RI 9758/01 2017

1 | Page Question 1 No. Suggested Solution Remarks for Student ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 23 2 3 23 2 22 2 32 23 e ln 1 221 2 ... ... , 1 12! 3! 2 3 22 ...23 2 4 36 ...23 x ax x x ax axx ax ax ax ax ax xax ax x ax aa a a aax x x +       = + + + + − + − −< ≤     = − ++ − + +   − −+= ++ +     2No term in x , we have 24 0 0 (rejected since 0) or 4.aa a a a− =⇒= ≠ = Raffles Institution H2 Mathematics (9758) Solution for 2017 A-Level Paper 1

2 | Page Question 2 No. Suggested Solution Remarks for Student (i) (ii) To solve 1 bx axa <−− , we consider the graphs 1y xa= − and y bx a= − drawn in (i), noting they intersect at a point where x > a. ( ) ( ) 2 1 ,. 11 1 , reject since bx a x axa x a xa xa xabb b = −>− − =⇒=+ =− >  1 bx axa <−− 1xa b>+ or xa< You need to refer to (i) if you are using “Hence”. x y ab x=a y=b|x – a|

3 | Page Question 3 No. Suggested Solution Remarks for Student (i) 22 2 5 10 0y xy x− + −= … (1) Differentiate (1) with respect to x: dd2 2 2 10 0dd yyy x yxxx− −+ = dd 50dd yyy x yxxx− −+ = … (2) When d 0d y x = , 5yx= Sub into (1): 2 22 2 1125 10 5 10 0 20 10. or 22 x xx x x− + − =⇒ = ∴= − (ii) Differentiate (2) with respect to x: 222 22 d d d dd 50d d d dd y y yyyyx x x x xx + − −−+ = …(3) When 1 5d, and 0 d22 yxy x= = = . Sub into (3): 22 2 22 2 5d 1d d 52 50 0dd d422 yy y xx x− +=⇒ = − < Thus, 15, 22   is a maximum turning point. Use 2 nd Derivative Test here. 1st derivative test cannot be used a s d d y x depends on both x and y. Note that exact values with working are required as no calculator is allowed for this question.

4 | Page Question 4 No. Suggested Solution Remarks for Student (i) ( )4 2149 1422 2 xxy xx x +++= = = +++ + Note that C is defined on all values of x except –2. ( ) 2 d1 0 for all 2d 2 y xx x =− < ≠− + Thus, gradient of C is negative for all points on C. (ii) 14 2y x= + + Asymptotes are 2, 4xy= −= (iii) A translation of 2 units in the positive x direction follow by a translation of 4 units in the negative y direction. Replace x by (x – 2) follow by replacing y by (y + 4) is not acceptable.

5 | Page Question 5 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) ( ) 32f f 1 8 1 8 7 ...(1) f 2 12 8 4 2 12 4 2 4 ...(2) f 3 25 27 9 3 25 9 3 2 ...(3) x x ax bx c abc abc a bc a bc a bc a bc =+ ++ = ⇒+++= ⇒++= = ⇒+ + += ⇒ + += = ⇒ + + += ⇒ + += − Solving (1), (2) and (3), 33, , 722a bc= −== (ii) ( ) 32 33f7 22xx x x= − ++ We know that in general a cubic function has 1, 2 or 3 x-intercepts. ( ) ( ) 2 2 2 22 3f ' 3 3 2 33 2 11 33 44 2 13 13 0 for all since 0 for all 24 2 x xx xx xx x xx x = −+ = −+ = −+− +  = − +> − ≥   ( ) ( )f as and f as . Since gradient of curve is always positive, curve is always increasing. x x xx→−∞ →−∞ →∞ →∞ (in particular, f is 1-1) ( )So, f 0 has only one root.x = Using GC, the root is –1.33 (3 s.f.) (iii) ( ) 2 2 13f ' 2 3 224 15 2 12 15 2 12 xx x x =⇒ − += ⇒− =  = ±

6 | Page Question 6 No. Suggested Solution Remarks for Student (i) It is a line passing through the point A with position vector a and is parallel to the vector .b (ii) It is a plane with normal vector n such that the dot product of the position vector of any point on the plane with the normal has the value d. Since n is a unit vector, the magnitude of d is the distance from the origin to the plane. (iii) ( ) . .. .. 0, . . . a tb n d a n tb n d d anbn t bn d anra b bn + =⇒+= −≠=  −∴= +             which is the position vector of the point of intersection of the line in (i) and the plane in (ii).

7 | Page Question 7 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) ( ) sin 2 sin 2 d 1 2sin 2 sin 2 d2 1 cos 2 2 cos 2 2 d2 11 sin 2 2 sin 2 244 44 mx nx x mx nx x m nx m nx x m nx m nx Cmn mn = −− = − +− − = − ++ −++− ∫ ∫ ∫ (ii) ( )( ) ( ) ( ) ( ) ( ) 2 0 2 0 22 0 0 00 0 fd sin 2 sin 2 d sin 2 sin 2 2sin 2 sin 2 d 1 cos 4 1 cos 4 2sin 2 sin 2 d22 11 11 sin 4 sin 428 28 11 sin 2 2 sin 2 222 22 sin xx mx nx x mx nx mx nx x mx nx mx nx x x mx x nxmn m nx m nxmn mn k π π π π ππ π π = + = ++ −−= ++   =− +−     +− + + − +− = ∫ ∫ ∫ ∫ ( )0 for kπ = ∈ 

8 | Page Question 8 No. Suggested Solution Remarks for Student (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 i 2 5 5i 0 2 2 41 i51 i 21 i 2 4 40 21 i 2 6i 21 i 2 6i 1 i 2 6i 1 ior21 i 1 i 21 i 1 i 2 2i 6i 6 2 2i 6i 6or 44 1 2i or z 2 i zz z z z zz zz z −− ++ = ± −− += − ±−= − ±= − ++ −+= ×= ×−+ − + ++ − +− += = = −+ = − You can also divide through by ( )1i− before applying the quadratic formula Note that detailed working is required as calculator is not allowed. (b) (i) ( ) ( ) ( ) ( ) 22 3 2 242 1i 1 i 2i 2i 1 i 2 2i 2i 4 w w w ww = − = −= − = − − = −− = = −= − ( ) ( ) ( ) 43 2 39 58 0 4 2 2i 39 2i 1 i 58 0 Compare Real and Imaginary parts: 4 2 58 0 ...(1) 2 78 0 ...(2) (1) (2) : 4 24 0 6 Sub into (2) : 12 78 0 66 w pw w qw pq pq pq pp qq + + ++= −+ −− + − + − + = −− ++ = − − −= + −− =⇒= − − −=⇒= − (ii) ( ) ( ) ( )( ) ( )( ) 43 2 2 6 39 66 58 0 Note that 1 i is a root and coefficients are real, thus 1 i is also a root One quaratic factor is 1 i 1 i 2 2 ww w w w w ww − + − += −+ −− −+ = − + ( )( ) 43 2 2 26 39 66 58 2 2w w w w w w w pw q∴− + − += −+ + + By comparing constant terms, 58 2 29qq= ⇒= By comparing linear term, 66 2(29) 2 4 pp−= − + ⇒= − ( )( ) 43 2 2 26 39 66 58 2 2 4 29ww w w ww ww∴− + − += −+ −+

9 | Page Alternatively ( ) ( ) ( ) 2 2 43 2 432 32 32 2 2 4 29 2 2 6 39 66 58 22 4 37 66 58 488 29 58 58 29 58 58 0 ww wwww w w www www www ww ww −+ −+ − + − + −−+ −+ −+ −− + − −+ − −+

10 | Page Question 9 No. Suggested Solution Remarks for Student (a) 2 1 n nr r S u An Bn = = = +∑ (i) ( ) ( ) ( ) 1 22 11 21 n nnu SS An Bn A n B n An B −= − =+ −− −− = −+ (ii) 10 17 48 19 48 90 33 90 Solving, 3, 9 u AB u AB AB = ⇒ += = ⇒ += = =− (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 222 22 2 2 3 2232 2 11 2 2 22 22 22 22 22 2 222 2222 22 11 11 11 11 22 4 1 114 1 12 014 23 12 34 23 ... 1 21 11 1 14 nn rr r rr r r rr rr r r r rr r r rr r r n nn n nn n n nn = = + −− = + −−  = +−− ++−     = = = + −− = − +− +− + +− −− − + +−−  = + ∑∑ (c) ( ) 1 1 Let . ! ! 1! 1 n n n n n n xa n a x nx a n xn + + = = ×=++ 1lim lim 0 1 1 n nn n xa an + →∞ →∞ = = <+ for each x. Therefore 0 ! r r x r ∞ = ∑ converges. 0 e! r x r x r ∞ = =∑ from MF26.