2024 ASRJC JC1 H2 Math Promo Exam Solutions

[Turn over 2024 JC1 Promotional Exam 1. Sub ( )7,10 => 25 121 0a b c− − = Line 1yx=− cuts the graph at 1x= Sub 1x= and 0y= into equation of hyperbola => 0abc− − = Using GC: 5 4ac= , 1 4bc= Sub into equation of hyperbola ( ) ( ) 2251 2144c x c y c− − + = Since ,,abc are positive integers ( ) ( ) 22 5 2 1 1 4xy− − + = is a possible equation of the hyperbola Where 5, 1, 4a b c= = =

2 © ASRJC 2024 2(a) 2d 3sec 3d y xx = ( ) 23 1 tan 3 x=+ ( ) 231 y=+ 2 2 dd 06dd yy yxx =+ d6 d yy x= , where A = 6 1st Alternative Method 2nd Alternative Method 2d 3sec 3d y xx = 2 2 d 6sec3 3sec3 tan3d y x x xx =   26(3sec 3 )tan3xx= d6 d yy x= , where A = 6 1 2 2 tan 3 1d 31d d 33d yx y yx y yx − = =+ =+ 2 2 dd 6dd yy yxx = , where A = 6 (b) 32 32 d d d d6 . 6 .d d d d y y y y yx x x x=+ 2 2 2 dd66 dd yy yxx =+   4 2 2 3 4 2 2 3 d d d d d d6 2 . 6. . 6d d d d d d y y y y y y yx x x x x x = + + 23 23 d d d18. . 6d d d y y y yx x x=+ , where B = 18 and C = 6 3(a) Area 1 2 0 11d (1)22xx =−  OR 11 2 10 2 d 2 1dx x x x= − − 13 0 1 34 x=− 11 34=− 1 12= units2 Alternative method

3 © ASRJC 2024 [Turn over x = 2 x = 0.5 y = 0 y x (0, 2) (4, 0) Area ( ) 1 0 1 1d2 y y y= + − 13 2 2 0 1 1 2 4 2 3 112 4 2 3 y y y= + −  = + − 1 12= units2 (b) Volume ( ) 21 22 0 11d (1)32xx =−  OR ( ) ( ) 2211 2 10 2 d 2 1 dx x x x= − − 15 0 1 56 x=−  56 =− 30 = units3 4(a) 2 4 2 3 2 xy xx −= +− = ( )( ) 4 2 1 2 x xx − − − +  Asymptotes are y = 0, x = 2, x = 1 2− (b) From the graph, 0.5 2x−   or 4x OR: 2 4 02 3 2 x xx − +−  ( )( )( )4 2 2 1x x x− − − + > 0  0.5 2x−   or 4x –0.5 2 4

4 © ASRJC 2024 (c) Replace x with x , 0.5 2 x−   or 4x  Since 0 0.5x  − for all real values of x, 2x  or 4x  22 x−   or 4x− or 4x 5(a) ca represents the perpendicular distance from the point R to the line PQ. Area of PQR = 1 2  base  height = 1 ˆ 2  a c a = 1 2   aac a = 1 2 ca units2 (Shown) (b) By replacing a with –a and c with b in part (a), area of PQR = ( )11 22  − = b a b a  1 2 ca 1 2= ba  ca =ba (Shown) OR: From the diagram, c = a + b so ( ) = + c a a b a =  + a a b a = ba since =a a 0 (Shown) ca =ba  sin sinQPR PQR = c a b a  sin sinPQR QPR= cb (Shown)

5 © ASRJC 2024 [Turn over 6(a) (b) 1. Translation of 4 units in the positive x-direction: ( ) 22 2 2 1 replace by 4 4 1y x x x y x− = − − − = 2. Scale by scale factor k parallel to the y-axis: ( ) ( ) 2 222 24 1 replace by 4 1 yyy x y x kk− − = − − = 3. Translation of 1 unit in the negative y-direction: ( ) ( ) ( ) 22 22 22 14 1 replace by 1 4 1 yy x y y xkk +− − = + − − = Alternative (manipulate y first then x) 1. Scale by scale factor k parallel to the y-axis 2. Translation of 1 unit in the negative y-direction 3. Translation of 4 units in the positive x-direction Alternative (translation before scaling) 1. Translation of 4 units in the positive x-direction 2. Translation of 1 k unit in the negative y-direction ( ) ( ) 2 222 114 1 replace by 4 1y x y y y x kk − − = + + − − =  3. Scale by scale factor k parallel to the y-axis ( ) ( ) ( ) ( ) 22 22 2 2 2 11 4 1 replace by 4 1 1 4 1 yyy x y xk k k k y xk    + − − = + − − =       + − − = (c) C1 has asymptotes 12 xy=− + and 4x= , which intersects at (4,-1) C2 is a hyperbola center at (4,-1) and has oblique asymptote 1 ( 4)y k x=−  − , For C1 and C2 to cut exactly 2 times, the gradient of oblique asymptote 1 ( 4)y k x=−  − must be greater or equals to than that of C1. 1 2k 4x= 1 2y=− y x

6 © ASRJC 2024 7(a) 2f ( ) 5a = f( 5 2 ) 5a+= 5 2 5 2 5 a+ + = 25 552 2a −+= 47.5a= (b) largest 0k= (c) Let 12 xy x= − 012 xyx x=− − 2y xy x− =− 21 yx y= − 1g ( ) 21 xx x − = − Domain of g−1 = 11,32   (d) g 11R, 32 =  f 5D, 2 = −  Since gfRD , so gf exists. (  gf 1 1 17, 1 , , 6 3 2 3 − − ⎯⎯ → ⎯⎯ →   So fg 17R , 6 3 =  

7 © ASRJC 2024 [Turn over 8(a) When 0x= 1 10 1 2cos2 2 cos 23  − = −  = =   6  = 3sin 1 132y = − = − The coordinate of the point where C cuts the y-axis is 30, 12  − . (b) d 4sin 2d x  = ; 2d cd os 2y  = s d d d 4 in 2cos 2d d 2d y y xx     = = 1 cot 22 = When 4  = , 1x= and 0y= . (1,0) When 4  = , d 0d y x = When 0→ , the gradient will tend towards + The tangent of C as 0→ will get steeper and steeper until it tends towards a vertical line. (c) When 0= , 1x=− and 1y=− . ( 1, 1) − − When 3 8  = , 31 2cos 1 24x = − = + 32sin 1 142y = − = − (d) 1 2cos2x =− and sin 2 1y =− 1cos2 2 x −= and sin 2 1 y =+ Since 22sin cos 1AA+= Then ( ) 2 21 112 x y− + + =

8 © ASRJC 2024 9(a) 2 2 d23 x xxx + +−  2 2 2 1 2 2 1 +d2 2 3 ( 1) (2) x xx x x +=  + − + −   21 1 ( 1) 2ln 2 3 + ln2 2(2) ( 1) 2 xx x C x +−= + − + ++ 21 1 1ln 2 3 + ln2 4 3 xx x C x −= + − + + Alternative Method 2 2 d23 x xxx + +−  1 3 1 d4 1 3 xxx=+ −+  31ln 1 ln 344 x x C= − + + + (b) 2 sin dx x x 2 ( cos ) 2( cos ) dx x x x= − − −  2 cos 2cos dx x x x=− +  2 cos 2sinx x x C=− + + (c) For 3sinx = d 3cosd x  = 3 22 0 9d xx− 6 2 0 d9 (3sin ) d d x    =−    ( ) 26 0 3 1 sin 3cos d    =− 26 0 9cos d  = 6 0 cos2 19d 2   +=  6 0 9 sin 2 22   =+  99sin 0 04 3 2 6  = + − −  9 3 3 4 2 4 =+  93 384 =+

9 © ASRJC 2024 [Turn over 10(i) The lines are coplanar  The lines are intersecting lines since they are not parallel. Let 1 2 9 7 5 3 9 a b    −+        =+      ++    By comparing rows, 2 = 9 + 7   = –1 1 –  = a +    = 2 – a …Eq(1) 5 + 3 = 9 + b   = 4 3 b− …Eq(2) Eq(1) = Eq(2): 42 3 ba −−=  6 – 3a = 4 – b  3a = b + 2 (Shown) (ii) Angle between the two lines is 1 11cos 660 −  1cos ab ab −=  11 11 07 311cos cos 11660 07 3 b b −− −                =               −     2 3111 660 1 9 1 49 b b −= + + + 2 2 11 1 6 9 6 50 bb b −+= + 43b2 – 36b – 544 = 0 b = 36 1296 93568 86 + = 4 (reject negative value of b)

10 © ASRJC 2024 (iii) A normal for 2 is 1 2 1 0 1 2 3 3 5            −          −      11 01 32        =  −      −−    33 11 11 −       = − =−      −    3 1 3 1 2 1 3 2 5 10 1 5 1             =  = + + =                r  2 : 3 1 10 1  = r (iv) Angle required = 1 33 11 11cos 11 11 −        −          = 1 9cos 11 − = 0.613 rad (3 sf) or 35.1 (nearest 0.1) (v) 1 : 3 1 1 6 11 11  −=r < 7 11 Since 1 and 3 are parallel, then 3 : 3 1 1 11 k  −=r where k is a real constant. •(2, 1, 3) • (1, 2, 5) • O k 1 0 3 −  