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Mathematics key –March 2017 Part A
Type A
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Option (2) (1)
Answer of order 1 and degree 2
Q. No 21 22
Option (2) (1)
2 -t2
Answer
23
(1)
9
3
(2)
1 ,8 2 ( , ]
4
(4)
b:a
24
(3)
5
(2)
An asymptote parallel to y-axis
25
(2)
6
(3)
26
(1)
9
7
(4)
wn k 3 r (i 3 j 5k ) t (i 5 j k ) 2
27
(1)
The straight line x =
8
(1)
Gradient of the velocity/time graph
28
(2)
9
(4)
5
29
(2)
10
(3)
(i),(iii),(iv)
30
(4)
11 12
(1) (4)
(6,6) Two rows can have same number of zeroes before the first non zero entry
31 32
(2) (1)
13
(2)
x ce my
33
(3)
14
(4)
34
(2)
*
15 16 17 18 19
(4) (3) (2) (4) (4)
1 5 2 60
35 36 37 38 39
(1) (2) (1) (1) (2)
[ ⃗ ⃗⃗⃗⃗ ⃗⃗ ⃗] cos x K3 det(A) (0,0,-4)
20
(3)
40
(4)
[ ]
2 1/3 1/2 (25,1/5)
f '' ( x0 ) 0
f ( x) 0
( 0
dy y dx
(
+
)
)
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[
]
[
]
[
]
( ) 42
| | [
]=[
]
Adj A = [ -1
A = 43
]
[
]
Equation of the line is Equation of the xy plane is z=0 X = 0, y = Point of Intersection is ( 0, , 0) ⃗⃗ (or) 0 ( ⃗ ⃗) ( ⃗⃗ ⃗)
44(i) (⃗ 44(ii)
⃗) and ( ⃗⃗
⃗) are parallel
Vector form of the line
r (2i 3 j k ) t (i 4 j 3k )
Direction cosines are
(
)
45
46
AB = BC = CA = ABC is a right angles triangle 47
(
) (
)
When t = 2, velocity = 0 Acceleration = -12 K.E = 0 48 (i)
( )
(
)
Critical numbers are 0, 48(ii)
The curve is concave upward everywhere (or) The curve is convex downward everywhere. 49 Maximum error in area Relative error in A
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The characteristic equation is (
CF = P.I = P.I =
)
(
51
)
p T T F F
[ 52
(
(
q T F T F
)
( T T F T
F T F T
(
p T T T T F F F F
)
54
)] is a tautology q T T F F T T F F
(
r T F T F T F T F
T T F F F F F F
F T F T F T F T
P(0 < Z < C) = 0.45 C = -1.65 p=1/6 q=5/6 n = 36 (
)
(
)
(
)
)]
T T T T
53(i)
53(ii)
(
[
)
Mean =1 ( ) 55(a) (or)
The equation of hyperbola is
(
)
(
55(b)
Volume
∫
(
)
cubic units
56
All minors of , , , y = s; z = t X = 4 – s – 2t (x, y, z) = (4 – s – 2t, s, t); s, t
are zero
R
)
) ( T T F T F T F T
)
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⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗ ( 58
⃗ ⃗⃗
⃗ ⃗ )
(
) a 3i 4 j 2k b 2i 2 j k c 7i k Vector form r (1 s t )(3i 4 j 2k ) s(2i 2 j k ) t (7i k ) r (3i 4 j 2k ) s(i 6 j 3k ) t (4i 4 j k ) Cartesian Form |
|
59 Solving
remove the root x =-1
( [
) (
)
(
)
( (
The values are c is , c is The roots are c is , c is
)] ) , c is , c is , c is
, c is
, c is
,
,
60
2
X = -4ay 2
X = -9y
At (-6, -4) Slope m = ( ) 61
a = 10 , b=6
R is (6, y1) y1 = 4.8 PR = 12 + 4.8 = 16.8 62
The other asymptote is 2x – y + k =0 Equation of R.H is ( x + 2y - 5) (2x – y + k) + c = 0 k + c = -12 -8k + c = -48 k = 4 ; c= - 16 Equation of R.H is ( x + 2y - 5) (2x – y + 4) - 16 = 0
(or)
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√( ( )
) (
(
) (
)
)
( ) y=2 When y < 2, ( ) and ( ) d is minimum when y = 2 The required point is ( 2, 2)
when y > 2
64
65
√ Area of Ellipse = ∫ sq.units 66
(
(
)
)
(
√( )
dx
)
( )
( ) =
( ) ( )
The required length = ∫ = 4a 67
(or) T = S + C ekt T = 15 + C ekt When t = 0, T = 100 ; T = 15 + 85 ekt When t = 5, T = 60, e
(
)
C = 85 5k
=
When t = 10, T =
68
. I A B C D E (i) (ii) (iii) (iv)
I A B C D E I A B C D E A B I E C D B I A D E C C D E I A B D E C B I A E C D A B I Closure Axiom : Closure axiom is true Associative Axiom: Multiplication is associative Identity Axiom: I is the identity element in G Inverse Axiom : I is the inverse of I, A and B are inverses of each other, C is the inverse of C, D is the inverse of D, E is the inverse of E.
(G,.) is a group
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( )
∫
( )
70(a)
∫
=1
∫
Slope of tangent
The points are (4, 6) and (-4, -6) The equations of tangents are 2x + 3y + 26 = 0 2x + 3y - 26 = 0 70(b)
Put x + y = z
∫(
)
∫
( ) (
)
R.DHAKSHINA MOORTHY M.Sc.,M.Ed.,M.Phil., P.G.Asst (Maths) BHARATHIYAR VIDYALAYAM H.S.S THOOTHUKUDI www.nammakalvi.weebly.com
