MT171: Calculus - Feedback on the January Test, 10 January 2018 The last section of the MT171 Moodle page holds detailed solutions for this test. I strongly recommend that you consult these solutions to understand the level of work that you are expected to produce. Overall comments (which are quite similar to the ones of last year): . Some of you explained what you were doing but many of you didn't. You are penalised for that. . It was pleasing to see that many of you used the correct notation for integrals; only a few didn't do it. As warned you lose marks for that. . In many cases you used incorrect mathematical notation: a mathematical expression cannot be written on its own and should always be written as part of an equation or of some statement. Feedback on individual questions: Q1: For the vertical asymptotes most students wrote the corresponding equations without explaining that they are found when the denominator is 0. They also missed the justification for it, using the limit of a function. For the slant asymptote almost all students used polynomial division to find the equation of the slant asymptote but many didn't give any argument to justify their claim (again the argument should be based on limits of functions leading to lim f(x)-(x-2) =0, or alternatively to state that , as x +00, f(x) - (x-2). x->+00 Q2: (a) This was one of the better answered questions, although solutions could often have been laid out more clearly and logically. Defining an intermediate variable often helped with that clarity. A few students didn't express their answer for the variable x and some lost marks because of algebraic mistakes. A significant number of students missed the justification showing why only one of the solutions was acceptable. Q3: (a) It was surprising to see that so many of you can't prove this result. From the ones who could, V ? 2 2 needed to consider the function as only one student considered the restriction ? V TT invertible. (b) In this part you needed to do a very straightforward use of the chain rule. However only a few students explained that they were using the chain rule and fewer did use it correctly! Q4: This question was well done. What could have been better? See below, . The auxiliary equation needs to be justified by considering the initial trial function and what follows. . The roots of the auxiliary equation were trivial but some got them wrong! Check your work. . Most students wrote that the general solution was the sum of a homogeneous part (complementary function) and inhomogeneous part (particular integral) but you need to remember that a general solution of a second order differential equation is written with two arbitrary constants!
. This was a very easy particular integral but some treated it as if the trial solution was repeated to one of the linearly independent solutions of the complementary equation ... Also you need to