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 41 74 4139 nn r r rr n r r n n p n n n = = = = − = −  −= − − = −− ∑∑ ∑ (b)(i) ( )1 11 1! n rn r uS n= = = − +∑ As n→∞ , ( ) 1 01!n →+ . Therefore, 1 n r r u = ∑ converges. 1 1r r u ∞ = =∑ . (b)(ii) ( ) ( ) ( ) ( ) 1For 2, 11 11 1! ! 11 ! 1! ( 1) 1 1! 1! n nnn uSS nn nn n n n n −≥= − = − −− +  = − + +−= + = + 11 1111 (1 1) ! 2 (1 1) ! for 1.( 1) ! n uS nun n = = −= = ++ ∴= ≥ +

5 | Page Question 7 No. Suggested Solution Remarks for Student (i) 1.88525 1.885 (3 d.p.)α = = At ( ),7β − , 64 3 77ββ− −= − ( ) 42 30ββ −= 0 or 3 or 3β = − . From diagram, 0β > , thus 3β = (ii) f ( ) d 0.597 (3 d.p.)xx α β =−∫ (iii) ( ) ( ) ( ) 3 3 64 7 5 0 0 75 133 7d 775 13 3 3 7375 27 273 3 7375 299 335 xx x x xx −− = −−  =−− =−− =− ∫ Required area 299 543 73 335 35= −= (iv) ( ) ( ) ( ) ( ) 6464f 3 7 3 7 f (Shown)xx x x x x= − −= − −− −= − Thus, f is symmetrical about the y-axis. Both α and α− are the real roots of f(x) = 0. The other 4 roots are non-real. Since the coefficients and constant term of f are real, the 4 roots will be 2 conjugate pairs. Question asked students to comment on the 6 roots. Thus concepts of complex numbers are requ ired for (iv) Question 8 No. Suggested Solution Remarks for Student (i) ( ) 1 2 1f d d sin 39 xxx x c x −= = + −∫∫

6 | Page (ii) ( ) 2 2 1 2 2 222 32 24 6 24 6 1f 9 1 31 9 1 139 13 11 2213 29 2 9 135 222 69 1 3513 18 648 11664 15 3 54 648 34992 x x x x xx x xx x xx x − = − = − = −   −−     ≈ +− − + −       −−−      +−     = ++ + = ++ + As stated in the question, the intended method is to apply Binomial Theorem on 1 2 2 1 9 x − − instead of finding the Maclaurin’s expansion by repeated differentiation. (iii) 1 2 24 6 35 7 35 7 1 1sin d3 9 15 d3 54 648 34992 15 3 162 3240 244944 Note that 0 when 0. 15Thus, sin 3 3 162 3240 244944 x cx x xx x x xx xx cx x xx xx − − += − ≈ ++ + = ++ + = = ≈+ + + ∫ ∫

7 | Page Question 9 No. Suggested Solution Remarks for Student 1 : 2 12, 3 p  ⋅=− r 12 : 1 1 , 34 l λλ    = −+ − ∈    r  (i) 12 5 1 2 1 10 5 2 34 5 1 −       ×− =− = −      −−    11 1 : 2 1 2 123 0 1 31 q −−   ⋅= − ⋅= − − + =    r Therefore the Cartesian equation of q is 20x yz−+ += (ii) Using GC, 64 : 3 1 , 02 m µµ    = +∈    r  (iii) 64 3 2 OB µ µ µ + = +   64 1 54 31 4 23 23 AB µµ µµ µµ +− +     = ++ = +    −−     ( ) ( ) ( ) 2 22 2 Let 5 4 4 2 3y AB µ µµ= =+ ++ + −   ( ) ( ) ( )d 85 4 24 42 3 0d 40 32 8 2 8 12 0 42 36 6 7 y µ µµµ µ µµ µ µ = + + ++ −= ⇒ + ++ + − = ⇒= − ∴= − 2 2 d6 32 2 8 42 0 gives the min value of .d7 y yµµ = ++= > ⇒ = − Coordinates of required point 6 6 6 18 15 126 4 ,3 ,2 , ,777 7 7 7    =+− − − = −     

8 | Page Question 10 No. Suggested Solution Remarks for Student ( ) 2d 1d x k xxt = +− (i) 1 2x= , d1 d4 x t =− when 0t = , we have 1 11 5 11 (S hown)4 24 4 5k kk−= +− = ⇒= − (ii) ( ) 22 2 2 11 11 144 51 42 xx x x xx x +− = − − + = − −+− + = −−  2 2 d 15 1 1 1 d dd 54 2 551 42 x x xtt x  = − −− ⇒ = −    −− ∫∫ 51 11 22ln5 5152 222 51 1 11122ln , since 0 0222551 22 x tc x x cx x x +−−= + −−    +− = + ≤ ≤ ⇔− ≤ − ≤  −+ Thus, 2 515ln 2 51 xtc x  +−=−+ −+ + 1 2x= , d1 d4 x t =− when 0t = , we have ( )1 510 5 ln 5 ln 1 0 1 51 cc +−= − +⇒= =−+ + 2 515ln 2 51 xt x  +−∴= − −+ +

9 | Page (iii) (a) 1 511 25 125 ln 5 ln14 2 5 1512 xt  +−  += ⇒= − =   −−+ + (iii) (b) 510 5 ln 2.152 min (3 d.p.) 51 xt  −=⇒= − = + (iv) ( ) ( ) ( ) 1 5 1 5 11 55 1 5 1 5 2 515 ln 2 51 2 51e 2 51 e 25 1 25 1 2 2e e 5 1 1 5 e 51 1 5 2 2e t t tt t t xt x x x xx x x − − −− − −  +−=− −+ + +−= −+ + − ++ =+−  + = + +− + +− ∴= + x 2.152 t 0

10 | Page Question 11 No. Suggested Solution Remarks for Student (i) 22 2 24 16hr h r+ = ⇒= − ( ) 23 2 23 22 1 14 3 23 12 1633 1 16 23 V rh r rrr r rr ππ ππ π = +  = −+ = −+ ( ) ( ) 22 2 d2 1 116 2 2 2d3 3 2 16 V r rr r rr r ππ = −+ + −+ − When 1 d, 0d Vrr r= = ( ) ( ) ( ) ( ) ( ) ( ) 22 2 2 2 2 2 22 22 2 24 2 2 44 2 42 2 11 16 2 2 2 033 2 16 2 16 4 2 0 16 2 16 6 16 0 6 16 3 32 36 16 9 192 1024 576 36 9 192 1024 45 768 1024 0 (Shown) r rr r r r rrr r r r r rr r rr r rr r r rr r rr ππ −+ + −+= − −+− += − − + − −= −=− −=− + −= − + − += (ii) 4245 768 1024 0rr− += From GC, 1.20742 or 3.950797 =1.207 (3dp) or 3.951 (3dp) r ≈≈ (iii) ( ) ( ) 22 2 d2 1 116 2 2 2d3 3 2 16