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STEP 2005, Paper 1 – Notes / Solutions (18 pages; 23/1/16)

1 2 3 4 5 6 7 8

Sol'n (N) Sol'n Sol'n Sol'n N N N

9 10 11 12 13 14

(N) (N) Sol'n Sol'n Sol'n Sol'n

STEP 1, 2005, Q1 (S)

(i) A 'case by case' approach can be adopted here. We could, for example,

consider the different possibilities for the 1st digit:

79999: 1 rearrangement of 9999

89998: 4 rearrangements of 9998

99997: 4 rearrangements of 9997

99988: 𝐶2 = 6 4 rearrangements of 9988

giving a total of 15

Alternatively, we could categorise the numbers according to the number of

9s:

4 9s: 99997: 5 rearrangements

3 9s: 99988: 5!

3!2!= 10 rearrangements

again, giving a total of 15

(ii) The 2nd approach in (i) seems to be shorter:

4 9s: 99993: 5 rearrangements

3 9s: 99984/99975/99966:

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5!

3!+

5!

3!+

5!

3!2!= 20 + 20 + 10 = 50 rearrangements

2 9s: 99885/99876/99777 (we can consider the number of 8s here)

5!

2!2!+

5!

2!+

5!

2!3!= 30 + 60 + 10 = 100 rearrangements

1 9: 98886/98877 (again, considering the number of 8s)

5!

3!+

5!

2!2!= 20 + 30 = 50 rearrangements

0 9s: 88887: 5 rearrangements

giving a total of 5 + 50 + 100 + 50 + 5 = 210 rearrangements

STEP 1, 2005, Q2

Straightforward application of straight line equations.

STEP 1, 2005, Q3 (S)

[The phrases "two distinct real solutions" and "exactly one real solution"

virtually guarantee that only a quadratic equation is involved - making this

question very attractive (especially as all the parts are of the 'show that'

type).]

(i) 𝑥

𝑥−𝑎+

𝑥

𝑥−𝑏= 1 ⇒ 𝑥(𝑥 − 𝑏) + 𝑥(𝑥 − 𝑎) = (𝑥 − 𝑎)(𝑥 − 𝑏)

[𝑥 ≠ 𝑎 𝑜𝑟 𝑏 for the original equation to make sense]

⇒ 𝑥2 + 𝑥(−𝑏 − 𝑎 + 𝑎 + 𝑏) − 𝑎𝑏 = 0

⇒ 𝑥2 = 𝑎𝑏

As 𝑎 & 𝑏 are either both +ve or both -ve (and real), 𝑎𝑏 > 0

Hence the two roots 𝑥 = √𝑎𝑏 & − √𝑎𝑏 are real and distinct.

(ii) 𝑥

𝑥−𝑎+

𝑥

𝑥−𝑏= 1 + 𝑐 ⇒ 𝑥(𝑥 − 𝑏) + 𝑥(𝑥 − 𝑎) = (1 + 𝑐)(𝑥 − 𝑎)(𝑥 − 𝑏)

⇒ 𝑥2(2 − [1 + 𝑐]) + 𝑥(−𝑏 − 𝑎 + (𝑎 + 𝑏)(1 + 𝑐)) − (1 + 𝑐)𝑎𝑏 = 0

⇒ (1 − 𝑐)𝑥2 + 𝑐(𝑎 + 𝑏)𝑥 − (1 + 𝑐)𝑎𝑏 = 0

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Exactly one real root ⇔ ∆ = 0

[The symbol ∆ for the discriminant tends not to be used in A level

textbooks (perhaps because it might encourage students to confuse

'discriminant' with 'determinant', which has the same symbol!)]

so that 𝑐2(𝑎 + 𝑏)2 + 4(1 − 𝑐)(1 + 𝑐)𝑎𝑏 = 0

⇔ 𝑐2(𝑎 + 𝑏)2 + 4𝑎𝑏(1 − 𝑐2) = 0

⇔ 𝑐2(𝑎 − 𝑏)2 + 4𝑎𝑏 = 0

⇔ 𝑐2 =−4𝑎𝑏

(𝑎−𝑏)2 (1)

1 − (𝑎+𝑏

𝑎−𝑏)

2=

1

(𝑎−𝑏)2((𝑎 − 𝑏)2 − (𝑎 + 𝑏)2)

=−4𝑎𝑏

(𝑎−𝑏)2 [as this is a 'show that' result, you might want to deliberately give

more working in the exam], so that 𝑐2 =−4𝑎𝑏

(𝑎−𝑏)2 ⇔ 𝑐2 = 1 − (

𝑎+𝑏

𝑎−𝑏)

2

As 𝑎 & 𝑏 are real, (𝑎+𝑏

𝑎−𝑏)

2≥ 0, so that 𝑐2 ≤ 1

Again, as 𝑎 & 𝑏 are real, 𝑐2 ≥ 0

[At this point, we could show very easily that 𝑐2 > 0 from 𝑐2 =−4𝑎𝑏

(𝑎−𝑏)2 , since

𝑎 & 𝑏 are non-zero (2); however, it isn't clear whether the instruction

'deduce' applies to every aspect of the last part. I would have thought that

(2) was acceptable, but the Official sol'ns don't do it this way.]

Suppose that 𝑐2 = 0, so that (𝑎+𝑏

𝑎−𝑏)

2= 1

Then 𝑎+𝑏

𝑎−𝑏= 1 𝑜𝑟 − 1,

so that either 𝑎 + 𝑏 = 𝑎 − 𝑏 ⇒ 𝑏 = −𝑏 ⇒ 𝑏 = 0 (contradiction)

or 𝑎 + 𝑏 = 𝑏 − 𝑎 ⇒ 𝑎 = −𝑎 ⇒ 𝑎 = 0 (contradiction)

Thus 𝑐2 ≠ 0, and so 0 < 𝑐2 ≤ 1, as required.

[Note that we have only proved that this is a necessary condition for

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𝑐2 =−4𝑎𝑏

(𝑎−𝑏)2 to hold; not (necessarily!) a sufficient condition. There could

conceivably be values of 𝑐2 in the range (0,1] for which the result was not

true.]

STEP 1, 2005, Q4 (S)

(a) The diagram below shows the two values of 𝜃 in the range 0 to 2𝜋 for

which 𝑐𝑜𝑠𝜃 =3

5 [note that, from the 3,4,5 triangle, the smaller 𝜃 𝑖𝑠

>𝜋

4 (just for the purpose of sketching]. For these two values, 𝑠𝑖𝑛𝜃 =

4

5 &

−4

5, respectively (from the 3,4,5 triangle; or 𝑐𝑜𝑠2𝜃 + 𝑠𝑖𝑛2𝜃 = 1)

So, as 3𝜋

2≤ 𝜃 ≤ 2𝜋, 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 = 2 (−

4

5) (

3

5) = −

24

25

𝑐𝑜𝑠3𝜃 = cos(2𝜃 + 𝜃) = 𝑐𝑜𝑠2𝜃𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛2𝜃𝑠𝑖𝑛𝜃

Now 𝑐𝑜𝑠2𝜃 = 𝑐𝑜𝑠2𝜃 − 𝑠𝑖𝑛2𝜃 =9

25−

16

25= −

7

25

So 𝑐𝑜𝑠3𝜃 = (−7

25) (

3

5) − (−

24

25) (−

4

5) =

−21−96

125= −

117

125

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(b) 𝑡𝑎𝑛3𝜃 = tan(2𝜃 + 𝜃) =𝑡𝑎𝑛2𝜃+𝑡𝑎𝑛𝜃

1−𝑡𝑎𝑛2𝜃𝑡𝑎𝑛𝜃=

2𝑡𝑎𝑛𝜃

1−𝑡𝑎𝑛2𝜃+𝑡𝑎𝑛𝜃

1−2𝑡𝑎𝑛𝜃

1−𝑡𝑎𝑛2𝜃.𝑡𝑎𝑛𝜃

=2𝑡𝑎𝑛𝜃+(𝑡𝑎𝑛𝜃−𝑡𝑎𝑛3𝜃)

(1−𝑡𝑎𝑛2𝜃)−2𝑡𝑎𝑛2𝜃=

3𝑡𝑎𝑛𝜃−𝑡𝑎𝑛3𝜃

1−3𝑡𝑎𝑛2𝜃

Hence 𝑡𝑎𝑛3𝜃 =11

2⇒

11

2−

33

2𝑡𝑎𝑛2𝜃 = 3𝑡𝑎𝑛𝜃 − 𝑡𝑎𝑛3𝜃.

Writing 𝑥 = 𝑡𝑎𝑛𝜃, 𝑓(𝑥) (𝑠𝑎𝑦) = 2𝑥3 − 33𝑥2 − 6𝑥 + 11 = 0

We have to find a root ≥ 𝑡𝑎𝑛 (𝜋

4) = 1

The question is: does this cubic factorise easily?

We can try applying the factor theorem, but unfortunately this doesn't yield

any integer roots. At this point you could be forgiven for wondering if a

mistake had been made somewhere.

In fact the cubic has a factor of 2𝑥 − 1. However there are a few other

possibilities (assuming of course that a convenient factorisation does in

fact exist). The full list is:

(1) (𝑥 + 1)(2𝑥2 + 𝑎𝑥 + 11)

(2) (𝑥 − 1)(2𝑥2 + 𝑎𝑥 − 11)

(3) (𝑥 + 11)(2𝑥2 + 𝑎𝑥 + 1)

(4) (𝑥 − 11)(2𝑥2 + 𝑎𝑥 − 1)

(5) (2𝑥 + 1)(𝑥2 + 𝑎𝑥 + 11)

(6) (2𝑥 − 1)(𝑥2 + 𝑎𝑥 − 11)

(7) (2𝑥 + 11)(𝑥2 + 𝑎𝑥 + 1)

(8) (2𝑥 − 11)(𝑥2 + 𝑎𝑥 − 1)

(1) & (2) will have already been eliminated by the Factor theorem. It

doesn't take that long to establish whether a consistent value can be found

for 𝑎 in each of the other cases (by equating coefficients), but the process

seems a bit pedestrian for a STEP question.

We could attempt a sketch by considering the location of the turning points.

𝑑𝑦

𝑑𝑥= 6𝑥2 − 66𝑥 − 6

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Then 𝑑𝑦

𝑑𝑥= 0 ⇒ 𝑥2 − 11𝑥 − 1 = 0

⇒ 𝑥 =11±√125

2 ; ie just over 11 and just less than 0

Because the coefficient of 𝑥3 is positive, the general shape of 𝑓(𝑥) indicates

that the maximum will be at just less than 0, whilst the minimim will be at

just over 11.

Also, 𝑓(0) = 11 & 𝑓(1) = −26, so that there is a root between 0 & 1 (since

𝑓(𝑥) is continuous).

Hence we can deduce that there are 3 roots, with the following positions:

< 0 , between 0 & 1, and >11

[For any cubic, there will be a point of inflexion halfway between the

turning points (if they exist), with the curve having rotational symmetry of

order 2 about the point of inflexion.]

From this we can eliminate (4) & (8), but this still leaves 4 options to

investigate.

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We find that only 2𝑥3 − 33𝑥2 − 6𝑥 + 11 = (2𝑥 − 1)(𝑥2 + 𝑎𝑥 − 11) gives a

consistent value for 𝑎:

Equating coefficients of 𝑥2: −33 = 2𝑎 − 1 ⇒ 𝑎 = −16

Equating coefficients of 𝑥: −6 = −22 − 𝑎 ⇒ 𝑎 = −16

So one root is 1

2 (the one we were expecting between 0 & 1). We are trying

to find the root that is ≥ 1, so we want the positive root of 𝑥2 − 16𝑥 − 11 =

0; namely 16+√162+44

2= 8 + √64 + 11 = 8 + √75

[Note that it wasn't necessary to work out 162]

[As a check, the negative root, 8 − √75 is clearly less than 11−√125

2 , the

position of the maximum - as expected.]

STEP 1, 2005, Q5 (S)

(i) For 𝑘 ≠ 0, ∫ (𝑥 + 1)𝑘−1𝑑𝑥 = [1

𝑘(𝑥 + 1)𝑘]

10

=1

𝑘(2𝑘 − 1)

1

0

For 𝑘 = 0, ∫1

𝑥+1𝑑𝑥 = [ln (𝑥 + 1)]

10

= 𝑙𝑛21

0

As (𝑥 + 1)𝑘−1 is a continuous function of 𝑘 around 𝑘 = 0, for a given value

of 𝑥 [ie it doesn't jump at 𝑘 = 0], it follows that the two integrals will get

closer as 𝑘 approaches 0, since they represent areas under curves of

functions that become progressively closer.

So 1

𝑘(2𝑘 − 1) ≈ 𝑙𝑛2 when 𝑘 ≈ 0.

[It isn't really possible at A Level to give a much more rigorous argument.

In this sort of situation, the examiners are likely to be just testing a

candidate's awareness of the issue. Having said all this, the official solutions

don't seem to address the issue at all - which is unusual, as STEP solutions

regularly discuss such refinements. But you can never be entirely sure what

the examiners are going to home in on.]

[Although it may seem a trivial point, it is necessary to explicitly state the

conclusion, as otherwise the examiner may not be sure that you know that

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the solution is complete! (Perhaps more applicable in more complicated

situations.)]

(ii) [The official solutions imply that candidates are likely to notice that

𝑥(𝑥 + 1)𝑚 can usefully be rewritten as (𝑥 + 1 − 1)(𝑥 + 1)𝑚 (though it's a

useful trick to recall).]

Parts is possible. In general though, Parts has a couple of drawbacks:

(i) You can't always be sure that it will lead anywhere. In the case of

definite integrals, it is possible to end up proving that 𝐼 = 𝐼 (where 𝐼 is the

original integral).

(ii) It can require more than one application of the Parts formula.

So it is usually a good idea to consider the other possible approaches first.

Apart from the possibility of the integral being in the formulae booklet,

these are:

(a) looking for a rearrangement of the integrand (as suggested in the

official solutions for this question); this includes breaking down into partial

fractions

(b) a substitution

(i) 𝑢 = 𝑓(𝑥), where the integrand can be written as 𝑓′(𝑥)𝑔(𝑓(𝑥)), provided

that 𝑔(𝑢) can be integrated

(eg 𝑢 = 𝑠𝑖𝑛𝑥, where the integrand is 𝑐𝑜𝑠𝑥𝑠𝑖𝑛7𝑥)

(ii) eg 𝑢 = 𝑥 + 1, to simplify 𝑥(𝑥 + 1)𝑚 (ie as in this question)

(iii) a specialist substitution such as 𝑡 = 𝑡𝑎𝑛 (𝜃

2), which can reduce an

integrand involving trig. functions to one involving powers of 𝑡

[See "Integration Methods" for further details.]

So in this case we can write 𝑢 = 𝑥 + 1, to give

∫ 𝑥(𝑥 + 1)𝑚𝑑𝑥 = ∫ (𝑢 − 1)𝑢𝑚𝑑𝑢 = ∫ 𝑢𝑚+1 − 𝑢𝑚𝑑𝑢2

1

2

1

1

0

The 3 cases to consider are 𝑚 = −1, 𝑚 = −2 and other 𝑚.

For 𝑚 = −1, ∫ 𝑢𝑚+1 − 𝑢𝑚𝑑𝑢2

1= ∫ 1 −

1

𝑢𝑑𝑢 = [𝑢 − 𝑙𝑛𝑢]2

12

1

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= (2 − 𝑙𝑛2) − (1 − 0) = 1 − 𝑙𝑛2

For 𝑚 = −2, ∫ 𝑢𝑚+1 − 𝑢𝑚𝑑𝑢2

1= ∫

1

𝑢− 𝑢−2𝑑𝑢 = [𝑙𝑛𝑢 + 𝑢−1]2

12

1

= (𝑙𝑛2 +1

2) − (0 + 1) = 𝑙𝑛2 −

1

2

For other 𝑚, ∫ 𝑢𝑚+1 − 𝑢𝑚𝑑𝑢2

1= [

1

𝑚+2𝑢𝑚+2 −

1

𝑚+1𝑢𝑚+1]

21

= (2𝑚+2

𝑚+2−

2𝑚+1

𝑚+1) − (

1

𝑚+2−

1

𝑚+1)

=2𝑚+1

(𝑚+1)(𝑚+2)[2(𝑚 + 1) − (𝑚 + 2)]

−1

(𝑚+1)(𝑚+2)[(𝑚 + 1) − (𝑚 + 2)]

=2𝑚+1(𝑚)+1

(𝑚+1)(𝑚+2)

Q6 For (ii), we have to eliminate one variable (𝑘2) from two equations

involving three variables. In order to minimise the risk of errors (and of

possibly going round in circles), a standard 'sledgehammer' approach is

simply to make 𝑘2 the subject of each equation, and equate the two

expressions for 𝑘2. In fact, this gives the required result straightaway, but

the method could have been applied even had this not been the case.

For the last part, the fact that 𝑎 ≠ 𝑏 virtually tells us to divide by 𝑎 − 𝑏 at

some point.

Note that each part is a 'show that' question.

Q7 Parts (i) & (ii) are straightforward. Once numbers are plugged in (ie

𝑟 = 1,2, … ), part (iii) is no harder (and doesn't use parts (i) & (ii)). Of

course it looks horrible, but it shows the value of starting to write things

out.

(Instead of rearranging the rth term - as done in the official sol'ns - it is

possible to first simplify the term for r=1, then r=2, and observe the

pattern.)

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Q8 Straightforward and short. Because the only method of solving

differential equations in the C1-C4 syllabus is Separation of Variables, there

is really no theory to know for these questions.

The Hints & Answers states that the solution in (i) is the pair of straight

lines 𝑦 = ±(𝑥 + 1), but this seems to overlook the fact that 𝑦 = 2 when

𝑥 = 1. So presumably it should just be the straight line 𝑦 = 𝑥 + 1.

Also, for (ii), 𝑦 = 1 when 𝑥 = 1, so presumably the solution of 𝑦2 =(𝑙𝑛𝑥)2+1

𝑥

stated in the Hints & Answers should be modified to 𝑦 = √(𝑙𝑛𝑥)2+1

𝑥 (ie

excluding the negative root).

Q9 Also straightforward and short.

Q10 Unusually, there are (apparently) no complications in this question.

STEP 1, 2005, Q11

[Note that the sketch in (iv) is likely to be fairly time-consuming.]

(i) 𝑠𝑖𝑛2𝑡 = 2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡 = 0 & 2𝑐𝑜𝑠𝑡 = 0 ⇒ 𝑐𝑜𝑠𝑡 = 0 ⇒ 𝑡 =𝜋

2 𝑜𝑟

3𝜋

2

(ii) [It's probably a good idea to use 𝑣 instead of �̇� , as the dot might get

overlooked (either by yourself or the examiner).]

𝑣 = 2𝑐𝑜𝑠2𝑡𝑖 − 2𝑠𝑖𝑛𝑡𝑗

velocity is perpendicular to displacement when 𝑟. 𝑣 = 0 , provided that 𝑟

and 𝑣 are not zero [this seems to have been overlooked in the official sol'ns;

normally they're quite particular about this sort of refinement]

𝑟. 𝑣 = 0 ⇒ 2𝑠𝑖𝑛2𝑡𝑐𝑜𝑠2𝑡 − 4𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡 = 0

⇒ 𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡𝑐𝑜𝑠2𝑡 − 𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡 = 0

⇒ 𝑒𝑖𝑡ℎ𝑒𝑟 𝑠𝑖𝑛𝑡 = 0 𝑜𝑟 𝑐𝑜𝑠𝑡 = 0 𝑜𝑟 𝑐𝑜𝑠2𝑡 = 1

⇒ 𝑡 = 0, 𝜋,𝜋

2 𝑜𝑟

3𝜋

2

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but the particle is at the Origin at 𝑡 =𝜋

2 𝑜𝑟

3𝜋

2 ; ie its displacement vector is

0, so it doesn't make sense to talk about the velocity being perpendicular to

it; so we are left with 𝑡 = 0 𝑜𝑟 𝜋.

(iii) 𝑣 parallel to 𝑟 ⇒2𝑐𝑜𝑠2𝑡

𝑠𝑖𝑛2𝑡=

−2𝑠𝑖𝑛𝑡

2𝑐𝑜𝑠𝑡 (1) , provided that 𝑠𝑖𝑛2𝑡 & 𝑐𝑜𝑠𝑡 ≠ 0

As particle is not at O, 𝑐𝑜𝑠𝑡 ≠ 0

If 𝑠𝑖𝑛2𝑡 = 0 & 𝑐𝑜𝑠𝑡 ≠ 0, 𝑡ℎ𝑒𝑛 𝑠𝑖𝑛𝑡 = 0, 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑡 = 0 𝑜𝑟 𝜋

𝑡 = 0 ⇒ 𝑟 = 2𝑗 & 𝑣 = 2𝑖, 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑟 & 𝑣 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙

𝑡 = 𝜋 ⇒ 𝑟 = −2𝑗 & 𝑣 = 2𝑖, 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑎𝑔𝑎𝑖𝑛 𝑟 & 𝑣 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙

Then (1) ⇒ 2𝑐𝑜𝑠2𝑡𝑐𝑜𝑠𝑡 = −𝑠𝑖𝑛𝑡(2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡)

⇒ 𝑐𝑜𝑠2𝑡 + 𝑠𝑖𝑛2𝑡 = 0 (𝑎𝑠 𝑐𝑜𝑠𝑡 ≠ 0)

⇒ (𝑐𝑜𝑠2𝑡 − 𝑠𝑖𝑛2𝑡) + 𝑠𝑖𝑛2𝑡 = 0

⇒ 𝑐𝑜𝑠2𝑡 = 0

⇒ 𝑐𝑜𝑠𝑡 = 0

But 𝑐𝑜𝑠𝑡 ≠ 0, so that there are no situations when 𝑣 is parallel to 𝑟.

(iv) |𝑟|2

= 𝑠𝑖𝑛2(2𝑡) + 4𝑐𝑜𝑠2𝑡 = 4𝑠𝑖𝑛2𝑡𝑐𝑜𝑠2𝑡 + 4𝑐𝑜𝑠2𝑡

= 4(1 − 𝑐𝑜𝑠2𝑡)𝑐𝑜𝑠2𝑡 + 4𝑐𝑜𝑠2𝑡

= 8𝑐𝑜𝑠2𝑡 − 4𝑐𝑜𝑠4𝑡

𝑑

𝑑𝑡(8𝑐𝑜𝑠2𝑡 − 4𝑐𝑜𝑠4𝑡 ) = 0

⇒ −16𝑐𝑜𝑠𝑡𝑠𝑖𝑛𝑡 + 16𝑐𝑜𝑠3𝑡𝑠𝑖𝑛𝑡 = 0

⇒ 𝑠𝑖𝑛𝑡 = 0 (𝐴), 𝑐𝑜𝑠𝑡 = 0 (𝐵) 𝑜𝑟 𝑐𝑜𝑠2𝑡 = 1 (𝐶)

(𝐴) ⇒ |𝑟|2

= 4

(𝐵) ⇒ |𝑟|2

= 0

(𝐶) ⇒ 𝑡 = 0 𝑜𝑟 𝜋

𝑡 = 0 ⇒ |𝑟|2

= 4 & 𝑡 = 𝜋 ⇒ |𝑟|2

= 4 also

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So the max. distance is 2

[A fairly safe way of sketching the curve is to establish 𝑟 & 𝑣 at intervals of 𝜋

4 for t, and draw arrows to represent the direction of motion (ie the

direction of 𝑣 ). As it's a bit time-consuming, you might want to save this for

the end of the exam, when corners can be cut if necessary.]

[ (1) is 𝑡 = 0, (2) is 𝑡 =𝜋

4 etc; (9) would be 𝑡 = 2𝜋, which is equivalent to

𝑡 = 0; since 𝑐𝑜𝑠𝑡 has a period of 2𝜋 and 𝑠𝑖𝑛2𝑡 has a period of 𝜋 (being 𝑠𝑖𝑛𝑡

stretched by a scale factor of 1

2)]

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STEP 1, 2005, Q12

In finding the range of possible values of q in (b), it is not easy to be sure

that all of the constraints have been taken into account. The official sol’ns

don’t explain the reason behind the substitution x = 0.6 – k.

An alternative sol’n for this part is as follows:

[Note that Prob(Hat only) etc. have been expressed in terms of x, y & z – in

order to keep the number of variables to a minimum.]

q = 𝑥

0.8

From the previous part, x+y+z = 0.75 (A).

[Although we are trying to find x, we can reduce the number of variables

further by substituting 0.75 – y - z for x (which can be retrieved at the

end).]

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All the constraints can then be summarised by:

0≤0.6-x-y = 0.6 – (0.75-y-z) – y = z – 0.15 (B)

0≤0.7-x-z = 0.7 - (0.75-y-z) – z = y – 0.05 (C)

0≤0.3-y-z (D)

together with x≥0, y≥0 & z≥0

(B) & (C) then give:

z ≥ 0.15 (which takes care of z≥0)

y ≥ 0.05 (which takes care of y≥0)

This then gives y+z ≥ 0.05+0.15 = 0.2

and (D) gives y+z ≤ 0.3

Then, from (A), 0.75 – 0.3 ≤ x ≤ 0.75 – 0.2; ie 0.45 ≤ x ≤ 0.55,

so that 45

80 ≤ x ≤

55

80 or

9

16 ≤ x ≤

11

16

STEP 1, 2005, Q13

[You can never be sure with this type of question whether the

earlier parts are intended to be used later, or are just there as a

warm-up exercise (as in this case, it seems).]

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15

(a)

𝑃 (𝜇 −𝜎

2≤ 𝑋 ≤ 𝜇 + 𝜎) = 1 − (1 − 𝑏) − (1 − 𝑎) = 𝑎 + 𝑏 − 1

(b) 𝑃 (𝑋 ≤ 𝜇 +𝜎

2|𝑋 ≥ 𝜇 −

𝜎

2) =

𝑃(𝜇−𝜎

2≤𝑋≤𝜇+

𝜎

2)

𝑃(𝑋≥𝜇−𝜎

2)

=1−2(1−𝑏)

𝑏=

2𝑏−1

𝑏

(b)(i) [Some parts of this question involve ≤ etc; others < ;

obviously it makes no difference.]

Let Y be the volume for the unknown type of milk.

Let SM denote skimmed milk and FF full fat.

Then 𝑃(𝑌 > 500|𝑌 < 505) =𝑃(500<𝑌<505)

𝑃(𝑌<505)

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𝑃(𝑌 < 505) = 𝑃(𝑆𝑀)𝑃(𝑌 < 505|𝑆𝑀) + 𝑃(𝐹𝐹)𝑃(𝑌 < 505|𝐹𝐹)

= 0.6𝑏 + 0.4𝑎

Similarly, 𝑃(500 < 𝑌 < 505)

= 0.6(𝑏 − 0.5) + 0.4𝑃(𝜇 +𝜎

2< 𝑋 < 𝜇 + 𝜎)

= 0.6(𝑏 − 0.5) + 0.4(1 − [𝑏 + 1 − 𝑎]) [see diagram above]

= 0.6(𝑏 − 0.5) + 0.4(𝑎 − 𝑏)

= 0.2𝑏 + 0.4𝑎 − 0.3

Hence 𝑃(𝑌 > 500|𝑌 < 505) =0.2𝑏+0.4𝑎−0.3

0.6𝑏+0.4𝑎=

2𝑏+4𝑎−3

6𝑏+4𝑎

(b)(ii) 𝑃(𝑌 ≤ 505) = 0.7 ⇒ 0.6𝑏 + 0.4𝑎 = 0.7 (from (b)(i)) (1)

𝑃(𝐹𝐹|𝑌 ≥ 495) =1

3

⇒𝑃(𝐹𝐹 & 𝑌≥495)

𝑃(𝑌≥495)=

1

3 (2)

𝑃(𝐹𝐹 & 𝑌 ≥ 495) = 0.4(0.5) = 0.2

𝑃(𝑌 ≥ 495) = 0.6𝑃 (𝑋 ≥ 𝜇 −𝜎

2) + 0.4(0.5)

= 0.6𝑏 + 0.2

Then (2) ⇒0.2

0.6𝑏+0.2=

1

3

⇒ 0.6 = 0.6𝑏 + 0.2

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⇒ 0.4 = 0.6𝑏 ⇒ 𝑏 =2

3

Then, from (1), 0.4𝑎 = 0.7 − 0.6 (2

3) = 0.3 ⇒ 𝑎 =

3

4

STEP 1, 2005, Q14

[According to the ER, there were no successful attempts at this

question; which indicates how under-rated these distribution

questions are: there is hardly any theory involved, and the

question often amounts to no more than a bit of integration.]

[Be careful not to use the cdf instead of the pdf, or vice-versa, for

distribution questions.]

(i) Total prob. =1, so

𝑚 + 𝑘(1 − 𝑒−∞) = 1 ⇒ 𝑚 + 𝑘 = 1 ⇒ 𝑘 = 1 − 𝑚

(ii) For 0 ≤ 𝑋 < ∞, pdf of X =𝑑

𝑑𝑥(𝑘(1 − 𝑒−𝑥))

= 𝑘𝑒−𝑥 = (1 − 𝑚)𝑒−𝑥

𝐸(𝑋) = 𝑃(𝑋 = −1)(−1) + ∫ 𝑥. (1 − 𝑚)𝑒−𝑥𝑑𝑥∞

0 (*)

= −𝑚 + (1 − 𝑚)[𝑥(−𝑒−𝑥)]∞0

− (1 − 𝑚) ∫ (−𝑒−𝑥)𝑑𝑥∞

0

[integrating by Parts]

since 𝑥𝑒−𝑥 → 0 𝑎𝑠 𝑥 → ∞

= −𝑚 + 0 + (1 − 𝑚)[−𝑒−𝑥]∞0

= −𝑚 + (1 − 𝑚)(0 + 1) = 1 − 2𝑚

(iii) 𝑉𝑎𝑟(𝑋) = 𝐸(𝑋2) − (𝐸(𝑋))2

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𝐸(𝑋2) = 𝑚(−1)2 + ∫ 𝑥2. (1 − 𝑚)𝑒−𝑥𝑑𝑥∞

0

= 𝑚 + (1 − 𝑚)[𝑥2(−𝑒−𝑥)]∞0

− (1 − 𝑚) ∫ 2𝑥(−𝑒−𝑥)𝑑𝑥∞

0

= 𝑚 + 0 + 2(1 − 𝑚) ∫ 𝑥𝑒−𝑥𝑑𝑥∞

0

= 𝑚 + 2[𝐸(𝑋) + 𝑚], 𝑓𝑟𝑜𝑚 (∗)

= 𝑚 + 2[1 − 2𝑚 + 𝑚] = 2 − 𝑚

Then 𝑉𝑎𝑟(𝑋) = 2 − 𝑚 − (1 − 2𝑚)2 = 2 − 𝑚 − 1 + 4𝑚 − 4𝑚2

= 1 + 3𝑚 − 4𝑚2

Let M be the median.

Then 𝑃(𝑋 < 𝑀) =1

2 ,

so that 𝑚 + 𝑘(1 − 𝑒−𝑀) = 1/2 (since 𝑚 < 1/2, so that 𝑀 > 0)

Then 1 − 𝑒−𝑀 =1

2−𝑚

1−𝑚=

1−2𝑚

2(1−𝑚)

⇒ 𝑒−𝑀 = 1 −1−2𝑚

2(1−𝑚)=

2−2𝑚−1+2𝑚

2(1−𝑚)=

1

2(1−𝑚)

⇒ 𝑒𝑀 = 2(1 − 𝑚) and 𝑀 = ln (2 − 2𝑚)

(iv) 𝐸(|𝑋|1/2) = 1(𝑚) + ∫ 𝑥1/2(1 − 𝑚)𝑒−𝑥𝑑𝑥∞

0

Let 𝑦2 = 𝑥, so that 2𝑦𝑑𝑦 = 𝑑𝑥

Then 𝐸(|𝑋|1/2) = 𝑚 + 2(1 − 𝑚) ∫ 𝑦2𝑒−𝑦2𝑑𝑦

∞

0

= 𝑚 + 2(1 − 𝑚).1

4√𝜋

= 𝑚 +1

2(1 − 𝑚)√𝜋