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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