–y2= –7-----------① 2xy=24---------②

Also x2+y2 = 222224)( yxyx

= 5764924)7(22

= 25625 -------------③

Solving ①and ③we get x2=9 and y2=16 ∴x=±3 and y=±4 But xy is positive ⇒x and y have the same sign. (x=3, y=4) or (x = – 3, y = – 4)

∴ iorii 4343247

47. (a) Evaluate x

e

xxLim 2

(b) Determine the domain the concavity of the curve y=2–x2

a)

x

e

xxLim

2 which is in indeterminate form

Applying L’Hopital’s Rule we get

x

e

xxLim 2 which is in indeterminate form

Again applying L’Hopital’s Rule we get

022

xex

Lim

b) y=2–x2

y’ = – 2x y’’ = –2, Here y’’ is always (–)ve The curve is concave downwards (–∞,∞) 48. At what angles θ do the curvesy=ax and y=bx intersect (a≠b)?




49. If u=log(ex+ey), Findyu

xu

.

Differentiate u partially w.r.t. x yx

x

ee

e

x

u

Differentiate u partially w.r.t. yyx

y

ee

e

y

u

1

yx

yx

yx

y

yx

x

ee

ee

ee

e

ee

eyu

xu

50. Evaluate :

3

6 1

/

/ tan x

dx

Soln:- Let I =

①

②

①+② 2I=

51. Solve: xyyxdx

dy 1

)()( xyx

dx

dy 11 ))(( yx

dx

dy 11

dxxy

dy)(

)(

1

1 Integrating we get

3

6tan1

1

dxx

3

6

3

6sincos

cos

1

1

cos

sin

dxxx

xdxI

x

x

3

6)sin()cos(

)cos(

6363

63

dxxx

x

3

6cossin

sin

dxxx

xI xx sin)cos(

2

3

6sincos

cos

cossin

sin

dxxx

x

xx

x

663

1cossin

cossin3

6

3

6

3

6

xdxdx

xx

xx

1262

II

12tan1

13

6

dxx




cx

xy 2

12

)log( is the solution.

52. Prepare a truth table for (ii) (p∧q)∨(∼(p∧q)) There are TWO statements p,q ∴ The no of rows in the table are 22 =4

p q p∧q ∼(p∧q) (p∧q)∨(∼(p∧q))

T T T F T

T F F T T

F T F T T

F F F T T

53. Find the mean and variance of the distribution :

elsewhere003 3

,,)( xexf

x

54. Marks in an aptitude test given to 800 students of a school was found to be normally distributed. 10% of the students scored below 40 marks and 10% of the students scored above 90 marks. Find the number of students scored between 40 & 90.

Let X be the marks of a student.

P(x<40)=10% and P(x>90)=10%

P(40<X<90)=P( <X< )–{P(x<40)+P(x>90)}

= 100% – 10% – 10% = 80%

Out of 800 students number of student scored between 40 & 90 = =640 students.

55. a) Find the equation of the hyperbola whose centre is (1, 2) the distance between the directices is 20/3 and the distance between the foci is 30 and the transvers axis is parallel to y-axis.

[OR]

(b) Show that the set of four matrices ,10

01

,10

01

,

10

01

,

10

01

form an abelian group

under multiplication of matrices.

0

3)3()()( dxexdxxfxxEMean

x

0

!

31

3

!1

0

312

33n

nxnxdxexGammadxxe

0

92

3

!232223

3)3()()( dxexdxxfxxEx

9

1

9

122

3

1

9

222)]([)()( XExExVar

91

31 ; VarianceMean

100

80800




III. Answer Any NINE AND 70TH 10×10=100




56. Solve by matrix determinant method 2x + y – z = 4, x + y – 2z = 0, 3x + 2y – 3z = 4

The matrix equation is

404

323211112

zyx

Let 0

132113112

323211112

)()()(

Let 0

484418114

324210114

)()()(

x

Let 0

41216413482

343201142

)()()(

y

Let 0

448144142

423011412

)()()(

z

Since Δ=0 & Δₓ= Δy= Δz=0 and one 2⨉2 minor ≠ 0

The given system is consistent and it has infinitely

many solutions. Take z=k

Now the system has been reduced to two

equations

They are 2x + y = 4+k and x + y = 2k

Let 1121112

Let kkkk

kx

424

1214

Let 434421

42

kkk

kk

By Crammer’s rule,

yx yx ,

∴ x = 4 – k, y = 3k – 4 and z = k

Solution {4 – k, 3k – 4, k}

57. Altitudes of a triangle are concurrent-Prove by vector method

In ∆ABC, let AD, BE be the two altitudes

intersecting at O. In order to prove that the altitudes are

concurrent, it is sufficient to prove that CO is ⊥lar to AB.

Taking O as the origin. let the position

vertors of A. B. C be cba

,, respectively.

Then

cOCbOBaOA ,,

Now AD ⊥ BC

BCOA

0

BCOA

0

)( bca

0

baca ----①

Now BE ⊥ CA

CAOB

0

CAOB

0

)( cab

0

cbab ----②

Adding ①&② we get

0

cbca

0

cba )(

0

OCBA

ABOC

Hence the three altitudes are congruent.

58. Find all the values of 43

23

21

i and hence prove that the product of the values is 1.

Let 23

21 i = r(cos θ +i sin θ) ⇒ r= 1

43

41 r

Equating the real & imaginary parts, we get

cos θ = 21 and sin θ =

2

3 (+)ve and (+)ve

∴ θ lies in the I Quadrant. ∴ θ = α ⟾ θ = 3

)]sin()[cos()(332

321 1 ii

43

43

3323

21 )]sin()[cos()( ii

41

)]sin()[cos( i

Add 2kπ with the argument, we get,

41

22 )]sin()[cos( kik

Apply De’Moivre’s Theorem, we get,

)sin()cos(4

24

2

ki

k k=0,1,2,3




The values are )(),(),(),(4

74

54

34

ciscisciscis

Now their Product is= )(4

74

54

34

cis

= )()( 44

16 ciscis

= 44 sincos i = 1+0

= 1 Hence proved

59. Find the axis, vertex, focus, directrix, length and equation of latusrectum and draw their graph for the parabolas y2 –8x–2y+17=0

(i) y2 –8x–2y+17=0 ⇒ y2–2y+ = 8x –17+ ⇒

⇒ y2–2y+1 = 8x–17+1⇒y

2–2(y)(1)+12= 8x–16

(y–1)2 = 8(x–2)

This is of the form Y2 = 4aX

Where X=x–2, Y=y–2 and 4a=8⇒ a 2

rightwardopenparabolaThis

Refferred to X,Y axes

x–2 = X, y–1=Y

Refferred to x,y axes

x=X+2, y =Y+1

Axis of the parabola

is X-axis, ie., Y=0

Y=0⇒ y =0+1= 1

y = 1

Vertex is (0,0)

∴X=0, Y=0

X=0⇒ x =0+2=2

Y=0⇒ y =0+1= 1

∴Vertex is (2,1)

Focus is (a,0)=(2,0)

∴X=2, Y=0

X=2⇒ x =2+2=4

Y=0⇒ y =0+1=1

∴Focus is (4,1)

Eqn. of Directrix X= –a ⇒ X= –2

X= –2⇒ x = –2+2=0

⇒ x = 0

Eqn. of

Latusrectum

X= a ⇒ X= 2

X= 2⇒ x = 2+2=4

⇒ x = 4

Length of Latus-

rectum 4a=8 4a=8

60. A kho-kho player in a practice session while running realises that the sum of the distances

from the two kho-kho poles from him is always 8m. Find the equation of the path traced by him if the distance between the poles is 6m.

From the given data, the two kho-kho poles

be taken as the foci F1 and F2.

Let P(x,y) be the position of the player.

∴ F1P + F2P = 8 ⟹ 2a = 8 ⟹ a = 4

& F1 F2 = 6 ⟹ 2ae = 6 ⟹ e = 43




Also b2 = a2 ( 1 – e2) ⟹ 167

1692 16116 b

⟹ b2 = 7

∴ The equation of the path is 12

2

2

2

b

y

a

x

1716

22

yxei .,.

61. Find the equation of the hyperbola if its asymptotes are parallel to x + 2y – 12 = 0 and

x–2y+8=0, (2,4) is the centre of the hyperbola and it passes through (2,0). The asymptotes are parallel to the lines

x+2y–12=0 & x–2y+8=0

∴The equations of asymptotes are of the form

x+2y+m=0 and x–2y+n=0

But these are passing through the centre (2,4)

⟹2+8+m=0 and 2–8+n=0 ⟹ m=–10 and n=6

∴The equations of asymptotes are

x+2y–10=0 and x–2y+6=0

The combined equation of asymptotes is

(x+2y–10)(x–2y+6)=0

∴The equation of the hyperbola is of the form

(x+2y–10)(x–2y+6)+k=0

This hyperbolas passes through (2,0)

∴ (2+0–10)(2–0+6)+k=0 ⟹ k = 64

∴The equation of the hyperbola is

(x+2y–10)(x–2y+6)+64=0

62. Find the local minimum and maximum values of f(x)= x4–3x3 +3x2–x Soln:- Let f(x) = x4–3x3 +3x2–x

f ‘(x) = 4x3 – 9x2 +6x– 1 f ‘’(x) = 12x2 –18x+6 = 6(2x2 –3x+1) = 6(x–1)(2x–1) At the critical point f ‘ (x)=0 4x3 –9x2+6x–1=0 (x–1)2(4x–1)=0x= 1, 1 and 1/4 At x = 1, f ‘’(1)= 6(x–1)(2x–1) = 0 Thus the II derivative test gives no information

about the extremum nature of f at x=1 x = 1 is not the maxi. or mini. point. f(x) = x4–3x3 +3x2–x

f(1) = 0 (1, 0) is not the extremum point.

At x = 1/4 , f ‘’(1/4)= + ve > 0 x = 1/4 is the minimum point. f(x) = x4–3x3 +3x2–x

f(1/3) =(1/4)4–3(1/4)3 +3(1/4)2 –(1/4) = 1/256 – 3/192 + 3/16 – 1/4 = – 27/256 is the local minimum value the minimum point is (1/4, – 27/256)

63. Use differentials to find 43 021021 .. y 43 02.102.1 y




64. Find the area between the line 2x – y = 4 and the curve y2 = 4x.

From the figure, use y-axis to find the area. To find the limits of y, solve the two equations.

x = ½ (y+4) ⟹ y2 = 2(y+4) ⟹ y2 – 2y–8=0

⟹ (y+2)(y–4)=0 ⟹ y = –2 and y = 4

The required area dyxleftxRight

4

2

)(

4

2

324

2

2

341

422

141

421

yy

ydyyy

128

824

21

1264

162

1621

32

33

1612 = 9 Sq. Units.

65. Find the surface area of the solid generated by revolving the cycloid ),tsint(ax

)tcos1(ay about its base (X-axis)

To get the intersecting point with x-axis,put y=0 ,1cos0)cos1(.,. tttei

Given that x=a(t + sin t) & y= a(1 + cos t)

⟹dt

dx= a(1 + cos t) &

dt

dy= a(0 – sin t)

tatadt

dy

dt

dx 222222

sin)cos1(

tatta2222

sin)coscos21(

)sincoscos21(222

ttta

)cos22(2

ta )cos1(22

ta

22

22cos2)(cos22 tt aa

∴Surface area dtdt

dy

dt

dxy

22

2

dtata t

2

cos2)cos1(2

dtaa tt

22

2cos2cos22




66. Solve: 0322 xydydxyx )(

This eqn.can be written as --①

Put y = v.x---② Diff.① ---③

Sub.②&③ in ① we get

①

Taking integration on both sides

67. The sum of Rs.1000 is compounded continuously at the nominal rate of interest 4% per

annum. In how many years will the amount be twice the original principle? (Loge 2 = 0.6931)

Let A be the Principal amount and varies at a particular time ‘t’ years. Rate of interest 4%

W.k.t

Taking integration on both sides

--------------------①

When t = 0, the amount A=1000

① Now we have to find t, when A=2000

①

years

After 17 years the amount will be doubled.

68. Show that the set G of all matrices of the form

xx

xx where }0{Rx is a group under

matrix multiplication.

Let we shall show

that G is a group under matrix multiplication. (i) Closure axiom:

Let A= B=

dta t

0

2

32cos28

dxdt

xTakedxxa

t

22cos16 2

0

322

1

3

23232

23

2aIa

UnitsSqa .3

64 2

xy

yx

dx

dy

3

22

dxdv

dx

dyxv

vxxxvx

dxdvxv

3

222

vxvv

x

xdxdv

31 2

2

2

v

v

dxdv

vvv

dxdv xx

3

)41(

331

222

xdx

v

v dv 241

3

8

2

83

41

3

33

80

412

dtx

dxt

dt

xdx

v

v

dvv

dtdvv

tvputdv

cxt logloglog83 2

41 vt

cxv loglog8)41(log32

cxv loglog)41(log832

cxv log)41(log832

cxv 832)41( cx

x

y

8

34

2

2

1

cxx

yx

83

4

2

22 cx

x

yx

84

6

322

cxyx 23224

AkAAdtdA

dtdA

dtdA 04.0

dtA

dA 04.0

ctA loglog 04.0

t

cA

cA et

04.004.0 log

tecA

04.0

100010000 cec

t

e04.0

10002000

te et 04.02log2

04.0

32.1704.0

6931.0

04.0

2log ttt e

}0{/ Rx

xx

xxG ,G

xx

xx

G

xx

xx

02

0,0

22

22

xy

yxG

xyxy

xyxyAB




Closure axiom is true. (ii) Associative axiom: Matrix multiplication is always associative.

Associative axiom is true.

(iii) Identity axiom: Let E=

By the def n. of e,

By the defn. of Mat ⨉, AE=

xexexexe

eeee

xxxx

2222

xxxx

xexexexe

2222

2xe=x21e

Thus E = G

21

21

21

21

such that AE = A for all AG

Identity axiom is true.

(iii) Inverse axiom: Let A–1= Gyy

yy

, AG

By the def n. of inverse,

AA–1=E

21

21

21

21

yy

yyxxxx

By the defn. of Matrix Multiplication,

AA–1=

xyxy

xyxy

yy

yyxxxx

2222

21

21

21

21

2222

xyxy

xyxy2xy=

21

xy

41

Thus A–1 = Gxx

xx

41

41

41

41

such that AA–1=E AG

Inverse axiom is true. G is group under matrix multiplication.

Hence proved. 70. (a) Derive the equation of plane in the intercept form.

Let a,b & c be the X,Y & Z intercepts of the

plane respectively. ∴ The plane passes through the points (a,0,0),

(0,b,0) & (0,0,c)

The vect.eqn. is ctbsatsr

..)1( ktcjsbiatskzjyixei

)1(,. Equating the Co-efficients of i, j and k vectors x=(1–s–t)a, y=sb and z= tc

tc

zands

b

yts

a

x ,,1

Adding these three we get

11 tstsc

z

b

y

a

x

Cartesian form: Here (x1,y1,z1)=(a,0,0), (x2,y2,z2)=(0,b,0) &

(x3,y3,z3)=(0,0,c)

The Car. eqn. is 0

131313

121212

111

zzyyxx

zzyyxx

zzyyxx

i.e., 0

0000

0000

00

ca

ba

zyax

⟾ (x – a) bc + y (ac) + z (ab) = 0 ⟾bcx – abc + acy + ab z = 0

⟾ 1c

z

b

y

a

xabc

GAGee

ee

,

xx

xx

ee

ee

xx

xxAAE




[OR] 69. A random variable X has the following probability mass function

(i) Find k, (ii) Evaluate p (X< 4), p (X5) & p (3< X 6), (iii) Find the smallest value of x

for which p (X x) > ½. (i) Since is a probability mass

function, ∑

⟹

⟹ ⟹

(ii)

(iii)

∴ The smallest value of is

X 0 1 2 3 4 5 6

P(X=x) k 3k 5k 7k 9k 11k 13k




(b) At noon, ship A is 100 km West of ship B. Ship A is sailing East at 35

km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4.00 p.m.

Let P & Q be the initial position of ships A

& B After 4 hours, from the given data, ships A

& B travelled 140 kms and 100 kms. From the diagram z2 = x2 + y2 Differentiating we get,

dt

dyy

dt

dxx

dt

dzz 222

Given that x=40, y=100.

Then 29201160010040 22 z

And

dt

dyy

dt

dxx

zdt

dz 1

39002920

1251003540

2920

1

dt

dz

./ hrKm29

195

The distance between the ships is changing

at the rate ./ hrKm29

195


preparred by a.immanuvel maduram thangiah, st. …...2015/03/12 · preparred by a. mmanuvel...

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