RI 9758 2023 Prelim P1 Solution

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 2023 Year 6 H2 Mathematics Preliminary Examination Paper 1: Solutions 1 Solution [4] ( )( ) 95 ,111 1 x xx xx +< ≠±−+ + ( )( ) ( )( ) 9 51 011 xx xx −+ − <−+ ( )( ) 2 44 011 xx xx ++ <−+ ( ) ( )( ) ( ) 2 21 1 0 *x xx+ − +<  Method 1: 2 or 2 1 or 1x xx∴ <− − < <− > Method 2: Since ( ) 2 2 0 for ,xx+≥ ∈  (*) is equivalent to ( )( )11 0xx− +< and 2x≠− 1 or 1 and 2xxx∴ <− > ≠− 2 Solution [4] Let $x, $y and $z denote the usual selling price of a small, medium and large bag of Griffles popcorn respectively. To receive a total of 3 small, 7 medium and 1 large bag of Griffles popcorn, Beatrice bought 2 small, 5 medium and 1 large bag of popcorn. ( )0.95 3 7 85.5 3 7 90 (1)x yz x yz++= ⇒++ =  2 5 85.5 18.50 2 5 67 (2)x yz x yz+ += − ⇒ + +=  2.4 2.4 0 (3)z x xz= ⇒ −=  On solving, 5, 9 and 12xy z= = = The usual selling price of a small, medium and large bag of Griffles popcorn is $5, $9 and $12 respectively. x

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 3 Solution (a) [2] We have 2 dd48 dd ArAr r ttππ= ⇒= . Hence dd100 8 (5) 0.796 cm/sdd rr ttπ−= ⇒= − Alternative 2 d48 dr AAr r ππ= ⇒= ddd d dd 1 100 8 (5) 5 2 r Ar t tA π π = × = −× =− Hence the radius is decreasing at 5 cm/s2π . (b) [2] 324d dVolume of meteorite, 43d d VrVr r ttππ= ⇒= Since V decreases with t, we have d d V kAt =− for proportionality constant 0k > . This means that ( ) 22dd44 dd rrr kr kttππ = − ⇒= − , which is a negative constant. Hence the radius is decreasing at a constant rate. Alternative We have 2d (4 )d V kA k rt π= −= − for proportionality constant 0k > . 324d 43d VVr r rππ= ⇒= 2 2 dd d d dd 1 (4 ) 4 0 rVr t tV kr r k π π = × = −× = −< Since k is a negative constant, thus the radius is decreasing at a constant rate.

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 4 Solution (a) [3] Method 1 Reflect the graph of ( )fyx= about the x-axis, followed by scaling of the resulting graph by a factor of 2 parallel to the y -axis, and translating the resulting graph by 1 unit in the positive y-direction. Method 2 Translate the graph of ( )fyx= by 1 2 unit in the negative y-direction, followed by reflecting the resulting graph about the x-axis, and scaling by a factor of 2 parallel to the y-axis. Method 3: Reflect the graph of ( )fyx= about the x-axis, followed by translating the resulting graph by 1 2 unit in the positive y-direction, and scaling by a factor of 2 parallel to the y-axis. Method 4: Scale the graph of ( )fyx= by a factor of 2 parallel to the y-axis, followed by translating the resulting graph by 1 unit in the negative y-direction and then reflecting about the x-axis. (b) [2] A point is R-invariant if ( ) 21,, 1 1ab a b b b = ⇔ =⇔= ± . Hence if there are no R-invariant points, the graph must not intersect the lines 1y=± . Since f() 0x′ < , the graph is strictly decreasing for all real values of x and some possible graphs are thus x y −1 0

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ x y 1 −1 x y 1 0 0

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 5 Solution (a) [3] 32 1 32 ( 3) 2 2 3 xy x yx y x yx y yx y += − −= + −= + += − ( ) 1 2f 3 xx x − += − ( )1 ffD R 3,− = = ∞ OR: 32 5 311 xy xx += = +−− 53 1 51 3 51 3 y x x y x y −= − −= − = + − ( ) 1 5f1 3x x − = + − , ( )1 ffD R 3,− = = ∞ (b) [5] ( ) ( ) 1From graph, f f intersects at x x yx−= = x y

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ ( ) 2 f 32 1 4 20 xx x xx xx = + =− − −= 4 24 2 26 x ±= = ± Since 3x> , 26x= + OR: ( ) ( ) ( )( ) ( )( ) 1 2 2 ff 32 2 13 32 3 2 1 2 8 40 4 20 xx xx xx xx xx xx xx −= ++ =−− +− = +− − −= − −=

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 6 Solution (a) [5] ( )d 32 for some constant 0d 1 d 1 d32 kkt kt θ θ θθ = −− > =−−∫∫ ( ) Since pie is cooling down, 32. ln 32 kt C θ θ ≥ − = −+ ( )ln 32 32 e , where ekt C kt C AA θ θ − − = −+ = += Alternative: ln 32 32 e 32 e , where e kt C kt C kt C AA θ θ θ −+ − − = −+ −= ± = += ± When 0, =200t θ= ⇒ 200 32 168AA= +⇒= When 15, =180t θ= ⇒ 15180 32 168e k−= + 115 15 37 37ee 42 42 kk−− =⇒=  153732 168 42 t θ ∴= +  (b) [2] (c) [2] From GC, solution of 153760 32 168 42 t = +  is t = 212.04 (5sf) = 212 mins (nearest min), equivalent to 3h 32 mins. Alternatively, ( ) ( ) 15 115ln37 660 32 168 212.043742 ln 42 t t= + ⇒= ≈ =212 mins (nearest min), equivalent to 3h 32 mins. To safely store the pie, t < 212 mins. Latest time to keep the pie is 4.32 pm.

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 7 Solution (a) [4] ( )( ) 1 21 21 21 21 AB rr r r = +−+ −+ 11 and 22AB= =− . ( )( )1 1 2 12 1 n r rr= −+∑ 1 111 2 2 12 1 n r rr= = − −+∑ 11 13 11 35 111 572 11 23 21 11 21 21 nn nn  −   +−   +−=   + +− −− +− −+  1112 21 n = − + (b) [2] Since 1 0 21n →+ when n→∞ , the series converges and the sum to infinity is 1 2. (c) [4] ( )( )4 1 23 25 n r rr= ++∑ ( ) ( ) 2 6 1 (replace with 2)2 23 2 25 n m rmmm + = = −−+ −+     ∑ [ ][ ] 2 6 1 2 12 1 n m mm + = = −+∑ [ ][ ] [ ][ ] 25 11 11 2 12 1 2 12 1 n mm mm mm + = = = −−+ −+∑∑ ( ) 1 1 1111222 1 22 ( 5 ) 1n  =− −− ++ +  11 22 4 10n= − +

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ 8 Solution (a) [4] d1e e d xx uu x= −⇒ = − ( ) 22 1e 3 3 1e1 e1 dd 1e x x xu u − − −= − ⌠ ⌠ ⌡⌡ ( ) ( ) 21e 2 1e 2 22 1 2 1 1e 1e2 u − − − − − = = − −− Otherwise, ( ) ( ) ( ) ( ) ( ) 2 2 3 3 1 1 22 1 2 22 e d ( e)1 e d 1e 1e 2 1 1e 1e2 x xx x x xx − − − − = −− − −  −= − − = − −− ⌠ ⌡ ∫ (b) [3] 11 111 200 0 1 2 0 tan d tan d1 1 = ln(1 )42 1 = ln 242 xxx x x xx xπ π −− = − + −+ − ⌠⌡∫ (c) [3] −2 −1 0 1 2 4 π 4 π−

Raffles Institution H2 Mathematics 2023 Year 6 _____________________________________________________________________________________________ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 11 1 3221 1 22 1 1 33 1 11 2 1 1 30 1 2 22 2 22 efd fd t a n d d 1e ee d tan d d 1e 1e e2 tan d 2 d 1e 112 l n 2 2 1e 1e42 2 l n 2 1e 1e2 x x xx xx x x xx xx x x x x xx x xx x π π − − −−− − − − − − − − = ++ − = ++ −− = − −   = − − − −−    =− −− +− ⌠   ⌡ ⌠⌠   ⌡⌡ ⌠ ⌡ ∫∫∫ ∫ ∫