Dunearn Prelim 2009 Am P1 Solutions

A.Maths Paper 1 – prelim 2009 – Answers and marking scheme Answer all the questions 1. (a) 5x + 2y = 3 2x – 3y = 5  5 2  x   3        ……(M1)  2  3  y   5   x 1   3  2  3         ……(M1) 19   2 5  5   y  x 1   19        ………(B1) 19  19   y  x  1       …….(A1)  y    1 x 2 (b) (3 ) – 10(3x . 3) + 81 = 0 a2 – 30a + 81 = 0 …… (M1) (a – 27)(a – 3) = 0 ………(B1) a = 27 or a = 3 ---------(A1) 3x = 33 or 3x = 31 x = 3 or x = 1 ….(A1) 2. (a)

Max.Mark: 80

Total marks = 4

Total marks = 4

lg(2x -1) = 1 + lgx – lg(x +3) lg(2x -1) = lg 10 + lgx – lg(x +3) ……..(M1) 10 x Hence, 2x – 1 = ………………… (M1) x3 2x2 – 5x – 3 = 0 .......................(M1) x = - ½ (NA) or x = 3 ................(A1) Total marks = 4

(b) using (0,5), we get, a = 5 ……….(A1) 2 5 2 5 using  ,  , we get  5  b 3 ……..(M1) 4 3 4 2

b3 

1 4 2

2  13   b    1  ---------(B1)   2   3

1 1 b   …….(A1) 8 2

6. (a) 2x2 -5x - 6 = 0 , ( and ),  +  = 5/2 and   = -3 ………(B1) (i) 2 + 2 = ( + )2 – 2   …….(M1) = (5/2)2 – 2(-3) = 49/4 ……(A1)

Total marks = 4

Total marks = 3

A.Maths Paper 1 – prelim 2009 – Answers and marking scheme

6 (a) (ii) Sum of roots =

2 2( 2   2 ) 49 2 + =  …..(B1)    6

2 2 x = 4 ……(B1)   Equation, x2 – (SR)x + PR = 0 ........(M1) 6x2 +49x + 24 = 0 ………(A1) Total marks = 4 2 (b) b – 4ac < 0 [ 1 M]  16 – 4(2)(3-k) < 0 [ 1 M ]  k < 1 [ 2A] Product of roots =

4. (a) (x + 2) is a common factor for both, hence we have, (-2)3 + a(-2)2 – (-2) + b = 0 and (-2)3 + b(-2)2 -5(-2) + 3a = 0 …..(M2) We get, 4a + b = 6 and 3a + 4b = -2, ……..(M2) Solving, a = 2 and b = -2 ………(A1) Total marks = 5

(b)

7. (a)

(b)

(3 + ax)5 = 35 +(5)(34)(ax) + (10)(33)(ax)2 + … ……(M1) = 243 + 405a x + 270 a2 x2 ………(B1) Compare with = 243 – 810x + bx2 405 a = - 810 and b = 270 a2 ........(M1) a = -2 and b = 1080 …..(A2) Total marks = 5 |2x – 7| = |x – 3| (2x – 7)2 = (x-3)2 3x2 – 22x + 40 = 0 …….(M2) (3x – 10) (x – 4) = 0 x = 10/3 or x = 4 ……(A2)

Total marks = 4

4  2x A Bx  c =  2 ……..(M1) 2 ( x  1)( x  7) x  1 x  7 4 – 2x = A(x2 + 7) + (Bx + C)(x -1)……….(M2) Solving for A, B & C: A = ¼, B = - ¼ & C = - 9/4 ……..(B1) 4  2x 1 x9 Hence, =  ….(A1) 2 ( x  1)( x  7) 4( x  1) 4( x 2  7)

Hence, change x to (-x), we get, …………(M1) 4  2x 1 x9 =  …(A1) 2 ( x  1)( x  7) 4( x  1) 4( x 2  7)

A.Maths Paper 1 – prelim 2009 – Answers and marking scheme 8. (a) (i) 2(cos x + 5 sin x ) = 3 sin x 2 cos x = -7 sin x ……..(M1) tan x = -2/7 x = 164.1º , 344.1º …….(A2) (ii)

6 sin x = 7 + cos 2x 6 sin x = 7 + ( 1 – 2 sin2 x)…….(M1) 2 sin2 x + 6 sin x – 8 = 0 sin2 x + 3 sin x – 4 = 0……(M1) Solving for sin x and then solve for x (sin x + 4) (sin x – 1) = 0……(B1) Sinx = -4 (NA) sinx = 1  x = 90º……..(A1)

(iii)

cos x + cos 3x = 0 2 cos 2x cos x = 0 ..........(M2) cos 2x = 0 or cos x = 0…..(B1) x = 45 º, 90 º, 135 º……….(A1)

8 (b)

tan (3x + 2) = 2.4 ……(M1) Using radian measure for angles selecting the angle after 12, we get x = 12.3…….(A2)

5 (a) y = 2x3 - 3x2 – 9x + 1 dy  6 x 2  6 x  9 …….(B1) dx Gradient of y – 3x = 4 is 3 …….(M1) Solving 6x2 – 6x – 9 = 3 …….(M1) (x-2)(x+1) = 0  x = 2 or x = -1 ……(B1) Coordinates (2, -13) and (-1, 5)……(A1) (b) (i) y = tan3(3x – 1) dy  3 tan 2 (3 x  1)  (sec 2 (3 x  1))(3) ………(M1) dx = 9 tan2(3x-1)sec2(3x-1) ……(A1) (ii) y = (x2 – 1) cos 2x dy  ( x 2  1)( sin 2 x )(2)  cos 2 x(2 x ) …..(B2) dx = 2[x cos 2x – (x2 -1) sin 2x].........(A1)

A.Maths Paper 1 – prelim 2009 – Answers and marking scheme

3.

y

m xn

y(x – n) = m  yx – ny = m  ny = xy – m  y 

1 m xy  n n

…..(M1)

let Y = y and X = xy X= xy Y=y

1.19 1.19

1.74 0.87

2.07 0.69

2.28 2.40 0.57 0.48 ...... (B1)

graph, ………………………………………….( 3marks) 1 m (i)  0.59 and   1.9 n n We get, n = -1.69 and m = 3.21 (A2) (ii) when y = Y = 0.75, from graph X = 1.93, xy = 1.93 x = 2.57 ………(A1)