MATH 102 Exam 1 With Solution, Spring 2012

MATH102 mid-term quiz 1 Saturday 17 March 2012 Name: ID: Circle your recitation section: 5 (8am) 6 (9am) 7 (11am) 8 (12am) Question: 1 2 3 4 5 6 7 Bonus Total Points: 18 12 12 12 19 12 15 5 100 Score: 1. (18 points) Evaluate the following indefinite integrals. Show your working. (a) � x dx x2 − 1 Solution: Substitute u = x2 − 1, so that du = 2x dx. Then � x dx x2 − 1 = 1 2 � 1 u du = 1 2 ln |u| + C = 1 2 ln |x2 + 1| + C. (b) � dx x2 − 2x + 2 Solution: Firstly, complete the square: � dx x2 − 2x + 2 = � dx (x − 1)2 + 1. Now substitute u = x − 1 to obtain � dx (x − 1)2 + 1 = � du u2 + 1 = tan−1(u) + C = tan−1(x − 1) + C. (c) � 10x dx Solution: Use the definition of 10x as e(ln 10)x: � 10x dx = � e(ln 10)x dx = 1 ln 10e(ln 10)x + C = 1 ln 1010x + C. (You may like to know that d dx(tan−1 x) = 1 1 + x2.) Page 1 of 6 MATH 102 Mid-term quiz 1 17 March, 2012 2. (12 points) Evaluate the following limits. Show your working. (a) lim x→0 ex − 1 x Solution: lim x→0 ex − 1 x = lim x→0 ex 1 by l’Hˆopital’s rule = 1. (b) lim x→∞ (ln x)2 x + 1 Solution: lim x→∞ (ln x)2 x + 1 = lim x→∞ 2 ln x(1/x) 1 = lim x→∞ 2 ln x x by l’Hˆopital’s rule = lim x→∞ 2/x 1 by l’Hˆopital’s rule = 0. 3. (a) (6 points) Indicate which of the following functions are one-to-one. For each function which is not one-to-one, find two different values of x for which the function gives the same result. i. f(x) = (x − 1)3 Solution: This is one-to-one. ii. g(x) = (x − 1)4 Solution: This is not one-to-one. For example, g(0) = g(2) = 1. iii. h(x) = 1/(x − 1) Solution: This is one-to-one. (b) (6 points) I am thinking of a function f, which satisfies f(2) = 3 and f ′(2) = −2. If I tell you that f has an inverse f −1, write numbers in the four blank spaces to show two facts you can deduce about f −1: f −1( 3 ) = 2 ; (f −1)′( 3 ) = −1/2 . Page 2 of 6 MATH 102 Mid-term quiz 1 17 March, 2012 4. (12 points) Write letters in the boxes to match the following functions to their graphs (which are not all to the same scale). Functions Graphs A. y = ln |x| C B. y = xe−x D C. y = e−x2 B D. y = tan−1 x A Page 3 of 6 MATH 102 Mid-term quiz 1 17 March, 2012 5. (a) (9 points) Suppose a hot object is placed in a cool environment. If we let H(t) denote the temperature of the object at time t, and HS the constant temperature of the environment, then Newton’s law of cooling states that dH dt = −k(H − HS), where k is some constant depending on the physical conditions. Show how to solve this differential equation to get the solution H(t) = HS + Ae−kt for some constant A. Solution: dH dt = −k(H − HS) 1 H − HS dH dt = −k (rearranging) � dH H − HS = � (−k) dt (integrating both sides with respect to t) ln(H − HS) = −kt + C (no need for absolute value since H − HS > 0) H − HS = e−kt+C (exponentiating both sides) H − HS = Ae−kt (letting A = eC) H = HS + Ae−kt. Question 5 continues ... Page 4 of 6 MATH 102 Mid-term quiz 1 17 March, 2012 Question 5 (continued) (b) (10 points) I like to drink my coffee at precisely 55 ℃. This morning I took my coffee at 95 ℃ and left it to cool in a room at 15 ℃. Unfortunately, I left it too long: after 6 minutes, the coffee had already cooled to 35 ℃. Assuming that my coffee obeys Newton’s law of cooling, how long should I have left the coffee before drinking it? Solution: From the question, HS = 15. We have two further pieces of infor- mation: when t = 0, H = 95; and when t = 6, H = 35. 1. Substituting t = 0 and H = 95 into the equation from part (a) gives 95 = 15 + A, so A = 80. 2. Substituting t = 6 and H = 35 gives 35 = 15+80e−6k, so e−6k = 20/80 = 1/4 and k = − ln(1/4)/6. 3. Now substitute H = 55 and solve for t: we get 55 = 15 + 80e−kt, so e−kt = 40/80 = 1/2 and therefore t = ln(1/2) −k = 6 × ln(1/2) ln(1/4) = 6 × 1 2 = 3. 6. (12 points) Using integration by parts, evaluate the definite integral � 1 0 x2e−x dx. Show your working. Solution: � 1 0 x2e−x dx = � −x2e−x�1 0 + � 1 0 2xe−x dx (integration by parts) = � −x2e−x�1 0 + � −2xe−x�1 0 + � 1 0 2e−x dx (integration by parts again) = � −x2e−x�1 0 − � 2xe−x�1 0 + � −2e−x�1 0 = � e−x(−x2 − 2x − 2) �1 0 = e−1(−1 − 2 − 2) − e0(−2) = −5e−1 + 2 ≈ 0.1606. Page 5 of 6 MATH 102 Mid-term quiz 1 17 March, 2012 7. (15 points) Indicate whether each of the following statements is true or false. You do not need to justify your answers. Each correct answer scores 3 points, but if you answer and get it wrong then you lose one point. T F 3x grows faster then 2x as x → ∞. T F log2 x grows faster than log3 x as x → ∞. T F For any value of x, we have sin−1(sin x) = x. T F For any value of x, we have sin(sin−1 x) = x. It could be argued that this is only true for −1 ≤ x ≤ 1, since it doesn’t even make sense for other values of x, so either answer to this question was counted as correct. T F The function y = cos 2x satisfies the differential equation d2y dx2 + y = 0. Bonus question (5 points) Find two functions f and g such that f ◦ g is one-to-one, but f is not one-to-one. Solution: One possible answer is g(x) = ex and f(x) = x2. Page 6 of 6