document13
TRANSCRIPT
�
13 Linear Law
1. (a) y
x01 2 3
3
2
1
(b) y
x01 2 3
3
2
1
2. (a) Gradient= 3–�——–3–�
=� y-intercept=0 Equationofthelineisy=x.
(b) Gradient=– 4—5
,y-intercept=3.4
Equationofthelineisy=– 4—5
x+3.4
3. (a) y=�.5
(b) x=�.7
4. (a) Whenx=�.5, y=x
y=�.5
(b) Wheny=2.0,
2=– 4—5
x+3.4
x=(2–3.4)×1– 5—4 2
=�.75
5. (a) Gradient= 6–3——–4–�
=�
EquationoflineABis y–3=�(x–�) y=x+2
(b) Gradient= 5–2——–�–4
=–�
EquationoflinePQis r–5 =–�(t–�) =–t+� r =–t+6
(c) Gradient=– 5—–�0
=– �—2
EquationoflineRSisp=– �—2
q+5.
6. (a) Wheny=2,2=x+2 x=0
(b) Whent=3.5,r=–3.5+6 =2.5
(c) Whenp=3,3=– �—2
q+5
�—2
q=5–3
q=2×2 =4
7. (a) yx=4x2+5x
yx
—–x
=4x2—––x +
5x—–x y=4x+5
2
OR yx=4x2+5x
yx
—–x2 =
4x2—––x2
+5x—–x2
y
—x =4+51 �—x 2 =51 �—x 2+4
(b) y= 5—x –3x
xy=5–3x2
=–3x2+5 OR
y= 5—x –3x
y
—x =5—–x2
–3
y
—x =51 �—–x2 2–3
(c) ABy—–
x=
�—–x2
+ 7—x
x1 ABy—–
x 2=x1 �—–x2
+ 7—x 2 ABy= �—x +7
OR
ABy—–
x=
�—–x2
+ 7—x
x21 ABy—–
x 2=x21 �—–x2
+ 7—x 2 xABy=�+7x xABy =7x+�
(d) y2=8x2–5x
y2
—–x=
8x2
—–x –5x—–x
y2
—–x=8x–5
OR y2=8x2–5x
y2
—–x2
=8x2
—–x2
–5x—–x2
1 y—x 2
2
=–51 �—x 2+8
(e) y=pkx
log�0
y=log�0
pkx
=log�0
p+log�0
kx
=(log�0
k)x+log�0
p
(f) y=kxn
log�0
y=log�0
k+log�0
xn
=log�0
k+nlog�0
x =nlog
�0x+log
�0k
(g) y=pk(x+2)
log�0
y=log�0
p+log�0
k(x+2)
=log�0
p+(x+2)log�0
k =log
�0p+xlog
�0k+2log
�0k
=(log�0
k)x+log�0
p+2log�0
k
(h) y=k——
px–�
log�0
y=log�0
k–log�0
px–�
=log�0
k–(x–�)log�0
p =log
�0k–xlog
�0p+log
�0p
=(–log�0
p)x+log�0
k+log�0
p
(i) y= k——pnx2
log�0
y=log�0
1 k——pnx2 2
=log�0
k–log�0
pnx2
=log�0
k–(log�0
p+log�0
nx2) =log
�0k–log
�0p–x2log
�0n
=(–log�0
n)x2+log�0
k–log�0
p
(j) y=abx
—–k
log�0
y=log�0
1abx
—–k 2
=log�0
abx–log�0
k =log
�0a+log
�0bx–log
�0k
=log�0
a+xlog�0
b–log�0
k =(log
�0b)x+log
�0a–log
�0k
(k) y–4=abx
log�0
(y–4)=log�0
(abx) =log
�0a+xlog
�0b
=(log�0
b)x+log�0
a
(l) y=5+axn
y–5=axn
log�0
(y–5)=log�0
a+nlog�0
x =nlog
�0x+log
�0a
(m) y
—2
= 4—x –6—–x2
2x21 y—2 2=2x21 4—x –
6—–x2 2
x2y=8x–�2
3
OR
y
—2
= 4—x –6—–x2
2x1 y—2 2=2x1 4—x –
6—–x2 2
xy=8–�2—–x
xy=–�21 �—x 2+8
(n) xy2=3ABx+5x
xy2
—–x
=3ABx——x +
5x—–x
y2= 3—–ABx
+5
=31 �—–ABx 2+5
OR
xy2=3ABx+5x
xy2
—–ABx
=3ABx——ABx
+5x
——ABx
y2ABx=3+5ABx =5ABx+3
(o) y= 2x——–3+x
y(3+x)=2x 3y+yx=2x
3y+yx
———–y
=2x—–y
3+x=2x—–y
x—y = �—2
x+ 3—2
OR
y= 2x——–3+x
y(3+x)=2x 3y+xy=2x
3y+yx
———–x
=2x—–x
3y—–x +y=2
3y—–x =–y+2
y—x =1– �—
3 2y+ 2—3
8. (a) y=ax2+bx
y—x =
ax2
–—x +bx–—x
y—x =ax+b.................................. 1
a=Gradient
= 6–2——–5–�
= 4—4
=�
Substitutey—x =2,x=�anda=�into1,
2=�(�)+b b=�
Therefore,a=�andb=�.
(b) y=a+ c—x x(y)=x1a+ c—x 2 xy=ax+c................................ 1 a=Gradient
= 5–�——–�–4
= 4—––3
Substitutex=�,xy=5anda=– 4—3
into1,
5=– 4—3
(�)+c
c=5+ 4—3
= �9—–3
Therefore,a=– 4—3
andc= �9—–3
.
(c) y=abx
log�0
y=log�0
(abx) =log
�0a+log
�0bx
=(log�0
b)x+log�0
a
log�0
b=Gradient
= 5–2——–3–0
=� b=�0
log�0
a=log�0
y-intercept =2 a=�02
=�00
Therefore,a=�00andb=�0.
4
(d) y=axn
log�0
y=log�0
a+nlog�0
x =nlog
�0x+log
�0a
n=Gradient
= �–3——–�–0
=–2
log�0
a=log�0
y-intercept =3 a=�03
=�000
Therefore,a=�000andn=–2.
(e) y=abx2
log�0
y=log�0
a+log�0
bx2
=log�0
a+x2log�0
b log
�0y=(log
�0b)x2+log
�0a............ 1
log�0
b=Gradient
= 7–3——–6–�
= 4—5
=0.8 b=�00.8
=6.3�0
Substitute log�0
y = 3, x2 = � and log�0
b = 4—5
into1,
3= 4—5
(�)+log�0
a
log�0
a=3– 4—5
= ��—–5
a=�58.5
Therefore,a=�58.5andb=6.3�0.
9. (a) (i)x 2 3 4 5 6
y 2.2 2.8 4.0 5.0 5.3
yABx 3.� 4.8 8.0 ��.2 �3.0
y=kABx+ h—–ABx
yABx=kx+h
x��y
0
4
6
8
10
12
14
2
–2
x1 2 3 4 5 6
(ii) h=yABx-intercept =–2 k =Gradient
= �3–(–2)——–—–6–0
= 5—2
(b) (i) y=axn
log�0
y=log�0
axn
=log�0
a+nlog�0
x log
�0y=nlog
�0x+log
�0a
log10x 0.3 0.40 0.48 0.54 0.60 0.65
log10y 0.68 0.94 �.�8 �.37 �.57 �.74
0
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0.2
–0.2
log10 x
log10 y
0.2 0.4 0.6 0.8
5
(ii) log�0
a=log�0
y-intercept =–0.�8 a=0.66�
n =Gradient
= �.08–0.4——–—––0.44–0.2
=2.833
10. (a)
0
2
3
4
6
5
1
x
F
2 4 6 8 10 12
(b) (i) WhenF=2.5,x=5.2 (ii) Whenx=�2.0,F=5.7
(c) x=2.5iswronglytakenwhenF=�.0. Thecorrectvalueisx=2.0.
11. (a)v2 �9.98 40.07 60.06 79.92 9�.20 ��9.90
s �.0 2.0 3.0 4.0 5.0 6.0
0
40
60
80
120
100
20
s
v 2
2 4 6
(b) v=9.55waswronglytaken. Thecorrectvalueisv2=�00,thatis,v=�0.
(c) (i) Whenv=8,v2=64,s=3.2 (ii) Whens=3.5,v2=70 v=8.37
12. (a) Gradient= �40–22——–—–4–2
=59
y
—x =59x2+c............................... 1
Substitutex2=2,y
—x =22into1,
22=59(2)+c c=–96
Therefore,y
—x =59x2–96
y=59x3–96x
(b) Whenx=2.4,y=59(2.4)3–96(2.4) =585.2
13. (a) y=axn.......................................... 1
Substitutex=�,y=�00into1, �00=a(�)n
a=�00
Substitutex=3,y=900into1, 900=�00(3)n
3n=9 =32
n=2
(b) y=axn
y=�00(x)2
Y=mX+c Y=y,X=x2,c=0,m=�00
Therefore,Y=�00X
WhenY=5,X=a, 5=�00a
a= 5—––�00
= �—–20
WhenY=b,X=4, b=�00(4) =400
Hence,a= �—–20
,b=400andc=0.
6
1. y=3x+ 4–––x2
x2(y)=x213x+ 4–––x2 2
x2y=3x3+4 Y=mX+c
Interceptatverticalaxis,c=4
Therefore,pointatverticalaxisis(0,4).
m=Gradient=3
h–4——–�–0
=3
h–4=3 h=7
m=344–4—–—–k–0
=3
3k=40
k= 40–––3
2. y=px3+qx2
�–––x2
(y)= �–––x2
(px3+qx2)
y
–––x2
=px+q................................... 1
Y=mX+c
p=Gradient
= 45–7—–—–3–�
= 38–––2
=�9
Substitutex=�,y
–––x2
=7andp=�9into1,
7=�9(�)+qq=–�2
3. (a) y=k·�0x
log�0
y=log�0
(k·�0x) =log
�0k+log
�0�0x
=log�0
k+xlog�0
�0 log
�0y=x+log
�0k
Y=mX+c
(b) (i) log�0
k=� k=�0
(ii) Gradient=�
p–�—–—2–0
=�
p–�=2 p=3
4. y=2x2–3y
–––x2
= 2x2
––––x2
– 3–––x2
y–––x2
=–31 �–––x2 2+2
Y=–3X+c
Therefore,X= �–––x2
andY= y–––x2
.
5. (a)y
—x�.9� �.84 �.73 �.63 �.58 �.55
1—–x2 0.2� 0.�6 0.�� 0.06 0.04 0.03
0
1.0
1.5
2.0
0.5
x
x 20.05 0.10 0.15 0.20–1
–y
(b) (i) y=p—x +
q—p x
y
—x =p
—–x2
+q—p
=p1 �—–x2 2+
q—p
Y=mX+c
p=Gradient
= �.6–�.5—–—–—0.05–0
=2
(ii)q—p =
y—x -intercept
q—2
=�.5
q=3
7
6. (a)log10y �.08 �.2� �.3� �.43 �.52 �.60
ABx � �.22 �.4� �.58 �.73 �.87
0
1.0
1.5
0.5
��x0.5 1.0 1.5 2.0
log10 y
(b) y=pkABx
log�0
y=log�0
pkABx
=log�0
p+log�0
kABx
log�0
y=(log�0
k)ABx+log�0
p Y=mX+c
(i) log�0
p=log�0
y-intercept =0.45 \p=2.82
(ii) log�0
k=Gradient
= �.5–0.6—–—–—–�.7–0.25
=0.6207 k=4.�8
7. (a)log10y 0.48 0.80 �.�2 �.44 �.77 2.09
x–1 0 � 2 3 4 5
0
1.0
1.5
2.0
0.5
x – 12 4 6
log10 y
(b) (i) y=pkx–�
log�0
y=log�0
p+(x–�)log�0
k log
�0y=(log
�0k)(x–�)+log
�0p
Y=mX+c
log�0
p=x–�-intercept =0.5 p=3.�6
(ii) log�0
k=Gradient
= 2.0–�.0—–—–—4.8–�.6
=0.3�25 k=2.05
8. (a)x � 2 3 4 5 6
log10y 0.30 0.60 0.90 �.20 �.5� �.8�
0
1.0
1.5
2.0
0.5
x2 4 6
log10 y
(b) (i) y=kx
—–p
log�0
y=log�0
1 kx
—–p 2 =log
�0kx–log
�0p
log�0
y=(log�0
k)x–log�0p
Y=mX+c
log�0
y-intercept=–log�0
p 0=–log
�0p
log�0
p=0 p=�
(ii) log�0
k=Gradient
= �.7–0—–—–5.6–0
=0.3036 k=2.0�
1. y=kxn–�
log�0
y=log�0
(kxn–�) =log
�0k+log
�0xn–�
=log�0
k+(n–�)log�0
x =(n–�)log
�0x+log
�0k
Gradient=n–�Verticalintercept=log
�0k
8
2. Gradient= �3–3—–—–6–�
= �0—–5
=2
Therefore,ABy=2x2+c...................... 1
Substitutex2=�,ABy=3into1,3=2(�)+cc=�
Hence,theequationisABy=2x2+�,thatis,y=(2x2+�)2.
3. y=px+qx3—2
y—x =
px—–x +
qx3—2
——x =p+qABx =qABx+p...................................... 1 Y=mX+c
q=Gradient
= 8–4—–—4–6
= 4—––2
=–2
SubstituteABx=4,y
—x =8,q=–2into1,8=(–2)(4)+pp=�6
Therefore,p=�6andq=–2.
4. y=�02x+3
log�0
y=log�0
�02x+3
=(2x+3)log�0
�0log
�0y=2x+3
Y=mX+c
log�0
y-intercept=3
Gradient=2
ForpointA,4–3—–—q–0
=2
�=2q
q= �—2
ForpointB,p–3—–—�–0
=2
p–3=2 p=5
Therefore,p=5,q= �—2
.
5. y=ax5
log�0
y=log�0
a+5log�0
xlog
�0y=5log
�0x+log
�0a
Y=mX+c
log�0
a=Interceptoflog�0
y =4 a=�04
=�0000
Gradient=5
k–4——–3–0
=5
k–4=�5 k=�9
Therefore,a=�0000andk=�9.
6. (a) Gradient= 5–2——–3–0
=� Verticalintercept=2
Theequationofthelineis Y=mX+c y=log
�0x+2
(b) x=a×�0y–�
log�0
x=log�0
a+(y–�)log�0
�0 log
�0x=log
�0a+y–�
y=�–log�0
a+log�0
x
\�–log�0
a=2 log
�0a=–�
a=0.�
7. (a) Gradient= �0–4——––3–�
=3 Equationofthelineislog
�0y=3log
�0(x+2)+c
Substitute log�0
(x+2)=�andlog�0
y=4 intotheequation,
log�0
y=3log�0
(x+2)+c 4=3(�)+c c=�
Therefore,log�0
y=3log�0
(x+2)+�
y=k(x+2)n
log�0
y=log�0
k+nlog�0
(x+2) log
�0y=nlog
�0(x+2)+log
�0k
Comparetolog�0
y=3log�0
(x+2)+� Hence,n=3andlog
�0k=�
k=�0
9
(b) Wheny=0.08, y=k(x+2)n
0.08=�0(x+2)3
0.008=(x+2)3
x+2=3 0.008ABBBB =0.2 x=0.2–2 =–�.8
8. x—y = 3—x +x
�—x 1 x—y 2 = �—x 1 3—x +x2 �—y = 3—–
x2+�
�—y =31 �—–x2 2+�.......................... 1
Substitute �—–x2
=3and �—y =kinto1,
�—y =31 �—–x2 2+�
k=3(3)+� =�0
Substitute �—–x2
=pand �—y =�6into1,
�6=3(p)+�3p=�5 p=5
9. (a) y=rx2+t.................................... 1
Substitutex=�,y=4into1, 4=r(�)2+t r+t=4....................................... 2
Substitutex=3,y=36into 1, 36=r(3)2+t 9r+t=36................................... 3
3–2, 8r=32 r=4
Substituter=4into2, 4+t=4 t=0
(b) y=rx2+tbecomesy=4x2
�—x (y)= �—x (4x2)
y
—x =4x
Therefore,gradientis4.
ForA(2,p),
p–0——–2–0
=4
p=8
ForB(q,�2),
�2–0—–—–q–0
=4
�2=4q q=3
10. (a) Thegradientofthestraightline= ��–7—–—–4–2
=2
Gradient= r–7—–—3–2
=2
r–7=2 r=9
(b) log�0
y=2x+c............................ 1
Substitutelog�0
y=7,x=2into1, 7=2(2)+c c=3
Therefore,log�0
y=2x+3 y=�02x+3
11. (a) y=px2+q................................... 1
Substitutex=�,y=4into1, 4=p(�)2+q p+q=4...................................... 2
Substitutex=3,y=�2into1, �2=p(3)2+q 9p+q=�2.................................. 3
3–2,8p=8 p=�
Substitutep=�into2, �+q=4 q=3
(b) y=px2+qbecomes y=x2+3
y
—–x2
=�+3—–x2
y
—–x2
=31 �—–x2 2+�
ForA,
y
—–x2
=4—–�2
=4
�—–x2
=�—–�2
=�
�0
ForB,
y
—–x2
=�2—–32
=�2—–9
=4—3
�—–x2
=�—–32
=�—9
0 x2
11
A(1, 4)
–
x2
y–
�–, –�19
43B
12. x2y=px....................................... 1�—–x2
(x2y)=�—–x2
(px)
y=p1 �—x 2InDiagram(b),gradient =p
=�4–7———�– �—
2 =�4
Substitutep=�4into1, x2y=�4xForA, �2·r=�4(�) r=�4
ForB, t2(7)=�4t 7t2=�4t 7t=�4 t=2
13. (a)x 2.5 4.5 6.5 7.0 8.5 9.5
y2 2.5 4.0 5.6 6.0 7.� 7.9
0
4
6
8
2
x2 4 6 8 10
y 2
(b) (i) y2=px+q p=Gradient
=6–�———–
7–0.5
=�0—–�3
q=y2-intercept =0.6
(ii) Whenx=�.5,y2=�.7 y=�.3
14. (a)x � 4 9 �6 20 25
yABx 4.0� �4.02 30.06 52.80 66.0 82.0
0
40
60
80
20
x10 20 30
y ��x
(b) (i) y=p
——ABx
+(q+�)ABx
ABx(y)=ABx1 p——ABx
+(q+�)ABx2 yABx =p+(q+�)x yABx=(q+�)x+p
p=verticalintercept =2
q+�=Gradient
= 82–34———–25–�0
=3.2 q =2.2
(ii) Whenx=�0,yABx=34
yABB�0 =34
y=34——
ABB�0 =�0.75
��
15. (a)log10x 0 0.30 0.54 0.65 0.699 0.778
log10y 0.6 �.�8 �.69 �.9� 2 2.26
0
1.0
1.5
2.0
2.5
0.5
lg x
lg y
0.2 0.4 0.6 0.8
(b) y=pxn—2
log�0
y=log�0
p+ n—2
log�0
x
log�0
y= n—2
log�0
x+log�0
p
log�0
p=Interceptoflog�0
y =0.6 p=3.98�
n—2
=Gradient
=2.2–�.0————–0.8–0.2
=2 n=4
(c) Incorrectvalueofyis�80. Thecorrectvalueislog
�0y=2.2,
thatis, y=�58.5
16. (a)x 3.0 3.5 4.0 4.5 5.0 5.6
log10y �.4 �.53 �.68 �.83 �.98 log�0
p
0
1.0
1.5
2.0
0.5
x
log10 y
2 4 6
(b) (i) y=k·ax–�
log�0
y=log�0
k+(x–�)log�0
a =log
�0k+xlog
�0a–log
�0a
log�0
y=(log�0
a)x+log�0
k–log�0
a
log�0
a=Gradient
= 2.�–0.8————5.4–�
=0.2955 a=�.975
log�0
k–log�0
a=Interceptoflog�0
y =0.5 log
�0k–0.2955=0.5
log�0
k=0.7955 k=6.24
(ii) Whenx=5.6, log
�0y=2.�5
y=�4�.3 Therefore,p=�4�.3
(iii) Whenx=2.0,log�0
y=�.� y=�2.6
17. (a)T 0 26.8 32.9 38.0 42.9 46.5
ABl 0 4.5 5.5 6.3 7.� 7.7
0
10
20
30
40
50
–10
��l
T
2 4
New
Old
6 8
(b) Gradient= 30–0———5–0
=6
Therefore, T=6ABl T2=36l
(c) (i) T=6ABl–6
(ii) WhenT=6.5,ABl =2.� l =2.�2
=4.4�
�2
18. (a) y =k(x–�)n
log�0
y =log�0
k+nlog�0
(x–�) log
�0y =nlog
�0(x–�)+log
�0k
Y =mX+c
log10y 0.7 �.3 �.66 �.79 �.95 2.26
log10(x–1) 0 0.3 0.48 0.54 0.62 0.78
0
0.5
1.0
1.5
2.0
2.5
lg y
0.2 0.4lg (x – 1)
0.6 0.8
(b) (i) Wheny=�00,log�0
�00=2 log
�0(x–�)=0.65
x–�=4.467 x=5.467
(ii) Whenx=8, log�0
7=0.85 log
�0y=2.4
y=25�.2