UNIVERSITI TUNKU ABDUL RAHMAN ACADEMIC YEAR 2012/2013 APRIL EXAMINATION FHMM1014 MATHEMATICS I THURSDAY, 25 APRIL 2013 TIME: 5.00 PM – 7.00 PM (2 HOURS)

FOUNDATION IN SCIENCE

SOLUTIONS

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Solutions to FHMM1014 Mathematics I (April 2013)

2 FHMM1014 MATHEMATICS I Q1. (a) (i) a + bi = +i 1 (a − 2) + bi ⇒ a + bi = a − 2) + bi )(1 + i ) (( a + bi = (a − 2 − b) + (a − 2 + b)i

a a = − 2 − b So   b = a−2+b

⇒ ⇒

− b=2 a= 2

z= 2 − 2i

(ii) (iii)

z=

4+4 = 2 2
⇒ −π θ= 4

−2 = tan θ =−1 2 = arg( z ) θ= −π 4

(b)

(i)

3log x x log 3 =
3log x =81

⇒
⇒

2 × 3log x 162 =
3log x =34

log x = 4

⇒

x = 104

(ii)

= log10 x + log(1 + 2 x ) log 5x + log 6 log10 x (1 + 2 x )= log 5 x × 6 10 x (1 + 2 x ) = x × 6 5 Let ⇒ 2 x (1 + 2 x ) = 6

y = 2 x then y 2 + y − 6 = 0

( y − 2)( y + 3) = 0

y − 2 = 0 ⇒ y = 2 ⇒ 2x = 2 ⇒ x = 1 y + 3 = ⇒ y = 3 ⇒ 2 x = 3 impossible − − 0

Solutions to FHMM1014 Mathematics I (April 2013)

3 FHMM1014 MATHEMATICS I Q1. (Continued) (c)
P ( x)= A( x − 2)( x − (2 − i ))( x − (2 + i ))

P( x) = A( x − 2)( x 2 − 4 x + 5) P( x) = A( x3 − 6 x 2 + 13 x − 10)
P(0) = −5 P( x) = ⇒ 1 A= 2

1 3 ( x − 6 x 2 + 13 x − 10) 2

(d)

A(B  ( A − ( B  C ) )′ = C )′

(

)′

= A′  ( B  C )
= ( A′  B )  ( A′  C )
= ( A  B′ )′  ( A  C ′ )′

=B )′  ( A − C )′ (A−

Q2.

(a)

x3 − 2 x 2 + 3x − 4 − x2 + 2x − 3 = 1+ 3 x3 − x 2 + x − 1 x − x2 + x −1

− x2 + 2x − 3 − x2 + 2x − 3 A Bx + C = = + 2 3 2 2 x − x + x − 1 ( x − 1)( x + 1) ( x − 1) ( x + 1)
− x 2 + 2 x −= A( x 2 + 1) + ( Bx + C )( x − 1) 3

x= 1 x= 0

⇒
⇒

−1 + 2 − 3 =2 A ⇒
−3 = A − C ⇒

A =−1
C= 2
⇒ B= 0

x =1 ⇒ −

2 −1 − 2 − 3 = A − 2(− B + C )

Solutions to FHMM1014 Mathematics I (April 2013)

4 FHMM1014 MATHEMATICS I Q2.(a). (Continued)

x3 − 2 x 2 + 3x − 4 1 2 = 1− + 2 3 2 x − x + x −1 x − 1 ( x + 1)
(b) (i)
−1 < x −1 1 4 −1 2 & x> 1 4

So the valid range is x >

(ii)

x2 + x + 2 −1 ≤ 0 x 2 − 3x + 2
0 + +

⇒
1

4x ≤0 ( x − 1)( x − 2)
2 + + + + + +

x x x −1 x−2 4x ( x − 1)( x − 2) x≤0 and

1< x < 2

(c)

(i)

2n − 1 an = n ` 3 lim an = lim n →∞

(ii)

2n − 1 2 1 = lim   − lim   = 0 n →∞ 3n n →∞ 3 n →∞ 3     n n

The series is convergent.

Solutions to FHMM1014 Mathematics I (April 2013)

5 FHMM1014 MATHEMATICS I Q3 (a) (i)

5 a + ar = ; 3 2 3 ar + ar = r 2 (a + ar ) = 60 r 2 = 36

r=6 ⇒ a=

5 21 1 3

r = −6 ⇒ a = −

(ii)

a = −13 , d = 3
S = S 20 − S 9
20 9 S 20 = ( −26 + 19(3) ) = , S9 = ( −26 + 8(3) ) =9 310 − 2 2
S = 310 − (−9) = 319

(b)

(i)

 1  1   1  1  3    −    −  −  2 2 1 (− x )2 +  2  2  2  (− x )3 +  (1 − x) = 1 +  (− x ) +   2! 3! 2
1 2

= 1−

1 1 1 x − x 2 − x 3 + ... 2 8 16

(ii)

 502 − 1  2 1 2  50 1 − 2  = 2  50   50   50 

1

1

= (50 − 1)(50 + 1)
= 7 51

Solutions to FHMM1014 Mathematics I (April 2013)

6 FHMM1014 MATHEMATICS I Q3(b). (Continued)

(iii)

1 2 1  1  4999  1− 2  ≈ 1−  2  =  2  50  5000  50 
 4999  7 51 ≈ 50   ⇒  5000  51 ≈ 4999 700

1

P (c) 40.3 Q 31.8

S

63.5

R

(i)

cos QPR =

^

40.32 + 31.82 − 63.52 = (−0.5450) 2(40.3)(31.8)

∠ QPR = 123.03° (accept 123.02°)

(ii)

sin PQR =

^

31.8sin123.030 (0.4199) 63.5

∠ PQR = 24.83°

(iii)

PS =

40.3sin 24.830 sin 93.650

PS = 16.96 cm

Solutions to FHMM1014 Mathematics I (April 2013)

7 FHMM1014 MATHEMATICS I Q4. (a) (i) 1 1 1 + cos A + 1 − cos A + = 1 − cos A 1 + cos A (1 − cos A)(1 + cos A)
=
=

2 1 − cos 2 A
2 sin 2 A

= 2 csc 2 A

(ii)

5sin θ cos θ = 2 sin 2θ = 4 5

2θ = 0.9273r , 2.2143r , 7.2105r , 8.4975r

θ = 0.4637 r , 1.107 r , 3.605r , 4.249r

(b)

(i)

Let 12sin θ − 5cos θ ≡ R sin(θ − α )
≡ R sin θ cos α − R cos θ sin α
∴ R cos α = 12
R sin α = 5

 (1)
 (2)

(2) : (1)

tan α =

5 12

⇒ α = 22.62° ⇒ R = 13

(1) 2 + (2) 2 :

R 2 = 122 + 52 = 169

Thus, 12sin θ − 5cos θ ≡ 13sin(θ − 22.62°)

(ii)

7 sin(θ − 22.62°) = 13

θ − 22.62° 32.58° , 147.42° =

θ = ° , 170.04° 55.2
Solutions to FHMM1014 Mathematics I (April 2013)

8 FHMM1014 MATHEMATICS I Q4.(b). (Continued) (iii) The maximum value is when 12sin θ − 5cos θ is minimum. ( − 13 ) The maximum value of
1 1 1 is = 12sin θ − 5cos θ + 16 −13 + 16 3

(c)

(i)

a =5
~

 3 − 5 a=  ~  4     5 
^

(ii)

 −3   1   32  p   + q   =  4  3   −13 
32 −3 p + q = 4 p + 3q = −13 p= − q= 109 13

89 13

(iii)

a⋅ b = 3(1) + 4(3) = − 9
~ ~

a⋅ b = b cos θ a
~ ~ ~ ~

∴ cos θ=

= 0.5692 (5)( 10)

9

= 55.31° θ

Solutions to FHMM1014 Mathematics I (April 2013)