This preview shows page 1. Sign up to view the full content.
Unformatted text preview: STEP Solutions
2010
Mathematics
STEP 9465/9470/9475
October 2010 The Cambridge Assessment Group is Europe's largest assessment agency
and plays a leading role in researching, developing and delivering
assessment across the globe. Our qualifications are delivered in over 150
countries through our three major exam boards.
Cambridge Assessment is the brand name of the University of Cambridge
Local Examinations Syndicate, a department of the University of Cambridge.
Cambridge Assessment is a notforprofit organisation.
This mark scheme is published as an aid to teachers and students, to indicate
the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners. It does not indicate the details of the discussions
which took place at an Examiners’ meeting before marking commenced.
All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the
relevant knowledge and skills demonstrated.
Mark schemes should be read in conjunction with the published question
papers and the Report on the Examination.
Cambridge Assessment will not enter into any discussion or correspondence
in connection with this mark scheme.
© UCLES 2010
More information about STEP can be found at:
http://www.atsts.org.uk Contents
STEP Mathematics (9465, 9470, 9475)
Report
STEP Mathematics I
STEP Mathematics II
STEP Mathematics III Page
4
44
54 Question 1 Given that
5x2 + 2y 2 − 6xy + 4x − 4y ≡ a(x − y + 2)2 + b(cx + y )2 + d,
ﬁnd the values of the constants a, b, c and d.
We expand the right hand side, and then equate coeﬃcients:
5x2 + 2y 2 −6xy + 4x − 4y
≡ a(x − y + 2)2 + b(cx + y )2 + d
≡ a(x2 − 2xy + y 2 + 4x − 4y + 4) + b(c2 x2 + 2cxy + y 2 ) + d
≡ (a + bc2 )x2 + (2bc − 2a)xy + (a + b)y 2 + 4ax − 4ay + 4a + d,
so we require
a + bc2 = 5
2bc − 2a = −6
a+b=2
4a = 4
−4a = −4
4a + d = 0.
The fourth and ﬁfth equations both give a = 1 immediately, giving b = 1 from the third
equation. Then the second equation gives c = −2 and the ﬁnal equation gives d = −4. We must also check that this solution is consistent with the ﬁrst equation. We have
a + bc2 = 1 + 1 × (−2)2 = 5, as required. (Why is this necessary? Well, if the second
equation had begun with 7x2 + · · · , then our method would still have given us a = 1, etc.,
but the coeﬃcients for the x2 term would not have matched, so we would not have been
able to write the second equation in the same way as the ﬁrst.)
We thus deduce that
5x2 + 2y 2 − 6xy + 4x − 4y ≡ (x − y + 2)2 + (−2x + y )2 − 4. Solve the simultaneous equations
5x2 + 2y 2 − 6xy + 4x − 4y = 9,
6x2 + 3y 2 − 8xy + 8x − 8y = 14. (1)
(2) Spurred on by our success in the ﬁrst part, we will rewrite the ﬁrst equation in the
suggested form:
(x − y + 2)2 + (y − 2x)2 − 4 = 9.
(3) We are led to wonder whether the same trick will work for the second equation, so let’s
try writing:
6x2 + 3y 2 − 8xy + 8x − 8y ≡ a(x − y + 2)2 + b(cx + y )2 + d.
As before, we get equations:
a + bc2 = 6
2bc − 2a = −8
a+b=3
4a = 8
−4a = −8
4a + d = 0.
(We can write these down as the right hand side is the same as before.)
This time, a = 2 from both the fourth and ﬁfth equations, so we get b = 1 from the
third equation. The second equation gives us c = −2. Finally, the sixth equation gives
us d = −8. We must now check that our solution is consistent with the ﬁrst equation, which we have
not yet used. The left hand side is a + bc2 = 2 + 1 × (−2)2 = 6, which works, so we can
write the second equation as
2(x − y + 2)2 + (y − 2x)2 − 8 = 14.
(If we had not checked for consistency, we might have wrongly concluded that 183x2 +
3y 2 − 8xy + 8x − 8y can also be written in the same way.) These two equations now look remarkably similar! In fact, let’s move the constants to the
right hand side and write them together:
(x − y + 2)2 + (y − 2x)2 = 13
2(x − y + 2)2 + (y − 2x)2 = 22.
We now have two simultaneous equations which look almost linear. In fact, if we write
u = (x − y + 2)2 and v = (y − 2x)2 , we get
u + v = 13
2u + v = 22
which we can easily solve to get u = 9 and v = 4.
Therefore, we now have to solve the two equations
(x − y + 2)2 = 9
(y − 2x)2 = 4. (4)
(5) We can take square roots, so that (4) gives x − y + 2 = ±3 and (5) gives y − 2x = ±2. Thus we now have four possibilities (two from equation (4), and for each of these, two from
equation (5)), and we solve each one, checking our results back in the original equations.
2x − y x − y + 2
2
3
2
−3
−2
3
−2
−3 x
1
7
−3
3 y LHS of (1) LHS of (2)
0
9
14
12
9
14
−4
9
14
8
9
14 Therefore we see that the four solutions are (x, y ) = (1, 0), (7, 12), (−3, −4) and (3, 8).
An alternative is to observe that equation (2) looks almost double equation (1), so we
consider 2 × (1) − (2):
4x2 + y 2 − 4xy = 4.
But the left hand side is simply (2x − y )2 , so we get 2x − y = ±2.
Substituting this into equation (3) gives us (x − y + 2)2 + 4 − 4 = 9,
so that x − y + 2 = ±3. Thus we have the four possibilities we found in the ﬁrst approach, and we continue as
above.
Yet another alternative approach is to subtract (2) − (1) to get
x2 + y 2 − 2xy + 4x − 4y = 5, so that
(x − y )2 + 4(x − y ) = 5. Writing z = x − y , we get the quadratic z 2 + 4z − 5 = 0, which we can then factorise to
give (z + 5)(z − 1) = 0, so either z = 1 or z = −5, which gives x − y = 1 or x − y = −5.
Substituting x − y = 1 into (3) now gives (1 + 2)2 + (y − 2x)2 − 4 = 9,
so that (y − 2x)2 = 4; substituting x − y = −5, on the other hand, would lead us to
(−5 + 2)2 + (y − 2x)2 − 4 = 9,
and again we deduce (y − 2x)2 = 4. We have again reached the same deductions as in the ﬁrst approach, so we continue from
there. Question 2 The curve y =
Show that x−a x
e , where a and b are constants, has two stationary points.
x−b
a−b<0 or a − b > 4. We begin by diﬀerentiating using ﬁrst the product rule and then the quotient rule:
dy
d x − a x (x − a) x
=
e+
e
dx
dx x − b
(x − b)
(x − b).1 − (x − a).1 x (x − a) x
=
e
e+
(x − b)2
(x − b)
(a − b) x (x − a)(x − b) x
=
e+
e
(x − b)2
(x − b)2
x2 − (a + b)x + (ab + a − b) x
e.
=
(x − b)2
dy
Now solving dx = 0 gives x2 − (a + b)x +(ab + a − b) = 0. Since the curve has two stationary
points, this quadratic must have two distinct real roots. Therefore the discriminant must
be positive, that is
(a + b)2 − 4(ab + a − b) > 0, and expanding gives a2 − 2ab + b2 − 4a + 4b > 0, so (a − b)2 − 4(a − b) > 0. Factorising
this last expression gives
(a − b)(a − b − 4) > 0, so (sketching a graph to help, possibly also replacing a − b with a variable like x), we see
that we must either have a − b < 0 or a − b > 4.
(i) Show that, in the case a = 0 and b = 1 , there is one stationary point on either
2
side of the curve’s vertical asymptote, and sketch the curve.
We are studying the curve y = x
x− 1
2 ex . We have a − b = − 1 < 0, so the curve has two stationary points by the ﬁrst part of the
2
question. The xcoordinates of the stationary points are found by solving the quadratic
x2 − (a + b)x + (ab + a − b) = 0,
as above.
1
Substituting in our values for a and b, we get x2 − 2 x − 1 = 0, so 2x2 − x − 1 = 0,
2
which factorises to (x − 1)(2x + 1) = 0. Thus there are stationary points at (1, 2e) and
11
(− 2 , 2 e−1/2 ). The vertical asymptote is at x = b, that is at x = 1 .
2 1
Therefore, since the two stationary points are at x = 1 and x = − 2 , there is one stationary
point on either side of the curve’s vertical asymptote. We note that the only time the curve crosses the xaxis is when x = a, so this is when
x = 0, and this is also the y intercept in this case.
As x → ±∞, y ∼ ex (meaning y is approximately equal to ex ; formally, we say that y is
asymptotically equal to ex ), as the fraction (x − a)/(x − b) tends to 1. We can also note where the curve is positive and negative: since ex is always positive,
y > 0 whenever both x − a > 0 and x − b > 0, or when both x − a < 0 and x − b < 0, so
y < 0 when x lies between a and b and is positive or zero otherwise.
Using all of this, we can now sketch the graph of the function. The nature of the stationary
points will become clear from the graphs. In the graph, the dotted lines are the asymptotes
(x = 1 and y = ex ) and the red line is the graph we want, with the stationary points
2
indicated.
y (1, 2e) (− 1 , 1 e−1/2 )
22
0 (ii) Sketch the curve in the case a = 9
2 1
2 x and b = 0. x− 9 x
2
e.
x
Proceeding as in (i), we have a − b = 9 > 4, so again, the curve has two stationary points.
2
The xcoordinates of the stationary points are given by solving the quadratic
This time, we are studying the curve y = x2 − (a + b)x + (ab + a − b) = 0,
as above.
9
9
Substituting our values, we get x2 − 2 x + 2 , so 2x2 − 9x + 9 = 0. Again, this factorises
3
nicely to (x − 3)(2x − 3) = 0, giving stationary points at ( 2 , −2e3/2 ) and (3, − 1 e3 ).
2 The vertical asymptote is at x = b, that is at x = 0. This time, therefore, the stationary
points are both to the right of the vertical asymptote.
9
The xintercept is at x = a, that is, at ( 2 , 0). There is no y intercept as x = 0 is an
asymptote. Again, as x → ±∞, y ∼ ex . As in (i), y < 0 when x lies between a and b and is positive or zero otherwise.
Using all of this, we can now sketch the graph of this function. Note that the asymptote
y = ex is much greater than y until x is greater than 20 or so, as even then (x − a)/(x − b) ≈
15/20, and only slowly approaches 1. We don’t even attempt to sketch the function for
such large values of x!
y 0 9
2 ( 3 , −2e3/2 )
2
(3, − 1 e3 )
2 x Question 3 Show that
sin(x + y ) − sin(x − y ) = 2 cos x sin y
and deduce that
sin A − sin B = 2 cos 1 (A + B ) sin 1 (A − B ).
2
2
We use the compound angle formulæ (also called the addition formulæ) to expand the
left hand side, getting:
sin(x + y ) − sin(x − y ) = (sin x cos y + cos x sin y ) − (sin x cos y − cos x sin y )
= 2 cos x sin y,
as required.
1
For the deduction, we want A = x + y and B = x − y , so x = 2 (A + B ) and y = 1 (A − B ),
2
solving these two equations simultaneously to ﬁnd x and y . Then we simply substitute
these values of x and y into our previous identity, and we reach the desired conclusion: sin A − sin B = 2 cos 1 (A + B ) sin 1 (A − B ).
2
2
(This identity is known as one of the factor formulæ.)
Show also that
1
cos A − cos B = −2 sin 1 (A + B ) sin 2 (A − B ).
2 Likewise, we have
cos(x + y ) − cos(x − y ) = (cos x cos y − sin x sin y ) − (cos x cos y + sin x sin y )
= −2 sin x sin y,
1
so again substituting x = 2 (A + B ) and y = 1 (A − B ) gives
2
1
cos A − cos B = −2 sin 1 (A + B ) sin 2 (A − B ).
2 The points P , Q, R and S have coordinates (a cos p, b sin p), (a cos q, b sin q ),
(a cos r, b sin r) and (a cos s, b sin s) respectively, where 0
p < q < r < s < 2π ,
and a and b are positive.
Given that neither of the lines P Q and SR is vertical, show that these lines are parallel
if and only if
r + s − p − q = 2π.
Remark: The points P , Q, R and S all lie on an ellipse, which can be thought of as a
stretched circle, as their coordinates all have x = cos θ and y = sin θ, so they satisfy the
a
b
y2
x2
equation a + b = 1.
The lines P Q and SR are parallel if and only if their gradients are equal (and neither are
vertical, so their gradients are welldeﬁned), thus
PQ b sin q − b sin p
b sin s − b sin r
=
a cos q − a cos p
a cos s − a cos r
sin s − sin r
sin q − sin p
=
⇐⇒
cos q − cos p
cos s − cos r
1
1
1
1
2 cos 2 (q + p) sin 2 (q − p)
2 cos 2 (s + r) sin 2 (s − r)
⇐⇒
=
1
1
−2 sin 1 (q + p) sin 2 (q − p)
−2 sin 1 (s + r) sin 2 (s − r)
2
2 RS ⇐⇒ cos 1 (q + p)
cos 1 (s + r)
2
2
⇐⇒
=
1
− sin 2 (q + p)
− sin 1 (s + r)
2 1
⇐⇒ cot 1 (q + p) = cot 2 (s + r)
2
1
⇐⇒ 1 (q + p) = 2 (s + r) + kπ
for some k ∈ Z
2
⇐⇒ q + p = s + r + 2kπ
for some k ∈ Z
⇐⇒ r + s − p − q = 2nπ
for some n ∈ Z. The last four lines could have also been replaced by the following:
PQ RS ⇐⇒ · · · 1
cos 1 (s + r)
cos 2 (q + p)
2
⇐⇒
=
1
− sin 2 (q + p)
− sin 1 (s + r)
2 ⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒ 1
1
cos 2 (q + p) sin 1 (s + r) = cos 2 (s + r) sin 1 (q + p)
2
2
1
1
1
1
sin 2 (s + r) cos 2 (q + p) − cos 2 (s + r) sin 2 (q + p) = 0
1
sin 2 ((q + p) − (s + r)) = 0
1
(q + p − s − r ) = π
for some k ∈ Z
2
r + s − p − q = 2nπ
for some n ∈ Z. We are almost there; we now only need to show n = 1 in the ﬁnal line. We know that
0 p < q < r < s < 2π , so r + s < 4π and 0 < p + q < r + s, so that 0 < r + s − p − q < 4π ,
which means that n must equal 1 if P Q and RS are parallel.
Thus P Q and RS are parallel if and only if r + s − p − q = 2π . Question 4 Use the substitution x = t2 1
, where t > 1, to show that, for x > 0,
−1
1 x(x + 1) dx = 2 ln √ [Note: You may use without proof the result x+ t2 √
x + 1 + c. t−a
1
1
ln
+ constant. ]
dt =
2
−a
2a
t+a Using the given substitution, we ﬁrst use the chain rule to calculate
dx
2t
= −(t2 − 1)−2 · 2t = − 2
.
dt
(t − 1)2
(We could alternatively have used the quotient rule to reach the same conclusion.)
We can now perform the requested substitution, simplifying the algebra as we go:
1
x(x + 1) 1 dx =
= 1
t2 −1 1 t
t2 −1 ·
· t2
t2 −1 · dx
dt
dt −2t
dt
(t2 − 1)2 −2
dt
−1
t−1
1
+c
= −2 × ln
2
t+1
t+1
+ c.
= ln
t−1 = t2 using the given result At this point, we wish to substitute t for x, so we rearrange the original substitution to
get
1
x+1
t=
+1=
.
x
x
This now yields:
x+1
+1
t+1
x
ln
+ c = ln
+ c.
x+1
t−1
−1
x
We note immediately that we can drop the absolute value signs, since both the numerator
and denominator of the fraction are positive (the denominator is positive as t > 1 or
(x + 1)/x√ 1). So we get, on multiplying the numerator and denominator of the
>
fraction by x to clear the fractions, √
√
x+1+ x
√
+c
√
x+1− x
√√
√
√
x+1+ x
x+1+ x
√√
√
= ln
√
x+1− x
x+1+ x
√2
√
x+1+ x
= ln
+c
(x + 1) − x
√2
√
= ln x + 1 + x + c
√
√
= 2 ln x + 1 + x + c, t+1
ln
+ c = ln
t−1 +c which is what we were after.
The section of the curve 1
1
y=√ −√
x+1
x 9
between x = 1 and x = 16 is rotated through 360◦ about the xaxis. Show that the
8
5
volume enclosed is 2π ln 4 . 9/16
1/8 To ﬁnd the volume of revolution, we need to calculate the deﬁnite integral πy 2 dx: 9/16 πy 2 dx
1/8
9/16 =π
1/8
9/16 1
1
√ −√
x+1
x 2 dx 1
1
1
−2
+
dx
x
x(x + 1) x + 1
1/8
√
√
= π ln x − 4 ln x + x + 1 + ln(x + 1) =π 9
= π ln 16 − 4 ln 9
16 + 25
16 9/16
1/8 using the above result 25
+ ln 16 − π ln 1 − 4 ln
8 1
= π 2 ln 3 − 4 ln( 3 + 5 ) + 2 ln 5 − π 2 ln 2√2 − 4 ln
4
4
4
4 = π (2 ln 3 − 2 ln 4) − 4 ln 2 + (2 ln 5 − 2 ln 4) −
√
√
√
π −2 ln(2 2) − 4 ln 2 + (2 ln 3 − 2 ln(2 2)
= π (2 ln 3 − 4 ln 2 − 4 ln 2 + 2 ln 5 − 4 ln 2)−
π (−3 ln 2 − 2 ln 2 + 2 ln 3 − 3 ln 2)
= π (−4 ln 2 + 2 ln 5)
= 2π (−2 ln 2 + ln 5)
5
= 2π ln 4 . 1
√
22 1
8 + + 3
√
22 9
8 + ln 9
8
3
+ 2 ln 2√2 Question 5 By considering the expansion of (1 + x)n where n is a positive integer, or otherwise,
show that:
(i) n
n
n
n
+
+
+ ··· +
0
1
2
n = 2n ; We take the advice and begin by writing out the expansion of (1 + x)n :
(1 + x)n = n0
n1
n2
nn
x+
x+
x + ··· +
x,
0
1
2
n (∗) where we have pedantically written in x0 and x1 in the ﬁrst two terms, as this may well
help us to understand what we are looking at.
Now comparing this expansion to the expression we are interested in, we see that the only
diﬀerence is the presence of the xs. If we substitute x = 1, we will get exactly what we
want:
n
n
n
n
(1 + 1)n =
+
+
+ ··· +
,
0
1
2
n
as all powers of 1 are just 1. (ii) n
n
n
n
+2
+3
+ ··· + n
1
2
3
n = n2n−1 ; For the rest of the question, there are two very distinct approaches, one via calculus and
one via properties of binomial coeﬃcients.
Approach 1: Use calculus
This one looks a little more challenging, and we must observe carefully that there is no
n
term. Comparing to the binomial expansion, we see that the term n xr has turned
0
r
into n .r. Now, setting x = 1 will again remove the x, but where are we to get the r
r
from? Calculus gives us the answer: if we diﬀerentiate with respect to x, then xr becomes
rxr−1 , and then setting x = 1 will complete the job. Now diﬀerentiating (∗) gives
n(1 + x)n−1 = n
n
n
n
.1x0 +
.2x1 +
.3x2 + · · · +
.nxn−1 ,
1
2
3
n so by setting x = 1, we get the desired result.
Approach 2: Use properties of binomial coeﬃcients
We know that n
r = n!
(n − r)!r! so we can manipulate this formula to pull out an r, using r! = r.(r − 1)! and similar
expressions. We get
n
r so that r n
r =n n−1
r−1 n!
(n − r)!r!
1
n!
=.
r (n − r)!(r − 1)!
n
(n − 1)!
=.
r (n − r)!(r − 1)!
n n−1
=
r r−1
= . This is true as long as r n
n
n
+2
+ ··· + n
1
2
n =n 1 and n 1, so we get n−1
n−1
n−1
+n
+ ··· + n
0
1
n−1 = n.2n−1 where we have used the result from part (i) with n − 1 in place of n to do the last step.
(iii) n
1n
n
1n
1
+
+
+ ··· +
0
21
32
n+1 n = 1
2n+1 − 1 ;
n+1 Approach 1: Use calculus
Spurred on by our previous success, we see that now the
think of integration instead. Integrating (∗) gives
1
(1 + x)n+1 =
n+1 n
2 x2 term gives us n
2 . 1 , so we
3 n
n 12
n 13
n
1
.x1 +
.x+
. x + ··· +
.
xn+1 + c .
0
12
23
n n+1 We do need to determine the constant of integration, so we put x = 0 to do this; this
gives
1 n+1
n
n1
n
1
1
=
.0 +
. .0 + + · · · +
.
.0 + c,
0
12
n n+1
n+1
so c = 1
.
n+1 Now substituting x = 1 gives 1
(1 + 1)n+1 =
n+1
Finally, subtracting 1
n+1 n
1n
1n
1
n
1
+
++
+ ··· +
+
.
0
21
32
n+1 n
n+1 from both sides gives us our required result. (An alternative way to think about this is to integrate both sides from x = 0 to x = 1.)
Approach 2: Use properties of binomial coeﬃcients
We can try rewriting our identity so that the
1n
nr = 1
r stays with the r − 1 term; this gives us 1 n−1
.
r r−1 n
r−1 Unfortunately, though, our expressions involve
we can ﬁx this by replacing n by n + 1 to get
1
n+1
n+1
r = terms rather than n−1
r −1 terms, but 1
n
.
r r−1 We substitute this in to get
1
n
n
1n
+··· +
+
21
n+1 n
0
= 1
n+1
1
n+1
1
n+1
+
+ ··· +
n+1
1
n+1
2
n+1 n+1 = 1
(2n+1 − 1),
n+1 where we have again used the result of part (i), this time with n + 1 replacing n. (iv) n
n
n
n
+ 22
+ 32
+ · · · + n2
1
2
3
n = n(n + 1)2n−2 . Approach 1: Use calculus
This looks similar to (ii), in that we have increasing multiples. So we try diﬀerentiating (∗)
twice, giving us:
n(n − 1)(1 + x)n−2 = n
n
n
n
.1.0 +
.2.1x0 +
.3.2x1 + · · · +
.n(n − 1)xn−2 .
1
2
3
n Unfortunately, though, the coeﬃcient of n is r(r − 1) rather than the r2 we actually
r
want. But no matter: we can just add r and we will be done, as r2 = r(r − 1) + r, and
we know from (ii) what terms like n .2 sum to give us. So we have, putting x = 1 in our
2
above expression:
n(n − 1)(1 + 1)n−2 = n
n
n
n
.1.0 +
.2.1 +
.3.2 + · · · +
.n(n − 1).
1
2
3
n Now adding the result of (ii) gives
n(n − 1)2n−2 + n2n−1 = n
n
n
.(1.0 + 1) +
.(2.1 + 2) +
.(3.2 + 3) + · · · +
1
2
3
n
.(n(n − 1) + n)
n so
(n(n − 1) + 2n)2n−2 = n
n2
n2
n
+
.2 +
.3 + · · · +
.n2 .
1
2
3
n The left side simpliﬁes to n(n + 1)2n−2 , and thus we are done. An alternative (calculusbased) method is as follows. The ﬁrst derivative of (∗), as we
have seen, is
n(1 + x)n−1 = n
n
n
n
.1x0 +
.2x1 +
.3x2 + · · · +
.nxn−1 .
1
2
3
n Now were we to diﬀerentiate again, we would end up with terms like n(n − 1)xn−2 , rather
than the desired n2 xk (for some k ). We can remedy this problem by multiplying the whole
identity by x before we diﬀerentiate, so that we are diﬀerentiating
nx(1 + x)n−1 = n
n
n
n
.1x1 +
.2x2 +
.3x3 + · · · +
.nxn .
1
2
3
n Diﬀerentiating this now gives (using the product rule for the left hand side):
n(1+ x)n−1 + n(n − 1)x(1+ x)n−2 = n 20
n 21
n 22
n
.1 x +
.2 x +
.3 x + · · · +
.n2 xn−1 .
1
2
3
n Substituting x = 1 into this gives our desired conclusion (after a small amount of algebra
on the left hand side).
Approach 2: Use properties of binomial coeﬃcients
As this looks similar to the result of part (ii), we can start with what we worked out there,
−1
namely r n = n n−1 , giving us
r
r
n
n
n
+ 22
+ · · · + n2
1
2
n =n n−1
n−1
n−1
n−1
+ n.2
+ n.3
+ · · · + n.n
0
1
2
n−1 Taking out the factor of n leaves us having to work out
n−1
n−1
n−1
n−1
+2
+3
+ ··· + n
0
1
2
n−1
This looks very similar to the problem of part (ii) with n replaced by n − 1, but now the
multiplier of n is r + 1 rather than r, and there is also an n term. We can get over
r
0
the ﬁrst problem by splitting up (r + 1) n as r n + n , so this expression becomes
r
r
r
n−1
n−1
n−1
+2
+ · · · + (n − 1)
+
1
2
n−1 n−1
n−1
+
+
0
1 n−1
+ ··· +
2 n−1
n−1 The ﬁrst line is just part (ii) with n replaced by n − 1, so that it sums to (n − 1).2n−2 , and
the second line is just 2n−1 = 2.2n−2 by part (i). So the answer to the original question
(remembering the factor of n we took out earlier) is
n (n − 1).2n−2 + 2.2n−2 = n(n − 1 + 2).2n−2 = n(n + 1).2n−2 . Question 6
Show that, if y = ex , then
d2 y
dy
(x − 1) 2 − x
+ y = 0.
dx
dx (∗) dy
d2 y
If y = ex , then
= ex and
= ex . Substituting these into the left hand side of (∗)
dx
dx2
gives
d2 y
dy
+ y = (x − 1)ex − xex + ex = 0,
(x − 1) 2 − x
dx
dx
so y = ex satisﬁes (∗).
In order to ﬁnd other solutions of this diﬀerential equation, now let y = uex , where u
is a function of x. By substituting this into (∗), show that
(x − 1) du
d2 u
+ (x − 2)
= 0.
2
dx
dx We have y = uex , so we apply the product rule to get:
dy
du x
=
e + uex
dx
dx
du
=
+ u ex
dx
d2 y
d2 u x du x
du x
=
e+
e+
e + uex
2
2
dx
dx
dx
dx
d2 u
du
= 2 ex + 2 ex + uex
dx
dx
d2 u
du
=
+2
+ u ex .
d x2
dx
(If you know Leibniz’s Theorem, then you could write down d2 u/dx2 directly.)
We now substitute these into (∗) to get
(x − 1) d2 u
du
du
+2
+ u ex − x
+ u ex + uex = 0.
2
dx
dx
dx Dividing by ex = 0 and collecting the derivatives of u then gives
(x − 1) d2 u
du
+ (2x − 2 − x)
+ (x − 1 − x + 1)u = 0,
2
dx
dx which gives (∗∗) on simplifying the brackets. (∗∗) du
= v in (∗∗) and solving the resulting ﬁrst order diﬀerential equation
dx
for v , ﬁnd u in terms of x. Hence show that y = Ax + B ex satisﬁes (∗), where A and B
are any constants.
By setting As instructed, we set d2 u
du
dv
= v , so that
, which gives us
=
2
dx
dx
dx
(x − 1) dv
+ (x − 2)v = 0.
dx This is a standard separable ﬁrstorder linear diﬀerential equation, so we separate the
variables to get
x−2
1 dv
=−
v dx
x−1
and then integrate with respect to x to get
1
dv =
v − x−2
dx.
x−1 Performing the integrations now gives us
1
dx
x−1
= −x + ln x − 1 + c, ln v  =
which we exponentiate to get − 1− v  = k e−x x − 1,
where k is some constant, so we ﬁnally arrive at
v = k e−x (x − 1).
We now recall that v = du/dx, so we need to integrate this last expression once more to
ﬁnd u. We use integration by parts to do this, integrating the e−x part and diﬀerentiating
(x − 1), to give us
u= k e−x (x − 1) dx = k (−e−x )(x − 1) − k (−e−x ).1 dx = k (−e−x )(x − 1) − k e−x + c
= −kxe−x + c,
which is the solution to (∗∗). Now recalling that y = uex gives us y = −kx + c ex as our general solution to (∗). In
particular, letting k = −A and c = B , where A and B are any constants, shows that
y = Ax + B ex satisﬁes (∗), as required. Question 7 Relative to a ﬁxed origin O, the points A and B have position vectors a and b, respectively. (The points O, A and B are not collinear.) The point C has position vector c
given by
c = αa + β b,
where α and β are positive constants with α + β < 1. The lines OA and BC meet
at the point P with position vector p, and the lines OB and AC meet at the point Q
with position vector q. Show that
p= αa
,
1−β and write down q in terms of α, β and b. The condition c = αa + β b with α + β < 1 and α and β both positive constants means
that C lies strictly inside the triangle OAB . Can you see why?
We start by sketching the setup so that we have something visual to help us with our
thinking.
S B
Q
R
C
A
O P The line OA has points with position vectors given by r1 = λa, and the line BC has
points with position vectors given by
−
−
→
−
−
→
r2 = OB + µBC = b + µ(c − b) = (1 − µ)b + µc.
The point P is where these two lines meet, so we must have
p = λa = (1 − µ)b + µc
= (1 − µ)b + µ(αa + β b)
= (1 − µ + βµ)b + αµa. Since a and b are not parallel, we must have 1 − µ + βµ = 0 and αµ = λ. The ﬁrst
equation gives (1 − β )µ = 1, so µ = 1/(1 − β ). This gives λ = α/(1 − β ), so that
p= αa
.
1−β Now swapping the roles of a and b (and hence also of α and β ) will give us the position
vector of Q:
βb
q=
.
1−α
Show further that the point R with position vector r given by
r= αa + β b
α+β lies on the lines OC and AB .
We could approach this question in two ways, either by ﬁnding the point of intersection
of OC and AB or by showing that the given point lies on both given lines. We give both
approaches.
Approach 1: Finding the point of intersection
We require r to lie on OC , so r = λc, and r to lie on AB , so r = (1 − µ)a + µb, as before.
Substituting for c and equating coeﬃcients gives
αλa + βλb = (1 − µ)a + µb,
so that
αλ = 1 − µ
βλ = µ.
Adding the two equations gives (α + β )λ = 1, so λ = 1/(α + β ) and hence
r= αa + β b
c
=
.
α+β
α+β Approach 2: Showing that the given point lies on both lines
The equation of line OC is r1 = λc, and
r= αa + β b
1
=
c
α+β
α+β is of the required form, so R lies on OC . −
→
The equation of the line AB can be written as r2 − a = µAB , so we want r − a = µ(b − a).
Now we have
αa + β b α + β
−
a
α+β
α+β
−β a + β b
=
α+β
β
=
(b − a),
α+β r−a= which is of the form µ(b − a), so R lies on both lines.
The lines OB and P R intersect at the point S . Prove that OS
OQ
=
.
BQ
BS S lies on both OB and P R, so we need to ﬁnd its position vector, s. Once again, we
require s = λb = (1 − µ)p + µr, so we substitute for p and r and compare coeﬃcients:
s = λb = (1 − µ)p + µr
αa + β b
αa
+µ
= (1 − µ)
1−β
α+β
(1 − µ)α(α + β )a + µ(1 − β )(αa + β b)
=
(1 − β )(α + β )
((1 − µ)α(α + β ) + µα(1 − β ))a + µ(1 − β )β b
=
(1 − β )(α + β )
α(α + β − µα − 2µβ + µ)a + µ(1 − β )β b
=
(1 − β )(α + β )
Since the coeﬃcient of a in this expression must be zero, we deduce that
µ= α+β
,
α + 2β − 1 so that
µ(1 − β )β
b
(1 − β )(α + β )
α+β
β
=
b
α + 2β − 1 (α + β )
β
=
b
α + 2β − 1 s= −
→
−
−
→
Now, since OQ and BQ are both multiples of the vector q , we can compare the lengths OQ
and BQ in terms of their multiples of q. This might come out to be negative, depending
on the relative directions, but at the end, we can just consider the magnitudes. We thus have, since q = βb
,
1−α
OQ
β
=
BQ
1−α
β
=
1−α
β
=
,
β−1+α β
−1
1−α
β−1+α
1−α while
OS
β
=
BS
α + 2β − 1
β
=
α + 2β − 1
β
.
=
−α − β + 1 β
−1
α + 2β − 1
−α − β + 1
α + 2β − 1 Thus these two ratios of lengths are equal, as the magnitude of both of these is
β
.
1 − (α + β ) Question 8 (i) Suppose that a, b and c are integers that satisfy the equation
a3 + 3b3 = 9c3 .
Explain why a must be divisible by 3, and show further that both b and c must
also be divisible by 3. Hence show that the only integer solution is a = b = c = 0.
We have a3 = 9c3 − 3b3 = 3(3c3 − b3 ), so a3 is a multiple of 3. But as 3 is prime, a itself
must be divisible by 3. (Why is this? If 3 divides a product rs, then 3 must divide either
r or s, as 3 is prime. Therefore since a3 = a.a.a, and 3 divides a3 , it follows that 3 must
divide one of the factors, that is, 3 must divide a.)
Now we can write a = 3d, where d is an integer. Therefore we have
(3d)3 + 3b3 = 9c3 ,
which, on dividing by 3, gives
9d3 + b3 = 3c3 .
By the same argument, as b3 = 3(c3 − 3d3 ), it follows that b3 , and hence also b, is divisible
by 3.
We repeat the same trick, writing b = 3e, where e is an integer, so that
9d3 + (3e)3 = 3c3 .
We again divide by 3 to get
3d3 + 9e3 = c3 ,
so that c3 , and hence also c, is divisible by 3.
We then write c = 3f , where f is an integer, giving
3d3 + 9e3 = (3f )3 .
Finally, we divide this equation by 3 to get
d3 + 3e3 = 9f 3 .
Note that this is the same equation that we started with, so if a, b, c are integers which
satisfy the equation, then so are d = a/3, e = b/3 and f = c/3. We can repeat this process
indeﬁnitely, so that a/3n , b/3n and c/3n are also integers which satisfy the equation. But
if a/3n is an integer for all n 0, we must have a = 0, and similarly for b and c.
Therefore the only integer solution is a = b = c = 0.
[In fact, we can say even more. If a, b and c are all rational, say a = d/r, b = e/s, c = f /t
(where d, e, f are integers and r, s, t are nonzero integers), then we have
d
r 3 +3 e
s 3 =9 f
t 3 . Now multiplying both sides by (rst)3 gives
(dst)3 + 3(ert)3 = 9(f rs)3 ,
with dst, ert and f rs all integers, and so they must all be zero, and hence d = e = f = 0.
Therefore, the only rational solution is also a = b = c = 0.]
(ii) Suppose that p, q and r are integers that satisfy the equation
p4 + 2q 4 = 5r4 .
By considering the possible ﬁnal digit of each term, or otherwise, show that p and q
are divisible by 5. Hence show that the only integer solution is p = q = r = 0.
We consider the ﬁnal digit of fourth powers:
a
0
1
2
3
4
5
6
7
8
9 a4
0
1
6
1
6
5
6
1
6
1 2a4
0
2
2
2
2
0
2
2
2
2 So the last digits of fourth powers are all either 0, 5, 1 or 6, and of twice fourth powers
are all either 0 or 2.
Also, 5r4 is a multiple of 5, so it must end in a 0 or a 5.
Therefore if 2q 4 ends in 0 (that is, when q is a multiple of 5), the possibilities for the ﬁnal
digit of p4 + 2q 4 are
(0 or 1 or 5 or 6) + 0 = 0 or 1 or 5 or 6,
so it can equal 5r4 (which ends in 0 or 5) only if p4 ends in 0 or 5, which is exactly when
p is a multiple of 5.
Similarly, if 2q 4 ends in 2 (so q is not a multiple of 5), the possibilities for the ﬁnal digit
of p4 + 2q 4 are
(0 or 1 or 5 or 6) + 2 = 2 or 3 or 7 or 8,
so it can not be equal to 5r4 (which ends in 0 or 5).
Therefore, if p4 + 2q 4 = 5r4 , we must have p and q both being multiples of 5.
Now as in part (i), we write p = 5a and q = 5b, where a and b are both integers, to get
(5a)4 + 2(5b)4 = 5r4 . Dividing both sides by 5 gives
53 a4 + 2.53 b4 = r4 ,
where we are using dot to mean multiplication, so as before, r4 must be a multiple of 5
as the left hand side is 5(52 a4 + 2.52 b4 ). Thus, since 5 is prime, r itself must be divisible
by 5. Then writing r = 5c gives
53 a4 + 2.53 b4 = (5c)4 ,
which yields
a4 + 2b4 = 5c4
on dividing by 53 .
So once again, if p, q , r give an integer solution to the equation, so do a = p/5, b = q/5
and c = r/5. Repeating this, so are p/5n , q/5n , r/5n , and as before, this shows that the
only integer solution is p = q = r = 0.
[Again, the same argument as before shows that this is also the only rational solution.]
This is an example of the use of Fermat’s Method of Descent, which he used to prove one
special case of his famous Last Theorem: he showed that x4 + y 4 = z 4 has no positive
integer solutions. In fact, he proved an even stronger result, namely that x4 + y 4 = z 2
has no positive integer solutions.
Another approach to solving the ﬁrst step of part (ii) of this problem is to use modular
arithmetic, where we only consider remainders when dividing by a certain ﬁxed number.
In this case, we would consider arithmetic modulo 5, so the only numbers to consider are
0, 1, 2, 3 and 4, and we want to solve p4 + 2q 4 ≡ 0 (mod 5), where ≡ means “leaves the
same remainder”. Now a quick calculation shows that p4 ≡ 1 unless p ≡ 0, while 2q 4 ≡ 2
unless q ≡ 0, so that
p4 + 2q 4 ≡ (0 or 1) + (0 or 2) ≡ 0, 1, 2 or 3 (mod 5) with p4 + 2q 4 ≡ 0 if and only if p ≡ q ≡ 0.
Incidentally, Fermat has another theorem relevant to this problem, which turns out to
be relatively easy to prove (Fermat himself claimed to have done so), and is known as
Fermat’s Little Theorem. This states that, if p is prime, then ap−1 ≡ 1 (mod p) unless
a ≡ 0 (mod p). In our case, p = 5 gives a4 ≡ 1 (mod 5) unless a ≡ 0 (mod 5), as we
wanted. Question 9 2b 2a
α The diagram shows a uniform rectangular lamina with sides of lengths 2a and 2b leaning
against a rough vertical wall, with one corner resting on a rough horizontal plane. The
plane of the lamina is vertical and perpendicular to the wall, and one edge makes an
angle of α with the horizontal plane. Show that the centre of mass of the lamina is a
distance a cos α + b sin α from the wall.
We start by redrawing the sketch, labelling the corners and indicating the centre of mass
as G, as well as showing various useful lengths.
D A a G C b
2b 2a
α
B It is now clear that the distance of G from the wall is a cos α (horizontal distance from
wall to midpoint of AB ) plus b sin α (horizontal distance from midpoint of AB to G), so
a total of a cos α + b sin α.
Also, in case it is useful later, we note that the vertical distance above the horizontal
plane is, by a similar argument from the same sketch, a sin α + b cos α.
The coeﬃcients of friction at the two points of contact are each µ and the friction is
limiting at both contacts. Show that
a cos(2λ + α) = b sin α,
where tan λ = µ.
There are two approaches to this. One is to indicate the reaction and friction forces
separately, while the other is to use the Three Forces Theorem. We show both of these. Approach 1: All forces separately
We start by sketching the lamina again, this time showing the forces on the lamina,
separating the normal reactions from the frictional forces.
D
F1
A R1 G
R2 C W
α
F2 B We now resolve and take moments:
R (↑)
R (→)
M (A) F1 + R2 − W = 0
R1 − F2 = 0
W (a cos α + b sin α) − R2 .2a cos α + F2 .2a sin α = 0 Since friction is limiting at both points of contact, we have F1 = µR1 and F2 = µR2 .
Substituting these gives:
R (↑)
R (→)
M (A) µR1 + R2 − W = 0
R1 − µR2 = 0
W (a cos α + b sin α) − 2aR2 cos α + 2aµR2 sin α = 0 (1)
(2)
(3) Equation (2) gives R1 = µR2 , so we can substitute this into (1) to get W = (1 + µ2 )R2 .
Substituting this into (3) now leads to
(1 + µ2 )R2 (a cos α + b sin α) = 2aR2 (cos α − µ sin α).
We can clearly divide both sides by R2 , and we are given that tan λ = µ, so we substitute
this in as well, to get
(1 + tan2 λ)(a cos α + b sin α) = 2a(cos α − tan λ sin α).
We spot 1 + tan2 λ = sec2 λ, and so multiply the whole equation through by cos2 λ, as the
form we are looking for does not involve sec λ:
a cos α + b sin α = 2a(cos2 λ cos α − sin λ cos λ sin α).
Since the form we are going for is b sin α = a cos(2λ + α), we make use of double angle
formulæ, after rearranging:
b sin α = a(2 cos2 λ cos α − 2 sin λ cos λ sin α) − a cos α
= a((2 cos2 λ − 1) cos α − 2 sin λ cos λ sin α)
= a(cos 2λ cos α − sin 2λ sin α)
= a cos(2λ + α), and we are done with this part.
Approach 2: Three Forces Theorem
The ‘Three Forces Theorem’ states that if three (nonzero) forces act on a large body in
equilibrium, and they are not all parallel, then they must pass through a single point.
(Why is this true? Let’s say two of the forces pass through point X . Taking moments
about X , the total moment must be zero, so the moment of the third force about X must
be zero. Therefore, the force itself is either zero or it passes through X . Since the forces
are nonzero, the third force must pass through X .)
In our case, we have a normal reaction and a friction force at each point of contact. We
can combine these into a single reaction force as shown in the sketch. Here we have
written N for the normal force, F for the friction and R for the resultant, which is at an
angle of θ to the normal.
R F θ
N We see from this sketch that tan θ = F/N = µN/N = µ. In our case, since tan λ = µ we
must have θ = λ.
We can now redraw our original diagram with the three (combined) forces shown:
R2
D A R1
X λ G C
λ 2a
α
O P B W We can now use the Three Forces Theorem is as follows. Looking at the diagram, we
know that the distance OB = 2a cos α = OP + P B . Now we know OP = a cos α + b sin α,
so we need only calculate P B .
But P B = P X tan λ (using the triangle P BX ), and
P X = OA + height of X above A
= 2a sin α + (a cos α + b sin α) tan λ. Putting these together gives
OB = 2a cos α = OP + P B
= a cos α + b sin α + (2a sin α + (a cos α + b sin α) tan λ) tan λ
= a cos α(1 + tan2 λ) + b sin α(1 + tan2 λ) + 2a sin α tan λ
= a cos α sec2 λ + b sin α sec2 λ + 2a sin α tan λ.
We can now rearrange to get
b sin α sec2 λ = 2a cos α − a cos α sec2 λ − 2a sin α tan λ.
Since we want an expression for b sin α, we now multiply by cos2 λ to get
b sin α = 2a cos α cos2 λ − a cos α − 2a sin α sin λ cos λ
= a((2 cos2 λ − 1) cos α − (2 sin λ cos λ) sin α)
= a(cos 2λ cos α − sin 2λ sin α)
= a cos(2λ + α).
An alternative argument using the Three Forces Theorem proceeds by considering the
distance XP . Using the left half of the diagram, we have
XP = OA + OP tan λ
= 2a sin α + (a cos α + b sin α) tan λ.
From the right half of the diagram, we also have XP = P B/ tan λ, and P B = 2a cos α −
OP = a cos α − b sin α, so that
2a sin α + (a cos α + b sin α) tan λ = (a cos α − b sin α)/ tan λ.
Now multiplying by tan λ and collecting like terms gives
b sin α(1 + tan2 λ) = a cos α(1 − tan2 λ) − 2a sin α tan λ.
Then using 1 + tan2 λ = sec2 λ and then multiplying by cos2 λ gives us
b sin α = a cos α(cos2 λ − sin2 λ) − 2a sin α sin λ cos λ
= a cos α cos 2λ − a sin α sin 2λ
= a cos(α + 2λ),
as we wanted.
Show also that if the lamina is square, then λ = π
4 − α. We have a = b as the lamina is square, so that our previous equation becomes
a sin α = a cos(2λ + α). Dividing by a gives
sin α = cos(2λ + α).
Now we can use the identity sin α = cos( π − α), so that
2
π
− α = 2λ + α.
2
(Being very careful, we should check that we can take inverse cosines of both sides to
deduce this equality. This will be the case if both 0 < π − α < π and 0 < 2λ + α < π .
2
2
2
But 0 < α < π so the ﬁrst inequality is clearly true. For the second inequality, we have
2
π
0 < λ < π so that 0 < 2λ + α < 32 . But since the cosine of this is positive (being sin α),
2
it must lie in the range 0 < 2λ + α < π as required.)
2
Subtracting α and dividing by 2 now gives our desired result:
π
− α = λ.
4 Question 10 A particle P moves so that, at time t, its displacement r from a ﬁxed origin is given by
r = et cos t i + et sin t j.
Show that the velocity of the particle always makes an angle of π with the particle’s
4
displacement, and that the acceleration of the particle is always perpendicular to its
displacement.
To ﬁnd the velocity, v, and acceleration, a, we diﬀerentiate with respect to t (using the
product rule).
We have
r = et cos t i + et sin t j
v = dr/dt = et cos t − et sin t i + et sin t + et cos t j
a = dv/dt = (et cos t − et sin t − (et sin t + et cos t))i+
(et sin t + et cos t) + (et cos t − et sin t j
= (−2et sin t)i + (2et cos t)j. We can rewrite these, if we wish, by taking out the common factors:
r = et (cos t)i + (sin t)j
v = et (cos t − sin t)i + (sin t + cos t)j
a = 2et (− sin t)i + (cos t)j) . From these, we can easily ﬁnd the magnitudes of the displacement, velocity and acceleration:
r = et v = et = et (cos t)2 + (sin t)2 = et
(cos t − sin t)2 + (sin t + cos t)2
2 cos2 t + 2 sin2 t √
= et 2
a = 2et (− sin t)2 + (cos t)2 = 2et . We can now ﬁnd the angles between these using a.b = 2ab cos θ; ﬁrstly, for displacement and velocity we have
r.v = e2t cos t(cos t − sin t) + sin t(sin t + cos t)
= e2t (cos2 t + sin2 t)
= e2t ,
while
√
tt
r.v = e .e 2 cos θ, √
so that cos θ = 1/ 2, so that θ = π
4 as required. Next, for displacement and acceleration we have
r.a = 2e2t cos t(− sin t) + sin t cos t = 0,
so they are perpendicular.
Geometrictrigonometric approach
There is another way to ﬁnd the angles involved which does not use the scalar (dot)
product.
Recall that the velocity is v = et (cos t − sin t)i + (sin t + cos t)j . We can use the
“R cos(θ + α)” technique, thinking of cos t − sin t as 1 cos t − 1 sin t, so that
√1
1
cos t − sin t = 2 √2 cos t − √2 sin t
√
= 2(cos t cos π − sin t sin π )
4
4
√
π
= 2 cos(t + 4 )
√1
1
sin t + cos t = 2 √2 sin t + √2 cos t
√
= 2(sin t cos π + cos t sin π )
4
4
√
π
= 2 sin(t + 4 )
√
Thus v = 2et cos(t + π )i + sin(t + π )j , so v is at an angle of π with r.
4
4
4
Likewise, a makes an angle of π
4 with v, and so an angle of Sketch the path of the particle for 0 t π
2 with r. π. One way of thinking about the path of the particle is that its displacement at time t is
given by r = et (cos t)i + (sin t)j , so that it is at distance et from the origin and at an
angle of t (in radians) to the xaxis (as (cos t)i +(sin t)j is a unit vector in this direction).
Thus its distance at time t = 0 is e0 = 1, and when it has gone a half circle, its distance
is eπ , which is approximately e3 ≈ 20. So the particle moves away from the origin very
quickly!
Another thing to bear in mind is that its velocity is always at an angle of π to its
4
displacement. Since it is moving away from the origin, its velocity is directed away from
the origin, so initially it is moving at an angle of π above the positive xaxis.
4
As we sketch the path, we also indicate the directions of the velocities at the times t = 0,
t = π and t = π .
2
y
eπ/2
−eπ 1x A second particle Q moves on the same path, passing through each point on the path
a ﬁxed time T after P does. Show that the distance between P and Q is proportional
to et .
We write rP = r for the position vector of P and rQ for the position vector of Q. We
therefore have
rP = et cos t i + et sin t j
rQ = et−T cos(t − T ) i + et−T sin(t − T ) j,
and so we can calculate rP − rQ 2 :
rP − rQ 2 = et cos t − et−T cos(t − T ) 2 + et sin t − et−T sin(t − T ) 2 = e2t cos2 t − 2et et−T cos t cos(t − T ) + 2e2(t−T ) cos2 (t − T ) +
e2t sin2 t − 2et et−T sin t sin(t − T ) + e2(t−T ) sin2 (t − T ) = e2t − 2e2t−T (cos t cos(t − T ) + sin t sin(t − T ) + e2(t−T )
= e2t − 2e2t−T cos(t − (t − T )) + e2t−2T
= e2t 1 − 2e−T cos T + e−2T , so that
rP − rQ  = et 1 − 2e−T cos T + e−2T , which is clearly proportional to et , as required, since T is a constant. Question 11 Two particles of masses m and M , with M > m, lie in a smooth circular groove
on a horizontal plane. The coeﬃcient of restitution between the particles is e. The
particles are initially projected round the groove with the same speed u but in opposite
directions. Find the speeds of the particles after they collide for the ﬁrst time and show
that they will both change direction if 2em > M − m.
This begins as a standard collision of particles question. ALWAYS draw a diagram
for collisions questions; you will do yourself (and the examiner) no favours if you try to
keep all of the directions in your head, and you are very likely to make a mistake. My
recommendation is to always have all of the velocity arrows pointing in the same direction.
In this way, there is no possibility of messing up the Law of Restitution; it always reads
v1 − v2
= e, and you only have to be careful with the signs of
v1 − v2 = e(u2 − u1 ) or
u2 − u1
the given velocities; the algebra will then keep track of the directions of the unknown
velocities for you. Before M After M u1 = u v1 m m u2 = −u
v2 Then Conservation of Momentum gives
M u1 + mu2 = M v1 + mv2
and Newton’s Law of Restitution gives
v2 − v1 = e(u1 − u2 ).
Substituting u1 = u and u2 = −u gives
M v1 + mv2 = (M − m)u
v2 − v1 = 2eu. (1)
(2) Then solving these equations (by (1) − m × (2) and (1) + M × (2)) gives
(M − m − 2em)u
M +m
(M − m + 2eM )u
v2 =
.
M +m
v1 = (3)
(4) The speeds are then (technically) the absolute values of these, but we will stick with these
formulæ as they are what are needed later. Now, the particles both change directions if v1 and v2 have the opposite signs from u1
and u2 , respectively, so v1 < 0 and v2 > 0. Thus we need
M − m − 2em < 0
M − m + 2eM > 0. and (5)
(6) But (6) is always true, as M > m, so we only need M − m < 2em from (5).
After a further 2n collisions, the speed of the particle of mass m is v and the speed of
the particle of mass M is V . Given that at each collision both particles change their
directions of motion, explain why
mv − M V = u(M − m),
and ﬁnd v and V in terms of m, M , e, u and n.
The fact that the particles both change their directions of motion at each collision means
that if they have velocities v1 and v2 after some collision, they will have velocities −v1 and
−v2 before the next collision. This is because they are moving around a circular track,
and therefore next meet on the opposite site, and hence are each moving in the opposite
direction from the one they were moving in. (We do not concern ourselves with precisely
where on the track they meet, and we are thinking of our velocities as onedimensional
directed speeds.)
Therefore, mvm + M vM is constant in value after each collision, where vm is the velocity
of the particle of mass m, and vM that of the particle of mass M , but it reverses in
sign before the next collision. So after the ﬁrst collision, it it M u − mu to the right (in
our above sketch), and hence after an even number of further collisions, it will still be
M vM + mvm = M u − mu to the right. But after an even number of further collisions,
the particle of mass M is moving to the left, so vM = −V , vm = v . Thus
mv − M V = (M − m)u.
Also, since there are a total of 2n +1 collisions, we have, by 2n +1 applications of Newton’s
Law of Restitution,
V + v = e2n+1 (u + u).
Solving these two equations simultaneously as before then yields
(2me2n+1 − M + m)u
M +m
2n+1
(2M e
+ M − m)u
v=
.
M +m V= Question 12 A discrete random variable X takes only positive integer values. Deﬁne E(X ) for this
case, and show that
E(X ) = ∞ P (X n). n=1 For the deﬁnition of E(X ), we simply plug the allowable values of X into the deﬁnition
of E(X ) for discrete random variables, to get
E(X ) = ∞ n P(X = n). n=1 Now, we can think of n, P(X = n) as the sum of n copies of P(X = n), so that we get
E(X ) = ∞ n P(X = n) n=1 = 1.P(X = 1) + 2.P(X = 2) + 3.P(X = 3) + 4.P(X = 4) + · · ·
= P(X = 1) +
P(X = 2) + P(X = 2) +
P(X = 3) + P(X = 3) + P(X = 3) +
P(X = 4) + P(X = 4) + P(X = 4) + P(X = 4) + · · ·
Adding each column now gives us something interesting: the ﬁrst column is P(X =
1) + P(X = 2) + P(X = 3) + · · · = P(X 1), the second column is P(X = 2) + P(X =
3) + · · · = P(X 2), the third column is P(X = 3) + P(X = 4) + · · · = P(X 3), and
so on. So we get
E(X ) = P(X
= ∞ 1) + P(X P(X 2) + P(X 3) + P(X 4) + · · · n), n=1 as we wanted.
An alternative, more formal, way of writing this proof is as follows, using what is sometimes called “summation algebra”: E(X ) =
= ∞
n=1
∞ nP (X = n)
n P(X = n) summing n copies of a constant n=1 m=1 =
=
= P(X = n)
1 m n<∞
∞
∞ P(X = n) m=1 n=m
∞ P(X writing it as one big sum
see below m), m=1 which is the sum we wanted. For the penultimate step, note the we are originally summing
all pairs of values (m, n) where n is any positive integer and m lies between 1 and n, so
we have 1
m
n < ∞, as written on the third line. This can also be thought of as
summing over all pairs of values (m, n) where m is any positive integer (i.e., 1 m < ∞),
and n is chosen so that m n < ∞, that is, we are summing on n from m to ∞.
[One ﬁnal technical note: we are allowed to reorder the terms of this inﬁnite sum because
all of the summands (the things we are adding) are nonnegative. If some were positive
and others were negative, we might get all sorts of weird things happening if we reordered
the terms. An undergraduate course in Analysis will usually explore such questions.]
I am collecting toy penguins from cereal boxes. Each box contains either one daddy
penguin or one mummy penguin. The probability that a given box contains a daddy
penguin is p and the probability that a given box contains a mummy penguin is q ,
where p = 0, q = 0 and p + q = 1.
Let X be the number of boxes that I need to open to get at least one of each kind of
penguin. Show that P(X 4) = p3 + q 3 , and that
E(X ) = 1
− 1.
pq We ask ourselves: what needs to happen to have X
4? This means that we need to
open at least 4 boxes to get both a daddy and a mummy penguin. In other words, we
can’t have had both a daddy and a mummy among the ﬁrst three boxes, so they must
have all had daddies or all had mummies. Therefore P(X 4) = p3 + q 3 .
This immediately generalises to give P(X
n = 1, P(X 1) = 1, and for n = 2, P(X n) = pn−1 + q n−1 , at least for n
3. For
1
1
2) = 1 = p + q , as we argued above. Therefore, we have
E(X ) = ∞ P(X n) n=1 = 1 + ( p1 + q 1 ) + ( p2 + q 2 ) + ( p3 + q 3 ) + · · ·
= (1 + p + p2 + p3 + · · · ) + (1 + q + q 2 + q 3 + · · · ) − 1
1
1
+
−1
adding the geometric series
=
1−p 1−q
11
= + −1
as p + q = 1
qp
p+q
−1
=
qp
1
=
−1
again using p + q = 1.
pq Hence show that E(X ) 3. To show that E(X )
3, we simply need to show that
1
showing that pq 4 , by taking reciprocals. 1
pq Now, recall that q = 1 − p, so we need to show that p(1 − p)
the quadratic in p by completing the square:
p(1 − p) = p − p2 = 1
4 4. But this is the same as
1
.
4 To do this, we rewrite − (p − 1 )2 .
2 1
Since (p − 2 )2 0 for all p (even outside the range 0 < p < 1), we have p(1 − p)
required, with equality only when p = q = 1 .
2 1
,
4 as This can also be proved using calculus, or using the AM–GM inequality, or by writing
1/pq = (p + q )2 /pq and then rearranging to get (p − q )2 0. Question 13 The number of texts that George receives on his mobile phone can be modelled by
a Poisson random variable with mean λ texts per hour. Given that the probability
George waits between 1 and 2 hours in the morning before he receives his ﬁrst text is p,
show that
pe2λ − eλ + 1 = 0.
Given that 4p < 1, show that there are two positive values of λ that satisfy this
equation.
Let X be the number of texts George receives in the ﬁrst hour of the morning and Y be
the number he receives in the second hour.
Then X ∼ Po(λ) and Y ∼ Po(λ), with X and Y independent random variables.
We thus have P(George waits between 1 and 2 hours for ﬁrst text) = P(X = 0 and Y > 0)
= P(X = 0).P(Y > 0)
= e−λ .(1 − e−λ )
= p,
so that e−λ − e−2λ = p. Multiplying this last equation by e2λ gives eλ − 1 = pe2λ ; a straightforward rearrangement
yields our desired equation.
(This equation can also be deduced by considering the waiting time until the ﬁrst text;
this is generally not studied until university, though.)
The solutions of the quadratic equation in eλ are given by
√
1 ± 1 − 4p
λ
.
e=
2p
But we are given that 4p < 1, so that 1 − 4p > 0 and there are real solutions. We need
to show, though, that the two values of eλ that we get are both greater than 1, so that
the resulting values of λ itself are both greater than 0.
We have
1± √ 1 − 4p
> 1 ⇐⇒ 1 ± 1 − 4p > 2p
2p
⇐⇒ ± 1 − 4p > 2p − 1. Now, since 4p < 1, we have 2p < 1 , so 2p − 1 < 0, from which it follows that for the
√2
positive sign in the inequality, 1 − 4p > 0 > 2p − 1. It therefore only remains to show
√
that − 1 − 4p > 2p − 1. But
− 1 − 4p > 2p − 1 ⇐⇒
1 − 4p < −(2p − 1)
⇐⇒ 1 − 4p < (2p − 1)2
⇐⇒ 1 − 4p < 4p2 − 4p + 1, which is clearly true as 4p2 > 0. (We were allowed to square between the ﬁrst and second
lines as both sides are positive.)
Thus the two solutions to our quadratic in eλ are both greater than 1, so there are two
positive values of λ which satisfy the equation.
The number of texts that Mildred receives on each of her two mobile phones can be
modelled by independent Poisson random variables but with diﬀerent means λ1 and λ2
texts per hour. Given that, for each phone, the probability that Mildred waits between
1 and 2 hours in the morning before she receives her ﬁrst text is also p, ﬁnd an expression
for λ1 + λ2 in terms of p.
Each phone behaves in the same way as George’s phone above, so the two possible values
of λ are those found above. That is, the values of eλ1 and eλ2 are the two roots of
pe2λ − eλ + 1 = 0. We know that the product of the roots of the equation ax2 + bx + c = 0 is c/a,1 so in our
case, eλ1 eλ2 = 1/p, so that eλ1 +λ2 = 1/p, giving
λ1 + λ2 = ln(1/p) = − ln p. Find the probability, in terms of p, that she waits between 1 and 2 hours in the morning
to receive her ﬁrst text.
Let X1 be the number of texts she receives on the ﬁrst phone during the ﬁrst hour and
Y1 be the number of texts that she receives on the ﬁrst phone during the second hour.
Then X1 and Y1 are both distributed as Po(λ1 ), so
P(X1 = 0) = e−λ1
P(Y1 = 0) = e−λ1 .
Now let X2 and Y2 be the corresponding random variables for the second phone, so we
have
P(X2 = 0) = e−λ2
P(Y2 = 0) = e−λ2 .
We must now consider the possible situations in which she receives her ﬁrst text between
1 and 2 hours in the morning. She must receive no texts on either phone in the ﬁrst hour,
and at least one text on one of the phones in the second hour. We use the above result
that λ1 + λ2 = − ln p, so that e−λ1 −λ2 = p.
1 Why is this? If the roots of ax2 + bx + c = 0 are α and β , then we can write the quadratic as
a(x − α)(x − β ) = a(x2 − (α + β )x + αβ ), so that c = aαβ , or αβ = c/a. Likewise, b = −a(α + β ) so that
α + β = −b/a. Thus
P(ﬁrst text between 1 and 2 hours)
= P(X1 = 0 and X2 = 0 and Y1 > 0 or Y2 > 0 or both)
= P(X1 = 0).P(X2 = 0). 1 − P(Y1 = 0 and Y2 = 0)
= P(X1 = 0).P(X2 = 0). 1 − P(Y1 = 0).P(Y2 = 0) = e−λ1 .e−λ2 . 1 − e−λ1 .e−λ2 = e−λ1 −λ2 .(1 − e−λ1 −λ2 )
= p(1 − p),
and we are done.
Alternative approach This approach uses a result which you may not have come across yet: the sum X + Y of
two independent Poisson random variables X ∼ Po(λ) and Y ∼ Po(µ) is itself a Poisson
variable with X + Y ∼ Po(λ + µ). Since the number of texts received on the two phones together is the sum of the number
of texts received on each one, the total can be modelled by a Poisson random variable
with mean Λ = λ1 + λ2 texts per hour.
Then the probability of waiting between 1 and 2 hours in the morning for the ﬁrst text is
given by q , where
q e2Λ − eΛ + 1 = 0, using the result from the very beginning of the question, replacing p with q and λ with Λ.
Since Λ = λ1 + λ2 = ln(1/p) from above, eΛ = 1/p.
Therefore
eΛ − 1
e2Λ
1/p − 1
=
(1/p)2
= p2 (1/p − 1)
= p(1 − p). q= Hints & Solutions for STEP II 2010
1 When two curves meet they share common coordinates; when they “touch” they also share a
common gradient. In the case of the osculating circle, they also have a common curvature at the
d2 y
dy
, the
and
point of contact. Since curvature (a further maths topic) is a function of both
dx
dx 2
question merely states that C and its osculating circle at P have equal rates of change of gradient.
It makes sense then to differentiate twice both the equation for C and that for a circle, with
equation of the form (x – a)2 + (y – b)2 = r2, and then equate them when x = 1 . The three
4
resulting equations in the three unknowns a, b and r then simply need to be solved simultaneously. For y = 1 – x + tan x , d2 y
dy
= 2 sec2x tan x .
= – 1 + sec2x and
2
dx
dx
2 d2 y
dy dy + 2 = 0 .
= 0 and 2 + 2(y – b)
For (x – a) + (y – b) = r , 2(x – a) + 2(y – b)
2
dx
dx dx 2
2
When x = 1 , y = 2 1 and so 1 a 2 1 b r 2 ;
4
4
4
4
2 2 2 ( x a)
dy 1 then gives a relationship between a and b;
=
dx
( y b)
4
d2 y
gives the value of b.
=4= 2
2( y b)
dx
Working back then gives a and r.
and Answers: The osculating circle to C at P has centre 2 1 1 ,
4
2 5
2 1 and radius
4 1
2 . The singlemaths approach to the very first part is to use the standard trig. “Addition” formulae for
sine and cosine, and then to use these results, twice, in (i); firstly, to rewrite sin3x in terms of
sin3x so that direct integration can be undertaken; then to express cos3x in terms of cos3x in
order to get the required “polynomial” in cosx. Using the given “misunderstanding” in (ii) then
leads to a second such polynomial which, when equated to the first, gives an equation for which a
couple of roots have already been flagged. Unfortunately, the several versions of the question that
were tried, in order to help candidates, ultimately led to the inadvertent disappearance of the
interval 0 to in which answers had originally been intended. This meant that there was a little bit
more work to be done at the end than was initially planned.
cos3x = cos(2x + x) = cos2x cos x – sin2x sin x = (2c2 – 1)c – 2sc.s = (2c2 – 1)c – 2c(1 – c2)
= 4c3 – 3c .
sin3x = sin(2x + x) = sin2x cos x + cos2x sin x = 2sc.c + (1 – 2s2)s = 2s(1 – s2) + s(1 – 2s2)
= 3s – 4s3
(i) I () = 7 sin x 8 sin x dx = sin x 2 sin 3x dx = cos x 3 0 0 = – cos – 2 (4cos3 – 3 cos) + 1 +
3
and I () = 0 when c = 1 ( = 0) 2
3 = – 8 c3 + c +
3 5
3 2
3 cos 3x 0 (ii) J () = sin
7
2 2 x 8 sin 4 x
4 0 = 7
2 (1 – cos2 ) – 2(1 – cos2)2 = – 2c4 + 1
2 c2 + 3
2 I () = J () 0 = 12c4 – 16c3 – 3c2 + 6c + 1 = (c – 1)2 (2c + 1)(6c + 1)
Thus cos = 1, = 0; cos = – 1 , = 2 ; and cos = – 1 , = – cos – 1 ( 1 ).
6
2
3
6
Answers: 2n , 2n 2 , (2n 1) cos 1
3 3 1
6 You don’t have to have too wide an experience of mathematics to be able to recognise the
Fibonacci Numbers in a modest disguise here. (However, this is of little help here, as you should
be looking to follow the guidance of the question.) In (i), you are clearly intended to begin by
substituting n = 0, 1, 2 and 3, in turn, into the given formula for F n , using the four given terms of
the sequence. You now have four equations in four unknowns, and the given result in (i) is
intended to help you make progress; with (ii) having you check the formula in a further case. In
the final part, you should split the summation into two parts, each of which is an infinite geometric
progression.
(i) F 0 = 0 0 = a + b or b = – a . Then F 1 = 1 1 = a( – ).
[F 2 = 1 1 = a(2 – 2) + = 1 is needed later] and F 3 = 2 2 = a(3 – 3) = a( – ) (2 + + 2) by the difference of two cubes
= 1.(2 + + 2) 2 + + 2 = 2
1
and + = 1, and
a
1
1
1
1
, b=–
, = 1 5 , = 1 5 .
solving simultaneously gives a =
2
2
5
5 Then, using any two suitable eqns., e.g. any two of = – 1, – = (ii) Using the formula F n = a n + b n = Theorem gives 1 5 Similarly, 1 5 6 1
2 n (1 5 5 ) n (1 5 ) n with n = 6 ; the Binomial 6 1 6 5 15.5 20.5 5 15.5 2 6.5 2 5 5 3 = 576 + 256 5 .
1 576 256 5 so that F 6 = 6
512 5 = 8.
25 n
n Fn
1
1
a a 1 (iii) n 1 = =
2 n0 2 2 n0 2 2 5 1 1 1 5 2 5 1 1
n0 2
4 4
S formula for the two GPs;
1 4 1 4 = . 2 5 3 5 2 5 3 5 2 3 5 2 3 5 =
Rationalising denominators then yields
5 95 5 95 1 using the
1 5 2 2 5 = 1.
5 4 4 Hopefully, the obvious choice is y = a – x for the initial substitution and, as with any given
result, you should make every effort to be clear in your working to establish it. Thereafter, the two
integrals that follow in (i) use this result with differing functions and for different choices of the
upper limit a. Since this may be thought an obvious way to proceed, it is (again) important that
your working is clear in identifying the roles of f(x) and f(a – x) in each case. In part (ii), however,
it is not the first result that is to be used, but rather the process that yielded it. The required
substitution should, again, be obvious, and then you should be trying to mimic the first process in
this second situation.
(i) Using the substn. y = a – x , dy = – dx and (0, a) (a, 0) so that
0
a
a
f ( x)
f (a y )
f (a y ) f ( x) f (a x) dx = f (a y) f ( y) .– dy = f (a y) f ( y) dy
0
a
0
f (a x) a = f ( x) f (a x) dx , since the x/y interchange here is nothing more than a relabelling.
0 a Then 2 I = 0 f ( x) f (a x)
dx = 1. dx = x a = a I =
0
f ( x) f (a x)
0
a 1
2 a. For f(x) = ln(1 + x) , ln(2 + x – x2) = ln[(1 + x)(2 – x)] = ln(1 + x) + ln(2 – x)
1 and ln(2 – x) = ln(1 + [1 – x]) = f(a – x) with a = 1 so that f ( x) f ( x) f (1 x) dx = 1
2 . 0 /2 0 sin x
dx =
sin x 1 4 /2 0 sin x
sin x. 12 cos x. 1
1
, du = 2 dx and
x
x
2
1
sin x
dx =
Then .
x sin x sin 1 x
0.5 2 0 sin x
dx = 1 2 .
4
1
sin x sin 2 x 1
x sin x
.5 x 2 . sin x sin 1x dx =
0
sin 1 1
x
.
.5 x sin x sin 1x dx
0 1
.sin u sin u1 sin u .– du
2 0.5 1
u 2 or x dx = ln x0.5 = 2 ln 2
1 2 2 Adding then gives 2 I = 1
2 2 1 , 2 2, 1 .
2
2 (ii) For u = 1
sin u 1
= .
du
1
u sin u sin u 0.5 /2 dx = 2 I = ln 2 . 0.5 5 The opener here is a standard bit of Alevel maths using the scalar product, and the following
parts use this method, but with a bit of additional imagination needed. In 3dimensions, there are
infinitely lines inclined at a given angle to another, specified line, and this is the key idea of the
final part of the question. Leading up to that, in (i), you need only realise that a line equally
inclined to two specified (nonskew) lines must lie in the plane that bisects them (and is
perpendicular to the plane that contains, in this case, the points O, A and B). One might argue that
the vector treatment of “planes” is further maths work, but these ideas are simple geometric ones. cos2 = (1, 1, 1) (5, 1, 1) 1 3
3. 27 (i) l 1 equally inclined to OA and OB iff (m, n, p) (1, 1, 1) m2 n2 p 2 . 3
i.e. 3(m + n + p) = 5m – n – p or m = 2(n + p). cos = 1
3 2
3 ( m, n, p) (5, 1, 1) m 2 n 2 p 2 . 27 mn p For l 1 to be the angle bisector, we also require (e.g.)
cos2 = 2 cos2 – 1 = m2 n2 p2 . 3 = cos , where m2 n2 p 2 . 2 . , so that m + n + p = Squaring both sides: m2 + n2 + p2 + 2mn + 2np + 2pm = 2(m2 + n2 + p2) 2mn + 2np + 2pm = m2 + n2 + p2
Setting m = 2n + 2p (or equivalent) then gives 2np + (2n + 2p)2 = (2n + 2p)2 + n2 + p2
which gives (n – p)2 = 0 p = n , m = 4n . m 4 Thus n 1 , or any nonzero multiple will suffice. p 1 (ii) If you used the above method then you already have this relationship; namely,
2uv + 2vw + 2wu = u2 + v2 + w2 .
Thus, 2xy + 2yz + 2zx = x2 + y2 + z2 gives all lines inclined at an angle cos – 1 2
3 to OA and hence describes the surface which is a doublecone, vertex at O, having central axis OA . 6 Although it seems that 3dimensional problems are not popular, this is actually a very, very easy
question indeed and requires little more than identifying an appropriate rightangled triangle and
using some basic trig. and/or Pythagoras. There are thus so many ways in which one can approach
the three parts to this question that it is difficult to put forward just the one.
D
(i) Taking the midpoint of AB as the origin, O,
with the xaxis along AB and the yaxis along
OC, we have a cartesian coordinate system to
help us organise our thoughts.
C
Then A = 1 , 0, 0 , B =
2 P = 0, C = 0, 3
2
3
6 1 ,
2 0, 0 , , 0 . The standard distance formula P , 0 by trig. or Pythagoras, and then gives PA (or PB) = 3
3 and PD = 6
3 or A
2
3 O
. B 1 3 in rightangled triangle
(ii) The angle between adjacent faces is (e.g.) DOC = cos – 1 6 1 3 2
–11
DOP, which gives the required answer, cos 3 . D (iii) 3
2
6
3 The centre, S, of the inscribed sphere must, by symmetry,
lie on PD, equidistant from each vertex. 3
6 r By Pythagoras, x2 =
r + Then r = x sin(90o – (ii)) = 2 6
9 1
3 6
3 x= x x2 6
12 x= 6
4 . . 3
6 S r Alternatively, if you know that the sphere’s centre is at
The centre of mass of the tetrahedron, the point (S)
with position vector 1 (a b c d) , then the answer
4 x
3
6 P 7 1
12 A is just 1
4 DP = 6
12 . The first two parts of the question begin, helpfully, by saying exactly what to consider in order to
proceed, and the material should certainly appear to be routine enough to make these parts very
accessible. Where things are going in (iii) may not immediately be obvious but, presumably, there
is a purpose to (i) and (ii) which should become clear in (iii).
(i) y = x3 – 3qx – q(1 + q) dy
= 3(x2 – q) = 0 for x q .
dx q , y = q q 1 < 0
2 When x = q , y = q q 1 < 0 since q > 0
When x = 2 since q > 0 and q 1 Since both TPs below xaxis, the curve crosses the xaxis once only (possibly with sketch)
q2 q3
q x3 = u 3 3uq 3 3
uu
u
q2 q3
q2
0 = x3 – 3qx – q(1+ q) = u 3 3uq 3 3 – 3qu – 3 q q 2
uu
u
3
2
q u3 + 3 q(1 q) = 0 or u 3 q(1 q) u 3 q 3 0
u (ii) x = u + q(1 q) q 2 (1 q) 2 4q 3
q
=
u=
1 q 1 2 q q 2 4q
2
2
q
q
2
1 q 1 q = 1 q (1 q ) = q or q2
=
2
2
3 1 1 2 2 giving u = q 3 or q 3 and x = q 3 + q 3
(iii) + = p , = q 3 3 3 ( ) = p3 – 3qp .
3 One root is the square of the other = 2 or = 2 0 2 2 .
Then 0 = 2 2 3 3 ( ) 2 p 3 3qp q(1 q) 1
3 2
3 p= q + q . 8 When asked to draw sketches of graphs, it is important to note the key features. The first curve is a
standard “exponential decay” curve; the second has the extra factor of sinx. Now sinx oscillates
between –1 and 1, and introduces zeroes at intervals of . Thus, C 2 oscillates between C 1 and –C 1 ,
with zeroes every units along the xaxis. This sketch of the two curves should then make it clear
that the x i that are then introduced are the xcoordinates of C 2 ’s maxima, when sinx = 1. [It is
important to be clear in your description of x n and x n+1 in terms of n as these are going to be
substituted as limits into the area integrals that follow.] The integration required to find one
representative area will involve the use of “parts”, and the final summation looks like it must be
that of an infinite GP. 1 y = e–x O 2 3 4 y = – e–x –1
The curves meet each time sin x = 1 when x = 2n +
(4n 3)
(4n 1)
and x n + 1 =
.
Thus x n =
2
2 e x 2 ( n = 0, 1, 2, …). sin x dx attempted by parts = e x . cos x e x . cos x dx or e x . sin x e x . sin x dx (depending on your choice of ‘1st’ and ‘2nd’ part) = e x . cos x e x . sin x e x . sin x dx . x
x
1
Then I = e (cos x sin x) – I (by “looping”) = 2 e (cos x sin x) e xn 1 An = x xn 1 x x e sin x dx = e 1 e
2 x cos x sin x x n or xn = 1e
2 1 ( 4 n 1)
2 0 1 2 1 e ( 4 n 3) 0 1 2 =
2
1
2 Note that A 1 = 1 e
2 = 1e
2 5
2 e 2 1 5
2 e 2 1 1
2 e 1 ( 4 n 1)
2 and A n + 1 = e 2 A n so that e
1
2 xn 1 x cos x sin x 2 x n 1 e 2 A
n 1 n = A1 1 e 2 e 2 ...
2 1
e 2 5 1
= 1 e 2 e 2 1 2
(using the S of a GP formula) = 1 e 2
2
2 2
1 e
e 1 9 Once you have written down all relevant possible equations of motion, this question is really quite
simple; the two results you are asked to prove arise from considering either times or distances to
the point of collision. There is, however, one crucial realisation to make in the process, without
which further progress is almost impossible; once noted, it seems terribly obvious, yet it probably
doesn’t usually fall within the remit of standard Alevel examination questions. For P 1 , x1 0 , x1 u cos , x1 ut cos , y1 g , y1 u sin gt , y1 ut sin 1 gt 2
2 For P 2 , x 2 0 , x 2 v cos , x 2 vt cos , y 2 g , y 2 v sin gt , y 2 vt sin 1 gt 2
2 u sin u 2 sin 2 Now P 1 is at its greatest height when y 2 0 t =
and it follows y1 = h =
g
2g that u sin = 2 gh Note that if the two particles are at the same height at any two distinct times (one of which is t = 0
here), then their vertical speeds are the same throughout their motions. Thus u sin = v sin .
2v sin . This is the time when P 2 would land. Also, the collision occurs
g
b
when x 2 = b t =
is the time of the collision.
v cos y 2 = 0, t 0 t = Then t P2 1
2 range < t(collision) < t P2 range (or by distances) v 2 sin cos v sin b
2v sin 2v 2 sin cos <b<
<
< g
v cos g
g
g (v sin ) 2
2(v sin ) 2
cot < b <
cot . Using u sin = v sin =
g
g
2 gh
4 gh
cot < b <
cot 2h cot < b < 4h cot .
g
g 2 gh then gives One could repeat all this work for P 1 , but this is not necessary. Since the particles are at their
maximum heights simultaneously (see the above reasoning) and would achieve their “ranges”
simultaneously also, we have 2h cot < a < 4h cot . 10 I always feel that collisions questions are very simple, since (as a rule) there are only the two main
principles – Conservation of Linear Momentum and Newton’s Experimental Law of Restitution –
to be applied. Such is the case here. Part (ii) is only rendered more difficult by the introduction of
a number of repetitions, and then the question concludes with some pure mathematical work using
logarithms.
u (i) B bm
vB A m
vA Using CLM: bmu = bmv B + mv A .
Using NEL:
u = vA – vB .
2bu
(b 1)u
Solving simultaneously: v A =
and v B =
.
b 1
b 1
2
Then v A = 1 u 2u – as b , and v A < 2u always.
1 b (ii) u = u1 un = vn u2 = v2 B 1 n m
v1 B 2 n – 1 m
v2 Bn m
vn ……… Am
v 2 2 2 Using the results of (i), v 2 = u 2 = u ; u 3 = u ; … etc. …
u 2 1 1 1
2 n 1 2 2 2 2 all the way down to u n = u and v = u.
u n 1 u n 1 1 1 1
2 1 , as > 1, it follows that v can be made as large as possible.
Since u n = 1 In the case when = 4, v = 8 n u
5 > 20u requires n log 8 > log 20 n 5 n log 20
.
log 8 5 Now log 2 = 0.30103 log 8 = 3log 2 = 0.90309
and log 5 = log 10 – log 2 = 1 – 0.30103 = 0.69897
so that log 8 = log 8 – log 5 = 0.20412.
5
1.30103
.
Also log 20 = log 10 + log 2 = 1 + 0.30103 = 1.30103, so we have n 0.20412
Since 6 0.20412 = 1.22472 and 7 0.20412 = 1.42884, n min = 7.
11 A few years ago, a standard “threeforce” problem such as this would have elicited responses
using Lami’s Theorem; since this tidy little result seems to have lapsed from the collective Alevel
consciousness, I shall run with the more popular, alternative Staticsquestion approach of
resolving twice and taking moments. In order to get started, however, it is important to have a
good, clear diagram suitably marked with correct angles. The later parts of the question consist
mostly of trignometric work.
R T
Res. T sin( ) + R sin( + ) = W B
C Res. T cos( ) = R cos( + ) l l A 2l W.2l cos = T.3l sin A W T cos( )
3T sin sin( + ) =
cos( )
2 cos 2 cos cos . cos sin . sin sin . cos cos . sin + 2 cos cos . cos sin . sin sin . cos cos . sin = 3 sin cos . cos sin . sin Substituting to eliminate T ’s (e.g.) T sin( ) + Dividing by cos cos cos 2(cos tan . sin )(tan . cos sin ) 2(cos tan . sin )(tan . cos sin )
= 3 tan (1 tan . tan )
Multiplying out, cancelling and collecting up terms, and then dividing by tan tan then gives
the required answer 2 cot + 3 tan = cot . = 30o, = 45o cot = 2.1 3. 1
= 2 3 ,
3 3 1 and tan15 = tan(60 – 45 ) =
1 3
o 12 o o 2 3 1
1 2 3 .
3 1
2 3 In some ways, the pdf f(x) couldn’t be much simpler, consisting of just two horizontal straightline segments (in the nonzero part). Part (i) is then relatively routine – use “total prob. = 1” to
find the value of k, before proceeding to find E(X); and the trickiest aspect of (ii) is in the
inequalities work. You also need to realise that the median could fall in either of the two nonzero
regions. For (iii), it is necessary only to follow through each possible value of M relative to E, the
expectation.
Since the pdf is only nonzero between 0 & 1, and the area under its graph = 1, if a, b are both <
(>) 1 then the total area will be < (>) 1. Since we are given that a > b, it must be the case that
a > 1 and b < 1.
1 k 1 k 1 0 0 k 0 k (i) 1 = f ( x) dx = a dx + b dx = ax + bx = ak + b – bk k = 1 b
.
a b k
1 bx 2 1 ax 2 k
ak 2 b bk 2
xf ( x) dx = ax dx + bx dx = =
+ 2
2
2 2 0 2 k
0
k
0 1 E(X) = ba b 2 1 2b b 2 1 2b ab
b ( a b) 1 b =
. = 2
2
2( a b )
2(a b)
a b
2 = (ii) If ak 1
2 and aM =
If ak (i.e. M (0, k)) then
1
2 or M = a ab ab 1
2 2a – 2ab a – b a + b 2ab 1
.
2a (i.e. M (k, 1)), and noting that this is equivalent to a + b 2ab ,
1
then ak + (M – k)b = 1 or (1 – M)b = 1 M = 1 –
2
2
2b
1
2 b(1 a ) 2
a 2ab a 2 b a b
1 2b ab 1
=
=
>0 2(a b)
2a
2a ( a b)
2a ( a b)
and the required result follows. (iii) If a + b 2ab , then – M = If a + b 2ab , then – M =
= 1
1 2b ab
b 2b 2 ab 2 2ab 2b 2 a b
=
1 2b
2(a b)
2b(a b)
a (1 b) 2
> 0 as required.
2b(a b) 13 This question is really little more than examining the various cases that arise for each outcome and
then doing a little bit of work algebraically. The result of part (i) is somewhat counterintuitive, in
that Rosalind should choose to play the more difficult opponent twice, while one intutively feels
she should be playing the easier opponent. The real issue, however, is that she needs to beat both
opponents (and not just win one game): examining the probabilities algebraically makes this very
obvious. Part (ii) is a nice adaptation, where there is a cutoff point separating the cases when one
strategy is always best from another situation when either strategy 1 or 2 can be best. Here, it is
most important to demonstrate that the various conditions hold, and not simply state a couple of
probabilities and hope they do the job. [It is perfectly possible to do (iii) by “trialanderror”, but I
have attempted to reproduce below an approach which incorporates a method for deciding the
matter.]
(i) P(W PPQ ) = P(W P W Q –) + P(L P W Q W P ) = p . q. 1 + (1 – p)qp = pq(2 – p).
Similarly, P(W PQQ ) = pq(2 – q) and P(W PPQ ) – P(W PQQ ) = pq(q – p) > 0 since q > p. Thus,
P(W PPQ ) > P(W PQQ ) for all p, q and “Ros plays Pardeep twice” is always her best strategy.
(ii) SI: P(W 1 ) = P(W Q W P – –) + P(W Q L P W P –) + P(W Q L P L P W P )
= pq + pq(1 – p) + pq(1 – p)2 or pq(3 – 3p + p2) SIII: P(W 3 ) = pq(3 – 3q + q2) similarly. SII: P(W 2 ) = P(W P W Q – –) + P(L P W P W Q –) + P(W P L Q W Q –) + P(L P W P L Q W Q )
= pq + pq(1 – p) + pq(1 – q) + pq(1 – p)(1 – q)
= pq(4 – 2p – 2q + pq) or pq(2 – p)(2 – q) . P(W 1 ) – P(W 3 ) = pq(q – p) 3 [ p q ] > 0 since q > p and p + q < 2 < 3 so that SI is
always better than S3 P(W 1 ) – P(W 2 ) = pq p 2 p 1 pq 2q = pq (2 p )(q p ) (1 p ) 1
1 p 1
> 0 whenever q – p >
.
2 p
2 p
1
< 1 , so that SI always better than SII
Now p + 1 < q < 1 0 < p < 1 1 < 1 2
2
3
2
2 p
when q – p > 1 .
2
P(W 1 ) – P(W 2 ) > < 0 q – p > < 1 p
.
2 p 1 p 3 7 > 1 so SII is better than SI.
4
2 p
1 p 3
Take p = 1 , q = 3 q – p = 1 < 1 and so choosing
4
4
2
2
2 p 7
1 p 3
3
11 so that < 7 1 114 (say 116 ) will give p = 1 , q = 16 and q – p = 176 >
2
4
2 p 7
SI is better than SII. Take p = 1
4 ,q= 1
2 q–p= 1
4 < 1
2 and [I believe that q – p > k has k = 1 as the least positive k which always gives SI better than SII,
2
but it is a long time ago that the problem was originally devised and I may be wrong.] STEP Mathematics III 2010: Solutions
Section A: Pure Mathematics 1. The first two parts are obtained by separating off the final term of the summation and
, and
expanding the brackets respectively giving
1
(the latter given in the question) .
By comparison with the expression for B,
1
1
which by substituting for
1
from the expression for B gives
1
1
.
Substituting for C from the initial result, the required expression can be obtained which can most
neatly be written 1
Thus 1 Also, 1
yielding the first inequality.
and this quadratic expression is only negative if and only if
.
Rearranging the inequality to make xn+1 the subject yields the required result.
2. The expression of cosh a in exponentials enables the integral to be written as
1
1
which can in turn can be expressed as
1
and so employing partial fractions this is
1 The evaluation of this with simplification of logarithms yields
1
1
1
2 sinh
giving the required result.
In part (ii), the same technique can be employed for both integrals giving, in the first case
1
1
1
2 cosh coth 2 and in the second
1
1 1 1 tan 1
2 sinh 2 2 sinh tan 2 or alternatively
4 cosh
3. The two primitive 4th roots of unity are 1
1
1 1,
1
1
1 0⇒
In part (ii),
8th roots of unity, n must be 8. 1 1
1 so
1
1⇒ 2
1 so
1, 1 so
1
1 so
1
1 1
1 so 1 1 so n is a multiple of 8, and as there are 4 primitive 1⇒
1 0⇒
1
⋯1
1 is the only nonprimitive root as no power of any other root less than the pth equals unity,
⋯1
because p is prime, so
No root of
0 is a root of
, there is no integer t such that
≡
, and if
Thus if
or
. 0 for any
1 when
0, then . (For if
, by the definition of
1. Similarly, if
.)
0 or
0 , so ≡
, and so
If
, then
likewise in the alternative case. ≡ 1 which is not possible for positive s, and 4. (i) As satisfies both equations,
these the desired result is simply found.
If
satisfies
0. But also, 0 and 0, so subtracting 0, then we may divide by , and find that and
so 0. satisfies On the other hand if there is a common root, then it is found at the start of the question and as it
0, the required result is found.
satisfies
If
0 and
, then
and so the two
equations are one and trivially have a common root. Alternatively, if there is a common root and
, then the initial subtraction yields
, and so the result is trivially true.
0, then
0 and
0 have a common root from (i), and so then do
0 and
0 which is the required result.
0
On the other hand, if the two equations have a common root , then
1
0 , and thus so does
and
1
0 which is a quadratic equation and we can use
the result from (i) again.
Using
,
,
, in the given condition, we obtain a cubic equation in b,
(ii) If 0, which has a solution
obtained as √ 1, meaning the other two can be simply . 5. The line CP can be shown to have equation 1 and so R is 0, So, similarly, S must be
,0 .
Thus RS has equation
1
1
and PQ has equation
As the coordinates of T satisfy both equations, they satisfy their difference which is
0. As RS and PQ intersect,
which yields . 0 and hence
0 implying that T’s coordinates
0,
satisfy
0 giving the desired result. (Alternatively,
0⇔
which is a contradiction.)
The construction can be achieved more than one way, but one is to label the given square ABCD
anticlockwise, choose points on AB and AD different distances from A, label them P and Q,
construct CP and CQ, and find their intersections with AD and AB, R and S, respectively, and
find the intersection of PQ and RS, label it T, then TA is perpendicular to AC. Rotating the
labelling through a right angle and repeating three more times achieves the desired square. is cos , sin
sin , cos , 0 ,
The scalar product , 0 , is cos cos , sin cos , sin ,
is sin , cos , 0 ,
is
is 0,0,1 and
is
cos sin , sin sin , cos .
∙
gives the quoted result immediately. The direction of the axis can
cos cos
1
giving the direction of the axis as
be found from the vector product 0
sin cos
0
sin
0
sin
.
sin cos
6. 7. The initial result can be obtained by differentiating y directly twice obtaining
sin sin
√ cos sin sin sin and substituting into the LHS.
sin (Slightly more elegant is to rearrange as cos
1 obtain 1 and then differentiate a second time.) The two similar results are 1
1 5 Using 3
4 2 1 1 0 and 0, which lead to the conjecture 1 0 , we find that , differentiate and then square to 0 which is proved simply by induction.
1, 0, and so the Maclaurin series commences ,
1 0, ! 4
⋯ ! Now replacing x by sin ,
⋯1
sin
cos
1
!
!
!
All the odd differentials are zero, and the even ones are 1
so if m is even all the terms are zero from a certain point (when
terminates and is a polynomial in sin , of degree m.
8. Substituting for ! sin
⋯
2…
2
2 ) and thus the series , , the desired integral is seen to be the reverse of the quotient rule, i.e. and
To choose a suitable function
in part (i), substitution of
12
3 in the given expression yields a quadratic equation, and equating the
coefficients of the powers of x gives 5
3
2, 2
3
,3
2
.
These three equations are linearly dependent and so their solution is not unique.
Choosing, for example
0,
3,
2 and then
1,
1,
1 gives solutions
which are related by
arbitrary constant. 1 i.e. the same bar the (ii) Rearranging the equation to be solved as , the integrating factor is
As a result, the RHS we require to integrate is
Repeating similar working to part (i), except with
1 cos
2 sin and
sin
cos , gives three linearly dependent equations,
5
2, 3
2 ,4
Choosing e.g.
4,
5,
0, the solution is
4 5 sin
1 cos
2 sin Section B: Mechanics 9. Resolving radially inwards for the mass P,
sin ,
where R is the normal reaction of the block on P, and v is the (common) speed of the masses
when OP makes an angle with the table.
Conserving energy,
sin
0 , and making
this formula to substitute in the first equation rearranged for R,
sin is found. Remaining in contact requires this expression to be nonnegative for all , 0
asin and Considering the graphs of
asin the subject of 0, ∀ , 0 if and only if asin so
0 for all , 0
required result. if and only if 3 ,
0 for sin 2 cos 10. Resolving perpendicularly to OB,
elastic string is for 0 . 0 which gives the , where the tension in the . The sine rule Putting these three results together gives the required expression.
Also from the sine rule,
, so for and small, yielding the desired result.
From this result, may be made the subject of the formula, so that the result
1 sin
, which for small angles becomes
1
and hence the period is can be written
2 . 11. If the acceleration of the block is
and
so the relative acceleration , and the acceleration of the bullet is
, , then The initial velocity of the bullet relative to the block is – and the final velocity of the bullet
relative to the block is 0. If the time between the bullet entering the block and stopping moving
through the block is T, then using”
“, 0
For the block, the initial velocity is 0 , the final velocity is , and again using
,
and so
as required.
If the distance moved by the block whilst the bullet is moving through the block is ,
2 using” 2 “, andso Once the bullet stops moving through the block, the next initial velocity of block/bullet is , the
final velocity is 0, the acceleration is – , so the distance moved using
2 “ ” is given by 0 2 Thus the total distance moved is If i.e. , then the block does not move, and the bullet penetrates to a depth Section C: . Probability and Statistics ⋯
⋯ which is 1 plus an infinite GP. Summing that GP
12.
1
and making S the subject produces the displayed result.
1 21 result with 1, 31
1 , ⋯ 1 ⋯ so making use of the first 1
1
both lead to the required result.
1 ⋯ which or alternatively,
⋯ or alternatively, The expected number of shots, S, is given by
3
4
5
1
2
5
⋯
2
12
13 ⋯
⋯
2 which using the initial result of the question
and can be shown to simplify to the required expression.
0 , 13. 1 1
1 0
1 1 1 0 and As 1, , , , 1 1
0 is given. ,
, 1 as seen before. 1 implies
implies
implies and hence , for
, 1
, 1 1,2,3 as
, ,
, and
as a linear transformation does not affect correlation. ...
View Full
Document
 Spring '12
 rotar
 Math

Click to edit the document details