TED (10)-1002 Reg. No. ………………………….

(REVISION-2010) Signature …….……………………

FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/

TECHNOLIGY- MARCH, 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 A = *

+ and B = *

+ find A – 3B

A – 3B = *

+ - 3*

+

= *

+ - *

+ = *

+

(b) If n = n find the value of n?

ncr = ncs ===>r + s = n or r = s

n = 7 + 2 = 9

(c) Find the value of sin30o.cos60

o + cos30

o.sin60

o

sinA.cosB + cosA.sinB = sin(A + B)

sin30o.cos60

o + cos30

o.sin60

o = sin(30 +60) = sin90 = 1

(d) Evaluate

= sin2 =

= sin30

o = ½

(e) Write down the equation of the line having slope ½ and y intercept -1

Slope-intercept form of a line is y = mx + c

Required equation is y =½ x-1

PART –B

Answer any five questions. Each question carries 6 marks

II.

(a) Solve using determinants:

x + y – z = 4

3x – y + z = 4

2x – 7y + 3z = -6

AX = B

[

] [ ] = [

]

x =

=

|

|

|

|

= ( ) ( ) ( )

( )– ( ) ( ) =

= 32/16 = 2

y=

=

|

|

= ( ) ( ) ( )

=

= 16/16 = 1

z=

=

|

|

= ( ) ( ) ( )

= -16/16 = -1

(b) If A =[

] and B =[

] be two matrices, find AB & BA

AB=[

] [

] = [

]

BA = [

] [

] = [

]

(c) Find the middle term in the expression of( ⁄ )

11

n = 11 n + 1 = 12

so 6th

and 7th

terms are middle terms

Tr+1= ncrxn-r

yr

T6= 11c5(2x)6(3/x)

5

= 11c526 x

6 3

5 x

-5

= 7185024x

T7= 11c6 (2x)5(3/x)

6

= 11c625 x

5 3

6 x

-6

= 10777536x-1

(d) Prove that

(i) ( ) + ( ) = 2cosec2A

( ) = cot2A – 2cotA + 1

( ) = cot2A + 2cotA + 1

( ) + ( ) = 2cot2A + 2 = 2(cot

2 + 1) = 2cosec

2A

(ii)

=

( )

( ) =

( ) =

( ) =

(e) Evaluate tan22½o without using tables

tan2 =

put = 22 ½o

then tan2(22 ½o) =

tan45o =

1 =

2tan = 1 – tan2

===> tan2 + 2tan - 1 = 0

Put tan = tan22 ½ o = x

Then becomes x2 + 2x – 1 = 0

===> x = √

=

√

x =tan22 ½ o = -1+√ is admissible

(f) Prove that

=tan3A

=

=

1

1

= ( )

( )

=tan3A

(g) The vertices of a triangle are A(3,4), B(5,6) and C(-1,-2). Find the equation to the median

through A

Coordinates of

D = (

,

)

= (2,2)

equation of AD =

⁄ ⁄

⁄ = ⁄

⁄ = ⁄

===>

⁄ = ⁄

===> y – 4 = 2x – 6

===> 2x – y = 2

PART –C

(maximum mark : 60)

A(3,4)

B(5,6) C(-1,-2) D(2,2)

Answer four full questions. Each question carries 15 marks.

III.

(a) If |

| = |

| find x

|

| = 2( -6 - 2) – (18 – 2 ) + x( 3 + 1 )

= -16 -16 + 4x

= 4x – 32

|

| = 8 – 3x

4x – 32 = 8 – 3x

7x = 40

x = 40/7

(b) If A is a square matrix, show that A + AT is symmetric and A - A

T is skew symmetric.

(A + AT)

T = A

T +(A

T)

T = A

T + A [ ( A + B )

T = A

T + B

T ]

A + AT

is symmetric

(A - AT)

T = A

T -(A

T)

T = A

T – A = -(A – A

T)

A - AT

is skew symmetric

(c) Solve the following system of equations using the inverse of coefficient matrix

x - y + z = 4

2x + y - 3z = 0

x + y + z = 2

AX = B

[

] = [ ] = [

]

First we find the inverse of A

| | = |

| = 1|

|+ 1|

|+ 1|

| = 10

Minors

= |

|= 4

=|

|= 5

= |

|= 1

= |

| = -2

= |

| = 0

= |

| = 2

= |

|= 2

=|

| = -5

= |

| = 3

Minor matrix = [

]

Cofactor matrix = [

]

Adjoint matrix = [

]

So inverse matrix =

| | =

[

]

Solution in X = A-1

B

[ ]=

[

][

]

[ ]=

[

]

= [

]

x = 2 y = -1 z = 1

IV.

(a) Solve for x if |

| = 0

|

| = 3(6 - 18) – (6x - 6x2) + 9(6x - 2x

2) = 0

= -36 -6x +6x2 + 54x – 18x

2 = 0

= -12x2 + 48x – 36 = 0

= 12(-x2 + 4x -3) = 0

= 12(x2 – 4x + 3) = 0

= x2 – 4x + 3 = 0

===> x = √

=

= 3,1

(b) If A = *

+ B = *

+ and C =[

]

Show that (A + B)C = AC + BC

(A + B)C = (*

+ *

+) [

]

= *

+ [

]

= *

+

AC + BC = *

+ [

] + *

+ [

]

1

= *

+ + *

+

= *

+

From and we have (A + B)C = AC + BC

(c) Find the inverse of A = [

]

| | = |

| = 3|

| + |

|+ |

|

= 3(2 - 3) + 2(4 + 4) + 3(-6 -4)

= -3 + 16 -30 = -17

| | = – 17

Minors

= |

| = -1

=|

| = 8

= |

| = -10

= |

| = 5

= |

| = -8

= |

| = -1

= |

| = -1

= |

| = -7

= |

| = 7

Minor matrix = [

]

2

1 2

Cofactor matrix = [

]

Adjoint matrix = [

]

So inverse matrix =

| | =

[

]

=

[

]

V.

(a) If nCn-2 = 210 find the value of ‘n’

nCn-2 = 210

we havenCn-r =nCr

===> nC2 = 210

===> ( )

⁄ = 210

===> n2 – n = 420

===> n2 – n – 420 = 0 n =

√

=

= 21, -20

n = 20 is admissible

n = 21

(b) Find the term independent of x in the expansion of (2x2 + 1/x

2)15

Tr+1 = ncr an-r

br

= 15cr (2x

2)15-r

(1/x)r

= 15cr (215-r

)(x30-2r

) x-r = 15cr (2

15-r (x

30-3r) = 15cr2

15-r x

30-3r

Then 30 – 3r = 0

===>3r = 30

===>r = 10

T11= 15c1025x

0

Term independent of x = T11= 96096

(c) Prove that

+

= 2cosecA

+

=

( )

( ) =

( )

=

( ) =

( )

( ) = 2cosecA

VI.

(a) Expand (x3 – 1/x

2)5 binomially

We know (a + b)n = a

n + nc1 a

n-1 b + nc2 a

n-2 b

2 + . . . . . . . + b

n

Thus (x3–

)5 = (x

3)5

+ 5c1( x3)4 (

) + 5c2(x3)3 (

)2

- 5c3 (x3)2 (

)3 + 5c4 (x

3)1 (

)4

+ (

)5

= x15

– 5 ×x12×

+ 10x9 x

-4 – 10x

6 x

-6 + 5x

3 x

-8 – x

-10

= x15

– 5x10

+ 10x5 – 10 + 5x

-5 – x

-10

(b) Find the coefficient of x10

in the expansion of (2x2 – 3/x)

11

Tr+1 = ncr an-r

br

Tr+1 = 11cr (2x2)11-r

(-3/x)r

= 11cr211-r

x22-2r(-

3)rx

-r = 11cr2

11-r(-3)

rx

22-3r

Now 22 – 3r = 10

-3r = -12

r = 4

T5 = 11c427(-3)

4 x

10 = 3421440x

10

Coefficient of x10

is 3421440

(c) If cotA= -

, 4

rd quadrant. Find all other trigonometric functions.

cotA= -

tanA = -

[ 4

rd quadrant.]

sinA = -

cosecA = -

cosA =

secA =

VII.

(a) Show that tan15o + cot15

o = 4 without using tables

tan15o = tan(45 – 30) =

=

√

√

=√

√ = 2-√

cot15o =

=

√ = 2+√

tan15o + cot15

o = 2-√ + 2+√ = 4

(b) Show that sin10o.sin50

o.sin70

o = 1/8

sin10 (sin50. sin70) = sin10 x

[cos120 – cos(-20) ]

=

sin10 [cos120 – cos20]

=

sin10 [-cos60 – cos20]

=

sin10 [-1/2 – cos20]

=

sin10 +

sin10 . cos20

=

sin10 +

.

[ sin30 + sin(-10)]

=

sin10 +

x

-

sin10

=

B C 8

17

15

(c) State and prove Napier’s formulae

Statement

in any ABC, tan(

)= (

).cotA/2

Proof:

Consider(

)cotA/2 =

.cotA/2

= ( )

( ) .cotA/2=

( )

( ). cotA/2

=

(

)

(

) (

). cotA/2

= cot( ) (

). cotA/2

= tan(

) . cot( –

) . cotA/2

= tan(

). . cotA/2

= tan(

)

VIII.

(a) Express √ cosx + sinx in the form Rsin(x+2)

√ 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

Squaring and adding &

3 + 1 = R2

sin2 + cos

2

4 = R2

===> R = 2

1

2

1 2

1 2

===>√ =

===> tan = √

===> = tan-1

(√ )

===> = 60o

(b) Derive expression for sin3A and cos3A

(i) Sin3A = sin (2A + A)

= sin2AcosA + cos2AsinA

= 2sinAcosA.cosA + (1 – 2sin2A)sinA

= 2sinAcos2A + sinA – 2sin

3A

= 2sinA (1 – sin2A) + sinA – 2sin

3A

= 3sinA – 4sin3A

(ii) 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

(c) Prove that bc.cosA + ca..cosB + .ab.cosC =

Using cosine rule,

a2 = b

2 + c

2 – 2bc.cosA

b2 = a

2 + c

2 – 2ac.cosB

c2 = a

2 + b

2 – 2ab.cosC

Adding , and we get

a2

+ b2 + c

2 = 2(b

2 + c

2+ a

2) – 2(bc.cosA +ac.cosB + ab.cosC)

===> 2(b2 + c

2+ a

2) – (a

2 + b

2 + c

2) = 2(bc.cosA + ac.cosb + ab.cosC)

1

2

3

1 2 3

===> 2(bc.cosA + ca.cosB + ab.cosC)

bc.cosA + ca..cosB + .ab.cosC =

IX.

(a) Solve the with a = 4cm, b = 5cm and c = 7cm

A = (

) = (

)

= (

)

= ( ) =34o03’

B = (

) = (

) = (

)

= ( ) = 44o25’

C = 180o – (A + B) = 180

o – (34

o03’ + 44

o25’) = 101

o32’

(b) The x intercept of a line is 3 times its y- intercept. The line passes through (-6,3). Find its

equation.

The intercept form of a line is

+

= 1

=1

Given that a = 3b

Passes through (-6,3) then,

= 1

= 1

===> 1/b = 1 ===> b = 1

Equation of the line is

+

= 1 or x + 3y = 1

(c) Find the value of ‘k’ for which the line :

1

1

3x + y – 2 = 0

kx + 2y – 3 = 0

2x – y – 3 = 0 is concurrent

It is concurrent

|

| = 0

= 3|

| - 1|

| - 2 |

| = 0

= 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0

= -27 + 3k – 6 + 2k + 8 = 0

5k – 25 = 0

5k = 25

K =

= 5

X.

(a) Solve ABC given a = 87cm , b = 53cm and C = 70o

tan(

) =

cot

A – B = 2tan-1[

cot

]

A – B = 2tan-1

[

cot35]

= 2tan-1

[

cot 35

o]

= 2tan-1

[ ]= 2 x 19o08

’ = 38

o16

’

A + B = 180 – 70 =110o

A = 148o16

’/2 = 74

o08

’

B = 110 - 74o08

’ = 35

o52

’

Now we have to find ‘c’, we have

=

=

===> c =

=

= 84.99cm

(b) Find the equation to the straight line passing through the point of intersection of the lines

2x – y – 3 = 0 and x – 2y + 1 = 0 and

(i) Parallel and

(ii) Perpendicular to the line x – y = 5

2x – y – 3 = 0

x – 2y + 1 = 0

x = |

|

|

| =

y = |

|

=

Case I

Equation of a line Parallel to the line x – y = 5 is x – y + k = 0

x – y + k passes through (

,

)

We have

+ k = 0

===> k = -2/3

Required equation is

x – y -

= 0

ie, 3x – 3y = 2

Case II: perpendicular to the line x – y = 15

Equation of the line perpendicular to x – y = 5 is –x – y + k = 0 which passing through (

,

)

We have

+ k = 0

===> k = 4

Required equation is

x – y + 4 = 0

or x + y = 4

(c) Find the point of intersection of the line 2x – 3y = 11 and 3x + 4y = 8

2x – 3y = 11

3x + 4y = 8

x = |

|

|

| =

=

= 4

y = |

|

=

=

= -1

1

2