JPJC 9758 2022 Promo Solutions

Jurong Pioneer Junior College H2 Mathematics JC1-2022 Year-End Exam Solution Q1 (i) (ii) 0.349 2 x or 2.15x Q2 (a) (i)   33 3 33 d 9e 3eln 9 3ed 9 3e 3 e xx x xxx     (ii)   4 2 3 2 2 3 2 2d sin 4sin cos 2 8 sin cosd x x x x x x xx  (iii) x = 2 y x 0.349

2 1d sec3 sin 2d xxx      1 2 1sec3 2 sin 2 3sec3 tan 3 12 x x x x x   1 2 2sec3 3tan 3 sin 2 14 x x x x    (b) 32 4 9 10 0y y x x     2 dd3 4 2 9 0dd yyyx xx    2 d 9 2 d 3 4 yx xy   Q3 32 d13 d uu t t t    When 0t  , 1u . When 2t  , 9u          22 325 3333 0 0 9 3 1 9 33 1 9 23 1 912 1 9 2 1 dd 11 11 d3 11 d3 1 d3 1 3 1 2 1 1 1 1 1 1 1 32 1 3 2 3 9 2 81 2 243 ttt tt tt u uu u uuu u u u uu uu                                    Q4 (i)

3 2 4 1 x kxy x          2 2 2 1 4d d 1 x k x x kxy x x            2 2 24 1 x x k x     d 0d y x     2 2 4 0x x k    For two stationary points:   22 4 4 0k   5k  Alternative: 2 45 ( 1)11 x kx ky x k xx            2 5d 1d 1 ky x x   d 0d y x     2 15xk   15xk   For two stationary points: 50 k 5k  (ii) When 4k  , 2 44 1 xxy x   (iii) 2 4 ( 3)( 1) 0x kx mx x      x = –1 y x y = x + 3 (0,4)

4 2 4 ( 3)( 1)x kx mx x     2 4 31 x kx mxx   , where 4k  . For graphs of 2 4 1 x kxy x   and 3y mx to have no intersection and hence 2 4 ( 3)( 1) 0x kx mx x      to have no real root, 01 m . Q5 (i) Using similar triangles, 0.6 0.7 6 7 x y xy   Let the volume of water in the tank be V.   21 6 12422 7 7V xy y y y     d 24 d7 V yy  Since d d d ,d d d V V y t y t     24 d0.0025 0.47d d 0.0018 m/s (4 d.p)d y t y t    (ii) Time taken =     21 120.6 (0.7)(4) 0.4 0.0025 226s27     (nearest second) Or Time taken =   2212 12(0.7) 0.4 0.0025 226s77     (nearest second) Q6 (i) 0.7 y x 0.6

5 Using GC, from graphs, the intersection points of C and l are (2, 4) and (5,1) . Alternative 2 2 or 6 8 16 7 10 0 (or use GC polynomial rootfinder) ( 5)( 2) 0 5 2 x x x xx xx xx               When 2x , 2 6 4y   When 5x , 5 6 1y   The intersection points of C and l are (2, 4) and (5,1) . (ii) Area of the region R   5 2 54 6 d 4 d a x x x x     or 5 2 54 6 d 8 16 d a x x x x x        532 5 4 4623 a xx x         or 52 3 2 54 86 162 3 2 a x x x xx                 322 5456 6(5)2 2 3 a a      2 103626 a a   units 2 Q7 (a) xa for 5 6d a xx , can also use area of trapezium   1 5 1 62 aa  

6 10e d5 2e x x x = 10 2e d2 5 2e x x x  = 2e5d 5 2e x x x  = 5ln 5 2e x c   (b) 2 d 18 x x x =   1 2 21 16 1 8 d16 x x x   =   1 2 2181 116 2 x c   =   1 2 21 188 xc (c) 2 (ln ) dx x x =     22 2 1ln 2 ln d22 xxx x x x             =     2 2 ln ln d 2 xx x x x   =   2 2 2 2 1ln (ln ) d2 2 2 x x xx x x x                    =   22 2 1ln (ln ) d2 2 2 xxx x x x           =   2 2 2 2 ln (ln ) 2 2 4 x x xx x c            Q8 (a) 12 11 5 62 1 2 1 xy xx      1 1 1 5 5 62 3 2 2 3 2 1 2 1 2 1y y y yx x x x x              Translate by 2 units in the negative x-direction. (This can be at any step.) Scale parallel to the y-axis by scale factor of 5. Translate by 6 units in the positive y-direction. (b) Let   2 lnux and d d v xx   d1 2 lnd u xxx   2 2 xv Let lnux and d d v xx  d1 d u xx 2 2 xv y

7 Step 1 : Translate by 2 units in the negative x-direction. Step 2 : Reflect about y-axis. Step 3 : Translate by 4 units in the positive y-direction.        0, 5 2, 5 2, 5 2, 1               6, 4 4, 4 4, 4 4,0        2 4 4 4x x x x       2 0 0 0x x x x       2 2 2 2y y y y       Q9 Let AP = y. 1tan 30 3 x y  3yx Since AP BQ y , 20 2 3PQ x cm. Let A denote the area of rectangle PQRS.   220 2 3 20 2 3A x x x x    d 20 4 3d A xx  . Step 1 Step 2 Step 3 y A P x S

8 To find maximum of minimum values, we let d 0d A x  . 20 4 3 0 x Hence, we get 55 333 x 2 2 d 4 3 0 d A x   . Area of PQRS is maximum when 5 33x . Hence, 55 3 cm, 20 2 3 3 10 cm33PS PQ      . Area of rectangle PQRS = 250 3 cm3 . Q10 (i) 2 3OC a , 4 3OD b (ii) 2 3BC ab , 4 3AD ba E is on Line BC: 2 , for a value of 3OE    b a b E is on Line AD: 4 , for a value of 3OE    a b a Consider 24 33OE                b a b a b a 24 (1 ) (1 )33      a b a b By comparing the coefficients: 2 13 ------------Equation (1) 41 3 ------------Equation (2) Solving: 3, 3  Thus 24( 3) 3 4 233b a b a b a b aOE                   (Shown) (ii)

9 42 33CD OD OC    ba , 84 3CE OE OC    ba Area of triangle CDE = 1 2 CD CE = 1 4 2 8 42 3 3 3             b a b a = 1 4 4 8 2 2 8442 3 3 3 3 3 3                             b b b a a b a a = 1 4 8 2 42 3 3 3              b a a b since aa = 0 and bb = 0 =      1 32 8 1 8 4 2 9 3 2 9 9      a b a b a b a b where 4 9k  Q11 (i) (ii) 2336x t y t 2dd 6 18dd xy tttt d 3d y tx  Since tangent is parallel to the line 42yx , d 32d y tx   2 3t  Hence, 23 2 4 2 163 , 63 3 3 9xy                4 16,39P  (iii) y x

10 At 4 16,39Q  , 3 16 26 93y t t    d2 3 3 2d3 y tx     Equation of tangent at Q : 16 4 293yx     82 9yx 9 18 8yx (shown) (iv) At R : 0y 40 18 8 9xx    R 4 ,09   Area of triangle PQR 1 4 4 16 16 ()2 3 9 9 9          128 81 units2 (v) 9 18 8yx     329 6 18 3 8tt  3227 27 4 0tt    21 (N.A as Point Q) or 33tt  23 1 1 1 23 , 63 3 3 9xy                12,39S  Q12 y x y