vidyalankar basic mathematics prelim question paper...

1013/FY/Pre_Pap/Maths_Soln 6

Vidyalankar F.Y. Diploma : Sem. I Basic Mathematics

Prelim Question Paper Solution

Given:

4 3 9

3 2 7

1 4 x

= 0

4 [2x 28] 3 [3x 7] + 9 [12 2] = 0 8x 112 9x + 21 + 108 18 = 0 x 1 = 0 x = 1

x =1 2

3 4

, y =4 5

1 3

3x + y = 1 2 4 5

33 4 1 3

= 3 6 4 5

9 12 1 3

=

3 4 6 5

9 1 12 3

3x + y = 7 11

8 9

A 2 = A A = 3 9

1 9

=3.3 (9)( 1) (9)(9) ( 9)( 9)

( 1)(3) ( 9)( 1) ( 1)(9) ( 9)( 9)

= 0 0

6 72

2A = 0 0

6 72 = 0

hence A2 is a singular matrix

We have, A =

3 2

3 2

1 1

0 4

B =

3 2

1 1

3 2

4 2

L.H.S. = A + B

L.H.S. =

3 2 3 2

3 2 1 1

1 1 3 2

0 4 4 2

1. (a)

1. (b)

1. (c)

1. (d) Vidyala

nkar



L.H.S. =

3 2

3 1 2 1

1 3 1 2

0 4 4 2

L.H.S. =

3 2

2 1

4 1

4 2

… (1)

R.H.S. = B + A

R.H.S. =

3 2 3 2

1 1 3 2

3 2 1 1

4 2 0 4

R.H.S. =

3 2

1 3 1 2

3 1 2 1

4 0 2 4

R.H.S. =

3 2

2 1

4 1

4 2

… (2)

From eq. (1) and eq. (2) we get, L.H.S. = R.H.S. (A + B) = (B + A) is verified.

Let 2

1

x x =

1

x x 1=

A Bx x 1

Multiplying throughout by x (x 1), we get, 1 = A (x 1) + Bx

When x = 1, 1 = 0 + B (1) B = 1

When x = 0, 1 = A (0 1) + 0 A = 1

2

1

x x =

1 1x x 1

We know cos(A + B) = cos A cos B sin A sin B put A = B = , cos 2 = cos cos sin sin

= cos2 sin2 But cos2 = 1 sin2 & sin2 = 1 cos2 cos 2 = (1 sin2 ) sin2 = 1 2sin2 cos 2 = cos2 (1 cos2 ) = 1 2 sin2 cos 2 = cos2 (1 cos2 ) = 2 cos2 1

1. (e)

1. (f) Vidyala

nkar

Vidyalankar : F.Y. Diploma Mathematics


Compound Angle The algebraic sum or difference of two or more angles is called compound angle.

Given : 2 sin 40 cos 10 = sin A + sin B

2 sin 40 cos 10 = 2 sin A B

2

cos A B

2

Equating both sides we get

40 = A B

2

A + B = 80 (1)

and 10 = A B

2

A B = 20 (2)

equation (1) + equation (2) 2A = 100 A = 50 equation (1) equation (2) 2B = 60 B = 30

L.H.S. = cos21 sin21cos21 sin21

Dividing numerator and denominator by cos 21°, we get,

= 1 tan211 tan21

= tan45 tan21

1 tan45 tan21

[ tan 45° = 1]

= tan (45° 21°) tanA tanBtan A B

1 tanA tanB

= tan 24° = R.H.S.

L.H.S. = tan117

+ tan1

113

In this case,

Let x = 17

> 0, y = 1

13> 0 and xy =

1 17 13 =

191

< 1

L.H.S. = tan1x y1 xy

= tan1

1 17 13

1 11

7 13

= tan1

13 791

11

91

= tan120 / 9190 / 91

= tan129

= cot192

1 11

tan cot xx

= R.H.S.

1. (g)

1. (h)

1. (i)

1. (j)

Vidyala

nkar



Let the lines be

1 : 3x 2y + 4 = 0 Its slope, m1 =32

=32

and

2 : 2x 3y 7 = 0 Its slope, m2 =23

=23

It is the a cute angle between lines, then

tan = 1 2

1 2

m m1 m .m

Putting, m1 = 3/2, m2 = 2/3, we get,

tan =

3 22 33 2

12 3

=

9 462

= 5

12

= tan15

12

Range = largest Quantity smallest Quantity = 31 1 = 30.

Then writing into D, Dx, Dy and Dz as required in Cramer’s Rule, we have

D =

3 3

3 3 1

2 1 2

4 3 2

= 3(2 6) 3(4 8) 1(6 + 4) = 3(8) 3(4) (10) = 24 + 12 10 = 22

Dx =

11 3 1

0 1 2

25 3 2

= 11(2 6) 3(0 50) 1(0 + 25) = 11(8) 3(50) 1(25) = 88 + 150 25 = 37

Dy =

3 11 1

2 0 2

4 25 2

= 3(0 50) 11(4 8) 1(50) = 150 11(4) 1(50) = 150 + 44 50 = 200 + 44 = 156

1. (k)

1. (l)

2. (a)

Vidyala

nkar



Dz =

3 3 11

2 1 0

4 3 25

= 3(25 0) 3(50 0) + 11(6 + 4) = 75 150 + 110 = 75 40 = 115

Then by Cramer’s rule,

x = xDD

= 3722

= 3722

y = yD

D =

15622

= 15622

z = zDD

= 11522

= 11522

x = 3722

, y = 7811

, and z = 11522

LHS = (x 4) (5 27) 7

(3x 4) (15 2y) 1

= 5 3 7

7 7 1

Comparing a11 and a12 x + 4 = 5 x = 1 5 + 2y = 3 y = (35) / 2 = 4

We have A = 2 2

2 3

1 5

and B = 2 3

3 1 2

1 0 0

Now L.H.S. = (AB)T … (1)

AB = 2 2 2 3

2 3 3 1 2

1 5 1 0 0

AB =

2 3 3 1 2 0 4 0

3 5 1 0 2 0

AB = 2 3

3 2 4

8 1 2

(AB)T =

3 2

3 8

2 1

4 2

= L.H.S. … (1)

We have,

A = 2 2

2 3

1 5

AT = 2

2 1

3 5

B = 2 3

3 1 2

1 0 0

BT =

2 3

3 1

1 0

2 0

R.H.S. = T TB A

2. (b)

2. (c)

Vidyala

nkar



R.H.S. = 2 2

3 2

3 12 1

1 03 5

2 0

R.H.S. =

6 3 3 5

2 0 1 0

4 0 2 0

R.H.S. =

3 2

3 8

2 1

4 2

… (2)

From eq. (1) and eq. (2) we get L.H.S. = R.H.S. (AB)T = T TB A is proved.

A = 2 3 1

4 5 0

B = 1 2 4

1 3 0

Now, A + B = 2 1 3 2 1 4

4 1 5 3 0 0

A + B = 1 5 3

5 8 0

(A + B)T =

1 5

5 8

3 0

… (i)

AT =

2 4

3 5

1 0

, BT =

1 1

2 3

4 0

AT + BT =

2 1 4 1

3 2 5 3

1 4 0 0

AT + BT =

1 5

5 8

3 0

… (ii)

From (i) and (ii) (A + B)T = AT + BT

3 2

x 5

x x 6x

= 2

x 5

x x x 6

= x 5

x(x 2)(x 3)

2. (d)

2. (e)

Vidyala

nkar



= A B Cx x 2 x 3

A = x 0

x 5(x 2)(x 3)

= 5

( 2)( 3)

=

56

B = x 2

x 5x(x 3)

= 3

(2)(5)

= 3

2 5

= 3

10

C = x 3

x 5x(x 2)

= 8

( 3)( 5)

=

83.5

= 8

15

Let 2

3x 2

x 1 x 1

=

3x 2

x 1 x 1 x 1

=

2

3x 2

x 1 x 1

= 2

A B Cx 1 x 1x 1

Multiplying throughout by (x + 1)2 (x 1), we get, 3x + 2 = A (x + 1) (x 1) + B (x 1) + C (x + 1)2

When x = 1, 3 + 2 = 0 + 0 + C (2)2 4C = 5 C = 5/4

When x = 1, 3 + 2 = 0 + B (1 1) + 0 2B = 1 B = 1/2

When x = 0, B = 1/2, C = 5/4

2 = A (1) (1) + 12

(1) + 54

(1)2

2 = A 1 52 4

A = 5 14 2 2 =

5 2 84

=

54

2

3x 2

x 1 x 1

=

2

5 1 54 x 1 4 x 12 x 1

Let A =

3 3

1 3 2

3 2 5

2 3 6

be the given matrix.

A =

3 3

1 3 2

3 2 5

2 3 6

A = 1[(6) (2) (5) (3)] 3[(3) (6) (2) (5)] + 2[(3) (3) (2)(2)]

A = [12 + 15] 3[18 10] + 2[9 + 4]

A = 3 3(8) + 2(5)

2. (f)

3. (a) Vidyala

nkar



A = 3 24 10

A = 31

A 0

A1 exists To find A1:

Let A =

3 3

1 3 2

3 2 5

2 3 6

= 1 1 1

2 2 2

3 3 3 3 3

a b c

a b c

a b c

Let A1, B1 and C1 ….. are cofactors of elements a1, b1, c1…. respectively.

A1 = 2 5

3 6

= 12 + 15 = 3

B1 = 3 5

2 6 = (18 10) = 8

C1 = 3 2

2 3

= 9 + 4 = 5

A2 = 3 2

3 6

= (18 (6)) = 24

B2 = 1 2

2 6 = 6 4 = 2

C2 = 1 3

2 3

= (3 6) = 9

A3 = +3 2

2 5 = 15 + 4 = 19

B3 = 1 2

3 5 = (5 6) = 1

C3 = +1 3

3 2 = 2 9 = 11

The matrix of cofactors = 1 1 1

2 2 2

3 3 3

A B C

A B C

A B C

=

3

3 8 5

24 2 9

19 1 11

adj A = Transpose of cofactors matrix

adj A =

3

3 24 19

8 2 1

5 9 11

Vidyala

nkar



A1 = 1

adj AA =

3 24 191

8 2 131

5 9 11

… (1)

We have x + 3y + 2z = 6 3x 2y + 5z = 5 2x 3y + 6z = 7 are given equations. Writing in matrix form, we get

1 3 2 x

3 2 5 y

2 3 6 z

=

6

5

7

A =

3 3

1 3 2

3 2 5

2 3 6

X =

3

x

y

z

B =

3

6

5

7

… (2)

We have AX = B X = A1B … (3) From eq. (1), eq. (2) and eq. (3), we get,

3 1

x

y

z

=

3 3 3 1

3 24 19 61

8 2 1 531

5 9 11 7

3 1

x

y

z

=

3 3

18 120 1331

48 10 731

30 45 77

3 1

x

y

z

=

3 1

311

3131

62

3 1

x

y

z

=

3 1

1

1

2

x = 1, y = 1, z = 2 which is required answer.

We have,

2

2

x 23x

x 3 x 1

=

2

A Bx Cx 3 x 1

2

2

x 23x

x 3 x 1

=

2

2

A x 1 Bx C x 3

x 3 x 1

x2 + 23x = A(x2) + A + Bx2 + 3Bx + Cx + 3C

3. (b) Vidyala

nkar



x2 + 23x = (A + B)x2 + x(3B + C) + A + 3C Equating corresponding coefficient (A + B) = 1 … (1) (3B + C) = 23 … (2) A + 3C = 0 … (3) From eq. (3), we get, A = 3C … (4)

From eq. (1) and eq. (4) we get 3C + B = 1 B = 1 + 3C … (5) From eq. (2) and eq. (5), we get 3(1 + 3C) + C = 23 3 + 9C + C = 23 10C = 20 C = 2 … (6) From eq. (4) and eq. (6), we get A = 6 … (7) From eq. (1) and eq. (7) we get B = 7 … (8)

2

2

x 23x

(x 3)(x 1)

= 2

6 7x 2x 3 (x 1)

2

2

x 23x

(x 3) (x 1)

= 2

7x 2 6x 3(x 1)

which are required partial fraction.

Let tan 1

(tan 2) (tan 3)

=

A Btan 2 tan 3

here A = tan 2

tan 1tan 3

= 1

1

= 1

B = tan 3

tan 1tan 2

= 21

= 2

hence,

tan 1

(tan 2) (tan 3)

=

1 2tan 2 tan 3

We know cos(A + B) = cos A cos B + sin A sin B

put A = 2

, = B

hence cos2

= cos cos sin sin

2 2

= 0 + 1 sin

cos2

= sin

3. (c)

3. (d) Vidyala

nkar



L.H.S. = sin A sin 2A sin 3A sin 4A

cos A cos 2A cos3A cos 4A

L.H.S. =

sin A sin 4A sin 2A sin 3A

cos A cos 4A cos2A cos 3A

L.H.S. =

4A A A 4A 3A 2A 2A 3A2sin cos 2sin cos

2 2 2 2A 4A A 4A 2A 3A 2A 3A

2cos cos 2cos cos2 2 2 2

L.H.S. =

5A 3A 5A A2 sin cos sin cos

2 2 2 2

5A 3A 5A A2 cos cos cos cos

2 2 2 2

L.H.S. =

5A 3A Asin cos cos

2 2 2

5A 3A Acos cos cos

2 2 2

L.H.S. = 5A

tan2

L.H.S. = R.H.S. Hence proved.

L.H.S. =1 sec2A

tan2A

=

11

cos2Asin2Acos2A

=cos2A 1

sin2A

= 1 cos2A

sin2A

=22cos A

2sinA.cosA =

cosAsinA

= cotA = R.H.S.

L.H.S. = 1 tan2A tanA1 tan2A tanA

=

sin2A sinA1

cos2A cos Asin2A sinA

1cos2A cos A

=

cos2A cosA sin2A sinAcos2A cos A

cos2A cosA sin2A sinAcos2A cos A

= cos2A cosA sin2A sinAcos2A cosA sin2A sinA

L.H.S. = sin 3A L.H.S. = sin(2A + A) L.H.S. = sin2A cos A cos2A sinA L.H.S. = (2 sin A cos A) cos A + [cos2 A – sin2 A] sin A

3. (e)

3. (f)

4. (a)

4. (b)

Vidyala

nkar



L.H.S. = 2sin A cos2 A + (cos2 A sin2 A) sin A L.H.S. = 2 sin A (1 – sin2 A) + (1 – sin2 A – sin2 A) sin A L.H.S. = 2 sin A – 2 sin3 A + (1 – 2 sin2 A) sin A L.H.S. = 2sin A 2sin3A + sin A 2sin3A L.H.S. = 3sin A 4 sin3A L.H.S. = R.H.S. Hence proved. L.H.S. = sin 20. sin 40. sin 60. sin 80

= sin 20. sin 40.3

2. sin 80

3sin60

2

= 3

4[2 sin 20. sin 40] .sin 80 [Multiplying and dividing by 2]

= 3

4[cos (20° 40°) cos (20° + 40°)] . sin80°

= 3

4[(cos 20° cos 60°) . sin80°]

= 3

4[cos 20° sin 80° cos 60° sin 80°]

= 3

8[2 cos 20° sin 80° 2.

12

sin 80°]

[Multiplying and dividing by 2 and cos 60° = 1/2]

= 3

8[sin (20° + 80°) sin (20° 80°) sin 80°]

= 3

8[sin 100° + sin 60° sin 80°] [ sin (60°) = sin 60°]

= 3

8[sin 80° + sin 60° sin 80°]

[ sin 100° = sin (2 90° 80°) = sin 80°]

= 3

8(sin 60°)

= 3 3

8 2

= 3

16 = R.H.S.

L.H.S. = cosec A cosec A

+cosec A-1 cosec A+1

= cosecA1 1

cosecA-1 cosecA+1

= cosecA2

cosecA+1+cosecA-1

cosec A 1

=

2

cosecA 2 cosecA

cot A

4. (c)

4. (d) Vidy

alank

ar



= 2

2

2 cosec A

cot A

=

2

2 2

2 sin A.

sin A cos A

= 2

2

cos A

= 2sec2A = R.H.S.

L.H.S. = cot coseccot cosec 1

L.H.S. =

cos 11

sin sincos 1

1sin sin

L.H.S. = cos 1 sincos 1 sin

L.H.S. =

2

2

2cos 2sin cos2 2 2

2sin 2sin cos2 2 2

L.H.S. = 2cos cos sin

2 2 2

2sin sin cos2 2 2

L.H.S. = 2cos cos sin

2 2 2

2sin cos sin2 2 2

L.H.S. = cos

2

sin2

L.H.S. = cot2


L.H.S. = cos145

sin1

513

Let cos145

= 1 cos 1 =

45

sin21 = 1 cos2 1 = 1 16

25

= 925

sin 1 = 35

4. (e)

4. (f) Vidyala

nkar



Also, Let sin1 513

= 2 sin 2 = 5

13

cos22 = 1 25

169 =

144169

cos 2 = 1213

Next cos (1 2) = cos 1 cos 2 + sin 1 sin 2

= 4 12 3 55 13 5 13

=48 1565 65

=6365

1 2 = cos1 6365

cos1 45

sin1 513

= cos1 6365

L.H.S. = sin A 1 cosA

1 cos A sinA

L.H.S. =

2 2sin A 1 cos A

sinA 1 cosA

L.H.S. =

2 2sin A sin AsinA 1 cosA

L.H.S. =

22sin AsinA 1 cosA

L.H.S. =

22 1 cos A

sinA 1 cosA

L.H.S. =

2 1 cosA 1 cosA

sinA 1 cosA

L.H.S. = 1 cosA

2sinA

L.H.S. = 2 [cosec A cot A] L.H.S. = R.H.S. Hence proved.

We know sin(A + B) = sin A cos B + cos A sin B A = C/2, B = D/2

C Dsin

2

= C D C D

sin cos cos sin2 2 2 2

... (1)

cos(A B) = cos A cos B + sin A sin B Similarly,

C Dcos

2

= C D C D

cos cos sin sin2 2 2 2 ... (2)

5. (a)

5. (b) Vidyala

nkar



Multiplying (1) and (2);

C D C D

sin cos2 2

=

C D C Dsin cos cos sin

2 2 2 2

C D C D

cos cos sin sin2 2 2 2

Multiplying by 2 throughout :

C D C D

2sin cos2 2

= 2 2C C D C D D2 sin cos cos cos cos sin

2 2 2 2 2 2

2 2C D D D C Csin cos sin sin cos sin

2 2 2 2 2 2

= 2 2C C D D2 sin cos cos sin

2 2 2 2

2 2D D C Csin cos cos sin

2 2 2 2

C D C D

2sin cos2 2

= D D C C

2 sin cos sin cos2 2 2 2

We know 2 sin cos = sin 2.

hence C D C D

2sin cos2 2

= [sin D + sin C] = RHS

hence proved.

L.H.S. = cot A cot 2Acot A cot 2A

L.H.S. =

cos A cos 2Asin A sin 2Acos A cos 2Asin A sin 2A

L.H.S. =

cos A sin 2A cos2A sinA

cos A sin2A cos2A sinA

L.H.S. =

sin 2A A

sin 2A A

( sin (A + B) = sin A . cos B + cos A sin B

and sin(A B) = sin A cos B cos A sin B)

L.H.S. = sin A

sin3A


We have, L1 : 2x + 3y = 13 … (1) L2 : 2x 5y = 7 … (2) are given equation of straight lines.

Let m1 and m2 be slopes of line given by eq. (1) and eq. (2)

5. (c)

5. (d) Vidy

alank

ar



We get m1 = 2

3

and m2 = 25

If ‘’ be the acute angle between lines, then

tan = 1 2

1 2

m m1 m m

tan =

2 23 5

2 21

3 5

tan =

10 615

15 415

tan =

16151115

tan = 1611

tan = 1611

tan = 1611

= 1 16tan

11

Consider P (x1, y1) = P (5, 4) and the line is 2x + y + 6 = 0. The perpendicular distance from P(x1, y1) on ax + by + c = 0 is

= 1 12 2

ax by c

a b

where a = 2, b = 1, c = 6 x1 = 5, y1 = 4

= 2 5 1 4 6

4 1

= 10 4 6

5

=

20

5

= 4 5 units. Let P(x, y) divide the join of A (+5, 4) and B (2, 3) in the ratio K: 1 m = K and n = 1 Using external division formula,

Px = 2 1mx nxm n

and Py = 2 1my nym n

5. (e)

5. (f)

Vidyala

nkar



Px = K 2 1 5

K 1

Py =

K 3 1 4K 1

Px = 2K 5K 1

Py =3K 4K 1

P (x, y) divides the x-axis, Py = 0 0 = 3K + 4 K = +4/3

Px =

42 5

34

13

=

85

313

= 23

The ratio is 4

:13

and the point is (23, 0).

Let for the line, x-intercept = a y- intercept = b Given : a = 2b

The equation of line in two intercept form is, x ya b = 1

Putting a = 2b, we get, x y

2b b = 1, multiplying throughout by 2b.

x + 2y = 2b (i) This line passes through the point (4, 1) Putting x = 4, y = 1, we get, 4 + 2 (1) = 2b 2b = 6 b = 3 a = 2b = 6

Putting b = 3 in equation (1), x + 2y = 2 (3) x + 2y = 6 This is the required line. Let P(x3, y3) and Q(x4, y4) be the points of trisection for the join A(3, 4) and B(2, 3) Then we observe from fig. that the point P(x3, y3) divides the join AB internally in the ratio 1 : 2. m = k, n = 2k k 0, k R Using internal division formula for x and y we get,

Px = 2 1mx nxm n

and Py = 2 1my nym n

A(3, 4) (x1, y1)

P(x3, y3) Q(x4, y4) B(2, 3) (x2, y2)

6. (a)

6. (b) Vidyala

nkar



Putting values of m, n, x1, y1, x2, y2 we get

Px = k 2 2k 3

k 2k

Py = k 3 2k 4

k 2k

Px = 2k 6k

3k

Py = 3k 8k

3k

Py = 4k3k

Py = 5k3k

P2 = 43

Py = 53

P(x3, y3) = 4 5

,3 3

Now point Q(x4, y4) is the midpoint of PB.

Qx =

42

32

Qy =

53

32

Qx =

4 632

Qy =

43

2

Qx =

23

2

Qy =

46

Qx = 26

Qy = 23

Qx = 13

Q = (x4, y4) = 1 2,

3 3

M = h N

L c.f.F 2

C.I. F CF 9.5 14.5 4 4 14.5 19.5 6 10 19.5 24.5 10 20 24.5 29.5 5 25 29.5 34.5 7 32 34.5 39.5 3 35 39.5 44.5 9 44 44.5 49.5 6 50 N = 50 Median class = 24.5 29.5 h = 5 F = freq. of median class = 5. C.f. = 20

6. (c)

Vidyala

nkar



Substituting

Median = 24.5 + 525 20

5

= 24.5 + 5 = 29.5

Range = L S

= 36 25 = 11

Coefficient of Range = L SL S

= 36 2536 25

= 1161

M D = ix x

n

x = fixifi

= 60 150 350 810 495 195

50

= 200050

= 40

MD = 25 15 5 5 15 25

50

= 9050

= 95

Variance = 21fi xi x

n

Coefficient of variance = 100| x |

, where = 5D = Variance

x = fixi

fi

= (9)(15) (17)(45) (75)(43) (105)(82) (195)(24) (135)(81) (165)(44)

300

= 118.7

Variance = 2i1

fi x xn

= 2 2 219(15 118.7) 17(45 118.7) 43(75 118.7)

300

2 2 282(105 118.7) 81(135 118.7) 24(195 118.7)

= 1492.90

Coefficient of variance = 100x =

Variance100

118.7

= 32.54

6. (e)

6. (d)

6. (f)

Vidyala

nkar

vidyalankar basic mathematics prelim question paper...

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