TN 11th Question Paper 2024 Maths

TN PUBLIC EXAM

QUESTION
PAPER
2024

No. of Printed Pages : 12
7612
A £vÄ Gs
Register Number

!7612IstYearMathematics!
PART - III
Pou® / MATHEMATICS
( uªÌ ©ØÖ® B[Q» ÁÈ / Tamil & English Version)

Põ» AÍÄ : 3.00 ©o ÷|µ® ] [ ö©õzu ©v¨ö£sPÒ : 90
Time Allowed : 3.00 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®,
AiU÷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 |õßS ©õØÖ ÂøhPÎÀ ªPÄ® Hئøh¯
Âøhø¯z ÷uº¢öukzxU SÔ±mkhß Âøh°øÚ²® ÷\ºzx GÊuÄ®.
Note : (i) Answer all the questions.

(ii) Choose the most appropriate answer from the given four alternatives and write
the option code and the corresponding answer.

[ v¸¨¦P / Turn over

7612 2

1. n §mkPЮ AuØS ö£õ¸zu©õP n \õÂPЮ EÒÍÚ. AøÚzx §mkPÐUS®
\õÂPÒ \›¯õP ö£õ¸¢xÁuØPõÚ AvP£m\ •¯Ø]PÎß GsoUøP :

n(n+1)
(A) n(n−1) (B) n(n+1) (C) n (D)
2

There are n locks and n matching keys. If all the locks and keys are to be perfectly matched,
then the maximum number of trials is :

n(n+1)
(a) n(n−1) (b) n(n+1) (c) n (d)
2

x2 y 2
2. − =k GßÓ {¯©¨£õøu°ß «x (8, −5) GßÓ ¦ÒÎ EÒÍx GÛÀ, k &ß
16 25
©v¨¦ :
(A) 2 (B) 0 (C) 3 (D) 1

x2 y 2
If the point (8, −5) lies on the locus − =k , then the value of k is :
16 25

(a) 2 (b) 0 (c) 3 (d) 1

3. x2−4y2=0 ©ØÖ® x=a GßÓ ÷PõkPÍõÀ E¸ÁõUP¨£k® •U÷Põnzvß £µ¨¦ :

1 2 2 3 2
(A) 2 a 2 (B) 2a 2 (C) 3
a (D) a
2

The area of the triangle formed by the lines x2−4y2=0 and x=a is :

1 2 2 2 3 2
(a) a (b) 2a 2 (c) a (d) a
2 3 2

4. nC +n C +......+n C =
0 1 n

(A) 2n+1 (B) 2 n (C) 2n−1 (D) 2n
nC +n C +......+n C =
0 1 n

(a) 2 n+1 (b) 2n (c) 2 n−1 (d) 2n

A

3 7612

5. A ©ØÖ® B BQ¯ C¸ {PÌa]PÒ A⊂B ©ØÖ® P(B) ≠ 0, GÚ C¸¨¤ß
¤ßÁ¸ÁÚÁØÖÒ Gx ö©´¯õÚx ?
P(A)
(A) P(A/B) / P(A) (B) P(A/B) = P(B)

(C) P(A/B) > P(B) (D) P(A/B) < P(A)
If A and B are two events such that A⊂B and P(B) ≠ 0, then which of the following is
correct ?

P(A)
(a) P(A/B) / P(A) (b) P(A/B) =
P(B)

(c) P(A/B) > P(B) (d) P(A/B) < P(A)

sinx
6. lim
x →∞ x

(A) ∞ (B) 1 (C) −∞ (D) 0
sinx
lim
x →∞ x

(a) ∞ (b) 1 (c) −∞ (d) 0

7. n((A×B)∩(A×C))=8 ©ØÖ® n(B∩C)=2 GÛÀ, n(A) Gߣx :

(A) 8 (B) 6 (C) 16 (D) 4
If n((A×B)∩(A×C))=8 and n(B∩C)=2 then n(A) is :
(a) 8 (b) 6 (c) 16 (d) 4

8. (x, −2), (5, 2), (8, 8) Gß£Ú J¸ ÷Põhø©¨ ¦ÒÎPÒ GÛÀ, x &ß ©v¨¦ :

1
(A) 1 (B) −3 (C) 3 (D) 3
If the points (x, −2), (5, 2), (8, 8) are collinear, then x is equal to :

1
(a) 1 (b) −3 (c) 3 (d)
3

A [ v¸¨¦P / Turn over

7612 4

3
9. A ©ØÖ® B GßÓ C¸ {PÌa]PÐUS P(A) = ©ØÖ® P(A ∩ B ) = 1 GÛÀ
10 2
P(A∩B) &ß ©v¨¦ :

1 1 1 1
(A) 4 (B) 2 (C) 5 (D) 3

3 1
If two events A and B are such that P(A) = and P(A ∩ B ) = then P(A∩B) is :
10 2

1 1 1 1
(a) (b) (c) (d)
4 2 5 3

→ ∧ ∧ ∧ → ∧ ∧ ∧ → ∧ ∧ ∧ → → →
10. a = i + j + k , b = 2 i+x j + k , c = i− j +4 k ©ØÖ® a ⋅ ( b × c ) = 70 GÛÀ x &ß
©v¨¦ :
(A) 26 (B) 5 (C) 10 (D) 7
→ ∧ ∧ ∧ → ∧ ∧ ∧ → ∧ ∧ ∧ → → →
If a = i + j + k , b = 2 i+x j + k , c = i− j +4 k and a ⋅ ( b × c ) = 70 then x is equal to :

(a) 26 (b) 5 (c) 10 (d) 7

dy
11. y=f (x2+2) ©ØÖ® f ’(3)=5 GÛÀ, x=1 &À Gߣx :
dx

(A) 15 (B) 5 (C) 10 (D) 25
dy
If y=f (x2+2) and f ’(3)=5, then at x=1 is :
dx
(a) 15 (b) 5 (c) 10 (d) 25

12. e−2x GßÓ öuõh›À x5 &ß öPÊ :

−4 2 4 3
(A) 15 (B) 3 (C) 15 (D) 2
The co-efficient of x5 in the series e−2x is :

−4 2 4 3
(a) (b) (c) (d)
15 3 15 2

A

5 7612

sin(A−B) sin(B−C) sin(C−A)
13. + + =
cosA cosB cosB cosC cosC cosA
(A) 0 (B) sinA+sinB+sinC
(C) cosA+cosB+cosC (D) 1
sin(A−B) sin(B−C) sin(C−A)
+ + is :
cosA cosB cosB cosC cosC cosA
(a) 0 (b) sinA+sinB+sinC
(c) cosA+cosB+cosC (d) 1

dy
14. y=esinx GÛÀ =
dx
(A) sinxesinx (B) esinx (C) cosxesinx (D) ecosx
dy
If y=esinx then =
dx
(a) sinxesinx (b) esinx (c) cosxesinx (d) ecosx

15. log 2 512 &ß ©v¨¦ :
(A) 9 (B) 16 (C) 12 (D) 18
The value of log 2 512 is :
(a) 9 (b) 16 (c) 12 (d) 18

→ ∧ ∧ ∧ ∧ ∧ ∧
16. BA = 3 i + 2 j + k ©ØÖ® B &ß {ø» öÁUhº i +3 j − k GÛÀ A &ß {ø» öÁUhº :
∧ ∧ ∧ ∧ ∧
(A) 4 ∧i (B) 4 i + 2 j + k (C) −4 ∧i (D) 4 i +5 j
→ ∧ ∧ ∧ ∧ ∧ ∧
If BA = 3 i + 2 j + k and the position vector of B is i +3 j − k , then the position vector A is :

∧ ∧ ∧ ∧ ∧ ∧ ∧
(a) 4i (b) 4 i +2 j + k (c) −4 i (d) 4 i +5 j

tan x
17. ∫ sin 2 x dx =
1 1
(A) 2 tan x + C (B) tan x + C (C) 4 tan x + C (D) 2 tan x + C

tan x
∫ sin 2 x dx is :
1 1
(a) tan x + C (b) tan x + C (c) tan x + C (d) 2 tan x + C
2 4

A [ v¸¨¦P / Turn over

7612 6

3−x − 6 3
18. −6 3− x 3 = 0 GßÓ \©ß£õmiß J¸ wºÄ :
3 3 −6−x

(A) 0 (B) 6 (C) −6 (D) 3
3−x − 6 3
A root of the equation − 6 3 − x 3 = 0 is :
3 3 −6−x
(a) 0 (b) 6 (c) −6 (d) 3

19. f : [−3, 3] → S GßÓ \õº¦ f (x)=x2 GÚ Áøµ¯ÖUP¨£mk ÷©Ø÷PõºzuÀ GÛÀ,
S Gߣx :
(A) [−3, 3] (B) [−9, 9] (C) [0, 9] (D) R
If the function f : [−3, 3] → S defined by f (x)=x2 is onto, then S is :
(a) [−3, 3] (b) [−9, 9] (c) [0, 9] (d) R

a x − bx
20. lim =
x →0 x

b a a
(A) log  a  (B) log ab (C) b (D) log  
b

a x − bx
lim =
x →0 x

b a a
(a) log   (b) log ab (c) (d) log  
a b b

£Sv - II / PART - II
SÔ¨¦ : GøÁ÷¯Ý® HÊ ÂÚõUPÐUS Âøh¯ÎUPÄ®. ÂÚõ Gs 30 &US
Pmhõ¯©õP Âøh¯ÎUPÄ®. 7x2=14
Note : Answer any seven questions. Question No. 30 is Compulsory.

1 1 A
21. + = GÛÀ A &ß ©v¨¦ GßÚ ?
7! 8! 9!
1 1 A
If + = then find the value of A.
7! 8! 9!

22. 3x+2y+9=0 ©ØÖ® 12x+8y−15=0 BQ¯øÁ Cøn÷PõkPÒ GÚU PõmkP.
Show that the lines are 3x+2y+9=0 and 12x+8y−15=0 are parallel lines.

A

7 7612

3 4 1
23. A = 0 −1 2 
 GÛÀ, ?A? &ß ©v¨ø£U PõsP.
 
 5 −2 6 

3 4 1
Compute ?A? if A = 0 −1 2 

 
 5 −2 6 

3
24. A ©ØÖ® B JßøÓö¯õßÖ Â»US® {PÌa]PÒ, P(A) = ©ØÖ® P(B) = 1 GÛÀ,
8 8
P(A ∪ B) &IU PõsP.

3 1
If A and B are mutually exclusive events P(A) = and P(B) = then find P(A ∪ B) .
8 8

→ ∧ ∧ → ∧ ∧ ∧ → →
25. a = 3 i + 4 j ©ØÖ® b = i + j + k GÛÀ, a × b &ß ©v¨ø£U PõsP.

→ → → ∧ ∧ → ∧ ∧ ∧
Find a × b , where a = 3 i + 4 j and b = i + j + k .

1
26.
1− 2 sin x
GßÓ \õº¤ß \õº£PzøuU PõsP.

1
Find the domain of
1− 2 sin x

1 − tanA
27. tan (45− A) = GÚU PõmkP.
1 + tanA

1 − tanA
Show that tan (45− A) =
1 + tanA

28. f (x)=x cosx GÛÀ, f '' PõsP.
Find f '' if f (x)=x cosx.

A [ v¸¨¦P / Turn over

7612 8

29. GßÓ Cµmøh ÷|º÷Põmiß uÛzuÛ ÷|º÷PõkPÎß
5x 2 +6xy+y 2 =0
\©ß£õmøhU PõsP.
Find the separate equation of the pair of straight lines 5x2+6xy+y2=0.

30. INDIA GßÓ Áõºzøu°À EÒÍ GÊzxPøÍ GzuøÚ ÁøPPÎÀ Á›ø\¨&
£kzu»õ® ?
Find the number of ways of arranging the letters of the word INDIA.

£Sv - III / PART - III
SÔ¨¦ : GøÁ÷¯Ý® HÊ ÂÚõUPÐUS Âøh¯ÎUPÄ®. ÂÚõ Gs 40 &US
Pmhõ¯©õP Âøh¯ÎUPÄ®. 7x3=21
Note : Answer any seven questions. Question No. 40 is Compulsory.

31. _¸USP :
1 1 1 1 1
− + − +
3− 8 8− 7 7− 6 6− 5 5 −2
Simplify :
1 1 1 1 1
− + − +
3− 8 8− 7 7− 6 6− 5 5 −2

32. ©v¨¤kP :

∫ xe dx
x

Evaluate :

∫ xe dx
x

33. ¤ßÁ¸® öuõh›À •uÀ n EÖ¨¦PÎß TkuÀ PõsP.
6+66+666+6666+...........
Compute the sum of first n terms of the series.
6+66+666+6666+...........

x 2 − 6x + 5
34. lim &ß ©v¨ø£U PõsP.
x → 3 x 3 − 8x + 7

x 2 − 6x + 5
Calculate : lim
x → 3 x 3 − 8x + 7

A

9 7612

35. x &I¨ ö£õÖzx ÁøPUöPÊøÁU PõsP.
y=xexlogx
Differentiate the following with respect to x.
y=xexlogx

36. nP =720 ©ØÖ® nC =120 GÛÀ, n, r &IU PõsP.
r r
If nPr=720 and nCr=120 find n, r.

37. ©v¨¦ PõsP : cos1058.
Find the value of cos1058.

 x 2+x−5 ; x  (−∞, 0)

 2
 x +3x−2 ; x  (3, ∞)
38. f (x) = 
2
 x ; x  (0, 2)
 2

 x −3 ; ©ØÓ Ch[PÎÀ

GÚ Áøµ¯ÖUP¨£iß −3, 5, 0 BQ¯ÁØÔÀ f &ß ©v¨¦PøÍU PõsP.

 x 2+x−5 if x  (−∞, 0)

 2
x +3x−2 if x  (3, ∞)
Write the values of f at −3, 5, 0 if f ( x ) =  2
 x if x  (0, 2)
 2

 x −3 otherwise

∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
39. 2 i − j + k , 3 i −4 j − 4k , i − 3 j −5 k BQ¯ öÁUhºPÒ J¸ ö\[÷Põn •U÷Põnzøu
Aø©US® GÚU PõmkP.
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
Show that the vectors 2 i − j + k , 3 i −4 j − 4k , i − 3 j −5 k form a right angled triangle.

cos11
+ sin11

40. = tan56
GÚ {ÖÄP.
cos11
− sin11

Prove that :
cos11
+ sin11

= tan56

cos11
− sin11


A [ v¸¨¦P / Turn over

7612 10

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

Note : Answer all the questions.

2x
41. (A) £Sv ¤ßÚ[PÍõP¨ ¤›UPÄ®. 2
( x + 1) (x − 1)

AÀ»x
(B) y=etan−1x GÛÀ, (1+x2)y ''+(2x−1)y '=0 GÚU PõmkP.
2x
(a) Resolve into partial fractions 2
.
( x + 1) (x − 1)

OR
−1x
(b) If y=etan , show that (1+x2)y ''+(2x−1)y '=0.

42. (A) ABCD GßÓ |õØPµzvÀ AC, BD &ß |k¨¦ÒÎPÒ E ©ØÖ® F &BP C¸¨¤ß,
GÚ {ÖÄP.
AÀ»x
(B) f ¤ßÁ¸©õÖ Áøµ¯ÖUP¨£mkÒÍx.

 sin x
 + cosx ; x≠0
f ( x ) = x
 2 ; x=0

GßÓ \õº¦ x=0 &CÀ öuõhºa]¯õÚx GÚU PõmkP.
(a) If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively,

then prove that .
OR

 sin x
 + cosx ; when x ≠ 0
(b) Let f ( x ) = x
 2 ; when x = 0

Show that f is continuous at x=0.

A

11 7612

43. (A) log 10 2+16 log 10 16 +12 log 10 25 +7 log 10 81 =1 GÚ {ÖÄP.
15 24 80
AÀ»x
(B) Jzu C¸ áõiPÎÀ, JßÔÀ 6 P¸¨¦ ©ØÖ® 4 ]Á¨¦ {Ó¨ £¢xPÒ EÒÍÚ.
©ØöÓõ¸ áõi°À 2 P¸¨¦ ©ØÖ® 2 ]Á¨¦ {Ó¨ £¢xPÒ EÒÍÚ. \©Áõ´¨¦
•øÓ°À J¸ áõi ÷uº¢öukUP¨£mk Av¼¸¢x J¸ £¢x GkUP¨£kQÓx.
(i) A¨£¢x P¸¨£õP C¸¨£uØPõÚ {PÌuPøÁU PõsP.
(ii) GkUP¨£mh £¢x P¸¨¦ GÛÀ •uÀ áõi°¼¸¢x GkUP¨£mhuØPõÚ
{PÌuPÄ ¯õx ?
16 25 81
(a) Prove that log 10 2+16 log 10 +12 log 10 +7 log 10 =1
15 24 80
OR
(b) There are two identical urns containing respectively 6 black and 4 red balls, 2 black
and 2 red balls. An urn is chosen at random and a ball is drawn from it.
(i) Find the probability that the ball is black.
(ii) If the ball is black, what is the probability that it is from the first urn ?

3x + 5
44. (A) ©v¨¤kP : ∫ 2 dx
x + 4x + 7
AÀ»x
 ᑻ
(B) cot  7 1  = 2 + 3 + 4 + 6 GÚU Põs¤UPÄ®.
 2 
3x + 5
(a) Evaluate : ∫ x2 + 4x + 7 dx
OR
 1ᑻ 
(b) Show that cot  7  = 2 + 3 + 4 + 6 .
 2 

45. (A) x J¸ ÷uøÁ¯õÚ AÍ»õÚ ö£›¯ Gs GÛÀ,
3 3 3 1
x + 6 − x 3 + 3 &ß ©v¨ø£z ÷uõµõ¯©õP GÚ {ÖÄP.
x2
AÀ»x
(B) f : R → R GßÓ \õº¦ f (x)=2x−3 GÚ Áøµ¯ÖUP¨£iß f J¸ C¸¦Óa\õº¦ GÚ
{¹¤zx, Auß ÷|º©õÔøÚU PõsP.
1
(a) Prove that 3 x 3 + 6 − 3 x 3 + 3 is approximately equal to when x is sufficiently
x2
large.
OR
(b) If f : R → R is defined by f (x)=2x−3, prove that f is a bijection and find its inverse.

A [ v¸¨¦P / Turn over

7612 12

46. (A) ÷|¨¤¯›ß `zvµzøu TÔ, {¹¤UPÄ®.
AÀ»x
(B) λx2−10xy+12y2+5x−16y−3=0 Gߣx J¸ Cµmøh ÷|ºU÷Põmøh SÔUS®
GÛÀ
(i) λ &ß ©v¨¦ ©ØÖ® uÛzuÛa \©ß£õkPøÍU PõsP.
(ii) CƸ ÷PõkPÒ öÁmk® ¦ÒÎø¯U PõsP.
(iii) C¸ ÷PõkPÐUS Cøh¨£mh ÷Põn® PõsP.
(a) State and prove Napier’s Formula.
OR
(b) If the equation λx2−10xy+12y2+5x−16y−3=0 represents a pair of straight lines,
find :
(i) The value of λ and the separate equations of the lines.
(ii) Point of intersection of the lines.
(iii) Angle between the lines.

47. (A) Põµoz ÷uØÓzøu £¯ß£kzv

b+ c a a2
c+ a b b 2 = (a + b + c) (a − b) (b − c) (c − a) GÚ {ÖÄP.
a+b c c2

AÀ»x
(B) n/1 &US 32n+2−8n−9 BÚx 8 &BÀ ÁS£k® Gߣøu {ÖÄP.

b+ c a a2
(a) Using Factor theorem, prove that c + a b b 2 = (a + b + c) (a − b) (b − c) (c − a)
a+b c c2

OR
(b) Prove that 32n+2−8n−9 is divisible by 8 for all n/1.

-o0o-

A

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