43
2015 Bored of Studies Trial Examinations Mathematics Advanced Solutions
2015 Bored of Studies Trial Examinations
Mathematics Advanced
Solutions
Section I
1. A
2. C
3. D
4. A
5. B
6. B
7. C
8. B
9. D
10. A
Working/Justification
Question 1
Substitute the values in to obtain the equation
16000 = 1000(
1 +r
100
)4
16 =(
1 +r
100
)4±2 = 1 +
r
100
r = 100,−300
Looking at the choices available, the answer is (A)
Question 2
(A) is not necessarily true because we could construct an isosceles triangle that has angles 45◦, 67.5◦
and 67.5◦, which is not a right angled triangle.
(B) is not necessarily true because from the requirement of SAS if the corresponding angle is not theincluded angle of the two equal corresponding sides, then the two triangles are not always congruent.
(D) is not necessarily true because if the hexagon is irregular the interior angle is not necessarily 120◦
(C) is always correct as any pairs of diagonals which bisect each other in a quadrilateral will definetwo sets of congruent triangles where we can deduce that equal sides must be parallel due to alternateangles.
1
Question 3
Note that f ′(x) = g′(x) is equivalent to f(x) = g(x) + c i.e. f(x) − g(x) = c. In other words if thedifference between f(x) and g(x) is a constant, then they both must have the same derivative.
For (A), f(x)− g(x) = (sinx+ cosx)2 − 2 sinx cosx = sin2 x+ cos2 x = 1
For (B), f(x)− g(x) = tan2 x− sec2 x = −1
For (C), f(x)− g(x) = loge x− loge 5x = − loge 5
For (D), f(x) − g(x) = sin2 x − cos2 x = 2 cos2 x − 1 which cannot be simplified further into a con-stant. Hence the answer is (D).
Question 4
First find the gradient of the secant between the points (x0, f(x0)) and (x0 + h, f(x0 + h)) then let h→ 0to find the derivative by first principles. Hence the expression is
f ′(x0) = limh→0
f(x0 + h)− f(x0)
x0 + h− x0
= limh→0
f(x0 + h)− f(x0)
h
So the correct answer is (A)
Question 5
The quadratic polynomial has the general equation f(x) = ax2 + bx + c. Stationary points occur whenf ′(x) = 0 which means
2ax+ b = 0
x =1
2×− b
a
=1
2(α + β)
Hence f ′(α + β
2
)= 0 so the answer is (B)
2
Question 6
Note that the gradients of the two lines are −a1b1
and −a2b2
.
Rearranging (A) suggests that −a1b1
= −a2b2
so the lines must be parallel which is consistent with the
conclusion.
Rearranging (C) suggests thata1a2b1b2
= −1 so the lines must intersect at right angles which is consis-
tent with the conclusion.
If the coefficients are exactly equal on both lines then the lines have the same equation and must co-incide, so (D) is consistent with the conclusion.
If the coefficients are not equal this does not necessarily mean the two lines cannot coincide. The lines cancoincide if the coefficients are in the same ratio but not necessarily equal. For example, x+ y + 1 = 0 and2x+ 2y + 2 = 0 have coefficients which are not equal, yet the lines coincide. Hence, (B) is not necessarilytrue.
Question 7
From sketching the region defined, the area of R is
∫ 1
0
((x− 2)2 − x2
)dx, which then simplifies to the
integral
∫ 1
0
(−4x + 4) dx. From sketching each of the choices, it can be seen that (A) and (B) have no
resemblance to the region R. From the remaining choices (C) and (D) we see that (C) is the correct
answer because the area of the region is given by
∫ 1
0
(−4x+ 4) dx.
Question 8
(A) is not necessarily correct because the integral
∫ t0
0
f ′(t) dt is actually the amount of water that has
leaked from the bucket, not the water remaining in the bucket.
(C) is not necessarily correct because whilst the rate is f ′(t) at time t, the volume that has leaked isf(t) + c for some constant c, which could possibly be non-zero.
(D) is not necessarily correct because when t→ T the value of f ′(T ) is not necessarily zero.
(B) is correct because the volume that has leaked out of the bucket completely is given by
∫ T
0
f ′(t) dt
which is f(T )− f(0)
3
Question 9
From a sketch it is can be seen that PS is in fact a vertical line and therefore has the same x-coordinateas S which is x = a substitute this back into the equation y2 = 4ax to get y = −2a or y = 2a, whichsuggests that (D) is correct.
Question 10
If A and B are mutually exclusive then it is impossible for outcomes A and B to occur together. Hencethe answer is (A). Note that (C) is actually the probability that either outcome A OR outcome B occurswhen they are mutually exclusive.
4
Section II
Question 11
(a) Using the change of base law to common base 4
log2 x log3 x = log4 x
log4 x
log4 2× log4 x
log4 3= log4 x
log4 x
log4 x1
2log4 3
− 1
= 0
log4 x = 0 or log4 x = log4
√3
x = 1,√
3
(b) Using the fact that cos(90◦ − x) = sinx
sin 0◦ + sin 1◦ + sin 2◦ + ...+ sin 90◦
cos 0◦ + cos 1◦ + cos 2◦ + ...+ cos 90◦=
sin 0◦ + sin 1◦ + sin 2◦ + ...+ sin 90◦
sin 90◦ + sin 89◦ + sin 88◦ + ...+ sin 0◦
=sin 0◦ + sin 1◦ + sin 2◦ + ...+ sin 90◦
sin 0◦ + sin 1◦ + sin 2◦ + ...+ sin 90◦
= 1
(c)
sin3 x− cos3 x = sinx− cosx
(sinx− cosx)(sin2 x+ sinx cosx+ cos2 x)− (sinx− cosx) = 0
(sinx− cosx)(sin2 x+ sinx cosx+ cos2 x− 1) = 0
sinx cosx(sinx− cosx) = 0
5
Consider the cases
(1) sinx = 0
x = 0, π, 2π
(2) cosx = 0
x =π
2,3π
2
(3) sinx = cosx
tanx = 1
x =π
4,5π
4
Combining cases (1), (2) and (3) to get full set of solutions: x = 0,π
4,π
2, π,
5π
4,3π
2, 2π
(d) (i) Given that
P = P0ekt
dP
dt= kP0e
kt
= kP
Hence P = P0ekt satisfies
dP
dt= kP
(ii) When t = t0, P = aP0 so
aP0 = P0ekt0
a = ekt0
k =ln a
t0
6
Substitute this back into P
P = P0ett0
ln a
= P0eln a
tt0
= P0att0
(e) The shortest distance to the lines ax+ by = 0 and bx+ ay = 0 for the point P (x, y) must be equal so
|ax+ by|√a2 + b2
=|bx+ ay|√a2 + b2
|ax+ by| = |bx+ ay|
ax+ by = bx+ ay
(a− b)x = (a− b)y
y = x
OR ax+ by =− bx− ay
(a+ b)x = −(a+ b)y
y = −x
Hence the locus is y = ±x which is shown in the sketch below
7
(f) If x2 + xy + y2 = 0 then (x− y)(x2 − xy − y2) = 0 or equivalently x3 = y3 (1). Also note that
x2
y2+x
y+ 1 = 0
x
y+ 1 = −x
2
y2(2)
(x
x+ y
)2015
+
(y
x+ y
)2015
=x2015 + y2015
(x+ y)2015
=
x2015+y2015
y2015
(x+y)2015
y2015
=
(xy
)2013× x2
y2+ 1(
xy
+ 1)2015
=
(x3
y3
)671× x2
y2+ 1(
xy
+ 1)2015 substitute results (1) and (2)
=
x2
y2+ 1(
−x2
y2
)2015
=
x2
y2+ 1
−x4030
y4030
=
x2
y2+ 1
−(x3
y3
)1343× x
y
=
x2
y2+ 1
−xy
= 1 using (2)
8
Question 12
(a)
alogb x
= aloga xloga b by change of base
= aloga x1
loga b
= x1
loga b
= xloga aloga b
= xlogb a
(b) By distance formula
OA =√
62 + 22
=√
40
Note that OB = BA as 4ABO is an isosceles triangle. Using Pythagoras’s Theorem:
OB2 +OA2 = 40
OB2 = OA2 = 20
Using distance formula for OB and BA:
OB2 = x2 + y2
BA2 = (x− 6)2 + (y − 2)2
But since OB2 = OA2 = 20
x2 + y2 = 20
x2 + y2 − 12x− 4y + 40 = 20
9
This implies that
−12x− 4y = −40
y = 10− 3x sub this into x2 + y2 = 20
x2 + (10− 3x)2 = 20
10x2 − 60x+ 80 = 0
x2 − 6x+ 8 = 0
(x− 4)(x− 2) = 0
x = 2, 4
When x = 2, y = 4
When x = 4, y = −2
However, from the diagram, the point B lies in the first quadrant, hence B has coordinates (2, 4).
(c) (i)
10
(ii) Let VC be the volume of the cylinder with radius 1 and height π2
and VX be the volume of the solid
formed by rotating the region bounded by y = tan(x
2
), the line x = π
2and the x-axis, about the x-axis.
VC = πr2h
= π × 12 × π
2
=π2
2units2
VX = π
∫ π2
0
tan2(x
2
)dx
= π
∫ π2
0
(sec2
(x2
)− 1)dx
= π[2 tan
(x2
)− x]π
2
0
= π[2 tan
(π4
)− π
2
]
=π
2(2− π) units2
Volume of solid = VC − VX
=π2
2− π
2(2− π)
= π(π − 1) units3
(d) (i) Substituting the equation of the tangent y = mx+ b into the parabola x2 = 4ay
x2 = 4a(mx+ b)
x2 − 4amx− 4ab = 0
Since the line y = mx+ b is a tangent to the parabola then ∆ = 0
11
∆ = 16a2m2 − 4(1)(−4ab)
= 16a(am2 + b)
∴ am2 + b = 0
(ii) For the y-intercept to lie on the directrix then b = −a
am2 − a = 0
a(m2 − 1) = 0
m = ±1
∴ Equations of possible tangents are y = ±x− a
12
Question 13
(a) (i)
|p+ q| = |p|+ |q|
(p+ q)2 = (|p|+ |q|)2
p2 + 2pq + q2 = p2 + 2 |p| |q|+ q2
|pq| = pq
This means that pq ≥ 0
(ii) If we let p = x− a and q = b− x then
|x− a|+ |x− b| = |a− b|
⇒ |p|+ |−q| = |p+ q|
|p|+ |q| = |p+ q|
From part (i), this implies that pq ≥ 0 hence
(x− a)(b− x) ≥ 0
(x− a)(x− b) ≤ 0
∴ a ≤ x ≤ b
(b) (i) The angle that the line joining (α, 0) and T (p, q) makes with the positive x-axis is π3.
So the gradient of that line is tan π3
=√
3
The equation of this line is
y − q =√
3(x− p) sub (α, 0)
−1 =√
3(α− p)
α = p− q√3
Similarly, the line joining (β, 0) and T (p, q) makes with the positive x-axis is 2π3
hence β = p+q√3
13
(ii) Noting that
α + β = 2p
αβ =
(p+
q√3
)(p− q√
3
)
= p2 − q2
3
Since the quadratic polynomial is monic then P (x) = x2 − 2px+
(p2 − q2
3
)
(c) (i)
Since ∆ABZ is equilateral then AZ = BZ (1)
Also AD = BC (opposite sides of parallelogram ABCD)
Since X and Y are the centres of equilateral triangles ∆PAD and ∆QBC respectively then
• The distance from the centres X and Y to the vertices of ∆PAD and ∆QBC respectively are equal
• XA bisects ∠PAD and Y B bisects ∠CBQ
From the first point it can be deduced that AX = BY (2).
From the second point, it can be deduced that
∠DAX = ∠CBY
=π
6
Let ∠ABC = x. Since ABCD is a parallelogram
∠BAD = π − x (co-interior angles AD ‖ BC)
In ∆ZBY
∠ZBY = ∠CBY + ∠ABC + ∠ZBA
=π
6+ x+
π
3
=π
2+ x
14
In ∆XAZ
∠XAZ = 2π − ∠BAD − ∠ZAB − ∠DAX (angles at a point A)
= 2π − (π − x)− π
3− π
6
=π
2+ x
Hence ∠XAZ = ∠ZBY (3).
From the results of (1), (2) and (3)
∆XAZ ≡ ∆Y BZ (SAS)
(ii) Using the result in part (i)
XZ = Y Z (corresponding sides of congruent triangles)
∠Y ZB = ∠XZA (corresponding angles of congruent triangles)
Noting that ∠AZB =π
3and that
∠AZB = ∠AZY + ∠Y ZB
= ∠AZY + ∠XZA
⇒ ∠AZY + ∠XZA =π
3
Hence ∆XY Z is equilateral due to two adjacent sides equal and the included angle being π3
15
Question 14
(a) (i) Using cosine rule on 4AOB
AB2 = OB2 +OA2 − 2(OA)(OB) cos∠AOB
= OB2 +OA2 − 2(OA)(OB) cos(180◦ − θ) but cos(180◦ − θ) = − cos θ
= OB2 +OB2 + 2(OA)(OB) cos θ
(ii) Similarly, using cosine rule
BC2 = OC2 +OB2 − 2(OB)(OC) cos θ
CD2 = OD2 +OC2 + 2(OC)(OD) cos θ
AD2 = OD2 +OA2 − 2(OA)(OD) cos θ
AB2 = OB2 +OB2 + 2(OA)(OB) cos θ from part (i)
But since diagonals AC and BD bisect each other then OA = OC and OB = OD. Hence the equationsbecome
BC2 = OC2 +OB2 − 2(OD)(OC) cos θ (1)
CD2 = OD2 +OC2 + 2(OC)(OD) cos θ (2)
AD2 = OD2 +OA2 − 2(OA)(OB) cos θ (3)
AB2 = OB2 +OB2 + 2(OA)(OB) cos θ (4)
16
Taking (1) + (2) + (3) + (4):
AB2 +BC2 + CD2 + AD2 = 2(OA2 +OB2 +OC2 +OD2)
= 2(2OA2 + 2OB2) but OA =AC
2and OB =
BD
2
= 2
(2AC2
4+
2BD2
4
)
= AC2 +BD2
(b) Since square roots can only apply to non-negative values
1− x2 ≥ 0
(1 + x)(1− x) ≥ 0
−1 ≤ x ≤ 1
Now consider the outer square root
x−√
1− x2 ≥ 0
x ≥√
1− x2 note that this implies x > 0 since√
1− x2 ≥ 0
x2 ≥ 1− x2
2x2 ≥ 1
(x√
2− 1)(x√
2 + 1) ≥ 0
x ≤ − 1√2
or x ≥ 1√2
But −1 ≤ x ≤ 1 and x > 0 so the domain is1√2≤ x ≤ 1
17
From the domain
1
2≤ x2 ≤ 1
−1 ≤ −x2 ≤ −1
2
0 ≤ 1− x2 ≤ 1
2
0 ≤√
1− x2 ≤ 1√2
− 1√2≤ −√
1− x2 ≤ 0
0 ≤ x−√
1− x2 ≤ 1 as
(1√2≤ x ≤ 1
)∴ Range is 0 ≤ f(x) ≤ 1
(c) (i)
tan θ =
√3
21
2
=√
3
θ =π
3
18
(ii)
Area of segment =1
2r2θ
=1
2× 12 × π
3
=π
6units2
Area of triangle =1
2bh
=1
2× 1
2×√
3
2
=
√3
8units2
∴∫ 1
12
√1− x2 dx = Area of segment− Area of triangle
=π
6−√
3
8
=4π − 3
√3
24
(iii) Using trapezoidal rule:
∫ 1
12
√1− x2 dx ≈ 1
8
[√3
2+ 0 + 2×
√7
4
]
=
√3 +√
7
16
19
Using the result from (ii)
4π − 3√
3
24≈√
3 +√
7
16
π ≈ 9√
3 + 3√
7
8
= 2.941 to four significant figures
(iv) Jennifer has the more accurate estimation of π due to the concavity of the function.For larger values of x, the function decreases at a faster rate than for smaller values of x.Therefore the trapezoidal areas compared to the actual area will be closer for 0 ≤ x ≤ 1
2than
for 12≤ x ≤ 1
20
Question 15
(a) (i)
f(−x) = ln(−x+
√x2 + 1
)
= ln
((−x+
√x2 + 1
)× x+
√x2 + 1
x+√x2 + 1
)
= ln
(−x2 + x2 + 1
x+√x2 + 1
)
= ln
(1
x+√x2 + 1
)
= ln(x+√x2 + 1
)−1= − ln
(x+√x2 + 1
)= −f(x)
∴ f(x) is an odd function.
(ii)
f(x) = ln(x+√x2 + 1
)
f ′(x) =
1 +x√x2 + 1
x+√x2 + 1
Stationary points occur when f ′(x) = 0
1 +x√x2 + 1
= 0
√x2 + 1 + x√x2 + 1
= 0
But since y = ln(x+√x2 + 1
)then x+
√x2 + 1 cannot be zero, hence f(x) has no stationary points.
21
(iii) From (ii):
f ′(x) =
1 +x√x2 + 1
x+√x2 + 1
=
√x2 + 1 + x√
x2 + 1(x+√x2 + 1
)=
1√x2 + 1
f ′′(x) = − x
(x2 + 1)32
Possible points of inflexion occur when f ′′(x) = 0 which occurs when x = 0, y = 0. Checking point ofinflexion
x −0.1 0 0.1f ′′(x) −0.0998 0 0.0998
There is a change in concavity hence the point of inflexion is (0, 0).
(iv)
22
(b) Note in each term, the denominator is an arithmetic series, so
1
1 + 2+
1
1 + 2 + 3+ ...+
1
1 + 2 + 3 + ...+ n=
122(1 + 2)
+1
32(1 + 3)
+ ...+1
n2(1 + n)
=n∑k=2
1k2(1 + k)
=n∑k=2
2
k(1 + k)
= 2n∑k=2
1 + k − kk(1 + k)
= 2n∑k=2
(1 + k
k(1 + k)− k
k(1 + k)
)
= 2n∑k=2
(1
k− 1
1 + k
)But note that
2n∑k=2
(1
k− 1
1 + k
)= 2
((1
2− 1
3
)+
(1
3− 1
4
)+
(1
4− 1
5
)+ ...+
(1
n− 1
1 + n
))
= 2
(1
2− 1
1 + n
)
= 1− 2
n+ 1
=n− 1
n+ 1
∴1
1 + 2+
1
1 + 2 + 3+ ...+
1
1 + 2 + 3 + ...+ n=n− 1
n+ 1
23
(c) Consider two cases
Case 1 - Player A wins first
For player A to win two games in a row, the following sequences of wins can be possible
{A,A}, {A,B,A,A}, {A,B,A,B,A,A}, ...
The last two games must be won by player A and any remainder must be alternating wins between playersA and B. The probability of this is given by
P (case 1) = p2 + (pq)p2 + (pq)2p2 + ...
= p2(1 + pq + (pq)2 + ...)
=p2
1− pq
Case 2 - Player B wins first
For player A to win two games in a row, the following sequences of wins can be possible
{B,A,A}, {B,A,B,A,A}, {B,A,B,A,B,A,A}, ...
The last two games must be won by player A and any remainder must be alternating wins between playersA and B. The probability of this is given by
P (case 2) = qp2 + q(pq)p2 + q(pq)2p2 + ...
= qp2(1 + pq + (pq)2 + ...)
=p2q
1− pq
Hence the total probability of player A winning the series is the sum of the two cases, which givesp2(1 + q)
1− pq.
24
Question 16
(a) (i) Let Ak be the amount still owing after k months where the loan grows with interest and theamount owing is reduced M each month
A1 = P (1 + r)−M
A2 = A1(1 + r)−M
= P (1 + r)2 −M(1 + r)−M
A3 = A2(1 + r)−M
= P (1 + r)3 −M(1 + r)2 −M(1 + r)−M
......
An = P (1 + r)n −M(1 + r)n−1 −M(1 + r)n−2 − ...−M(1 + r)−M
= P (1 + r)n −M(1 + (1 + r) + (1 + r)2 + ...+ (1 + r)n−1)
= P (1 + r)n − M((1 + r)n − 1)
r
But An = 0, which is when the loan is completely repaid
M((1 + r)n − 1)
r= P (1 + r)n
M =(1 + r)nrP
(1 + r)n − 1
(ii) The loan is repaid off in n2
months. After m months the amount owing is given by
Am = P (1 + r)m −M(1 + r)m−1 −M(1 + r)n−2 − ...−M(1 + r)−M −B
Am+1 = Am(1 + r)−M
= P (1 + r)m+1 −M(1 + r)m −M(1 + r)m−1 − ...−M(1 + r)−M −B(1 + r)
25
Am+2 = Am+1(1 + r)−M
= P (1 + r)m+2 −M(1 + r)m+1 −M(1 + r)m − ...−M(1 + r)−M −B(1 + r)2
......
An2
= P (1 + r)n2 −M(1 + r)
n2−1 −M(1 + r)
n2−2 − ...−M(1 + r)−M −B(1 + r)
n2−m
= P (1 + r)n2 − M((1 + r)
n2 − 1)
r−B(1 + r)
n2−m
But An2
= 0 where the loan is completely repaid
B(1 + r)n2−m = P (1 + r)
n2 − M((1 + r)
n2 − 1)
r
= P (1 + r)n2 − P (1 + r)n((1 + r)
n2 − 1)
(1 + r)n − 1using part (i)
B
P(1 + r)−m = 1− (1 + r)
n2 ((1 + r)
n2 − 1)
((1 + r)n2 − 1)((1 + r)
n2 + 1)
= 1− (1 + r)n2
(1 + r)n2 + 1
=(1 + r)
n2 + 1− (1 + r)
n2
(1 + r)n2 + 1
=1
(1 + r)n2 + 1
(1 + r)m =B
P
((1 + r)
n2 + 1
)
m = log1+m
(B
P
((1 + r)
n2 + 1
))
26
(b)
y = ln(f(x))
dy
dx=f ′(x)
f(x)
d2y
dx2=f(x)f ′′(x)− [f ′(x)]2
[f(x)]2
The local maximum occurs whendy
dx= 0 and
d2y
dx2< 0. This occurs at x = α so
f ′(α)
f(α)= 0
f ′(α) = 0
Also
f(α)f ′′(α)− [f ′(α)]2
[f(α)]2< 0 but f ′(α) = 0
f(α)f ′′(α)
[f(α)]2< 0
f ′′(α)
f(α)< 0
But f(α) > 0 for y = ln(f(x)) then f ′′(α) < 0.
Since f ′(x) = 0 and f ′′(x) < 0 at x = α then f(x) also has a local maximum at x = α.
27
(c) (i)
By trigonometry on the right triangle
cos θ =x1 − x0
d
x1 = x0 + d cos θ
sin θ =x21 − x20
d
=(x0 + d cos θ)2 − x20
d
=2x0d cos θ + d2 cos2 θ
d
= d cos2 θ + 2x0 cos θ
x0 =sin θ − d cos2 θ
2 cos θ
28
By trigonometry on the left triangle
sin θ =x0 − x2
d
x2 = x0 − d sin θ
cos θ =x22 − x20
d
=(x0 − d sin θ)2 − x20
d
=−2x0d sin θ + d2 sin2 θ
d
= d sin2 θ − 2x0 sin θ
x0 =d sin2 θ − cos θ
2 sin θ
Equating the x0
sin θ − d cos2 θ
2 cos θ=d sin2 θ − cos θ
2 sin θ
sin2 θ − d sin θ cos2 θ = d sin2 θ cos θ − cos2 θ
d sin θ cos2 θ + d sin2 θ cos θ = sin2 θ + cos2 θ
d sin θ cos θ(sin θ + cos θ) = 1
d =1
sin θ cos θ(sin θ + cos θ)
29
(ii)
Note that to minimise the area, d needs to be minimised which is equivalent to maximising the de-nominator sin θ cos θ(sin θ + cos θ)
Let y = ln (sin θ cos θ(sin θ + cos θ))
= ln(sin θ) + ln(cos θ) + ln(sin θ + cos θ)
dy
dx=
cos θ
sin θ− sin θ
cos θ+
cos θ − sin θ
sin θ + cos θ
A local maximum occurs whendy
dx= 0
cos2 θ(sin θ + cos θ)− sin2 θ(sin θ + cos θ) + sin θ cos θ(cos θ − sin θ)
sin θ cos θ(sin θ + cos θ)= 0
(sin θ + cos θ)(cos2 θ − sin2 θ) + sin θ cos θ(cos θ − sin θ) = 0
(sin θ + cos θ)2(cos θ − sin θ) + sin θ cos θ(cos θ − sin θ) = 0
(cos θ − sin θ)((sin θ + cos θ)2 + sin θ cos θ) = 0
But (sin θ + cos θ)2 + sin θ cos θ 6= 09 because sin θ > 0 and cos θ > 0 for 0 < θ <π
2. Hence
sin θ = cos θ
tan θ = 1
θ =π
4for 0 < θ <
π
2
Since there are no other stationary points then the local maximum of y is also the absolute maximum.From part (b) this means that when θ = π
4, the expression sin θ cos θ(sin θ + cos θ) is maximised so d is
minimised and hence the area is minimised.
The minimum value of d is
d =1
sin(π4
)cos(π4
) (sin(π4
)+ cos
(π4
))=√
2
Hence the minimum area is d2 which is 2 square units.
30
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 1 –
2015 Bored Of Studies
Mathematics Advanced Trial Examination
Marking Guidelines
Section I
Multiple-Choice Answer Key
Question Answer
1 A
2 C
3 D
4 A
5 B
6 B
7 C
8 B
9 D
10 A
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 2 –
Section II
Question 11 (a)
Criteria Marks
All correct answers. 2
One correct answer. 1
Question 11 (b)
Criteria Marks
Correct solution. 1
Identifies that sin 90 cos .
Question 11 (c) (i)
Criteria Marks
Correct solution. 1
Question 11 (c) (ii)
Criteria Marks
Correct solution. 1
Question 11 (e)
Criteria Marks
Finds both locus equations. 3
Finds one of the locus equations. 2
Uses the perpendicular distance formula on both lines. 1
Question 11 (c)
Criteria Marks
Finds all solutions. 3
Finds part of the solution set. 2
Identifies that 3 3cos sin cos 1 sin cosi sn x x x x x x . 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 3 –
Question 11 (f)
Criteria Marks
Correct simplification and answer. 3
Makes substantial progress. 2
Identifies that 2
21 0
x x
y y , or equivalent merit. 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 4 –
Question 12 (a)
Criteria Marks
Correct solution. 2
Attempts to use change of base, or equivalent merit. 1
Question 12 (b)
Criteria Marks
Finds the correct coordinates of B. 4
Makes substantial progress. 3
Makes some progress using distance formula or gradient formula to
obtain simultaneous equations, or equivalent merit. 2
Attempts to use properties of OBA , or equivalent merit. 1
Question 12 (c) (i)
Criteria Marks
Correct sketch.
Partially correct sketch. 1
Question 12 (c) (ii)
Criteria Marks
Finds correct volume. 3
Makes substantial progress. 2
Uses an appropriate integral of 2tan2
x
. 1
Question 12 (d) (i)
Criteria Marks
Correct solution. 2
Uses discriminant, equates gradients, or equivalent merit. 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 5 –
Question 12 (d) (ii)
Criteria Marks
Finds equations of all possible tangents. 2
Finds the gradient of possible tangents, or equivalent merit. 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 6 –
Question 13 (a) (i)
Criteria Marks
Correct solution. 2
Makes some progress. 1
Question 13 (a) (ii)
Criteria Marks
Correct solution. 2
Attempts to use part (i), or equivalent merit. 1
Question 13 (b) (i)
Criteria Marks
Correct solution. 3
Shows that 3
qp but does not find the correct value of .
OR
Finds the correct value of and makes substantial progress towards
showing that 3
qp .
2
Finds the equation of the line that connects ,0 to T, or equivalent
merit.
OR
States the correct value of .
1
Question 13 (b) (ii)
Criteria Marks
Correct solution. 2
Attempts to use relationship between roots and coefficients, or
equivalent merit. 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 7 –
Question 13 (c) (i)
Criteria Marks
Correct proof. 4
Makes substantial progress. 3
Shows at least two sides equal, or equivalent merit. 2
Shows at least one side equal, or equivalent merit. 1
Question 13 (c) (ii)
Criteria Marks
Correct deduction. 2
Makes some progress using part (i). 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 8 –
Question 14 (a) (i)
Criteria Marks
Correct solution. 1
Question 14 (a) (ii)
Criteria Marks
Correct proof . 3
Makes substantial progress. 2
Uses the cosine rule to find an expression for AB, or equivalent merit. 1
Question 14 (b)
Criteria Marks
Finds correct domain AND range. 3
Finds correct domain
OR
Finds the correct range
OR
Domain and range are both partially correct
2
Domain is partially correct
OR
Range is partially correct
1
Question 14 (c) (i)
Criteria Marks
Finds correct value of . 1
Question 14 (c) (ii)
Criteria Marks
Correct solution. 3
Makes substantial progress. 2
Finds the area of a relevant segment, or equivalent merit. 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 9 –
Question 14 (c) (iii)
Criteria Marks
Correct approximation for , to four significant figures. 3
Correct approximation for
1
1
2
21 , or equivalent merit.x dx
2
Attempts to use trapezoidal rule to approximate
1
1
2
21 .x dx
1
Question 14 (c) (iv)
Criteria Marks
Correct explanation involving concavity. 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 10 –
Question 15 (a) (i)
Criteria Marks
Correct solution. 3
Makes substantial progress. 2
Attempts to rationalise 2 1x x , or equivalent merit. 1
Question 15 (a) (ii)
Criteria Marks
Correct solution 2
Finds f x , or equivalent merit. 1
Question 15 (a) (iii)
Criteria Marks
Finds the correct point (0,0) 2
Finds f x , or equivalent merit. 1
Question 15 (a) (iv)
Criteria Marks
Correct sketch of the graph. 2
Partially correct sketch of the graph. 1
Question 15 (b)
Criteria Marks
Correct solution. 3
Makes substantial progress. 2
Uses properties of arithmetic series, or equivalent merit 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 11 –
Question 15 (c)
Criteria Marks
Correct solution. 3
Makes substantial progress. 2
Finds the correct probability in part of the series. 1
2015 Bored Of Studies Mathematics Advanced Marking Guidelines
– 12 –
Question 16 (a) (i)
Criteria Marks
Correct solution. 2
Shows correct sequence of repayments, or equivalent merit. 1
Question 16 (a) (ii)
Criteria Marks
Correct solution. 4
Makes substantial progress. 3
Shows correct equation for full repayment in terms of B, or equivalent
merit. 2
Shows correct sequence of payments in terms of B, or equivalent
merit. 1
Question 16 (b)
Criteria Marks
Correct proof. 3
Shows that 0f when ln f is maximised. 2
Obtains correct derivative of ln f x , or equivalent merit. 1
Question 16 (c) (i)
Criteria Marks
Correct solution. 3
Makes substantial progress. 2
Obtains valid relationship between coordinates and . 1
Question 16 (c) (ii)
Criteria Marks
Correct solution. 3
Makes substantial progress. 2
Obtains correct derivative of d or ln d . 1
