RI 9740/01 2014

1 | Page Question 1 No. Suggested Solution Remarks for Student (i) ( ) 1Let f 1yx x= = − 1 1111 yxx y yy −−=⇒=−= ( ) 1 1f xx x − −∴= ( ) ( ) 21 11 1 1ff f11 11 1 11 xxx xxx xx −−= = = = = −−− − −− (ii) ( ) ( )( ) ( )( ) 32 1f ff ffx x xx −= = = Question 2 No. Suggested Solution Remarks for Student 2 22 54 0x y xy+ += … (1) Differentiate with respect to x: 22dd 22 0dd yyx xy xy yxx+ + += When d 1d y x =− , we have 22 22 0x xy xy y−+ − + = 22yx y x= ⇒= ± Sub yx= into (1): 33 354 0 27 3xx x x y+ + =⇒ = − ⇒ = −= Sub yx=− into (1): 33 54 0 54 0 (no solution)xx−++ =⇒ = Therefore there is only one point on C with gradient −1, and coordinates of that point is ( )3, 3−− . It is not necessary to make d d y x the subject before substituting d 1d y x =− . Raffles Institution H2 Mathematics Solution for 2014 A-Level Paper 1

2 | Page Question 3 No. Suggested Solution Remarks for Student (i) ×=ab 0 Either =a0 or =b0 or a and b are parallel to each other. =a0 or =b0 are trivial cases where students might discard. (ii) ( ) 222 11 1 1 122 22 3 3 3312 2 22    = = =+− + +−  −−  n i jk Alternative answer: 122 333 = − +− n i jk Read question carefully and note that they asked for unit vector. (iii) ( ) 222 10 12cosine of acute angle 2 0 312 2 21    = ⋅= + +−  −  Question 4 No. Suggested Solution Remarks for Student (i) (ii) They are the vertical lines xa=− , xb= and .xc= Alternative: Tangents at these points are parallel to the y axis. Not sufficient to say “tangents are undefined” (–a,0) (b,0) (c,0) y x y 2 = f(x) O

3 | Page Question 5 No. Suggested Solution Remarks for Student (i) 1 2iz= + ( ) 22 1 2i 1 4i 4 3 4iz = + =+ − = −+ ( )( ) 3 3 4i 1 2i 3 4i 6i 8 11 2iz =−+ + = −+ − −= − − ( )3 1 1 1 11 211 2i i11 2i 125 125 125z = = −+ = − +−− Read question carefully, we need 3 1 z , not just 3z (ii) ( ) 2 3 11 23 4i i125 125 qpz p qz + = −+ + − +  is real Therefore, 24 0 250125pq q p+ =⇒= − ( ) 2 3 11 23 4i 250 i125 125 3 4i 2 2 4i 19 qpz p pz pp pp p + = −+ − − +  = −+ + − = Note that 2 3 qpz z+ is real. This is a “hint” to “check” your answer. Question 6 No. Suggested Solution Remarks for Student (a)(i) Let nP be the statement ( )1 743 n np = − for .n +∈  When 1,n= ( ) ( ) 1 1 LHS 1 11RHS 7 4 3 133 p= = = −= = LHS = RHS, so 1P is true. Now assume that kP is true for some ,k +∈  i.e. ( )1 743 k kp = − We need to show that 1kP+ is true, i.e. ( ) 1 1 1 743 k kp + + = − We have ( ) 1 47 14 74 73 kk k pp+ = − = −−

4 | Page ( ) ( ) ( ) 1 1 4 74 73 1 28 4 213 1 743 k k k + + = −− = −− = − Thus 1is true kkP P +⇒ is true. Since 1P is also true, nP is true for all n +∈  by Mathematical Induction. (a)(ii) ( ) ( ) ( ) 11 1 1 743 1 743 44 11 73 4