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) ( ) 22 22 1 2 xy a a += 2 22 241 xya a =−  ( ) 2 2 2 4y a x=− 222y a x= − Since 0, 0xy for coordinates of P, Area of rectangle PQRS, ( )( )22A x y= 4xy= 224 (2 )x a x=− 228x a x=− (shown) (iii) 22 22 d2 88d 2 Ax x a xx ax  −= + − − ( ) 2 2 2 22 88x a x ax − + − = − 22 22 8 16ax ax −= − When A is a maximum, d 0d A x = . Alternatively ( ) 2 2 2 2 2 2 4 23 22 64 ( ) 64( ) d2 64(2 4 )d dPut 0, 2 2 0d 0( ) or ( 0) 2 A x a x a x x AA a x xx A x a xx ax NA x x = − = − =− = − = = = 

6 22 22 8 16 0ax ax − = − 22 1 2xa= 1 2 xa= Since 0x , 1 2 xa= (iv) 228A x a x=− 1 , 100 2 x a A== 2 2118 100 22 a a a    −=       118 100 22 aa   =     2 25a = 5a= Since 0a , 5a= Q7 (i) 2 53 22 x xx − −+− 2 2 2 5 3 2( 2) 02 21 0( 2)( 1) x x x xx xx xx − + + − +− −+ +− 2 2 22 2 2 21 121 2 1 1 121 2 4 4 172 48 xx xx xx x −+ = − +     = − + − − − +         = − + Since 2 1 04x − , 2 1720 48x − +  for all real values of x. OR: Alternatively 22 22 2 2 4 100 64 22 100 4 5( 0) aa a a aa   =−      = =

7 Consider 22 1 0xx− + = Discriminant = ( ) ( )( ) 2 1 4 2 1 7 0− − =−  and coefficient of x2 = 2 > 0 Hence 22 1 0xx− +  for all real values of x. Thus, consider 1 0( 2)( 1)xx +− . ( 2)( 1) 0xx+ −  Therefore 2x− or 1x (ii) ( ) ( ) ( ) ( ) 2 2 2 2 53 22 53 22 53 22 53 2 (Note that the inequality sign is different from that in (i)) 2 x xx x xx x xx x xx −− −− −+ −− + −−− −− − − + − − Replace x by x− : 21 12 x x −  −  −   Q8 (a)(i) ( )fy x a a= − + ( )' 0,Aa ( )' ,0Ba ( )' 3 ,C a a − x y x O B’ C’ A’

8 (a)(ii) ( ) 1 fy x= 1'' 0,B a  − 1'' 2 , 2Ca a  − (b) ( )g siny x x== After A, ( ) ( )g siny x x = + = + After B, g sin33 xxy    = + = +       After C, g 2 sin 233 xxy    = + + = + +       Q9 (a) ( ) 22 22 2 2sind cos d 1 1 sin x x x    = −− ( ) 2 2 2sin cos d cos    = ( ) 2 1 12 2sin d 1 cos 2 d 1 sin 22 sin cos sin sin 1 c xc x x x c    − − = =− = − + = − + = − − +   sinx = d cosd x  = Note : To obtain cos in terms of x, use trigo identities or triangle. 22cos sin 1+= Or x 1 y x O B’’ C’’

9 (b)(i) d1sin(ln ) cos(ln )d xxxx =  (b)(ii) 1sin(ln ) d sin(ln ) cos(ln ) dx x x x x x x x =−   sin(ln ) cos(ln )d 1sin(ln ) cos(ln ) sin(ln )d sin(ln ) cos(ln ) sin(ln )d x x x x x x x x x x x x x x x x x x =−  = − − −   = − −    12 sin(ln )d sin(ln ) cos(ln )x x x x x x c= − +  1sin(ln )d sin(ln ) cos(ln )2x x x x x c = − + Q10 (i) (ii) 2 d 21d x t t x tt =+ =+ 2 d 21d y t t y tt =− =− d 2 1 d 2 1 yt xt −= + sin(ln )ux= d 1d v x = d1 cos(ln )d u xxx =  vx= cos(ln )ux= d 1d v x = d1 sin(ln )d u xxx =−  vx= y x (0,2) Note : Must include negative values of t in the window settings. ( ) 2When 0, 0 10 0 or 1 0 or 2 x t t tt t y = + = += =− = ( ) 2When 0, 0 10 0 or 1 0 or 2 y t t tt t x = − = −= = =

10 Gradient of line 5 4 20yx=− is 4 5 . d 2 1 4 d 2 1 5 yt xt −== + 10 5 8 4tt− = + 9 2t = 2 99 24.7522x = + = 2 99 15.7522y = − = Tangent to C is parallel to 5 4 20yx=− at point ( )24.75,15.75 (iii) From (i), at ( )0, 2 , 1t=− (Can also obtain 1t=− from graph on GC) ( ) ( ) 2 1 1d 3d 2 1 1 y x −−== −+ Equation of tangent at ( )0, 2 is 2 3( 0)yx− = − 32yx=+ ( ) 22 32t t t t− = + + 22 4 2 0tt + + = 2 2 1 0tt + + = ( ) 2 10t+= 1t=− Since 1t=− is the only solution, the tangent at ( )0, 2 does not cut C again. Q11 (i) Given aOA= , bOB= , 1 2 aOC= . By Ratio Theorem, ( )2 3 1 1 1 2 3 35 5 2 5 c + b a b a bOD = = + = +  (ii) Area of triangle OCD ( ) ( ) ( ) 1 2 1 1 1 32 2 5 1 1 3 2 10 10 a a b a a a b OC OD= =  + =  + 