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