1a em.pdf
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Mathematics - Paper I(A)Model Paper (English Version)
Time: 3 Hours Max. Marks: 75
Note: The question paper consists of 3 sections A, B and C.
SECTION - A
I. (i) Very short answer type questions.
(ii) Answer all questions.
(iii) Each question carrier TWO marks. (10 × 2 = 20)
1. If A = {−2, −1, 0, 1, 2} and f: A → B is a surjection defined by
f(x) = x2 + x +1, then find B.
2. Find the domain of the real valued function f(x) = log (x2-
4x + 3).
2 43. If A = [ ] and A2 = 0, then find the value of 'k'.-1 k
1 2 0 -1
4. Find the rank of the matrix [ 3 4 1 2 ]-2 3 2 5
5. Find the vector equation of the plane passing through the points i
- 2 j
+ 5 k
,
-
5 j
-
k
and-
3 i
+ 5 j
.6. If the vectors 2 i
+λ j
- k
and 4 i
- 2 j
+ 2 k
are perpendicular to each other,
find 'λ'.
7. If a
× b
= b
× c ≠ 0, then show that a + c = p b
where 'p' is some scalar.
8. Find a cosine function whose period is '7'.
9. Simplify: sin2 42° - sin2 12°
510. If cos hx = , find (i) cos h 2x and (ii) sinh 2x
2SECTION - B
II. (i) Short answer type questions.
(ii) Answer any 5 questions.
(iii) Each question carries 4 marks. (5 × 4 = 20)
1 2 211. If A = [2 1 2] then show that A2 - 4A - 5I = 0.2 2 112. Show that the points A (2 i
-
j
+ k
), B ( i
-
3 j
-
5 k
) and C (3 i
-
4 j
-
4 k
)
form a right angled triangle.
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N E W
S Y L L A B U S
N E W
S Y L L
A B U S
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13. If a + b + c = 0, then prove that a × b = b × c = c × a .
3 514. If cos α = and cos β = and α, β are acute angles, then prove that
5 13
α - β 1 α + β 16
(i) sin2
=
and (ii) cos2
=
.
2 65 2 65
15. Solve the equation: sinx + √ 3cosx = √
2.
√ 41 ∏
16. Prove: cot-19 + cosec-1 = .4 4
B - C B + C17. Prove in ∆ ABC: a cos = (b + c) cos .
2 2
SECTION - C
III. (i) Long answer type questions.
(ii) Answer any 5 questions.
(iii) Each question carries 7 marks. (5 × 7 = 35)
18. If f: A → B is a bijection, then prove that fof -1 = IB and f -1 of = IA
19. Using the principle of mathematical induction, prove that
n(n + 1)2 (n + 2)12 + (12 + 22) + (12 + 22 + 32) + ..... upto n brackets =
12a1 b1 c1
20. If A = [a2 b2 c2 ] is a non-singular matrix, then prove that A is invertible anda3 b3 c3Adj.A
A-1 = det.A
21. Solve the equations by using matrix-inversion method:
3x + 4y + 5z = 18; 2x-
y + 8z = 13, 5x-
2y + 7z = 20.22. Find the shortest distance between the skew lines.
r = (6 i
+ 2j
+ 2k
) + t(i
- 2j
+ 2k
) and r = (-4 i
- k
) + s(3 i
- 2 j
- 2k
).
A B C π -A π - B π - C23. In ∆ ABC, prove: cos + cos + cos = 4 cos cos cos .
2 2 2 4 4 4
r1 r2 r3 1 124. Prove: + + = - .
bc ca ab r 2R
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SOLUTIONSI 1. f: A → B is a surjection
A = {−2, −1, 0, 1, 2}
f(x) = x2
+ x + 1f(−2) = (−2)2 + (−2) + 1 = 3
f(−1) = (−1)2 + (−1) + 1 = 1
f(0) = 0 + 0 + 1 = 1
f(1) = 1 + 1 + 1 = 3
f(2) = (2)2 + 2 + 1 = 7
∴ B = {1, 3, 7}
2. f(x) = log (x2 − 4x + 3) is a real valued function
Hence, x2 − 4x + 3 > 0
⇒ (x−1) (x−3) > 0
⇒ x ∈ (−∞, 1) ∪ (3, ∞)
∴ domain is R-[1, 3]
2 4 2 4 0 0
3. A2 = 0 ⇒ ( )( )=
( )−1 k −1 k 0 0
0 8 + 4k 0 0⇒ ( ) = ( )−2−k −4 + k 2 0 0⇒ 8 + 4k = 0 (or) −2 − k = 0 (or) −4 + k 2 = 0
⇒ k = −2
1 2 0
4. Consider the sub-matrix
(3 4 1
)−2 3 2determinant = 1(8 − 3) − 2(6 + 2)
= 5 − 16 = −11 ≠ 0
∴ Rank of the given matrix = 3
5. The vector equation of the plane passing through the points a,
b and c is
r = a + s(
b − a) + t( c − a ) where s and t are scalars.
Here a =
i − 2 j + 5
k;
b = −5
j −
k; c = −3 i + 5
j
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∴ The required equation is: r = (
i − 2 j + 5 k) + s(−5 j − k − i + 2 j − 5 k) + t(−3 i + 5 j − i + 2 j − 5 k )
⇒ r = ( i − 2 j + 5 k) + s(− i − 3 j − 6 k) + t(−4 i + 7 j − 5 k)
⇒
r = (1 − s − 4t)
i + (−2 − 3s + 7t)
j + (5 − 6s − 5t)
k 6. For mutually perpendicular vectors, (2
i + λ j − k) . (4 i − 2 j + 2 k) = 0
⇒ 8( i . i) − 2λ( j . j) − 2( k . k) = 0
⇒ 8 − 2λ −2 = 0 ⇒ λ = 3
7.
a × b = b × c ≠ 0
⇒ a × b ≠ 0; b × c ≠ 0
also
a × b = b × c
⇒ ( a × b ) − ( b × c ) = 0
⇒ ( a × b ) + ( c × b) = 0
⇒ ( a + c ) × b = 0
⇒ ( a + c) and b are collinear vectors.
Hence
a +
c = p
b for some scalar p
8. Let cos px be the required function
2Π∴ Period is .p
2Π 2ΠGiven period is 7 ∴ = 7 (or) p=
p 72Π
∴ Cos x is the required function.7
9. Sin2 42° - Sin2 12° = Sin (42° + 12° ). Sin (42° − 12°)
= Sin 54°. Sin 30°√ 5 +1 1 √
5 +1
= ( ) ( ) = .4 2 810. Cosh2x − Sinh2x =1
⇒ Sinh2 x = Cosh2x−1
5 2= ( ) −12
21=
4
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√ 21
∴ Sinhx = ± 2
Cosh 2x = Cosh2x + Sinh2x
5 2 21= ( ) + 2 423
∴ Cosh 2x = 2
also, Sinh 2x = 2.Sinhx.Coshx
± √ 21 5
= 2( ) ( )2 2
5√ 21= ±
2
1 2 2 1 2 2 1 2 2 9 8 8
11. A = ( 2 1 2 ) ⇒ A2 = ( 2 1 2 ) ( 2 1 2 ) = ( 8 9 8 )2 2 1 2 2 1 2 2 1 8 8 9
1 2 2 4 8 8
4A = 4
(2 1 2
)=
(8 4 8
)2 2 1 8 8 41 0 0 5 0 0
5I = 5 ( 0 1 0 ) = ( 0 5 0 )0 0 1 0 0 5
9 8 8 4 8 8 5 0 0
Now, A2 − 4A − 5I = ( 8 9 8 ) − ( 8 4 8 )− ( 0 5 0 )8 8 9 8 8 4 0 0 5
9 − 4 − 5 8 − 8 − 0 8 − 8 − 0 0 0 0
= ( 8 − 8 − 0 9 − 4 − 5 8 − 8 − 0 ) = ( 0 0 0 )8 − 8 − 0 8 − 8 − 0 9 − 4 − 5 0 0 0
= O
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12. A = 2
i −
j +
k; B =
i − 3
j − 5
k; C = 3
i − 4
j − 4
k
are given points
AB = (
i − 3
j − 5
k) − (2
i −
j +
k) = −
i −2
j − 6
k
√ √ AB= 1 + 4 + 36 = 41
∴ AB2 = AB 2 = 41 → (1)
BC = (3
i − 4
j − 4
k) − (
i − 3
j − 5
k) = 2
i −
j +
k
√
√ BC= 4 + 1 + 1 = 6
∴ BC2 = BC 2 = 6 → (2)
AC = (3
i − 4
j − 4
k) − (2
i −
j +
k) =
i − 3
j − 5
k
√
√ AC= 1 + 9 + 25 = 35
∴ AC2 = AC2 = 35 → (3)
(1) = (2) + (3) ⇒ AB2 = BC2 + AC2
∴ ABC is a right angled triangle
13.
a +
b +
c =
0 given
∴
a +
b = −
c
a × (
a +
b) =
a × (−
c)
⇒ (
a ×
a) + (
a ×
b) = −
a ×
c
⇒
a ×
b =
c ×
a → (1)
b +
c = −
a
⇒
b × (
b +
c) =
b ×
(−
a)
⇒ (
b ×
b) + (
b ×
c) = −(
b ×
a)
⇒
b ×
c =
a ×
b → (2)
c +
a = −
b
c × (
c ×
a) =
c × (−
b)
⇒ ( c × c) + ( c × a) = −( c × b)
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⇒
c ×
a =
b ×
c → (3)
(1), (2), (3) ⇒
a ×
b =
b ×
c =
c ×
a
14. cosα = 3 5
and α is acute angle ∴ sinα = 4 5
5 12cosβ = and β is acute angle ∴sinβ = 13 13
3 5 4 12 63Now, cos(α - β) = cosα cosβ + sinα sinβ = ( )( ) + ( ) ( ) = 5 13 5 13 65
3 5 4 12 -33also cos(α + β) = cosα cosβ - sinα sinβ = ( )( ) - ( ) ( ) = 5 13 5 13 65
(α - β)(i) 2sin2 = 1 - cos(α - β)
2
63 2= 1 - =
65 65(α - β) 1
∴ sin2 = 2 65
(α + β)(ii) 2 cos2 = 1 + cos(α + β)
2
33 32= 1 - =
65 65
(α + β) 16∴ cos2 =
2 65
15. sinx + √
3 cosx = √
2
dividing both sides by √
1 + 3 = 2 we get
1 √
3 1 sinx + cosx = 2 2 √
2
Π Π 1cosx cos + sinx sin = 6 6 √
2
Π 1 Πcos(x - ) = = cos 6 √ 2 4
Π Π⇒ x - = 2nΠ ± , n ∈ z6 4
Π Π Π Π⇒ x = 2nΠ + + (or) x = 2nΠ - + 4 6 4 6
5Π Π∴Solution : x = 2nΠ + (or) x = 2nΠ - , n ∈ z
12 12
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√
4116. Put A = cot-19; B = cosec-1
4
√
41∴cotA = 9; cosec B =
4
Π 1 40 < A, B < ⇒ tanA = ; tanB = 2 9 5
(tanA + tanB)tan (A + B) =
(1 - tanA tanB)
1 4 + 9 5
= 1 4
1 - ( )( )9 541
45 Π Π= = 1 = tan ∴ A + B =
41 4 4
45
∴ The result proved
b + c 2R SinB + 2R SinC17. Consider =
a 2RSinA
SinB + SinC= SinA
B+C B−C2Sin
2Cos
2=
A2sin
2Cos
A
2B+C B−C
Sin 2
Cos 2
=
(...A+B+C = 180°)
B+C B+CCos
2Sin
2
b+c Cos B−C
2 =
aCos
B+C2
∴ By cross multiplication, we get
B−C B+Ca Cos = (b+c) Cos
2 2
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18. f: A →B is a bijection
⇒ f −1: B →A is a bijection
∴ fof −1: B →B and f −1 of: A →A are also bijections
Also, IA
: A →A; IB
: B →B are bijections
Hence, IA and f −1 of are defined on the same domain A
Also, IB and fof −1 are defined on the same domain B
Let b ∈ B... f: A →B, we have for a ∈ A, f(a) = b (or) a = f −1(b)
(fof −1) (b) = f [f −1(b)] = f(a) = b = IB(b)
This implies that fof −1 = IB →(1)
Similarly (f −1of) (a) = f −1[f(a)] = f −1(b) = a = IA(a)
∴ f −1of = IA →(2)
(1) and (2) imply that the result proved.
19. 1st Bracket (12) = 1
2nd Bracket (12 + 22) = 5
3rd Bracket (12 + 22 + 32) = 14
n(n+1) (2n+1)nth Bracket= (12 + 22 + 32 +... + n2) =
6
K(K+1) (2K+1)Kth Bracket = (12 + 22 + 32 +... +K2) =
6
1(1+1)2 (1+2)n = 1 ⇒ 1 =
12
12
⇒ 1 = = 112∴ S(1) is true
2(2+1)2 (2+2)n = 2 ⇒ 1 + 5 =
12
(2)(9)(4)6 = = 6
12
∴ S(2) is trueLet S(K) be true
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∴ We have
K(K+1) (2K+1) K(K+1)2 (K+2)1 + 5 + 14 + ........ + =
6 12
Adding (K + 1)th Bracket = (12 + 22 + 32 +.... + (K+1)2)
(K+1) (K+2) (2K+3)= both sides, we have
6
K(K+1) (2K+1) (K+1) (K+2) (2K+3)1 + 5 + 14 +... + + =
6 6
K (K+1)2 (K+2) (K+1) (K+2) (2K+3) +
12 6
(K+1)(K+2) K(K+1) 2K+3= [ + ]6 2 1
(K+1)(K+2) (K+2) (K+3)= [ ]6 2
(K+1)(K+2)2 (K+3)=
12
n(n+1+1)2 (n+1+2)
=
form12
This implies that S(K+1) is true. And hence the result proved by using the prin-
ciple of mathematical induction.
20. A = a1 b1 c1[a2 b2 c2]a3 b3 c3A ≠ 0 Since A is non-singular matrix (given)
Let matrix of cofactors of A be taken as = A1 B1 C1[A2 B2 C2]A3 B3 C3This given that Adjoint A = A1 A2 A3[B1 B2 B3]C1 C2 C3
a1 b1 c1 A1 A2 A3Consider A (Adj A) =
(a2 b2 c2
) (B1 B2 B3
)a3 b3 c3 C1 C2 C3
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a1A1+b1B1+c1C1 a1A2+b1B2+c1C2 a1A3+b1B3+c1C3
= [a2A1+b2B1+c2C1 a2A2+b2B2+c2C2 a2A3+b2B3+c2C3 ]a3A1+b3B1+c3C1 a3A2+b3B2+c3C2 a3A3+b3B3+c3C3
det.A 0 0
= ( 0 det.A 0 )0 0 det.A(by using the definition and properties of determinant)
1 0 0
= det.A(0 1 0)0 0 1= det A(I)
Adj.A∴ A( ) = I → (1)det.A
Adj.ASimilarly it can be proved that ( ) A = I → (2)det.A
Adj.A Adj.A∴ (1), (2) ⇒ A ( ) = ( ) A = Idet.A det.A
Adj.AThis implies that A−1 is invertible and A−1 =
det.A
∴ The Proof
21. 3x + 4y + 5z = 18
2x − y + 8z = 13
5x− 2y + 7z = 20
are written as AX = B where3 4 5 x 18
A =(2 −1 8 ) ; X = (y); B =(13)5 −2 7 z 20∴ X = A−1B ------(1)
A = 3(−7+16) −4(14-40) + 5(−4+5)
= 27 + 104 + 5
= 136 ≠ 0∴ A is invertible and A−1 exists
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(−7+16) −(14−40) +(−4+5)
Matrix of co-factors of A = (−(28+10) +(21−25) −(−6−20))+(32+5) −(24−10) +(−3-8)]
9 26 1
= (−38 −4 26)37 −14 −119 −38 37
∴ Adj A = (26 −4 −14)1 26 −11
Adj A 1 9 −38 37
∴ A−1 =
= [
26 −4 −14
]det.A 136
1 26 −11
x1
9 −38 37 18
Now, X = A−1 B ⇒ [y]= [26 −4 −14] [13]z 136 1 26 −11 201 162 −494 +740
=
[468 −52 −280
]136 18 +338 −220
1 408
= [136]136 1363
= [1]1⇒ x = 3, y = 1, z = 1 is the solution
22. r = (6 i + 2 j + 2 k) + t( i −2 j + 2 k) represents the vector equation of the linepassing through the point A(6, 2, 2) and parallel to the vector (1, -2, 2) = b (say)
r = (−4 i − k) + s(3 i − 2 j − 2 k) represents the vector equation of the line pass-ing through the point C(−4, 0, −1) and parallel to the vector (3, −2, −2) = d (say)
[ AC b d]Now, shortest distance between these lines = --------- (1)
b × d AC = (−4, 0, −1) − (6, 2, 2) = (−10, −2, −3)
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b = (1, −2, 2) ; d = (3, −2, −2)
−10 −2 −3
[
AC
b
d] = 1 −2 23 −2 −2
= −10(8) + 2(−8) − 3(4)
= − 80 − 16 − 12 = −108
[ AC b d] = −108 = 108
i
j
k
b × d =
1 −2 2
3 −2 −2
=
i(8) − j(−8) + k(4)
= 8
i + 8
j + 4
k
b × d = √
64 + 64 + 16 = 12
108∴ (1) ⇒ Shortest distance = = 9
12
A B C Π23. A + B + C = Π ⇒ + + = 2 2 2 2
A B CL.H.S. = Cos + Cos + Cos
2 2 2
A+B A−B C= 2 Cos Cos + Cos
4 4 2
A+B A−B Π A + B= 2 Cos
Cos
+ Cos
( − )4 4 2 2A+B A−B A+B
= 2 Cos Cos + Sin 4 4 2
A+B A−B A+B A+B A+B A+B= 2 Cos [Cos + Sin ](... Sin = 2 Sin Cos )4 4 4 2 4 4
Π−C A−B Π−C= 2 Cos [Cos + Sin ]4 4 4
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Π−C A−B Π Π−C= 2 Cos [Cos + Cos( − )]4 4 2 4
Π−C A−B Π+C= 2 Cos
[Cos + Cos
(
)]4 4 4
Π−C A−B+Π+C A−B−Π−C= 2 Cos [2Cos Cos ]4 8 8
Π−C 2Π−2B −2B−2C= 4 Cos [Cos Cos ]4 8 8
Π−C Π−B Β+C= 4 Cos
[Cos Cos
]4 4 4
Π−C Π−B Π−A= 4 Cos [Cos Cos ]4 4 4
Π−A Π−B Π−C= 4 Cos Cos Cos
4 4 4
= R.H.S.
r1 r2 r324. LHS = + + bc ca ab
ar1 + br2 + cr3=
abc
a.s. TanA
2+ b.s.Tan
B
2+ c.s.Tan
C
2=
abc
s
= [
a.Tan A 2 + b.Tan B 2 + c.Tan C 2 ]abc
s Sin A / 2 SinB / 2 Sin
C / 2= [2R SinA + 2R Sin B + + 2R SinC ]
abc Cos A / 2 CosB / 2 Cos
C / 2
s 2R. 2Sin A / 2 CosA / 2 Sin
A / 2 2R. 2SinB / 2 Cos
B / 2 sinB / 2
= [ + +abc CosA / 2 Cos
B / 2
2R. 2Sin C / 2
Cos C / 2. Sin C /
2
]Cos C / 2
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s= [4R Sin2 A / 2 + 4 R Sin2 B / 2 + 4R Sin2 C / 2]
abc
4R.s= [Sin2 A 2 + Sin2 B 2 + Sin2 C 2 ]abc
s 1−cos A 1−cosB 1−cosC abc= [ + + ] (∴ ∆ = )∆ 2 2 2 4R
s 3 1 ∆= [ − (CosA + CosB + CosC)] (∴ r = )∆ 2 2 s
1 3 1 A+B A−B= [ − (2Cos Cos + CosC)]r 2 2 2 2
1 C A−B C= [3 − (
2Sin Cos
+ 1−2 Sin2 )]2r 2 2 21 C A−B C
= [2 − 2Sin (Cos − sin )]2r 2 2 22 C A−B A+B
= [1 − Sin (Cos − Cos )]2r 2 2 21 C A B
=
[1 − Sin
(2 Sin Sin
)]r 2 2 2
1 2 A B C= − [Sin Sin Sin ]
r r 2 2 2
1 2 Sin A / 2 SinB / 2 Sin
C / 2= −
r 4R Sin A / 2 SinB / 2 Sin
C / 2
1 1= −
r 2R= R.H.S.
(Prepared by C. Sadasiva Sastry)
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