FHMM1124 General Mathematics II (Past Year Paper Answer) September 2010 1 (a) Z=92 , x =20 , y =16 (b) Laspeyres price index = 108.43 2 (a) Year 2006 2007 2008 Total Deviation Average deviation Adjustment Adjusted seasonal factor,S Q1 -2.00 -3.13 -5.13 -2.57 0.06 -2.63 Deviation (Y-T) Q2 Q3 0.87 4.62 1.37 6.50 11.12 2.24 5.56 1.12 0.06 0.06 5.50 1.06
Q4 -3.50 -4.25 -7.75 -3.88 0.05 -3.93
(b)(i)
(ii)
3
(a)(i)
(ii) Limit is exist (iii) f(x) is continuous at x =-2 (b)(ii) Area = (c)
4
(a)
Equation of normal :
(b)(i) f(x) is increasing at f(x) is decreasing at (ii) Local maximum point = Local minimum point =
1
FHMM1124 General Mathematics II (Past Year Paper Answer) December 2010 1 (a) g=34 , u =2 , v =6 (b)(i) Aggregate price Index for 2008=90.48 Aggregate price Index for 2009=112.38
(ii) Laspeyres price index = 89.53 2 (a) Week/ Day 2006 2007 2008 Total Deviation Average deviation Adjustment Adjusted seasonal factor,S Mon -12.20 -13.80 -26.00 -13.00 0.06 -13.06 Deviation (Y-T) Tue Wed 2.00 14.80 1.40 17.20 0.80 32.00 4.20 16.00 1.40 0.06 0.06 15.94 1.34
Thu 27.6 29.00 56.60 28.30 0.06 28.24
Fri -31.60 -33.20 -64.80 32.40 0.06 -32.46
(a)(iv) Yp=98 (b)(i)
(ii)
3
(a)(i)
(ii) Limit does not exist. (iii) f(x) is discontinuous at x = 0 (iv) c = 8 (b)(i) f(x) is increasing at f(x) is decreasing at (ii) Local maximum point = Local minimum point = (iv) Absolute maximum point =155 Absolute minimum point = -53
2
FHMM1124 General Mathematics II (Past Year Paper Answer) 4 (a) (b) Equation of tangent :
(c)
Area =
April 2011 1 (a) Z=25 , x =5 , y =5 (b) Paasche price index = 110.07 Laspeyres quantity index = 127.47 2 (a) Year 2005 2006 2007 Total Deviation Average deviation Adjustment Adjusted seasonal factor,S (ii) Q1 15.50 56.88 72.38 36.19 0.13 36.06 Deviation (Y-T) Q2 Q3 -35.75 -51.13 -53.88 -72.25 -123.38 -89.63 -61.69 -44.82 0.13 0.13 -61.82 -44.95
Q4 68.50 73.13 141.63 70.82 0.13 70.69
3
(a)(i)
(ii) Limit is exist (iii) f(x) is discontinuous at x =1. (b) (c) 4 (a)(i) Absolute maximum point = 0 Absolute minimum point = -9 Equation of normal : f(x) is increasing at f(x) is decreasing at
(ii) Local maximum point = Local minimum point =
3
FHMM1124 General Mathematics II (Past Year Paper Answer) 4 (b)
(c)(ii) Volume = September 2011 1 (a) g=560 , u =26 , v =16 (b)(i) Laspeyres price index = 123.65 (ii) Paasche quantity index = 103.59 2 (a) Year 2005 2006 2007 Total Deviation Average deviation Adjustment Adjusted seasonal factor, S (iii) Yp=RM25886 (b)(i) (ii) (iii) (iv) x=2, x=5 Q1 -2.50 0.50 -2.00 -1.00 0.07813 -1.07813 Deviation (Y-T) Q2 Q3 9.25 0.125 8.25 -1.50 -1.375 17.50 -0.6875 8.75 0.07813 0.07813 -0.76563 8.67187
Q4 -5.75 -7.75 -13.50 -6.75 0.07813 -6.82813
3
(a)(i)
(ii) Limit is exist (iii) f(x) is continuous at x =0. (iv) c = 2
(b)(i)
f(x) is increasing at f(x) is decreasing at
4
FHMM1124 General Mathematics II (Past Year Paper Answer) (ii) Local maximum point = Local minimum point = 4 (a) (b) Equation of tangent :
(c)(ii) Area =
December 2011 1 (a) g=320 , u =8 , v =8 (b)(i) Laspeyres price index = 126.79 The price has increase 26.79% in year 2011 compare to year 2010. Laspeyres quantity index = 91.96 The quantity has decrease 8.04% in year 2011 compare to year 2010. (ii) Paasche quantity index = 94.37 The quantity has decrease 5.63% in year 2011 compare to year 2010.
2
(a) Year 2009 2010 2011 Total Deviation Average deviation Adjustment Adjusted seasonal factor, S (iii) Yp=14.53 ≈RM15 million (b)(i) ∞ (ii) ∪ Q1 77.1513 77.5281 154.6794 77.3397 0.9996 77.3088
Deviation (Y-T) Q2 Q3 117.9941 54.0382 112.9683 55.3250 109.3632 230.9624 54.6816 115.4812 0.9996 0.9996 54.6597 115.4350
Q4 150.2959 154.9858 305.2817 152.6409 0.9996 152.5798
(c)
is exist because
is a one to one function.
5
FHMM1124 General Mathematics II (Past Year Paper Answer)
3
(a)(i)
(ii) Limit is exist.
(iii) f(1)=2 f(x) is discontinuous at x =1.
(b)(i)
f(x) is increasing at f(x) is decreasing at
(ii) Local maximum point = Local minimum point = 4 (a)
(b)
*Change the question become: Use implicit differentiation to find the tangent line to the curve Equation of tangent :
for
. Hence, find an equation of at the point (-2, -2).
(c)(ii) Area =
April 2012 1 (a) C=160 , u =1 , v =4 (b)(i) Laspeyres quantity index =83.33 The quantity has decrease 16.67% in July 2011 compare to January 2011. (ii) Paasche price index = 116.30 The price has increase 16.30% in July 2011 compare to January 2011.
6
FHMM1124 General Mathematics II (Past Year Paper Answer) 2 (a) Year 2007 2008 2009 Total Deviation Average deviation Adjustment Adjusted seasonal factor, S (iii) Yp=41.15 ≈RM42000 (b)(i) Q1 -29 -31.75 -60.75 -30.375 -0.3438 -30.0312 Deviation (Y-T) Q2 Q3 45.25 -8 55.25 -8.25 -16.25 100.50 -8.125 50.25 -0.3438 -0.3438 -7.7812 50.5938
Q4 -11 -15.25 -26.25 -13.125 -0.3438 -12.7812
(ii) (iii) 3 (a)(i) (ii) Limit does not exist.
(iii) f(-3)=0 f(x) is discontinuous at x =-3. (b)(ii) (c) Area =
4
(a) (b)(i)
Equation of normal : f(x) is increasing at f(x) is decreasing at (ii) Local maximum point = Local minimum point =
7
FHMM1124 General Mathematics II (Past Year Paper Answer) September 2012 1 (a)(i) Maximize C = 45x + 50y + 35z Subject to 4x + 5y + 3z ≤ 7 3x + 2y + z ≤ 4 x,y,z≥0 (ii) Standard form: Maximize C - 45x - 50y -35z = 0 Subject to 4x + 5y +3z + u =7 3x + 2y + z + v = 4 x,y,z,u,v≥0 Not optimized. Z=75, x = 10 , y =5 (c)(i) Laspeyres price index =126.53 The price has increase 26.53% from year 2005 to year 2011.
(b)
(ii) Paasche price index = 126.38 The price has increase 26.38% from year 2005 to year 2011.
2
(a)(i) Year 2009 2010 2011 Total Deviation Average deviation Adjustment Adjusted seasonal factor, S (ii) Quarter 3: Yp= 812.58 thousands Quarter 4: Yp= 902.94 thousands (b)(i) (ii) Q1 -38.75 -46.25 -85.00 -42.50 0.47 -42.97
Deviation (Y-T) Q2 Q3 20.00 -76.25 17.50 -75.00 -151.25 37.50 -75.63 18.75 0.47 0.47 -76.10 18.28
Q4 101.25 101.25 202.50 101.25 0.47 100.78
8
FHMM1124 General Mathematics II (Past Year Paper Answer) 3 (a)(i)
(ii) k = 9 (iii)
(b) (c)(i)
Equation of normal :
(ii)
4
(a)(i)
f(x) is increasing at f(x) is decreasing at
(ii) Local maximum point = Local minimum point = (iii) Absolute maximum value =33 Absolute minimum value =-51 (b) Area =
9 