Name: AK-Nummer: Ergänzungsprüfung January 29, 2016

Transcription

1 INSTRUCTIONS: The test has a total of 32 pages including this title page and 9 questions which are marked out of 10 points; ensure that you do not omit a page by mistake. Please write your name and AK-Nummer on every page. You may use scrap paper for rough work, but only the answers written on the test paper will be considered for marks. Therefore, show all of your work on the test paper itself; you may use the back of the paper if you need extra space, but be sure to indicate this clearly if you do. Write as neatly and clearly as possible. Calculators are not needed and are not allowed. Likewise, the use of cell phones or any other electronic devices is not allowed during the test. You may not consult your notes, textbooks, or any other pre-prepared written material during the test. You have 3 hours to complete all of the questions; you are free to leave as soon as you are finished. 1/32

2 1. a) (5 Marks) Solve the following system of equations for (x, y, z) by your preferred method. A. x + y + z =4 B. x + 3y + 3z =10 C. 2x + y z =3 2/32

3 b) (5 Marks) Does the following linear system have: a) no solutions; b) exactly one solution; c) infinitely many solutions? Support your answer with a graph. If the linear system has one solution, derive that solution. If it has infinitely many solution give a particular solution. A. y 3x = 2 B. 3y 9x =20 3/32

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5 Solution: 1. a) Subtract equation A. from B. and 2 times B. from C. to get the following system of two equations and two unknowns: I. 2y + 2z =6 II. 5y 7z = 17 Now add 5/2 times I. to II. to get 2z = 2 or equivalently z = 1. Plugging this into I. gives y = 2, and then plugging these results into A. gives x = 1. We may check that (x, y, z) = (1, 2, 1) satisfies all equations and is therefore our solution. b) We write the equations in familiar slope-intercept form A. y =3x + 2 B. y =3x which shows that the equations represent parallel lines as shown in Figure 1 and therefore we have no solution. 5/32

6 40 y 30 y = 3x y = 3x x Figure 1: The lines do not intersect, they are parallel 6/32

7 2. a) (5 Marks) Give the formula, in the form y = c + bx + ax 2, of the quadratic function which has zeros x = 3 and x = 1 and takes value y = 1 when x = 1. Sketch the graph of the function and 2 give the coordinates of the vertex. 7/32

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9 b) (5 Marks) Solve the following cubic equation x 3 7x x 9 = 0 9/32

10 Solution: 2. a) We set up an equation based on the information we have ( y = c(x 3) x + 1 ) 2 (0.1) where c is unknown. Now we plug in (x, y) = (1, 1) and solve for c. This gives c = 1 3. Expanding (0.1) gives To graph the function, we factor out 1 3 y = x 1 3 x2 y = 1 3 again and then complete the square to get ( x 5 ) Plotting is then just a matter of applying the transformations a) scale by 1/3 (blue in Figure 2 ) b) shift right by 5/4 (red) c) shift up by 49/48 (green, final) to the graph of x 2 (black). We can check that the resulting graph (green line) passes through the points (3, 0), ( 1/2, 0), and (1, 1). We also know that the vertex, or maximum in this case, must lie at (5/4, 49/48) since this is how much we have shifted the vertex of the graph of x 2 which lies at (0, 0). b) We write the equation in the form x(x 2 7x + 15) = 9 which shows that any integer solution must divide 9 exactly. Trying the divisors of 9 we see that 1 is a solution. Now we do polynomial division: x 2 6x + 9 x 1 ) x 3 7x x 9 x 3 + x 2 6x x 6x 2 6x 9x 9 9x + 9 Factoring, we get x 2 6x + 9 = (x 3) 2 so our solution set is {1, 3}. 0 10/32

11 ( 5 4, ) ( 1 2, 0) (3, 0) Figure 2: Graph of y = x 1 3 x2 11/32

12 3. a) (5 Marks) Fill in the coordinates on the unit circle corresponding to the given angle. (0, 1) (, ) (, ) 3π 4 π 6 ( 1, 0) (1, 0) 4π 3 7π 4 (, ) (, ) (0, 1) 12/32

13 b) (5 Marks) Draw the graph of y = 3 cos ( 2 ( x π 2 )) so that one can see one entire period. State the period of the function, the location of all zeros, and the value and location of all extreme points. 13/32

14 Solution: 3. a) See Figure 3 b) Apply the transformations: a) reflect in x (blue in Figure 4); b) scale vertically by 3 and horizontally by 1/2 c) shift right by π/2 (green) to the function cos(x) (black). The resulting function has maximum value 3 at points {kπ} k Z and minimum value -3 at points {(2k + 1)π/2} k Z, zeros at points {(2k + 1)π/4} k Z, and period π. 14/32

15 Figure 3: The unit circle 3 Period: π (0, 3) (π, 3) π 4 3π ( π 3, 3) Figure 4: Graph of y = 3 cos ( 2 ( x π 2 )) 15/32

16 4. a) (5 Marks) State whether or not the following limits exist as real numbers. If they exist as real numbers, find the limits. If they do not exist, state whether or not the limits diverge to ±. 2 ln(x) (i) lim x 10 log(10x) + e (ii) lim x 1 x 2 3x+2 x 2 +4x 5 16/32

17 b) (5 Marks) Use the squeeze theorem to show that the following limit is zero. lim x 0 x 2 csc ( 1 x ) 17/32

18 Solution: 4. a) (i) By continuity of the two functions which are being summed, we know that the individual limits exist, and that the limit is just the sum of the functions evaluated at x = 10. Plugging this in gives b) Since log(100) + e 2 ln(10) = log(10 2 ) = (ii) We factor the top and bottom polynomials to get x 2 3x + 2 lim x 1 x 2 + 4x 5 = lim (x 1)(x 2) x 1 (x 1)(x + 5) = lim (x 2) x 1 (x + 5) = 1 6 x 2 x 2 sin(1/x) = x2 csc ( ) x 2 1 x and lim x 0 x 2 = lim x 0 ( x 2 ) = 0 the result follows by the squeeze theorem. 18/32

19 5. Find y. In part c) your answer may contain x and y terms. a) (3 Marks) y = tan(x) 1+cos(x) b) (3 Marks) y = 1 x /32

20 c) (4 Marks) xe y = y sin(x) 20/32

21 Solution: 5. a) Just use the quotient rule to get y = (1 + cos(x)) sec2 (x) + tan(x) sin(x) (1 + cos(x)) 2. Although it is possible to perhaps simplify further by applying the identities cos 2 (x) + sin 2 (x) = 1 and sec 2 tan 2 (x) = 1 I would accept this as an answer. b) The power rule gives y = x 1 1 6/5 2 x, 3/2 and further simplification is not possible, so we can stop here. c) We differentiate both sides with respect to x to get and then we solve for y which gives e y + xe y y = y sin(x) + y cos(x), y = y cos(x) ey xe y sin(x). 21/32

22 6. (10 Marks) Analyse the function y = x2 x 2. x + 3 Find: a) The location of maxima and minima (you do not need to compute the maximal or minimal values though, just the location i.e. the x-coordinate); b) The location of inflection points; c) The location of saddle points; d) The formulas of any asymptotes (vertical, horizontal, or slant). 22/32

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24 Solution: 6. First we factor the numerator to see if any factors which might cancel. This shows that y = (x + 1)(x 2) x + 3 so that there is no cancellation of factors. We can see, however, that there is a vertical asymptote at x = 3. Differentiating y gives, after some work, y = x2 + 6x 1 (x + 3) 2. Setting this equal to 0 and using the quadratic formula shows that we have critical points x = 3 ± 40/2. Differentiating again, gives y = 20 (x + 3) 3 which shows that there are no inflection or saddle points since y will never be zero. Checking the values of y at the critical points shows that ) ( ) y ( 3 < 0 and y 3 + > 0, 2 2 meaning that x = 3 40/2 is a relative or local maximum, and x = /2 is a relative or local minimum. Last, we observe that the degree of the numerator of y is one bigger than the degree of the denominator. Therefore, we expect linear behaviour as x grows in the positive and negative direction, i.e. we expect a slant asymptote. We know that we can get the formula of the asymptote by polynomial division, i.e. x 4 x + 3 ) x 2 x 2 x 2 3x 4x 2 4x + 12 so that we have a slant asymptote with formula y = x 4. Summarizing, we have a) a local maximum at x = 3 40/2 and a local minimum at x = /2. b) no inflection points c) no saddle points d) A vertical asymptote with formula x = 3 and a slant asymptote with formula y = x /32

25 7. Evaluate the given integrals. a) (3 Marks) 1 0 (1 + x) 4 dx b) (3 Marks) s2 s ds 25/32

26 c) (4 Marks) e 2x 1+e 4x dx 26/32

27 Solution: 7. a) We let u = 1 + x so that x = (u 1) 2 and dx = 2(u 1)du. Then 1 0 ( ) x dx = 2 u 4 (u 1)du = ( u 5 u 4) du = 1 4 u4 1 3 u3 u=2 b) We use integration by parts with u = s, du = dt, dv = 2 s ds, and v = 2 s / ln(2). Then s2 2 ds = 1 ln(2) s2s 1 2 s ds = 1 1 ln(2) ln(2) s2s (ln(2)) 2 2s + c. c) We make the substitution u = e 2x, du = 2e 2x dx to obtain e 2x 1 + e dx = 1 1 4x u du = tan 1 (u) + c = 1 2 tan 1 (e 2x ) + c u=1 = /32

28 8. (10 Marks) Evaluate the integral x 2 2x 1 (x 1) 2 (x 2 + 1) dx. 28/32

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30 Solution: 8. We proceed by the standard technique for integration of rational functions. Here we have a repeated linear factor and an irreducible quadratic factor in the denominator. Therefore the set-up we want looks like x 2 2x 1 (x 1) 2 (x 2 + 1) = A x 1 + B (x 1) + Cx + D 2 x Solving for A, B, C and D gives x 2 ( 2x 1 1 (x 1) 2 (x 2 + 1) dx = x 1 1 (x 1) x 1 2 = ln( x 1 ) 1 x 1 ) dx x x x dx + 1 x dx where we have used the substitution u = x 1 for the first two integrals. For the remaining integrals we make the substitution u = x in the first, and recognize the second integrand as the derivative of tan 1 (x). This gives x 2 2x 1 1 dx = ln( x 1 ) (x 1) 2 (x 2 + 1) x ln(x2 + 1) + tan 1 (x) + c. 30/32

31 9. a) (5 Marks) For a probability measure P one of the following is always true for two events A and B P(A B) P(A) + P(B) 1, P(A B) P(A) + P(B) 1. or Which inequality is always true? Why? b) (5 Marks) There are two boxes, Box O and Box E. Box O contains 1 black ball and 3 white balls and Box E contains 2 black balls and 4 white balls. Suppose we select a box with equal probability 1/2, say by flipping a coin, and that we then blindly draw a ball out of that box. What is the probability that the ball we draw is black? 31/32

32 Solution: 9. a) We know that in general P(A B) = P(A) + P(B) P(A B). But, for any A and B we always have P(A B) 1 so that P(A B) 1P(A) + P(B) P(A B). b) Call the event that the ball is black B and let O and E denote the events of picking box O and E respectively. P( pick a black ball ) = P(B) = P(E B) + P(O B) = P(B E)P(E) + P(B O)P(O) = = /32