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− + ++ =  ++ ++  + +++ ++   ++ + +  