CJC 2020 Prelims P1 Ans

1 2020 JC2 H2 PRELIMS Paper 1’s Suggested Solutions 1: (i) 23f( ) 35 21 3 3(3 5) 1 2 3 33 5 xx x x x += + = − + − = + + (ii) y = 1 x → 11 1 99y xx = = : scale by a factor of 1 9 parallel to the y-axis Or scale by a factor of 1 9 parallel to the x-axis → 1 59 3 y x =  + : translate by 5 3 in the negative x-direction → 1 59 3 y x =−  + : reflect about the x-axis → 21 53 9 3 y x = −  + : translate by 2 3 in the positive y-direction

2 2. (i) i 41 i 2e w π =+= i 8 i 4 1 i 842 2e * 2e 2e n nn n n z w π π ππ   −   +−   = = (ii) For * nz w to be real and negative, arg 2 , where * nz kkw ππ= ±∈  2 84 1 1284 6 16 Smallest 6, 22 n k n k nk n ππ ππ+=± += ± = ± =

3 Q3: Since all coefficients are real, 1iz= −− is also a root. A quadratic factor ( ) ( ) [ ][ ] ( ) ( ) 2 1i 1i 1i 1i 1i zz zz z = −−+ −−−     = +− ++ =+− 2 22zz=++ ( )( ) 32 22 5 2 22z z p zq z z zd+ + += + + + ( ) ( ) 32 32 2 32 25 244 22 2 4 42 2 z z pz q z z z dz dz d z dz dz d + + += + + + + + = ++ ++ + Comparing the coefficients of 2z , z and constant terms, 45 1dd+=⇒= 42 6dp p+= ⇔ = 2 2dq q= ⇔= Alternative: ( ) ( ) ( ) ( ) ( ) 32 2 1i 5 1i 1i 0 4 6i i 0 4 i 6 0 i0 pq pp q pq p −+ + −+ + −+ + = −−+ += − ++ −= + Comparing the real and imaginary parts, 40 pq−+= 6p= 4qp= − 2q= Last factor is ( )2zd+ = ( )21z+ . Therefore, the real root is 1 2− . All the other roots are 11 i, 2z= −± − .

4 Q4: (i) Horizontal asymptote (y = 0) and y-intercept 50, 2   (ii) Since every horizontal line cuts the graph at most once. Therefore, f is a 1-1 function and hence f has an inverse function. When x < 2, let y = 5 2 x− 5 2xy= − 1 5f () 2xx− = − , 1 2x> When x ≥ 2, let 1y x= 1x y= 1 1f () x x − = , 10 2x<≤ Hence, 1 1 1, ,0 2f () 15 ,, 22 xxxx xxx −  ∈ <≤=  ∈>−   (iii) Since the range of f -1 =  or ( ),−∞ ∞ ⊄ of Domain of g =[2, )∞ , gf -1 does not exist.

5 Q5: 2 32 3 81002 2 16200 2 8100 3 1 8100 3 A r rh h r r V rr r r rr ππ π ππ π π = + = ⇒= − = +−  = −+ 2d 8100 d V rr π= −+ At maximum V, d 0d V r = . 2 2 2 8100 0 8100 8100 8100, since 0 r r r rr π π π π −+ = = = = > Maximum ( ) 1 8100 8100 810081003 8100 1 8100 81003 90 5400 486000 V π ππ π πππ π π =−+ =−+  = = 2 2 d 20d V r π= −< Hence, V is maximum.

6 Q6: (i) 22 22 2 2 2 125 9 125 9 25 1 9 51 9 yx yx xy xy −= = + = + = ±+ Asymptotes 552, 2, 2, and 33x x y yx y x= −=== = − Turning point/Intercept ( )9 40,− Vertices ( ) ( )0,5 , 0, 5− ( ) ( ) ( ) ( )1.31, 5.46 , 1.31, 5.46 , 2.75,6.78 , 2.75,6.78−− − − (ii) From the graph, 2.75x≤− or 22 x−<< or 2.75x≥ O x y

7 Q7: (i) 32 2 23y y yx x++=− 2 d dd3 2 2 23d dd y yyyy xx xx+ += − 2 22 2 2 22 ddd dd dd3 6 2 22 2ddd dd dd yy y yy y yy yyxxx xx x x+ + + += 222 22 2 2 22 d d ddd3 6 2 2 22d d ddd y y yy yyy yx x xxx  + + + +=   When 0x= , 32 20yy y++= ( ) 2 2 2 20 11 2024 17 024 yy y yy yy ++ =  + −+ =  + += 2 170 since 024yy ⇒= + + ≠ When 0x= , 2 d dd3 223d dd y yyyy x xx+ += − d23d d3 d2 y x y x =− =− When 0x= , 222 22 2 2 22 d d ddd3 6 2 2 22d d ddd y y yy yyy yx x xxx  + + + +=   2 2 2 2 2 2 2 2 2 2 dd2 22dd 3d 12d d9 1d4 d5 d4 yy xx y x y x y x  += −+=  = − =− d3 d2 y x =− and 2 2 d5 d4 y x =− .

8 2 5 3 40 ...2 2!y xx −= +− + + 235 ...28y xx= −− + (ii) 1 11 22 1 2 yy =+ + 11 111122 2 2 yy −−   + = −−       1 2 1 1 ...2 22 1 122 y yy −   = +− +− +      +    1 2 .1 12 ..82 1 24 y yy − =+ ++  − 2 22 1 .1 1 135 135 ... ... ..24 22 8 82812 xx xxy −  = ++ −−− + −− + +   22 1 22 1 1 3 5 19 ... ...2 4 2 8 84 13 5 9 ...1 122 2 8 32 32 xx x xx xy −   = −−− ++ +     = +++  + + 2 1 13 7 ...1 2 8 16122 y xx − = ++ +  + 

9 Q8: (i) 23xt= , 6yt=− d d 6x t t= , d d 6y t =− , d d d ddd 6 1 6 y y x ttx tt − = = =− At point P, gradient of tangent = 1 p− . Equation of the tangent to curve C at P : 21(6) ( 3 ) py p xp−− = − − , 1 36py xpp= −+− , 1 3py xp= −− . At point P, gradient of normal = 11 gradient of tangent 1 p p − − = −= . Equation of the normal to curve C at P : 2(6) ( 3 )y p px p−− = − , 336y px p p= −− . (ii) When the tangent at P meets the x-axis, y = 0 103 p xp= −− , 1 3p xp=− , 23xp=− 2 ( 3 , 0)Tp⇒− When the normal at P meets the x-axis, y = 0 30 36px p p= −− , 336px p p= + 0p≠ , 236xp= + 2 (3 6, 0)Np⇒+ ( )( ) 221 2 3 3 6 , 0 (3, 0), which is independent of (sho wn) M pp p ∴=− + + = (iii) Proof : For any value of p where 01 p<< , let the acute angle PN makes with the x-axis be θ, and let the acute angle PM makes with the x-axis be α.

10 2 2 Length (3 6) 3 33 MN p p = +− = + ( ) ( ) ( ) ( ) ( ) 2 22 24 2 24 22 22 Length 3 3 0 ( 6 ) 9 18 9 36 9 18 9 33 33 33 0 PM p p pp p pp p pp = − + −− = −++ = ++ = + = + +≥  Length Length PM MN=∴ MPN∆ is an isosceles triangle. ⇒ MPN MNP θ= = The acute angle α that PM makes with the x-axis (ext. of = sum of opp. int. s) 2 (proven). MPN MNP θθ θ = +∆ = + =   