1a em.pdf

8/17/2019 1a em.pdf

1/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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.

www.eenadupratibha.net


N E W

S Y L L A B U S

N E W

S Y L L

A B U S

8/17/2019 1a em.pdf

2/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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



8/17/2019 1a em.pdf

3/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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



8/17/2019 1a em.pdf

4/15

w w w . e

e n a d u p r

a t i b h

a . n e t

∴ 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



8/17/2019 1a em.pdf

5/15

w w w . e

e n a d u p r

a t i b h

a . n e t

√ 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



8/17/2019 1a em.pdf

6/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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)



8/17/2019 1a em.pdf

7/15

w w w . e

e n a d u p r

a t i b h

a . n e t

⇒

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



8/17/2019 1a em.pdf

8/15

w w w . e

e n a d u p r

a t i b h

a . n e t

√

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



8/17/2019 1a em.pdf

9/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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



8/17/2019 1a em.pdf

10/15

w w w . e

e n a d u p r

a t i b h

a . n e t

∴ 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



8/17/2019 1a em.pdf

11/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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



8/17/2019 1a em.pdf

12/15

w w w . e

e n a d u p r

a t i b h

a . n e t

(−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)



8/17/2019 1a em.pdf

13/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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



8/17/2019 1a em.pdf

14/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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



8/17/2019 1a em.pdf

15/15

w w w . e

e n a d u p r

a t i b h

a . n e t

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)



1a em.pdf

Documents

Transcript of 1a em.pdf