STEP Support Programme. Pure STEP 3 Solutions

Size: px

Start display at page:

Download "STEP Support Programme. Pure STEP 3 Solutions"

Clyde West
5 years ago
Views:

1 STEP Support Programme Pure STEP 3 Solutions S3 Q6 Preparation Completing the square on gives + + y, so the centre is at, and the radius is. First draw a sketch of y 4 3. This has roots at and, and you can show that that there is a minimum point when 3 4. Using the graph we have 4 when or. For the second part, sketch y. This crosses the -ais at only, is asymptotic to the y-ais and has a stationary point at when y is negative. Using the sketch we have when <. iii You should have a diagram looking something like the one below: Note that y includes both y and y. You can find some more on sketching graphs of y f in the STEP Curve Sketching module. For the second and third graphs the gradient tends to infinity as tends to from above. Pure STEP 3 Solutions

2 The STEP 3 question Substituting in z + iy gives: + iy + p + iy + y + iy + p + piy + Real Part: y + p + Imaginary Part: y + py + py Taking the imaginary part then either y or p. Taking y in the real part gives: p + p + for Therefore we either have p or p +. If p then the real part equation becomes y + which becomes + y, i.e. a circle radius centre,. If p + then we have y, which is the -ais with the point, ecluded since we cannot have. Here we have: p + iy + + iy + p py + piy + + iy + Real Part: p py + + Imaginary Part: py + y p + y The imaginary equation gives y or p. Taking these in turn: Substituting y into the real part equation gives p + + i.e. p +. This means that we cannot have, so the locus of this case is the real ais with the origin ecluded. Substituting p into the real part equation gives: + y y y + which is the equation of a circle centre, and radius. Since p to eclude the point, which again is the origin. we need Don t forget to sketch the loci! Pure STEP 3 Solutions

3 iii In this last case we have: p + iy + p + iy + p py + piy + p + p iy + Real Part: p py + p + Imaginary Part: py + p y + ppy The first thing to note is that the real part equation implies that we cannot have p. Hence the imaginary equation implies that either y or p. If y then the real part equation becomes: p + p + Solving for p gives p ± 4 8. p is real if and only if 4 8 and, i.e. iff < or. This locus is the real ais with the segment < ecluded. If p then we have: This gives y y y + The picture looks something like this: To make it clearer, you could put an empty circle on the origin to show that this point is ecluded. Pure STEP 3 Solutions 3

4 S3 Q 3 Preparation d sinln cosln d d e sinln cosln d sinln cosln e e sinln Base case When n the conjecture becomes d, which is true sketch y d if you must! Induction step Assume the conjecture is true when n k, i.e. we have k k. Now consider the case n k +. We have: d d k+ d k d d d k + k d d k k + k k k + k k + k d d k Which is the result when n k +. Hence if it is true for n k then it is true for n k + and as it is true for n it is true for all integers n. iii If f sin then we have: f cos f f sin f f 3 cos f 3 f 4 sin f 4 f 5 cos f 5 This means that the first three non-zero terms in the Maclaurin s epansion of sin are: sin 3 3! + 5 5! +... Pure STEP 3 Solutions 4

5 Hence: sin...3 3! +.5 5!. 6 +,, So to seven decimal places we have sin Pure STEP 3 Solutions 5

6 4 The STEP 3 question d ln + d as required. + Since we are going to need the second derivative as well for the base case, lets work it out here to give: d y d + + 3/ Base case When n we have: + y + + y + y + + 3/ / + + Hence the conjecture is true when n. Assume now that the conjecture is true when n k, so we have: + y k+ + k + y k+ + k y k Differentiating this with respect to gives: [ + y k+3 + y k+] [ + k + y k+ + k + y k+] + k y k+ + [ ] [ y k k + y k+ + k + + k ] y k+ + [ ] [ ] y k y k+3 + [ k k + y k+ + k + k + y k+ ] [ y k+ + k + ] y k+ Which is the conjecture with n k +. Hence if it is true for n k then it is true for n k + and as it is true for n it is true for all integers n. Pure STEP 3 Solutions 6

7 This result can be rearranged to give: + y n+ n + y n+ n y n and so, if we have: y n+ n y n. We now have: y ln y + y + 3/ y 3 y y 4 y y 5 3 y 3 3 y 6 4 y 4 y 7 5 y Hence the Maclaurin s epansion is: y y + y +! y + 3 3! y ! + 5 5! 3 7 7! Pure STEP 3 Solutions 7

8 9 S3 Q8 5 Preparation There are different ways you can approach these. Below is one possible set of solutions. a Limit is, by iv. Note that the limits of f and g must be finite. b Limit is 5, by and. c cos is continuous, so using v: lim t cos t cos lim t d e is continuous, so we can use v to get a limit of. t cos. If we consider t, both the numerator and denominator tend to infinity as t, so et we cannot deduce anything about the limit like this. Instead, try: e t t t e t t + t + t! + t3 3! +... t + + t! + t 3! +... Now as t, The numerator stays as, but the denominator tends to infinity hence lim t e t t. A useful rule of thumb is that eponentials always beat polynomials eventually. iii If e t then as t,. We have: lim ln lim e t ln e t t lim te t t by part. iv a r ln d r ln d [ ln r ln r ] r r [ ] r ln r r r ln r r r d d b Taking the limit as r, and using the result from part, we have: ln d. Pure STEP 3 Solutions 8

9 6 The STEP 3 question Don t forget to do the first bit! This is called the Stem of the question and anything proved or given here is very likely to be useful in the rest of the question. Setting e t gives us: lim m ln n lim e mt t n t n lim t e mt t n using the given result. lim lim e ln e lim ln e Here we have used the Stem result with m n. I n+ m ln n+ d [ m+ ln n+ m + n + m + n + m + I n ] m+ m + n + ln n d m ln n d Note that the third line used the result from the Stem of the question. Now we have: I n n m + I n n m + n m + I n n m + n m + n m +... m + I n n! m + n I Since I [ m m+ d m + ], we have: m + I n n n! m + n+. Pure STEP 3 Solutions 9

10 iii d e ln d ln ln 3 + ln d! 3! [ ] + ln d + ln d + 3 ln 3 d +...! 3! +! + + +!! ! 3! Pure STEP 3 Solutions

11 6 S3 Q7 7 Preparation d d cosh e e sinh. d d sinh e e cosh. cosh is an even function i.e. cosh cosh and as such it is symmetrical about the y-ais. The minimum value is when, y, and it does not cross the -ais. sinh is an odd function, and so it had rotational symmetry about the origin. crosses the aes at, ad the gradient is always positive. It Try using Desmos to check your sketches. iii cosh + sinh e + e + e e e cosh sinh e + e e e e iv cosh sinh e + e e e 4 e + + e 4 e + e v y cosh + c, and y when implies that c and so y cosh +. vi dy d sinh y cosh y cosh y sinh y dy d ln sinh y + c sinh y Ae Pure STEP 3 Solutions

12 8 The STEP 3 question Using the quadratic formula gives: u sinh ± 4 sinh + 4 sinh ± cosh e or e Taking each of these solutions in turn gives: dy d dy e or d e y e + a or y e + b Since we want y and dy d y e +. > which ecludes dy d e at, we have Using the quadratic formula gives: dy d ± sinh y sinh y ± cosh y sinh y Hence we have: sinh y ± cosh y dy d ± lncosh y + c We can only set y into lncosh y +, so doing this gives: and so: ln + c lncosh y + ln lncosh y + ln cosh y + e cosh y e Since cosh y, y is only defined for. The asymptotes are what happens as when y ±. y + ey e y ln 4 y e y e y ln 4 + Therefore the asymptotes are y ± + ln 4. Pure STEP 3 Solutions

13 988 S Q5 9 Preparation a Base case When n we have cos θ + i sin θ cos θ + i sin θ, which is true. You could instead show it is true when n. Inductive step Assume the conjecture is true when n k, so cos θ + i sin θ k cos kθ + i sin kθ. Now consider n k + : cos θ + i sin θ k+ cos θ + i sin θ k cos θ + i sin θ cos kθ + i sin kθ cos θ + i sin θ cos kθ cos θ + i cos kθ sin θ + i sin kθ cos θ sin kθ sin θ cos kθ cos θ sin kθ sin θ + i cos kθ sin θ + sin kθ cos θ cosk + θ + i sink + θ Therefore if it is true when n k, it is true when n k +, and since it is true when n it is true for all integers n. b z 5 r 5 cos θ + i sin θ 5 r 5 cos 5θ + i sin 5θ Then equating real parts gives r 5 cos 5θ hence r and equating imaginary parts gives r 5 sin 5θ sin 5θ. This means that cos 5θ ±. Since r, we must have r and cos 5θ θ, π 5, 4π 5, 6π 5 and 8π 5. We therefore have z cos kπ 5 + i sin kπ 5 for k,,, 3, 4. These are the 5 fifth roots of unity. c We have: cos 5θ + i sin 5θ cos θ + i sin θ 5 Equating real parts gives: cos θ cos θ 4 i sin θ + cos θ 3 i sin θ +... cos 5θ cos θ 5 cos θ 3 sin θ + 5 cos θ sin θ 4 You can then use sin θ cos θ to write this just in terms of cos θ if you needed/wanted to. Considering α α... α n n + a n +... a n + a n gives us α α... α n n a n. Pure STEP 3 Solutions 3

14 The STEP 3 question Let z + iy, so we have + iy 7. The imaginary part of this gives: 7 6 y 35 4 y 3 + y 5 y 7 which means that either y or y + y 4 y 6. It we divide the second equation by 6 and set t y where we get 7 35t + t 4 t 6 i.e. t 6 t t 7. Solving z 7 gives z e πki 7 for k 3,,,,,, 3. Here we are taking the argument as lying between π and π rather than between and π. The reason for this should become clear in the net few lines. When k we have z, and so y. The other 6 solutions relate to the roots of t 6 t t 7. The roots are therefore t y sin πk 7 cos tan πk πk 7. 7 Using α α... α n n a n gives: tan 6π 7 tan 4π 7 tan π 7 tan π 7 tan 4π 7 tan 6π tan 6π 7 tan 4π 7 tan π 7 tan π 7 tan 4π 7 tan 6π 7 7 tan π 7 tan 4π 7 tan 6π 7 7 tan π 7 tan 4π 7 tan 6π 7 ± 7 Then, since tan π 7 is positive, whilst the other two are negative we have: tan π 7 tan 4π 7 tan 6π 7 7 When n 9, start by considering z 9, and the imaginary part of this, which gives: y 6 y y 5 y 7 + y Dividing by 9 and setting t y gives either y or t 8 t 6 + t 4 t The ninth roots of unity are z e πki 9 for k 4, 3,,,,,, 3, 4 where k corresponds to y. We have: tan 8π 9 tan 6π 9... tan 6π 9 tan 8π tan 8π 9 tan 6π 9... tan 6π 9 tan 8π 9 9 tan π 9 tan 4π 9 tan 6π 9 tan 8π 9 9 tan π 9 tan 4π 9 tan 6π 9 tan 8π 9 ± 9 Two of tan values are positive and two are negative, so we have: tan π 9 tan 4π 9 tan 6π 9 tan 8π 9 9 For the final part, follow eactly the same idea, but you will find that tan π are positive whilst tan 6π, tan 8π and tan π are negative, so we have: tan π tan 4π tan 6π tan 8π tan π. and tan 4π Pure STEP 3 Solutions 4

15 3 S3 Q3 Preparation This is quite hard. One way of thinking about it is to start by assuming that p + p + p 3. Now if we rotate everything by about the origin then p p etc, so that p + p + p 3 is unchanged. This means that stays in the same place after a rotation of about the origin, therefore must be at the origin and we have p + p + p 3. p.p i.e. the radius of the circle p + p + p 3.p.p p + p + p 3.p p.p + p.p + p 3.p p.p + p 3.p p.p p 3.p by symmetry p.p iii p. p..p + p.p k.p p i. p i k.p i + i i 3k 3.p i + 3 i 3k. p + p + p k + 3 iv Let p a, b. Then p.p b. We also have p.p a + b a ± 3. So p and p 3 are given by ± 3,. Pure STEP 3 Solutions 5

16 The STEP 3 question Similarly to before we have p + p + p 3 + p 4.p and so p.p + p 3.p + p 4.p. By symmetry we have p i.p 3 for i, 3, 4 and by etending the argument we have p i.p j 3. XP i p i.p i p i. +. i i p i. + i 4 p i. 4 i 8 p + p + p 3 + p p.p 3 b 3 and p.p a + b a 8 9 and since we are told that a > we have a 3. Letting P 3 and P 4 be of the form P j p, q, r we have: p.p j 3 r 3 p.p j 3 3 p p p j.p j 9 + q + 9 q 3 So P 3 and P 4 are given by 3, ±,. 3 3 iii XP i 4 p i.p i i 4 i p i.p i p i. +. i p i. i 4 p i. i p i. i Pure STEP 3 Solutions 6

17 Simplifying further gives: XP i 4 4 i 4 p i. i pi. + p i. i 4 4 p + p + p 3 + p p i. Substituting X, y, z and the values of P i found we have: p i. z + i i p i. 3 3 z y 3 z y 3 z z z + 9 z + z z + 9 z + i 3 3 y + 3 z y + 3 z z z 3 y z z z z + 9 z z + 9 z + 3 y z y y + z since + y + z z 3 y + 3 y + 3 y Hence: XP i p i. i i which is independent of, y and z. Pure STEP 3 Solutions 7

18 8 S3 Q4 3 Preparation The gradient is given by: dy d cosh sinh cosh cosh Therefore the gradient is always positive and tends to as ±. We also have: tanh e e e + e As we have tanh and as we have tanh. This should be enough for you to sketch your graph and you can use Desmos to check your sketch. First note that cosh sinh get: cosh cosh + cosh + sinh cosh sinh cosh + sinh + cosh sinh sinh cosh tanh tanh and cosh + sinh cosh. We then If < we have cosh cosh + tanh. iii If y arcosh then cosh y. Differentiating this second equation with respect to gives us: sinh y dy d dy d sinh y dy d cosh y dy d Pure STEP 3 Solutions 8

19 iv + d sinh y sinh cosh y dy y + sinh y dy cosh y + k + + k To help us integrate directly first note that Hence: + d + + k. d + d +. v A quick sketch of y and y sin remember that is in radians shows that > sin for >. We then have: t dt > sin t dt [ t] [ ] > cos t > cos Pure STEP 3 Solutions 9

20 4 The STEP 3 question Differentiating tanh y with respect to y gives cosh y is less than or equal to. The graphs look like:, so the gradient of tanh y So when y > we have y > tanh y. Letting y arcosh which needs > we have cosh y and: cosh y cosh y cosh y sinh y sinh y + sinh y cosh y sinh y cosh y tanh y Then since we have y > tanh y for y > we have arcosh > for >. Pure STEP 3 Solutions

21 Integrating the LHS of gives: arcosht dt arcosh [ θ cosh θ θ sinh θ dθ ] arcosh [ arcosh sinh θ arcosh arcosh ] arcosh [ cosh θ arcosh cosh θ dθ ] arcosh using t cosh θ integration by parts Integrating the RHS of gives: t arcosh t dt [ sinh θ θ cosh θ sinh θ dθ using t cosh θ sinh θ ] arcosh arcosh Since f > g f d > g d we have: arcosh > arcosh + arcosh > arcosh > + + arcosh > + arcosh > + arcosh > + arcosh > iii Now integrate the result from part. The LHS is as before and the RHS is twice what it was before. This gives us: arcosh > arcosh + arcosh > 3 arcosh > 3 + Pure STEP 3 Solutions

22 3 S3 Q7 5 Preparation If f then f in increasing and so we have f f and for this function this means f for. a cos e sin b sinh c f cosh f d f + f e f f iii Integrating by parts for the first term gives: A sin + B cos d [ A cos ] + A cos d + B sin A sin cos + B sin + k iv If we let f y then dy d f. We have: fe f f d ye y dy ye y e y + k fe f e f + k v Considering cosh we have: and hence we have cosh cosh e + e e / e / When asked to show an inequality it is often easier to rearrange and show greater than or less than. There are other ways you can approach this question, including differentiation to find the minimum point. Pure STEP 3 Solutions

23 6 The STEP 3 question Differentiating E gives: dy d E y dy d d + y 3 d dy d y d d + y 3 Hence E is constant. Since y and dy d when we have E y 4 E dy. d and so Differentiating E gives: de dv d d v dv d d + sinh v d dv d v d d + sinh v dv dv d d dv d de for d We now know that E is decreasing for, so we have E E + cosh ln 3 Since cosh ln 3 e ln 3 + e ln 3 3 we have: E dv + cosh v d 3 and hence cosh v 5 3 as required. iii Here you are left on your own. Comparing to the previous parts suggests that it would be a good idea to consider an E. dw Let E + w sinh w + cosh w. We can find this by comparing to the d other two functions the first term is the square of the relevant derivative and the last term is twice the integral of the last term in the given differential equation. Pure STEP 3 Solutions 3

24 Differentiating E gives us: de dw d d dw d dw d d w d d w + sinh w + w cosh w + sinh w + w cosh w d 5 cosh 4 sinh 3 dw d dw d 5 cosh 4 sinh 3 Considering the last set of brackets here we have: dw d 5 cosh 4 sinh 3 5e + e 4e e 6 e + 9e 6 e e 6e + 9 e e 3 Hence we have E. We also have E + 5, and so E 5 for. This gives: dw + w sinh w + cosh w 5 d w sinh w + cosh w 5 4 and as w sinh w since if w <, sinh w < etc we have cosh w 5 4 as required. Pure STEP 3 Solutions 4

STEP Support Programme. STEP 2 Complex Numbers: Solutions

STEP Support Programme. STEP 2 Complex Numbers: Solutions STEP Support Programme STEP Complex Numbers: Solutions i Rewriting the given relationship gives arg = arg arg = α. We can then draw a picture as below: The loci is therefore a section of the circle between