Practice Paper A Sec 3 Math Solutions for Students

1 Practice Paper A Sec 3 Solutions for Students Qn Solutions 1(i) 5 2 and 1 2  1(ii) 22 22 22 11      2 2 ( ) 2 ()      21 22 11 4   Hence equation is 2 21 4 0xx   2(a) 222 4 4 4( ) 2x x x x     2 143 2x   Since 2 1 02x  , 2 14 3 3 2x    so 22 4 4xx cannot be greater than 3. 2(b) Need 40p and 2[ 4( 2)] 4(4 )(6 5) 0p p p     i.e. 0p and 25 4 0 (5 4)( 1) 0 41 or 5 pp pp pp           Hence 4 5p

2 Qn Solutions 3(a) 2 2 22 4(2 1) 84 ps u su p     2 2 4 8 sup  2 4 8 sup   3(b) 3 3 2 3 2 2 2 22 2 8 16 4 4 (4 2 ) ( 2 )( 2 4 ) 4 (2 )(2 ) ( 2 )( 2 ) 4(2 ) x y xy x x y y x x y x xy y x y x y x x y x y yx           2 2 22 4 2 22 x xy y xy x x y y x     2224 2 xy xy   4(a) 22 4 4 ( )( ) 4( )x x y y x y x y x y        Since xy , 04x y x y     Hence 3 3 4xy 6 4(b) 2432(27 ) 45(3 ) 2(81 ) 45(9 ) 2433 x x x x x     Let 9xy , equation becomes 222 45 243 2 45 243 0y y y y      Solving, 27y ( 9 2y is rejected since 9x cannot be negative) Hence 39 27 2 x x  

3 Qn Solutions 5(i) Largest possible value of k is 4. 5(ii) (a) 12f ( ) 4 (2 )xx    or (4 )xx  , domain of 1f is 1x 5(ii) (b) 1 1 2 f ( ) f ( ) f ( ) 4 ( 3) 0 xx xx x x x xx           hence 0x since 3x domain of 1f 6(a) ln 2 3log ln 2 3 2.14 (3 s.f .) x x   6(b) 221 log ( 1) log (22 7 )yy    2 2 2 2log 2 log ( 1) log (22 7 )yy     Hence 222( 1) 22 7 2 3 20 0y y y y       Solving, 5 or 4 (rej.)2y

4 Qn Solutions 7(i) 87 49  C the numbers represent the cost of producing one table and one chair ($87 and $49 respectively). 7(ii)  (20 50) 4190D DC = , it represents the total cost ($4190) of producing 20 tables and 50 chairs. 8(i) [See Graph Paper] Incorrect value of y recorded is 6.81. 8(ii) [See Graph Paper] The gradient represents a, so 2.8a (accept 2.73 2.87a ) The vertical intercept represents b, so 1.5b (accept 1.3 1.7b ) 2 12.7 7.13y yx    (accept 7.07 7.18y )

5 Qn Solutions 9(a) 3 3 22(e 2) e (7e 5) 2 7 5 4 0x x x yyy        where exy 1y is a root so 1( 1)(2 1)( 4) 0 e 1 (rej.) or or 4 2 xy y y      Hence ln 2 or ln 4x 9(b) f (2) 2 8 8 0 pq     and 3f 9 27 8 3002 pq     Solving, 4p and 24q Hence 4 3 2f ( ) 8 4 30 24x x x x x     By long division, remainder is 62 x 10(i) Midpoint of AB 1 ,12M  and gradient of AB 2 Hence equation of 2l : 11( 1) 4 5 222y x y x        or 15 24yx 10(ii) Equation of 1l : 18 ( 5) 2 112y x y x      or 1 11 22yx

6 Qn Solutions 10(iii) Equation of perpendicular to 1l at 110, 2   : 112 2yx  Intersecting with 2l : 11 1 52 2 2 4xx    2.7x and 0.1y Hence distance between the two lines 222.7 (5.5 0.1)   units 2.7 5 or 6.04 (3 sf.) units OR Equation of AB:  4 2 ( 2) 2y x y x      Intersecting with 1l : 1 11 222xx  2.2x  and 4.4y Hence distance between the two lines 22(0.5 2.2) (4.4 1)    units 2.7 5 or 6.04 (3 sf.) units 11(i) Angle of elevation 1 25tan 80  17.4 (1 d.p.) 11(ii) 50 (200 180 ) DMP     30 Distance he walks 80cos30 m 40 3 or 69.3 (3 s.f.) m

7 Qn Solutions 11(iii) By Sine Rule, 80 m sin 30 sin(180 30 110 ) DP     80 m2sin 40DP  62.2 (3 s.f.) mDP 12 24 3 ( 1)( 4)y x x y x x       22 412 9 3 4 3 2 4 3 3 2 3x x x x x x         By symmetry, 03 x . 13 22sec tan 0 sec sec 1 0x x x x      Hence 1 5 2sec cos 2 15 xx     But 1 cos 1x   hence 2cos 15 x  13(i) cosec sec cosec sec cosec 1 cosec 1 x x x x xx   2 2 2cosec 2sec cosec 1 xx x   x y 4 323yx 0 4 3 2 x y 4 323yx 0 4 3 2 243y x x  

8 2 2 2cosec 2sec cot xx x  2 2 22cosec tan 2sec tanx x x x 222sec 2sec tanx x x 13(ii) cosec sec cosec sec 1 cosec 1 cosec x x x x xx   cosec sec cosec sec 0cosec 1 cosec 1 x x x x xx     Hence 222sec 2sec tan 0x x x 2sec (sec tan ) 0x x x   Since sec 0x , 2 2sec tan 0 cos 15 x x x     Hence 180 51.8 (1 d.p.) or 180 51.8 (1 d.p.)x     128.2 (1 d.p.) or 231.8 (1 d.p.)  

9 Qn Solutions 14 tan 2 sin2 x x By including the graph of 2 sinyx it can be seen that there are three solutions between 0 and 360 inclusive. x y 1 1 180 360 tan 2 xy 2 sinyx 0 2