MATENB1 Sick test 1memofinal

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MEMODEPARTMENT OF MATHEMATICS AND APPLIED MATHEMATICSMODULEMATENB1APPLICATIONS OF CALCULUS FOR ENGINEERSCAMPUSAPKASSESSMENTSEMESTER SICK TEST 1 memorandumDATE 11/09/2023TIME 08:00ASSESSORSDR S SINGHDR K SEBOGODIDURATION 90 MINUTESMARKS 50SURNAME AND INITIALSSTUDENT NUMBERCONTACT NUMBERNUMBER OF PAGES: 1 + 11 PAGESINSTRUCTIONS: 1. ANSWER ALL THE QUESTIONS ON THE PAPER IN PEN.2. NO CALCULATORS ARE ALLOWED.3. SHOW ALL CALCULATIONS AND MOTIVATE ALL ANSWERS.4. IF YOU REQUIRE EXTRA SPACE, CONTINUE ON THEADJACENTBLANK PAGE AND INDICATE THIS CLEARLY.

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20231/11Question 1[10 marks]For questions 1.1 – 1.10, chooseonecorrect answer, and make a cross (X) in the correct block.Questionabcde1.1X1.2X1.3X1.4X1.5X1.6X1.7X1.8X1.9X1.10XFor 1.1 and 1.2 below, considerP(x)Q(x)=x3−7x2+ 4xx2−5x+ 1=S(x) +R(x)Q(x).After long division:1.1S(x) =(1)(a)x−2(b)x+ 2(c)x+ 1(d)x−1(e) None of the above.1.2R(x) =(1)(a) 7x+ 2(b) 2x−7(c)−2x+ 7(d)−7x+ 2(e) None of the above.

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20232/111.3 When we use integration by parts to evaluate the integralZx·2xdx=, we chooseuanddvsuch that(1)(a)du=x·2x−1dxandv=x22.(b)du=dxandv=2xln 2.(c)du= 2xln 2dxandx2.(d)du=dxandv= 2x+1.(e) None of the above.1.4 To evaluate the integralZcot4xcsc6x dx=, we use(1)(a) the identity 1 + csc2x= cot2xand the substitutionu= cotx.(b) the identity 1 + cot2x= csc2xand the substitutionu= cscx.(c) the identity 1 + cot2x= csc2xand the substitutionu= cotx.(d) the identity 1 + cot2x=−csc2xand the substitutionu= cscx.(e) None of the above.1.5 Choose the correct trigonometric substitution for the given integralZ√a2−b2x2dx(1)(a)x=abtan(θ),−π2≤θ≤π2(b)x=abcos(θ),−π2≤θ≤π2(c)x=absin(θ),−π2≤θ≤π2(d)x=absec(θ),−π2≤θ≤π2(e) None of the above.1.6 Why does the Mean Value Theorem not apply tof(x) =|x−3|on [1,4]?(1)(a)f(x) is not continuous on [1,4](b)f(x) is not differentiable on (1,4)(c)f(1)̸=f(4)(d)f(1)> f(4)(e) None of the above.

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20233/111.7 Choose the correct partial fraction decomposition for the given rational function:(1)t3−2t2+ 2t−5t4+ 4t2+ 3(a)At+B(t2+ 1)(t2+ 3)(b)At+B(t2+ 1)+C(t2+ 3)(c)At+B(t2+ 1)+Ct+D(t2+ 3)(d)A(t2+ 1)+C(t2+ 3)(e) None of the above.1.8 Consider the following integrals:(1)I1=Z104dxex−10I2=Z8−8dx((x+ 1)−7)((x+ 1) + 7)Select the correct option:(a) Only the integralI1is improper.(b) Only the integralI2is improper.(c) IntegralsI1andI2are improper.(d) NeitherI1norI2are improper.(e) None of the above..1.9 Letfbe a differentiable function and suppose thatf(1) =f(7). Select the statement thatis true:(1)(a) There exists a numberc∈(1,7) such thatf′(c) = 0.(b) By Rolle’s Theorem there exists a numberc∈(1,7) such thatf′(c) = 0.(c) By the Intermediate Value Theorem there exists a numberc∈(1,7) such thatf(c) = 0.(d) By Fermat’s Theoremfhas a local minimum or local maximum on (1,7).(e) None of the above.1.10 Letfbe a twice-differentiable function ofxsuch that, whenx=c,fis decreasing, concaveup, and has anx-intercept. Which of the following is true?(1)(a)f′(c)< f′′(c)< f(c)(b)f(c)< f′(c)< f′′(c)(c)f′′(c)< f(c)< f′(c)(d)f′(c)< f(c)< f′′(c)(e) None of the above.

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20234/11Question 2[23 marks]Evaluate the following integrals:(a)Zπ20sin5xcos3x dx(3)SolutionZπ20sin5xcos3x dx=Zπ20sin5xcos2xcosx dx=Zπ20sin5x(1−sin2xcosx dx=Zπ20(sin5x−sin7x) cosx dx=16sin6x−18sin8xπ20=16[sin6π2−sin60]−18[sin8π2−sin80]=16−18=124(b)Z∞−∞e−|x|dx(4)Solution:By definition we have:Z∞−∞e−|x|dx=Z0−∞e−|x|dx+Z∞0e−|x|dx=limt→−∞Z0te−|x|dx+ limt→∞Zt0e−|x|dx=limt→−∞Z0texdx+ limt→∞Zt0e−xdx=limt→−∞[ex]x=0x=t+ limt→∞[−e−x]x=tx=0=limt→−∞[1−et] + limt→∞[−e−t+ 1]= 2

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20235/11(c) Use trigonometric substitution to show thatZdx√x2+b2= lnx+√x2+b2+C(4)Solution:Letx=btanθ, then√x2+b2=bsecθ, thusZdx√x2+b2=Zbsec2θbsecθdθ=Zsecθ dθ= ln|secθ+ tanθ|+C1= ln√x2+b2b+xb+C1= ln(x+√x2+b2) +CwhereC=C1−ln|b|(d) EvaluateZ10x2−2x(2x+ 1)(x2+ 1)dx(4)Solution:The denominator is factorized so write the partial fractionsx2−2x(2x+ 1)(x2+ 1)=A2x+ 1+Bx+Cx2+ 1Form an identity; multiply by (2x+ 1)(x2+ 1), thusx2−2x=A(x2+ 1) + (Bx+C)(2x+ 1)Choose values ofxand substitute into the identityLetx=−12A= 1Expand the identity in order to solve for the rest of the constantsx2−2x=Ax2+A+ 2Bx2+ 2Cx+Bx+CCompare coefficients ofx21 =A+ 2BSinceA= 1, thereforeB= 0. Compare constants0 =A+C, henceC=−1 So thatZ10x2−2x(2x+ 1)(x2+ 1)dx=Z1012x+ 1−1x2+ 1dx=12ln|2x+ 1| −tan−1x10=12ln 3−tan−11 ==12ln 3−π4

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20236/11(e)Zx3√1 +x2dx.Compute this integral by integrating by parts. Letu=x2.(4)SolutionWe use integration by parts (IBP):Letu=x2;dv=x√1 +x2dx=⇒du= 2xdxandv=13(1 +x2)32Then,Zx3√1 +x2dx=x23(1 +x2)32−23Zx(1 +x2)32dx=x23(1 +x2)32−23·15(1 +x2)52+C=p(1 +x2)3153x2−2+CQuestion 3[2 marks]Letfandgbe functions that are strictly positive and increasing(i.e. For eachx∈R, f(x)>0, f′(x)>0 andg(x)>0, g′(x)>0).Consider the functionFdefined byF(x) =f(e3x)g(2x+ 1).Prove thatFhas no critical points.Solution:Using the product and chain rules we obtain:F′(x) = 3e3xf′(e3x)g(2x+ 1) + 2f(e3x)g′(2x+ 1)Using the assumptions imposed onfandgit’s clear thatf′(e3x), g(2x+1), f(e3x) andg′(2x+1)are all strictly larger than zero. It’s also clear thate3xis strictly larger than zero.ThereforeF′(x) is strictly larger than zero and hence there is no pointxwhich makesF′(x) = 0. This, together with the fact thatF′(x) is defined for all values ofximplies thatFhas no critical points.

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20237/11Question 4[4 marks]Prove the Mean Value Theorem, that is, prove that:Iffis a function that satisfies the following hypotheses:1.fis continuous on the closed interval [a, b].2.fis differentiable on the open interval (a, b).Then there is a numbercin (a, b) such thatf′(c) =f(b)−f(a)b−aSolution:We prove Mean Value Theorem with an aid of a graph.We apply Rolle’s Theorem to a newfunctionhdefined as the difference betweenfand the function whose graph is the secant lineAB. Using point-slope equation of a line, we see that the equation of the line AB can be writtenasy−f(a) =f(b)−f(a)b−a(x−a)As shown in the below figure, we leth(x) to beh(x) =f(x)−f(a)−f(b)−f(a)b−a(x−a)First we must verify thath(x) =f(x)−f(a)−f(b)−f(a)b−a(x−a) satisfies the three hypotheses ofRolle’s Theorem.1. The functionhis continuous on [a, b] because it is the sum offand a first-degree polynomial(straight line), both of which are continuous.2. The functionhis differentiable on (a, b) because bothfand the first-degree polynomial aredifferentiable. Soh′(x) =f′(x)−f(b)−f(a)b−a

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20238/113.h(a)=f(a)−f(a)−f(b)−f(a)b−a(a−a) = 0h(b)=f(b)−f(a)−f(b)−f(a)b−a(b−a)=f(b)−f(a)−[f(b)−f(a)] = 0Thereforeh(a) =h(b)Sincehsatisfies the hypotheses of Rolle’s Theorem, there is a numbercin (a, b) such thath′(c) = 0.Therefore0 =h′(c) =f′(c)−f(b)−f(a)b−aand sof′(c) =f(b)−f(a)b−aQuestion 5[4 marks]Use Rolle’s Theorem to show thatf(x) =x36+x22+x+ 1 has exactlyone real root.Solution:First we use the Intermediate Value Theorem to show that a root exists.f(0) = 1>0 andf(−2) =−13<0.Sincefis a polynomial, it is continuous, hence there is a numbercbetween−2 and 0 such thatf(c) = 0. Thus the given equation has a root.To show that the equation has no other real root, we use Rolle’s Theorem and argue by contradic-tion. Suppose that it had two rootsaandb. Thenf(a) = 0 =f(b) and, sincefis a polynomial,it is differentiable on (a, b) and continuous on [a, b]. Thus, by Rolle’s Theorem, there is a numbercbetweenaandbsuch thatf′(c) = 0.Butf′(x) =12x2+x+ 1The discriminant forf′(x) isb2−4ac=−1. Hencef′(c)̸= 0∀x.This gives a contradiction. Therefore the equation must have exactly one real root.Question 6[11 marks]Letf(x) =x√2−x2.(a) Find the domain off.(1)Solution:(−2; 2)

MEMOMATENB1 SEMESTER SICK TEST 1 memorandum– 11 SEPTEMBER 20239/11(b) Find the intercepts off(x).(1)Solution:They−intercept is (0,0) and thex−intercepts are (0,0), (−√2,0) and (√2,0).(c) Determine the intervals of increase and decrease and find the local maximum and/or mini-mum values off.(4)Solution:f′(x) =√2−x2−x2√2−x2=2(1−x2)√2−x2.The critical numbers are−1 and 1. The signs of the derivative are determined in the tablebelow.Interval1 +x1−xf′(x)f(x)x <−1−+−Decreasing−1< x <1+++Increasingx >1+−−DecreasingThe local maximum value isf(1) = 1 and the local maximum value isf(−1) =−1.(d) Given:f′′(x) =2x(x2−3)(2−x2)3/2.Discuss the concavity of the graph and find the inflection point(s).(3)Solution:f′′(x) =−2x(4−x2)3/2. So,f′′(x)>0 whenx <0 andf′′(x)<0 whenx >0, meaning thatfisconcave upward on (−∞,0) and concave downward on (0,∞) Point of inflection is (0,0).(e) Sketch the graph off.(2)Solution: