SAJC P2 ans

1 Section A: Pure Mathematics (40 marks) 1(i) ( ) 2 19 22 2 1244 1 xxy xx =−++ += + Vertical Asymptote: 1 2x=− Oblique asymptote: 1 24 xy= − Intersection with axes: x = 0, y = 2  (0 , 2) No intersection with x-axis From G.C. Turning points: Min (1 , 1), Max ( )22,−− y x (0 , 2) (1 , 1) O

2 (ii) Substitute x = 0 into 1 24 xy= − 1 4y=− P 4, 10 −   For 41y mx= − , 14 4 1 LHS RHS −=  =− = when x = 0 Therefore, P lies on 41y mx= − for all real values of m. (iii) ( )( ) 2 2 4 8 12 1 21 214 4 x mx x xm xx += − + + = −+ Since 02 m<< , 10 42 m<< , 1 44 myx= − will not intersect with the graph of 2 2 21 xy x= + + . Hence, there is no real root for the equation ( )( ) 24 8 12 1x mx x+= − + . (iv) Since there are no real roots for the equation, ( )( ) 24 8 12 1x mx x+= − + , the two roots are complex roots. Since all the coefficients in the equation are real, the roots must occur in complex conjugate pairs, i.e. ix ab= + or ix ab= − , ,a b∈  .

3 2(a) Equation of the line : λλ= +∈r a u,  is parallel to λλ−= ∈ ⇒− r a u, r a u  ( )−×=ra u 0 (b) ( ) ( ) 2 2 22 22 2 2 2 22 7 () () 2 || 2 || 2 || 742 2 3 3 02 3 − =−⋅− =⋅− ⋅+⋅ − = − ⋅+ = − = ⋅+ = −⋅ ⋅ − − ≠⋅= ba ba ba bb ab aa b a b ab a ab a a a ab ab a aa a b a is not perpendicular to OA OB⇒     Hence, AOB∠ is not a right angle. Therefore, points O, A and B are not points that lie on the circumference of a circle with diameter AB. 3(i) d d1 y xy x xy −= −+ --- (1) Let 1uxy=−+ Differentiating with respect to x: dd 1dd dd 1dd uy xx yu xx = − ⇒= −

4 From equation (1): d1 d u x− = 1u u − d1 d u xu= --- (2) ( ) ( ) 2 2 2 dd 2 ( 1) 2 1 22 1 2 ,where =2 uu x u xA xy xA xy x A y x xC C A = = + −+ = + −+ = + =+± + ∫∫ A and C are arbitrary constants.

5 (ii) (iii) When 1.yx>+ 2 23 d1 d ( 1) y x xy= −+ < 0. The part of the solution graph above the line y = x + 1 is concave downwards. 4 (i) 4 22 2 22 4 ( 2) 4 = ( 2 2 )( 2 2 ) (Shown) xx x x xx x += + − ++ +− Using partial fractions 22 1 ( 2 2)( 2 2)xx xx++ −+ = 22 22 22 ax b cx d xx xx ++ +++ −+ Multiplying across by 22( 2 2) ( 2 2)x xx x++ +− , y x O y = x + 1 (0 ,1)

6 22 032 1 ( )( 2 2) ( )( 2 2) Comparing coefficients of , , and respec tively in sequence, 2 2 1 (1) 0 (2) 2 2 0 (3) 2 2 2 2 0 (4) ax b x x cx d x x xxx x bd ac ab cd abcd = + −++ + ++ + = −−− + = −−− − + + + = −−− − + + = −−− Solving using GC, 11 1 1,, ,84 8 4abc d=== −= . 224 11 ( 2 2)( 2 2)4x xx xx= ++ −++ = 22 22 8( 2 2) 8( 2 2) xx xx xx +− −++ −+ (ii) ∴ 1 4 0 1 d4 xx   +∫ = 1 220 12 2 d8 22 22 xx xxx xx +−   −  ++ −+  ∫ 11 220 0 11 22 00 1121 00 1 2 11 2 22 d d8 22 8 2 22 1 22 1 2 =d d16 2 2 16 ( 1) 1 11 = ln( 2 2) tan ( 1)16 8 xx xxxx xx x xxxx x xx x− + ++ =  ++ ++  + +++ ++   ++ + +   ∫∫ ∫∫ ( ) 111 = ln5 ln2 tan 216 8 32 −−+ − π

7 11 220 0 11 22 00 1121 00 1 2 11 2 2 2 d d8 22 8 2 22 1 22 1 2 =d d16 2 2 16 ( 1) 1 11 = ln( 2 2) tan ( 1)16 8 xx xxxx xx x xxxx x xx x− − −− =  −+ −+  − −−+ −+   −+ − −   ∫∫ ∫∫ ( )1 = ln 216 32−− π 1 220 12 2 d8 ( 1) 1 2 2 xx xx xx  +− − ++ −+ ∫ = ( ) 111 ln5 ln2 tan 216 8 32 π− −+ − - ( )1 ln216 32 π −− ( ) ( ) 111ln 5 tan 216 8 −= + (Ans)

8 5 (i) Let 2 2yx x= +− Since 2 20xx+−≤ when 01 x<< , ( ) 2 2 2 19 24 y xx x = − +− = −+ + 2 19 24 19 24 91 42 xy xy xy ± ±  += − += − = −− y = f(x) y x O (0, 2) (1 , 0)

9 91 42xy= −− since 01 x<< 1 91f () 42xx−∴ = −− (ii) ( ) ( ) ( ) -1 -1 -1 -1 -1 fg f ff g ff gfDR RD R 0 ,01 0, 2 , =  → =  →  →  → ∞ ( )1gfR0 ,− = ∞ . y O y = g(x) (6, ln 3)− x (0, ln 3)− (3, ln 3)− 3− 3 6

10 (iii) Area required [ ] ( ) 5.5 4 2.5 1 g( ) d ln d 0.79073 0.791 (to 3 sf, by GC) xx xx =− = −− = = ∫ ∫ Section B: Statistics (60 marks) 6 (a) The researchers obtained the sampling frame from the 30 000 patients who volunteered for the study. Step 1: Number the members of the volunteers, say from 1 to 30,000. Step 2: Randomly select 500 of these members by generating 500 distinct random numbers and then selecting the corresponding members to form the required sample (b) Random selection means that the subjects in the sample were assigned to a treatment at random, without bias. Or The selection of patients to assign treatment to is free of bias. 7(i) 1. The condition of one semi-conductor chip is independent of the condition of any other semi-conductor chips.