Oxford MAT Prep: Multiple Choice Questions Dr J Frost ([email protected]) www.drfrostmaths.com Copyright Notice: This resource is free-to-use for all NOT FOR PROFIT contexts only. I do not give permission for them to be used in any context involving financial gain, notably by private tutors or Oxbridge preparation agencies. Last Updated: 28th January 2016 Index Click to go to the corresponding section. Comparing Values Sequences
Trapezium Rule Number Theory Area/Perimeter Remainder Theorem Logarithms Circles
Graph Sketching Calculus Reasoning about Solutions Trigonometry General Points 1. 2. 3.
4. 5. 6. 7. The Oxford MAT paper is the admissions test used for applicants applying to Oxford for Mathematics and/or Computer Science, or to Mathematics at Imperial. It consists of two sections. The first is multiple choice, consisting of 10 questions each worth 4% each (for a total of 40%). The second consists of 4 longer questions, each worth 15% (for a total of 60%). We deal with the first section here. The paper is non-calculator. You need roughly 50% to be invited for interviews. However, successful maths
applicants have an average of around 75%. The questions only test knowledge from C1 and C2. You must ensure you know the content of these two modules inside out. You should also keep in mind that the MAT wont test you on theory you wouldnt have covered, so should think in the context of what you can be expected to do. The multiple choice questions become progressively easier (and quicker) the more you practise. So practise these papers regularly. Even redoing a paper youve done before has value. Ive grouped some of the questions from these papers here by topic, to help you spot some of the common strategies you can use. Comparing Values Preliminary Tips You often have to compare logs. If one gives you a rational number and the other not,
then form an inequality and rearrange to see if one is bigger than the other. E.g.: Which is bigger: or ? Suppose . Taking 5 to the power of each side: Since this is true, we must have that . Make approximations where appropriate. E.g. is roughly 2. Similarly, form inequalities by considering approximations. E.g. is approximately 2, but slightly less than 2, i.e. we know Remember that
Remember that when . Comparing values A B C D
Key Points 1. Make sure you know your sin/cos/tan of 30/45/60 degrees. 2. Be comfortable with making estimates. is somewhere between 1.5 and 2. Similarly is just less than 4. 3. Know your laws of logs like the back of your hand! Comparing values A
B C D Key Points 1. Changing bases occasionally allows us to evaluate less obvious logs. For example 2. We can see that
3. Forming inequalities and manipulating them often helps us compare logs. If were trying to find which of and are bigger, then: If , (by taking 5 to the power of each side) , so we were right. We can use the same technique to show that is bigger. Comparing values A
B C D Key Points 1. Again, realise that is less than 1 when is larger than . 2. The root inside outside and outside the log function has different effects. In (d), we can put on front of the log instead. With (b), we cant
do this. 3. Realise that squaring a number between 0 and 1 makes it smaller. Alternatively: is approximately , because . We can use this value to approximate (b), (c) and (d). Comparing values A B
C D Sequences Preliminary Tips Know your formulae for the following and be able to apply them quickly: Sum of an arithmetic series: Sum of a geometric series: Infinite sum of a geometric series: Two series might be interleaved, e.g. Find the sum of each in turn. In some MAT questions (as well as Senior Maths Challenge) it sometimes helps
to consider the running total when you have a series that oscillates between negative and positive values. Youll see an example of this. Sequences A 2011 B C
D Key Points 1. With sequences, remember that for the (n+1) th term, we can just replace n with n+1. So just substitute these expressions into the inequality. 2. We can sometimes exploit the fact the questions are multiple choice. Once we simplify to , subbing the four different values in tells us (a) must be the answer. 3. Although of course, we could use a C1 approach to solve a quadratic inequality.
Sequences A 2010 B C
D Key Points 1. If you have two interweaved sequences, find formulae for them separately. This means we just want the sum of the first n terms from each of the two. 2. Know your formulae for the sum/infinite sum of a geometric series like the back of your hand. Sequences A
2009 B C D Key Points 1. Sometimes it helps to think about the running total as we progress along
the sum. Our cumulative totals here are 1, -1, 2, -2, 3, -3, 2. Alternatively, try to spot when you can pair off terms such that things either cancel or become the same. In this case, 1-2 = -1, 3-4=-1, and so on. Although this makes it harder to spot exactly when we hit 100 in this case. 3. Think carefully about what happens at the end. Looking at the running totals, if the 1st is 1, the 3rd is 2, the 5th 3, then the (2n-1)th gives us n. So when our running total was 100, 2n-1 = 199. Trapezium Rule Preliminary Tips The Trapezium Rule is: where w is the width of each strip. Know when the trapezium rule overestimates or underestimates area.
Overestimates when line curves upwards. Underestimates when line curves downwards. Trapezium Rule A 2010 B
C D Key Points 1. Remember that that the trapezium rule uses equal length intervals. This suggests that the boundaries of the strips have to coincide with the dots in the diagram, otherwise our
trapeziums wouldnt exactly match the function. 2. (a) we can eliminate because multiples of wont include a . We can eliminate (b) and (c) in a similar way. 3. In general, itll be the LCM of the denominators. Trapezium Rule A
2009 B C D Key Points 1. This is all about being proficient with the sum of a geometric series. The question is more a test
of your dexterity, not your problem solving skills. 2. By the trapezium rule, we get . So using and and , for the summation in the middle we get . Simplifying, we end up with (b). 3. Occasionally we can absorb terms into the geometric series to make our life easier. Since the first length inside the outer brackets was , it makes much more sense to make and . 4. Be careful in considering the number of points you use in the trapezium rule. If you have regions, then you actually have points, of which are duplicated. 5. Again, we could exploit the fact the question is multiple choice. What if we had 1 trapezium! Trapezium Rule A
2008 B C D Key Points 1. Thinking about the question visually helps. A function which curves
upwards will give an overestimate, and a function which curves downwards gives an underestimate. 2. (d) is the only transform which changes the shape of the curve, giving us a reflection on the y-axis (in the line y=1). A curve for example curving up will now curve down, giving us an underestimate. Number Theory Preliminary Tips For many problems it helps to find the prime factorisation of a number: A number is square if and only if the powers in its prime factorisation are even. Similarly, a number is a cube if and only if its powers are divisible by 3, and son on. We can get the number of factors a number has by adding 1 to each
power. E.g. , so it has factors. Diophantine Equations are equations where youre trying to identify integer solutions. There will never be anything too difficult on this front relative to the Maths Olympiad for example, but it may be worth seeing the RZC Number Theory slides on this topic. Know your divisibility rules (also on the RZC Number Theory slides): A number is divisible by 3 if its digits add up to a multiple of 3, divisible by 4 if its last two digits are divisible by 4, divisible by 6 if its both divisible by 2 and 3, and so on. Be sure to check out the Riemann Zeta Club resources on Number Theory on my website. Number Theory
A 2008 B C D
Key Points 1. These kinds of questions are quite common in Maths Challenge papers. The key is to systematically consider how many times each digit appears for the units digit, and then the tens digit (rather than considering each full number in turn). 2. For the units digit, 1 to 9 is each seen 10 times (0 is seen 9 times, but this doesnt matter because it doesnt contribute to the sum). And . 3. Each tens digit, 1 to 9, occurs 10 times. This again gives 450, so our total is 900. Number Theory A
2008 B C D Key Points
1. From the RZC lecture notes, remember that a number is square if all the powers in the prime factorisation are even. 2. The prime factorisation is . 3. Be adept with recognising odd/evenness when considering sums and products. If is even, then consider which of the four statements supports this. E.g. If is even, is even (because ), and since , is even. Number Theory A 2009
B C D Key Points 1. This clearly looks like a binomial expansion! As per the RZC algebra slides, you should always try to spot potential factorisations in number theory
problems. Factorising gives 2. Then so . Since and must be positive (which does not include 0), can have any value between and so that remains at least 1. Number Theory A 2007 B
C D Number Theory A 2010 B
C D Area/Perimeter Preliminary Tips
Remember the trick: The diagonal of square is times longer than the side. Similarly, the side is times shorter than the diagonal. Another quick trick: The height of an equilateral triangle is times longer than half the base. For circles, add key radii at strategic places. Split up the shape into manageable chunks (e.g. right-angled triangles). Youre likely to be able to use either simple trigonometry or Pythagoras. You MUST memorise sin 30/cos 30/sin 60/cos 60. Remember also that and , which helps if youre trying to work out without a calculator. Dont forget your circle theorems (although they dont really feature very prominently in MAT). Sometimes you can come up with two expressions for the same length. Example:
1 1 r The radius of the big circle is 1. What is the radius of the small circle? 2 So ?
Area/Perimeter A 2012 B C
D Key Points 10 2 120 10 2 As per usual, draw in the radius of the circle at strategic places (in this case, where the triangle
touches the circle). This allows us to divide up the triangle into manageable chunks. The area of one third of the triangle is By splitting each isosceles triangle into two right angle ones, then each half of the side of the triangle is . Theres 6 of these half lengths. Using these A and P, we find (a) is true. Area/Perimeter A
2012 B C D Key Points 1. I initially tried to think how the orientation of the spotlight affected the area covered. I
realised that pointing it symmetrically at the opposite end maximised the area. 2. My approach was to draw radii from Q and R to the centre (where angle QOR = using circle theorems) and P to the centre. This gave me a sector and two triangles, and a bit of simple geometry led me to (b). 3. However, we could again exploit the fact we have multiple choice to try a specific case. If we were to choose , we have a semicircle and a right-angled triangle, which quickly gives us . This gives us (b). Area/Perimeter A 2011
B C D Key Points 1. A sensible first step is to form equations for the perimeter and area. Say and . 2. Often we can introduce an inequality when equations are involved by using the
discriminant. 3. Using substitution: . Then using the discriminant: There are multiple alternative approaches: We could have exploited the fact that the question is multiple choice. Choosing a few possible widths and heights will eliminate the incorrect possibilities. The area of a rectangle is maximal when we have a square. We can then generate appropriate inequalities from this, which matches the answer. A more advanced method is using the AM-GM inequality (see my Algebra RZC slides), initially giving Area/Perimeter A
2006 B C D Key Points
1. If the sides of the triangle are , and , start with by Pythagoras. 2. We could always find the areas of the triangles, but its quicker to realise that the area of each triangle is proportional to the area of each implied square around the triangle, say , and . Then , so . Remainder Theorem Preliminary Tips The remainder when is divided by is . If is a factor then clearly because theres no remainder. You might have to factorise the factor first! E.g. If is divisible by , then and . In C2 you only use the remainder theorem for division by linear divisors, i.e. . However, keep in mind the principle of the remainder theorem is that we evaluate for whatever makes the divisor 0. If for example we wanted to find the remainder when dividing by the nonlinear , since and would make this divisor 0, we can evaluate or to find the
remainder. Remainder Theorem A 2008 B C
D Remainder Theorem A 2006 B C
D Remainder Theorem A 2009 B C
D Key Points If is a factor then both and is a factor. By the remainder theorem, and . For the latter, we get or . However, for the former, the middle term will be positive or negative depending on whether is odd or even. can only be 10 or -15. is the only case when both are satisfied. Logarithms Preliminary Tips
If you see numbers like 2 and 8 together for example, you should be able to spot that and somehow use that to simplify. The related groups youll see are usually powers of 2: and powers of 3: . You need to know how to change the base: and are the inverse of each other. This means for example that: Logarithms A 2011 B
C D Key Points 1. Some log tips that frequently crop up: and . This is because the exponential and logarithm functions are inverses of each other, so applying one then the other to gets us back to . 2. Change bases of the exponential to be consistent with the log, so we can cancel in this way.
3. Whenever you have cubics (in this case ), MAT papers are generous, usually allowing you to factorise the first two terms and the last two terms. i.e. . Logarithms A 2007 B C
D Key Points 1. The principle that squared terms are always comes up a LOT in MAT papers (as well as BMO/SMC), so its worth building up a sixth sense for spotting when we can apply this principle. 2. Note that the squaring is outside the log function, so we cant move the 2s to the coefficients. 3. To make a maximum, we need to make b minimum. When then its log will be 0. Then solving, we end up with (c). Note that choosing would make this term larger, as the log of a number in the range is negative, but this would be squared.
Circles Preliminary Tips As per C2, if you have a mixture of , , and , complete the square. When considering nearest points on a circles circumference, draw a line from the centre of the circle to the point of interest/centre of another circle (see below). Consider the conditions for which the equation of a circle are valid: The radius must clearly be positive, and similarly must be positive. The nearest point to this dot on the circumference of the circle can be found by
drawing this straight line from the centre. Nearest points to each other. Circles A 2012 B
C D Key Points 1. Drawing a quick sketch will often massively help (and immediately eliminates a number of possibilities). 2. (a) goes through the top-most and right-most point of the circle. 3. For (c), the x value occurs before the rightmost point where x=2.
4. Similarly, (d) the y-intercept of occurs before the top-most point where y=2. 5. This leaves (b). We neednt do any geometry, but it wouldnt be too difficult to do so if we wanted to verify it. Circles A 2011 B C
D Key Points 1. A general principle is to reflect on the valid range of values for different standard formulae. 1. With a quadratic for example, the discriminant must be if it has a solution. 2. Similarly here, for , then must clearly be positive, since a squared number is always positive. 2. As you would usually do in a C2 exam, complete the square. This gives us . 3. So and so or .
4. Sketching out a sin curve will help visualise the solutions. We can see that in the required range, (b) is the correct answer. Circles A 2009 B C
D Key Points (-3, -4) 1. As per usual, complete the square, and then a sketch may help. 2. By inspection, we can see the nearest point to the origin must be on the line that goes from the centre and through the origin. 3. Since the radius is 10 and the distance from
the centre to the origin is 5 (by Pythagoras), then the answer must be 5. Circles A 2007 B C
D Key Points 1. Again, draw a diagram! 2. Drawing a line between the centres of the circles often helps for questions like these. We can see visually that the nearest point must lie on this line. Circles A
2006 B C D Graph Sketching Preliminary Tips The RZC Graph Theory slides go through a lot of detail on this!
Since this is multiple choice, you should look out for the following features: Is the y-intercept correct? What happens as ? Do the turning points look right? Do the roots (i.e. x-intercepts) look right? Think about whether a transformation is inside some other function or outside. Compare and for example: in the first, as increases, the input we use for sin increases more rapidly, so we accelerate across the sin graph more rapidly. The result is that is gradually oscillates more quickly. Squaring a function always results in a positive value. In polynomial equations: A single factor of means the curve crosses at . A repeated factor of means the curve touches the x-axis at . Graph Sketching
A 2011 B C D
Key Points 1. Consider the y-intercept (by using x = 0) 2. Consider the roots (by using y = 0) 3. Cubics given in MAT exams are often conveniently factorisable. In this case, we can see we can factorise it to . 4. But we didnt even need to do this in this case, because we can eliminate (a) because its in the wrong direction, eliminate (b) because the y-intercept is wrong, and eliminate (d) because . Graph Sketching A
2008 B C D Key Points 1. We could again immediately eliminate (a) and (b) by considering the y-intercept.
2. We could eliminate (d) by considering a small value of x just above 0, and seeing if the value increases or decreases relative to the y-intercept of . E.g. If , then . This is more negative, which eliminates (d) where y increases. 3. But this is an inelegant method which shouldnt be used on anything other than multiple choice, but we cant guarantee our small value occurs before the turning point. You could instead complete the square to get . We can then see that the turning point occurs when , more easily eliminating (d). 4. In general, completing the square (or differentiation) allows us to find turning points. The fact weve taken the reciprocal doesnt change the at which they occur. Graph Sketching A 2007
B C D Key Points 1. Its quite common in MAT (and in interviews) to consider the effects both inside and outside of trigonometric functions.
2. Inside the sin, increases more rapidly than would, so our periodicity decreases. 3. Remember that a squared value is always positive! 4. Consider the shape of . This gradually decreases (although is always positive). Since were multiplying this by our trigonometric function, the peaks will gradually get smaller. 5. This evidence so far gives us (a) or (d). But , which eliminates (d). Graph Sketching A 2010 B
C D Key Points 1. This is similar to the previous question! 2. Inside the sin, increases more slowly than . So were going across a sin graph increasingly slowly, reducing the periodicity. This eliminates (a) and (d). 3. Were squaring the value, so it must be positive (although weve already eliminated (a)).
4. Now consider what happens to the peaks. The peaks will all still be 1, since squaring the value or changing the periodicity wont change this maximum. That eliminates (c). Graph Sketching A 2012 B C
D Key Points 1. This question tests whether youre comfortable with recognising that repeated factors lead to the x-axis being touched (straight from C1). 2. It cant be (b) because the origin is not a root in the graph. 3. It cant be (a) because the curve should touch at -3 and +3 but cross at +1, which is not consistent with the graph. 4. Equation (d) factorises to . So both (c) and (d) initially seem feasible. But considering the y-intercept, we see that only (d) is positive.
Calculus Preliminary Tips Remember that definite integration represents the area under the graph. For some questions, you neednt actually perform the integration, you just need to consider graphically if the area is positive or negative. It may sometimes help to sketch the graph (particularly in light of the above). MAT only expects you to know how to differentiate and integrate polynomials. They cant expect you to know how to deal with trigonometric functions or exponential functions in such instances you need to more generally reason. Whenever you see , you should think to yourself how does A change as B changes?. Thinking this statement may help you which conceptually more difficult questions, or where you wouldnt actually be expected to work out what is (e.g. because you dont know how to differentiate such a function). If you want to integrate a function where the gradient suddenly changes,
e.g. , then you have to split it up into pieces based on where the breaks occur and integrate each. e.g. Whenever you see the words minimum or maximum, think differentiation! Calculus A 2009 B C
D Key Points 1. It might be helpful to first find what I(a) is. If we expand out the bracket, we get 2. Were asked for the smallest value of as varies, so differentiate w.r.t. and set to 0 (as we have a stationary point) and we get . 3. So the question is asking us for 4. A key thing to reflect on here is that the question is purposely trying to bamboozle you by combining integration and differentiation, as well as the fact that youre integrating with respect to , but differentiating with respect to (i.e. different variables). But as
long as you carefully consider what youre trying to do at each step, and with respect to what variable youre differentiating/integrating, then youll be OK! Calculus A 2012 B C
D Key Points 1. You dont need to actually do any integration here (and you wont be able to unless youve done C3!) 2. Looking at the multiple choice options, we only care if T is positive/negative/0. 3. Thus for each of the integrals, we only care whether the area under the graph is above the x-axis or below the x-axis. 4. By sketching the 3 graphs, we find the first area is positive, the second negative and the third positive. Thus T is negative.
Calculus A 2011 B C D
Key Points This one is pretty tricky! The complication obviously comes from the fact that were transforming the input were using for the function were integrating. A good place to start is to note were integrating between 0 and 1, i.e. . Then , so after is transformed before being used in the function, the input to the function is going to vary between only and 0. The equation of the function in this range is . But . So we have Calculus A
2010 B C D Key Points 1. It helps that you know that the paper cant expect you to know how to integrate ! It
suggests were going to have consider a graphical method. 2. If in doubt, sketch your function. 3 = 4 2 2 2 3. I(a) gives us the area up to .
4. Now consider what , actually means. Its asking when the area doesnt change as changes. Notice that as goes past , the area starts to decrease because theres a negative area beyond . Thus the area stops increasing and is about to decrease when . Calculus A 2008 B
C D Calculus A 2007
B C D Key Points 1. The key part here it that divide an area up into smaller chunks, and sum the areas of these smaller chunks. For example 2. Also note that we can take the constant factor outside the integral.
3. So if we let for example and , then were trying to find , when we know that and . These are obviously just simultaneous equations. Calculus Important Note: This question is sufficiently old that it was before the C modules existed at A Level (instead of C1-4 and FP1-3, there was P1-6). This kind of content would now appear in C3, and thus a question like this would no longer appear in a MAT. A 2006 B
C D Reasoning about Solutions Preliminary Tips Theres three ways to consider the number of solutions: METHOD 1: Factorise (when possible!) METHOD 2: Reason about the graph
e.g. This cubic conveniently factorises to: (Example from STEP) Sketch . Thus state the values of for which the equation has: (a) 0 solutions (b) 1 solution (c) 2 solutions (d) 3 solutions We can see it has three solutions (two of them equal). Look out for the difference of two
squares!!! METHOD 3: Consider the discriminant Remember that if there are real solutions, = 4 6 2+9 3 + 3 As changes, the graph slides up and down, we can see the
number of roots change. Clearly if then the curve wont touch the -axis for example. Reasoning about Solutions Preliminary Tips More on METHOD 2: Reason about the graph Cubics will always have at least one solution, because the y-value goes from to . In general, this is true when the greatest power is an odd number. Polynomials where the greatest power is even however (e.g. quadratic/quartic) have a global minimum or maximum. Considering the turning points often allows us to reason about solutions.
Cubic (where coefficient of is positive) Quartic (where coefficient of is positive) Quintic (where coefficient of is positive)
Reasoning about Solutions A 2009 B C D
Key Points Spot when you can use the difference of two squares (also useful for the SMC/BMO!) Make use of the discriminant. Thus If , and using the discriminant on the first, .
Using the discriminant on the second Reasoning about Solutions A 2009 B C
D Key Points By differentiating to find the turning points: So the turning points occur at . Then considering the graph of the quartic: If the x-axis is anywhere in the horizontal trip between the maximum and the greater of the two minimums (whichever it is), well have four solutions because the line will cross the axis 4 times. The y-values of the turning points are and respectively. So so the maximum is above the x-axis, and so that the greater of the two minimums
occurs below the x-axis. Reasoning about Solutions A 2008 B C D
Key Points If we could form a quadratic, then when can clearly use the discriminant: This is very much in the style of a C2 log question. The only trickier thing here is realising that , whereas in C2 papers, you only needed to be able to spot that . The quadratic we want is therefore , where . This gives However, if , we know must be positive (because exponential functions always give positive values). The larger root will always be positive, so this is therefore not an issue. Reasoning about Solutions A
2007 B C D Key Points
Its the same deal! The equation factorises to . As per usual, we were given a cubic which was nice and easy to factorise, and we could factorise the difference of two squares. So or or . The first one doesnt give a real solution, so theres two real solutions. Reasoning about Solutions A 2006 B C
D Key Points When you have , its often its useful to consider the graphs separately and then add them: Now its clear theres no solutions.
But its even easier when we realise that because the modulus function always leaves a positive value, the sum of and must be positive. Thus to sum to zero, both have to be 0. From the first, . But then would be 1. So there cant be any solutions. Trigonometry Preliminary Tips 1. Often when considering the number of possible solutions to an equation which involves trigonometric functions, we need only consider the range of the trig functions, as this may well constrain our solutions considerably: a) and obviously vary between -1 and 1. b) and vary between 0 and 1. 2. In particularly, as before, remember that things that are squared are always positive, i.e. their minimum value is 0. 3. Sometimes dividing the whole equation by // etc. puts our equation just in terms
of one trigonometric function (e.g. leading to a quadratic equation). Trigonometry A 2007 B C
D Key Points We know varies between -1 and 1 and thus between 0 and 1. The has just been put there to throw you: we can still get any value to input into the sin. therefore varies between 16 and 49. Thus the answer is (c). Trigonometry A 2008
B C D Key Points Our usual trick is to consider how each expression can vary as sin/cos varies. We can see that must be at least 4. Similarly, must be at most 4 since is positive. Thus
both expressions must be equal to 4. This happens when and . This gives the one solution . Trigonometry A 2009 B C
D Key Points You should build up a sixth sense of thinking POSITIVE!!! whenever you see something squared. This means that and must be at least 0. So the minimum value of the LHS is . Thus both have to be 0. But and arent both 0 at the same time, so theres no solutions. Trigonometry A
2011 B C D Key Points
I personally drew a quick sketch of and , which are similar to and in that the peaks are still at 1 (with the same x values), but negative values are now positive, and the values (except 1) will be much closer to the -axis since were doing the power of a value . That means well have solutions when (i.e. ) and (i.e. ) or vice versa, which gives . But you can sort see from the sketch that theyll be no solutions in between, due to the lack of symmetry in the line , given that the powers of and are different. But heres a better solution: The equation looks similar to , and thus unless and or vice versa, because taking the power of a number less than 1 reduces the value. Thus except in these circumstances, and there are no other solutions. Remember you have to consider when both and ,
(and the same for sin) otherwise youll miss a solution. Its also possible to use and end up with a difficult factorisation in terms of ! Trigonometry A 2010 B C
D Key Points The key here is dividing by . This gives us a quadratic in terms of . Factorising we get . Alternatively, you could have spotted that you can factorise the original equation as Each gives us 2 solutions.
In general, think about what you might be able to divide by to simplify the equation. Trigonometry A 2008 B C
D Key Points The key here is to get expressions for and in terms of . The only reason we might not have a solution for is if we had some division by 0. By suitable simplification we find theres no division at all, so the answer must be (a). To avoid a notational mess, its helpful to replace with just and so on, i.e. and . Then substitute one into the other! Trigonometry A
2011 B C D Key Points The fact I saw the tangent
suggested to me that I could form a right-angled triangle if I drew in the radius again. I just let the radius be 1. I could then see the two triangles share a common length (say ). For the bottom triangle . For the top triangle, using the sin rule: . Putting these two together, we get (b). Trigonometry
A 2011 B C D Key Points
Its helpful to draw out the two graphs on the same axis, and then shade the appropriate regions. Trigonometry A 2011 B C
D Key Points Clearly if , then , which rules out (b) and (d), as no line goes through the original. But note that since cos/sin repeats every , in general we have for any integer , thus . This gives us all the lines in (a). But note also that . So , i.e.. This gives us the lines with gradient -1 in (c).