Active Maths 4 Book 1 Additional Questions and...

Active Maths 4 Book 1:Additional Questions

and Solutions

2 A C T I V E M AT H S 4 B O O K 1 : A D D I T I O N A L Q U E S T I O N S A N D S O LU T I O N S

Question 1

(a) Find the vertex of y = f(x) where f(x) = 3x2 – 6x – 7, stating whether it is a minimum or a maximum.

(b) Find where the graph of y = f(x) crosses the axes and hence sketch the graph.

Question 2

The line l has equation y = 2x – 3 and the curve C has equation y = (x + 2)(x – 4).(a) Sketch the line l and the curve C on the same axes, showing the coordinates of the x- and y-intercepts.

(b) Show that the x-coordinates of the points of intersection of l and C satisfy the equation x2 – 4x – 5 = 0.

(c) Hence, or otherwise, find the points of intersection of l and C.

Question 3

The height of a roller coaster h metres above ground level is given by the equation

h = 1 ____ 100 ( x4 _____ 1200 – 11 ___ 90 x3 + 4x2 )

where x is the distance in metres along the track. The track is 97.5 m long and goes underground at some points. A ride photo of the roller coaster can only be taken when h is stationary.

(a) Find dh ___ dx

(b) Find d2h ____

dx2

The ride starts at x = 0.

(c) Verify that this belongs to a stationary point.

(d) At which other distances along the track can ride photos be taken?

It is easier to take a photo when the ride is at a minimum point because the track is more accessible.

(e) At what distance along the track should the ride photo be taken?

Question 4

(a) Rewrite the following equation in the form f(x) = 0, where f(x) is of the form f(x) = ax3 + bx2 + cx + d:

(x − 1)(x2 + x + 1) = 2x2 − 17

(b) Use the Factor Theorem to show that (x + 2) is a factor of f(x).

(c) Express f(x) in the form f(x) = (x + 2)(x2 + px + q), where p and q are constants.

(d) Show that f(x) = 0 has only one real root.

(e) Consider the equation y = f(x).

(i) Find dy

___ dx

.

(ii) Find the tangent to the curve y = f(x) at x = 1.

(iii) Show that f(x) is increasing when 3x2 − 4x > 0.

(iv) Solve the quadratic inequality 3x2 − 4x > 0.

3A C T I V E M AT H S 4 B O O K 1 : A D D I T I O N A L Q U E S T I O N S A N D S O LU T I O N S

Question 5

(a) A curve, C, has equation y = −x2 + x + 6. Show that C has a stationary point at (0.5, 6.25).

(b) The diagram shows the graphs of C and the line L, given by the equation y = x + 2.

(i) Determine the coordinates of the points of intersection of C and L.

(ii) Hence find the area of the shaded region

Ly

C

x

bound by C and L.

Question 6

The distance travelled by an object dropped from a height close to the earth’s surface is closely approximated by the function

d(t) = 1 __ 2 gt2

where t is the time in seconds that have elapsed since the object was dropped, and g is the constant acceleration of the object due to gravity. (a) A stone is dropped from a height of 50 metres.

(i) Find in terms of g the distance travelled by the stone in 3 seconds.

(ii) How long does it take the stone to fall a distance of 2g metres?

(b) One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard.

(i) If d is the depth of the well and t1 is the time it takes the stone to hit the water, write d in terms of g and t1.

(ii) Given that the speed of sound in air is approximately 330 m/s and t2 is the time taken for the sound to travel back up the well, write d in terms of t2.

(iii) T, the total time elapsed between dropping the stone and hearing the splash, can be written as

T = pd + q √__

d

Find the value of p and the value of q.

(iv) How deep is the well, if the total time elapsed is 4 seconds? Let g = 9.8 m/s2.

Question 7

The McCarthy family bought a house from the Moran family for €389,400. In lieu of a 20% down payment, the Moran’s accepted a 10% down payment at the time of the sale and a promissory note from the McCarthy’s for the remaining 10%, due in 4 years. The McCarthy’s also agreed to make monthly interest payments to the Moran’s at 11% per annum simple interest until the note expires.

The McCarthy’s obtained a 20 year mortgage, for the remaining 80% of the purchase price, from their bank at an APR of 4.55%. The bank in turn paid the sellers the remaining 80% of the purchase price, less a sales commission of 3% of the sales price paid to the sellers’ and buyers’ estate agents.

Note: A promissory note is a financial instrument, in which one party promises in writing to pay a determinate sum of money to the other (the payee), either at a fixed or determinable future time or on demand of the payee, under specific terms.


(a) Find the McCarthy’s down payment and the amount they borrowed from their bank.

(b) What amount did the McCarthy’s borrow from the Moran family?

(c) Find the McCarthy’s monthly interest-only payment to the Moran’s.

(d) Find the Moran family’s total income from all aspects of the down payment.

(e) How much did the Moran’s receive from the McCarthy’s bank?

(f) Find the Moran’s total income from all aspects of the sale.

(g) Calculate to the nearest euro the McCarthy’s monthly repayments.

Question 8

A is the closed interval [−5, 5]

The function f is defined on A by:

f: A → R: x � 1 __ x2

A graph of the function f is shown.

00

1–1

–1

1

2

3

4

5

6

7

8

–2

A B

DC

–3–4–5–6–7 2 3 4 5 6 7

(i) Find the co-ordinates of A, B, C and D

(ii) State whether f is injective. Give a reason for your answer.

(iii) State whether f is surjective. Give a reason for your answer.

(iv) For what positive value of x ∈ A is f(x) a minimum?

(v) Find the minimum value of f(x).

(vi) On the plane above, graph the function

g(x) = 1 ______ (x – 2)2

by transforming the graph of f(x).

(vii) Use long division and factoring to show that the function

h(x) = 2x2 + 4x + 5 ____________ x2 + 2x + 1


can be written as

h(x) = 2 + 3 _______ (x + 1)2

Then, on the plane above, graph h by transforming the graph of f(x).

(viii) Use integration to find the area of the region bounded by the graphs of f(x), g(x), the x and y axes and the line x = 2.

Question 9

A bungee jumper plummets from a high bridge to the river below and then bounces back over and over again. At time t seconds after her jump, her height H (in metres) above the river is given by

H(t) = 100 + 75ept ( cos π __ 4 t ) where p ∈ �.

(i) If at t = 12s, the jumper’s height above the river is 100 – 75 ____ 5 √

___ e3 m, find the value of p.

(ii) Find her height, to two decimal places, at the times indicated in the table.

t H(t)

0

1

2

4

6

8

12

(iii) If the jumper’s height can also be represented by the function

L(t) = 100 + 75ept sin ( π __ 4 t + q ) where q ∈ �, find the smallest positive value of q.

(iv) Find H’(t), the derivative of H(t).

(v) Using the fact that a sin q + b cos q = √_______

a2 + b2 sin (q + a), where a = sin–1 ( b ________ √

_______ a2 + b2 )

write H’(t) in the form

a √__

b ept sin ( π __ 4 t + a ) , where a, b ∈ � and a ∈ �.


(vi) The graphs of H(t) and H’(t) in the domain 0 ≤ t ≤ 120 are shown below.

–2010 20 30 40 50 60 70 80 90 100 110 120 t

00

20C

40

60

80

100

120

140

160

180H(t)

H'(t)

200

–40

–60

(a) From the graph estimate the shortest distance between the jumper and the river.

(b) Find the co-ordinates of the point C and hence find to two decimal places the shortest distance between the jumper and the river.

(c) Find the percentage error in your estimate from part (a).

(d) Use the graph to estimate lim (H(t)) as t → ∞


Solutions

Question 1

f(x) = 3x2 − 6x − 7(a) Function is a quadratic and the coefficient of x2 is positive, therefore vertex will be a minimum.

To find the vertex, we need to complete the square

f(x) = 3(x2 − 2x) − 7

= 3(x2 − 2x + 1) − 3 − 7

= 3(x − 1)2 − 10

To find x-value of vertex, put x − 1 = 0

⇒ x = 1

∴ Vertex is (1, −10) … Minimum

(b) f(x) crosses y-axis at x = 0 which gives y = −7

(0, −7)

f(x) crosses x-axis at y = 0

∴ 3x2 − 6x − 7 = 0

Using the completed square form: 3(x − 1)2 − 10 = 0

(x − 1)2 = 10 ___ 3

x − 1 = ± √___

10 ___ 3

x = 1 ± √___

10 ___ 3

( 1 + √___

10 ___ 3 , 0 ) , ( 1 − √___

10 ___ 3 , 0 )

–2.5 –2 –1.5 –1

1 – 10/3 1 + 10/3

–0.5–2

2468

101214

0

–4–6–8

–10

0.5 1 1.5

(1, –10)

2 2.5 3 3.5 4 4.5

x

y

0


Question 2

(a)

–8 –6 –4 –2 2 4 6(4, 0)

8 1000

–5

5

10

15

20

25

30

35

x

y

–10 (0, –8)

(0, –3)

(1.5, 0)(–2, 0)

(b) At points of intersection:

2x − 3 = (x + 2)(x − 4)

2x − 3 = x2 − 2x − 8

0 = x2 − 4x − 5 … Shown

(c) x2 − 4x − 5 = 0

(x − 5)(x + 1) = 0

x = 5, x = −1

x = 5 ⇒ y = 2(5) − 3

y = 7 … (5, 7)

x = −1 ⇒ y = 2(−1) − 3

y = −5 … (−1, −5)

Question 3

h = 1 ____ 100 ( x4 _____ 1200 − 11 ___ 90 x3 + 4x2 )

(a) dh ___ dx

= 1 ____ 100 ( x3 ____ 300 − 11 ___ 30 x2 + 8x )

(b) d2h ____

dx2 = 1 ____ 100 ( x2

____ 100 − 11 ___ 15 x + 8 ) (c) For x = 0 to belong to a stationary point, dh ___

dx must = 0 when x = 0.

∴ Substitute x = 0 into dh ___ dx

:

1 ____ 100 ( 03 ____ 300 − 11 ___ 30 (0)2 + 8(0) ) = 0,

so x = 0 is at a stationary point.

(d) Put dh ___ dx

= 0 & solve for x:

1 ____ 100 ( x3 ____ 300 − 11 ___ 30 x2 + 8x ) = 0


⇒ x3 ____ 300 − 11 ___ 30 x2 + 8x = 0

⇒ x3 − 110x2 + 2400x = 0

x(x2 − 110x + 2400) = 0

x(x − 80)(x − 30) = 0

x = 0, x = 80, x = 30

So ride photos can be taken at x = 30 m and at x = 80 m.

(e) At x = 30, d2h ____

dx2 = 1 ____ 100 ( (30)2 _____ 100 − 11 ___ 15 (30) + 8 )

= −0.05, (< 0 ⇒ Maximum)

At x = 80, d2h ____

dx2 = 1 ____ 100 ( (80)2 _____ 100 − 11 ___ 15 (80) + 8 )

≈ 0.13 (> 0 ⇒ Minimum)

∴ The ride photo should be taken 80 m along the track.

Question 4

(a) (x − 1)(x2 + x + 1) = 2x2 − 17

x3 + x2 + x − x2 − x − 1 = 2x2 − 17

⇒ x3 − 2x2 + 16 = 0

(b) If (x + 2) is a factor then, by the Factor Theorem, f(−2) will be equal to zero. Using the equation you found in part (a):

f(−2) = (−2)3 −2(−2)2 + 16

= −8 − 2(4) + 16 = 0

⇒ (x + 2) is a factor of f(x).

(c) Factorise the cubic by equating coefficients:

x3 − 2x2 + 16 = (x + 2)(px2 + qx + r)

= px3 + (q + 2p)x2 + (r + 2q)x + 2r

Equating coefficients: p = 1, 2r = 16 ⇒ r = 8

q + 2p = −2 ⇒ q + 2 = −2 ⇒ q = −4.

So x3 − 2x2 + 16 = (x + 2)(x2 − 4x + 8)

(d) f(x) = 0 has at least one real root, x = −2, as shown in part (b).

As shown in part (c), f(x) = (x + 2)(x2 − 4x + 8), so if f(x) = 0 then (x + 2)(x2 − 4x + 8) = 0, and so any other roots of f(x) = 0 will be the roots of (x2 − 4x + 8) = 0.

The discriminant of (x2 − 4x + 8) is:

(−4)2 − (4 × 1 × 8) = −16

−16 < 0, so (x2 − 4x + 8) = 0 has no real roots

Therefore f(x) = (x + 2)(x2 − 4x + 8) only has one real root, x = −2

(e) (i) dy

___ dx

= f ́ (x) = 3x2 − 4x.

(ii) The gradient of the tangent is given by:

m = f ́ (1) = 3(1)2 − 4(1) = −1

When x = 1, y = (1)3 − 2(1)2 + 16 = 15


Now use y − y1 = m(x − x1) to find the equation of the tangent:

y − 15 = −(x − 1)

⇒ y = −x + 16

(iii) f(x) is increasing where f ́ (x) > 0,

i.e. when 3x2 − 4x > 0.

(iv) The graph of y = 3x2 − 4x is drawn below.

0 4–3

3x2 − 4x = 0 ⇒ x(3x − 4) = 0, so the graph crosses the x-axis at x = 0 and x = 4 __ 3 .

The graph is positive (i.e. above the x-axis) when x < 0 and when x > 4 __ 3 .

Therefore, 3x2 − 4x > 0 when x < 0 and when x > 4 __ 3 .

Question 5

(a) dy

___ dx

= –2x + 1

Stationary points occur when dy

___ dx

= 0,

⇒ −2x + 1 = 0

⇒ x = 0.5

x = 0.5 ⇒ y = −(0.5)2 + 0.5 + 6

= −0.25 + 0.5 + 6 = 6.25

So there is a stationary point at (0.5, 6.25).

(b) (i) Finding the points of intersection:

−x2 + x + 6 = x + 2

⇒ x2 = 4

⇒ x = ± 2

Use the equation for L to find the corresponding y-values:

When x = 2, y = 2 + 2 = 4

When x = −2, y = −2 + 2 = 0

So the coordinates of intersection are (2, 4) and (−2, 0).

(ii)

Ly

C

(2, 4)

(–2, 0) x


The area between L and C is equal to the area under C between x = −2 and x = 2 minus the area under L between x = −2 and x = 2.

The area under C between x = −2 and x = 2 is given by:

∫ 2

–2

(−x2 + x + 6)dx = [ − x3 __ 3 + x

2 __ 2 + 6x ] –2

2

= ( − 23 __ 3 + 2

2 __ 2 + 6(2) ) − ( −

(−2)3 _____ 3 +

(−2)2 _____ 2 + 6(−2) )

= ( − 8 __ 3 + 14 ) − ( 8 __ 3 − 10 ) = 56 ___ 3

The area under L between x = −2 and x = 2 is equal to the area of a triangle of base 4 and height 4 [the vertices of the triangle are (−2, 0), (2, 0) and (2, 4).]

So, the area beneath L is 1 __ 2 × 4 × 4 = 8

The area between C and L is therefore:

56 ___ 3 − 8 = 56 ___ 3 − 24 ___ 3 = 32 ___ 3

Question 6

(a) (i) d(3) = 1 __ 2 g(3)2

= 9g

___ 2 metres.

(ii) 1 __ 2 gt2 = 2g

t2 = 4

t = 2 seconds.

(b) (i) d = 1 __ 2 g t 1 2

(ii) d = 330t2

(iii) t 1 2 = 2d ___ g T = t1 + t2

t1 = √___

2d ___ g T = √___

2d ___ g + d ____ 330

t2 = d ____ 330 T = √__

2 __ g √__

d + 1 ____ 330 d

T = 1 ____ 330 d + √__

2 __ g √__

d p = 1 ____ 330 q = √__

2 __ g

(iv) 1 ____ 330 d + √__

2 __ g √__

d = 4

Let √__

d = x

⇒ d = x2,

1 ____ 330 x2 + √__

2 __ g x − 4 = 0

0.00303x2 + 0.451754x − 4 = 0

x = −0.451754 ± √

___________________________ (0.451754)2 − 4(0.00303)(−4) _________________________________________ 2(0.00303)

x = 8.38 m.

The well has a depth of 8.38 m.


Question 7

(a) Down payment: 389,400 × 10% = €38,940

(b) €38,940

(c) Monthly Interest: 38940 × 11% _____________ 12 = €356.95

(d) Total Income = €38,940 + €38,940 + €356.95 × 48

= €95013.60

(e) 80% of 389,400 = €311,520

3% of 389,400 = € 11,682

Moran's received € 299,838

(f) Total Income: €299,838 + €95,013.60

= €394,851.60

(g) Monthly interest rate

F = P(1 + i)t

104.55 = 100(1 + i)12

(1 + i)12 = 1.0455

i = 12

√_______

1.0455 − 1

= 0.003714819

A = 311520 [ i (1 + i)240 ____________

(1 + i)240 − 1 ]

where i = 0.003714819

A = €1,963.73

Question 8

(i) A (−1, 1) C ( −2, 1 __ 4 ) B (1, 1) D ( 2, 1 __ 4 ) (ii) f is not injective, as f(−1) = 1 = f(1)

(iii) f is not surjective, as there does not exist y ∈ � such that f(0) = y

( iv) 5

(v) f(5) = 1 __ 52 = 1 ___ 25


(vi)

00

1–1

–1

1

2

3

4

5

g(x)

h(x)

6

7

8

–2

A B

DC

–3–4–5–6–7 2 3 4 5 6 7

(vii) 2

x2 + 2x + 1 2x2 + 4x + 5

2x2 + 4x + 2

3

⇒ h(x) = 2 + 3 ___________ x2 + 2x + 1

h(x) = 2 + 3 _______ (x + 1)2

(viii) Area = 2 ∫ 1 2

1 __ x2 dx

= 2 ∫ 1

2

x−2 dx

= 2 [ − 1 __ x ] 1 2 = 2 [ − 1 __ 2 + 1 ] = 1

Question 9

(i) 100 + 75e12p cos (3π) = 100 − 75 e − 3 __ 5

−75e12p = −75 e − 3 __ 5

12p = − 3 __ 5

p = − 1 ___ 20


(ii) t H(t)

0 175.00

1 150.45

2 100.00

4 38.60

6 100.00

8 150.27

12 58.84

(iii) 100 + 75ept sin ( π __ 4 t + q ) = 100 + 75ept cos ( π __ 4 t ) sin ( π __ 4 t + q ) = cos ( π __ 4 t ) ⇒ q = π __ 2

or

sin π __ 4 t cos q + sin q cos π __ 4 t = cos π __ 4 t

⇒ sin q = 1 and cos q = 0

smallest value for which this is true is q = π __ 2

(iv) H‘(t) = 75ept ( − π __ 4 sin ( π __ 4 t ) ) + cos ( π __ 4 t ) 75 pept

= 75 e − t ___ 20 ( − π __ 4 sin ( π __ 4 t ) ) − 15 ___ 4 e − t ___ 20 cos ( π __ 4 t ) = − 15 ___ 4 e − t ___ 20 [ 5π sin ( π __ 4 t ) + cos ( π __ 4 t ) ] (v) H’(t) = − 15 ___ 4 e − t ___ 20 [ √_________

25π2 + 12 sin ( π __ 4 t + a ) ], where a = sin−1 ( 1 _________ √

________ 25π2 + 1

) = 0.063576181

= − 15 ___ 4 e − t ___ 20 √________

25π2 + 1 sin ( π __ 4 t + a )

= − 15 ___ 4 √________

25π2 + 1 e − t ___ 20 sin ( π __ 4 t + a ) (vi) (a) 39 m.

(b) − 15 ___ 4 e − t ___ 20 [ √________

25π2 + 1 ] sin ( π __ 4 t + a ) = 0

e − t ___ 20 sin ( π __ 4 t + a ) = 0

⇒ sin ( π __ 4 t + a ) = 0

for t > 0

π __ 4 t + a = π

π __ 4 t + 0.063576181 = π

t = 3.919s

∴ C = (3.919, 0)

H(3.919) = 100 + 75 e − 3.919 ______ 20 cos ( π __ 4 × 3.919 ) = 38.471 m.


(c) Error = 39 − 38.471

= 0.529

% Error = 0.529 _______ 38.471 × 100

= 1.38% (2 d.p.) (d) lim H(t) = 100

t → ∞

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