Midterm2009_markingscheme

MID TERM EXAM 2009 –Matematics T paper 1 (MARKING SCHEME) No Marking Scheme 1 (A – B)’ ∩ A = (A ∩ B’)’ ∩ A = (A’ ∪ B) ∩ A = (A’ ∩ A) ∪ (B ∩ A) = 0 ∪ (B ∩ A) = B ∩ A = A∩ B 2

Marks B1 B1 B1

Total

B1

4M

1

− dy 1 = x. (2 − x) 2 (−1) + 2 − x dx 2

= 7 4

∫ 1

4 − 3x 2 2− x

M1 A1

7 4

4 − 3x 4 − 3x dx = 2∫ dx 2− x 1 2 2− x

M1

7

1 4   = 2 x 2 − x 2     1

A1

1

2 = 14  1  − 2(1) 4 4

A1

=-0.25 3

( x + yi) 2 = 5 − 12i x 2 − y 2 + 2 xyi = 5 − 12i x 2 − y 2 = 5, 2 xy = −12 6 y=− x

2

 6 x 2 −  −  = 5,  x 4 2 x − 5 x − 36 = 0 ( x 2 − 9)( x 2 + 4) = 0 x 2 = 9, x 2 = −4(reject ) x = ±3, y = 2 5 − 12i = ±(3 − 2i ) 4

y = e 2 x cos x dy = e 2 x ( − sin x) + cos x(2e 2 x ) dx

5M

M1 M1

M1 A1 A1 A1

6M

M1

= e 2 x (2 cos x − sin x)

d2y = e 2 x ( −2 sin x − cos x) + (2 cos x − sin x)(2e 2 x ) 2 dx = e 2 x (3 cos x − 4 sin x)

d2y dy =k + hy 2 dx dx e 2 x (3 cos x − 4 sin x) = ke 2 x (2 cos x − sin x) + h(e 2 x cos x) Compare coefficient of cos x : 3 = 2k + h Compare coefficient of sin x : − 4 = −k Therefore, k = 4 and h = −3

M1

M1 M1 A1, A1

6M

5

f ( x) = x 2 + 5 − f (0.5) = −0.75 f (0.6) = 0.36

3 x M1

since f(0.5) < 0 and f(0.6) > 0, there is a root between 0.5 and 0.6

3 f ′( x) = 2 x + 2 x x0 = 0.5

B1

3   2  0.5 + 5 −  0.5  x1 = 0.5 −  3    2 x 0.5 + 2  0.5   = 0.5577

M1

x2 = 0.5640 x3 = 0.5641

A1 M1 A1

The root is 0.564 correct to 3 s.f. 6

1 A B = + (1 + x) ( 2 + x) 1 + x 2+ x A(2 + x) + B(1 + x) ≡ 1 A= 1

B = -1

A1

(1 + x)-1 = 1 – x + x2 + …

B1

1 x (1 + )-1 2 2 (−1) (−2)  x  2 1 x = [ 1 + (-1) +   +…] 1− 2 2 2 2 1 1 1 2 = x + x + …. 2 4 8 1 1 1 1 2 = (1 – x + x2 + …) - ( x + x …) (1 + x) ( 2 + x) 2 4 8 1 3 7 − x + x 2 + .... 2 4 8

=

1 y = x+ x x2 +

M1

A1 M1 8M

M1

1 = y2 − 2 x2 3

B1

A1

1 ⇒ y2 = x2 + 2 + 2 x

A1

x − 2x − 6x − 2x + 1 = 0 2 1 x 2 − 2x − 6 − + 2 = 0 x x 4

7M

M1

1 1 1 = (1 + x) ( 2 + x) 1 + x 2 + x (2 + x)-1 =

7

A1

2

1 1 x 2 + 2 − 2( x + ) − 6 = 0 x x

( y − 4)( y + 2) = 0 y = 4, y = −2 1 When y = 4 , x + = 4 x When y = −2 , x +

1 = −2 x

M1

y2 − 2y − 8 = 0 M1 A1

⇒ x 2 − 4x + 1 = 0

M1

x = 2± 3

A1

⇒ x 2 + 2x + 1 = 0

x =1

A1

8M

8

f ( x) = g ( x)( x − 1)( x + 1)( x − 2) + ( ax 2 + bx + c) f (1) = a + b + c = 3 (1) f (−1) = a − b + c = 1 (2) f (2) = 4a + 2b + c = −2 (3) (1)+(2), 2a+2c=4, a+c=2 2(2)+(3), 6a+3c=0, 2a+c=0 a = -2, c = 4, b = 1 Remainder = f ( x) = −2 x 2 + x + 4

9

B1 B1 B1 M1 A1A1A1 M1

0 0 1 0  1 0 a 0      (a) P =  − 1 − 2 0  = m b − 5 0  + n 0 1  c − 3 − 2  0 0  −1 − 3 1      0 0   ma 0 0  n 0 1      0  + 0 n  − 1 − 2 0  =  mb − 5m  − 1 − 3 1   mc − 3m − 2m   0 0     

0  0 1 

0  0 n  0 0   ma + n 0 0 1      − 5m + n 0  − 1 − 2 0  =  mb   − 1 − 3 1   mc  − 3 m − 2 m + n    

8M

M1

M1

ma + n = 1

mb = −1 − 5m + n = −2 − 3m = −3

M1

mc = −1 − 2m + n = 1

m = 1, n = 3, a = −2, b = −1, c = −1 2

0 0  1 0 0  1     2 (b) P =  − 1 − 2 0  = 1 4 0   − 1 − 3 1  1 3 1     

B1

0 0 1 0 0 1 0 0   1       1 4 0  = s  − 1 − 2 0  + t  0 1 0  1 3 1   − 1 − 3 1   0 0 1        0 0  1 0 0   s + t     0  1 4 0  =  − s − 2 s + t 1 3 1   − s − 3s s + t     s = -1, t = 2 P 2 = − P + 2I P 4 = P 2 P 2 = (− P + 2 I ) 2 = P 2 − 4 P + 4 I = (− P + 2 I ) − 4 P + 4 I = −5P + 6 I 10

 1 1 − 1  − 5 − 5 5  10 0 0       6 − 4  =  0 10 0  AB =  2 3 2   8 5 4 1  − 7 1 1   0 0 10    = 10 I 1  A B  = I  10 

 5 −  10 1 8 −1 A = B=  10 10  7 −  10

5 10 6 10 1 10

−

5   10   − 0.5  4 −  =  0.8 10  1   − 0.7  10 

A1

− 0.5 0.5   0.6 − 0.4  0.1 0.1 

M1

M1 A1 M1 A1

M1 A1

M1 A1

10M

x+ y−z =0 2 x + 3 y + 2 z = 16 5 x + 4 y + z = 19

B1 B1 B1

 1 1 − 1 x   0        2 3 2  y  = 16   5 4 1  z  19      

M1

 x  − 5 − 5 5  0    1   6 − 4 16   y =  8  z  10  − 7 1 1 19    

M1

 1.5    = =  2.0   3.5   

A1

x = RM 1.50, y = RM 2.00, z = RM 3.50

11

S n = a + ar + ar + ar + ... + ar 2

3

n −1

A1

→ (1)

rS n = ar + ar 2 + ar 3 + ar 4 + ... + ar n → (2)

M1

(1) − (2), S n − rS n = a − ar

M1

n

12M

S n (1 − r ) = a(1 − r n ) a (1 − r n ) 1− r 1 − 1 r= 2 =− , a=2 2 4 Sn =

S∞ =

2

=

1 1+ 4

A1 B1

8 5

B1

  1 n  21 −  −   n   4   8   1   Sn =  = 1 −  −   1 5   4   1+ 4

M1

n

S∞ − S n =

8 8 8 1 − +  −  < 10− 5 5 5 5 4

M1

n

8 1 − < 10− 5 5 4 1 5  n lg < lg10− 5 ×  4 8 

M1 M1

5  lg10− 5 ×  8  n> 1 lg 4

M1

n > 8.64 ∴n = 9

A1 A1

12 y 12

y=12-x3 x

0

y=12-4x

D1 D1 D1

12 M

M1

12 − 4 x = 12 − x 3 x( x 2 − 4) = 0 x = 0, 2, − 2 (0,12) ( 2,4) (−2,20)

A1 A1

0

2

−2

0

3 3 A= ∫ [(12 − 4 x) − (12 − x )]dx + ∫ [(12 − x ) − (12 − 4 x )]dx 0

M1

2

∫

∫

3 = ( x − 4 x)]dx + (4 x − x )dx 3

−2

0

0

2

 x4   x4  =  − 2 x 2  + 2 x 2 −  4 0 4 −2    16   16   = 0 −  − 8  +  8 −  − 0 4     4

A1 M1

=8

A1 2

0

∫

∫

3 2 2 2 3 2 V= π [(12 − 4 x) − (12 − x ) ]dx + π [(12 − x ) − (12 − 4 x) ]dx −2

=π

0

2

0

∫ [24 x

−2

M1,M1

3

+ 16 x 2 − 96 x − x 6 ]dx + π ∫ [−24 x 3 − 16 x 2 + 96 x + x 6 ]dx 0

0

2

  16 x 3 x7  16 x 3 x7  = π 6 x 4 + − 48 x 2 −  + π − 6 x 4 − + 48 x 2 +  3 7 −2 3 7 0     2528   1504   = π 0 −  −  + π   − 0   21   21   = 192π

A1 M1 A1

15M