Paper 1 Section A. Question Answer 1 A 2 C 3 D 4 A 5 B 6 D 7 C 8 B 9 C 10 B 11 D 12 A 13 B 14 C 15 C 16 A 17 B 18 B 19 C 20 A. Summary A 5 B 6 C 6 D 3

Transcription

1 Paper Section A Detailed Marking Instructions : Higher Mathemats 0 Vs Question Answer A C D A B D C B C B D A B C C A B B C 0 A Summar A B C D Page

2 Paper Section B Triangle ABC has vertes A(, 0), B(,) and C(, 0),as shown in the diagram opposite. Medians AP and CR intersect at the point T(,). B R P T Q C (a) Find the equation of median BQ. O A x (b) Verif that T lies on BQ. (a) know and find midpoint of AC pd calculate gradient of BQ state equation (, ) or equivalent x or x ( ( )) ( ). Candidates who do not use a midpoint lose. There is no need to simplif m for BQ and Do not award for x + 0, although then x If m cannot be simplified, due to an error, then BQ. is available for using. Accept 0 x +... It must, however, be simplified before can be awarded. would be awarded for x + 0 is still available. and mx + c where m c.. Candidates who find the equations of AP or CR can onl gain mark. AP : 0 ( x ) or ( x ) CR : 0 ( x ) or ( x ) (b) substitute in for T and complete + + e.g. Substitution : () () 0 Gradients : m BT m Vectors : BT, TQ and BT TQ BQ. is available as follow through with an appropriate communation statement, e.g. 'T does not lie onbq'.. Statements such as PA, RC and BQ are all medians and therefore all share the same point T do not gain. Since onl mark is available here, do not penalise the omiion of an reference to a common point or parallel. Regularl occurring responses Gradient approach : (b) m m leading to : in (c), without further working, gains and BT BQ but loses. but (b) m and m leading to m m so T lies on BQ leading to : in (c), BQ TQ BQ TQ without further working, loses and but gains.. Page

3 Detailed Marking Instructions : Higher Mathemats 0 Vs Triangle ABC has vertes A(, 0), B(,) and C(, 0),as shown in the diagram opposite. Medians AP and CR intersect at the point T(,). B R P T Q C O (c) Find the ratio in whh T divides BQ. A x (c) valid method for finding the ratio complete to simplified ratio Method : Vector approach e.g. BT and TQ : Method : Stepping out approach For : without working, onl is awarded. Be aware that the working ma appear in (b). Some candidates obtain : from erroneous working thus losing. e.g. : Method : Distance Formula approach BT TQ e.g. d and d : B T Q or B T Q. An evidence of appropriate steps, e.g. and or and but not and, can be awarded to e.g., B T Q is not suffient on its own and so loses. : with no further simplifation ma be awarded but not but gains. In this question working for (c) ma appear in (b), where the working appears for Regularl occurring responses... leading Response Response Response (b) BT TQ (b) QT BQ QT (b) QT TB BT TQ (c) : (c) : (c) so : marks out of marks out of marks out of Response (b) BT TQ so BT TQ x (c) : x but : would have gained Page

4 (a) (i) Show that ( x ) is a factor of f x x x x ( ) + +. (ii) Hence factorise f( x ) full. (b) Solve x x x (a) Method : Using snthet division x know to use complete evaluation 0 state conclusion pd find quadrat factor pd factorise completel ( x ) is a factor see note x + x stated, or implied b + ( x )( x )( x ) stated explitl Method : Using substitution and inspection x + + know to use 0 ( x ) is a factor see note x x + x ( )( ) + ( x )( x )(x ) stated explitl. Communation at For. At must be consistent with working at. i.e. candidate s working must arrive legitimatel at zero before is awarded. If the remainder is not 0 then an appropriate statement would be '( x ) is not a factor'., minimum acceptable statement is factor. Unacceptable statements : x is a factor, ( x + ) the expreion ma be written as ( x ) (x ) +. is a factor, x is a root, ( x ) is a root etc. (b) state solutions x x and or or These ma appear in the working at (a). + leading to x, x then (, 0 ) and (, 0 ) x x x + leading to (, 0 ) and (, 0 ). From (a) ( x )( x )(x ) However, ( )( )( ). From (a) ( x )(x + ) onl leading to ( x )( x + )(x ) leading to gains x, x onl does not gain does not gain x, x and x gains.. as equation solved is not a cub, but as follow through from a cub equation. Page

5 Detailed Marking Instructions : Higher Mathemats 0 Vs (c) The line with equation x is a tangent to the curve with equation x x x + + at the point G. Find the coordinates of G. (d) This tangent meets the curve again at the point H. Write down the coordinates of H. (c) Method : Equating curve and line set CURVE G G LINE expre in standard form compare with (a) or factorise identif x pd evaluate Method : Differentiation know to and differentiate curve set derivative to gradient of line pd solve quadrat equation proce to identif x complete to CURVE G LINE Method : Equating curve and line x x x x stated explitl x x x ( x )( x )(x ) x Method : Differentiation x + x x + x x and at x evaluate 0 see note CURVE and LINE from both curve and line In method :.. is onl available if 0 appears at either the, and or are onl available via the working from. If ( x )( x )(x + ) does not appear at. At a quadrat used from (a) ma gain. If G and H are interchanged then stage. and. stage, it can be implied b, is lost but. Candidates who obtain three distinct factors at and.. A repeated factor at In both methods: or and stage is required for and can gain and but a quadrat from are still available.. ma gain and for evaluating all values, but lose to be awarded without justifation. onl.. All marks in (c) are available as a result of differentiating x + x x + and solving this equal to (from method ). Onl marks and x x x (from method ) are available to those candidates who choose to differentiate + + and solve this equal to 0.. Candidates ma choose a combination of making equations equal and differentiation. (d) pd state solution ( ), ma appear in (c). Method from (c) would not ield a value for H and so is not available. Page

6 (a) Diagram shows a right angled triangle, where the line OA has equation x 0. (i) Show that tan a. (ii) Find the value of sin a. a O x (b) A second right angled triangle is added Diagram as shown in Diagram. The line OB has equation x 0. Find the values of sinb andcos b. O a b A A Diagram B x (a) write in slope/intercept form connect gradient and tana pd calculate hpotenuse state value of sine ratio x or x stated explitl m and tan a or m tan a and tana stated, or implied b or ma not appear until (c). is onl available if sina.. Onl numeral answers are acceptable for and. Regularl occurring responses Response a marks out of x 0 () () 0 tana sina (b) determine tanb know to complete triangle pd determine hpotenuse state values of sine and cosine ratios tan b stated, or implied b right angled triangle with and correctl shown stated, or implied b sin b and cos b ma not appear until (c).. is onl available if sinb b and cos b. sin b and cos without working is awarded marks onl.. Onl numeral answers are acceptable for and. Page

7 (c) (i) Find the value of sin( a b). Detailed Marking Instructions : Higher Mathemats 0 Vs (ii) State the value of sin( b a). (c) know to use addition formula substitute into expansion pd evaluate sine of compound angle use sin( x) sinx a b sin cos cosasinb. sin(a B) sinacosb cosasinb, or just sinacosb cosasinb, with no further working does not gain. Candidates should not be penalised further at to.. Candidates who use sin( a b) sina sinb lose,, and and. for values of sine and cosine outside the range but can gain a non-zero answer whh is obtained from the result sin( x) sin x.. Treat sin cos cos sin as bad form onl if sin and cos subsequentl disappear.. It is acceptable to work through the whole expansion again for Regularl occurring responses sin( b a) 0 x., as follow through, onl for Response Response sin( a b) sina sinb x x sina cosa sinb cosb Marks lost in (a) or (b) 0 x sin( a b) sinacosb cosasinb 0 marks out of sin( b a) marks out of Eased - not dealing with fraction containing a surd. Response Response From (a) and (b) sina (i) sin( a b) sinasinb cosacos x b (c) (i) (ii) cosa cosb sinb sin( a b) sinacosb cosasinb sin( b a) sinbcosa cosbsina marks out of (ii) sin( b a) Here the working was not necear; the answer would gain marks out of, provided it is non zero. Page