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MATH 101 Quiz #5 (v.A1) Last Name: Friday, March 18 First Name: Grade: Student-No: Section: Very short answer question 1. 1 mark Does the series 1 X n=2 n p converge? (Answer “yes” or “no” in the box.) 2n + n Answer: No Solution: We have n 1 1 p = lim p = 6= 0, n!1 2n + 2 n n!1 2 + 1/ n lim so the series diverges by the Test for Divergence. Marking scheme: 1 for a correct answer in the box Short answer questions—you must show your work 2. 2 marks Relate the number 0.86̄ = 0.866666 . . . to the sum of a geometric series, and use that to represent it as a rational number (a fraction or combination of fractions, with no decimals). A calculator-ready answer is acceptable. Answer: 6 6 Solution: The number is 0.8 + 100 + 1000 + 1064 + · · · = 45 + 1062 series sums to 6 1 6 1 = , 1 = 2 10 1 10 10(10 1) 15 so the fraction is 13 15 P1 n=0 10 n . The geometric 4 1 13 + = . 5 15 15 Marking scheme: • 1 mark for writing the geometric series (either with • 1 mark for getting the right answer P or · · · notation) 3. 2 marks Show that the series 1 X n=3 5 converges. n(log n)3/2 5 . Then f (x) is positive and decreasing, so that the sum x(log x)3/2 Z 1 1 X f (n) and the integral f (x) dx either both converge or both diverge, by the Integral Solution: Let f (x) = 3 3 Test. For the integral, we use the substitution u = log x, du = Z 1 Z 1 5 dx 5 du = , 3/2 x(log x)3/2 3 log 3 u dx x to get which converges by the p-test (3/2 > 1). Marking scheme: • 1 point for correctly using the Integral Test. (For this quiz, students can earn this mark even if they don’t state that f (x) is positive and decreasing; however, on the final exam, students must show that they know these hypotheses are important.) • 1 point for showing the integral converges (citing the p-test is OK) Long answer question—you must show your work 4. 5 marks Find the solution to the di↵erential equation Solve completely for y as a function of x. yy 0 ex + e x = Solution: Cross-multiplying, we rewrite the equation as dy = ex + e x dx y 2 dy = (ex + e x ) dx. y2 Integrating both sides, we find 1 3 y = ex 3 Setting x = 0 and y = 5, we find C = 125 ; 3 e x + C. therefore the solution is 1 3 125 y = ex e x + 3 3 x x y = (3e 3e + 125)1/3 . Marking scheme: 1 that satisfies y(0) = 5. y • 1 mark for separating the variables • 2 marks for integrating both sides of the equation • 1 mark for finding the value of the constant • 1 mark for solving for y