06 as pure mathematics practice paper c mark scheme

Advanced Subsidiary l

PAPER C

Mark Scheme

Paper 1: Pure Mathematics

1 M ar k Sc he m e

Paper

States or implies the formula for differentiation from first principles.

f ( x) = 5 x

B1

3

f ( x + h) − f ( x) h →0 h

f ʹ′( x) = lim

Correctly applies the formula to the specific formula and expands and simplifies the formula.

M1

3

f ʹ′( x) = lim

5 ( x + h ) − 5 x3 h

h →0

f ʹ′( x) = lim

(

)

3

5 x + 3 x 2 h + 3xh 2 + h3 − 5 x 3

h 15 x h + 15 xh + 5h3 f ʹ′( x) = lim h →0 h h →0

2

2

Factorises the ‘h’ out of the numerator and then divides by h to simplify.

f ʹ′( x) = lim

(

h 15 x 2 + 15 xh + 5h 2

)

h

h →0

(

f ʹ′( x) = lim 15 x + 15 xh + 5h 2 h →0

A1

2

)

States that as h → 0, 15x2 + 15xh + 5h2 → 15x2 o.e. so derivative = 15x2 *

A1* (4 marks)

NOTES: Use of δx also acceptable.

Students must show a complete proof (without wrong working) to achieve all 4 marks.

Not all steps need to be present, and additional steps are also acceptable.

Graph has correct shape and does not touch

2 M ar k Sc he m e

M1

x-axis.

The point (0, 1) is given or labelled.

A1

(2 marks)

⎛ 1 ⎞ Translation 1 unit right (or positive x direction) or by ⎜ ⎟ ⎝ 0 ⎠

B1

⎛ 0 ⎞ Translation 5 units up (or positive y direction) or by ⎜ ⎟ ⎝ 5 ⎠

B1

(2 marks) Total 4 marks

3a M ar k Sc he me

uuur

States that AB = −a + b

M1

uuur uuur uuur uuur 3 3 States PQ = PO + OQ or PQ = − a + b 5 5

M1

uuur 3 uuur 3 uuur States PQ = ( −a + b ) or PQ = AB 5 5

A1

uuur

uuur

Draws the conclusion that as PQ is a multiple of AB the two lines PQ and AB must be parallel.

A1 (4 marks)

3b M ar k Sc he me

3 PQ = × 10 cm = 6 cm cao 5

B1

(1 mark) Total 5 marks

4 M ar k Sc he me

Asymptote drawn at x = 6

B1

Asymptote drawn at y = 5

B1

13 ⎛ 13 ⎞ Point ⎜ 0, ⎟ labelled. Condone clearly on y 3 ⎝ 3 ⎠ axis.

B1

⎛ 26 ⎞ Point ⎜ ,0 ⎟ labelled. ⎝ 5 ⎠

B1

Condone

26 clearly on x axis.

5

Correctly shaped graph drawn in the correct quadrants formed by the asymptotes.

B1 (5 marks)

5 Correctly shows that either M ⎛ 1 ⎞ f(3) = 0, f(−2) = 0 or f ⎜ − ⎟ = 0 ar ⎝ 2 ⎠ k Sc Draws the conclusion that (x – 3), (x + 2) or (2x + 1) must therefore be a factor. he Either makes an attempt at long division by setting up the long division, or makes an attempt to me find the remaining factors by matching coefficients. For example, stating

M1

M1 M1

( x − 3) ( ax2 + bx + c ) = 2x3 − x2 −13x − 6 or

( x + 2) ( rx2 + px + q ) = 2x3 − x2 −13x − 6 or

( 2x +1) (ux2 + vx + w) = 2x3 − x2 −13x − 6 For the long division, correctly finds the the first two coefficients.

A1

For the matching coefficients method, correctly deduces that a = 2 and c = 2 or correctly deduces that r = 2 and q = −3 or correctly deduces that u = 1 and w = –6 For the long division, correctly completes all steps in the division.

A1

For the matching coefficients method, correctly deduces that b = 5 or correctly deduces that p = −5 or correctly deduces that v = –1 States a fully correct, fully factorised final answer:

A1

(x – 3)(2x + 1)(x + 2) (6 marks)

NOTES: Other algebraic methods can be used to factorise h(x).

For example, if (x – 3) is known to be a factor then

2 x3 − x2 − 13x − 6 = 2 x2 ( x − 3) + 5 x( x − 3) + 2( x − 3) by balancing (M1) = (2 x2 + 5 x + 2)( x − 3) by factorising (M1)

= (2 x + 1)( x + 2)( x − 3) by factorising (A1)

6a M ar k Sc he me

5

Attempt is made at expanding ( p + q ) . Accept seeing the coefficients 1, 5, 10, 10, 5, 1

M1

or seeing

( p + q )5 = 5C0 p5 + 5C1 p 4 q + 5C2 p3q 2 + 5C3 p 2 q 3 + 5C4 pq 4 + 5C5 q 5

o.e.

Fully correct answer is stated:

A1

( p + q )5 = p5 + 5 p4 q + 10 p3q2 + 10 p2q3 + 5 pq4 + q5 (2 marks)

6b M ar k Sc he me

States that p, or the probability of rolling a 4, is

B1

1 4

States that q, or the probability of not rolling a 4, is

3 4

States or implies that the sum of the first 3 terms (or 1 − the sum of the last 3 terms) is the required probability.

B1

M1

For example,

p5 + 5 p 4 q + 10 p3q 2 or 1 − (10 p 2 q3 + 5 pq 4 + q5 ) 5

4

3

⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ + 5 ⎜ ⎟ ⎜ ⎟ + 10 ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠

2

or

1 15 90 + + 1024 1024 1024

or

4 5 ⎛ ⎛ 1 ⎞2 ⎛ 3 ⎞3 ⎛ 1 ⎞⎛ 3 ⎞ ⎛ 3 ⎞ ⎞ 1 − ⎜10 ⎜ ⎟ ⎜ ⎟ + 5 ⎜ ⎟⎜ ⎟ + ⎜ ⎟ ⎟ ⎜ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠⎝ 4 ⎠ ⎝ 4 ⎠ ⎟⎠ ⎝

or

405 243 ⎞ ⎛ 270 1 − ⎜ + + ⎟ ⎝ 1024 1024 1024 ⎠

Either

53 o.e. or awrt 0.104 512

M1

A1

(5 marks) Total 7 marks

uuur 7a States or implies that BC = 13i − 8 j o.e.

M ar Recognises that the cosine rule is needed to solve for ∠BAC by stating k a 2 = b2 + c 2 − 2bc × cos A Sc Makes correct substitutions into the cosine rule. he 2 2 2 me 233 = 45 + 104 − 2 45 104 × cos A o.e.

(

) (

cos A = −

) (

)

(

)(

M1 M1

M1

)

7 or awrt −0.614 (seen or implied by correct answer).

130

A = 127.9° cao

M1

A1 (5 marks)

7b M ar k Sc he me

States formula for the area of a triangle.

M1

1 Area = ab sin C 2 Makes correct substitutions using their values from above.

Area =

1 2

(

45

)(

Area = 27 (units2)

M1ft

)

104 sin127.9...o A1ft (3 marks) Total 8 marks

8a Use of the gradient formula to begin attempt to find k.

M k + 1 − (−2) 3 −2 − (k + 1) 3 = − or =− ar 1 − (3k − 4) 2 3k − 4 − 1 2 k Sc (i.e. correct substitution into gradient formula and equating to he me 2k + 6 = −15 + 9k

M1

3 − ).

2

A1*

21 = 7k k = 3*

(must show sufficient, convincing and correct working).

(2 marks)

8b M ar k Sc he me

Student identifies the coordinates of either A or B. Can be seen or implied, for example, in the subsequent step when student attempts to find the equation of the line.

B1

A(5, −2) or B(1, 4).

Correct substitution of their coordinates into y = mx + b or y − y1 = m(x − x1) o.e. to find the equation of the line.

For example,

M1

⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ −2 = ⎜ − ⎟ ( 5 ) + b or y + 2 = ⎜ − ⎟ ( x − 5 ) or 4 = ⎜ − ⎟ (1) + b or y − 4 = ⎜ − ⎟ ( x − 1) ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ A1

3 11 or 3x + 2 y − 11 = 0 y=− x+ 2 2

(3 marks)

8c M ar k Sc he me

Midpoint of AB is (3, 1) seen or implied.

B1

Slope of line perpendicular to AB is , seen or implied.

B1

Attempt to find the equation of the line (i.e. substituting their midpoint and gradient into a correct equation). For example,

M1

2 3

2 ⎛ 2 ⎞ 1 = ⎜ ⎟ ( 3) + b or y − 1 = ( x − 3) 3 ⎝ 3 ⎠

2x − 3 y − 3 = 0 or 3 y − 2x + 3 = 0 .

A1

Also accept any multiple of 2 x − 3 y − 3 = 0 providing a, b and c are still integers.

(4 marks) Total 9 marks

9a M ar k Sc he 9b me M ar k Sc he me

115 (m) is the height of the cliff (as this is the height of the ball when t = 0). Accept answer that states 115 (m) is the height of the cliff plus the height of the person who is ready to throw the stone or similar sensible comment.

B1

(1 mark) Attempt to factorise the – 4.9 out of the first two (or all) terms.

M1

5 ⎞ ⎛ h(t ) = −4.9 t 2 − 2.5t + 115 or h(t ) = −4.9 ⎜ t 2 − t ⎟ + 115 2 ⎠ ⎝

(

)

2

M1

2

h(t ) = −4.9 (t − 1.25) − (−4.9) (1.25) + 115 2

or

2

⎛ 5 ⎞ ⎛ 5 ⎞ h(t ) = −4.9 ⎜ t − ⎟ − (−4.9) ⎜ ⎟ + 115 ⎝ 4 ⎠ ⎝ 4 ⎠ 2

h(t ) = 122.65625 − 4.9 (t − 1.25) o.e.

(N.B. 122.65625 =

A1

3925 ) 32

Accept the first term written to 1, 2, 3 or 4 d.p. or the full answer as shown.

(3 marks)

9ci M ar k Sc he me

Statement that the stone will reach ground level when h(t) = 0, or −4.9t 2 + 12.25t + 115 = 0 is seen.

M1

Valid attempt to solve quadratic equation (could be using completed square form from part b, calculator or formula).

M1

Clearly states that t = 6.25 s (accept t = 6.3 s) is the answer, or circles that answer and crosses out the other answer, or explains that t must be positive as you cannot have a negative value for time.

A1

(3 marks)

9cii hmax = awrt 123 Ma rk t = 5 or t = 1.25 Sch 4 eme

ft A from part b.

B1ft

ft C from part b.

B1ft

(2 marks) Total 9 marks

NOTES: c: Award 4 marks for correct final answer, with some working missing. If not correct B1 for each of A, B and C correct.

If the student answered part b by completing the square, award full marks for part c, providing their answer to their part b was fully correct.

∠A = 45o seen or implied in later working.

10a a Ma Makes an attempt to use the sine rule, rk Sch States or implies that sin120o = 3 and 2 eme

B1

for example, writing

sin 45o =

sin120o sin 45o = 8x − 3 4x −1

M1

A1

2 2

NOTE: Award ft marks for correct work following incorrect values for sin 120° and sin 45° Makes an attempt to solve the equation for x.

Possible steps could include:

M1ft

3 2 6 1 3 6 or or = = = 16 x − 6 8 x − 2 16 x − 6 4 x − 1 16 x − 6 8 x − 2

(8 3 ) x − 2

(

)

(

)

(

)

3 = 16 2 x − 6 2 or 4 6 x − 6 = 16 x − 6 or 24 x − 6 = 16 6 x − 6 6

(

) (

)

( )

6 2 − 2 3 = x 16 2 − 8 3 or 4 6 x − 6 = 16 x − 6 or 12 x − 3 = 8 6 x − 3 6 x=

6 2 −2 3 16 2 − 8 3

or

x=

6− 6 16 − 4 6

or

x=

A1ft

3 6 − 3 o.e.

8 6 − 12

Makes an attempt to rationalise the denominator by multiplying top and bottom by the conjugate.

Possible steps could include: x=

( 3 2 − 3 ) × (8 (8 2 − 4 3 ) (8

) 3)

2 +4 3 2 +4

x=

48 + 12 6 − 8 6 − 12 128 − 48

States the fully correct simplifed version for x.

x=

x=

36 + 4 6 80

9+ 6 * 20

NOTE: Award ft marks for correct work following incorrect values for sin 120° and sin 45°

10b 1 1 1 States or implies that the formula for the area of a triangle is ab sin C or ac sin B or bc sin A a 2 2 2 Ma rk 1 ⎛ ⎛ 9 + 6 ⎞ ⎞⎛ ⎛ 9 + 6 ⎞ ⎞ ⎟ − 1⎟⎜ 8 ⎜⎜ ⎟ − 3 ⎟ (sin15 or awrt 0.259 ) Sch ⎜⎜ 4 ⎜⎜ 2 ⎝ ⎝ 20 ⎟⎠ ⎟⎜ 20 ⎟⎠ ⎟⎠ ⎝ ⎠⎝ eme or

M1ft

A1*

(7 marks)

M1

M1

1 ( awrt1.29 )( awrt1.58)(sin15 or awrt 0.259 ) .

2

Finds the correct answer to 2 decimal places. 0.26

A1

1 24 + 11 6 6− 2 .

200 If 0.26 not given, award M1M1A0 if exact value seen.

NOTE: Exact value of area is

(

)(

)

(3 marks) Total 10 marks

M1

11a Makes an attempt to find (10 − 6 x )dx ∫ a Ma Raising x powers by 1 would constitute an attempt. rk 2 2a Sch Shows a fully correct integral with limits. ⎡⎣10 x − 3x ⎤⎦ a = 1 eme

A1

Makes an attempt to substitute the limits into their expression. For example,

M1ft

(10(2a) − 3(2a) ) − (10(a) − 3(a) ) or (20a −12a ) − (10a − 3a ) is seen.

2

2

2

2

Rearranges to a 3-term quadratic equation (with = 0). 9a 2 − 10a + 1 = 0

M1ft

Correctly factorises the LHS: (9a – 1)(a – 1) = 0 or uses a valid method for solving a quadratic equation (can be implied by correct answers).

M1ft

States the two fully correct answers a =

A1

1 or a = 1 9

For the first solution accept awrt 0.111 (6 marks)

11b a Ma rk Sch eme

Figure 1

Straight line sloping downwards with positive x and y intercepts. Ignore portions of graph outside 0 ⩽ x ⩽ 2

M1

Fully correct sketch with points (0, 10), and ( 5 ,

A1

3

0) labelled. Ignore portions of graph outside 0 ⩽ x ⩽ 2 (2 marks)

11c a Ma rk Sch eme

Statements to the effect that the (definite) integral will only equal the area (1) if the function is above the x-axis (between the limits)

B1

AND when a = 1, 2a = 2, so part of the area will be above the x-axis and part will be below the xaxis.

Greater than 1.

B1 (2 marks) Total 10 marks

12a a Ma rk Sch eme

States that the perimeter of the track is 2πr + 2 x = 300 The choice of the variable x is not important, but there should be a variable other than r.

M1

Correctly solves for x. Award method mark if this is seen in a subsequent step.

A1

x=

300 − 2πr = 150 − πr 2

States that the area of the shape is A = πr 2 + 2rx

B1

Attempts to simplify this by substituting their expression for x.

M1

A = πr 2 + 2r (150 − πr ) A = πr 2 + 300r − 2πr 2

A1*

States that the area is A = 300r − πr 2 *

(5 marks)

12b Attempts to differentiate A with respect to r a Ma Finds dA = 300 − 2πr dr rk Sch Shows or implies that a maximum value will occur when 300 − 2πr = 0 eme Solves the equation for r, stating r =

A1 M1 A1

150 π

⎛ 150 ⎞ ⎛ 150 ⎞ Attempts to substitute for r in A = 300r − πr 2 , for example writing A = 300 ⎜ ⎟ − π ⎜ ⎟ ⎝ π ⎠ ⎝ π ⎠ Solves for A, stating A =

M1

2

22 500 π

M1

A1 (6 marks) Total 11 marks

NOTES: 12b: Ignore any attempts at deriving second derivative and related calculations.

Uses the equation of a straight line in the form Makes correct substitution. log 4 V = −

log 4 V = mt + c or log 4 V − k = m(t − t0 ) o.e.

M1 A1

1 t + log 4 40000 o.e.

10

(2 marks) Either correctly rearranges their equation by exponentiation −

1

M1

t + log 4 40000

For example, V = 4 10 or takes the log of both sides of the equation V = abt .

For example, log 4 V = log 4 (abt ) .

Completes rearrangement so that both equations are in directly comparable form

V=

⎛ − 1 40000 × 4 10 ⎜ ⎜ ⎝

⎞ 1 t ⎟ and V = ab or log 4 V = − t + log 4 40000 and log 4 V = log 4 a + t log 4 b . ⎟ 10 ⎠

States that a = 40 000 States that b = 4 NOTE:

−

M1

t

A1 A1

1 10

2nd M mark can be implied by correct values of a and b.

(4 marks)

a is the initial value of the car o.e.

B1

b is the annual proportional decrease in the value of the car o.e. (allow if explained in figures using their b. For example, (since b is ≈0.87) the car loses 13% of its value each year.)

NOTE: Accept answers that are the equivalent mathematically.

For example, for b. the value of the car in 87% of the value the previous year.

Substitutes 7 into their formula from part b. Correct answer is £15 157, accept awrt £15 000

B1 (2 marks) B1ft (1 mark)

Uses 10000 = abt with their values of a and b or writes log 4 10000 = −

1 t + log 4 40000 10

M1

(could be inequality).

Solves to find t = 10 years.

A1ft (2 marks)

Acceptable answers include.

B1

The model is not necessarily valid for larger values of t.

Value of the car is not necessarily just related to age.

Mileage (or other factors) will affect the value of the car.

(1 mark) Total 12 marks