C3 June 05 Ms

6665 Core C3 Mark Scheme (Post standardisation) Question Number 1

(b)

(a)

Scheme

Marks

sin 2 θ cos 2 θ 1 + ≡ 2 2 cos θ cos θ cos 2 θ 2 2 Completion : 1 + tan θ ≡ sec θ (no errors seen)

M1

Use of 1 + tan 2 θ = sec 2 θ :

M1

Dividing by cos 2 θ :

2 (sec 2 θ − 1) + sec θ = 1

A1

(2)

[ 2 sec 2 θ + sec θ − 3 = 0 ] Factorising or solving: (2 sec θ + 3)(sec θ − 1) = 0 3

[ sec θ = –

or sec θ = 1 ]

2

θ =0 cos θ = –

M1

B1 2 3

;

M1 A1

θ1 = 131.8 °

A1√

θ 2 = 228.2 °

(6)

[A1ft for θ 2 = 360° – θ1 ]

[8]

1

6665 Core C3 Mark Scheme (Post standardisation) Question Number 2

Scheme

Marks

(a) (i) 6 sin x cos x + 2 sec 2 x tan 2 x

or 3 sin 2 x (ii)

M1A1A1 (3)

+ 2sec 2 x tan 2 x

[M1 for 6 sin x]

B1M1A1 (3)

1

3( x + ln 2 x) 2 (1 + ) x

[ B1 for 3( x + ln 2 x)2 ] (b)

Differentiating numerator to obtain 10x – 10 Differentiating denominator to obtain 2(x-1) Using quotient rule formula correctly: To obtain

dy ( x − 1) 2 (10 x − 10) − (5 x 2 − 10 x + 9)2( x − 1) = dx ( x − 1) 4 2( x − 1)[5( x − 1) 2 − (5 x 2 − 10 x + 9) ( x − 1) 4

Simplifying to form =

3

(a)

−

8 ( x − 1) 3

*

(c.s.o.)

5x + 1 3 − ( x + 2)( x − 1) x + 2 5 x + 1 − 3( x − 1) = ( x + 2)( x − 1) M1 for combining fractions even if the denominator is not lowest common =

2x + 4 = ( x + 2)( x − 1)

2( x + 2) 2 = ( x + 2)( x − 1) x −1

B1 M1

M1 A1 cso (4)

* M1 must have linear numerator

(b)

2 ⇒ xy − y = 2 ⇒ xy = 2 + y x −1 2+x f –1(x) = o.e. x 2 2 fg(x) = 2 (attempt) [ ] " g" − 1 x +4 1 2 and finding x2 = …; x =± 2 Setting 2 = 4 x +4 y=

M1A1 A1

(3)

M1 M1; A1 (3)

[10] 2

6665 Core C3 Mark Scheme (Post standardisation) Question Number

4

(a)

Scheme

f ′ (x) = 3 ex – 3e x

−

(c)

M1A1A1

1 2x

(3) M1

1 =0 2x

⇒ 6α e α = 1

Marks

⇒α =

1 6

e

−α

A1 cso

(*)

x1 = 0.0613..., x 2 = 0.1568.., x3 = 0.1425..., x 4 = 0.1445....

M1 A1

(2) (2)

[M1 at least x1 correct , A1 all correct to 4 d.p.] (d)

1 with suitable interval 2x e.g. f ′ (0.14425) = – 0.0007 Using f ′ (x) = 3 ex –

M1

f ′ (0.14435) = + 0.002(1) Accuracy (change of sign and correct values)

A1

(2)

[9]

3

6665 Core C3 Mark Scheme (Post standardisation) Question Number 5

Scheme

Marks

(a) cos 2A = cos2 A – sin2 A ( + use of cos2 A + sin2 A ≡ 1)

= (1 – sin2 A); – sin2 A = 1 – 2 sin2 A (b)

(*)

A1

2sin 2θ − 3cos 2θ − 3sin θ + 3 ≡ 4sin θ cos θ ; −3(1 − 2sin 2 θ ) − 3sin θ + 3

(2)

B1; M1

M1

≡ 4 sin θ cos θ + 6 sin 2 θ − 3 sin θ (*)

≡ sin θ (4 cos θ + 6 sin θ − 3) (c)

M1

A1

(4)

4 cos θ + 6 sin θ ≡ R sin θ cos α + R cos θ sin α Complete method for R (may be implied by correct answer) [ R 2 = 4 2 + 6 2 , R sin α = 4 , R cos α = 6 ] R=

A1

52 or 7.21

Complete method for α ; (d)

α = 0.588 (allow 33.7º)

sin θ ( 4 cos θ + 6 sin θ − 3 ) = 0

3

M1 A1

(4)

M1

θ =0 sin( θ + 0.588) =

M1

B1 = 0.4160..

(24.6º)

M1

52 θ + 0.588 = (0.4291), 2.7125 [or θ + 33.7 ° = (24.6º), 155.4°]

dM1

θ = 2.12

A1

cao

(5)

[15]

4

6665 Core C3 Mark Scheme (Post standardisation) Question Number 6.

Scheme

(a)

Marks

Translation ← by 1

M1

Intercepts correct

A1

x ≥ 0, correct “shape”

B1

y

0

–2

x

2

(2)

a

(b)

y

[provided not just original] 0

–3

3

x

b

(c)

a = – 2,

Reflection in y-axis

B1√

Intercepts correct

B1

(3)

B1 B1

(2)

b=–1

(d) Intersection of y = 5x with y = – x – 1 Solving to give x = –

M1A1 M1A1

1

(4)

6

[11] [Notes: (i) If both values found for 5x = – x – 1 and 5x = x – 3, or solved algebraically, can score 3 out of 4 for x = –

1 6

and x = – ¾;

required to eliminate x = – ¾ for final mark. (ii) Squaring approach: M1 correct method, 24x2 + 22x + 3 = 0 ( correct 3 term quadratic, any form) A1 Solving M1, Final correct answer A1.]

5

6665 Core C3 Mark Scheme (Post standardisation) Question Number 7

(a)

Scheme

(300 = 2500a); (b)

2800a 1+ a a = 0.12 (c.s.o) *

Setting p = 300 at t = 0 ⇒ 300 =

Marks M1 dM1A1 (3)

2800(0.12)e 0.2 t ; e 0.2 t = 16.2... 0.2 t 1 + 0.12e Correctly taking logs to 0.2 t = ln k

M1A1

t = 14

A1

(4)

B1

(1)

1850 =

M1

(13.9..)

(c) Correct derivation: (Showing division of num. and den. by e 0.2 t ; using a)

M1

(d) Using t → ∞ , e − 0.2t → 0, p→

A1

336 = 2800 0.12

(2)

[10]

6