2013 Bored of Studies Trial Examinations

Mathematics

SOLUTIONS

Section I

1. B

2. D

3. B

4. B

5. A

6. A

7. B

8. D

9. C

10. C

Working/Justification

Question 1

We can eliminate (A) and (C), since they are not to 4 significant figures.

By direct calculation eπ = 23.14069... ≈ 23.14

Question 2

Sn =n

2(n+ 1) as Sn is an arithmetic series where n is an integer.

By trial and error (and taking positive solutions of n), if

Sn = 21⇒ n = 6

Sn = 55⇒ n = 10

Sn = 91⇒ n = 13

Sn = 155⇒ n =−1 +

√1241

2which is not an integer

Question 3

Solving directly using log laws:

ln

(36− x2

x

)= ln 5

36− x2 = 5x

(x+ 9)(x− 4) = 0

x = 4,−9

BUT from the original equation, we can only take the logarithms of positive values. In other words wehave the restriction 36− x2 > 0 and x > 0. Only x = 4 satisfies this condition.

1

Question 4∫ π

0

sin kx dx = −1

k[cos kx]π0

= −1

k(cos kπ − 1)

We require

−1

k(cos kπ − 1) = 0

cos kπ = 1

This is satified, when k = 0,±2,±4,±6,±8, ... (i.e. the even numbers)

Question 5

Let B, G and R represent blue, green and red respectively. Calculating the relevant probabilities:

P (BB) =4

9× 3

8

P (RR) =3

9× 2

8

P (GG) =2

9× 1

8

P (BG ∪GB) =4

9× 2

8+

2

9× 4

8

P (BR ∪RB) =4

9× 3

8+

3

9× 4

8

Note that

• (A) is true since P (BB) = 2P (RR)

• (B) is false since P (BB) 6= 2P (GG)

• (C) is false since P (BG ∪GB) 6= 2P (RR)

• (D) is false since P (BR ∪RB) 6= 2P (GG)

2

Question 6

dI

dt= 1− t

5

I(t) = t− t2

10+ c

Assuming I(t) ≥ 0, note that

• (A) is true since I(10) = I(0) = c

• (B) is false sincedI

dtactually decreases when t increases

• (C) is false since I(5) =5

2+ c =

5

2+ I(0) > 0

• (D) is false since at t = 5,dI

dt= 0 but

d2I

dt2= −1

5< 0 (i.e. local maximum occurs at t = 5)

Question 7

Since the circle contains dotted lines, then we can eliminate (A) and (C), leaving us with (B) and (D).Since the absolute value graph has solid lines, we can eliminate (D).

Question 8

Note that

• (A) is true as per ratio of intercepts of parallel lines theorem

• (B) is true as per the converse of the ratio of intercepts of parallel lines theorem (or can use similartriangles argument)

• (C) is true since this leads to ∆ABC|||∆ADE with ratio of corresponding sides 1:2

• (D) is not always true as it is possible to construct DE to be twice of BC whilst not being parallelto BC which immediately violates the statement that B and C are midpoints of AD and AE.

3

Question 9

Deceleration impliesdv

dt< 0 so there should be negative slope at first. By changing direction, the velocity

changes sign (thus eliminating (A) and (B)). Accelerating quickly to the original speed and remaining atthat speed implies that the velocity must be relatively stable, which eliminates (D).

Question 10

Note that

• (A) is not always true since f(x) = g(x) + c for some non-zero constant c satisfies f ′(x) = g′(x)

• (B) is not always true since we can have f(x) = g(x)+c for some non-zero constant c at the stationarypoint

• (C) is always true since differentiating both sides of f ′(x) = g′(x) leads to f ′′(x) = g′′(x) for all real x

• (D) is not always true since at the stationary points f ′(x)× g′(x) = 0

4

Section II

Question 11

(a)

|ax− 1| ≤ |bx− 1| square both sides

a2x2 − 2ax+ 1 ≤ b2x2 − 2bx+ 1

(b2 − a2)x2 − 2x(b− a) ≥ 0

x(b− a)((b+ a)x− 2) ≥ 0

∴ x ≤ 0 or x ≥ 2

a+ b

Note that 0 < a < b then b− a > 0 and a+ b > 0

(b)

d

dx

(ln

(√x− 1√x+ 1

))=

d

dx

(ln(√x− 1)− ln(

√x+ 1)

)

=1

2√x(√x− 1)

− 1

2√x(√x+ 1)

=(√x+ 1)− (

√x− 1)

2√x(√x− 1)(

√x+ 1)

×√x√x

=

√x

x(x− 1)

(c)

y

x O – k 2k

– k²

k

3k²

Consider the sketch of y = x2 − k2 for −k ≤ x ≤ 2k

∴ Range is −k2 ≤ f(x) ≤ 3k2

5

(d)(i)

y

x O 2π

2

–2

1

π

3π

2

π

2

(d)(ii)First solve for 2 sin 2x = 1 for 0 ≤ x ≤ 2π or 0 ≤ 2x ≤ 4π

sin 2x =1

2

2x =π

6,5π

6,13π

6,17π

6

x =π

12,5π

12,13π

12,17π

12

Reading from the graph, solutions are: 0 ≤ x ≤ π

12or

5π

12≤ x ≤ 13π

12or

17π

12≤ x ≤ 2π

(e)(i)

y

x O b

3b

b + a k

The general equation is (y − y0)2 = −4a(x − x0)

where (x0, y0) is the vertex and a is the focal length.The y-coordinate of the vertex must also be the y-coordinate of the focus (b, 3b) hence y0 = 3b. Also,the horizontal distance between the focus and thevertex is the focal a then x0 = b + a. So the equa-tion of the parabola is (y− 3b)2 = −4a(x− (b+ a)).

6

(e)(ii) The horizontal distance between the vertex and the directrix is the focal length a.

Hence k = b+ 2a, but b < k ≤ 3b.

Substituting k = b+ 2a we get b < b+ 2a ≤ 3b which gives 0 < a ≤ b

(f)

S =∞∑k=0

sin2k θ Note that 0 < sin θ < 1 since 0 < θ <π

2

= 1 + sin2 θ + sin4 θ + ..... which is a limiting sum

=1

1− sin2 θ

=1

cos2 θ

Similarly,

C =∞∑k=0

cos2k θ Note that 0 < cos θ < 1 since 0 < θ <π

2

= 1 + cos2 θ + cos4 θ + ..... which is a limiting sum

=1

1− cos2 θ

=1

sin2 θ

LHS =1

S+

1

C

= cos2 θ + sin2 θ

= 1

= RHS

7

(g) Let A1 be the area bounded by y =1

x, x = a1, x = b1 and the positive x− axis.

A1 =

∫ b1

a1

1

xdx

= [lnx]b1a1

= ln b1 − ln a1

= ln

(b1a1

)

If we similarly define A2 then we have A2 = ln

(b2a2

).

But a1b2 = a2b1 ⇒ b1a1

=b2a2

.

Hence A2 = ln

(b1a1

)so A1 = A2.

8

Question 12

(a)(i) Let the volume of the solid be V .

y

x O π

6 π

2

π

2

V = π

∫ π6

0

tan2 x dx

= π

∫ π6

0

(sec2 x− 1) dx

= π[tanx− x]π60

= π

(1√3− π

6

)cubic units

(a)(ii) Let the volume of the cylinder which has radius1√3

and heightπ

6be denoted VC . We can see that

VC > V

π ×(

1√3

)2

× π

6> π

(1√3− π

6

)

π

18+π

6>

1√3×√

3√3

π >3√

3

2

Also, we know that V > 0

π

(1√3− π

6

)> 0

π

6<

1√3×√

3√3

π < 2√

3

∴3√

3

2< π < 2

√3

9

(b)

(ax)ln a = (bx)ln b

ln((ax)ln a

)= ln

((bx)ln b

)ln a ln ax = ln b ln bx

ln a(ln a+ lnx) = ln b(ln b+ lnx)

lnx(ln a− ln b) = (ln b)2 − (ln a)2

lnx(ln a− ln b) = (ln b− ln a)(ln b+ ln a) but a 6= b⇒ ln a 6= ln b

− lnx = ln b+ ln a

ln

(1

x

)= ln ab

1

x= ab

∴ x =1

ab

(c)

y =1

f(x)

dy

dx= − f ′(x)

[f(x)]2

d2y

dx2=−f ′′(x)[f(x)]2 + 2f(x)[f ′(x)]2

[f(x)]4

Since f(x) has a minimum stationary point at x = α then f ′(α) = 0 and f ′′(α) > 0, provided f(α) 6= 0.

10

So when x = α for y =1

f(x)then

d2y

dx2=−f ′′(α)[f(α)]2 + 2f(α)[f ′(α)]2

[f(α)]4

= − f ′′(α)

[f(α)]2< 0 since f ′′(α) > 0 and [f(α)]2 > 0

Hence a maximum occurs at x = α for y =1

f(x)

(d)(i)

LHS =cos θ

1 + sin θ+

cos θ

1− sin θ

= cos θ

((1− sin θ) + (1 + sin θ)

(1 + sin θ)(1− sin θ)

)

=2 cos θ

1− sin2 θ

=2 cos θ

cos2 θ

=2

cos θ

= 2 sec θ

= RHS

11

(d)(ii)∫ π4

0

sec θ dθ =1

2

∫ π4

0

(cos θ

1 + sin θ+

cos θ

1− sin θ

)dθ

=1

2[ln(1 + sin θ)− ln(1− sin θ)]

π40

=1

2

[ln

(1 +

1√2

)− ln

(1− 1√

2

)]

=1

2

[ln

(√2 + 1√

2

)− ln

(√2− 1√

2

)]

=1

2

[ln(√

2 + 1)− ln√

2− ln(√

2− 1)

+ ln√

2]

=1

2ln

(√2 + 1√2− 1

×√

2 + 1√2 + 1

)

=1

2ln(√

2 + 1)2

= ln(√

2 + 1)

12

Question 13

(a) In order to win Game A, the person must pull any card other than the target number withoutreplacement, in each turn for all (n− 1) turns.

P (win Game A) =n− 1

n× n− 2

n− 1× n− 3

n− 2× ...× 2

3× 1

2× 1

=1

n

In order to win Game B, the person only needs to select the target number once so

P (win Game B) =1

n

= P (win Game A)

(b)(i)

y = x2

dy

dx= 2x

= 2r at point R

Gradient of PQ

mPQ =p2 − q2

p− q

= p+ q

But PQ is parallel to the tangent at R hence

2r = p+ q

r =p+ q

2

13

(b)(ii) First obtain the equation of PQ.

y − p2 = (p+ q)(x− p)

(p+ q)x− y − pq = 0

Let d be the perpendicular distance from PQ to R.

d =|r(p+ q)− r2 − pq|√

(p+ q)2 + 1but r =

p+ q

2

=| (p+q)

2

2− (p+q)2

4− pq|√

(p+ q)2 + 1

=|(p+ q)2 − 4pq|4√

(p+ q)2 + 1

=|p2 − 2pq + q2|4√

(p+ q)2 + 1

=(q − p)2

4√

(p+ q)2 + 1

We also need the distance PQ.

PQ =√

(q − p)2 + (q2 − p2)2

=√

(q − p)2 + (q − p)2(q + p)2

= (q − p)√

1 + (p+ q)2 noting that p < r < q ⇒ q − p > 0

Let A be the area of the triangle PQR

A =1

2× d× PQ

=1

2× (q − p)2

4√

(p+ q)2 + 1× (q − p)

√1 + (p+ q)2

=1

8(q − p)3 square units

14

(c) Let the two roots be α and β. Since the two roots are real then the disciminant is non-negative.

a2 − 4b ≥ 0

a2 ≥ 4b

a ≥ 2√b noting that a > 0 and b > 0

Since the roots are no more than 1 unit apart then |α− β| ≤ 1.

|α− β|2 ≤ 1

|α2 + β2 − 2αβ| ≤ 1

|(α + β)2 − 4αβ| ≤ 1

|a2 − 4b| ≤ 1

−1 ≤ a2 − 4b ≤ 1

−1 + 4b ≤ a2 ≤ 1 + 4b

Extracting the right hand side of the inequality

a2 ≤ 1 + 4b

a ≤√

1 + 4b noting that a > 0 and b > 0

∴ 2√b ≤ a ≤

√1 + 4b

(d)(i)

A C D

B

X

tanα =BD

AD

1

tanα=AD

BD

Similarly,1

tan β=CD

BD

15

LHS =1

tanα+

1

tan β

=AC + CD

BD

=AC

BDbut AC = BD

= 1

= RHS

(d)(ii) Using the similar approach to part (i) then

1

tan(90◦ − β)+

1

tan β= 1 but tan(90◦ − β) = cot β

1

cot β+ cot β = 1

tan β + cot β = 1

(d)(iii) In part (ii) we showed that if we can construct DX such that DX = BC and DX⊥BC thentan β + cot β = 1. But

tan β +1

tan β=1

tan2 β − tan β + 1 = 0

The disciminant of this quadratic equation in tan β is −3 which is negative so there are no real roots fortan β. This means there cannot exist such a β where the condition in (ii) is satisfied. Hence the initialassumption that we find a point X where we can construct DX such that DX = BC and DX⊥BC isfalse.

16

Question 14

(a) At some particular time say t = t0 we have 2A(t0) = B(t0) and

dB

dt= k

dA

dt

bP0ebt0 = akP0e

at0 but A(t0) = P0eat0 and B(t0) = P0e

bt0

bB(t0) = akA(t0)

2bA(t0) = akA(t0)

k =2b

anoting that A(t) 6= 0

(a)(ii) If b = 2a then k = 4. Given the condition 2A(t0) = B(t0)

2P0eat0 = P0e

bt0

2eat0 = e2at0

eat0 = 2

t0 =ln 2

a

(b)(i)

f(x) =x

x2 + 1

f ′(x) =(x2 + 1)− 2x2

(x2 + 1)2

=1− x2

(x2 + 1)2

Stationary points occur when f ′(x) = 0

17

1− x2

(x2 + 1)2= 0

x = 1 noting that x ≥ 0

f(1) =1

2

Using the first derivative test around the neighbourhood of the stationary point

x 0.9 1.0 1.1f ′(x) 0.06 0 -0.04

∴ Maximum turning point at(1, 1

2

)

(b)(ii)

f(x) =x

x2 + 1×

1x2

1x2

=1x

1 + 1x2

as x→ 0 then f(x)→ 0 since1

x→ 0 and

1

x2→ 0

(b)(iii)

y

x O 1

½

18

(b)(iv) y = mx and y = f(x) enclose a region if they intersect more than once (assuming continuity ofthe curves across the region). This means we require multiple solutions to the equation:

mx =x

x2 + 1

x(mx2 − (1−m)) = 0 ∗

The quadratic factor needs to have real roots for the original equation to have multiple solutions. So werequire the discriminant to be non-negative.

02 + 4m(1−m) ≥ 0

m(m− 1) ≤ 0

0 ≤ m ≤ 1

BUT consider the boundary cases, when m = 0 or m = 1 we only get the solution x = 0 in ∗. This also isconsistent graphically, as m = 0 for y = mx gives a flat horizontal line which will never enclose a boundedarea with the curve and m = 1 is actually where the line y = mx is tangent to the curve. Hence thecondition is actually 0 < m < 1.

(b)(v) Let the area be A. First find the points of intersection by solving ∗.

x(mx2 − (1−m)) = 0

x = 0,

√1−mm

noting the given domain x ≥ 0

A =

∫ √ 1−mm

0

(x

x2 + 1−mx

)dx

=1

2

[ln(x2 + 1)−mx2

]√ 1−mm

0

=

ln

(1

m

)− 1 +m

2

=m− 1− lnm

2square units

19

(b)(vi) Note that

A > 0

m− 1− lnm

2> 0

m− 1 > lnm

em−1 > m

Given m is some variable between 0 and 1 we can say that ex−1 > x where some variable x is used in placeof m.

(c) Let I =

∫ a

−af(x) dx and IT be the trapezoidal rule approximation to I.

IT =a

2(f(a) + f(−a) + 2f(0))

=a

2(g(a) + k + g(−a) + k + 2g(0) + 2k) but since g(x) is an odd function g(−a) = −g(a)

= a(g(0) + 2k) but g(−0) = −g(0)⇒ g(0) = 0

= 2ak

But

I =

∫ a

−a(g(x) + k) dx

=

∫ a

−ag(x) dx+ k

∫ a

−adx but since g(x) is an odd function

∫ a

−ag(x) dx = 0

= k[x]a−a

= 2ak

= IT

∴ the trapezoial rule approximation gives the exact value of the integral

20

Question 15

(a)(i) We know that ∠ABC is common.

BQ2 = BP ×BC but AB = BQ

AB2 = BP ×BC

AB

BC=BC

BP

∴ ∆APB|||∆ABC (two sides in proportion and included angle equal)

(a)(ii) Let ∠BAP = x and ∠PAQ = y

Since AB = BQ then ∆BAQ is isosceles hence

∠BQA = ∠BAQ

= x+ y (equal angles opposite equal sides)

∠BPA = ∠PAQ+ ∠AQP

= x+ 2y (exterior angle equals sum of opposite interior angles of a triangle)

Since ∠ABC is common and ∆APB|||∆ABC, then we have either ∠BAP = ∠BAC or ∠BAP = ∠ACBfor corresponding angles to be equal. However, ∠BAC is clearly greater than ∠BAP (as y > 0) so therefore

∠BAP = ∠ACB

= x

∠QAC + ∠ACQ = ∠AQP (exterior angle equals sum of opposite interior angles of a triangle)

∠QAC = ∠AQP − ∠ACQ

= x+ y − x

= y

= ∠PAQ

∴ AQ bisects ∠PAC

21

(b)(i)

x =1

t+ 1

x =

∫1

t+ 1dt

= ln(1 + t) + c

When t = 0 then x = 0⇒ c = 0

x = ln(1 + t)

The distance travelled from time zero to time t = T where the displacement is x = d is

d =

∫ T

0

ln(1 + t) dt

(b)(ii) Consider y = ex− 1. Let A1 be the area bounded by the curve, the y-axis and the line y = T . LetA2 be the area bounded by the curve y = ex − 1, the x-axis and the line x = ln(T + 1).

y

x O

A1

A2

T

ln (T + 1)

y = ex − 1

ex = 1 + y

x = ln(1 + y)

A1 =

∫ T

0

ln(1 + y) dy

But we can also evaluate A1 by considering A2.

A2 =

∫ ln(T+1)

0

(ex − 1) dx

= [ex − x]ln(T+1)0

= eln(T+1) − ln(T + 1)− 1

= T − ln(T + 1)

22

A1 + A2 = Area of rectangle

= T ln(T + 1)

⇒ A1 = T ln(T + 1)− T + ln(T + 1)

= (T + 1) ln(T + 1)− T

∴ d =

∫ T

0

ln(1 + t) dt

= (T + 1) ln(T + 1)− T

(c)(i) and (ii)

y

x O A’(– r, 0) A (r, 0)

P (rcos α, rsin α)

(0, –r)

(0, r)

mAP =r sinα− 0

r cosα− r

=sinα

cosα− 1

mA′P =r sinα− 0

r cosα + r

=sinα

cosα + 1

mAP ×mA′P =sinα

cosα− 1× sinα

cosα + 1

=sin2 α

cos2 α− 1

= −sin2 α

sin2 α

= −1

∴ AP⊥A′P

23

(d)(i) From part (c)(ii), ∠OAR = 90◦

cos θ =OA

ORbut OR = 2a

OA = 2a cos θ

(d)(ii) The x-value of P is equal to the length of OD. The y-value of P is equal to the length of AC.Note that BD = OR = 2a and ∠ROB = ∠OAC = ∠OBD = θ (alternate angles as OR||AC||DB). In∆OCA:

cos θ =AC

OA

AC = OA cos θ

y = 2a cos2 θ from part (i)

In ∆BDO:

tan θ =OD

BD

OD = BD tan θ

x = 2a tan θ

(d)(iii)

y = 2a cos2 θ

=2a

sec2 θ

=2a

1 + tan2 θbut x = 2a tan θ

=2a

1 + x2

4a2

=8a3

4a2 + x2

24

Question 16

(a)(i)

s

s h

r

First note that the circumference of the base of the cone equals the arc length of the sector.

2πr = sθ

r =sθ

2π

Also, by Pythagoras’ theorem:

r2 + h2 = s2

h =√s2 − r2

=

√s2 − s2θ2

4π2

=s

2π

√4π2 − θ2

Let V be volume of the cone.

V =πr2h

3

=π

3× s2θ2

4π2× s

2π

√4π2 − θ2

=s3θ2

24π2

√4π2 − θ2

25

dV

dθ=

s3

24π2

(2θ√

4π2 − θ2 − θ3√4π2 − θ2

)

=s3

24π2

(2θ(4π2 − θ2)− θ3√

4π2 − θ2

)

=s3

24π2

(8θπ2 − 3θ3√

4π2 − θ2

)

Stationary points occur whendV

dθ= 0

s3

24π2

(8θπ2 − 3θ3√

4π2 − θ2

)= 0

θ(8π2 − 3θ2) = 0

θ = 0 or θ2 =8π2

3

but clearly θ = 0 does not maximise the volume so consider

θ2 =8π2

3

θ =2√

2π√3×√

3√3

as 0 < θ < 2π

=2π√

6

3

Since we are given thatd2V

dθ2< 0 at this value, and there are no other stationary points in the domain,

then the volume is maximised at this value of θ.

26

(a)(ii) We know that A =1

2s2θ and B = πr2 but volume is maximised at θ =

2π√

6

3and also r =

sθ

2π.

3B2 = 3π2

(s

2π× 2π

√6

3

)4

=4π2s4

3

2A2 =1

4s4

(2π√

6

3

)2

=4π2s4

3

∴ 3B2 = 2A2

(b)(i) In ∆ABC:

A

B C

By the sine rule

sin∠ABCAC

=sin∠BAC

BC

∴ sin∠ABC =AC sin∠CAB

BC

(b)(ii) Similarly for ∆PQR

sin∠PQR =PR sin∠RPQ

QR

⇒ sin∠ABCsin∠PQR

=AC ×QR× sin∠CABPR×BC × sin∠RPQ

= 1

since AC = PR, BC = QR and ∠CAB = ∠RPQ⇒ sin∠CAB = sin∠RPQ

27

∴ sin∠ABC = sin∠PQR

⇒ ∠ABC = ∠PQR or ∠ABC + ∠PQR = 180◦

since sin(180◦ − x) = sinx

(c)(i) Let Aj be the value of the j-th monthly deposit n months from present.

A1 = MRn

A2 = MRn−1

A3 = MRn−2

......

......

An = MR

∴ P = A1 + A2 + A3 + ...+ An

= MR(1 +R +R2 + ...+Rn−1) which is a geometric series

= MR

(Rn − 1

R− 1

)but S =

Rn − 1

R− 1

= MRS

(c)(ii) Let Bj be the amount remaining in the account j months after retirement.

B1 = RP − kM

B2 = RB1 − kM

= R2P − kMR− kM

B3 = RB2 − kM

= R3P − kMR2 − kMR− kM......

......

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Bm = RmP − kMRm−1 − kMRm−2 − ...− kMR− kM

Bm = RmP − kM(1 +R +R2 + ...+Rm−1)

= RmP − kM(Rm − 1)

R− 1

We expect Bm = 0 where the account is exhausted

RmP − kM(Rm − 1)

R− 1= 0

RmP =kM(Rm − 1)

R− 1

RmP (R− 1) = kMRm − kM

Rm(kM − P (R− 1)) = kM

Rm =kM

(kM − P (R− 1))

ln(Rm) = ln

(kM

(kM − P (R− 1))

)but P = MRS

m lnR = ln

(kM

(kM −MRS(R− 1))

)but S =

Rn − 1

R− 1

m =1

lnR× ln

(k

k −R(Rn − 1)

)

(c)(iii) When k ≤ R(Rn−1) then m is undefined which means there does not exist an m such that accountgets reduced to zero. In other words, the account actually grows over time where the rate of interest gainedexceeds the rate of monthly withdrawals (as an aside, one can formally show that Bj > Bj−1).

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