H2_Math_Paper_2_Solution

1 Section A: Pure Mathematics [40 marks] 1 Let x be the distance between A and P. 22 286 4PW x x=+ =+ Cost of laying the pipe, () 260000 64 45000 12Cxx= ++ − d 0d C x = 2 60000 45000 0 64 x x −= + () 2 2 22 2 60000 45000 64 43 6 4 16 9 64 576 7 576 576 (reject as 0)77 xx xx xx x xx =+ =+ =+ = =− > To show that C is minimum when 576 7x= : Method 1: First Derivative Test x 576 7x − ⎛⎞=⎜⎟⎜⎟⎝⎠ 576 7x= 576 7x + ⎛⎞=⎜⎟⎜⎟⎝⎠ 2 d 60000 45000d 64 Cx x x =− + < 0 0 > 0 ∴C is a minimum when 576 7x= . Land W P B A 8 Ocean x 12 – x

2 Method 2: Second Derivative Test () () () 2 2 2 2 22 22 33 22 22 6000060000 64 d 64 d6 4 60000 64 60000 3840000 0 for all 64 64 xx C x xx xx x xx +− += + +− == > ++ ∴C is a minimum when 576 7x= . 2 (i) t y d d = k(10 – y) 1 d d10 yk ty =−∫∫ – ln |10 – y| = kt + c |10 – y| = e kt c−− 10 – y = e kt c−−± 10 – y = ee eck t k t A−− −±= , where A = ± e c− y = 10 – ktA −e When t = 0, y = 0 ⇒ A = 10 When t = 2, y = 5 ⇒ 225 10 10e 10e 5kk−−=− ⇒ = k = 1 ln 22 ∴ y = 10 – 1 ln 2210e t⎛⎞−⎜⎟⎝⎠ (ii) The amount of material me morised tends to 10 units. (iii) t y d d = k(10 – y) – αy, α > 0 y t 10 0 y = 10 – 1 ln 2210e t⎛⎞−⎜⎟⎝⎠ y = 10

3 3 (i) ()02 i 4z−+ < and () ( )46 i 02 izz−−+ ≤ − + (ii) 1 4tan 1.10712AOC − ⎛⎞∠= = ⎜⎟⎝⎠ rad Least possible arg z = π 2 Largest possible arg z = π 2+1.1071 = 2.68 π arg 2.682 z∴ << (iii) ()() () 3arg 2 4i arg 1 i π4w−+ = − −= − ()04 26,2 , 422MM −+⎛⎞ ⇒−⎜⎟⎝⎠ Least value of zw− = CM () ( ) 22 20 42=− − +− = 22 4 (i) 2 14 1 11 4 10 5 n −−⎛⎞ ⎛ ⎞⎛⎞ ⎜⎟ ⎜ ⎟⎜⎟=× =⎜⎟ ⎜ ⎟⎜⎟⎜⎟ ⎜ ⎟⎜⎟ −⎝⎠ ⎝ ⎠⎝⎠ Acute angle between 1Π and 2Π 1 11 24 15cos 11 2 4 15 − −⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−•⎜⎟ ⎜⎟⎜⎟ ⎜⎟ −⎝⎠ ⎝⎠= −⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−⎜⎟ ⎜⎟⎜⎟ ⎜⎟ −⎝⎠ ⎝⎠ 1 14cos 64 2 − ⎛⎞= ⎜⎟⎝⎠ 28.1= D (to 1 dp) D C(0,2) y 3 4 π− 6 -2 (4 , 6 )− 4 O A (2,4) M x

4 (ii) 31 14 1 2 15 −⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟−= −⎜⎟ ⎜⎟⎜⎟ ⎜⎟ −⎝⎠ ⎝⎠ i 2Π∴ : 1 41 2 4 51 2 5 xyz −⎛⎞ ⎜⎟ =− ⇒− + − =−⎜⎟⎜⎟−⎝⎠ ri 1Π : 2 4xy z−+ = 2Π : 45 1 2xyz−+ − = − 1 43 Using GC, eqn of is 4 2 , where 01 l αα −⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟=− + ∈⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ \r . (iii) Given 6 3 5 OA ⎛⎞ ⎜⎟=⎜⎟⎜⎟−⎝⎠ JJJG and 2 3 1 OB ⎛⎞ ⎜⎟=⎜⎟⎜⎟⎝⎠ JJJG , 26 4 33 0 15 6 AB −⎛⎞⎛ ⎞⎛ ⎞ ⎜⎟⎜ ⎟⎜ ⎟=− =⎜⎟⎜ ⎟⎜ ⎟⎜⎟⎜ ⎟⎜ ⎟−⎝⎠⎝ ⎠⎝ ⎠ JJJG Length of projection of AB JJJG onto the line 1l 22 2 43 02 61 321 −⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟•⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠= ++ 6 14 −= 31 4 7= 1Π : 1 24 1 ⎛⎞ ⎜⎟−=⎜⎟⎜⎟⎝⎠ ri l2 : 3 21 3 3 p q pt q ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟=+ + −⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟−⎝⎠ ⎝ ⎠ r Let D be a point on the plane 1Π . Since 41 02 4 01 ⎛⎞ ⎛ ⎞ ⎜⎟ ⎜ ⎟−=⎜⎟ ⎜ ⎟⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ i , D is (4, 0, 0). ∴ 44 21 0 21 30 3 pp DC p p −⎛⎞ ⎛ ⎞ ⎛⎞ ⎜⎟ ⎜ ⎟ ⎜⎟=+ −=+⎜⎟ ⎜ ⎟ ⎜⎟⎜⎟ ⎜ ⎟ ⎜⎟−−⎝⎠ ⎝ ⎠ ⎝⎠ JJJG

5 Perpendicular distance from C to 1Π = Length of projection of DC JJJG onto the normal of 1Π 14 1 22 1 2 13 1 15 1 6