YORK UNIVERSITY. Faculty of Science Department of Mathematics and Statistics. MATH C Test #1. October 23, 2013.

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1 YORK UNIVERSITY Faculty of Science Department of Mathematics and Statistics MATH C Test #1 October 3, 013 Solutions Family Name (print): Student No: Given Name: Signature: INSTRUCTIONS: 1. Please write your name, student number and final answers in ink.. This is a closed-book test, duration 50 minutes. 3. Non-programmable, non-graphing calculator is permitted. NO INTERNET CON- NECTED DEVICES OR OTHER AIDS ARE PERMITTED. 4. There are six questions on seven pages. Answer all questions. Fill in answers in designated spaces. Your work must justify the answer you give. Show your work on the space provided. If you need more space, use the back of a page and clearly indicate this fact on an original page each time when you use the back of a page for your work. 5. Remain seated until we collect all the test papers. 6. Do the easiest questions first, GOOD LUCK!

2 1. ( pts) (a) Solve the following equation for x: ln(8 x) ln(4 + x) = ln( + x). ln(8 x) ln(4 + x) ln( + x) = 0 ln (8 x) (4 + x)( + x) = ln 1 (8 x) (4 + x)( + x) = 1 8 x = (4 + x)( + x) x + 8x = 0 x(x + 8) = 0. Hence, x = 0 or x = 8. However, if x = 8, then 4 + x < 0 and + x < 0. Therefore, x = 0 is the only solution. (b) Solve the following equation for x on the interval [0, π): sec x = 3 tan x + 1. sec x 1 = 3 tan x tan x 3 tan x = 0 tan x(tan x 3) = 0. So, tan x = 0 or tan x = 3. Therefore, the solutions are x = 0 and x = π 3. (c) Solve the following inequality for x: 4 5x <. Be sure to write your answer in interval notation. Using the property of the absolute value, < 4 5x < 4 < 5x < 4 6 < 5x < Hence, x ( 5, 6 5 ). < 5x < 6 5 < x < 6 5.

3 . (4 + pts) (a) Determine the equation in standard form for the line passing through the point ( 1, 1) and perpendicular to the line x 3y 6 = 0. x 3y 6 = 0 y = 1 3 x. Whence, the slope of the line x 3y 6 = 0, m 1 = 1. Then the line perpendicular 3 to the given line will have a slope m = 1 = 3. So, the point-slope form of an m 1 equation of the line passing through the point ( 1, 1) and perpendicular to the given line will be y ( 1) = 3(x ( 1)) or in the standard form y +3x+4 = 0. (b) Find the amplitude and period of the function f(x) = 3 cos( πx ). The amplitude, a = 3 = 3. Let p be the period of f(x). Then since the period of cos x is π, 3. ( pts) π p = π = πp = 4π = p = 4. (a) Find the largest domain on which the function f(x) = Dom (f) = {x R x x 3 > 0}. 1 x x 3 is defined. Whence, x x 3 > 0 (x 3)(x + 1) > 0. x x 3 > 0 either (x 3 > 0 and x+1 > 0) or (x 3 < 0 and x+1 < 0). either (x > 3 and x > 1) or (x < 3 and x < 1). But x > 3 and x > 1 = x > 3 = x (3, ), and x < 3 and x < 1 = x < 1 = x (, 1). Therefore, Dom (f) = (, 1) (3, ). 3

4 (b) Show that the function g(x) = 1 + is one-to-one. x 1 x 1, x Dom g(x), f(x 1 ) = f(x ) = 1 x = 1 x 1 + = 1 x 1 1 = 1 x 1 = x 1 1 = x 1 = x 1 = x. (c) Find the inverse of g(x). What is the range of g 1 (x)? y = 1 x 1 + = y = 1 x 1 = 1 y = x 1 = x = 1 y + 1. Therefore, g 1 (x) = 1 y + 1 = y 1 y. Range (g 1 ) = Dom (g) = (, 1) (1, ). 4. (5 + 3 pts) The radioactive isotope gallium-67 (Ga 67 ), which is used in diagnosis of malignant tumors, decay exponentially and has a half-life of 46.5 hours. (a) Determine the formula for the amount of isotope at any time t > 0. If we start with 100 milligrams of the isotope, how many milligrams (to two decimal places) will be left in 4 hours? Let W (t) be the amount of the isotope after t hours. Then, W (t) = W 0 e λt, t 0. We are given the half-life, T h = So, 0.5W 0 = W 0 e λ(46.5) = 0.5 = e λ(46.5). Taking the natural logarithm of the both sides, we obtain ln 0.5 = λ(46.5) or λ = ln = ln Thus, W (t) = W 0 e ln 46.5 t, t > 0. Now, when W 0 = 100 and t = 4, we will obtain W (4) = 100e ln = 69.9 (milligrams). (b) In how many hours will there be only 5 milligrams of the isotope left? Use the formula from part (a) to find t such that 5 = 100e ln 1 4 = e ln 4

5 ln 1 4 = ln ln = ln t = 46.5 = ( pts) Explain how the following functions can be obtained from y = cos x by basic transformations: (a) y = 1 + cos x; Stretch y = cos x vertically by the factor, then shift up one unit. (b) y = cos(x + π 4 ); Shift y = cos x to the left π 4 (c) y = cos( π x). units, then reflect about the x-axis. Since cos is an even function, cos( π x) = cos[ (x π )] = cos(x π ). Hence, it is sufficient to consider the equivalent function y = cos(x π ). To obtain the latter shift y = cos x to the right π units, then reflect about the x-axis. Note: cos( π x) = sinx. 6. (3 + 6 pts) (a) Consider the following system of linear equation { x 3y = r x + 6y = 4 where r is a constant. Determine values of r such that the system has: i. Infinitely many solutions, r =. ii. No solution, r. iii. Exactly one solution. No such value of r exist. 5

6 (b) Find all solutions to the following system of linear equation: 5x 3y + z = 17 x + 4y z = 7 x 11y + 4z = 3 Use Gaussian elimination method to reduce the augmented coefficient matrix of the system: Hence, the given system of linear equations is equivalent to the following system: x 11y + 4z = 3 y 9 z = = 0 which has infinitely many solutions. Let z = t, where t is any real number. Then from the second equation and from the first equation y = z = t, x = y 4z = t. Therefore, for any real number t, an ordered triple ( t, t, t) is a 6 solution to the system. The end. 6

Math 110 Test # 1. The set of real numbers in both of the intervals [0, 2) and ( 1, 0] is equal to. Question 1. (F) [ 1, 2) (G) (2, ) (H) [ 1, 2]

Math 110 Test # 1. The set of real numbers in both of the intervals [0, 2) and ( 1, 0] is equal to. Question 1. (F) [ 1, 2) (G) (2, ) (H) [ 1, 2] Friday July 8, 00 Jacek Szmigielski Math 0 Test # Fill in the bubbles that correspond to the correct answers. No aids: no calculators, closed book. You are not permitted to consult with your fellow students