Am Mock Solutions

Marking Scheme 1.

A man buys a new car at its pre-launch price. After t months, its value $V is given by V = 85988e − pt , where p is a constant.

(i)

What is the value of the new car at its pre-launch price?

[1]

The value of the car after 1 year is expected to be $74011. Calculate (ii)

the expected value, to the nearest dollar, of the car after 4 years,

(iii) the age of the car, to the nearest month, when its expected value will be less than $30000.

(i)

$85988 – A1

(ii)

85988e −12 p = 74011 - M1 p = 0.0125(3sf ) Expected value = 85988e −48×0.0125 - M1 (using t = 48) =$47191 – A1 (nearest dollar)

(iii)

85988e −0.0125t < 30000 ⎛ 30000 ⎞ −0.0125t < ln ⎜ ⎟ - M1;M1 (V<30000; t > 84.2 ) ⎝ 85988 ⎠ t > 84.2

85 months – A1

1

[3] [3]

2.

The roots of the quadratic equation −3 x 2 + 6 x − 7 = 0 are α and β . Find the quadratic equation [7]

whose roots are α 2 + 3 and β 2 + 3 . −3 x 2 + 6 x − 7 = 0

x2 − 2x +

7 =0 3

α +β =2 αβ =

- A1 (both answers are correct for sum and product of roots)

7 3

α2 +3+ β 2 +3 =α2 + β2 +6 = (α + β ) − 2αβ + 6 - M1 2

=

16 - A1 (sum of roots) 3

(α

2

+ 3)( β 2 + 3)

= (αβ ) + 3 (α 2 + β 2 ) + 9 2

(

)

= (αβ ) + 3 (α + β ) − 2αβ + 9 - M1 2

=

2

112 - A1 (product of roots) 9

New Equation: x 2 −

16 112 x+ = 0 - A1 3 9

9 x 2 − 48 x + 112 = 0 - A1

2

3

(i)

Prove the identity

sin x sin x + ≡ 2 cot x . sec x − 1 sec x + 1

[5]

Find all the angles between 0° and 360° which satisfy the equation (ii)

sin x sin x + = 3. sec x − 1 sec x + 1

(i)

sin x sin x + sec x − 1 sec x + 1

[5]

=

sin x(sec x + 1) + sin x(sec x − 1) -M1 (common denominator) (sec x − 1)(sec x + 1)

=

2 sin x sec x - M1 (simplifying after common denominator) sec 2 x − 1

=

2 tan x - M1;M1 ( sin x sec x = tan x;sec 2 x − 1 = tan 2 x ) 2 tan x

=

2 1 - M1 ( = cot x ) tan x tan x

= 2cot x (shown)

(ii)

sin x sin x + =3 sec x − 1 sec x + 1

2 cot x = 3 3 - M1; M1 (using (i) 2cotx = 3;getting tan x = 2/3) 2 2 tan x = 3

cot x =

basic angle = 33.69° - M1 (correct basic angle) x = 33.7°, 213.7° - A2 (1 mark for each answer)

3

4.

Solve the equation (i)

3log x 2 + log x 18 = 2

[3]

(ii)

log 7 y − 15log y 7 = −2

[5]

(i)

3log x 2 + log x 18 = 2

log x (144 ) = 2 - M1 (apply laws of log) 144 = x 2 - M1 (index form) x = 12 or − 12( NA) - A1

(ii)

log 7 y − 15log y 7 = −2

⎛ 1 log 7 y − 15 ⎜ ⎝ log 7

⎞ ⎟ = −2 - M1 (apply laws of log) y⎠

Let x = log 7 y ⎛1⎞ x − 15 ⎜ ⎟ = −2 - M1 ⎝ x⎠

x 2 + 2 x − 15 = 0 - M1 x = 3 or − 5 3 = log 7 y or − 5 = log 7 y - M1 y = 343 or

1 - A1 75

5

3x 2 + 12 x − 27 3x 2 + 12 x − 27 = x( x − 3)( x + 3) x 3 − 9x A B C = + + x x−3 x+3

M1

3 x 2 + 12 x − 27 = A( x − 3)( x + 3) + B( x)( x + 3) + C ( x)( x − 3) Let x = 0, − 27 = −9 A ⇒ A = 3

4

let x = 3, 36 = 18B ⇒ B = 2

M1

let x = –3, –36 = 18C ⇒ C = −2 3x 2 + 12 x − 27 3 2 2 = + − There fore, 3 x x−3 x+3 x − 9x

A1

d ⎛ 3 x 2 + 12 x − 27 ⎞ d ⎛ 3 2 2 ⎞ ⎜⎜ ⎟⎟ = − ⎜ + ⎟ 3 dx ⎝ x − 9x ⎠ dx ⎝ x x − 3 x + 3 ⎠ ⎛ 3 2 2 + = ⎜⎜ − 2 − 2 (x − 3) (x + 3)2 ⎝ x

⎞ ⎟ ⎟ ⎠

M2

⎛ 3 2 2 + Gradient of the curve when x = 1 is ⎜⎜ − 2 − 2 (1 + 3)2 ⎝ 1 (1 − 3)

6(a)

(b)

f ( x) = ( x − 1)( x − (1 + k ))( x − k 2 ) − 12 = (0 − 1)(0 − (1 + k ))(− k 2 ) k 3 + k − 12 = 0

M2 M1

5

(c)

⎞ 3 ⎟ = −3 . ⎟ 8 ⎠

A1

Let k = e y 2 = ey ln 2 = y ln e y = 0.693

A1

Total

9m

M1

Trial error :

g(2) = k 3 + k 2 − 12 =0 k − 2 is a factor of g By synthetic division g(k) = (k-2)( k 2 + 3k + 6 ) Consider k 2 + 3k + 6 = 0 − 3 ± 9 − 24 k= 2 Since D < 0, there are no other real roots i.e k = 2

M1

M1 M1

A1

5

7(a)

∠ABE = ∠DBE (given, BE ∠bisector) BE common side ∠BED =∠BEA = 90° (∠ in semicircle) ΔABE≡ΔDBE (ASA) ∴AE = DE

(b)

8(ai)

∠ADE = ∠CDE common angle ∠CED = ∠ABD (or ∠DCE = ∠DAB) (ext ∠ of cyclic quad) ΔABD is similar ΔCED AAAtest

∠OSP = 90 0 − θ ∠RST = θ

both M1 M1

M2

AD BD = CD ED

but AE = DE, AD = 2AE 2 AE BD = CD AE Proven AE2 = 12 BD×CD

M1

M1

M1 M1 M1

Total

(bi) M1 M1 A1

1.5 sin θ + 0.8 cos θ = R sin(θ + α )

(bii)

R = 1.5 2 + 0.8 2 = 1.7

M1

0 .8 α = 28.07 0 1 .5 OT = 1.7 sin(θ + 28.10 )

M1

tan α =

(c)

Proven

OS = 1.5 sin θ ST = 0.8 cosθ OT = 1.5 sin θ + 0.8 cosθ (aii)

7

9m

OT is max when sin(θ + 28.10 ) = 1 0 0 < θ < 90 0 (θ + 28.10 ) = 90 0 28.07 0 < θ + 28.07 < 118.07 0 θ = 61.9 0 1.7 sin(θ + 28.10 ) = 1.2 1 .2 sin(θ + 28.10 ) = 1 .7 0 Since 28.07 < θ + 28.07 < 118.07 0 (θ + 28.10 ) = 44.90 0 θ = 16.8 0

A1

Total

6

M1

A1

M1

A1 10m

Evaluate

9

∫

1 0

⎛ ⎜3 x − ⎝

integers.

⎞ ⎟ dx , giving your answer in the form a + b 3 , where a and b are x+3⎠ [5] 1

⎛ 1 ⎞ ∫0 ⎜⎝ 3 x − x + 3 ⎟⎠ dx 1

1

=

⎡ 3 1 ⎢ 3 x 2 ( x + 3) 2 ⎢ 3 − 1 ⎢ ⎣ 2 2

⎤ ⎥ ⎥ ⎥ ⎦0

=

1 ⎤ ⎡ 32 2 − + 2 x 2 x 3 ( ) ⎢ ⎥ ⎣ ⎦0

3

1

( x + 3) 2 3x 2 B1 for , B1 for 3 1 2 2 1

= = =

3 1 ⎤ 1 ⎡ ⎡ 2 2 − + − − + 2(1) 2 1 3 2(0) 2 0 3 ( ) ⎥ ⎢ ( ) 2 ⎤⎥ ⎢ ⎦ ⎣ ⎦ ⎣ [ 2 − 2(2)] − ⎡⎣ −2 3 ⎤⎦

−2 + 2 3

M2

A1

10 A curve has the equation y = 2 cos 2 x − sin 2 x , where 0 ≤ x ≤ π .

d2 y dy (i) Obtain expressions for and . dx dx 2 function.

Express each answer as a single trigonometric [4]

(ii) Find the x-coordinate of the stationary point of the curve, and determine the nature of this stationary point. [4] 10 (i)

dy dx

B1 for 4 cos x ( − sin x ) , B1 for −2cos 2x

= 4 cos x ( − sin x ) − 2 cos 2 x

= −4sin x cos x − 2cos2x = −2sin 2x − 2cos2x = −2 2 sin ( 2 x + 0.7854 ) or equivalent

d2 y = −4cos 2x + 4sin2x dx 2

B1 for −4cos 2x , B1 for 4sin2x

= −4 2 cos ( 2 x + 0.7854 ) or equivalent dy = 0. dx −2sin 2x − 2cos2x = 0 tan2x = −1

(ii) For stationary point,

M1

7

2 x = π − 0.7854, 2π − 0.7854, , 3π − 0.7854 A1 - arrive at both answers x = 1.1781, 2.7489

d2 y When x = 1.1781, = −4 2 cos ( 2(1.1781) + 0.7854 ) = 5.657 > 0 dx 2 The stationary point at x = 1.1781 is a minimum point.

M1 for method of determining nature of stationary pt

d2 y When x = 2.7489, = −4 2 cos ( 2(2.7489) + 0.7854 ) = −5.657 < 0 dx 2 The stationary point at x = 2.7489 is a maximum point.

A1 for determining nature of both stationary pts

11 (a)(i)

(a)(ii)

dy dx

=

x(−1)e− x + e − x (1)

=

(1 − x ) e− x

(shown)

∫ (1 − x ) e dx = xe + c ∫ e − x e d x = xe + c ∫ e dx − ∫ xe dx = xe + c − e − ∫ xe dx = xe + c ∫ xe dx = − e − xe + C −x −x

−x

−x

−x

−x

−x

−x

−x

−x

(a)(iii)

B1 for x(−1)e − x , B1 for e− x (1)

M1

−x

−x

−x

−x

A1

dy >0 dx (1 − x ) e− x > 0

1 − x > 0, since e− x is always positive x<1 B1

11 (b) A circle passes through the points A (0, 8) and B (7, 1). The centre of the circle lies on the 4 line y = x . Find the equation of the circle. [6] 3

8

11 (b) Let the equation by ( x − p ) + ( y − q ) = r 2 ----- (1) 2

Since centre (p, q) lies on the line y =

2

4 x, 3

4 p 3 Subst into (1): q=

M1 2

4p ( x − p ) + ⎛⎜ y − ⎞⎟ = r 2 ----- (2) 3 ⎠ ⎝ Subst (0, 8) into (2): 2

2

4p 2 ( 0 − p ) + ⎛⎜ 8 − ⎞⎟ = r 2 3 ⎠ ⎝ 64 p 16 p 2 p 2 + 64 − + = r2 3 9 2 25 p 64 p − + 64 = r 2 ----- (3) 9 3 Subst (7, 1) into (2):

M1

2

4p ( 7 − p ) + ⎛⎜1 − ⎞⎟ = r 2 3 ⎠ ⎝ 8 p 16 p 2 49 − 14 p + p 2 + 1 − + = r2 3 9 2 25 p 50 p − + 50 = r 2 ----- (4) 9 3 (3) – (4): 14 p − + 14 = 0 3 p=3 q=4 Subst into (3): 25(32 ) 64 ( 3) − + 64 = r 2 9 3 r=5 2

M1

A1

A1

Subst p, q, r into (1): Equation of the circle is: 2 2 ( x − 3 ) + ( y − 4 ) = 52 x2 − 6 x + y 2 − 8 y = 0

A1

9

11 (b) Alternative Method

Centre of circle lies on perpendicular bisector of AB Mid-point of AB = (3.5, 4.5) Gradient of AB = −1 Gradient of perpendicular = 1 Equation of perpendicular bisector of AB: y – 4.5 = 1 (x − 3.5) y = x + 1 ----- (1) 4 y = x ----- (2) 3 Solving (1) and (2), x = 3, y = 4 Centre of circle = (3, 4)

M1 A1

M1 A1

Let the equation of the circle be ( x − 3) + ( y − 4 ) = r 2 Subst A (0, 8): 2 2 ( 0 − 3) + ( 8 − 4 ) = r 2 2

25 = r 2 r=5

2

M1

Equation of the circle is: 2 2 ( x − 3 ) + ( y − 4 ) = 52 x2 − 6 x + y 2 − 8 y = 0

A1

10

12(a)(i) Area of A 3 18 = ∫ dx 2 6− x

M1

= ⎡⎣ − 18ln ( 6 − x ) ⎤⎦ 2 A1 – correct integration = ⎣⎡ − 18ln ( 6 − 3) ⎦⎤ − ⎡⎣ − 18ln ( 6 − 2 ) ⎦⎤ 3

= −18ln3 + 18ln4 = 5.1783 = 5.18 sq units

A1

12(a)(ii) Subst x = 2: y = 4.5

Subst x = 3: y=6

M1 for obtaining y = 4.5 and y = 6

Area of B = Total – Area of A = 1.5×3 + 4.5×1 – 5.1783 = 3.8217 = 3.82 sq units

A1

π⎞ ⎛ (b) The diagram shows part of the curve y = 4 sin ⎜ 2 x + ⎟ . Find the area of the shaded region, 6⎠ ⎝ which is bounded by the curve, the line y = − 2 and the line x =

π

6

.

[5]

y

π⎞ ⎛ y = 4sin ⎜ 2 x + ⎟ 6⎠ ⎝ A

O

x B

x=

12(b)

C

π 6

π⎞ ⎛ Subst 4sin ⎜ 2 x + ⎟ = 0: 6⎠ ⎝

11

y=−2

2x + x=

π 6

= 0, π , 2π ,...

5π [or 1.309] 12

π⎞ ⎛ At intersection point, subst 4sin ⎜ 2 x + ⎟ = −2: 6⎠ ⎝ π⎞ 1 ⎛ sin ⎜ 2 x + ⎟ = − 6⎠ 2 ⎝ π 7π 11π 2x + = , ,... 6 6 6 π x= 2

M1 for obtaining either x-coordinate

Area A =

5π 12

∫π

6

π⎞ ⎛ 4sin ⎜ 2 x + ⎟ dx 6⎠ ⎝

M1

5π

⎡− 4 π ⎞ ⎤ 12 ⎛ = ⎢ cos ⎜ 2 x + ⎟ ⎥ 6 ⎠⎦ π ⎝ ⎣ 2 6 5π 12

⎡− 4 π ⎞⎤ ⎛ =⎢ cos ⎜ 2 x + ⎟ ⎥ 6 ⎠⎦ π ⎝ ⎣ 2

6

⎛ ⎛ 5π ⎞ π ⎞ ⎛ ⎛π ⎞ π ⎞ = − 2 cos ⎜ 2 ⎜ ⎟ + ⎟ + 2 cos ⎜ 2 ⎜ ⎟ + ⎟ ⎝ ⎝ 12 ⎠ 6 ⎠ ⎝ ⎝6⎠ 6⎠ = 2 sq units Area of B ⎛ 5π π ⎞ =2× ⎜ − ⎟ ⎝ 12 6 ⎠ π = sq units 2

M1 [or 1.571]

Area of C ⎛ π 5π ⎞ =2× ⎜ − ⎟− ⎝ 2 12 ⎠

π

⎛

π⎞

∫ π 4sin ⎜⎝ 2 x + 6 ⎟⎠ dx 2 5 12

M1

π

⎡ π ⎞⎤ 2 ⎛ = + ⎢ 2 cos ⎜ 2 x + ⎟ ⎥ 6 ⎠ ⎦ 5π 6 ⎣ ⎝

π

12

12

⎛ ⎛π ⎞ π ⎞ ⎛ ⎛ 5π ⎞ π ⎞ + 2 cos ⎜ 2 ⎜ ⎟ + ⎟ − 2 cos ⎜ 2 ⎜ ⎟+ ⎟ 6 ⎝ ⎝2⎠ 6⎠ ⎝ ⎝ 12 ⎠ 6 ⎠ π = + ( −1.732 ) − ( −2 ) 6 = 0.7916 sq units =

π

Total shaded area = 2 + 1.571 + 0.7916 = 4.36 sq units

A1

13