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Mark Scheme (Results)Summer 2009

AEA

AEA Mathematics (9801/01)

June 20099801 Advanced Extension Award Mathematics

Mark Scheme

9801 AEA Mathematics Summer 2009 2

Question

NumberScheme Marks Notes

Q1 (a) B1

B1

B1(3)

Don’t insist on labels

(b) One intersection at = 2 B1

Second at ( +1) (2- ) = ( +2) M1 Attempt correct equationMust be + 2 on RHS

(0 = ) 2 2 + - 2 A1 Correct 3TQ

= , since root is in (-2, -1)

=

M1

A1 cso(5)

[8]

Solving

Must choose -


-2, -1, 2, (0,2)

Question


Q2 (a) = so when = = = B1

lny = sin ln M1 Use of logs (o.e)

= cos ln + M1A1

Use of product rule

at gradient = M1

Some correct sub in their y'

Equation of tangent is A1 cso

(6)(b) If it touches again then = sin = 1 M1 Method → sin x = 1

= + 2 A1 May be listed…

Gradient at is = 1

at points = is a tangent.

A1(3)

[9]

Check points satisfy = 1 plus comment

Q3 (a) M1 Use of sin ( - )

(o.e)M1

Use of sin , cos

and collect terms

A1 oe.

A1, B1

(5)(b)

sin = sin

M1 Use of sin

1 – 2 = 2 -

1 [2 – 3 = 2 ]

4 – 12 + 9 2 = 3 – 3 2

M1,B1

M1

B1 for cos[arcsinx] =

Simplify to quadratic in x


Question


12 2 – 12 + 1 (=0) A1 correct 3TQ

= M1Attempt to solve if at least one previous M scored in (b)

=

0 < < 0.5 = (o.e.) A1 Must choose '_'

(7)[12]

Q4 (a) '' ( ) = M1 Use of Quotient rule

' ( ) = M1 Sub ( ) = 0

" (accept ) A1 cso (3) Insist on not

(b) (i) (0, -3) B1 (1) Accept

(ii) Asymptotes and

B1

B1 (2)

M1

A1 cso (2)

Both

Any correct exp. for T in terms of a and b or complete 2nd line

(iv)= M1

Use of quotient

rule to find

= = M1Solving =0 -->

=… or =...

= 0 or (or = 0 but > 0) A1 (S+) Condone =

= compare 22

3

1

3020

a

aaM1

Full method e.g. " ( ) attempted

T" ( ) = 4

330360 > 0 min A1 Full accuracy + comment

Minimum area = = B1 (6)Must come from T(

) not T" ( )

[14]

N.B or SuggestS1 > 12S2 for S+ and 13 or 14.


Area, =

(*)

Question


ALT for (iv) Attempt … M1 No value of needed.

Correct and comment. A1 Fully correct and full comment.

Q5 (a)

(i)

= 180 = 60°

M1

A1

Equate S =180

Show = 60o

Area = M1 Use of

= = A1(4)

(ii)Sine Rule = OR sin = M1

Correct use of sine rule or and (a)

= 523

1552 A1

(2)(iii) Cosine Rule 2 =

5 =

M1 Use of cos rule where all terms are known, except c.

0 = c - 2 – 1 OR c - M1 Sub & simplify -> 3TQ

c = M1 Solving

c = OR A1 (4)(b)

= [2 143 + 2 ( – 1)] = M1For use of needn’t be simplified.

Sum of internal angles = 180 ( - 2 ) B1 (142 + ) = 180 ( – 2) 0 = 2 – 38 + 360 A1 Correct 3TQ.

0 = ( – 19)2 – 192 + 360 – 19 = 1 ( = 20 or 18)

M1 Attempt to solve relevant 3TQ

Internal angles all <180 20 = 143 + 19 2 > 180 18 = 143 + 17 2 < 180 ] S+

= 18 A1 (5)[15]

ALT for c If get sin = or method to find this M1

= use of M1

= 1 + M1 A1

If get cos = or method to find this M1

Then 2 = 2 + 2 – 2 cos use of M19801 AEA Mathematics Summer 2009 6

Question


c2 = 3 + 2 c = (3 + 2 M1 A1

Look out for similar variations of cosine rule with cos A

Pythagoras height = a sin 60 + Pythagoras once M1 2nd Pythagoras M1

cos 60 = 1 + other bit M11 + A1

Q6 (a)

is B1 Score anywhere.

M1 A1 M1 attempt

A1 correct

When = = A1

Equation of tangent at is: = M1 Attempt tangent at P.√ their P and m

is where = 0 A1 cso(6) Allow

(b)Area under curve = M1

Attempt

condone missing 2

= [2 sin ln sec t] - 2 sin tan M1A1

Attempt parts.Both parts correct.

= [ ] - M1 Use of s = 1 - c

= [ ] - 2 sec t + 2 cos M1 Split = [ 2 sin ln sec t] - A1, A1 Accept cos t tan t

= ln 2 - (2 ln [2 + ] – 0) + (2 M1 Use of correct limits on all 3 integrals

Area of = B1 Any correct expression.

Area of = are under curve – area of M1 Strategy must be or area

= A1 cso

(11)[17]

ALT Area = - ln (1 - o.e. M1 Condone missing -

= + M1A1

Parts correct


Question


= [ ] + 1 - M1 Split

= [ ] + M1 Partial Fractions

= + A1, A1 o.e.

Then use of limits etc as before.

Q7 (a) =

Attempt bothM1 Allow

. = - 10 = Use of .M1

Use of . to form equation for cos

cos = o.e. A1 cso

(3)(b) Area of K = 2 Area of ABC

= M1Use of

sin ( M1 Attempt √ their (a)

Area = = 30 A1

(3)(c) Identify to BC and to

Area = 2 [Area of + Area of ]

30 = 2

=

B1

M1

A1

M1

Method equation in r

Correct equation in r

Attempt = with rational denom.

= A1(5)

(d) AC =

= + t =

But

= 0

-35 + 49 + 16 – 25 + 25 = 0

M1

M1

M1

M1

Attempt AC

Expression for in terms of

Use of … … = 0

Linear equation in based on ‾

90 = 609801 AEA Mathematics Summer 2009 8

Question


= A1

= + 2 M1Method for in terms of known vectors

= A1

(7)[18]

Marks for Style Clarity and Presentation (up to max of 7)

S1 or S2 For a fully correct (or nearly fully correct) solution that is neat and succinct in question 2 to question 7

T1 For a good attempt at the whole paper. Progress in all questions.

Pick best 3 S1/S2 scores to form total.


june 2008 · web viewmark scheme (results) summer 2009-2, -1, 2, (0,2) title june 2008 author...

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