TED (10)-1002 Reg. No. ………………………….

(REVISION-2010) Signature …….……………………

FIRST SEMESTER DIPLOMA EXAMINATION IN ENGINEERING/

TECHNOLIGY- MARCH, 2013

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 |

| = 0 find the value of x.

|

| = 0

===> 24x – 14 = 0

===> 24x = 14

===> x =

=

(b) If A = *

+ B = *

+ find 2A – 3B

2A – 3B = 2*

+ + 3 *

+ = *

+ - *

+ = *

+

(c) = find n

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

Then n = r + s

n= 10 + 15 = 25

(d) Evaluate cos and tan if sin = ½

Sin2 + cos

2 = 1

Cos2 = 1 – (1/2)

2

Cos2 = 1 – (1/4)

cos = √

tan =

√

(e) Find the slope of the line whose inclination to the x axis is 45o

Slope = tan = tan45o = 1

PART –B

Answer any five questions. Each question carries 6 marks

II.

(a) If A = *

+ show that AAT is symmetric.

A = *

+ AT = [

]

AAT = *

+ [

] = *

+

Clearly (AAT)

T = *

+

= *

+

AAT is symmetric.

(b) Solve using determinants:

3x + y – z = 3

-x + y + z = 1

x + y + z = 3

AX = B

[

] [ ] = [

]

x =

=

|

|

|

|

= ( ) ( ) ( )

( ) ( ) ( )

=

=

= 1

y=

=

|

|

= ( ) ( ) ( )

=

=

= 1

z=

=

|

|

= ( ) ( ) ( )

=

=

= 1

(c) Find the term independent of x in the expression of( )5

Tr+1= ncran-r

br

= 5cr(3x)5-r

(-y2)r

We have to find

T4= 5c3(3x)2(-y

2)3

= 5c3 .32 x

2 (-1)

3 y

6

= -90x2y

6

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

sin + sin + sin + sin [sinC + sinD =2.sin

]

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

= 2sin4 .cos3 + 2sin4 .cos

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

]

= 2sin4 . 2(cos2 . cos )

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

(e) Prove that

+

= 4cos2

+

=

+

= = ( )

=

(f) Find the equation of the line passing through the point (2 , -1) and (-6 , 3). Also find the

slope of the line.

Two points of a line is given by,

=

(x1, y1) = (2, -1)

(x2, y2) = (-6, 3)

=

=

-2(y+1) = x – 2

-2y – 2 = x – 2

x + 2y = 0

Slope of the line x + 2y = 0 is m =

=

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

5x + 2y – 4 = 0

2x + ky + 11 = 0

3x – 4y – 18 = 0

5x + 2y – 4 = 0

2x + ky + 11 = 0

3x – 4y – 18 = 0

Since the lines are concurrent

|

| = 0

5|

| – 2|

| – 4|

|= 0

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

-90k + 220 – 2 × -69 + 32 + 12k = 0

-78k + 220 + 138 + 32 = 0

390 = 78k

K =

= 5

PART –C

(Maximummark: 60)

Answer four full questions. Each question carries 15 marks.

III.

(a) If A=*

+and I unit matrix of same order, then find A3 -3A

2 + 2A + I

A3 = A

2.A

A3

= A2. A = *

+ *

+A

= *

+ *

+ = *

+

A3 -3A

2 + 2A + I = *

+ - *

+ + *

++*

+

= *

+

(b) If A = *

+ , B = *

+ show that (AB)-1

= B-1

A-1

AB = *

+ *

+ *

+

| | = |

| = 4

Cofactor matrix = *

+

Adj. (AB) = *

+

Inverse of AB =

| |= *

+

= [

]

A = *

+ | | = 4

Cofactor matrix of A = *

+

Adj. (A) = *

+

A-1

=

| |=

*

+

B = *

+ | | = 1

Cofactor matrix of B = *

+

Adj. (B) = *

+

B-1

=

| |= *

+

B-1

A-1

= *

+[

]

= [

]

(AB)-1

= B-1

A-1

(c) Find x if |

| = |

|

|

| = 2|

| - x|

| + 3|

|

=2(1 – 2) - x( 4 )+ 3( 4 )

= -2 - 4x + 12 = 10 -4x

|

| = 3x – 4

3x – 4 = 10 – 4x

===> 7x = 14

===>x = 2

IV.

(a) If A = *

+, B = *

+, show that( ) = +

=*

+ + *

+= *

+

( ) = *

+ 1

= *

+

= *

+

= *

+

= *

+

+ = *

+

From & it is clear that ( ) = +

(b) Find k, if the following system of equation are consistent

x + y + 1 = 0, x + 2y + 1 = 0, 2x+3y + k = 0

If the system is consistent then,

|

| = 0

1|

| - 1 |

| + 1|

| = 0

1(2k – 3) – (k – 2) + (3 - 4) = 0

2k – 3 – k + 2 – 1 = 0

k – 2 = 0

k = 2

(c) Solve using inverse of the coefficient matrix

x + y + z = 1,

2x + 2y + 3z = 6,

x + 4y + 9z = 3

AX = B

[

] = [ ] = [

]

Calculation for A-1

= |

| = 6

= |

| = 15

= |

| = 6

2

1 2

= |

| = 5

= |

| = 8

= |

| = 3

= |

| = 1

= |

| = 1

= |

| = 0

| | = -3

Minor matrix = [

]

Cofactor matrix = [

]

Adjoint matrix = [

]

So inverse matrix, A-1

=

| | =

[

]

X = A-1

B =

[

][ ]

=

[

]

x =

= 7

y =

-10

z =

= 4

V.

(a) If 20Cr = 20Cr+2find r

nCr =nCs ===> r = s or r + s = n

ie, r + r + 2 = 20

2r + 2 = 20

2r = 18, r = 9

(b) Find middle term in the expansion of (x2 +3/x)

20

Tr+1 = ncr an-r

br , n= 20

n + 1 = 21, odd.

(

)

= 22/2 = 11th

term is the middle term

T11 =20c10 (x2)10

(3/x)10

= 20c10x20

310

x-10

=20c10x10

310

= 20c10310

x20

(c) Prove that

= 2 - √

=

(

√ )

(

√ )=

√

√

(√ )

(√ )

(√ )

(√ )= (√ )

=

√

= 2 - √

VI.

(a) Expand (3x –

)4 binomially

(a + b)n = a

n + nc1 a

n-1 b + nc2 a

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

n

(3x –

)4 = (3x)

4 – 4c1 (3x)

3 (

) + 4c2 (3x)

2 (

) - 4c3 (3x)

1 (

) + 4c4(

)

= 81x4 – 4 ×27x

3(

) + 6 ×9x

2×(

) – 4 × 3 ×x×(

) + (

)

=81x4 – 54x

3y + (

)x

2y

2 – (

) +

(b) Find the constant term in the expansion of (√x +

)10

Tr+1 = ncr an-r

br

Tr+1 = 10cr (√x)10-r

(

)r

= 10cr

2rx

-2r

= 10cr

x-2r

2r

= 10cr

2r

= 10cr

2r

Then

= 0 ===> 10 – 5r = 0

===> r = 2

Therefore T3 = 10c222

= 45 x 4 = 180 is the constant term.

(c) Prove that

=

=

=

VII. (a)Prove that in ABC, a(sinB - sinC) = 0

in ABC, we know that sinB =

(by sine rule)

andsinC =

a(sinB - sinC) = a (

)

=

( a (b – c))

=

[a (b – c) + b (c – a) +c (a – b)]

=

[ab – ac + bc – ba +ca – cb]

=

x 0 = 0

(b)Prove that cos20.cos40.cos60.cos80 =

We have cos60 =

ie,

cos20.cos40.cos80 =

=

cos20

[cos120 – cos(-40)]

= ⁄ . cos20 [ + cos40]

= cos20 +

cos20.cos40

= . cos20 +

x (cos60 + cos20)

= . cos20 +

cos60 +

cos20

=

. cos60 =

x

=

(c )Show that sin120.cos330 + cos240.sin330 = 1

Sin120 = sin(1 x 90 + 30) = cos30 = √

Cos330 = cos(360 – 30) = cos(4 x 90 – 30) = cos30 = √

Cos240 = cos(270 – 30) = cos(3 x 90 – 30) = -sin30 =

Sin330 = sin(360 – 30) = sin(4 x 90 – 30) = -sin30 =

sin120.cos330 + cos240.sin330

= √

x

√

+

x

=

+

= 1

VIII.

(a) Express 3cos + 4sin 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

Squaring and adding &

3 + 1 = R2

sin2 + cos

2

4 = R2

===> R = 2

===>√ =

===> tan = √

===> = tan-1

(√ )

===> = 60o

(b) Prove that sin( A + B).sin(A – B) = sin2A – sin

2B

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

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

1

2

1 2

1 2

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

= sin2A.cos

2B – sinA.cosAsinB.cosB

+ sinAcosA.sinBcosB - cos2A.sin

2B

= sin2A.cos

2B - cos

2A.sins

2B

= sin2A(1 - sin

2B) – (1 - sin

2A)sin

2B

= sin2A-sin

2A.sin

2B -sin

2B - sin

2A.sin2B

= sin2A - sin

2B

(c) In any ABC, show that (b + c)sinA/2 = a.cos(

)

LHS = (a + b).sin

= (2RsinA + 2RsinB)sin

*

+

= 2R(sinA + sinB)sin

= 2R.2.sin

.cos

.sin

= cos

.4R.sin

.sin

= cos

.4R.sin (90 -

).sin

= cos

.2R.(2cos

.sin

)

= cos

.2R.sinC

= cos

.c = RHS

IX.

(a) Solve ABC if a = 5cm, B = 30o& c = 8cm

tan(

) =

cot

=tan

-1[

cot

]

=tan

-1[

cot

]

=tan

-1[

cot 15

o]

=tan

-1[ ] = -40.736

A – B = -81.473

A + B = 180 – 30 =150

Solving +

A = 34.2635 = 34O16’ B = 150 - 34.2635 = 115

O44’

Now we have to find ‘C’

We have

=

=

c =

× sin30o

= 4.44cm

(b) Find the slope and intercept of the line 3x + 4y = 12

Slope of 3x + 4y = 12 is

=

Intercept form of a line is

= 1

3x + 4y = 12

= 1

===>

= 1

X intercept = 4

Y intercept = 3

(c) Find k so that the lines kx + 2y – 10 = 0, 2x – 4y + 15 = 0 are

(i). Perpendicular to each other.

1

2

1 2

(ii). Parallel to each other.

(i). m1 x m2 = -1

x

= -1-1

= ===> 2k = 8 ===> k =4

(ii). m1 = m2

=

=

X.

(a) Solve using Napier’s formula, given a = 87cm, b = 53cm and C = 70o

tan(

) =

cot

A – B =2tan-1[

cot

]

A – B =2tan-1

[

cot35]

=2tan-1

[

cot 15

o]

=2tan-1

[ ]= 2 x 19o08

’ = 38

o16

’

A + B = 180 – 70 =110o

2A = 148o16

’/2 = 74

o08

’

A = B = 110

B = 110 - 74o08

’ = 35

o52

’

Now we have to find ‘c’, we have

=

=

===>c =

=

= 84.99cm

(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 = *

+

x = |

|

|

|

=

=

= -1

Y = |

|

|

|

=

=

= 0

Point of intersection = (-1, 0)

Given the line is,

x + y – 6 = 0

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

Perpendicular line is,

bx - ay + k = 0

x – y + k = 0

Passes through (-1, 0)

===> -1 – 0 + k = 0

K = 1

===> x – y + 1 = 0

1

1

1

1

(c ) A line passes through (-6 , 3 ). The X-intrecept of the line is 3 times its Y-intercept. Find the

. equation of the line.

a = 3 b (given)

The equation of a line is

+

= 1

Ie,

+

= 1……………….(1)

(1) Pass through (-6 , 3 )

(1) Implies,

+

= 1

+

= 1

= 1 , b = 1

(1) Implies

+

= 1