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# S3A - SOLUTIONS TO EXAM 3A 1[20 Marks Evaluate each of the...

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Unformatted text preview: SOLUTIONS TO EXAM 3A 1. [20 Marks] Evaluate each of the following limits. SHOW ALL OF YOUR WORK. . 1 — sec 9 (a) \$1310 92 Solution Note 1—55“) = 9. Hence we can apply L’Hopital’s Rule to this limit. 0 o 1, 1 — sec0 _ 1_ D(1 — sec6) _1, —sec0tan€ _ _ lim sec61im Sing/C086 913i) 62 T 913) D(62) T 913%) 26 T «9—»0 9—0 26 1 sin 9 1 1 : — 1 l' l' 2 1 : ()013b2cos0613b 0 2() 2 . tanG u sec 0 (b) .3122. T Solution Substitute (9 = 0 to evaluate this limit: , tanQeseCQ tanO—secO 0e1 11m = : z 00. Geo 92 02 02 2. [20 Marks] A ball is kicked upwards from the ground, rises for 5 seconds and then falls downwards until it lands on the roof of a building 8 seconds after it was kicked. How many feet high is the building? Solution Let the yeaxis have its origin on the ground with units measured in feet. Let time t be measured in seconds. Then 5(0) 2 0 because the object starts on the ground. Hence not) = 7322: + 22(0) and 30:) = 716752 + 22(0):: + 3(0) 2 716252 + v(0)t. At its highest point t = 5 the ball stops: 0 2 21(5) = 732(5) + 11(0). Hence 11(0) 2 160 ft/sec. When the ball is caught at t = 8: 3(8) = 716(82) + (160)(8) = 71024 + 1280 = 256 feet. 2 3. [20 Marks] Find the values of :r where the function f(x) : 4: 16 with domain [1,00) has its as minimum and maximum values (if any). J USTIFY YOUR ANSWER. Solution By the quotient rule: f’(x) _ 233(334 + 16) — 332(4333) _ 23:5 + 323: — 4335 _ 322: e 23:5 _ 233(16 e 334) (x4 + 16)2 (91:4 + 16)2 (x4 + 16)2 (x4 + 16)2 Thus f has one crictical point: at a: : 2 where the numerator of f’(a:) vanishes and f’(:t) equals zero. Note f(1) = %, f(2) 2 % and 2 . x hm = 0. x—>oo 1:4 +16 Since 0 is the smallest of these numbers, the function f has no absolute minimum. Since % is the largest of these numbers, f has an absolute maximum at m : 2. 4. [20 Marks] Find the lower and upper Riemann sums of ﬂm) : {C2 for the partition P = {73, *1, 0, 2, 5}. Solution The lower Riemann sum is sum of the areas of the rectangles in the left ﬁgure below. L(P,f) = f(*1)l(*1)*(i3)l+f(0)10*(*1)l ~f(0)[2*01+f(2)l5*2] = (1)(2)+(0)(1)+(0)(2)+(4)(3) = 2~0+0+12=14 The upper Riemann sum is sum of the areas of the rectangles in the right ﬁgure below. Uﬂiﬂ = f(3H(1) (3ﬂif(1ﬂ0 (lﬂifOHQ—[email protected]—ﬂ = @xm+«nuy+mxm+r%x\$ =18+1+8+75=1m y y 25 y =x 9 4 4 I x a.” x -3 -I 0 2 5 -3 -1 0 2 5 5. [20 Marks] JUSTIFY EACH OF YOUR ANSWERS. 'rr/4 (a) Evaluate / tanO d0. —7r/4 'rt/4 Solution Since tan0 is an odd function, / tanQ d0 — O. —'rr/4 d0 exists. 7r - 0 (b) Explain Why / 813 0 Solution The integrand is f(6) = “20 for 0 < 0 < 1 and f(0) = 1. This function is continuous. For 0 < «9 < 1 it is the quotient of two continuous functions. At 6 : 0: sin 0 1. 012% 0 :1=fml A theorem in the text says that the deﬁnite integral of a continuous function exists. 2 (0) Assume that f is a continuous function with average value 12 on [—1,2]. Find / f(x) dx. 71 Solution The average value of 12 of f on [—1,2] is the value of the integral j: f(a:) dw divided by the length of the interval [71, 2]. That is 2 f: f(w) day 12 3 Hence f: f(:c) d1: 2 3(12) 2 36. ((1) Which is larger: foﬁ sinm dm or L? sin2 1: dx? Solution Note 0 E sina: E 1 for 0 E a: E 71'. Hence Note 0 E singx E sina: for 0 E a: E 71'. Therefore f0” sin2 a: dzz: E f0” sing; dds, and f0” sing; d2; is the larger integral. (6) Assume Find /1 3 f(:z:) ([17. Solution By the additive property of integrals 4 4 f(x)dx : /0f(x)dx—/3 f(x)dx:15—2:13 k5 /_\ i R. x; | l 5 | [ [02m M) “a /_\ Q‘ g [l 3 1 " d" — :1: (1:1: : 3 — : 8 0 m l f f( > 1 5 ...
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S3A - SOLUTIONS TO EXAM 3A 1[20 Marks Evaluate each of the...

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