= 5 0 cos 0 = 1. = lim = 1. = 5 1/ 1 2 cos. QUESTION 1 Compute the following limits: (iii) lim 5 1/x cos. (10 marks) Solution Let y = 1 x.

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1 QUESTION 1 Compute the following its:. (i 5 1/x cos 6x (ii (iii 5 1/x cos x 2 1. x 5 1/x cos + 6x. 6x Solution Let y = 1 x. (i 5 1/x cos 6x = y + 5 y cos (ii Since 1 cos 6x 1 and 5 1/x >, we get Further, y = 5 cos = /x 5 1/x cos 5 1/x 6x x 5 1/x = 5 y = + y x 5 1/x = =. + By the Squeeze Theorem, we conclude that x 5 1/x cos = 6x (iii 5 1/x cos x 1 6x 2 = 5 1/ 1 2 cos ( π 6 12 (1 marks = 1 25 cos π = 1 5. : 1

2 QUESTION 2 You are celebrating the end of the semester with your favourite cocktail. The ice cubes in your drink are melting so that the surface area of each cube is decreasing at the constant rate of.16 cm 2 /s. At a certain moment, the volume of each ice cube is.125 cm. Find the rate at which the volume of each of the ice cubes is decreasing at that moment. Give the answer in cm /s. (1 marks Solution Let s(t be the surface area of an ice cube, let v(t be the volume, and let t be certain moment. We are given that s (t =.16 and that v(t =.125. Let s find a relation on v and s. If the volume is v, then the side length of the cube is x = v 1/ and the surface area is s = 6x 2 = 6v 2/. Differentiating it, we get which implies s (t = 6 2 v 1/ v (t = 4v 1/ v, v (t = 1 4 s (t [v(t ] 1/ = =.2. Thus the answer is that the volume of the ice cubes is decreasing at the rate of.2 cm /s. : 2

3 QUESTION Let m(t be a transmitter s signal at time t. We will look at what happens near the moment that the transmitter is switched on. When the transmitter is off, the signal is, naturally, zero. When it is switched on, it starts transmitting the idle signal a sinusoidal wave of a particular frequency ω >, amplitude A >, and initial phase P. Let s for simplicity assume that A = 1 and that the idle signal is given by the function sin [ω(t P ]. Given that the transmitter is switched on at t = 5, what is the condition on ω and P making the function m(t continuous near the point t = 5? Is it possible to find values of ω and P making m(t differentiable at t = 5? Solution The function representing the transmitter s signal is {, t < 5, m(t = sin [ω(t P ], t 5, (1 marks and the objective of the problem is to check whether this function is continuous / differentiable at t = 5. For m to be continuous at 5, we must have = m(t = m(5 = m(t = sin [ω(5 P ], t 5 t 5 + which is true if and only if ω(5 P is a multiple of π. Thus the answer to the first part of the question is ω(5 P = πn, where n is an arbitrary integer. For m to be differentiable at 5, it has to be continuous and satisfy the additional condition that the tangent line on the left of t = 5 is the same as the tangent line on the right, that is, = [ ] t=5 = [ sin ω(t P ] t=5 = ω cos[ω(5 P ]. Thus we need to have cos[ω(5 P ] = and sin[ω(5 P ] = at the same time, which is, of course, impossible. Thus m cannot be differentiable at t = 5. :

4 QUESTION 4 Figure 1 shows the graph of a function f : (, R. Sketch the graph of its derivative. Your graph should clearly show where the value of f is positive, negative, or zero and give a rough estimate of the magnitude of f. (1 marks Figure 1: The graph of the function f (x in Question 4. y O 1 2 x Figure 2: The graph of f (x in Question 4. 4

5 QUESTION 5 Figure shows the graph of a function f : [ 4, 4] R. (i At which points of the open interval ( 4, 4 is f not differentiable? (ii Find 4 4 f (xdx. (iii Let g(x = x f (tdt. Evaluate g ( 2. (1 marks Figure : The graph of the function f (x in Question 5. Solution (i 2, 1, 2. (ii 5.5 = 2.5. (iii g ( 2 = f ( 2 = 2. : 5

6 QUESTION 6 Santa Claus prepared 5 kg of candy for good children and put it in a huge bag. He loaded everything onto his sleigh and took off. The unladen weight of the sleigh plus the mass of Santa himself and all the reindeer is 15 kg. Unfortunately, the bag has a hole and candies are now dropping off the sleigh. To make things even worse, candies falling through the hole tear its edges further apart, causing the hole to grow bigger. As a result, the rate at which the sweets are being released into the air is growing linearly. Specifically, t minutes after the take-off, the mass of the candy bag is decreasing at the rate of 4t kg/min. It is also given that Santa s sleigh is cbing at the constant rate of 1.2 km/min (the growth rate of the altitude. (a Write down the explicit formula expressing the mass of the candy bag as a function of t. (b How long will it take until the bag is empty? (c What is the sleigh s altitude A at the moment the candy is gone? (d Find the amount of work the reindeer have done to lift the sleigh to the altitude A. You may assume that the gravitational acceleration is constant and equals 1 m/s 2. Recall that the gravitational force acting on a body of mass m is gm, where g is the gravitational acceleration. Also, recall that if the force F was constant, then its work on displacing an object would be F s, where s is the displacement. (2 marks 6

7 Solution (a Let m(t be the mass of the candy bag. Then we are given that m (t = 4t. In means that m(t = 2t 2 + C, where C can be found from the condition m( = 5 = C = C. Thus m(t = 2t (b The bag is empty when m(t = = 2t 2 + 5, that is 2t 2 = 5, which means t 2 = 25 and t = ±5. Since the time must be positive, we see that the bag empties in 5 minutes. (c Since the speed of cbing is constant 1.2 km/min, we see that the sleigh s altitude after 5 min is A = = 6 km. (d The work is the integral of the gravitational force with respect to the altitude. Thus we need to express everything via the altitude. Also, let s convert kilometres to metres so that the final answer is in Joules. Let x(t be the altitude of the sleigh in metres. Then x = 12t and t = 12 x for x 6. The mass of the candy is m = 2t2 + 5 = 2 x2 + 5 = x The total mass of the sleigh is therefore M(x = 6 12 x2 x = kg. The gravitational force is F(x = gm(x, where g is the gravitational acceleration. The work done W is the integral of the force, that is, 6 W = g ( = g [ ( x dx x 72 ] = g ( = g ( = g Thus the answer is 11 MJ. : 7

8 QUESTION 7 Compute the following integrals: (i 5 e x dx 4 + e x. (ii 2π x cos xdx. (1 marks Solution get (i Let s substitute u = e x + 4. Then u (x = e x and hence we 5 (ii Integrating by parts, we get 2π x cos xdx = e x dx e e x = 1 e +4 u du = [ ln u ] e5 +4 = ln e = [ ] 2π x sin x [ x sin x + cos x 2π ] 2π sin xdx = = 2π [ ] 2π x sin x [ ] 2π cos x = π 2 : 8

9 QUESTION 8 Compute the it x e t dt 1 + e x2. Solution First, notice that x e t dt = 1 + e x2 =. (1 marks Thus the it in question is an indeterminate form of the type and we can apply l Hôpital s Rule. Further, by Chain Rule and Fundamental Theorem of Calculus, we get for the numerator and d dx d x 2 dx for the denominator. Thus, x e t dt = 2x 1 + e x2 1 + e x2 = 2xex e 1 + e t dt 1 + e x2 = 2x xex2 = x2 1 + e x2 1 + e x2 1 + e = x2 2 xex2 e x2 1+e x2 = 2 e x = = 2. : 9

10 QUESTION 9 Find the maximal value of the integral b (4 + x x 2 dx for all a b. a (1 marks Figure 4: The parabola y = 4 + x x 2 and the areas under it. Solution Let f (x = 4 + x x 2 = (x 2 x 4 = (x + 1(x 4 and notice that we have f (x < for 1 < x < 4, f (x > for x < 1 or x > 4. The integral b a (4 + x x2 dx is the sum of areas in Figure 4 dark green with negative sign and dark red with positive sign (we have a = 2 and b = 6 specifically in the illustration, but it doesn t matter. Obviously, the maximum is attained when there is no green part and the red part is as large as possible, that is, for a = 1 and b = 4. Thus the answer is 4 [ x (4 + x x 2 2 ] 4 [ x ] 4 dx = = : 1 1 1