Paper H Sec 3 Solutions for students

1 Paper H Sec 3 IP Solutions for students Qn Solutions 1. (a) Substitute x = a into 2() 5f xxx  to get 2 2 51 60 (3 ) (2 ) 0 aa aa aa      3a or 2a (b) 3 2 27 2 29 (3 ) (4 ) (5 ) 2 0 xx xxx x x     = 2(3 ) ( 39 ) 22 9 ( 3)( 4)( 5) ( 4)( 5) xx x x xxx xx      = 2 39 22 9 (4 ) (5 ) xx x xx    2 20 (4 ) (5 ) 1 xx xx     2. (a) Substitute 31yx into 2 33 4xx y  .  2 2 31 4 45 0 xx x x x     (1 ) (5 ) 0 1 or 5 0 or 2 xx xx yy        1, 0 or  5, 2    22 1( 5 ) 02 2 10 or 6.32 (3 s.f.) or 40 units AB    Alternatively, students can make x the subject and substitute into 2 33 4xx y to find y first. The equation will be 291 8 0yy . (b) 22 (2 ) 9xx c 2254 9 0xc x c   Does not intersect means no real solution of x  discriminant < 0 22(4 ) 4(5)( 9) 0cc   2 45c 35c or 35c

2 3. (i) Using the graph of 32 27 9yx x x  To solve 32 25 6 0xx x   32 25 6 2 3 2 3x xx x x     32 27 9 2 3x xx x    = y Hence 23yx  is the graph to be drawn. (ii) 32 27 9 2 3x xx x     From the graph, 21 o r 3xx  . Alternatively, students can obtain (x + 2)(x – 1)(x – 3) > 0, then solve it by regional test on a number line. 4. (i) 91 27 35 75 4 4   (ii) 4 9 4    (iii) 491 27 35 747 975 4 4 352 4            747(1 1) (1099)352   (or by simple addition) Total energy = 1099 calories 5. (i) f is a 1-1 function and hence its inverse function exists. Alternatively, f passes the horizontal line test, hence its inverse function exists.

3 (ii)   1 2fl o gx x  or lg lg 2 x or ln ln 2 x (iii) f( )x and 1f( )x are reflection of each other about the line y = x. 6. (a) 1 332 28 ...(1) xy   1 336 2 51 ...(2)xy    1 3 2(1) (2), 10 5 ...(2)x    11 33 1 2 82xx x     Subst 1 3 1 into (1)2x   gives 33122 8 2 7 32 yyy    (b) 22 5 1xx  21 25xx   2 21 25x x   244 1 2 5x xx  223 2 0xx   (2 ) ( 21 ) 0xx   2x or 1 2x (na) Overal P if does show rejection. (c) Since 30x , 3x 34 3 3 4 2x     That is 34 2 3 2xx  for all real value of x. Hence, no real solution. 7. (i) ( )f x 2 2 2 2 24 7 2( 2 ) 7 2( 2 1 1) 7 2( 1) 9 xx xx xx x           7. (ii)

4 (iii) () 7fx  2 2 2( 1) 9 7 2( 1) 9 7 x x      2 2 2( 1) 9 7 (1 )8 22 1 x x x      x = 3.83 or -1.83 (out of domain) or 2 2 2( 1) 9 7 (1 )1 11 x x x       x = 0 or 2 Hence, x = 0 or 2 or 3.83 8. (i) Using cosine rule, 22 2 820 970 2(820)(970) cos56 723735AC    851 mAC  (nearest metre) (ii) tan18 970 BD 970 tan18 315 mBD   (nearest metre) (iii) DTB is larger. 9. (a) 1y x Stretch/scale the graph along the x-axis by a factor of 3 to obtain 1 3 y x i.e. 3y x . Translate the graph obtained downwards by 1 unit to obtain 3 1y x . (–1, 1) (1, 9) (4, 9) (3.12, 0) 7

5 Alternative answer 2 Stretch/scale the graph along the y-axis by a factor of 3 to obtain 1 3 y x i.e. 3y x . Translate the graph obtained downwards by 1 unit to obtain 3 1y x . Alternative answer 3 Translate the graph obtained downwards by 1 unit to obtain 1 1y x . Stretch/scale the graph along the x-axis by a factor of 3 to obtain 1 1 3 y x i.e. 3 1y x . Alternative answer 4 Translate the graph obtained downwards by 1 3 unit to obtain 11 3y x . Stretch/scale the graph along the y-axis by a factor of 3 to obtain 11 3 3 y x i.e. 3 1y x . (b)  243 6xy  223 3xy (Compress along y axis by factor 2) 10. (i) Gradient of BC = 1 2 Using B (0, 3), equation of BC is 13( 0 )2yx   1 32yx   When y = 0, 1103 3 622xx x    6  Alternatively,

6 Gradient of BC = 30 0    30 210     6  (shown) (ii) Area of quadrilateral ABCD, units2 061 8 6 01 30 61 532 1 (36 270 18 (18 36))2 135       (iii) Midpoint of CD = 61 8 06,1 2 , 322   Gradient of CD = 60 1 18 6 2   Gradient of perpendicular bisector of CD = 2 Equation of perpendicular bisector is 3 2( 12) 2 27yx y x      (iv) Yes point A lies on the perpendicular bisector. When x = 6,  26 2 7 1 5y   which is consistent with the y- coordiante of A. Alternatively, students can show AC = AD = 15. 11. (a) 5 75log 2(0)3  = 5log 25 2 11 (b)   22log 2 log 2 logxx x x x 2 1 2log 12 12 or x u uu   (c) log ( ) loglog ( ) loglog ( ) log n n n aa ana aa bnbbb ana    2 3 4 2021 2 3 4 2021 32 222 2log 2 log 3 log 3 log 3 log 3 ... log 3   32 2 2 2 2 2021 terms l o g2 l o g3 l o g3 l o g3 l o g3 . . . l o g3        

7       32 2 2 log 2 2021 (log 3) 1 2021 log 3log 3 2021     12. (a) 2138 1xy        21ln 3 ln 81 2 ln ( 1) ln 3 4 ln 3 2l n l n3 1 4l n3 xy yx yx        ln 3ln 1 2 ln 32yx    or ln 3 ln 81ln 1 22yx    (b) 6( 2 ) 8Gradient 4 53 2   Method 1: Equation: 264 ( 5 )yx x   2 31 22 41 4 41 4 yx x yx x     Method 2: Equation: 24yx x c Substituting (5, 6), 64 ( 5 ) 1 4cc   241 4yx x  31 2241 4yx x   Hence, a = 4 and b = –14 Note: students can manipulate 31 22y ax bx   to get 2y xa x b first. 13. (i) P(–12, 4) Equation of C1 is    22 12 4 16xy   . (ii)(a) ( 4) unitsQR r (ii)(b) ( 4) unitsQM r Or  2 224 12 = 8 128 unitsQM r r r   (iii) 22 2(4 )(4 ) + 1 2rr  22 81 6 81 6 1 4 4rr rr 16 144 9 r r   Alternatively, 2 8 128 4rr r   can be formed.

8 (iv) 5tan 0.395 12QPM SPM   (shown) (v) Area of minor sector, APS, units2 = 21 (4) 0.39479 15.72522   Area of minor sector, SQO, units2 = 21 (9) 0.39479 47.62822   Area of polygon, APQRC, units2 = 1 24 5 24 4 1562   Area of shaded regions, units2 =156 2(47.628) 2(15.725) 29.294 29.3(3 ) sf 