12TH STD - MATHEMATICS
HALF YEARLY EXAMINATION β 2017
ANSWER KEY (18 - 12- 17)
ONE MARK QUESTIONS : 40 X 1 = 40
1 b) 2
2 d) 1( )
nk a d j I
3 d) no solution
4 d) has only trivial solution only if rank of the coefficient matrix is equal to the number of unknowns
5 d) 2
6 d) 3 3 0
7 b) 2
8 b) 9 1 1 0x y z
9 c) skew
10 d) .
s inb n
b n
11 d) collinear
12 c) 2 s in ( )i
13 d) (1,0)
14 * c) 1 2 1 2
a rg ( ) a rg ( ) a rg ( )z z z z (m.a)
15 d) 9 0o
16 a)
2 2
11 4 4 4 3 2
x y
17 c) 3
, 2 48
18 b) 2 2 2
x y a
19 b) 1
, / m in5 0
c m
20 a) 2
8 / seccm
21 d) everywhere concave downward
22 a) vertical tangent 1
x x
23 b) 1
1 1
24 d) , ,x y
u u u are continuous
25 a) 0
26 d) 9
27 a) 2 0
28 b) (2 ) ( )f a x f x
29 b) lo g x
30 c) '' '
2 2 0y y y
31 c) 2
( 2 )3
x x
32 c) 2, 1
33 c) ( )p q
34 b) 3
35 b) 1
36 d) {0}C
37 a) all
38 d) 8
39 b) 5
40 c) 80
SIX MARK QUESTIONS : 10 X 6 = 60
41 π΄ = 1 21 1
= 1 β 2 = β1 β 0 A is invertible.
π΅ = 0 β11 2
= 1 β 0 B is invertible.
AB = 2 31 1
π΄π΅ = β1 β 0 AB is invertible.
adj A = 1 β2β1 1
; π΄β1 = β1 21 β 1
adj B = 2 1β1 0
; π΅β1 = 2 1
β1 0
adj(AB) = 1 β3β1 2
; π΄π΅ β1 =1
π΄π΅
πππ(π΄π΅) = β1 3 1 β2
β βββ (1)
π΅β1 π΄β1 = β1 3 1 β2
β β β β (2)
From (1) and (2) ,
π΄π΅ β1 = π΅β1π΄β1 42 The corresponding matrix equation is
1 1 11 2 30 1 2
π₯π¦π§ =
7186
Augmented matrix is π΄, π΅ = 1 1 1 71 2 3 180 1 2 6
~ 1 1 1 70 1 2 110 1 2 6
π 2 β π 2 β π 1
~ 1 1 1 70 1 2 110 0 0 β 5
π 3 β π 3 β π 2
It is in echelon form and it has three non-zero rows.
π π΄, π΅ = 3 and π π΄ = 2.
π π΄, π΅ β π π΄ 43 Cartesian form :
π₯β2
2=
π¦β1
β1=
π§+3
β1= π(π ππ¦)
Any point : 2π + 2, βπ + 1, βπ β 3
This point lie on the plane π₯ β 2π¦ + 3π§ + 7 = 0
2π + 2 β 2 βπ + 1 + 3 βπ β 3 + 7 = 0
2π + 2 + 2π β 2 β 3π β 9 + 7 = 0 β π = 2
β΄ The required point is (6 , β1 , β5) 44 (i)
(ii)
π₯ β π = 0 βΉ π₯ β₯ π , π₯ β π = 0 βΉ π₯ β₯ π
π₯ is β₯ to both π πππ π
π₯ is to π Γ π βΉ π₯ = π(π Γ π )
π₯ β π = 0
βΉ π π Γ π β π = 0
βΉ π , π , π = 0
π , π , π are coplanar.
The angle between the line and plane is π πππ =π βπ
π π
π β π = 6 β 2 + 12 = 16
π = 3; π = 7
π = π ππβ1 π β π
π π = π ππβ1
16
21
45 Let π§1 = π1 πππ π1 + ππ πππ1
π§2 = π2 πππ π2 + ππ πππ2 . π§1 = π1, πππ π§1 = π1;
π§2 = π2, πππ π§2 = π2
1 1
1 2 1 2
2 2
[c o s ( ) s in ( )]z r
iz r
1 1
2 2
z r
z r 1
1 2 1 2
2
a rg a rg a rgz
z zz
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46 AB= β21 + 23π = 970
BC= 9 β 27π = 810
CA = 12 + 4π = 144 + 16 = 160
π΄π΅2 = π΅πΆ2 + πΆπ΄2 = 970 Hence the given points represent
a right angled triangle on the Argand plane.
47 (i)
(ii)
π¦ =1
π₯ ππ π₯ β β, π¦ β 0
limπ₯β+β
π ππ2π₯
1π₯
= limπ¦β0
sin 2π¦
π¦
0
0ππππ
= limπ¦β0
2 cos2 π¦
1 = 2 1 = 2
Continuous on [0,2]
Not differential on (0,2)
Rolls theorem cannot applied to the given function. 48 ' 2 2
' '
( ) s e c c o s
0 , 02 6
f x x e c x
f f
f is not monotonic on 0 ,2
49 Let π¦ = π π₯ =π₯1
2
ππ¦ = π β² π₯ ππ₯ =1
2 (π₯)
β12 ππ₯
Let π₯ = 36 , ππ₯ = βπ₯ = 0.1, π 36 = 6
ππ¦ =1
2 (36)
β12 Γ 0.1 =
0.1
12= 0.0083
π π₯ + βπ₯ β π¦ + ππ¦
β΄ 36.1 β π 36 + 0.008 = 6.008
36.1 β 6.01 (two decimal) 50 . ( )
xC F A x B e
2
1. 7 4P I x x ,
2. 2P I x ,
3. 1P I
General Solution : 2
. . ( ) 3 6x
y C F P I A x B e x x
51
The last columns are identical.
β΄ ~(π β§ π) β‘ (~π)(~π)
p q p β§ q ~(π β§ π)
T T T F
T F F T
F T F T
F F F T
p q ~π ~π (~π)(~π)
T T F F F
T F F T T
F T T F T
F F T T T
52 STATEMENT:Let G be a group. β π, π,c β G, Then
(i) π*π = π*π β π = π (left cancellation law)
(ii) π*π = π*π β π = π (Right cancellation law)
Proof : (i) π*π = π β π β πβ1 β π β π = πβ1 β (π β π)
β (πβ1 β π) β π = (πβ1 β π) β π β π β π = π β π β π = π (ii)π β π = π β π
β π β π β πβ1 = π β π β πβ1 β π β (π β πβ1) = π β (π β πβ1) β π β π = π β π β π = π
53 π π < 40 = 0.1 ; π π > 90 = 0.1
β΄π 40 < π < 90
= 1 β [ π π < 40 + π π > 90 ] = 1 β 0.1 β 0.1 = 0.8 Out of 800 Students, number of
students scored between 40 and 90 is
800 Γ 0.8 = 640 students 54 (i)
(ii)
p = 9/10, q= 1/10 Mean = 450, variance = 45
S.D = 3 5
p = 2/100, n = 200
4
( 3) 0 .5 6 6 9P X
55 (a) Equation of tangent at any point β²π‘β² ππ π₯π¦ =π2 ππ π₯ + π¦π‘2 = 2ππ‘ π₯
2ππ‘+
π¦
2π π‘ = 1
β΄ Intercept on the axes are π = 2ππ‘ , π =2π
π‘
Equation of normal at Μπ‘β² on π₯π¦ = π2 ππ
π¦ β π₯π‘2 =π
π‘β ππ‘3
π₯
ππ‘βππ‘ 3
βπ‘2
+π¦
π
π‘βππ‘ 3
= 1
β΄ Intercept on axes are
π =β1
π‘2 π
π‘β ππ‘3 ; π =
π
π‘β ππ‘3
β΄ππ + ππ = 2ππ‘ β1
π‘2 π
π‘β ππ‘3 +
2π
π‘
π
π‘β ππ‘3
= β2π
π‘
π
π‘β ππ‘3 +
2π
π‘
π
π‘β ππ‘3 = 0
55 (b)
2
4 2
0
s in (1 s in )x x d x
2 2
4 6
0 0
s in s inx d x x d x
3 2
TEN MARK QUESTIONS : 10 X 10 = 100
56 SOLUTION:
Let π₯, π¦ πππ π§ be the number of coins in each
category Re.1 , Rs.2 and Rs.5 respectively.
β΄ π₯ + π¦ + π§ = 30
π₯ + 2π¦ + 5π§ = 100
Put π§ = π, π β π .
π₯ + π¦ = 30 β π π₯ + 2π¦ = 100 β 5π, π β π
β= 1 11 2
=1
βπ₯=3π β 40; βπ¦=70 β 4π
By Cramerβs rule ,
π₯= βπ₯
β = 3π β 40 ; π¦=
βπ¦
β = 70 β 4π
The Solution set (π, π, π) = ( ππ β ππ, ππ β ππ, π) ,
1π β€ π β€ ππ
possible solutions are (2,14,14) , (5,10,15) , (8,6,16)
and (11,2,17)
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57 π Γ π = π + π β 2π
π Γ π = π β 3π + π
π Γ π Γ π Γ π = β5π β 3π β 4π ββ (1)
π π π = β2
π π π π = β4 π β 2π β 2π
π π π = 1
π π π π = π + π + 2π
π π π π β π π π π
= β5 π β 3π β 4π β ββ (2)
From (1) and (2),
π Γ π Γ π Γ π
= π π π π β π π π π 58 VECTOR FORM:
π = 2π + 2π + π , π’ = 2π + 3π + 3π and
π£ = 3π + 2π + π The equation of the plane is
π = π + π π’ + π‘π£ π = ππ + ππ + π + π ππ + ππ + ππ + π(ππ
+ ππ + π ) CARTESIAN FORM:
π₯1, π¦1, π§1 = 2, 2, 1 , π1, π1, π1 = 2, 3, 3 and π2, π2, π2 = 3, 2, 1
The equation of the plane is
π₯ β π₯1 π¦ β π¦1 π§ β π§1
π1 π1 π1
π2 π2 π2
= 0
π₯ β 2 π¦ β 2 π§ β 1
2 3 3 3 2 1
= 0
ππ β ππ + ππ + π = π 59 The roots of the equation are pΒ±ππ
Let πΌ = π + ππ and π½ = π β ππ ; πΌ β π½ = 2ππ
π¦ = π cot π β π
π¦ + πΌ = π πππ‘ π + π
= π πππ π
π ππ π+ π =
π
π ππ π πππ π + ππ ππ π
(y + Ξ±)π =ππ
sinn π(cos π + isin π)n
(y + Ξ±)π =ππ
sinn π(cos ππ + isin ππ)
Similarly,
(y + Ξ²)π =ππ
sin n π(cos ππ β isin ππ)
(y+ Ξ±)π β(y+ Ξ²)π
Ξ±β Ξ²=
ππ
(2ππ) sin n π(2isin ππ)
(π²+ π)πβ(π²+ π)π
πβ π=
ππβπ πππππ½
π¬π’π§π§ π½
60 Consider the parabola is open downward.
Equation of the parabola is π₯2 = β4ππ¦.
It passes through the point B(3, -2.5)
32 = β4π(β2.5)
π =9
10
π₯2 = β49
10π¦ β β β β β β β 1
The point π₯1, β7.5 lies on 1
π₯12 = β4
9
10 β7.5 = 27 β π₯1 = 3 3
The water strikes the ground π π m beyond
this vertical line.
61 Let F1 be the position of the Sun.
Given e = 0.206 and a = 36 million miles.
The shortest distance
πΉ1π΄ = πΆπ΄ β πΆπΉ1 = π β ππ = π 1 β π = 36 1 β 0.206 = 28.584 million miles
The longest distance πΉ1π΄β² = πΆπΉ1 + πΆπ΄β² = π + ππ = π 1 + π = 36 1 + 0.206
= 36 Γ 1.206
= 43.416 million miles
62 Center of ellipse = centre of hyperbola
Length of LR = 9/2 2
2 9
2
b
a ie., 4a
5
4e 3b
Conjugate axis x =1 Conjugate axis parallel to Y axis Transverse axis is X axis
2 2( 1) ( 1)
11 6 9
x y
2 2 2 2
2 2
2 2
( 1) ( 1) 1 6 9
2 2 5 0
x y a b
x y
x y x y
63 π₯=ππππ 3π, π¦ = ππ ππ3π ππ₯
ππ= β3ππππ 2ππ πππ ;
ππ¦
ππ= 3ππ ππ2ππππ π;
ππ¦
ππ₯= β
3ππ ππ 2ππππ π
3ππππ 2ππ πππ= β
π πππ
πππ π
Slope of the tangent at βπβis βπ πππ
πππ π
Slope of the normal at βπβis πππ π
π πππ
Equation of the normal at βπβis
π¦ β ππ ππ3π = πππ π
π πππ( π₯ β ππππ 3π)
π¦π πππ β ππ ππ4π = π₯πππ π β ππππ 4π
π₯πππ π β π¦π πππ = π πππ 4π β π ππ4π
π₯πππ π β π¦π ππ
= π πππ 2π + π ππ2π πππ 2π β π ππ2π
πππππ½ β πππππ½ = ππππππ½
64 Let π₯ and π¦ be the length and breadth of the printed
area. π₯π¦ = 384 βΉ π¦ =384
π₯
Dimensions of the poster area are
π₯ + 8 and π¦ + 12.
Area of the poster π΄ = π₯ + 8 π¦ + 12
π΄ π₯ = 12π₯ + 8 384
π₯ + 480
π΄β² (π₯) = 12 β 8 384
π₯2
π΄β²β² π₯ = 16 384
π₯3
For max/min, π΄β² π₯ = 0 βΉ π₯ = Β±16
β΄ π₯ = 16 [β΅ π₯ cannot be β π£π]
When π₯ = 16, π΄β²β²(π₯) =3
2> 0
When π₯ = 16, π΄ is minimum
When π₯ = 16, π¦ = 24. Dimensions of the poster are 24 cm and 36 cm.
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65 R.H.S. is not a homogeneous function.
Define π = π‘πππ’ =π₯3+π¦3
π₯βπ¦
f is a homogeneous function of degree 2.
By Eulerβs theorem,
π₯ππ
ππ₯+ π¦
ππ
ππ¦= 2π
π₯π π‘πππ’
ππ₯+ π¦
π π‘πππ’
ππ¦= 2 π‘πππ’
π₯π ππ2π’ππ’
ππ₯+ π¦π ππ2π’
ππ’
ππ¦= 2
π πππ’
πππ π’
πππ
ππ+ π
ππ
ππ= πππππ
66 π₯ = π π‘ + π πππ‘ ; π¦ = π(1 + πππ π‘)
Put π¦ = 0 β 1 + πππ π‘ = 0 β πππ π‘ = β1 β π‘ =
βπ, π ππ₯
ππ‘= π(1 + πππ π‘);
ππ¦
ππ‘= βππ ππt
ππ₯
ππ‘
2+
ππ¦
ππ‘
2= 2ππππ
π‘
2
Surface area = 2ππ¦π
π
ππ₯
ππ‘
2+
ππ¦
ππ‘
2ππ‘
= 2ππ 1 + πππ π‘ 2ππππ π‘
2ππ‘
π
βπ
= 16ππ2 πππ 3π‘
2ππ‘
π
0
= 16ππ2 2πππ 3π’ππ’π
20
Put π‘
2= π’, ππ‘ = 2ππ’
= 32ππ2 2πππ 3π’ππ’π
20
= 32ππ2 2
3 β΅ by reduction formula.
=πππ ππ
π ππππππ πππππ.
67 The point of intersection of 4y2 = 9x and 3x
2 =16y
are (0,0) and (4, 3).
Required area can be solved about x-axis.
Required area = (π¦1 β π¦2)ππ₯4
0
π¦1 means π¦ from 4y2 = 9x
π¦2 means π¦ from 3x2
=16y
R.A.= 3
2 π₯ β
3
16π₯2
4
0ππ₯ = π₯
3
2 βπ₯3
16
0
4
= 8 β 4 = 4 square units.
68 Let T be the temperature of the coffee at any time
t.
By Newtonβs law of cooling, ππ
ππ‘ β π β 15 since S =15Β°πΆ β
ππ
ππ‘= π(π β 15) β π β 15 =
ππππ‘
When π‘ = 0, π = 100 β 100 β 15 =
ππ0 β π = 85
β΄ π β 15 = 85πππ‘
t T S
0 100 15
5 60 15
10 ? 15
When t = 5, T = 60
β 60 β 15 = 85π5π β 45 = 85π5π
β π5π =45
85
When t = 10, T β15 = 85π10π
π = 15 + 85 π5π 2
= 15 + 23.82β = 38.82β
π = 38.82β
The required temperature after a further
interval of 5 minutes is ππ. ππβ
69 π π < 40 = 0.1 ; π π > 90 = 0.1
β΄π 40 < π < 90
= 1 β [ π π < 40 + π π > 90 ]
= 1 β 0.1 β 0.1 = 0.8
Out of 800 Students, number of
students scored between 40 and 90 is
800 Γ 0.8 = 640 students
(i) WITH REPLACEMENT:
X can take the values as 0, 1, 2, 3.
P(success)=π π = 3
7 ; P(Failure) = π πΉ =
4
7
π π = 0 =64
343 π π = 1 =
144
343
π π = 2 =108
343 π π = 3 =
27
343
The required probability distribution is
X 0 1 2 3
π π= π₯
64
343
144
343
108
343
27
343
ππ = 1, β ππ β₯ 0
(ii) WITHOUT REPLACEMENT:
π π = 0 =3πΆ0 β 4πΆ3
7πΆ3=
1 β 4
35=
4
35
π π = 1 =3πΆ1 β 4πΆ2
7πΆ3=
3 β 6
35=
18
35
π π = 2 =3πΆ2 β 4πΆ1
7πΆ3=
3 β 4
35=
12
35
π π = 3 =3πΆ3 β 4πΆ0
7πΆ3=
1 β 1
35=
1
35
70 (a) Let the cubic polynomial be π¦ = π π₯ .
Since it attains maximum at π₯ = β1 and a
minimum at π₯ = 1.
ππ¦
ππ₯= 0 at π₯ = β1 and π₯ = +1.
ππ¦
ππ₯= π π₯ + 1 π₯ β 1
ππ¦ = π π₯2 β 1 ππ₯
β ππ¦ = π (π₯2 β 1) ππ₯ β π¦ = π π₯3
3β π₯ + π
When x= -1 and y = 4 β 2π + 3π = 12 ----(1)
When x= +1 and y = 0 β β2π + 3π = 0 -----(2)
Solving (1) and (2) we get, k=3 and c=2
The cubic polynomial is π = ππ β ππ + π.
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70 (b) Let π = π§ β πΆ / z = 1 (i) Closure axiom:
Let π§1 , π§2 β π
π§1π§2 = π§1 π§2 = 1 β 1 = 1, β π§1 , π§2 β π
β΄ π§1 π§2 β π
β΄ Closure axiom is true.
(ii) Associative axiom:
Complex multiplication is always associative . β΄ Associative axiom is true.
(iii) Identity axiom:
Let π§ β π and 1 = 1
such that 1 β π§ = π§ β 1 = 1 β 1 β π. β΄ The identity element 1 β π. β΄ The identity axiom is true.
(iv) Inverse axiom:
Let z β π , π€ππππ z = 1. Also 1
π§ =
1
π§ =
1
1= 1 β
1
π§β π
such that zβ1
π§=
1
π§β π§ = 1
β΄ The inverse element of z is 1
π§β π
β΄ The inverse axiom is true.
β΄ π΄,β is a group.
PREPARED BY
Tr. S SELVARAJ
MSc., BEd., MA., COA.,
VICE PRINCIPAL
MGM MATRIC. HR. SEC. SCHOOL,
KRISHNAGIRI DISTRICT.
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