IJC JC2 H2 Maths 2012 Solutions Paper 1

Qn 1(i) Let the number of units of sand, stone and brick re quired by the company be x, y and z respectively. 15 10.5 8.1 205.2 x y z + + = 11 17.3 7 229.4 x y z + + = 12 13 10 208 x y z + + = From GC, 7, 8, 2 x y z = = = . The number of units of sand, stone and brick requir ed is 7, 8 and 2 units respectively. 1(ii) Total amount that the company must pay ( ) ( ) ( )$0.9 11 7 10.5 8 7 2 $157.50 = + +     = 2(ii) 2 2 sin d y x x dx = ( cos ) cos dy x x x dx dx = − − − ∫ cos sin x x x C = − + + cos sin x x x C = − + + sin ( 1)(sin ) cos y x x x dx x Cx D = − − − − + + ∫ sin 2 cos y x x x Cx D = − − + + Given (0) 0, '(0) 3 f f = = ( )f passes through the origin (0) 0 2 y x f D = ⇒ = ⇒ = '(0) 3; 3 f C = = sin 2 cos 3 2 y x x x x = − − + + 3 Let Pn be the statement 3 2 2 1 1 ( 1) 4 n r r n n = = + ∑ for n + ∈ Z . Prove that P1 is true, i.e. LHS: 1 3 3 1 (1) 1 r r = = = ∑ RHS: 2 2 1 (1) (1 1) 1 4 + = P1 is true. Assume Pk is true, i.e. 3 2 2 1 1 ( 1) 4 k r r k k = = + ∑ for k + ∈ Z . Prove that Pk+1 is true, i.e. 1 3 2 2 1 1 ( 1) ( 2) 4 k r r k k + = = + + ∑ for k + ∈ Z . LHS: 1 3 3 3 1 1 ( 1) k k r r r r k + = = = + + ∑ ∑

2 2 3 2 2 2 2 2 2 1 ( 1) ( 1) 4 1 ( 1) 4( 1) 4 1 ( 1) 4 4) 4 1 ( 1) ( 2) 4 k k k k k k k k k k k RHS = + + + = + + +     = + + +     = + + = Thus, kP is true ⇒ 1kP + is true Since 1P is true and kP is true ⇒ 1kP + is true, by mathematical induction, nP is true for all n + ∈ /Zbb. (Shown) 3 3 ln(2 ) ln 2 ln r rv a r a = = + ( ) ( ) ( ) 1 3 1 3 3 3 3 4 2 2 2( 1) 2( 1) ln 2 ln ln 2 (ln )(1 2 3 ... 4 ) 1ln 2 (ln ) ( 1) 4 4 ln16 ln( ) 4 ln16 ( ) 4 n n r r n r n n n n S v r a n a n a n n n a n a proven = = + + = ∑ = + ∑ = + + + + +   = + +     = + = 4(i) ( ) 2 2 d d 2 4 66 d d xy y x x x − + = d d 4 8 0 d d y y x y y x x x + − + = ( )d 4 8 d y x y x y x − = − − d 8 d 4 y x y x y x += − For tangent parallel to y-axis, 4 0 y x − = 4x y =

Substitute 4x y = into equation of curve, ( ) ( ) 224 2 4 4 66 y y y y − + = 266 66 y = 2 1y = 1y = ± . When 1y = , 4x = When 1y = − , 4x = − Coordinates are ( ) ( )4,1 , 4,1 − 4(ii) Substitute y k = into equation of the curve, 2 2 2 4 66 kx k x − + = ( ) 2 2 4 2 66 0 x kx k + + − − = Considering the discriminant, ( ) 2 2 2 4(4) 2 66 33 1056 0 for all real values of k k k k − − − = + > The line y k = cuts the curve for all real values of k. 5 From the GC, ln(2 9) y x = + intersects 2 10 y x = − at x = – 2.9539, 1.8760 Hence for 2 ln(2 9) 10 x x + ≥ − , 10 2.9539 1.8760 10 3.16 2.95 1.88 3.16 x or x x or x − ≤ ≤ − ≤ ≤ − ≤ ≤ − ≤ ≤ Using previous result to solve for 2 ln(2 9) 10 x x + = − , 10 2.9539( ) 1.8760 10 x rej or x − ≤ ≤ − ≤ ≤ 3.16 1.88 1.88 3.16 x or x − ≤ ≤ − ≤ ≤

6(i) Since a and b are the radiuses of a circle, | a| = | b|=2 2 10 2a b r   ⋅ = = −     o a b cos12 Let AN:NB = k:(1 – k), hence (1 ) a b ON k k = + − /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp Since ON is perpendicular to OB , ( ) 2 2 2 . 0 (1 ) 0 . (1 ) 0 1 (1 ) 0 2 1 (1 ) 0 2 2 3 ON OB k k k k kr k r k k k = ⋅ + − = + − =   − + − =     − + − = = /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp b a b a b b Hence, 1 (2 ) 3 a+b ON = /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp Alternatively, Use of geometry (various method) Area of OAN △ 1 1 (2 ) 3 3 1 26 1 6 = × + = × + × = × a a b a a a b a b 6(ii) tan 30 3 ON r rON = = ∘ 3 3 (2 ) 3 ON OC r r = −       = − /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp /combarrowextender/combarrowextender/combarrowextender /arrowrightnosp a+b

7(a) 7(i) R z ≡ − 7(ii) (2 2cos )(2 2sin ) Area θ θ = × × 16sin cos θ θ = 8sin 2 θ= 7(b) 3 3 3(1 ) i i e e θ θ + = + 2 2 2 3 ( ) i i i e e e θ θ θ − = + 23 (2cos ) 2 i e θ θ= 26 cos 2 i e θ θ= 26 cos (1 ) 2 2 i ae i θ θ = + 2 4 6 cos 2 2 2 i i ae e θ π θ   =     2 4 6 cos 2 i i e ae θ π θ = 2 4 6cos , 2 i i a e e θ π θ = = 2 πθ = , 3 2 a = 8(i) ( )d 2 e e ln 1 2 d 1 2 x x y xx x − − −  = − −   −  ( ) 1d 2e 1 2 d xy x y x −−= − − − ( ) ( )d1 2 2e 1 2 d xyx y x x −− = − − − (shown) Differentiating with respect to x, ( ) ( ) ( ) ( )( ) 2 2 2 2 d d d 1 2 2 2e 2 1 2 d d d d d 1 2 2 2e 2e 1 2 2 d d x x x y y y x y x x x x y y x x y y x x − − − − − = + − − − = + − − − − +

( ) ( ) 2 2 d d 1 2 2 4e 3 2 d d xy y x x y x x −− = + + − (shown) 8(ii) Differentiating with respect to x, ( ) ( ) 3 2 2 3 2 2 d d d d 1 2 2 2 4e 3 2 2 d d d d xy y y y x x y x x x x −− − = − + − − When 2 3 2 3 d d d 0, 0, 2, 0, 10 d d d y y y x y x x x = = = − = = − . 352 ... 3y x x ∴ = − − + 8(iii) 2 3 3 2 81 ... 2 2 ... 2 6 3 x x x y x x x    = − + − + − − − +       2 3 2 3 3 82 2 2 2 ... 3y x x x x x x = − − − + + − + 352 ... 3y x x = − − + (verified). 9 (i) (ii) 3 0 2 3 1 d 2 sin d 8.38 (3 s.f.) y x t t π π π − = − = ∫ ∫ (iii) 1 cos 2 x t− = ( ) 2 2 2 1cos 2 xt −= , 2 2 sin t y = Since 2 2 sin cos 1 t t + = ( ) 2 2 2 1 12 x y− + = (iv) Stretch parallel to the y axis with stretch factor 2. Translate -1 unit in the direction of the x axis. 10 (a)(i) ( , ) hR a = ∞ gD = [ , ) a ∞ Since h g R D ⊆ , gh exists.

(ii) 2gh( ) x x a a = + − 2gh( ) x x = gh( ) x x = gh( ) since 0 x x x = − < gh : 0 x x x − < /arrowbarright 10(b) (i) 1f ( ) , 1 2sin 2 x x x= < < + π π ( ) 2 2cos f '( ) 1 2sin xx x −= + ( ) 2 Since 1 2sin 0 x+ > and 0 2cos 2 x< − < for 2 xπ π< < , ( ) 2 2cos f '( ) 0 1 2sin xx x −= > + for 2 xπ π< < Thus f ( ) x is an increasing function. (Shown) (ii) 1 1 2sin y x= + 11 2sin x y+ = 1 1 1 sin 1 2x y −     = −         -1 1 1 1 f ( ) sin 1 2x x −     = −         1 ff 1D R ,1 3 −   = =     11(a) (i) M 1 = {5}, M2 = {10, 15} and M3 = {20, 25, 30}… Mn has n elements Hence, the number of elements in each set up to and including M n is an AP with a = 1, d = 1 and n number of terms [ ]2(1) ( 1)(1) ( 1) 2 2 n n n S n n = + − = + As the elements of the sets are multiples of 5, Last element of M n = 55 ( 1) ( 1) 2 2 n n n n × + = + (Proven) 1 f

(a)(ii) The elements in Mn+1 follow an AP with first term = 5 ( 1) 5 2 n n + + , common difference = 5, and n+1 number of terms. Hence, sum of all the elements in Mn+1 ( ) 2 2 1 5 ( 1) 5( 1)( 2) 52 2 2 1 5( 1) 2 2 5 2 2 5 5 ( 1)( 2 2) or ( 1) ( 1) 1 2 2 n n n n n n n n n n n n n + + + +   = + +     + +   = + +     = + + + + + +     (b)(i) It is a GP with r = 1.15 3 3 6 6 (1.15) 150 150 98.627 1.15 98.627(1.15 1) 863.35 1.15 1 a a S = = = −= = − Hence, David will take 863s (or14 mins 23s) to run 2.4km. (b)(ii) The time taken for Tommy to run each round is equivalent to a GP with a = 100, r = 1.1 and n = 6. Time taken for Tommy to run 2.4km = 6 110(1.1 1) 848.72 1.1 1 − =− Since 848.72 + 60 > 863.35 , Tommy will not complete the 2.4km run before David. 12(i) &(ii) 1 12 5 iz i − − = 12+ 5 (12- ) 5 i z i z i − = ⇒ − =

5tan , 0.39491 12 θ θ = ⇒ = 12sin(0.39491) 4.6167 BC BC = ⇒ = 12 cos(0.39491) 11.0764 AC AC = ⇒ = 1 (12 ) (12 ) z BC i AC = − + − 1 7.38 0.924 z i = + 2 (12 ) (12 ) z BC i AC = + + − 2 16.6 0.924 z i = + Alternatively, ( ) ( ) ( ) ( ) 2 2 2 2 2 2 12 12 12 (1) 12 1 5 (2) x y x y − + − = − − − + + = − − ( ) ( ) 2 2 (1) (2) : 12 1 144 25 12 13 y y y − − − + = − = 12 su