JPJC_9758_2024_Promo_Solution (JCMTC)

Jurong Pioneer Junior College H2 Mathematics JC1-2024 Year-End Exam Solution Q1 (a) ( ) ( ) 12 22 4 d6cos 3d 13 6 19 xxx x x x − =− − =− − (b) ( ) ( ) ( ) 3 d 2 1 d 1ln ln 2 1 3lnd d 2 23 2 2 1 13 21 53 21 x xxxx x xx xx x xx  + = + −   =− + =− + −−= + Alternative (not recommended) ( ) ( ) 32 3 36 5 6 2 3 2 1d 2 1 2 2 1lnd 21 3 2 1 21 21 53 21 x x xxx x x xxx xxx x xx x xx  −+ + += + −+= + + −−= +

2 Q2 Let V be the volume of the water and h be the depth of the water at time t minutes after the start. Given 21 3V r h = and 3d 0.1m / mind V t = . Consider o 3tan 30 3 r rhh=  = Consider 2 31 3 1 3 3 9V h h h==  Then 2d1 d3 V hh = When 33mV = , 1 3331 27 273 9 h h h  =  =  =  Consider d d d d d d V V h t h t= 2 31 27 d0.1 3d h t  =  d 0.0228m/min (nearest to 3 s.f.)d h t= or 1 3 1 30 m/min Q3 2f ( ) ax bx cx= + + At ( )1, 4− , f ( 1) 4−= 2 ( 1) 4( 1) 4 ------ (1) a bc a b c + − + =− − + = At ( )2, 11− , f (2) 11=− 2 (2) 11(2) 2 11 ------ (2)4 a bc a bc + + =− + + =− 1 f ( )y x= has a vertical asymptote with equation 1x= implies C has an x-intercept at 1x= . 2 f (1) 0 (1) 0(1) 0 ------ (3) a bc abc = + + = + + = Or

3 32 2f ( ) a bx cxx x ++= 2 32 1 f ( ) x x a bx cx= ++ 1 f ( )y x= has a vertical asymptote with equation 1x= , ( ) ( ) 32 1 1 0a b c+ + = 0 ------ (3)abc+ + = Solving (1), (2) & (3) using GC: 12, 2, 10a b c= =− =− . Hence, 2 12f ( ) 2 10xx x= − − . Q4 Consider ( )88x A x B= − + By comparing coefficient of x and constant term: 1 8A= , 1B= . 1 2 0 d 4 8 5 x xxx−+ = 11 2200 1 8 8 1 d d8 4 8 5 4 8 5 x xxx x x x − +− + − + 12 0 1 ln 4 8 58 xx= − + + ( ) ( ) 1 2202 1 d 4 2 1 1 5 x xx − + − − − +   1 ln1 ln58=− + ( ) 1 20 1 d 4 1 1 x x−+ = 1 ln58− + ( ) 1 20 2 1 d 141 2 x x  −+   = 1 ln58− + 1 1 0 1 1 1 tan 114 22 x−    −    = 1111ln5 tan (0) tan ( 2)82 −−− + − −  111ln5 tan ( 2)82 −=− − −

4 Q5 (i) 2 1 x ax by x ++= + (2,0) is a turning point on C, 20 2 2 ab= + + 24ab+ =− Eqn (1) ( )( ) ( ) ( ) ( ) 2 2 2 2 21d d 1 2 1 x a x x ax by x x x x a b x + + − + + = + + + −= + At ( )2,0 , d 0d y x = 22 2(2) 0 ab+ + − = 8ab− =− Eqn (2) (1) + (2) 3 12a=− 4a=− From (2) 8 4 8 4ba= + =− + = (ii) 2 4 4 9 511 xxyx xx −+= = − +++ x = –1 y x (0,4) 5

5 (iii) ( ) 32 1 1 0x ax b x+ + + + = 32 1x ax bx x+ + =− − ( ) ( ) 2 1x x ax b x+ + =− + 2 1 1 x ax b xx ++ =−+ Since the graphs of 2 1 x ax by x ++= + and 1y x=− intersect only once , the equation ( ) 32 1 1 0x ax b x+ + + + = has only one real root. Q6 (i) Cartesian equation of ellipse is ( ) 22 22 1 2 xy a a += (ii