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STEP Prep: Step I

STEP Prep: Step I. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Last modified: 27 th December 2013. General points about STEP questions.

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STEP Prep: Step I

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  1. STEP Prep: Step I Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 27th December 2013

  2. General points about STEP questions • STEP questions are in some ways similar to the longer questions in MAT papers. STEP questions are probably, on average, slightly harder, and much more ‘fiddly’, in the sense that a greater degree of working is often required. • Mercifully, you only have to choose 6 of the 13 questions to do. Any extra questions you answer will be marked, but only your 6 best questions will count. • Each question is out of 20 marks, although the breakdown of marks within a question is not made known to you. Most marks however are method marks. • For 2013 the STEP Paper 1 grade boundaries were: Cambridge requires 1 in STEP II and III.

  3. General points about STEP questions • This sounds obvious, by you really need to know your A Level syllabi inside out. You should be well practiced at using trigonometric identities, know sin/cos/tan of 30/45/60 off-hand, etc. • The paper is non-calculator. • You can often answer later parts of questions and pick up marks, even if you weren’t able to finish an earlier part. • Early, possibly guided part of the question (i.e. where a method is recommended) are often there to serve a later unguided part of the question. It’s therefore sensible to try and use a similar/identical approach. • What marks STEP questions out is that you’re given no indication of method. It’s often a case at making an ‘intelligent guess’ at a method which perhaps: • tries to transform you expression/equation into something which resembles a previous part. • reflects on how this part is different to an earlier part, and what would be the most common method to deal with this change. General pointers:

  4. Preparing for STEP • Many STEP questions might seem insurmountable at first, but get easier with practice as you gradually become adept with common strategies/tricks and become quicker and less error prone with your manipulation. • You need to practice regularly and from early on. • One realistic target is to work on 1 problem a day, say for 5 days each week. • This TheStudentRoom thread is excellent: http://www.thestudentroom.co.uk/showthread.php?t=2403970 • All past papers since 1987 with solutions. • There’s a series of support papers by Stephen Siklos (the STEP Chief Coordinator).

  5. Integrating Expressions A Level practice: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

  6. Integrating Expressions A Level practice: ? ? ? ? ? ?

  7. STEP 1 Prep: Integrating Expressions Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 27th December 2013

  8. Integrating Expressions Here we deal with integrating single expressions – we’ll deal with differential equations later. Remember your 3 methods of integration: 1. By inspection (IBI) 2. By substitution (IBS) 3. By parts (IBP) From experience, I tend to find, when given the choice, a substitution often works much better to start off, and IBP often leads to a dead-end. However, a substitution may allow for IBP to then be used. You should be able to spot off hand say that: You’re rarely given the substitution, but the most obvious guess is usually the right one. Don’t forget to change your bounds for definite integration. Sometimes you have to use a further substitution! • For ‘hence integrate ...’ questions, sometimes the bounds have changed. In which case, use a substitution! E.g. If you previously had \int[1,0] and now need to determine \int[2,1], use the substitution u = x – 1 . • It often helps to simplify an expression first. If you have sin(2x), it may or may not be worth replacing it with 2sin(x)cos(x). With fractions, can you simplify using polynomial division?

  9. Integrating Expressions 2011 • Discussion: • IBP: Messy. In fact, you may be tempted to use u = x and du/dx = e^x/(1+x) thinking the latter will give you E (as I initially tried!), but you’ll get stuck. • Substitution: u = 1 + x would seem sensible. • Alternatively, could simplify x/(1+x) using polynomial division. ? • Discussion: • Now polynomial division really would be helpful. • Eventually will get 2 – e + E ? • Discussion: • The substitution u = (1-x)/(1+x) would seem sensible here. • Will eventually reduce to the expression at the start of the question, so answer is E. ? Discussion: Limits have changed! Using u = x2 seems sensible given the power of e, which changes out limits to 1 and 2. Further substitution v = u – 1 needed to change limits to 0 and 1. Answer eE/2 ? Full solutions can be found here: http://www.mathshelper.co.uk/oxb.htm I only give some overarching discussion of approaches here.

  10. Integrating Expressions 2009 • Hint: • Again, either use polynomial division to simplify, or the substitution u = x+1. ? • Hint: • Notice that powered term, xn, has moved from the denominator to the numerator. This suggest a substitution of 1/x. ?

  11. Integrating Expressions 2009 We earlier showed that: • Hints: • We have something similar to the previous result (when m = 2), except we have the x5 + 3 in the numerator. • We can split into two fractions (a common strategy). After simplifying, this gives two integrals in the form we want. • Ignoring the previous parts and using partial fractions also works. Answer: 3/2 + 4ln 2 ? ? • Hints: • The bounds make this one a bit harder, because they’re not in the form m and 1/m. • Again, we could split up into 3 fractions. The second fraction is easy. For the third, we could use result from (ii) to change 2 and 1 to ½ and 1 (making the result negative), before swapping the ½ and 1 so the limits are in the right order, cancelling out the minus. • This combines nicely with the first fraction, with limits 1 and 2, since the two integrals combined have limits of ½ and 2, which is precisely what we want. Answer: 3/8 + ln 3 ? ?

  12. Integrating Expressions 2009 ? • Hints: • This is a classic ‘IBP twice’ question. The fact in the result we’re dividing by an expression involving m suggest we make the rather than the u. • Be careful with signs. • Solution • cos(A + B) + cos(A – B) = 2 cos A cos B • Thus we have • After expanding out the brackets, we have two integrals of the same form as the first part of the question. ?

  13. Integrating Expressions 2009 • Hints • This is a classic “it looks like I have to use a similar approach to the previous question, but I have to work out myself how to adapt this method”. Clearly, we need some way to get sin A sin B as a sum of two cos expressions. • Note that cos(A + B) – cos(A – B) = -2sin A sin B. • Subbing this in and expanding the brackets, we get two integrals, the first exactly the same as the previous part, and the other which we can integrate in the same way. ? ? Solution

  14. STEP 1 Prep: Differential Equations Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 27th December 2013

  15. Differential Equations Some general tips: Be very careful in reflecting on the textual information given to construct your initial equation. Don’t get confused between constants and variables when integrating. Use implicit or explicitly given starting conditions (e.g. “the water level initially has height H” means “h = H when t = 0”) to work out constants of integration. Recall that ‘rate’ means /dt Recall that when we have ln(x) + c, we can simplify this to ln(Ax) by letting A = ln C. These questions tend to be unpopular (the next question was only answered by a third of candidates, and the median mark was 3/20!). They have the potential to go very wrong, but also give the potential for easier marks.

  16. Differential Equations 2011 • Solution • Note first that the water level is gradually increasing, not decreasing. • Water leaks at a rate . Suppose the rate flowing in is F. Then . But at this point, so , i.e. Thus • Solving this differential equation involves noting that h = H when t = 0 (allowing us to work out our constant of integration ). A bit of log manipulation and we’re done. • Using the result in the first sentence of the question, is approximately . But as becomes larger, the -1 in the numerator becomes inconsequential, so we can ignore it, and the fraction simplifies to . ?

  17. Differential Equations 2011 • Solution • We solve in the same way as the first part. We initially find that • This rearranges to • A sensible substitution at this point would be • As before we use t = 0, h = H to find our constant of integration. • The once we substitute t with T’ and h with , it’s just a bit of manipulation to simplify to the desired result. ?

  18. Differential Equations 2003

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