Some Integrals which cannot be found
1) sinx/x dx
2) ∫ cosx /x dx
3) ∫ dx/log x
4) ∫ √(1 - k²sin²x) dx
5) ∫ √(sinx) dx
6) ∫ sin(x²) dx
7) ∫ cos(x²) dx
8) ∫ xtan x dx
9) ∫ ₑ-x² dx
10) ∫ ₑx² dx
11) ∫ x²/(1+x⁵) dx
12) ³√(1+x²) dx
13) ∫ √(1+x³) dx
TIME SERIES ANALYSIS
TIME SERIES
A) Find the moving average of
1) 1, 0, -1, 0, 1,0, -1, 0,1 4yrs..
2) Year: 1 2 3 4 5 6 7 8 9 Sales: 36 43 43 34 44 54 34 24 15
4 years centred moving average.
3) 5 yrly moving average:
Year: 1 2 3 4 5 6 7 8 9 10
Sales:32 17 57 92 2 5 10 27 5 31
4) 3 year moving average:
Year: 1 2 3 4 5 6 7
Values: 2 4 5 7 8 10 13
5) Using 3-year moving average method determine the trend and short-term fluctuations::
Year: 1 2 3 4 5 6 7
Values: 21 30 40 25 40 55 70
6) Find the 3-year weighted moving average with weights 1,4,1 for the following series:
Year : 1 2 3 4 5 6 7
Values: 2 5 6 5 8 3 3
7) Year production ('0000 tonnes)
1995 506
1996 620
1997 1036
1998 673
1999 588
2000 699
2001 1116
2002 783
2003 663
2004 773
2005 1189
Find the 4 years moving average.
B) Use the method of least square
1) Year: 01 02 03 04 05
Prod: 10 12 8 10 14
Use the method of least squares to fit a straight line to the data. Also find the trend values for different years. 10.8+ 0.6x; 9.6, 10.2, 10.8, 11.4, 12.0
2) Year: 00 01 02 03 04
Insured: 11.3 13 9.7 10.6 10.7Find the insuranced people in 1997? Y=11.06 - 0.4x; 8.18.
3) Yr: 1991 92 93 94 95 96 97
Sales:125 128 133 135 140 141 143
Find the trend values:. Y=135+ 3.107x, 125.679, 128.786, 131.893, 135, 138.107, 141.214, 144.321.
4) Yr: 1980 81 82 83 84 85 1986
Profit: 60 72 75 65 80 85 95
Estimate the profit for 1987. Y= 76+ 4.857x; 95.428 lacs.
5) Obtain the trend Equation and tabulate against each year after estimation of the trend and short-term fluctuations.
Year Value
1997 380
1998 400
1999 650
2000 720
2001 690
2002 620
2003 670
2004 950
2005 1040 y= 680 + 70.5x; -18.0, -68.5, 111.0, 110.5, 10.0, -130.5, - 151.0, 58.5, 78.0.
6) Determine the equation of a straight line which best fits the following data:
year:. 01 02 03 04 05
Sales: 35 56 79 80 40
Compute the trend values for all the years from 2001 to 2005. Y= 58 + 3.4x, 51.2, 54.6, 58.0, 61.4, 64.8
7) Fit a suitable straight line to the following data by the method of least squares and estimate the percentage of insured people in 1997:
Year: 1989 1990 1991 1992 1993
Ins: 11.3 13.0 9.7 10.6 10.7
Y= 11.06- 0.36x, 8.9
8) The following table gives the annual profits(in thousand ₹) in a factory:
Year profit ('000 ₹)
1991 60
1992 72
1993 75
1994 65
1995 80
1996 85
1997 90
a) fit a straight line trend by the method of least squares. Y=75.29 + 4.32x
b) find the gradient of the fitted trend line. 4.32
c) calculate the projected profit for 1998. 92570
C)
1) Year: 1994 95 96 97 98 99
Sale : 12 15 17 22 24 30
Estimate the volume of sale for 2000. Y=20+1.74x; 32.2 units.
2) yr: 1997 98 99 00 01 02
Price: 250 207 228 240 281 300
Find the trend value of 2004: y= 266.33 + 13.49x; 387.74.
3) Year. Value
1996 380
1997 400
1998 650
1999 720
2000 690
2001 600
2002 870
2003 930
Estimate the value of 2005. Y= 655+ 35.83x referred to midpoint of 1983-84 as origin and unit of x as 6 months; 1049.13.
4) Fit a straight line trend equation by the method of least squares and estimate the trend values:
Year values
1991 80
1992 90
1993 92
1994 83
1995 95
1996 99
1997 92
1998 104
Y= 91.75 + 1.25x, 83, 85.5, 88, 90.5, 93, 95.5, 98, 100.5
5) The weight of a calf taken at weekly intervals are given:
Age in weeks(x) Weight (y)
1 52.5
2 58.7
3 65
4 70.2
5 75.4
6 81.1
7 87.2
8 95.5
9 102.2
10 108.4 Fit a straight line using the method of least squares and estimate the average rate of growth per week. Y= 45.73 + 6.162x, 6.162.
D)
1) year: 2005 10 15 20 25 2030
Profit: 10 13 15 20 22 28
Find the profit for 2035. Y= 18.1.743x; x=2.5yrs; 30.20 lakh.
2) Yr: 1995 97 99 2001 2003
Sales: 18 21 23 27 16
Estimate the sales in 2000, and 2005. Y= 21+ 0.1x; 21000, 21600
E) Find the Seasonal indices and seasonal Index of following using the method of average:
1) Year: Quarters
I II III IV
2003 37 38 37 40
2004 41 34 25 31
2005 35 37 35 41
1.75, 0.42, -3.59, 1.42
2) Year: Quarters
I II III IV
2002 39 21 52 81
2003 45 23 63 76
2004 44 26 69 75
2005 53 23 64 84
86.4, 44.4, 118.4, 150.8
3) year I II III IV
2001 72 68 80 70
2002 76 70 82 74
2003 74 66 84 80
2004 76 74 84 78
2005 78 74 86 82
96.4, 92.1, 106.9 100.5
4) Year 1st 2nd 3rd 4th
2002 16.00 13.50 14.70 17.00
2003 15.90 12.20 15.60 18.10
2004 16.30 11.90 16.90 19.20
2005 17.10 13.20 15.00 18.70
0.77, -3.31, 0.11, 2.43
5) Year I II III IV
2001 71 68 79 71
2002 76 69 82 74
2003 74 66 84 80
2004 76 73 84 78
2005 78 74 86 82
-1.25, -6.25, 6.75, 0.75
6) year. I II III IV
2000 75 60 54 59
2001 86 65 63 80
2002 90 72 66 85
2003 100 78 72 93
117.20, 91.82, 85.14, 105.84
F) Find the seasonal Indices by the method of moving averages from the following:
1) Year: Quarters
I II III IV
2002 97 100 106 110
2003 88 93 96 101
2004 76 79 83 88
2005 94 98 103 106
10.12,0.13, -14.08, 3.83
2) Year: Quarters
I II III IV
1990 34 32 31 36
1991 37 34 33 41
1992 43 40 38 48
33.625, 34.25, 34.75, 35.625, 37, 38.5, 39.875, 41.375
3) Year: Quarters
I II III IV
2002 101 93 79 98
2003 106 96 83 103
2004 110 101 88 106
110.9,99.9, 84.9, 104.3
4) Year: Quarters
I II III IV
2002 65 58 56 61
2003 58 63 63 67
2004 70 59 56 52
2005 60 55 51 58
2.83, - 0.50, -2.17, -0.17
5) Year: Quarters
I II III IV
2002 75 60 54 59
2003 86 65 63 80
2004 90 72 66 85
2005 100 78 72 93
122.36, 92.43, 84.70, 100.51
G) 1) A company estimates it's average monthly sales in a particular year to be ₹200000. The seasonal Indices (SI) of the sales data are as follows:
Month SI
Jan 76
Feb 77
Mar 98
Apr 128
May 137
June. 122
2) A company estimates it's average monthly sales in a particular year to be ₹2000000. The seasonal Indices (SI) of the sales data are as follows:
Month SI
Jan 78
Feb 75
Mar 100
Apr 126
May 138
June. 121
July 101
Aug 104
Sept. 99
Oct 103
Nov 80
Dec 75
Ignoring the possibile existence of a trend, use the above information to draw up a monthly sales budget for the company. 15.6, 15.0, 20.0, 25.2, 27.6, 24.2, 20.2, 20.8, 19.8, 20.6, 16.0, 15.0
3) A company estimates its average monthly sales in a particular year to be ₹2000000. The seasonal Indices (SI) of the sales data are as follows:
Month SI
Jan 76
Feb 77
Mar 98
Apr 128
May 137
June. 122
July 101
Aug 104
Sept. 100
Oct 102
Nov 82
Dec 73
Using this information, draw up a monthly sales budget for the company. (Assume that there is no trend) 152000, 154000, 196000, 256000, 274000, 244000, 202000, 208000, 200000, 204000, 164000, 146000.
1) a) Find the inverse of. 2 -2
4 3
(2)
b) If f: N -> R be a function defined
as f(x) = 4x²+12x+15, show that
f: N -> S, where S is the range of f
is invertible.
Find the inverse of f. (2)
c) eˣ+eʸ=eˣ⁺ʸ prove y₁= - eʸ⁻ˣ. (2)
d) prove. (2)
tan⁻¹ 1/2+ tan⁻¹1/5 + tan⁻¹1/8=π/4
e)Evaluate limₓ-₀ tan8x/sin 2x. (2)
2) using the properties of Determinant find
1 a a²-bc
1 b b²-ca
1 c c²-ab (4)
3) show sin⁻¹12/13+cos⁻¹4/5 +
tan⁻¹63/16 =π (4)
4) If eʸ(x +1) =1 show y₂ =( y₁)² (4)
5) i) ∫ x²eᵃˣ dx. ii) ∫ xlogx dx. OR
∫ dx/(x³+x²+x+1) (4)
6) Find the point on the curve
y= x³-11x+5 at which the
equation of the tangent is
y= x -11 ( 4)
OR
If f(x) = (4x+3)/(6x -4) , x≠2/3. what is the inverse of f.
7) If A= 2 -3 5
3 2 -4
1 1 -2 , find A⁻¹ . Using A⁻¹, solve the following system of equations. 2x-3y+5z=16, 3x+2y-4z=-4, x+y-2z=-3 (6)
OR
If A = 1 2 3
2 3 1
-1 1 1 using elementary transfirmation, find A⁻¹ and verify A⁻¹A = I = AA⁻¹.
8) Show that the semi-vertical angle of the cone of maximum volume and of given slant height is tan⁻¹√2
OR (6)
Find the area of the great rectangle that can be inscribed in the ellipse
x²/a² + y²/b² = 1.
9) ∫ (tanθ +tan³θ)/(1+tan³θ)dθ. (6)
10) a) The demand functiin of a
monopolist is given by
p= 100 -x-x².
find
i) the revenue function.
ii) marginal revenue function. (2)
11) A furniture dealer deals in only two items: tables and chairs. He has Rs20000 to invest and a space to store at most 80 pieces. A table costs him Rs500 and a chair costs him Rs200. He can sell a table for Rs950 and a chair for Rs280. Assume that he can sell all the items that he buys. Formulate this problem as an LPP so that he can maximize his profit. (6)
SUBMULTIPLE ANGLE
+--++++++----------------
A)Prove
1) cot(x/2) -tan(x/2) = 2cotx
2) cos⁴(x/2) - sin⁴(x/2) = cos x
3) (1+cosθ)/sinθ = cot(θ/2)
4) (sinα/2 ⊕ cosα/2)² = 1⊕ sinα
5) sin2α/(1-cos2α).(1-cosα)/cosα
= tanα/2
6) (1+sinα -cosα)/(1+sinα +cosα)
= tan(α/2)
7) 2cscα = tan(α/2) + cot(α/2)
8) sin2θ/(1+cos2θ). cosθ/
(1+cos2θ)
= tan(θ/2)
9) (tanx+secx -1)/(tanx -secx +1)
= tan(π/4 +x/2)
10) secα+tanα =tan(π/4+ α/2)
11) secα - tanα = tan(π/4 -α/2)
12) tan(π/4 +α/2)
= √{(1+sinα)/(1-sinα)
= secα -tanα
13) tan(α+β)/2 + tan(α-β)/2 =
2sinα/(cosα +cosβ)
14) cotx = ½(cot(x/2) - tan(x/2))
15) 8sin⁴(α/2) - 8 sin²(α/2)+1
= cos2α
16){sinα/2-√(1+sinα)}/
{cosα/2-√(1+sinα)
= cot(α/2)
17) (2sinθ -sin2θ)/(2sinθ +sin2θ)
= tan²(θ/2)
18) (1+tan(α/2))/(1-tan(α/2)) =
(1+sinα)/cosα
19) sinα =2tan(α/2)/(1+tan(α/2))
20) cosα + cosβ)²+(sinα +sinβ)² =
4cos²(α-β)/2
21) 2cos(π/6)=√[2+{√2+√(2)}]
22) tan6 tan42 tan66 tan78 =1
23) sin(15/2) =√6 - √3 + √2 -2
24) 2sinx/2= ⊕√(1+sinx) ⊕
√(1 -sinx)
B) Evaluate
1) sin²72 - cos²30
2) cosπ/5 + cos 3π/5
3) sin 45/2
4) tan 44/2
5) sin15
6) cos 15
7) sin 105
8) 4cos9
9) sin54 + cos72
10) sin18 + cos 36
11) sin²36 + cos² 18
12) cos² 66 - sin² 6
13) cos²48 - sin²12
14) 8cos²20 - 6 cos20
15) 3sin40 - 4 sin³ 40
16) sin²24 - sin²6
C) If tan(α/2) = √{(1-e)/(1+e)} tan(α/2)
then show cosα= (cosα -e)/
(1-ecosα)
D) tanα = sinαsinβ/(cosα+cosβ) then show one of the values of tan(α/2) is
tan(α/2) tan(β/2)
E) If sinα +sinβ =a and cosα+cosβ=b
then find the value if cos(α+β)
F) If A= 320 prove
tanA/2={-1 + √(tan²A)}/tanA
FORMULAE:
1) sin 2x= 2sin x cos x
2) Sin 3x = 3 sin x - 4 sin³x
3) Cos 2x = cos²x - sin²x
= 2 cos²x - 1
= 1 - 2 sin²x
= (1- tan²x)/(1+ tan²x)
4) 1 + cos 2x = 2 cos²x OR
Cos x= ± √{(1+ cos 2x)/2}
5) 1 - cos 2x= 2 sin²x OR
Cos x= ± √{(1- cos 2x)/2}
6) Cos 3x = 4 cos²x - 3 cos x
7) Tan x= (1- cos 2x)/sin 2x
= Sin 2x/(1+ cos 2x)
=± √{(1- cos 2x)/(1+ cos 2x)
8) Tan 2x = 2 tanx/(1- tan²x)
9) tan 3x=(3tan x -tan³x)/(1-3 tan²x)
10) sin18°=(√5 -1)/4=cos72°= sin π/10
11) cos 36°= (√5+2)/4= sin54°= cos π/5
12) sin15°= (√3-1)/2√2 = cos75°= sin π/2
13) cos15°= (√3+1)/2√2 = sin75°= cos π/2
14) tan π/12 = 2 -√3 =(√3-1)/(√3+1) = cot 5π/12
15) tan 5π/12 = 2 + √3 =(√3-1)/(√3+1) = cot π/12
16) tan 225°= √2 - 1 = cot 67.5°= cot 3π/8 = tan π/8
17) tan 67.5°= √2+ 1 = cot 22.5°
========================
EXERCISE -1
___________
1) Find Sin 2θ, cos2θ, tan2θ if
a) cosθ=⅗. 24/25, -7/25, -24/7
b) sinθ=⅘. 24/25, -7/25, -24/7
c) tanθ=½. 4/5, 3/5, 2
d) if cosA=⅗ find Sin2A
e) if sinA=⅗ find tan3A
f) If cos2a =24/25 find tan3A
2) EXPRESS:
a) cos4A in terms of sinA. 1 - 8 sin²A + 8 sin⁴A
b) tan4A in terms of tanA. (4 tanA - 4 tan⁴A)/(1+tan⁴A-6 tan²A)
c) sin4A in terms of tanA.
3) Prove:
a) Sin2α/(1-cos2α)= cotα
b) sin2θ/(1+cos2θ) = tanθ
c) cotβ + tanβ = 2csc2β
d) 1+ tanα tan2α = sec2α
e) tan²α = (sec2α -1)/(sec2α +1)
f) sec²θ(1+sec2θ) = 2sec2θ
g) (cosα+sinα)/(cosα-sinα) = tan2α + sec2α
h) tanβ(1+sec2β)=tan2β
i) tan2α/tanα = 1+sec2α
j) (cotθ+tanθ)/(cotθ-tanθ)=sec2θ
k) 2/(cotα-tanα) = tan2α
l) {sinα - √(1+sin2α)}/{cosα-√(1+sin2α)= cotα
m) tan(π/4 + α)=(1+sin2α)/cos2α
n) cos²(α+π/4)+cos²(α-π/4)=1
o) {(1-tanθ)/(1+tanθ)}² = (1-sin2θ)/(1+sin2θ)
p) tan(45+α)- tan(45-α) =2tan2α
q) tan(45+α)+ tan(45-α) =2sec2α
r) sec(45+α)sec(45-α) = 2sec2α
s) 4cosβcos(2π/3 +β) cos(4π/3+β) = cos(3β)
t) tan(π/4 +α/2)= secα +tanα = √{(1+sinα)/(1-sinα)}
u) (cotθ-tanθ)/(1-2sin2θ) = secθ cosec 2θ
v) cot15 - tan15 = 2tan60
w) tan15 + cot15 =4
x) (2cosα+1)(2cosα -1)=2cos2α +1
y) cos⁴α - sin⁴α = cos2α
z) cos⁶β -sin⁶β = cos2β(1- ¼ sin² 2β)
A) (sin²α-sin²β)/(sinα cosα-sinβcosβ) = tan(α+β)
B) (1-cos2α+sin2α) /(1+cos2α+sin2α) = tanα
C) cos³xcos3x+sin³xsin3x=cos³2x
D) {(cosα+sinα)/(cosα-sinα)} -
{(cosα-sinα)/(cosα+sinα)}=2tan2α
E) (sinα+sin2α)/(1+cosα+cos2α) = tanα
F*) (sinα/cosα) (1-cos2α)/(1-cos4α) = tanα
G) (cos³α+sin³α)/(cosα +sinα) =1 - ½ sin2α
H) 1/sin10 - √3/cos10 = 4
I) cotα/(cotα -cot3α) = tanα/(tan3α - tanα))
J) 1/(tan3α-tanα)- 1/(cot3α -cotα) = cot2α
K) sin8β=8sinβcosβcos2βcos4β
L) cos5θ = 16cos⁵θ - 20cos³θ+ 5cosθ
M) sin5α=16sin⁵α-20sin³α+5sinα
N) cos(120-α)+cosα+cos(120+α)=0
O) cos²(α-120) +cos²α +cos²(α+12) = 3/2
P) tan4α=(4tanα-4tan³α)/(1-6 tan²α+tan⁴α)
Q) (2cos2ⁿθ+1)/(2cosθ+1)=
(2cosθ-1)(2cos2θ-1)........ (2cos2ⁿ⁻¹θ-1)
R) tan2ⁿθ/tanθ = (1+sec2θ)(1+sec2²θ) ...(1+secⁿθ)
S) sin⁴α=⅜ - ½ cos2α +⅛ cos4α
T) cos⁸α+sin⁸α= 1-sin²2α+⅛ sin⁴2α
U) tan(π/4 +α)+tan(π/4 -α) =2sec2α
V) cos³α (sin3α)/3 +sin³α(cos³α)/3= (sin4α)/4
W) cos4α - cos4β= (cosα-cosβ)
(cosα+cosβ)(cosα-sinβ(cosα+sinβ)
X) tanθ+ 2tan2θ +4tan4θ +8tan8θ= cotθ
Y) (2cos8α+1)/(2cosα +1) =
(2cosα-1)(2cos2α-1)(2cos4α-1)
Z) cos²(α-β) - sin²(α+β) = cos2αcos2β
a') 4(cos³7+sin³23) =3(cos7 +sin23)
EXERCISE-2
-------------------
1) if tanx =b/a, find acos2x+bsin2x
2) If cosx +sinx =√2 cosy show that tan2y =1
3) If sin²x = siny cosy then show cos2x = 2cos(π/4 + y)
4) If cos2a=x+1/x show 2cosa =(x³+ 1/x³)
5) If tan²x 2tanx tan2y =
tan²y+2tanytan2x prove that each side =1 or tanx = ± tany
6) If tan²θ = 1+ 2 tan²α show cos2α = 1+ 2cos2θ
7) If 2 tanα = 3tanβ show tan(α-β) =sin2β/(5 -cos2β)
8) If tan(α+β-θ)/tan(α-β+θ) =
tanθ/tanβ then show that either
sin(β-θ)=0 or sin2α+sin2β+sin2θ=0
9) If α and β are acute angles and cos2α = (3cos2α-1)/(3-cos2β) then show that tanα =√2 tanβ
10) If cosα = ½(a +1/a) show that
a)cos2θ= ½(a²+ 1/a²)
b) cos3θ= ½ (a³ +1/a³)
11) If tanα = sec2α, then prove that
sin2α = (1-tan⁴α)/(1+tan⁴α)
12) If csc2α+csc2β +csc2θ = 0 show that tanα + tanβ+ tanθ + cotα + cotβ + cotθ=0
13) If tan³x=2tan²y+1 then show that cos²2x + sin²y = 0
COMPOUND INTEREST
1) Find the difference between Simple interest and compound interest on Rs5000 invested for 4 years at 5% p.a., compounded yearly.
2) What rate of interest p.a. does a person get who is paid at the rate of 5% CI payable half yearly.
3) In 2 years a sum of money double itself in compound interest. Find the rate percentage
4) What principal will amount to Rs551.25 in 2 years at 5% compound interest ?
5) A sum becomes 1323 in 2 years at compound interest compounded yearly. Find the rate percentage.
5) If the population of a Town increases every year by 2% of the population at the beginning of that year, in how many years will the total increase of population be 10%
6) In how many years will the population of a village change from 2500 to 2601, if the rate of increase is 2% p.a ?
7) The difference between Simple and compound interest on a sum for 3 years at 5% p.a is Rs76.25. Find the sum.
8) The sum of 956 amounts to Rs1099.40 in 3 years, at simple interest What sum of money invested at compound interest, payable yearly, will amount to TmRs1481.97 in 3 years, the rate of interest p.a. in both the cases being the same ?
9) A sum of money invested at compound interest, payable yearly, amounts to Rs 2704 at the end of the second year and to Rs2812.16 at the end of the third year. Find the rate of interest and the sum.
10) The difference between the Simple and compound interest at the same rate for 2 years on a certain amount is 1/100th of the amount. Find the rate of interest.
11) A man deposits Rs5000 in a Savings account which pays compound interest at the rate of 4.5% for the first 2 years and at the rate of 5% for 1 year. Find the amount after 3 years.
12) Find the compound interest on Rs 6950 for 3 years, if interest is payable half-yearly, the rate for the first two years being 6% and for the third year 9%p.a.
13) A sum of money invested at compound interest amount to Rs10816 at the end of the second year and to Rs11248 at the end of third year, find the rate of interest and the sum invested.
14) A sum of money invested at compound interest amounts to Rs 17640 at the end of the second year and to Rs18522 at the end of third year. Find the sum originally invested and the rate of interest.
15) A sum of money is lent at 8% p.a compound interest. If the interest for the second year exceeds that for the first year by Rs32, find the original principal.
16) What is the present value of a reversion to Rs3000 after 7/2 years reckoning 4% C. I payable yearly.
17) A loan earns interest at a certain rate(compound) percent p.a. Four years ago amount was Rs81; now it is Rs144. What will the amount be two years hence ?
