STPM 2014 Maths T Paper 1 Trial Sem 1 SSIJB

SEK KEB MEN SULTAN ISMAIL, JOHOR BAHRU 954/3 MATHEMATICS T Pre-U 1 TERM 1 TRIAL EXAMINATION 2013 Section A [45, Marks] Answer all questions in this section.

1

A functions f is defined as f : x a x 2 + 2, x ∈ [ −5, −1] (a) Find f −1 . (b) Sketch f and f −1 on the same axes.

[4 marks] [3 marks]

2 The second and the fifth terms of a geometric series are 24 and −3 respectively. Find the smallest value of n for which the difference between the sum of the first n terms and the sum to infinity is less than 0.005. [5 marks]  1 2 2  −5 a 4   ÷  ÷ 3 Given that P =  1 1 3 ÷, Q =  8 −5 c ÷ and PQ = kI , where k is a constant and I is  3 b 1÷  c 4 c÷     a , b the 3 × 3 identity matrix. Find the values of and c . [4 marks] −1 Deduce Q and hence, solve the following system of linear equations. −5 x + ay + 4 z = 11 8 x − 5 y + cz = −5 cx + 4 y + cz = 4 [6 marks]

2π . 3

4

The complex number z is such that zz* = 64 and arg( z ) =

(a) (b)

Find z in polar form and deduce z 5 in polar form. [3 marks] 3 w Find the complex numbers such that w = −4 + 4 3i in the form of a + bi . [5 marks]

x2 y 2 5 The equation of an ellipse is given by + =1. 9 25 (a) Find the centre, vertices and the foci of the ellipse and sketch the curve. [6 marks] 2 (b) If the distances of a point on the ellipse from the foci are d1 and d 2 , show that (d1 + d 2 ) is constant. [4 marks]

6

0  2  4   −1   ÷  ÷  ÷  ÷ Find the acute angle between the lines r =  1 ÷+ m  3 ÷ and r =  7 ÷+ t  2 ÷.  −1÷  ÷  7 ÷  −2 ÷    4    

[5 marks]

Section B [15 marks] Answer any one question in this section 7.

(c)

x3 + 2 x 2 + 2 x + 3 in partial fractions. x2 + 2x − 3 If a = b + c , show that (a − b − c ) 2 = 4bc . 2 Solve the equation log 3 x + log 3 x = log 27 9 .

(d)

Solve the inequality cos( x −17°13') < −0.5, −180° ≤ x ≤ 180° .

(a) (b)

Express

[4 marks] [3 marks] [3 marks] [5 marks]

8. (a) The position vectors of the points A , B and C are i + 3 j + 5k , 4i − 9 j − k and pi − j + 3k respectively. (i) find the unit vector parallel to AB . [3 marks] (ii) find the value of p if A, B and C are collinear. [3 marks] (iii) if p = 2 , find the position vectors of the points D such that DABC is a parallelogram. [3 marks] (b) Solve the following system of linear equations using Gaussian elimination.

4x + 2 y − 3z = 5 x + y − 2z = 2 2 x − y + 3z = 1

[6 marks]

MATHEMATICAL FORMULAE sin 2 A = 2sin A cos B Trigonometrical identities sin( A ± B) = sin A cos B ± cos A sin B ; cos 2 A = cos 2 A − sin 2 A = 2cos 2 A − 1 = 1 − 2sin 2 A cos( A ± B) = cos A cos B mcos A cos B 2 tan A tan 2 A = tan A ± tan B 1 − tan 2 A tan( A ± B) = 1 mtan A tan B Sum of series: 1 1 For an arithmetic series: Sn = n(a + l ) = n[2a + ( n − 1) d ] 2 2 a (r n − 1) For a geometric series: S n = , r ≠1 r −1 Binomial expressions n n n (a + b) n = a n +  ÷a n −1b +  ÷a n − 2b 2 + L +  ÷a n −r b r + L + b n , n ∈ ¢ + 1 2 r n( n − 1) 2 n(n − 1)K (n − r + 1) r (1 + x) n = 1 + nx + x +L + x +L , n∈¤, x log(0.5) n > 12.64 the smallest value of n = 13

B1

M1 A1 M1 A1

5

 1 2 2  −5 a 4   ÷  ÷ 3 Given that P =  1 1 3 ÷, Q =  8 −5 c ÷ and PQ = kI , where k is a constant and I is  3 b 1÷  c 4 c÷     the 3 × 3 identity matrix. Find the values of a, b and c . [4 marks] −1 Deduce Q and hence, solve the following system of linear equations. −5 x + ay + 4 z = 11 8 x − 5 y + cz = −5 cx + 4 y + cz = 4 [6 marks]

3

a12 : a32 : a31 : a11 :

PQ = k I  1 2 2  −5 a 4   1 0 0  ÷ ÷  ÷  1 1 3 ÷ 8 −5 c ÷ = k  0 1 0 ÷  3 b 1÷ c 4 c ÷  0 0 1÷   a − 10 + 8 = 0 - - - -(1) 3a − 5b + 4 = 0 - - - -(2) a =2, b=2 −15 + 8b + c = 0 c = −1 k = −5 + 16 − 2 = 9  1 2 2 1 ÷ Q =  1 1 3÷ 9 ÷  3 b 1 1/ 9 2 / 9 2 / 9   ÷ = 1/ 9 1/ 9 1/ 3 ÷  1/ 3 2 / 9 1/ 9 ÷    −5 a 4  x   11  ÷ ÷  ÷  8 −5 c ÷ y ÷ =  −5 ÷,  c ÷  ÷ 4 c ÷   z   4   x   1  ÷  ÷  y ÷=  2 ÷  z ÷  3÷     x = 1, y = 2, z = 3

B1

M1 A1 A1

4

B1

−1

M1

A1

 x  1 2 2  11  ÷ 1 ÷ ÷ ⇒  y ÷ =  1 1 3 ÷ −5 ÷ 9  z÷  3 b 1÷ 4 ÷     

M1

A1 A1

2π . 3

6

4

The complex number z is such that zz = 64 and arg( z ) =

(a) (b)

Find z in polar form and deduce z 5 in polar form. [3 marks] 3 Find the complex numbers w such that w = −4 + 4 3i in the form of a + bi . [5 marks] 2

4(a) zz = z = 64 z =8

2π 2π   z = 8  cos + i sin ÷ 3 3  

B1

 10π  z 5 = 85 cis  ÷  3   2π  = 32768cis  − ÷  3  2π 2π   = 32768  cos − i sin ÷ 3 3  

M1

A1

3

4(b) −4 + 4 3i = 42 + (4 3) 2 = 8  4 3  2π arg(−4 + 4 3i ) = π − tan −1  ÷ ÷=  4  3  2π  w3 = 8cis  ÷  3   2π / 3 + 2kπ  w = 2cis  ÷, k = 0,1, 2 3    2π   8π   14π  w1 = 2cis  ÷ , w2 = 2cis  ÷, w3 = 2cis  ÷  9   9   9   2π   8π   4π  w1 = 2cis  ÷ , w2 = 2cis  ÷, w3 = 2cis  − ÷  9   9   9  w = 1.5321 + 1.2856i, −1.8794 + 0.6840i, 0.3473 − 1.9696i

M1

A1 M1 M1

A1

5

x2 y 2 5 The equation of an ellipse is given by + =1. 9 25 (a) Find the centre, vertices and the foci of the ellipse and sketch the curve. [6 marks] 2 (b) If the distances of a point on the ellipse from the foci are d1 and d 2 , show that (d1 + d 2 ) is constant. [4 marks] 5(a) Centre = (0, 0) Vertices = (0, −5), (0,5) c 2 = 25 − 9, ⇒ c = ±4 Foci = (0, −4), (0, 4) y 4 3 2 1

(b)

B1 B1 M1 A1

x

−5 −4 −3 −2 −1 −1 1 2 3 4 5 6 −2 −3 −4 −5 Let (a, b) be a point on the ellipse a 2 b2 25 + = 1, ⇒ b 2 = 25 − a 2 - - - (1) 9 25 9

D1 D1

B1

(d1 + d 2 ) 2 = ( a 2 + (b + 4) 2 + a 2 + (b − 4) 2 ) 2 = a 2 + (b + 4) 2 + a 2 + (b − 4) 2 + 2 a 2 + (b − 4) 2 a 2 + (b − 4) 2 = 2a + 2b + 32 + 2 (a + b + 16) − (8b) - - - - (2) Substitute (1) into (2), … … … 50 16 = 2a 2 + 50 − a 2 + 32 + 2 (9 + a 2 ) 2 9 9 50 32 = 2a 2 + 50 − a 2 + 32 + 18 + a 2 9 9 = 100 2 ∴ (d1 + d 2 ) is constant. 2

2

2

2

2

M1

2

M1 A1

6

0  2  4   −1   ÷  ÷  ÷  ÷ Find the acute angle between the lines r =  1 ÷+ m  3 ÷ and r =  7 ÷+ t  2 ÷.  −1÷  ÷  7 ÷  −2 ÷    4    

6

 2   −1   ÷ ÷ 2 2 2 2 2 2  3 ÷g 2 ÷ = 2 + 3 + 4 1 + 2 + 2 cos θ  4 ÷  −2 ÷    −2 + 6 − 8 = 3 29 cos θ 4 cos θ = − 29 87 θ = 104.34° The acute angle is θ = 75.66°

[5 marks]

B1M1 M1

A1 A1

Section B [15 marks] ( MARK ONLY ONE question in this section – the FIRST ONE AFTER QUESTION 6 ) 7.

(a) (b) (c) (d)

x3 + 2 x 2 + 2 x + 3 in partial fractions. x2 + 2x − 3 If a = b + c , show that (a − b − c ) 2 = 4bc . 2 Solve the equation log 3 x + log 3 x = log 27 9 . Solve the inequality cos( x −17°13') < −0.5, −180° ≤ x ≤ 180° . Express

x3 + 2 x 2 + 2 x + 3 x 3 + 2 x 2 − 3x + 5 x + 3 ≡ x2 + 2x − 3 x2 + 2 x − 3 5x + 3 ≡ x+ 2 x + 2x − 3 5x + 3 5x + 3 A B ≡ ≡ + 2 x + 2 x − 3 ( x − 1)( x + 3) x − 1 x + 3 5 x + 3 ≡ A( x + 3) + B ( x − 1) x = 1, 8 = 4 A ⇒ A=2 x = −3, − 12 = −4 B ⇒ B=3

[4 marks] [3 marks] [3 marks] [5 marks]

7(a)

B1 (or long division)

M1 A1

x3 + 2 x 2 + 2 x + 3 2 3 ≡ x+ + 2 x + 2x − 3 x −1 x + 3 7(b) a = b+ c ( a )2 = ( b + c )2 a =b+c+2 b c a −b −c = 2 b c

7(c)

4

B1 M1

(a − b − c ) 2 = (2 b c ) 2 = 4bc 2 log 3 x + log 3 x = log 27 9 log 3 9 log 3 x 3 = log 3 27 1 2 3log 3 x = log 3 9 or 3log 3 x = 3 3 1 2 log3 x = log 3 9 or log 3 x = 9 9 1 9

7(d)

A1

2

x = 9 = 1.277 or x = 39 = 1.277 cos( x − 17°13') < −0.5 . y

A1

3

B1

M1 A1

3

2 1 x −3

−2

−1

−1

1

2

M1

3

−2 −3 cos( x −17°13') = −0.5 = − cos 60° x − 17°13' = −120°,120° x = −102°47 ',137°13' Solution set = {x : −180° ≤ x < −102°47 ',137°13' < x ≤ 180°}

M1 A1 A1A1

8. (a) The position vectors of the points A , B and C are i + 3 j + 5k , 4i − 9 j − k and pi − j + 3k respectively. (i) find the unit vector parallel to AB . [3 marks] p A , B (ii) find the value of if and C are collinear. [3 marks] (iii) if p = 2 , find the position vectors of the points D such that DABC is a parallelogram. [3 marks] (b) Solve the following system of linear equations using Gaussian elimination.

4x + 2 y − 3z = 5 x + y − 2z = 2 2 x − y + 3z = 1

8(a) (i)

AB = 4i − 9 j − k − (i + 3j + 5k ) = 3i − 12 j − 6k

[6 marks] B1

The unit vector =

=

32 + 122 + 62 21 4 21 2 21 i− j− k 21 21 21

AC = kAB  p   1  3  ÷  ÷  ÷  −1÷−  3 ÷ = k  −12 ÷  3 ÷  5 ÷  −6 ÷       −4 = −12k or −2 = −6k 1 k= 3 1 p − 1 = (3) 3 p=2

(ii)

(iii)

(b)

3i − 12 j − 6k

M1 A1

3

M1

M1

A1

3

DABC is a parallelogram,

AD = BC  1  2   4   ÷  ÷  ÷ OD −  3 ÷ =  −1÷−  −9 ÷  5 ÷  3 ÷  −1÷        −1  ÷ OD =  11÷ = −i + 11j + 9k  9÷  

 4 2 −3   1 1 −2  2 −1 3 

5 ÷ 2÷ 1÷ 

 4 2 −3  4 R2 − R1 → R2   →  0 2 −5 2 R3 − R1 → R3  0 −4 9   4 2 −3   →  0 2 −5  0 0 −1  −z = 3 2 y − 5z = 3 4 x + 2 y − 3z = 5 x = 2, y = −6, z = −3 R3 + 2 R2 → R3

B1

M1

A1

3

B1

5 ÷ 3÷ −3 ÷  5 ÷ 3÷ 3÷ 

M1 M1

A1

M1 A1

6