TN 12th Model Question Paper Maths

A A

No. of Printed Pages : 16
1312 (NP)
£vÄ Gs
A Register Number

!1312NPMathematics!
PART - III
Pou® / MATHEMATICS
( uªÌ ©ØÖ® B[Q» ÁÈ / Tamil & English Versions)

Põ» AÍÄ : 2.30 ©o ÷|µ® ] [ ö©õzu ©v¨ö£sPÒ : 90
Time Allowed : 2.30 Hours ] [Maximum Marks : 90

AÔÄøµPÒ : (1) AøÚzx ÂÚõUPЮ \›¯õP¨ £vÁõQ EÒÍuõ GߣuøÚa
\›£õºzxU öPõÒÍÄ®. Aa_¨£vÂÀ SøÓ°¸¨¤ß, AøÓU
PsPõo¨£õÍ›h® EhÚi¯õPz öu›ÂUPÄ®.
(2) }»® AÀ»x P¸¨¦ ø©°øÚ ©mk÷© GÊxÁuØS®,
Ai÷PõikÁuØS® £¯ß£kzu ÷Ásk®. £h[PÒ ÁøµÁuØS
ö£ß]À £¯ß£kzuÄ®.

Instructions : (1) Check the question paper for fairness of printing. If there is any lack of
fairness, inform the Hall Supervisor immediately.
(2) Use Blue or Black ink to write and underline and pencil to draw diagrams.

£Sv & I / PART - I
SÔ¨¦ : (i) AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®. 20x1=20

(ii) öPõkUP¨£mkÒÍ ©õØÖ ÂøhPÎÀ ªPÄ® Hئøh¯
Âøhø¯z ÷uº¢öukzxU SÔ±mkhß Âøh°øÚ²® ÷\ºzx
GÊuÄ®.
Note : (i) All questions are compulsory.
(ii) Choose the most appropriate answer from the given four alternatives and
write the option code and the corresponding answer.

[ v¸¨¦P / Turn over

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1312 (NP) 2

1. 28 Cß 11 &B® £i‰» \uÂQu¨ ¤øÇ ÷uõµõ¯©õP 28 Cß \uÂQu¨
¤øÇø¯¨ ÷£õÀ __________ ©h[PõS®.
1 1
(1) 11 (2) 28 (3) (4)
28 11
th
The percentage error in the 11 root of the number 28 is approximately __________
times the percentage error in 28.
1 1
(1) 11 (2) 28 (3) (4)
28 11

2. 4x2−y2=36 US 5x−2y+4k=0 GßÓ ÷Põk J¸ öuõk÷Põk GÛÀ k Cß ©v¨¦ :
9 81 4 2
(1) (2) (3) (4)
4 16 9 3
2 2
The line 5x−2y+4k=0 is a tangent to 4x −y =36, then k is :
9 81 4 2
(1) (2) (3) (4)
4 16 9 3

3. ö£¸UPø»¨ ö£õÖzx S»©õQ¯ JßÔß •¨£i ‰»[PÎÀ, ω2 Cß Á›ø\.
(C[S ω Gߣx (1)1/3 -ß P»¨ö£s ‰»®)
(1) 2 (2) 1 (3) 4 (4) 3
2
In the multiplicative group of cube root of unity, the order of ω is : [ω is a complex cube
root of unity]
(1) 2 (2) 1 (3) 4 (4) 3

4. ©ØÖ® g(x) BQ¯ \õº¦PÒ ö£õxÁiÁ Cøh©v¨¦ Âv°À
f(x)
Áøµ¯ÖUP¨£mhøÁ ÷£õÀ Aø©²® GÛÀ, ö£õx ÁiÁ Cøh©v¨¦
Âv°ß G¢u SÔ¨¤mh {ø»°À Ax ö»Uµõg]°ß Cøh ©v¨¦ Âv¯õP
©õÖ® ?
(1) f 9(x)=0
(2) g9(x)=0
(3) g(x) Gߣx J¸ \©Ûa\õº¦
(4) f(x) Gߣx J¸ \©Ûa\õº¦
If f(x) and g(x) are two functions as defined in Generalized law of mean then Lagrange’s
law of mean is a particular case of Generalised law of mean for :
(1) f 9(x)=0
(2) g9(x)=0
(3) g(x) is an identity function
(4) f(x) is an identity function

A

A A

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3 1312 (NP)

5. −x−iy •uÀ PõÀ£Sv°À Aø©¢uõÀ −ix+y Aø©²® PõÀ £Sv :
(1) ‰ßÓõ® PõÀ £Sv (2) |õßPõ® PõÀ £Sv
(3) •uÀ PõÀ £Sv (4) Cµshõ® PõÀ £Sv
If −x−iy lies in the first quadrant, then −ix+y lies in the :
(1) third quadrant (2) fourth quadrant
(3) first quadrant (4) second quadrant

6. ¤ßÁ¸ÁÚÁØÖÒ Gx ö©´ø©¯õS® ?
(1) p∨(~p) (2) p∧(~p) (3) p∨q (4) p∧q
Which of the following is a tautology ?
(1) p∨(~p) (2) p∧(~p) (3) p∨q (4) p∧q

7. X GßÓ \©Áõ´¨¦ ©õÔ°ß £µÁØ£i 4 ÷©¾® \µõ\› 2 GÛÀ E(X2) Cß
©v¨¦ :
(1) 6 (2) 8 (3) 2 (4) 4
Variance of the random variable X is 4. Its mean is 2. Then E(X2) is :
(1) 6 (2) 8 (3) 2 (4) 4

→ → →
8. r = s i −t k GßÓ \©ß£õk SÔ¨£x :
(1) yz - uÍ®

(2) xz - uÍ®
→ →
(3) i ©ØÖ® k ¦ÒÎPøÍ CønUS® ÷|ºU÷Põk
(4) xy - uÍ®

→ → →
r = s i − t k is the equation of :
(1) yz - plane
(2) xz - plane
→ →
(3) a straight line joining the points i and k
(4) xy - plane

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1312 (NP) 4

1
9. y = x 3 GßÓ ÁøÍÁøµUS RÌUPõq® TØÖPÎÀ Gx ö©´¯õÚx ?

(1) ÁøÍÁøµUS J¸ ÁøÍÄ ©õØÖ¨ ¦ÒÎ EÒÍx. ÷©¾® A¨¦ÒΰÀ
y99 QøhUPõx

(2) ÁøÍÁøµUS JßÖUS ÷©»õÚ ÁøÍÄ ©õØÖ¨ ¦ÒÎPÒ EÒÍx
(3) ÁøÍÁøµUS ÁøÍÄ ©õØÖ¨ ¦ÒÎ Qøh¯õx
(4) ÁøÍÁøµUS J¸ ÁøÍÄ ©õØÖ¨ ¦ÒÎ EÒÍx. ÷©¾® A¨¦ÒΰÀ
y99=0 BS®

1
Which one of the following statements is true about the curve y = x 3 ?

(1) The curve has a point of inflection in which y99 does not exist

(2) The curve has more than one point of inflection

(3) The curve has no point of inflection

(4) The curve has a point of inflection in which y99=0

10. z 1=1+2i, z 2=1−3i ©ØÖ® z 3 =2+4i GÛÀ, z 1z 2z 3, 2z 1z 2z 3 ©ØÖ® −7z 1z 2z 3
Gß£Ú J¸ BºPß uÍzvÀ :
(1) C¸ \©£UP •U÷Põnzvß •øÚ¨¦ÒÎPÒ
(2) J÷µ ÷Põhø©ÁÚ
(3) ö\[÷Põn •U÷Põnzvß •øÚ¨¦ÒÎPÒ
(4) \©£UP •U÷Põnzvß •øÚ¨¦ÒÎPÒ
If z 1 =1+2i, z 2 =1−3i and z 3 =2+4i then, the points on the Argand diagram
representing z1z2z3, 2z1z2z3, −7z1z2z3 are :

(1) Vertices of an isosceles triangle

(2) Collinear

(3) Vertices of a right angled triangle

(4) Vertices of an equilateral triangle

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5 1312 (NP)

11. \©£izuõÚ ÷|›¯a \©ß£õkPÎß öuõS¨¤À ρ(A) Gߣx ©õÔPÎß
GsoUøPø¯ Âh SøÓÁõÚx GÛÀ öuõS¨£õÚx :
(1) öÁΨ£øh¯ØÓ wºÄPÒ ©mk÷© ö£ØÔ¸US®
(2) wºÄPÒ ö£ØÔ¸UPõx
(3) öÁΨ£øhz wºÄ ©mk÷© ö£ØÔ¸US®
(4) öÁΨ£øhz wºÄ ©ØÖ® GsoUøP¯ØÓ öÁΨ£øh¯ØÓ wºÄPÒ
ö£ØÔ¸US®
In the homogeneous system ρ(A) is less than the number of unknowns, then the system
has :

(1) only non-trivial solutions

(2) no solution

(3) only trivial solution

(4) trivial solution and infinitely many non-trivial solutions

12. y=cx−c2 GߣuøÚ¨ ö£õxz wºÁõP¨ ö£ØÓ ÁøPUöPÊ \©ß£õk :

(1) y9=c (2) (y9) 2+xy9+y=0

(3) (y9)2−xy9+y=0 (4) y99=0

y=cx−c2 is the general solution of the differential equation :

(1) y9=c (2) (y9) 2+xy9+y=0

(3) (y9)2−xy9+y=0 (4) y99=0

13. y9+(y99) 2 =x(x+y99) 2 GßÓ ÁøPUöPÊa \©ß£õmiß Á›ø\ ©ØÖ® £i
•øÓ÷¯ :
(1) 1, 2 (2) 1, 1 (3) 2, 2 (4) 2, 1

The order and degree of the differential equation y9+(y99)2=x(x+y99)2 are :

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

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1312 (NP) 6

π
2
tan x − cot x
14. ∫ 1 + tan x cot x dx Cß ©v¨¦ :
0

π π
(1) (2) π (3) (4) 0
4 2

π
2
tan x − cot x
The value of ∫ 1 + tan x cot x dx is :
0

π π
(1) (2) π (3) (4) 0
4 2

15. J¸ £õ´éõß £µÁ¼À P(X=2)=P(X=3) GÛÀ, £s£ÍøÁ λ Cß ©v¨¦ :
(1) 3 (2) 0 (3) 6 (4) 2
In a Poisson distribution if P(X=2)=P(X=3) then, the value of its parameter λ is :
(1) 3 (2) 0 (3) 6 (4) 2

16. x 2 +y 2 =4, ©ØÖ® x=2 CÁØÔØS Cøh÷¯ HØ£k® £µ¨¤øÚ
x=−2
x- Aaø\ ö£õÖzxa _ÇØÓ¨£k® ÷£õx QøhUS® vh¨ö£õ¸Îß
ÁøÍ£µ¨¦ :
(1) 64π (2) 32π (3) 8π (4) 16π
The surface area of the solid of revolution of the region bounded by x2+y2=4, x=−2
and x=2 about x-axis is :
(1) 64π (2) 32π (3) 8π (4) 16π

→ → → → → → → → →
17. a + b + c = 0 , a = 3, b = 4, c = 5 GÛÀ, a &US® b &US®
Cøh¨£mh ÷Põn® :
5π π π 2π
(1) (2) (3) (4)
3 2 6 3
→ → → → → → → → →
If a + b + c = 0 , a = 3, b = 4, c = 5 then, the angle between a and b is :

5π π π 2π
(1) (2) (3) (4)
3 2 6 3

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7 1312 (NP)

18. y2=12x GßÓ £µÁøͯzvß SÂ|õoß CÖv¨¦ÒÎPÎÀ Áøµ¯¨£k®
öuõk÷PõkPÒ \¢vUS® ¦ÒÎ Aø©²® ÷Põk :

(1) y+3=0 (2) y−3=0 (3) x−3=0 (4) x+3=0

The tangents at the end of any focal chord to the parabola y2=12x intersect on the
line :

(1) y+3=0 (2) y−3=0 (3) x−3=0 (4) x+3=0

19. A GßÓ vø\°¼ Ao°ß Á›ø\ 3, vø\°¼ k ≠ 0 GÛÀ A−1 Gߣx :

1 1 1
(1) I (2) kI (3) 2
I (4) I
k k k3

If A is a scalar matrix with scalar k ≠ 0, of order 3, then A−1 is :

1 1 1
(1) I (2) kI (3) 2
I (4) I
k k k3

20. J¸ ÷PõÍzvß PÚ AÍÄ ©ØÖ® BµzvÀ HØ£k® ©õÖÃu[PÒ
GsnÍÂÀ \©©õP C¸US® ÷£õx ÷PõÍzvß ÁøÍ£µ¨¦ :

4π 1
(1) 4π (2) (3) 1 (4)
3 2π

The surface area of a sphere when the volume is increasing at the same rate as its
radius, is :

4π 1
(1) 4π (2) (3) 1 (4)
3 2π

A [ v¸¨¦P / Turn over

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1312 (NP) 8

£Sv & II / PART - II

SÔ¨¦ : (i) H÷uÝ® HÊ ÂÚõUPÐUS Âøh¯ÎUPÄ®. 7x2=14

(ii) ÂÚõ Gs 30 &US Psi¨£õP Âøh¯ÎUPÄ®.
Note : (i) Answer any seven questions.
(ii) Question number 30 is compulsory.

21. JÆöÁõ¸ ÁøP |õn¯[PÎß GsoUøPø¯ Põs£uØPõÚ ÷|›¯a
\©ß£õmkz öuõS¨¤øÚ R÷Ç öPõkUP¨£mkÒÍ {PÌa]US HØÓÁõÖ
GÊxP.
""J¸ ø£°À ` 1 ©ØÖ® ` 2 ©ØÖ® ` 5 |õn¯[PÒ EÒÍÚ. ¹£õ´ 100
©v¨¤ØS ö©õzu® 30 |õn¯[PÒ EÒÍÚ.''
To find the number of coins, in each category, write the suitable system of equations for
the given situation :
“A bag contains 3 types of coins namely ` 1, ` 2 and ` 5. There are 30 coins amounting
to ` 100 in total.”

→ → → → → →
22. 3 i + 2 j +9 k ©ØÖ® i +m j +3 k Gß£Ú JßÖUöPõßÖ Cøn
2
öÁUhºPÒ GÛÀ m = 3 GÚ {ÖÄP.

→ → → → → →
If the two vectors 3 i + 2 j + 9 k and i + m j + 3 k are parallel, then prove that

2
m= .
3

n
1+ i 
23.   = 1 GÛÀ n &Cß «a]Ö ªøP •Ê Gs ©v¨ø£U PõsP.
1− i 

n
Find the least positive integer n such that 
1+ i 
 = 1.
1− i 

A

A A

A A

9 1312 (NP)

24. RÌUPsh {PÌa]US HØÓ Áøµ£hzøu ÁøµP.
""J¸ ÁõÀ Âs«ß (Comet) BÚx `›¯øÚa (Sun) _ØÔ
£µÁøͯ¨£õøu°À ö\ÀQÓx ©ØÖ® `›¯ß £µÁøͯzvß S¯zvÀ
Aø©QÓx. ÁõÀ Âs«ß `›¯Û¼¸¢x 80 ªÀ¼¯ß Q.«. öuõø»ÂÀ
Aø©¢x C¸US® ÷£õx ÁõÀ Âs«øÚ²®, `›¯øÚ²® CønUS®
π
÷Põk, £õøu°ß Aa_hß 3 GßÓ ÷PõnzvøÚ HØ£kzx®.''
Draw the diagram for the given situation :
“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 kms from the sun, the line segment from the
π
sun to the comet makes an angle of radians with the axis of the orbit.”
3

25. f(x)=sinx &ß ©õÖ{ø» GsPøÍU PõsP.
Find the critical numbers of f(x)=sinx.

26. f(x)=x3+1 GßQÓ ÁøÍÁøµ°ß \õº£P® ©ØÖ® }mi¨¦ BQ¯ÁØøÓ
PõsP.
Write the domain and extent of the function f(x)=x3+1.

π π
3 3
dx dx
27. {ÖÄP : ∫ 1 + cot x ∫ 1 + tanx
=
π π
6 6

π π
3 3
dx dx
Prove that ∫ = ∫
π 1 + cot x π 1 + tanx
6 6

28. §a]¯©ØÓ ÂQu•Ö GsPÎß Pn®, ÁÇUP©õÚ Tmh¼ß RÌ AøhÄ
AØÓx GÚ {ÖÄP.
Show that the set of all non-zero rational numbers is not closed under addition.

A [ v¸¨¦P / Turn over

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A A

1312 (NP) 10

 −3x ,
29. J¸ \©Áõ´¨¦ ©õÔ x&ß {PÌuPÄ Ahºzva \õº¦ f ( x ) = 3e x >0

 0 , x ≤0

GÛÀ £µÁÀ \õº¦ F(3)=1−e−9 GÚ {ÖÄP.
Prove that F(3)=1−e −9 if the probability density function f(x) is defined as

3e−3x , x >0
f (x) = 
 0 , x ≤0

30. f(x)=?x−2?+?x−5? GßÓ \õº¦US [1, 6] GßÓ CøhöÁΰÀ ÷µõ¼ß
÷uØÓzøua \› £õºUP.
Verify Rolle’s theorem for the function f(x)=?x−2?+?x−5? in [1, 6].

£Sv & III / PART - III
SÔ¨¦ : (i) H÷uÝ® HÊ ÂÚõUPÐUS Âøh¯ÎUPÄ®. 7x3=21

(ii) ÂÚõ Gs 40 &US Psi¨£õP Âøh¯ÎUPÄ®.
Note : (i) Answer any seven questions.
(ii) Question number 40 is compulsory.

31. A ©ØÖ® B BQ¯ H÷uÝ® C¸ ‰ßÓõ® Á›ø\²ÒÍ ö£õ¸zu©õÚ
AoPøÍU öPõsk ρ(A)+ρ(B) ≠ ρ(A+B) GߣuøÚ {¹¤UP.
Prove that ρ(A)+ρ(B) ≠ ρ(A+B) by giving the suitable matrices A and B of order 3.

→ → → → → →
32. 4 i − j +3 k , − 2 i + j −2 k GÝ® öÁUhºPÐUS ö\[SzuõÚx® Gs
AÍÄ 6 Eøh¯x©õÚ öÁUhºPøÍU PõsP.
Find the vectors of magnitude 6 which are perpendicular to both the vectors
→ → → → → →
4 i − j + 3 k and − 2 i + j − 2 k .

A

A A

A A

11 1312 (NP)

33. n Gߣx J¸ ªøP •Ê Gs GÛÀ :
n
 1 + sinθ − i cosθ  π  π 
  = cos n  − θ  − i sin n  − θ 
 1 + sinθ + i cosθ  2  2 
GÚ {¹¤UP.
If n is a positive integer, prove that
n
 1 + sinθ − i cosθ  π  π 
  = cos n  − θ  − i sin n  − θ 
 1 + sinθ + i cosθ  2  2 

34. J¸ ö\ÆÁP Av£µÁøͯzvØS Áøµ¯¨£mh öuõk÷Põmiß öuõk¦ÒÎ
öuõø»zöuõk ÷PõkPÐUS Cøh¨£mh £õPzvøÚ C¸ \©©õP¨ ¤›US®
GÚU PõmkP.
Show that the tangent to a rectangular hyperbola terminated by its asymptotes is bisected
at the point of contact.

 π
35. f(x)=tan−1(sinx+cosx), x > 0 GßÓ \õº¦  0,  GßÓ CøhöÁΰÀ vmh©õP
 4
HÖ® \õº¦ GÚU Põs¤UP.
Show that the function f(x)=tan−1(sinx+cosx), x > 0 is strictly increasing in the interval

 π
 0,  .
 4

1 ∂f ∂f
36. f (x , y ) = GÛÀ, x +y =−f GÚU PõmkP.
x2 + y 2 ∂x ∂y

1 ∂f ∂f
If f ( x , y ) = then, prove that x +y =−f .
2 2 ∂x ∂y
x +y

37. Bµ® ‘r’ , Szx¯µ® ‘h’ Eøh¯ E¸øÍ°ß PÚ AÍøÁ öuõøP±mk
•øÓ°À PõsP.
Derive the formula for the volume of a cylinder with radius ‘r’ and height ‘h’ by using
integration.

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A A

A A

1312 (NP) 12

38. (p ∧ q) → (p ∨ q) Gߣx J¸ ö©´ø© GÚU PõmkP.
Show that (p ∧ q) → (p ∨ q) is a tautology.

39. J¸ £Pøh 120 •øÓ E¸mh¨£kQÓx. £Pøh°ß ÷©À 1 AÀ»x 5
Qøh¨£x öÁØÔö¯ÚU öPõÒͨ£kQÓx. QøhUS® öÁØÔ°ß
GsoUøP°ß \µõ\› ©ØÖ® £µÁØ£iø¯U PõsP.
A die is thrown 120 times and getting 1 or 5 is considered a success. Find the mean and
variance of the number of successes.

40. yx 3 dx+e −x dy=0 GßÓ ÁøPUöPÊa \©ß£õmiß wºÄ
( x 3−3x 2+6 x−6) ex+log y=c GÚ {ÖÄP.
Show that the solution of the differential equation yx 3 dx+e −x dy=0 is

( x 3−3x 2+6 x−6) ex+log y=c.

£Sv & IV / PART - IV
SÔ¨¦ : AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®. 7x5=35
Note : Answer all the questions.

41. (a) µ &Cß G®©v¨¤ØS x+y+3z=0; 4x+3y+µz=0; 2x+y+2z=0 GßÓ
\©¨£izuõÚ öuõS¨¤ØS
(i) öÁΨ£øhz wºÄ ©mk®
(ii) JßÖUS ÷©Ø£mh wºÄPÒ QøhUS®, GÚU PõsP.
AÀ»x
(b) sin (A+B)=sinA cosB+cosA sinB Gߣøu öÁUhº •øÓ°À {ÖÄP.
(a) For what values of µ the system of homogeneous equations x+y+3z=0;
4x+3y+µz=0; 2x+y+2z=0 have :
(i) only trivial solution
(ii) infinitely many solutions
OR
(b) Prove by vector method that
sin (A+B)=sinA cosB+cosA sinB

A

A A

A A

13 1312 (NP)

x− 2 y− 2 z− 1
42. (a) = = GßÓ ÷Põmøh EÒÍhUQ¯x® (−1, 1, −1) GßÓ
2 3 −2
¦ÒÎ ÁÈ÷¯a ö\À»U Ti¯x©õÚ uÍzvß Põºj]¯ß \©ß£õmøhU
PõsP.
AÀ»x
(b) wºUP : x11−x6+x5−1=0
(a) Find the cartesian equation of the plane containing the line
x− 2 y− 2 z− 1
= = and passing through the point (−1, 1, −1).
2 3 −2
OR
(b) Solve : x11−x6+x5−1=0.

43. (a) ""}ÒÁmhzvß «xÒÍ H÷uÝ® J¸ ¦Ò롧 SÂzöuõø»ÄPÎß
TkuÀ Auß ö|mha]ß }ÍzvØSa \©®'' GÚ {ÖÄP. ÷©¾®, J¸
¦ÒίõÚx A¨¦ÒÎUS® (3, 0) ©ØÖ® (−3, 0) GßÓ ¦ÒÎPÐUS®
Cøh÷¯¯õÚ yµ[PÎß TkuÀ 9 BP C¸US©õÖ |P¸©õÚõÀ

x2 y2
A¨¦Ò롧 C¯[SÁøµ + = 1 GÚ {ÖÄP.
 81   45 
   
 4   4 

AÀ»x
(b) ‘r’ Bµ•ÒÍ ÁmhzvÝÒ ö£¸® AÍÄ öPõÒЩõÖ Áøµ¯¨£k®
ö\ÆÁPzvß £µ¨¦ 2r2 GÚ {ÖÄP.

(a) Show that the sum of the focal distances of any point on an ellipse is equal to the
length of the major axis and also prove that the locus of a point which moves so

x2 y2
that the sum of its distances from (3, 0) and (−3, 0) is 9, is + =1.
 81   45 
   
 4   4 

OR
(b) Prove that the area of the largest rectangle that can be inscribed in a circle of
radius ‘r’ is 2r2.

A [ v¸¨¦P / Turn over

A A

A A

1312 (NP) 14

44. (a) J¸ HÄPøn, uøµ°¼¸¢x ö\[SzuõP ÷©À÷|õUQa ö\¾zx® ÷£õx
25 2
t ÷|µzvÀ ö\À¾® E¯µ® x GßP. Auß \©ß£õk x = 100t − t
2
GÛÀ :

(i) HÄPøn°ß öuõhUP vø\÷ÁP®

(ii) HÄPøn Ea\ E¯µzøu Aøh²® ÷£õx Auß ÷|µ®

(iii) HÄPøn Aøh²® Ea\ E¯µ®

(iv) HÄPøn uøµø¯ Aøh²® ÷£õx Auß vø\÷ÁP®

BQ¯ÁØøÓU PõsP.

AÀ»x

(b) GßÓ Av£µÁøͯzvß ø©¯®,
16x 2 −9y 2 −32x−18y+151=0
S¯[PÒ ©ØÖ® Ea]PÒ BQ¯ÁØøÓU PõsP. ÷©¾® Auß
ÁøÍÁøµø¯ ÁøµP.

(a) A missile fired from ground level rises x metres vertically upwards in t seconds
25 2
and x = 100t − t . Find :
2

(i) the initial velocity of the missile

(ii) the time when the height of the missile is a maximum

(iii) the maximum height reached

(iv) the velocity with which the missile strikes the ground

OR

(b) Find the centre, foci and vertices of the hyperbola 16x2−9y2−32x−18y+151=0
and draw the diagram.

A

A A

A A

15 1312 (NP)

45. (a) J¸ ÷uºÂÀ 1000 ©õnÁºPÎß \µõ\› ©v¨ö£s 34 ©ØÖ® vmh
»UP® 16 BS®. ©v¨ö£s C¯À{ø»¨ £µÁø» ö£ØÔ¸¨¤ß
©zv¯ 70% ©õnÁºPÒ ö£Ö® ©v¨ö£sPÎß GÀø»PøÍU PõsP.
P[0 < Z < 1.04]=0.35
AÀ»x
(b) y=sinx ©ØÖ® y=cosx GßÓ ÁøÍÁøµPÒ x=0 ©ØÖ® x=π GßÓ
÷PõkPÒ BQ¯ÁØÖUS Cøh÷¯ EÒÍ Aµ[Pzvß £µ¨ø£U PõsP.
(a) The mean score of 1000 students for an examination is 34 and the standard
deviation is 16. Determine the limit of the marks of the central 70% of the
candidates by assuming the distribution is normal.
P[0 < Z < 1.04]=0.35
OR
(b) Compute the area between the curve y=sinx and y=cosx and the lines x=0 and
x=π.

46. (a) w=x+2y+z2 GßÓ \õº¤À x=cos t; y=sin t; z=t GÛÀ \[Q¼ Âvø¯¨

£¯ß£kzv dw &IU PõsP. ÷©¾® x, y ©ØÖ® z &ß ©v¨¦PøÍ w &À
dt

dw
¤µv°mk &ß ©v¨ø£U Psk Âøhø¯ \› £õºUP.
dt
AÀ»x
(b) öÁ¨£{ø» 158C EÒÍ J¸ AøÓ°À øÁUP¨£mkÒÍ ÷u}›ß
öÁ¨£{ø» 1008C BS®. Ax 5 {ªh[PÎÀ 608C BP SøÓ¢x
ÂkQÓx. ÷©¾® 5 {ªh® PÈzx ÷u}›ß öÁ¨£ {ø»°øÚU PõsP.
dw
(a) If w=x+2y+z2 and x=cos t; y=sin t; z=t find by using chain rule. Also
dt

dw
find by substitution of x, y and z in w and hence verify the result.
dt
OR
(b) A cup of tea at temperature 1008C is placed in a room whose temperature is
158C and it cools to 608C in 5 minutes. Find its temperature after further interval
of 5 minutes.

A [ v¸¨¦P / Turn over

A A

A A

1312 (NP) 16

47. (a) S»[PÎß I¢x £s¦PøͲ® GÊxP.
AÀ»x
−2
x
x
(b) (5D2−8D−4) y = 5 e 5 + 2e + 3 GßÓ ÁøPUöPÊa \©ß£õmiß wºÄ
−2 −2
x 5 x 2 3
y = Ae 2x + Be 5 − xe 5 − e x − GÚ {ÖÄP.
12 7 4
(a) State all the five properties of groups.
OR
(b) Prove that the solution of the differential equation :

−2 −2 −2
x x 5 x 2 3
2
(5D −8D−4) y = 5 e 5 + 2e x + 3 is y = Ae 2x + Be 5 − xe 5 − e x − .
12 7 4

-oOo-

A

A A