# Mat 1581

Topics: Academic term, Logarithm, Natural logarithm Pages: 31 (3877 words) Published: March 5, 2013
MAT1581/201/2/2012

Tutorial Letter 201/2/2012
Mathematics I

MAT1581 Semester 2
Department of Mathematical Sciences
This tutorial letter contains solutions to your assignments and guidelines for the examination.

Bar code

CONTENTS
1 2 3 4 5 6 SOLUTIONS ASSIGNMENT 1 ....................................................................................................... 3 SOLUTIONS ASSIGNMENT 2 ....................................................................................................... 7 ANSWERS PREPARATION PAPER 1 (MAY 2010) .................................................................... 7 ANSWERS PREPARATION PAPER 2 (OCTOBER 2011) ........................................................... 8 SOLUTIONS PREPARATION PAPER 3 (MAY 2011) ................................................................. 11 GUIDELINES TO WRITING THE EXAMINATION ....................................................................... 17

Please note: 1) Your study guides have the codes MAT181Q and WIM131U. Always use the new code MAT1581. 2) Register on myUNISA and activate your mylife e-mail to receive new information immediately. 3) Solutions for semester 1 assignments on myUNISA additional resources. 4) More past examination papers on myUNISA official study material. (Should solutions be made available it will be on additional resources). Students may send an answer to the lecturer for comments. 5) No past examination papers or solutions are sent to students upon request. 6) Check on myUNISA before the examination for any corrections or alternative solutions to assignment questions provided in this letter.

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MAT1581/201

1

SOLUTIONS ASSIGNMENT 1

Question 1 Solve for x without the use of a calculator: 1.1

x = log3 9 − 6log 4 0.5
log 2−1 log 4 log 2 = log 3 3 + 6 2 log 2 6 = 1+ 2 =4 = log 3 32

( )

1 2

−6

(6)

1.2

Solve for x: e

2x

− 3e x − 4 = 0

(e x − 4)(e x + 1) = 0 (e x − 4) = 0 ex = 4 x = ln 4 = 1,386 or (e x + 1) = 0 impossible

(4)

[10] Question 2 Use the Binomial expansion to find the first three terms of (16 − 3x ) Simplify each term. (Note the change of marks from 3 to 6) −1 4

.

T1 = (16 )

−1

4

= ( 24 )

−1

4

= 2 −1 =

1 2

3x −5  −1  T2 =   (16 ) 4 ( −3 x ) = 128  4   −1  −5  2     4  4  16 − 9 4 −3 x 2 = 45 x T3 = ( ) ( ) 2! 16384 Thus (16 − 3 x ) −1 4

[6]

1 3x 45 x 2 = + + + .... 2 128 16384 = 0,5 + 0,0234 x + 0,00275 x 2 + ..... 3

Question 3 Resolve into partial fractions: Answer: Remember the constants A, B and C goes with the factor in the denominator and can be chosen differently each the student.

(x − 1)(x − 1)
2

2x

(x

2x
2

− 1 ( x − 1)

)

=

2x

( x − 1) ( x + 1)
2

=
2

A B C + + ( x − 1) ( x + 1) ( x − 1)2 ∴C = 1 ∴B = − 1 2 1 2 (7)

∴ 2 x = A ( x + 1)( x − 1) + B ( x − 1) + C ( x + 1) Put x = 1: Put x = −1: Put x = 0 : 2x 2 = 2C −2 = 4 B 0 = − A + B + C∴ A =

( x − 1) ( x + 1)
2

=

1 1 1 − + 2 ( x − 1) 2 ( x + 1) ( x − 1)2
[7]

Question 4 The following equations represent observations made by a surveyor in the adjustments of a level of networks: 2 xa − 3 xb = 1,313 4 xb − 3 xc = 1,906 xa − xc = 3, 784

Use Cramer’s rule to determine the value of xa, xb and xc. Answer:

Rearrange equations: = 1,313 2 xa − 3xb 4 xb − 3xc = 1,906 − xc = 3, 784 xa 2 −3 0 ∆ = 0 4 −3 = 1 1 0 −1 1,313 −3 ∆ xa = 1,906 3, 784 4

0 −3 = 23, 086 −1

4 0

MAT1581/201

xa =

∆ xa ∆

= 23, 086

∆ xb

2 1,313 0 = 0 1,906 −3 = 14,953 1 3, 784 −1 ∆ xb ∆ = 14,953 (10) [10]

xb =

∆ xc

2 −3 1,313 = 0 4 1,906 = 19,302 1 0 3, 784 ∆ xc ∆ = 19,302

xc =

Question 5 Solve for θ if 0 < θ < 2π : 9 cos 2 θ − 2cosθ − 3 = 0 Give your answers in terms of π. Answer: − ( −2 ) ± (−2) 2 − 4 ( 9 )( −3) 2 (9)

cos θ

= =

2 ± 112 18 cos θ = 0, 699 ref ∠ =45,65° θ = 45, 65°(= 0, 796rad) =45, 65°× =0,253 π or θ

or cos θ = −0, 47 ref ∠ =61,52° θ...