# MATH 123 Fall 1998 Final Exam Solutions - UNIVERSITY OF...

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UNIVERSITY OF SASKATCHEWANDepartment of Mathematics & StatisticsMathematics 123.3 Final Examination9:00 A.M., Friday, December 11, 1998Time:3 hoursInstructors:Browne, Chigogidze, MacLean, SongCLOSED BOOK — NO CALCULATORS PERMITTEDThe first part of this examination consists of 12 multiple choice questions worth 5% each. Submit youranswers to these on the blue opscan sheet provided.The second part of this examination consists of 4 questions worth 10% each. Show all your work: nocredit will be given for unsubstantiated correct answers. Submit your answers to these on one of thebooklets provided.PART I.(Multiple Choice)The possible answers to the following two questions are in:ANSWER SET I.(A) (-6,6)(B) (4,6](C)[-6,)(D)[-4,)(E)(-∞,4](F)(-∞,6](G)[0,)(H)[4,)(I)[6,)(J) None of these1. Find the range of the functionf (x)= |x-1| +x-1.
2. The interval on whichf (x)=x6-xis decreasing is:
1
The possible answers to the following six questions are in:ANSWER SET II.(A) 0(B) 1(C) 2(D) 3(E) 4(F) 5(G) 6(H) 7(I) 8(J) None of these3. They-intercept of the line passing through the point (6,5) and perpendicular to the liney=2x+17is:Solution:The required line has slope-12, so its equation isy-5= -12(x-6).Whenx=0,y-5= -12(0-6)=3, soy=8.Answer: (I) 84. Find limx→∞2s2x2-1x+8x2.Solution:limx→∞2s2x2-1x+8x2=limx0+2vuuuut21x2-11x+81x2=limx0+2vuuut2-x2x2x+8x2=limx0+2s2-x2x+8=2s2-020+8=2s14Answer: (B) 15. The absolute maximum value off (x)=x2-4x+1 over the interval [0,4] is:Solution:The graph offis a parabola which is concave up.Thus the maxima must occur at the endpoints of [0,4]. The value off (x)atx=0 and atx=4 is 1.