MATH 101 Midterm Quiz 1 with Solution, Fall 2013

AMERICAN UNIVERSITY OF BEIRUT MATH 101 — Calculus & Analytic Geometry I Fall Semester 2013–14 Mid-term Exam 1 Saturday 5 October, 2013 Time: 1 hour Name: ID: Circle your recitation section: 1 2 3 4 5 Fri 9am Fri 10am Fri 4pm Fri 11am Thurs 2pm 6 7 8 9 10 Thurs 11am Thurs 12:30pm Mon 8am Mon 9am Mon 2pm The backs of the pages may be used as scratch paper. Please do not open this paper until you are instructed to do so. Question: 1 2 3 4 5 6 Total Points: 6 15 10 6 9 14 60 Score: MATH 101 Mid-term exam 1 5 October, 2013 1. Here is the graph of a piecewise-defined function f(x). −4 −3 −2 −1 1 2 3 4 −3 −2 −1 1 2 3 4 (a) (4 points) Fill in the blanks: lim x→−1− f(x) = 3 lim x→−1+ f(x) = 1 lim x→4− f(x) = −3 lim x→0 f(x) = 0 (b) (2 points) Is f continuous at x = 0? Justify your answer. Solution: We have limx→0 f(x) = 0, but f(0) = −1. So f is not continuous at x = 0. Page 1 of 5 MATH 101 Mid-term exam 1 5 October, 2013 2. (15 points) Calculate the following limits. Show your working. (a) lim x→0 cos x + 1 5x − 2 Solution: lim x→0 cos x + 1 5x − 2 = limx→0 cos x + limx→0 1 limx→0 5x − limx→0 2 = 1 + 1 0 − 2 = −1. (b) lim x→2 � x2 − 3x + 2 x2 − 4 Solution: lim x→2 � x2 − 3x + 2 x2 − 4 = lim x→2 � (x − 2)(x − 1) (x − 2)(x + 2) = lim x→2 � x − 1 x + 2 = 1 2. (c) lim x→∞ 2x + 1 x2 − 9 Solution: lim x→∞ 2x + 1 x2 − 9 = lim x→∞ 2/x2 + 1/x 1 − 9/x2 = 0 1 = 0. (d) lim x→7 √x + 2 − 3 x − 7 Solution: lim x→7 √x + 2 − 3 x − 7 = lim x→7 (√x + 2 − 3)(√x + 2 + 3) (x − 7)(√x + 2 + 3) = lim x→7 x − 7 (x − 7)(√x + 2 + 3) = lim x→7 1 √x + 2 + 3 = 1 6. (e) lim x→0 sin(3x) x4 + x Solution: lim x→0 sin(3x) x4 + x = lim x→0 sin(3x) 3x 3x x4 + x = lim x→0 sin(3x) 3x lim x→0 3 x3 + 1 = 1 × 3 = 3. Page 2 of 5 MATH 101 Mid-term exam 1 5 October, 2013 3. (a) (8 points) Complete the following table by putting a tick (check mark) ✓ or a cross × in each empty box to indicate whether each function is: even; odd; increasing; decreasing. Function even odd increasing decreasing 4 sin x × ✓ × × 7 ✓ × × × x + ⌊x⌋ × × ✓ × |2x + 1| × × × × (b) (2 points) If f is an increasing function and g is a decreasing function, does the composition f ◦ g have to be increasing or decreasing? Justify your answer. Solution: For any numbers x1 and x2 with x1 > x2, we have g(x1) < g(x2) because g is decreasing, and then f(g(x1)) < f(g(x2)) because f is increasing. So f ◦ g is decreasing. 4. (6 points) Use the Intermediate Value Theorem to show that the equation cos x = x has a solution between 0.7 and 0.8. Solution: The equation can be re-arranged to f(x) = cos x − x = 0. Now f(0.7) = cos(0.7) − 0.7 = 0.065 . . . and f(0.8) = cos(0.8) − 0.8 = −0.103 . . . . Since f is a continuous function, and 0 lies between −0.103 and 0.065, the Inter- mediate Value Theorem implies that there is some c ∈ [0.7, 0.8] satisfying f(c) = 0, that is, cos c = c. Page 3 of 5 MATH 101 Mid-term exam 1 5 October, 2013 5. (a) (5 points) Consider the function f defined as follows. f(x) = � ⌊x⌋ + 1 if x < 2; x2 − x if x ≥ 2. Is the function f(x) continuous at x = 2? Justify your answer. Solution: If x is slightly smaller than 2, then we have f(x) = ⌊x⌋ + 1 = 1 + 1 = 2. Therefore limx→2− f(x) = 2. On the other hand, lim x→2+ f(x) = lim x→2+(x2 − x) = 2 because x2 − x is continuous. So we have lim x→2− f(x) = lim x→2+ f(x) = f(2) = 2 and therefore the function is continuous at x = 2. (b) (4 points) Suppose that f and g are continuous functions with f(2) = 5 and lim x→2 � 4f(x) − g(x) � = 11. Find g(2), and justify your answer. Solution: Because f and g are continuous, we have lim x→2 f(x) = f(2) = 5 and lim x→2 g(x) = g(2). The limit laws give lim x→2 � 4f(x) − g(x) � = 4 lim x→2 f(x) − lim x→2 g(x) = 20 − lim x→2 g(x). So 20 − limx→2 g(x) = 11, and therefore limx→2 g(x) = 9. Page 4 of 5 MATH 101 Mid-term exam 1 5 October, 2013 6. (14 points) Indicate whether each of the following statements is true or false. You receive two points for each correct answer. T F The line y = 3 is a horizontal asymptote to the graph y = 6x2 − x4 − 5 2x4 + 3 . T F The function f(x) = ⌊|x − 2|⌋ is continuous at x = 2. T F The function f(x) = sin(x − π/2) is an even function. T F If f(3) = 5 and g(5) = 2, then (f ◦ g)(3) = 2. T F If f is a function satisfying |f(x)| ≤ 2 for all x, then lim x→∞ f(x) x = 0. T F The formula sin(3θ) = sin3 θ − 3 sin θ cos2 θ is true for all θ. T F The domain of the function f(x) = √cos x is [−π/2, π/2]. Page 5 of 5