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)