2013 test 1

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Department
Mathematics and Statistics
Course
MATH 1510
Professor
Iouldouz Raguimov
Semester
Fall

Description
YORK UNIVERSITY Faculty of Science Department of Mathematics and Statistics MATH 1510 6.00 A Test #1 October 18, 2013 Solutions Surname (print): Given Name: Student No: Signature: INSTRUCTIONS: 1. Please write your name, student number and ▯nal answers in ink. 2. This is a closed-book test, duration { 50 minutes. 3. NO CALCULATORS, INTERNET CONNECTED DEVICES OR OTHER AIDS ARE PERMITTED. 4. There are 7 questions on 8 pages. Fill in answers in designated spaces, or in multiple choice questions, circle the correct answer(s). Your work must justify the answer you give. To get part marks show your work on the space provided. Clearly indicate each time when you use the back of a page for your work. 5. Remain seated until we collect all the test papers. 6. Please note that there are 60 points in the test, however your test grade will be counted out of 50. Do the easiest questions ▯rst, GOOD LUCK! 1. (4 + 4 pts) 2 ▯ 4 (a) Simplify the expression (y )▯4: Write your answer without negative exponents. ▯ 5 y 4 Assume x and y denote any positive real numbers. ANSWER: 1 7 5 x y▯4 y▯ 4 ( 5 )4 = ( 1 7)4 y▯4 x y▯ 4 ▯ (4) ▯5 ▯5▯(▯7) 2 y 4 y y y = 1(4) ▯ (4)= x y▯7 = x2 = x2: x2 y 4 √ √ (b) Simplify the expression54xy − y 3128x: Assume that x and y are any real numbers. ANSWER: 3 3 √3 √3 54xy − y 128x = y 27 · 2x − y64 · 2x √3 3 √ √ √ √ = y 3 · 2x − y 4 · 2x = 3y2x − 4 2x = y(3 − 4) 2x = −y 3 2x: 2. (4 + 4 pts) 3 2 (a) Factor the expression x + 4x − 3x − 12 completely by grouping terms. ANSWER: 3 2 2 x + 4x − 3x − 12 = x (x + 4) − 3(x + 4) √ √ = (x + 4)(x − 3) = (x + 4)(x −3)(x + 3): 1 1 (b) Factor the expression x(x + 1) (x + 1)2 and simplify your answer as much as pos- sible. Write your answer in radical form without negative exponents. ANSWER: 1 1 1 x(x + 1)▯2− (x + 1) = (x + 1)▯2[x − (x + 1)] 1 1 = (x + 1)▯2[x − x − 1] = − : x + 1 3. (4 + 4 pts) (a) Simplify the expression and express your answer in a form that is free of radicals in the x + 8 2 denominator (i.e., rationalize the denominato√): √ − √ √ : 5( x − 2) x − 2 ANSWER: x + 8 2 x + 8 − 2(5) x − 2 √ √ − √ √ = √ √ = √ √ 5( x − 2) x − 2 5( x − 2) 5( x − 2) √ √ √ √ (x − 2)( x + 2) (x − 2)( x + 2) 1 √ √ = √ √ √ √ = = ( x + 2): 5( x − 2)( x + 2) 5(x − 2) 5 1 3x 2x (b) Perform the subtractions and simpl2fy: − 2 : x + 3x − 10 x + x − 6 ANSWER: 3x 2x 3x 2x x + 3x − 10− x + x − 6= (x − 2)(x + 5) (x − 2)(x + 3) 2 2 3x(x + 3) − 2x(x + 5)3x + 9x − 2x − 10x = (x − 2)(x + 5)(x + 3)(x − 2)(x + 5)(x + 3) x − x = ; ∀x ̸= −5;−3;2: (x − 2)(x + 5)(x + 3) 4. (4 + 4 pts) 2 1 2x (a) Solve the equation: − = 2 : x + 1 x − 1 x − 1 ANSWER: 2 1 2x x + 1− x − 1= x − 1 2 1 2x ⇐⇒ x + 1− x − 1= (x − 1)(x + 1) ⇐⇒ 2(x − 1) − (x + 1) = 2x ⇐⇒ −x − 3 = 0 ⇐⇒ x = −3: 1 1 1 (b) Solve the equation+ = : x x + 3 4 ANSWER: 1 + 1 = 1 x x + 3 4 1 ⇐⇒ (x + 3) + x 4x(x + 3) ⇐⇒ 4(2x + 3) = x + 3x ⇐⇒ x − 5x − 12 = 0: Hence, √ 2
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