TED (10)-1002 Reg. No. ………………………….

(REVISION-2010) Signature …….……………………

FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/

TECHNOLIGY- OCTOBER, 2011

TECHNICAL MATHEMATICS- I

(Common – Except DCP and CABM)

[Time: 3 hours

(Maximum marks: 100)

Marks

PART –A

(Maximum marks: 10)

(Answer all questions. Each question carries 2 marks)

I.

(a) If |

| = |

| find x.

|

| = |

|

===>9x – 14 = 8 – 4

===>9x – 14 = 4

===>9x = 18

===> x = 18/9 = 2

(b) If = find n

= ===> r = s or r + s = n

Then n = r + s = 20 + 23 = 43

(c) If cos = √ find sin &tan

sin = 1/2 tan = 1/√

√

1 2

(d) If sin =a find sin3

Sin3 = 3sin – 4sin3

Sin3 = 3a – 4a3

(e) Find the slope of the linejoining the vertices (2, 6), (4, 0)

Slope =

=

=

= -3

PART –B

Answer any five questions. Each question carries 6 marks

II.

(a) Solve using determinants:

x + y + z = 3

2x + 3y + z = –6

x + y - z = –3

AX = B

[

] [ ] = [

]

x =

=

|

|

|

|

=( ) ( )

=

= 9

y=

=

|

|

=

= – 9

z =

=

|

|

=

= 3

(b) If A = [

] B = [

] evaluate AB & BA

AB = [

] [

]

=[

] = [

]

BA = [

] [

]

= [

] = [

]

(c) Find the middle term of (x2 + 1/x)

12

Tr+1 = ncr an-r

br , n= 12

Tr+1 = 12cr(x2)12-r

(1/x)r

n + 1 = 13, odd.

(

)

=

= 6 + 1= 7

th term is the middle term

T7 = 12c6 (x2)6(1/x)

6

= 12c10x12

x-6

=12c6x6

= 924x6

(d) Prove that sin + sin + sin + sin = 4 cos . cos . sin

sin + sin + sin + sin [sinC + sinD =2 ( ) ⁄ ). ( )

⁄ )]

= (sin + sin7 ) + (sin + sin )

= 2sin4 .cos3 + 2sin4 .cos

= 2sin4 (cos3 + cos ) [cosC + cosD = 2 ( ) ⁄ ). ( )

⁄ )]

= 2sin4 . 2(cos2 . cos )

= 2sin4 . cos2 . cos , hence the result.

(e) Find the equation of the line parallel and perpendicular to 2x – 3y + 10 = 0 and passing

through (1, 1).

2x – 3y + 10 = 0

a = 2, b = -3, c = 10

Case I

Equation of a line parallel is ax + by + k = 0

2x – 3y +k = 0

Passes through (1, 1)

===> 2 x 1 – 3 x 1 + k = 0

2 – 3 + k = 0

-1 + k = 0

K = 1

So, ===>2x– 3y + 1 = 0

Case II

Equation of a line perpendicular is, bx- ay + k = 0

-3x – 2y +k = 0

Passing through (1, 1)

===> -3 x 1 – 2 x 1 + k = 0

-3 – 2 + k = 0

-5 + k = 0

K = 5

So, ===>-3x– 2y + 5 = 0

1

1

1

1

2

2

1

2

(f) Prove thatcos3A = 4cos3A – 3cosA

Cos3A = cos(2A + A)

= cos2A.cosA – sin2A.sinA

= (2cos2A – 1)cosA – 2sinA.cosA.sinA

= 2cos3A – cosA - 2sin

2A.cosA

= 2cos3A – cosA – 2(1 - cos

2A)cosA

= 2cos3A – cosA – 2cosA + 2cos

3A

= 4cos3A – 3cosA

(g) The straight line through (4, 3) makes intercepts of 4a and 3a on the X axis and Y axis

respectively. Find a

Equation of a line in intercept form:

+

= 1

+

= 1

Passing through (4, 3) then

+

= 1

===>

+

= 1

===> a = 2

Then becomes,

+

= 1

1

1

1

PART –C

(Maximum mark: 60)

Answer four full questions. Each question carries 15 marks.

III.

(a) Find x if |

| = 0

|

| = 0

===> 2(2x + 5) – 3(4 - 15) + 5(-2 - 3x) = 0

===> 4x + 10 – 12 + 45 + - 10 – 15x = 0

===> -11x = -33

x = -33/-11 = 3

(b) Express the matrix A = [

] as the sum of a symmetric and skew symmetric matrices.

A = [

]

=

[

] [

]

=

[

]

=[

] a symmetric matrix

=

[

] [

]

=

[

]

= [

] a skew symmetric matrix

Adding &&

1

2

2 1

[

] + [

] = [

] = A

Hence the result.

(c) Find the inverse of [

]

A = [

]

Minors

= |

| = 15

= |

| = 0

= |

| = -10

= |

| = 6

= |

| = 3-6=-3

= |

| = 0

= |

| = -15

= |

| = 0

= |

| = 5

Minor matrix of A= [

]

Cofactor matrix of A= [

]

Adj A = [

]

| | = |

|= 1|

| – 2|

| + 3|

|

= 15 x 2 x 0 + 3 x -10 = 15 – 30 = -15

So inverse matrix =

| | =

[

]

IV.

(a) Find the value of ‘p’

2x + 3y + 9 = 0

4x + py + 13 = 0

px – 2y – 25 = 0

Since the system is consistent,

The eliminant, |

| = 0

2(-25p + 26) – 3(-100 – 13p) + 9(-8 – p2) = 0

-50p + 52 + 300 +39p – 72 – 9p2 = 0

-9p2 – 11p +280 = 0

P = √

= √

=

P =

= or p =

= 5

(b) If A = [

] show that A2

– 4A – 5I= 0

A2 = [

] [

] =[

]

A2

– 4A – 5I = [

] - [

] - [

]

= [

] - [

] = [

]

(c) If A = [

] and B = [

] show that (AB)-1

= B-1

A-1

AB = [

] [

]

AB = [

]

To find (AB)-1

Minors

m11=16

m12 = 22

m21 = 34

m22 = 47

Minor matrix = [

]

Cofactor of AB = [

]

Adj AB = [

]

| |= |

|

= 752 – 748

= 4

(AB)-1

=

| |

= [

]

To find B-1

B = [

]

Minors

m11=3

m12 = 4

m21 = 5

m22 = 7

Minor matrix = [

]

Cofactor of B = [

]

AdjB = [

]

| |= |

|

= 1

B-1

=

| |

= [

]

To find A1

A = [

]

Minors

m11=2

m12 = 2

m21 = 3

m22 = 5

Minor matrix = [

]

Cofactor of A = [

]

AdjA = [

]

| |= |

|

= 10 – 6

= 4

A-1

=

| |

= [

]

B-1

x A-1

= [

]

x [

]

B-1

x A-1

= [

]

(AB)-1

= B-1

A-1

V.

(a) Expand (x + 1/√ )5 binomially

(a + b)n = a

n + nc1 a

n-1 b + nc2 a

n-2 b + . . . . . . . + ncnb

n

(x + 1/√ )5= x

5 + 5c1 x

4(1/√ ) + 5c2 x

3(1/√ )

2 + 5c3 x

2(1/√ )

3 + 5c4x (1/√ )

4 + (1/√ )

5

= x5 + 5x

4 (1/√ ) + 10x

3 (1/ ) + 10x

2 (1/ √ )+ 5x (1/ 2

) + (1/ √ )

= x5+ 5x

4 x

-1/2 + 10x

2+ 10x.x

-1/2+ 5x

-1 + x

-5/2

= x5 + 5x

7/2 + 10x

2 + 10x

1/2 + 5x

-1 + x

-5/2

(b) Find the 10th

term in the expansion of (x2 +1/x)

20

Tr+1 = ncr an-r

br , n= 20

Tr+1 = 20cr (x2)20-r

(1/x2)r

= 20cr x40 – 2r

x-2r

= 20cr x40 – 4r

T10= 20c9 x40 – 36

= 20c9 x4

(c) Prove that

+

=

+

=

( )

( )

=

( )

=

( ) =

( )

= ( )

( ) =

VI.

(a) Find the middle term of (2x + 3/x)9

Tr+1 = ncr an-r

br , n= 9

Tr+1 = 9cr(2x)9-r

(3/x)r

n + 1 = 10, even.

5th

term and 6th

term are the middle term.

T5 = 9c4 (2x)5(3/x)

4

= 9c425

x53

4 x

-4

=9c425x3

4

= 9c425

34

x

T6 = 9c5 (2x)4(3/x)

5

= 9c524

x43

5 x

-5

=9c524

x-1

35

= 9c524

35

x-1

(b) Find the coefficient of x4 in the expansion of (x

4 - 1/x

3)15

Tr+1 = ncr an-r

br , n = 11

Tr+1 = 15cr (x4)15-r

(-1/x3)r

= 15crx60-4r

(-1)rx

-3r

= 15crx60-7r

(-1)r

Now 60 – 7r = 4

-7r =-56

r = 56/7 = 8

T9 = 15c8x4

(-1)8

Coefficient of x4 is 15c8

(c) If sinA = , A lies in first quadrant. Find all t- functions.

AB = √ = √

sinA = ½

cosA = √ secA = √

tanA = √ cotA = √

cosecA = 2

VII.

(a) If tanA = 18/17, tanB 1/35, prove that A – B = 45o

tan(A + B) =

( )

=

=

=

= 1

(b) Prove that ( ) ( ) ( )

( ) ( ) ( ) = 1

Cos(90 + A) = -sinA

Sec(360 + A) = 1/cos(360 + A) = 1/cosA = secA

Tan(180 – A) = -tanA

Sec(A – 720) = sec –(720 – A) = secA

1

√

2 A

B C

Sin(540 + A) = sin(360 + 180 + A) = sin(180 + A) = -sinA

Cot(A – 90) = cot(90 – A) = -tanA

Substituting the above values

( ) ( ) ( )

( ) ( ) ( ) =

= 1

(c) Prove that in , (a + b)sinC/2 = c.cos(

).

LHS = (a + b).sinC/2

= (2RsinA + 2RsinB)sinC/2 [

]

=2R(sinA + sinB)sinC/2

= 2R.2.sin(

).cos(

).sinC/2

= cos(

).4R.sin(

).sinC/2

= cos(

).4R.sin (90 -

).sinC/2

= cos(

).2R.(2cosC/2.sinC/2) [sinA = 2 sinA/2.cosA/2]

= cos(

).2R.sinC

= cos(

).c = RHS

VIII.

(a) Express √ cosx + sinx in the form of Rsin( ) where is acute.

√ cosx + sinx = R.sin( )

= R.sinx.cos + Rcosx.sin

Equating the similar terms on both sides,

√ cosx = Rsin .cos

Sinx = Rsin .cos

===>√ = Rsin

===> 1 = Rcos

1

2

Squaring and adding &

3 + 1 = R2

sin2 + cos

2

4 = R2

===> R = 2

===>√ =

===> tan = √

===> = tan-1

(√ )

===> = 60o

(b) Show that cos20.cos40.cos60.cos80 = 1/16

We have cos60 = ½

ie, ½cos20.cos40.cos80 =

= ⁄ cos20 ⁄ [cos120 – cos(-40)] [ cosA.cosB = 1/2[cos(A+B)-cos(A-B)]

= ⁄ .cos20[ ⁄ + cos40]

= ⁄ cos20 + ⁄ cos20.cos40

= ⁄ . cos20 + ⁄ x (cos60 + cos20)

= ⁄ . cos20 + ⁄ cos60 + ⁄ cos20

= ⁄ . cos60 = ⁄ x ⁄ = ⁄

(c) In a , A = 30o,C = 45

o, a = 2cm find c

A = 30o,C = 45

o B = 180 – (A + C)

= 180 – 75

= 105o

Using sine rule we have

===>

===>

===> c =

x ½

c= 4 x

√ =

√ = 2.828

1 2

1 2

IX.

(a) Solve ABC if a = 24.5cm, b = 18.6cm & c = 26.4cm

We have tan

=

√(( )( ))

( )where s =

=

= 34.75

=

√(( )( ))

( )

= 0.6153

tan

= 0.6153

===> A/2 = tan-1

(0.6153) = 31o36

’

A = 2 x 31o36

’ =63

o12

’

tanB/2 =

√(( )( ))

( )=

√(( )( ))

( )

= 0.3905

(b) The x – intercept of a line is 3times its y – intercept. The line passes through (-2, 3). Find its

equation.

Intercept form of a line is

+

= 1

+

= 1

The equation passes through (-2, 3)

+

= 1

-2b + 9b = 3b2

7b = 3b2

===> b = 7/3

Equation of the line is

+

= 1

(c) Find the value of k so that the following lines are concurrent.

5x + 2y – 4 = 0

2x + ky + 11 = 0

3x – 4y – 18 = 0

Eliminant = 0

[

]= 0

5|

| - 2|

| - 4|

| = 0

5(-18k + 44) – 2(-36 – 33) – 4(-8 – 3k) = 0

-90k + 220 + 72 + 66 + 32 +12k = 0

-78k + 390 = 0

-78k = -390

K =

= 5

X.

(a) Solve a = 4, b = 5 and C = 50o

tan(

) =

cot

A – B =2tan-1[

cot

]

A – B =2tan-1

[

cot50/2]

=2tan-1

[

cot 25

o]

=2tan-1

[ 2.145]

A – B = -26.7849

A + B = 180 – 50 = 130

Solving &

A – B = -26.7849

A + B = 130

1

2

1 2

2A = 103.215

A = 51.6075 = 51o36

’

B = 130 – A=130 - 51o36

’= 78

o24

’

Now we have to find ‘C’, we have

=

=

=

===>c =

= 3.909

(b) Find the equation to the line passing through the point of intersection of x – y + 1 = 0 and

2x +3y +2 = 0 and perpendicular to the line x + y – 6 = 0

Given x – y + 1 = 0

2x – 3y + 2 = 0

x – y = -1

2x – 3y = -2

A = [

]

B = [

]

= |

|

|

|

=

=

= -1

= |

|

|

|

=

=

= 0

Point of intersection = (-1, 0)

Given line is

x + y – 6 = 0

a = 1, b = 1, c = -6

Perpendicular line is

bx – ay + k = 0

x – y + k = 0

Pass through (-1, 0)

===> -1 – 0 + k = 0

K = 1

===> x – y + 1 = 0

(c) Find the foot of the perpendicular from the origin to the line 3x – 2y – 13 = 0

3x – 2y – 13 = 0

Slope of the line perpendicular to the line 3x – 2y = 13 is -2/3

Equation of the line perpendicular to 3x – 2y = 13 is

y – 0 =-2/3 (x – 0)

y = -2/3 x

3y = -2x

2x + 3y = 0

Foot of the perpendicular is obtained by solutions

3x – 2y = 13

2x + 3y = 0

6x – 4y = 26

6x + 9y = 0

-13y = 26

y =

= -2

3 ×x – 2 ×-2 = 13

3x = 13 – 4

x = 9/3 = 3

1

(0, 0)

m =

1

2

1

1

1

Foot perpendicularis (3, -2)