SIT190 Assignment 2 – Solutions Total marks: 80
(i)
(ii)
(iii)
Simplify, and express in terms of positive indices:
= 3−4 = 3
4−3 −3−1
3
24 �6−1�2 = 24 × 36−2
4
3−2 = 829−4 −4
= 824 −3 = 3�−3�
2−1+−2 3(2−1+−2) =1
2� 22+
� 12 1�2 3�2
(iv) �34�3= � = −3�2 −3�2
= 2
(v) √�2√ − √16� = 2 − 4
= −2
2. Sketch =2 and=2− onthesamesetofaxes. y = 2e−x y 2 x 3.
(a) Simplify (by writing as a single exponential):
(i) 3+2 = 3+2−(4−3) 4−3 = 5−1
(ii) �+3�2 = 2+6 4−3 4−3
= 2+6−(4−3)
= 5+2
(iii) (3−1)2 × 3−6 = 6−23−6
= 1 =
4.
(a) Simplify (by expressing as a single natural logarithm): (i) 3ln4−5ln2+ln3=ln43 −ln25 +ln3
(b) Expand and simplify:
(i) (3 − −3)2 = (3)2 − 23−3 + (−3)2 (perfect square)
= 6 − 2 + −6
(ii) 2(2 + −2)(2 − −2) = 2((2)2 − (−2)2) = 2(4 − −4)
(difference of squares)
= ln�43×3� 64×3
=ln�25 �
= 6 − −2
= ln 6 32
(ii) 2ln(3) − ln(34) + 2 ln(23) = ln(92) − ln(34) + ln(46) = ln �92×46�
34 = ln(124)
(iii) 2ln(2) + ln �3� − 4 ln = ln(42) + ln �3� − ln(4) = ln�42×3�
4× = ln(3)
(b) Simplify (without using a calculator): (i) log 1 =log 1
4 16 4 42 = log4 4−2
5.
Solve for :
(i) = 23−1
= −2 �1
(ii)ln�12�= �2
(iii) ln(3+1) = 3 + 1 3 − 1 = 2 ln(3−1) = ln �2�
3 − 1 = l n � 2 � 3 = l n � 2 � + 1 = l n � 2 � + 1
3

(ii) = 1 ln �5+2� 22 ln �5+2� = 2 ln�5+2� 2
2 2=
5+2 = 2
2
5 = 22 − 2 = 22−2
5
(iii) + 3 = 4 − 2+3 2+3 + + 3 = 4
2+3 = 1 − ln(2+3) = ln(1 − ) 2 + 3 = ln(1 − ) = ln(1 − ) − 3
2
6.
7.
(i) sin 60° = 0.866
(ii) tan 2.33 = −1.054
(iii) cos 2 + cos 20° = −0.41615 + 0.93969
Use a calculator to evaluate (answer correct to 3 decimal places):
= 0.524
Find the unknown length and angle for each of the following triangles.
(i) (ii)

47°
35
52° 46° = 90 − 52 = 38° =90−46=44°
10
sin52°= cos46°=10 35 = 35 × sin 52° = 27.580
= 10 =14.369
(iii)
(iv) cos 46°
50°

10

5.2
=√100−64=6

 = 90 − 50 = 40° tan50°=
2 28 2 +8=10
5.2
= 5.2 × tan 50° cos=8=0.8  = 6.197
−1
= cos 10(0.8) = 36.870°

8.
9.
Express in degrees:
(i) 2.5 = 2.5 × 180° = 143.2°
5 5 180°
(ii) =  ×   = 300°
33
(iii) 3 = 3 × 180° = 540°

Express in radians in terms of : (i) 25° =25× =25
= 5 180 180 36
(ii) 75° =75× =75 = 5 180 180
12
10. If cos = −√11 and is in the 2 (i) sin 5 nd quadrant, find the exact values of:
cos2 + sin2 = 1
11 + sin2 = 1
25 14 sin2 = 25
sin = ±√14 5
Since is in the 2 sin = √14 nd quadrant,
sin (ii) tan = 5
= cos √14�5
−√11�5
= −�14 11