2015 CCHY Midyear 4E5N Additional Maths P2 Marking Scheme [M1] [M1] [A1] [M1] [A1]

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1 05 CCHY Midear 4E5N Additional Maths P Marking Scheme (i) Side [M] [M] [A] (ii) B Pthagoras Theorem, Longest side 0 6 [M] [A] f ( ) (i) Divisible b f () [M] 6 4 0, + 0 ( )( ) 0 or [M] f when when f + () 6 f + () 0 [M] f, therefore [A],, since 0 05 CCHY 4E5N MYE AMP Solution Page of 4

2 (ii)(a) Method Let ( )( + b 5) b + ( b ) + Comaring terms, b b ( )( + 5) 0 [M] 0 or Hence,,.79 or [M].79,.79 (sf) Method (long division) ) 8 ) ) Method (snthetic division) ( )( + 5) 0 0 or Hence,, 4 5 or [A] 6 4 (ii)(b) 0 Let,. 79 or. 79,. 79 (N.A.) Hence, the euation has 4 real roots [B] 05 CCHY 4E5N MYE AMP Solution Page of 4

3 (a) lg lg 6 4lg lg 4lg 6 4 lg lg 6 [M] lg 6 lg 6 lg [M] [A] (b)(i) and z z + z [M] z [A] + [A] b(ii) Method From, z [M] z z Method From z, z [M] [A] z z From, + z + 05 CCHY 4E5N MYE AMP Solution Page of 4

4 05 CCHY 4E5N MYE AMP Solution Page 4 of 4 Taking on both sides, (roven) Method From, () From z z z () subst () into (), (roven) Method 4 z From b(i) z

5 z () () subst () into (), z (roven) 4(i) kt V Ae When t 0, V A A [B] (ii) e 4k e 4k [M] 4 e 4k 89 4 ln 4k [M] 89 k 4 4 ln (sf) [A] (iii) e t [M] t e t ln ln e ln t Year to scra car ears after 04 [A] 06 e t 05 CCHY 4E5N MYE AMP Solution Page 5 of 4

6 (iv) V Asmtote at V [B] Shae of eonential grah starting from [B] V e kt 0 t 5(i) Sum of roots, [M] [M] Product of roots, ( + )( + ) ( + ) [M] (shown) Mmcd 0 5(ii) from (i),. 5 Since and are of oosite signs, (.5) 9 6 (Reject.5) [M] [M] [M] 4 Quadratic euations: or [A] 05 CCHY 4E5N MYE AMP Solution Page 6 of 4

7 6(i) Coordinates of A (-, 0) [M] Coordinates of B (,0) [M] d d 8 [M] Gradient of tangent at A ( ) 8( ) + Euation of tangent at A: 0 ( ) ---- () [M] Gradient of tangent at B () 8() + 6 Euation of tangent at B: () [M] ()(), subst into (), + 4 Therefore coordinates of T,4 [M] 6(ii) Area of shaded region d [M] 4 [M] area of shaded region units [A] 05 CCHY 4E5N MYE AMP Solution Page 7 of 4

8 7(i) A B 5 5 A B let 5 B comaring constant, 5A A 5 B comaring coefficient of, 0 A + B B A [M] [M] B 4 5 [A] (ii) ln 5 Let ( )( ) ln 5 d 5 [M] d ln 5 [A] 5 (iii) From (ii) d 5 ln 5 d ln 5 c ln 5 d ln 5 d c 5 4 ln 5 d d ln 5 ln ln [A] [A] + c c [M] [M], where c, c are arbitrar constants 8(i) Method RHS tan tan( ) tan tan tan tan tan tan tan tan tan tan [M] [M] 05 CCHY 4E5N MYE AMP Solution Page 8 of 4

9 tan tan tan tan tan tan tan tan tan ( tan ) [M] tan tan tan tan tan tan tan tan [A] tan LHS tan tan tan (shown) tan Method tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan tan + tan tan tan tan [ tan + tan ] [ (tan) tan ] tan tan tan tan tan ( + ) tan (shown) tan tan tan 8(ii) tan tan d d sec cos when, d d cos 05 CCHY 4E5N MYE AMP Solution Page 9 of 4

10 tan tan 4 6 [M] gradient of normal is 6 Euation of normal: ( ) 6 [A] 6 7 8(iii) d d sec [M] cos when 6, cos cos 0 when 0, cos cos 0 when, cos cos For 0 and, 6 cos > 0 cos > 0 [R] d 0 d the curve is increasing for 0 and 6 9(i) DEF EBD (tangent-chord theorem or s in alt segment) [M] EDB EBD (base angle of isosceles EBD, BE DE) Since DEF EDB, BD//AF b roert of alternate angles [M] 05 CCHY 4E5N MYE AMP Solution Page 0 of 4

11 9(ii) AEB ACE (tangent-chord theorem or s in alt segment) [M] BAE EAC (common angle) [M] Since two air of corresonding angles are eual, AEB is similar to ACE. Since ratio of corresonding sides of similar triangles are eual, AE AB [M] AC AE AE AB AC (roven) 9(iii) Method AC Given AE AC AE AE AC () [M] Substitute () into results from (ii) AC AB AC [M] AB AC [M] B is the midoint of AC (roven) Method ACE is similar to AEB AC AE CE EB AE AB CE EB AE AB AE AB AE AB Suaring both sides, AE AB () from (ii), AE AB AC () () (), AB AB AC AB AC i.e. B is the midoint of AC (roven) 0. (i) [M] Centre (, 6) [A] 05 CCHY 4E5N MYE AMP Solution Page of 4

12 Radius 5 units (ii) Substitute [A] 6 () 0 0 [M] [A] Therefore is the tangent of the circle C. Method Discriminant Method ( - ) 0 Since the euation has onl real root, is a tangent to circle C (iii) Distance from A to P Since distance AP is greater than radius, Therefore A lies outside the circle C [M] (iv) To find the coordinates of M and N Substitute 0 into ( )( 0 ) 0 and 0 Therefore M (0,) and N(0,0) or (reverse order) [M] The -coordinate of Q is the same as the -coordinate of P Therefore Q (, 6) [M] 00 0 (reject) or 0 Therefore coordinates of Q 0, 6 [A]. (i) BAC 90 ( in a semi-circle) AB cos DE cos 9cos 4 [M] BD 4 AE sin sin [M] 4 P cos 9cos sin sin cos ( shown) (ii) P sin cos 05 CCHY 4E5N MYE AMP Solution Page of 4

13 R [M] tan [M] 8.0 P 5 cos 8. [A] (iii) Maimum value of P 4.(sf) or +5 [B] Corresonding angle 8. [B] (iv) 8 5 cos 8. 0 cos 8. 0 Basic angle, ( d) [A] [M]. (i) Height of triangle face Area of triangle 0 (ii) Area 4 h h 40 [M] h 48 [M] 5 0 h 4 [M] 60 [A] [M] (iii) Volume [M] (iv) dv d [M] At stationar value, dv d [M].9. [A] 05 CCHY 4E5N MYE AMP Solution Page of 4

14 (v) d V d 06 d V When. 9, [M] d Therefore this value of gives the maimum volume. Maimum V 4500(.9) 6(.9) ~ 640 (sf) Hence, the maimum volume 640 cm [A] 05 CCHY 4E5N MYE AMP Solution Page 4 of 4