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−