TN 12th Question Paper 2026 Mathematics

FOR TN 12TH EXAM PREPARATION

TN 12th 2026
Question Paper ·
Mathematics
EXAM YEAR TYPE SUBJECT

TN 12th 2026 Question Paper Mathematics

Notes · Sample Papers · Previous Year Papers · Mock Tests

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No. of Printed Pages : 12
a
9012
£vÄ Gs
M Register Number

!9012Mathematics!
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c. Pou®
PART - III

m .co
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/ MATHEMATICS

s l a
g la uªÌ ©ØÖ® B[Q» ÁÈ ag
a ( / 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®.
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(2) }»® AÀ»x P¸¨¦ ø©°øÚ ©mk÷© GÊxÁuØS®,
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AiU÷PõikÁuØS® £¯ß£kzu ÷Ásk®. £h[PÒ ÁøµÁuØS
ö£ß]À £¯ß£kzuÄ®.
g la
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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 &
m
I / PART - I

m .co
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SÔ¨¦ AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®.
em
: (i) 20x1=20

em l as
s
(ii) öPõkUP¨£mkÒÍ |õßS ©õØÖ ÂøhPÎÀ ªPÄ® Hئøh¯

g la Âøhø¯z ÷uº¢öukzxU SÔ±mkhß Âøh°øÚ²® ÷\ºzx ag
a 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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9012 2

1. AGßÓ 3 §a]¯©ØÓU ÷PõøÁ AoUS
3× AA
T
= A A ©ØÖ® B = A
T −1
A
T
GßÓÁõÖ
C¸¨¤ß, = T
BB

(A) I
3
(B) (C) A B
T
(D) B

T T −1 T T
If A is a 3 × 3 non-singular matrix such that AA = A A and B = A A , then BB =

T
(a) I (b) A (c) B (d) B
3

2. ρ(A) = ρ([A ? B]) GÛÀ, AX = B GßÓ ÷|›¯a \©ß£õkPÎß öuõS¨£õÚx :
(A) J¸[Pø©Äøh¯x ©ØÖ® GsnØÓ wºÄPÒ ö£ØÔ¸US®.
(B) J¸[Pø©Äøh¯x ©ØÖ® J÷µ J¸ wºÄ ö£ØÔ¸US®.
(C) J¸[Pø©ÁØÓx.
(D) J¸[Pø©Äøh¯x.
If ρ(A) = ρ([A ? B]), then the system AX = B of linear equations is :
(a) consistent and has infinitely many solutions

(b) consistent and has a unique solution

(c) inconsistent

(d) consistent

1
3. z GßÓ P»¨ö£snõÚx z ∈ C \ R BPÄ® z + ∈ R GÚÄ® C¸¢uõÀ, z &ß ©v¨¦ :
z

(A) 2 (B) 0 (C) 3 (D) 1

1
If z is a complex number such that z ∈ C \ R and z + ∈ R , then z is :
z

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

3

 π π 4
4.  cos + i sin  &ß GÀ»õ |õßS ©v¨¦PÎß ö£¸USz öuõøP :
 3 3 

(A) 1 (B) −2 (C) 2 (D) −1

3

 π π 4
The product of all four values of  cos + i sin  is :
 3 3 

(a) 1 (b) −2 (c) 2 (d) −1

5. f ©ØÖ® g Gß£Ú •øÓ÷¯ ©ØÖ® m n £i²ÒÍ £À¾Ö¨¦U÷PõøÁPÒ ©ØÖ®
h(x)=(fog)(x) GÛÀ, &ß £i¯õÚx :h

(A) m
n
(B) mn (C) n
m
(D) m+n

If f and g are polynomials of degrees m and n respectively and if h(x)=(fog)(x), then the

degree of h is :
n m
(a) m (b) mn (c) n (d) m+n

M

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−1  1 
6. x<0 GÛÀ, tan   &ß ©v¨¦ :
 x 

(A) −π +cot
−1
(x) (B) tan
−1
(x)

(C) −π + tan
−1
x (D) cot
−1
(x)

m
.co
−1  1 
If x < 0, then tan 
m
 is equal to :

.co
 x 

e m
s
−1 −1

em
(a) −π + cot (x) (b) tan (x)

(c)

s
−π + tan
−1
x (d) cot
−1
(x)
l a
g laø©¯zöuõø»zuPÄ : ag
7.
a
Ámhzvß
1
(A) 2
(B) 0 (C) 2 (D) 1

The eccentricity of the circle is :

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

mAø©¢xÒÍx GÛÀ,
→ → →
8. β ©ØÖ® γ BQ¯øÁ Aø©US® uÍzvÀ
.c
α
o
 

em  
→ → → → → →
(A) (B)
 α , β , γ  =0

as
l(D)
 α , β , γ  =1

(C) 
→ → → 
 α , β , γ =2 ag 
→ → → 
 α , β , γ  =−1

→ → →
If a vector α lies in the plane β and γ , then

   
→ → → → → →
(a) (b)
 α , β , γ  =0  α , β , γ  =1

   
→ → → → → →
(c) (d)
 α , β , γ =2  α , β , γ  =−1

o m
m c
GßÓ. ¦Òΰß
c. o ⋅
→ ∧ ∧ ∧
9. r GßÓ uÍzøu¨ ö£õÖzx
 i + 2 j + 4 k  = 38

e m A(1, 2, 3)

em ¤®£¨¦ÒÎ,
 

l as
las g
GÛÀ, GßÓ ¦Òΰ¼¸¢x öPõkUP¨£mh
A'(3, 6, 11)

a
uÍzvØS A

ag Áøµ¯¨£k® ö\[Szvß Ai
(A) (B) (2, 5, 7)(C) (D) (2, 3, 7)
:

(2, −4, 7) (2, 4, 7)

→ ∧ ∧ ∧
If the image of the point A(1, 2, 3) with respect to the plane r ⋅  i + 2 j + 4 k  = 38 is
 

A'(3, 6, 11), then the foot of the perpendicular from the point A to the given plane is :

(a) (2, 5, 7) (b) (2, 3, 7) (c) (2, −4, 7) (d) (2, 4, 7)

M [ v¸¨¦P / Turn over

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9012 4

10. GßÓ ¦ÒÎUS®
(6, 0) GßÓ ÁøÍÁøµ «xÒÍ ¦ÒÎUS® EÒÍ 2
x − y =4
2

öuõø»Ä SøÓ¢u£m\® GÛÀ, A¨¦ÒÎ :
(A) ( 3, 5 ) (B) (2, 0) (C) ( 13 , − 3 ) (D) ( 5, 1 )
2 2
One of the closest points on the curve x −y =4 to the point (6, 0) is :

(a) ( 3, 5 ) (b) (2, 0) (c) ( 13 , − 3 ) (d) ( 5, 1 )
11.
3
f(x)=x −3x , x
2
∈ [0, 3] GßÓ \õº¤ØS ÷µõ¼ß ÷uØÓzøu {øÓÄ ö\´²® ‘c’ &°ß
©v¨¦ :

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

3 2
The value of ‘c’ satisfied by the Rolle’s theorem for the function f(x)=x −3x , x ∈ [0, 3] is :
3
(a) (b) 1 (c) 2 (d) 2
2

12. 31&ß B® £i ‰» \uÃu¨ ¤øÇ ÷uõµõ¯©õP,
5 31 &ß \uÃu¨ ¤øÇø¯¨ ÷£õÀ
GzuøÚ ©h[PõS® ?
1 1
(A) 5 (B) 31
(C) 31 (D) 5

The percentage error of fifth root of 31 is approximately how many times the percentage

error in 31 ?

1 1
(a) 5 (b) (c) 31 (d)
31 5

2 2
13. A = {(x, y) * a<x<b, c<y<d} ⊂ R GßP. \õº¦ u : A → R BÚx A &À ^µõÚx GÛÀ,
2 2 2 2
∂ u ∂ u ∂ u ∂ u
(A) ∂ 2
+
2
=0 ∀ ( x, y )  A (B) ∂ 2
+
2
=1 ∀ ( x, y )  A
x ∂y x ∂y

2 2 2 2
∂ u ∂ u ∂ u ∂ u
(C) ∂ 2
−
2
=0 ∀ ( x, y )  A (D) 2
−
2
=1 ∀ ( x, y )  A
x ∂y ∂x ∂y
2 2
Let A = {(x, y) * a < x < b, c < y < d} ⊂ R . If the function u : A → R is harmonic in A, then :

2 2 2 2
∂ u ∂ u ∂ u ∂ u
(a) 2
+
2
=0 ∀ ( x, y )  A (b) 2
+
2
=1 ∀ ( x, y )  A
∂x ∂y ∂x ∂y

2 2 2 2
∂ u ∂ u ∂ u ∂ u
(c) 2
−
2
=0 ∀ ( x, y )  A (d) 2
−
2
=1 ∀ ( x, y )  A
∂x ∂y ∂x ∂y

M

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x
df
14. f ( x ) = ∫ t cos t dt , GÛÀ =
dx
0

(A) x cos x (B) cos x−x sin x (C) x sin x (D) sin x+x cos x

m
.co
x
df
If f ( x ) = ∫ t cos t dt , then
m
.co
=

0
dx

e m
em l as
s
(a) x cos x (b) cos x−x sin x (c) x sin x (d) sin x+x cos x

g la ag
a x
e
sinu 3
e
sinx
2

1
15. f ( )
x =
∫ > ©ØÖ® du , x 1
∫ dx =  f

( a ) − f ( 1)  GÛÀ, a ö£ÓUTi¯
u x 2
1 1

J¸ ©v¨¦ :
(A) 9 (B) 3 (C) 5 (D) 6

2
x sinu 3 sinx
e e 1
If f (x) = ∫ du , x > 1 and
∫ dx =  f

( a ) − f ( 1)  , then one of the possible value
u x 2
m
.co
1 1

of a is :

(a) 9 (b)

s em 3 (c) 5 (d) 6

g la
16. 2x
dy
− y= 3
a
GÝ® ÁøPUöPÊa \©ß£õmiß wºÄ SÔ¨¤kÁx :
dx

(A) £µÁøͯ® (B) ÷|ºU÷PõkPÒ
(C) }ÒÁmh® (D) Ámh[PÒ
dy
The solution of the differential equation 2 x − y = 3 represents :
dx

m
.co
(a) Parabola (b) Straight lines

m
.co
(c) Ellipse (d) Circles

e m
em GÝ® ÷|µzvØS¨ ¤ÓS «u•ÒÍ J¸ ö£õ¸Îß AÍÄ l s
a BS®. ö£õ¸Ò
las 17. t

g
a AÍÂØS ÂQu©õP
P

ag B¯õS® Ãu©õÚx A¢÷|µzvÀ «uª¸US® ö£õ¸Îß
Aø©¢xÒÍx GÛÀ, ¤ßÚº :
kt
(A) P=Ckt (B) P=Ce (C) Pt=C (D) P=Ce
−kt

P is the amount of certain substance left in after time t. If the rate of evaporation of the

substance is proportional to the amount remaining, then :

kt −kt
(a) P=Ckt (b) P=Ce (c) Pt=C (d) P=Ce

M [ v¸¨¦P / Turn over

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9012 6

1
18. f (x) = , a<x<b GÝ® \õº¦ J¸ öuõhºa]¯õÚ \©Áõ´¨¦ ©õÔ X &ß {PÌuPÄ
12

Ahºzv \õº¤øÚU SÔUQÓx GÛÀ, ¤ßÁ¸ÁÚÁØÖÒ Gx a ©ØÖ® b &Cß
©v¨¦PÍõP Cµõx ?
(A) 7 ©ØÖ® 19 (B) ©ØÖ®0 12

(C) 16 ©ØÖ® 24 (D) 5 ©ØÖ® 17

1
If the function f (x) = for a<x<b, represents a probability density function of a continuous
12

random variable X, then which of the following cannot be the value of a and b ?

(a) 7 and 19 (b) 0 and 12

(c) 16 and 24 (d) 5 and 17

19. 2l }Í•ÒÍ J¸ P®¤ \©Áõ´¨¦ •øÓ°À C¸ xshõP EøhUP¨£kQÓx. C¸
1
 0 < x < l
xskPÎÀ Smøh¯õÚuØPõÚ {PÌuPÄ Ahºzv \õº¦ f (x) = l

0 l ≤ x < 2l


GÛÀ, Smøh¯õÚ¨ £SvUPõÚ \µõ\› ©ØÖ® £µÁØ£i •øÓ÷¯ :
2 2
2 2
l l l l l l l
(A) l, (B) , (C) ,
(D) ,
12 2 3 2 12 2 6

A rod of length 2l is broken into two pieces at random. The probability density function of

1
 0 < x < l
the shorter of the two pieces is f (x) = l . The mean and variance of the shorter

0 l ≤ x < 2l


of the two pieces are respectively :

2 2
2 2
l l l l l l l
(a) l, (b) , (c) (d)
, ,
12 2 3 2 12 2 6

20. ¬(p∨q)∨[p∨(p∧¬r)] &ß C¸©® :
(A) ¬ (p ∧q)∧[p∧(p∧r)] (B) ¬ (p ∧q)∧[p∨(p∧¬r)]
(C) ¬ (p ∧q)∧[p∧(p∨¬r)] (D) (p ∧q)∧[p∧(p∨¬r)]
The dual of ¬(p∨q)∨[p∨(p∧¬r)] is :
(a) ¬(p∧q)∧[p∧(p∧r)] (b) ¬(p∧q)∧[p∨(p∧¬r)]
(c) ¬(p∧q)∧[p∧(p∨¬r)] (d) (p ∧q)∧[p∧(p∨¬r)]

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£Sv & II / PART - II

SÔ¨¦ : GøÁ÷¯Ý® HÊ ÂÚõUPÐUS Âøh¯ÎUPÄ®. ÂÚõ Gs 30 &US
Pmhõ¯©õP Âøh¯ÎUPÄ®. 7x2=14

m
.co
Note : Answer any seven questions. Question No. 30 is compulsory.

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21. AGߣx JØøÓ Á›ø\²øh¯ §a]¯©ØÓ ÷PõøÁ Ao GÛÀ,
m .co adj A Gߣx
s e m
ªøP Gs GÚ {ÖÄP.
s e l a
g l a ag
a
If A is a non-singular matrix of odd order, prove that adj A is positive.

12
n
22. _¸USP : ∑ i

n=1

12
n
Simplify : ∑ i

n=1

o m
c
. \©® GÛÀ, &ß ©v¨¦ PõsP.
m
2
GÝ® \©ß£õmiß ‰»[PÒ
e
23. x +2(k+2)x+9k=0 k

2
If x +2(k+2)x+9k=0 has equal roots, find k.

las
ag
24. _¸USP : sin
−1
[sin10]

−1
Simplify : sin [sin10]

25. Bµ® ö\.« Eøh¯x®, Aaø\ Bv¨¦ÒΰÀ öuõmka ö\ÀÁx©õÚ
5 x -

Ámh[PÎß \©ß£õmøhz u¸ÂUP.
Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in

m
general form.

m c o
.GÚ {ÖÄP.
.co e m
2
26. f(x)=x −2x−3 GßÓ \õº¦ (2, ∞) GßÓ CøhöÁΰÀ vmh©õP HÖ®
em
2
Prove that the function f(x)=x −2x−3 is strictly increasing in the interval (2, ∞).

l as
las ag
ag −1
 x  x
27. f ( x, y ) = cos   GÛÀ, f
y
= GÚ {ÖÄP.
y
 
2 2
y y −x

 x  x
−1
If f ( x, y ) = cos   , then show that f
y
= .
y
 
2 2
y y −x

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9012 8

π

2
10
28. ©v¨¤kP : ∫ sin x dx

0

π

2
10
Evaluate :
∫ sin x dx

0

1

 
2 2 2

2
d y  dy 
+ 1 +  =0 GßÓ ÁøPUöPÊ \©ß£õmiß Á›ø\ ©ØÖ® £iø¯
29.  
x
2
dx   dx  
 

(C¸US©õÚõÀ) wº©õÛUP.
Determine the order and degree (if exists) of the differential equation

1

 
2 2 2

2
d y  dy 
+ 1 +  =0
 
x
2
dx   dx  
 

1
30. Var (X) = GÛÀ, Var (2X+3) &ß ©v¨¦ PõsP.
2

1
If Var (X) = then, find the value of Var (2X+3).
2

£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.

 2 −2 4 3 
 
31.

−3 4 −2 −1

GßÓ Aoø¯ HÖ£i ÁiÂÀ ©õØÔ Aozuµ® PõsP.
 6 2 −1 7 
 

 2 −2 4 3 
 
Find the rank of the matrix −3 4 −2 −1 by reducing it to an echelon form.
 
 6 2 −1 7 
 

x+2 5
32. 4
x
−3(2GÝ® )+2 =0 \©ß£õmøh {øÓÄ ö\´²® AøÚzx
ö©´ö¯sPøͲ® PõsP.
x x+2 5
Find all real numbers satisfying the equation : 4 −3(2 )+2 =0.

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−1 1 −1 1 −1 31
33. {ÖÄP : 2 tan + tan = tan
2 7 17

−1 1 −1 1 −1 31
Prove that 2 tan + tan = tan
2 7 17

m
34. ©ØÖ®
(6, 7, 4) GßÓ ¦ÒÎPÒ ÁȯõPa ö\À¾® ÷|ºU÷Põk
(8, 4, 9)
m xz ©ØÖ® yz
.co
uÍ[PøÍ öÁmk® ¦ÒÎPøÍU PõsP.
m .co s e m
e
Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz

s l a
planes.

g l a ag
a
 1 − cos m θ 
35. lim   =1 GÛÀ, m=±n GÚ {ÖÄP.
θ →0  1 − cos n θ 

 1 − cos m θ 
If lim   = 1 , then prove that m=±n
θ →0  1 − cos n θ 

36. ÷|›¯À ÷uõµõ¯ ©v¨¥mk •øÓ°À 4
15 &ß ÷uõµõ¯ ©v¨¤øÚU PõsP.
m
.co
4
Use the linear approximation to find approximate value of 15

s em
∞

GÛÀ, ∫
∞
n−1

g la&ß ©v¨¤øÚU PõsP.
a
n

∫e
−x −x
37. x dx = 5! e x dx

0 0

∞ ∞
n−1

∫e
n

∫e
−x −x
If x dx = 5! , then find the value of x dx

0 0

38. ©ØÖ®
4P (X=4)=P(X=2) GÝ®£i EÒÍ n=6 X~B (n, p) &ß £µÁÀ, \µõ\› ©ØÖ®
vmh »UP® BQ¯ÁØøÓU PõsP.
m
m .co
If X~B (n, p) such that 4P (X=4)=P(X=2) and n=6, find the distribution, mean and Standard

.co m
Deviation of X.

s e
s em öPõkUP¨£mh Pnzvß «x ¤ßÁ¸® ö\¯»õÚx gla AøhĨ £s¦
39. (i)

g la ÷\º¨¦¨ £s¦ ©ØÖ®
(ii) a ö£ØÔ¸US©õ GÚa
\©Û¨ £s¦ BQ¯øÁPøͨ (iii)

a \›£õºUP.
m *n=m+n−mn; m, n ∈ Z
Verify (i) Closure property (ii) Associative property and (iii) Existence of identity for the

following operation on the given set :

m *n=m+n−mn; m, n ∈ Z

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9012 10

2026
 z 
1
40. z
1
= 1+i ©ØÖ® z
2
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 
z
2

2026
 z 
1
If z1 = 1 + i and z2 = 1 − i , find the inverse of  
 
z
2

£Sv & IV / PART - IV

SÔ¨¦ : AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®. 7x5=35

Note : Answer all the questions.

41. (A) ¤ßÁ¸® ÷|›¯a \©ß£õmkz öuõS¨ø£ ÷|º©õÖ Ao PõnÀ •øÓø¯
£¯ß£kzv wºUP :
2x +3x +3x =5
1 2 3

x −2x +x =−4
1 2 3

3x −x −2x =3
1 2 3

AÀ»x
3
(B) z + 2z = 0 GßÓ \©ß£õmiØS I¢x wºÄPÒ C¸US® GÚ {ÖÄP.
(a) Solve the following system of equations, using matrix inversion method.

2x +3x +3x =5
1 2 3

x −2x +x =−4
1 2 3

3x −x −2x =3
1 2 3

OR

3
(b) Show that the equation z + 2z = 0 has five solutions.

42. (A) p →(¬ q ∨ r ) ≡¬ p ∨ ( ¬ q ∨ r ) Gߣøu ö©´ø© AmhÁønø¯¨ £¯ß£kzv
{ÖÄP.
AÀ»x
π

4
π
(B) ∫ log ( 1 + tan x ) dx = log2 GÚ {ÖÄP.
8
0

(a) Prove that p →(¬q∨r)≡¬p∨(¬q∨r) using truth table.
OR

π

4
π
(b) Prove that
∫ log ( 1 + tan x ) dx = log2
8
0

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m .co s e m
se g l a
11
a 9012

43. (A) \©ß£õmøh wºUP : (x+1)(x+3)(x−2)(x−4)+21=0

AÀ»x
(B) GßÓ ÁøÍÁøµø¯ ÁøµP.
y=log(1+x)

(a) Solve the equation

(x+1)(x+3)(x−2)(x−4)+21=0
m
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OR

(b)

c o
Sketch the curve y=log(1+x).

. ÷|ºU÷Põk e m
44. (A) x−y+4=0
e m
GßÓ GßÓ }ÒÁmhzvß öuõk÷Põk GÚ
2
x +3y =12
2

l as
l as
{ÖÄP. ÷©¾® öuõk® ¦ÒÎø¯U PõsP.
ag
ag AÀ»x
(B) J¸ ©õv›°À Põn¨£k® Pv›¯UP AqUP¸UPÒ ]øuÄÖ® Ãu©õÚx
A¢÷|µzvÀ A¢u ©õv›°À Põn¨£k® AqUP¸UPÎß GsoUøPUS
ÂQu©õP Aø©¢xÒÍx. Bsk Põ» CøhöÁΰÀ J¸ ©õv›°À100

Bµ®£zvÀ Põn¨£k® Pv›¯UP AqUP¸UPÎß GsoUøP°À 10%

]øuÄÖQÓx. BskPÒ •iÂÀ Bµ®£zvÀ Põn¨£k® Pv›¯UP
1000

AqUP¸UPÎß GsoUøP°À GÆÁÍÄ «uª¸US® ?

om
2 2
(a) Show that the line x−y+4=0 is a tangent to the ellipse x +3y =12. Also find the

co-ordinates of the point of contact.

. c
e
OR
m
(b)

la
of such nuclei that are present in a given sample.
s
Assume that the rate at which radioactive nuclei decay is proportional to the number

In a certain sample 10% of the

ag
original number of radioactive nuclei have undergone disintegration in a period of

100 years. What percentage of the original radioactive nuclei will remain after

1000 years ?

45. (A) öÁUhº •øÓ°À, cos(α+β)=cosα cosβ−sinα sinβ GÚ {ÖÄP.
AÀ»x
(B) 2 2
(x +y )dy=xydx. y(1)=1 ©ØÖ® GÚU öPõkUP¨£mkÒÍx.
y(x )=e
0
x
0
&ß
©v¨ø£U PõsP.
(a) By Vector method prove that :

m
m .co
cos(α+β)=cosα cosβ−sinα sinβ

.co m
OR

m (A) J¸ uÛ{ø» \õº¦ &ß {PÌuPÄ {øÓ \õº£õÚx : lase
2 2
(b) (x +y )dy=xydx. It is given that y(1)=1 and y(x )=e. Find the value of x .

e
0 0

l as 46.

ag X

ag f(x)
x 1

k
2

2k 6k
3 4

5k
5

6k
6

10k

GÛÀ, (i) P(2<X<6) (ii) P(2≤X<5)

(iii) P(X≤4) (iv) P(3<X)

GߣÁØøÓU PõsP.
AÀ»x
M [ v¸¨¦P / Turn over

m .
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s em l a
g la ag
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9012 12

(B) J¸ }¸ØÔÀ, Bv°¼¸¢x « Qøh©mhz yµzvÀ }›ß AvP£m\ E¯µ® 0.5

«, }›ß £õøu J¸ £µÁøͯ® GÛÀ, Bv°¼¸¢x
4 « Qøh©mhz 0.75

yµzvÀ }›ß E¯µzøuU PõsP.
(a) A random variable X has the following probability mass function.

x 1 2 3 4 5 6

f(x) k 2k 6k 5k 6k 10k

Find (i) P(2<X<6) (ii) P(2≤X<5)

(iii) P(X≤4) (iv) P(3<X)

OR

(b) At a water fountain, water attains a maximum height of 4 m at horizontal distance of

0.5 m from its origin. If the path of water is a parabola, find the height of water at a

horizontal distance of 0.75 m from the point of origin.

47. (A) 3
s(t)=2t −9t +12t−4,
2
C[S t/0 GÝ® Âv¨£i J¸ ÷PõmiÀ J¸ xPÒ
|PºQÓx.
G¢÷|µ[PÎÀ xPÎß vø\ ©õÖQÓx ?
(i)

•uÀ ÂÚõiPÎÀ xPÒ £¯ozu ö©õzu yµzøuU PõsP.
(ii) 4

vø\÷ÁP® §a]¯ ©v¨ø£ Aøh²® ÷|µ[PÎÀ GÀ»õ® xPÎß
(iii)

•kUP® PõsP.
AÀ»x
(B) (1, GßÓ ¦ÒÎ ÁÈa ö\ÀÁx®
−2, 4) GßÓ uÍzvØS x+2y−3z=11

x +7 y +3 z
ö\[SzuõPÄ® = = GßÓ ÷PõmiØS Cøn¯õPÄ® Aø©²®
3 −1 1

uÍzvß xøn¯»S AÀ»õu öÁUhº \©ß£õk ©ØÖ® Põºj]¯ß
\©ß£õkPøÍU TÖP.
3 2
(a) A particle moves along a line according to the law s(t)=2t −9t +12t−4, where t/0.

(i) At what times the particle changes direction ?

(ii) Find the total distance travelled by the particle in the first 4 seconds.

(iii) Find the particle’s acceleration each time the velocity is zero.

OR

(b) Find the non-parametric form of Vector equation and Cartesian equation of the plane

passing through the point (1, −2, 4) and perpendicular to the plane x+2y−3z=11

x +7 y +3 z
and parallel to the line = =
3 −1 1

- o O o -

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