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1. (a)y
x01 2 3
3
2
1
(b) y
x01 2 3
3
2
1
2. (a) Gradient =3 13 1
= 1
y-intercept = 0
Equation of the line isy= x.
(b) Gradient = 45
,y-intercept = 3.4
Equation of the line isy= 45x + 3.4
3. (a) y= 1.5
(b) x= 1.7
4. (a) Whenx= 1.5,
y= x
y= 1.5
(b) Wheny= 2.0,
2 = 45x+ 3.4
x= (2 3.4) 54
= 1.75
5. (a) Gradient =6 34 1
= 1
Equation of lineABis
y 3 = 1(x 1)
y=x+ 2
(b) Gradient =5 21 4
= 1
Equation of line PQis
r 5 = 1(t 1)
= t+ 1
r = t+ 6
(c) Gradient = 5
10
= 12
Equation of lineRSis p= 12q+ 5.
6. (a) Wheny= 2, 2 =x+ 2
x= 0
(b) When t= 3.5, r= 3.5 + 6
= 2.5
(c) Whenp= 3, 3 = 12q+ 5
12q= 5 3
q= 2 2 = 4
7. (a) yx= 4x2+ 5x
yx
x
=4x 2
x
+5x
x
y= 4x+ 5
CHAPTER
13 Linear Law
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OR
yx= 4x2 + 5x
yx
x2
=4x 2
x2
+5x
x2
yx
= 4 + 5 1x
= 5 1x+ 4
(b) y=5x
3x
xy= 5 3x2
= 3x2+ 5
OR
y=5x
3x
yx
=5
x2
3
y
x
= 5
1
x2
3
(c)y
x
=1
x2
+7x
xyx=x1
x2
+7x
y = 1x
+ 7
OR
y
x
=1
x2
+7x
x2
yx
= x2
1x2
+7x xy = 1 + 7x
xy = 7x+ 1
(d) y2= 8x2 5x
y2
x
=8x2
x
5x
x
y2
x
= 8x 5
OR
y2= 8x2 5x
y2
x2 =8x2
x2 5x
x2
yx2
= 5 1x+ 8
(e) y=pkx
log10
y= log10
pkx
= log10
p+ log10
kx
= (log10k)x+ log
10p
(f) y= kxn
log10
y= log10
k+ log10
x n
= log10
k+ nlog10
x
= nlog10
x+ log10
k
(g) y=pk(x+ 2)
log10
y= log10
p+ log10
k(x+ 2)
= log10p+ (x+ 2) log10k = log
10p+ xlog
10k+ 2 log
10k
= (log10
k)x+ log10
p+ 2 log10
k
(h) y=k
px 1
log10
y= log10
k log10
px 1
= log10
k (x 1) log10
p
= log10
k xlog10
p+ log10
p
= (log10
p)x+ log10
k+ log10
p
(i) y=k
pnx
2
log10
y= log10
kpnx2 = log
10k log
10pnx
2
= log10
k (log10
p+ log10
nx2)
= log10
k log10
p x2log10
n
= (log10
n)x 2+ log10
k log10
p
(j) y=abx
k
log10
y= log10
abx
k
= log10
abx log10
k
= log10
a+ log10
bx log10
k
= log10
a+ xlog10
b log10
k
= (log10
b)x+ log10
a log10
k
(k) y 4 = abx
log10
(y 4) = log10
(abx)
= log10
a+ xlog10
b
= (log10
b)x+ log10
a
(l) y= 5 + axn
y 5 = axn
log10
(y 5) = log10
a+ nlog10
x
= nlog10
x+ log10
a
(m) y2
= 4x
6x2
2x2y2= 2x24x
6
x2
x2y= 8x 12
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OR
y2
=4x
6
x2
2xy2= 2x4x
6
x2
xy= 8
12
x xy= 12 1x+ 8
(n) xy2= 3x + 5x
xy2x
=3x
x
+5x
x
y2=3
x
+ 5
= 3 1x
+ 5 OR
xy2= 3x + 5x
xy2x
=3x
x
+5x
x
y2x = 3 + 5x = 5x + 3
(o) y=2x
3 + x
y(3 +x) = 2x
3y+ yx= 2x
3y+ yx
y
=2x
y
3 + x=2x
y
xy
=12x+
32
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
yx
= 13y+ 2
3
8. (a) y= ax2+ bx
yx
=ax2
x
+bx
x
yx
= ax+ b..................................
a= Gradient
=6 2
5 1
=44
= 1
Substituteyx
= 2,x= 1 and a= 1 into ,
2 = 1(1) + b
b= 1
Therefore, a= 1 and b= 1.
(b) y= a+c
x
x(y) =x
a+c
x xy= ax+ c................................
a= Gradient
=5 1
1 4
=4
3
Substitutex= 1,xy= 5 and a= 43
into ,
5 = 43
(1) + c
c= 5 +43
=19
3
Therefore, a= 43
and c=19
3
.
(c) y= abx
log10
y= log10
(abx)
= log10
a+ log10
bx
= (log10
b)x+ log10
a
log10
b= Gradient
=5 2
3 0
= 1 b= 10
log10
a= log10
y-intercept
= 2
a= 102
= 100
Therefore, a= 100 and b= 10.
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(d) y= axn
log10
y= log10
a+ nlog10
x
= nlog10
x+ log10a
n= Gradient
=1 3
1 0
= 2 log
10a= log
10y-intercept
= 3
a= 103
= 1000
Therefore, a= 1000 and n= 2.
(e) y= abx2
log10
y= log10
a+ log10
bx2
= log10
a+ x2log10
b
log10
y= (log10
b)x2+ log10
a............
log10
b= Gradient
= 7 36 1
=45
= 0.8
b= 100.8
= 6.310
Substitute log10
y = 3, x2 = 1 and log10
b =45
into ,
3 =45
(1) + log10
a
log10
a= 3 45
=11
5
a= 158.5
Therefore, a= 158.5 and b= 6.310.
9. (a) (i)x 2 3 4 5 6
y 2.2 2.8 4.0 5.0 5.3
yx 3.1 4.8 8.0 11.2 13.0
y= kx + hx
yx = kx+ h
xy
0
4
6
8
10
12
14
2
2
x
1 2 3 4 5 6
(ii) h=yx -intercept = 2
k= Gradient
=13 (2)
6 0
=52
(b) (i) y= axn
log10
y= log10
axn
= log10
a+ nlog10
x
log10
y= nlog10
x+ log10
a
log10
x 0.3 0.40 0.48 0.54 0.60 0.65
log10
y 0.68 0.94 1.18 1.37 1.57 1.74
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
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(ii) log10
a= log10
y-intercept
= 0.18
a= 0.661
n= Gradient
=1.08 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) When F= 2.5,x= 5.2
(ii) Whenx= 12.0, F= 5.7
(c) x= 2.5 is wrongly taken when F= 1.0.
The correct value isx= 2.0.
11. (a)v2 19.98 40.07 60.06 79.92 91.20 119.90
s 1.0 2.0 3.0 4.0 5.0 6.0
0
40
60
80
120
100
20
s
v2
2 4 6
(b) v= 9.55 was wrongly taken.
The correct value is v2= 100, that is, v= 10.
(c) (i) When v= 8, v2= 64, s= 3.2
(ii) When s= 3.5, v2= 70
v= 8.37
12. (a) Gradient = 140 224 2
= 59
yx
= 59x2+ c...............................
Substitutex2= 2,yx
= 22 into ,
22 = 59(2) + c
c= 96
Therefore,yx
= 59x2 96
y= 59x3 96x
(b) Whenx= 2.4,y= 59(2.4)3 96(2.4) = 585.2
13. (a) y= axn..........................................
Substitutex= 1,y= 100 into , 100 = a(1)n
a= 100
Substitutex= 3,y= 900 into , 900 = 100(3)n
3n= 9
= 32
n= 2
(b) y= axn
y= 100(x)2
Y= mX+ c
Y=y, X= x2, c= 0, m= 100
Therefore, Y= 100X
When Y= 5,X= a,
5 = 100a
a=5
100
=
1
20
When Y= b,X= 4,
b= 100(4)
= 400
Hence, a=1
20
, b= 400 and c= 0.
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1. py= kx3+ p2x
p,pyp
=kx3
p+p2xp
y=kp x
3
+ px
x,yx
=kp
x2+ p
Compare to Y= mX+ c,
Gradient, m=kp
h 0
0 3=kp
y
x-intercept =p p= h
h
3=k
h
k= 13
h2
2. (a) y= 5x
2
log10
y= log10
5x log10
2
log10
y= (log10
5)x log10
2
(b) Y= mX+ c
c= log10
2
k= log10
2
= 0.301
m= 0 kh 0
log10
5 =0.301
h
h=0.301
log
105
= 0.4306
3. (a) y= 100xp
log10
y= log10
100xp
= log10
100 + log10xp
= 2 +plog10
x
log10y=plog10x+ 2
(b) k= 2 (log10
y-intercept)
p= gradient
=8 2
3 0
=63
p= 2
4. y= 2x2 3
yx2
=2x2
x2
3
x2
yx2
= 3 1x2
+ 2 Y= 3X+ c
Therefore,X= 1x2
and Y= yx2
.
5. (a)yx
1.91 1.84 1.73 1.63 1.58 1.55
1x2
0.21 0.16 0.11 0.06 0.04 0.03
0
1.0
1.5
2.0
0.5
x
x20.05 0.10 0.15 0.20
1
y
(b) (i) y=p
x
+q
p
x
yx
=p
x2
+qp
=p 1x2+qp
Y= mX+ c
p= Gradient
=1.6 1.5
0.05 0
= 2
(ii)qp
=yx
-intercept
q2
= 1.5
q= 3
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6. (a)log
10y 1.08 1.21 1.31 1.43 1.52 1.60
x 1 1.22 1.41 1.58 1.73 1.87
0
1.0
1.5
0.5
x0.5 1.0 1.5 2.0
log10
y
(b) y=pkx
log10
y= log10
pkx
= log10
p+ log10
kx
log10y= (log10k)x + log10p Y= mX+ c
(i) log10
p= log10
y-intercept
= 0.45
p= 2.82
(ii) log10
k= Gradient
=1.5 0.6
1.7 0.25
= 0.6207
k = 4.18
7. (a) log10
y 0.48 0.80 1.12 1.44 1.77 2.09
x 1 0 1 2 3 4 5
0
1.0
1.5
2.0
0.5
x 12 4 6
log10
y
(b) (i) y=pkx 1
log10
y= log10
p+ (x 1) log10
k
log10
y= (log10
k)(x 1) + log10
p
Y= mX+ c
log10
p= (x 1)-intercept
= 0.5
p= 3.16
(ii) log10
k= Gradient
=2.0 1.0
4.8 1.6
= 0.3125
k= 2.05
8. (a)x 1 2 3 4 5 6
log10
y 0.30 0.60 0.90 1.20 1.51 1.81
0
1.0
1.5
2.0
0.5
x
2 4 6
log10
y
(b) (i) y=kx
p
log10
y= log10
kx
p
= log10
kx log10
p
log10
y= (log10
k)x log10p
Y= mX+ c
log10y
-intercept = log10p
0 = log10
p
log10
p= 0
p= 1
(ii) log10
k= Gradient
=1.7 0
5.6 0
= 0.3036
k= 2.01
1. y= kxn 1
log10
y= log10
(kxn 1)
= log10
k+ log10
xn 1
= log10
k+ (n 1) log10x
= (n 1) log10
x+ log10
k
Gradient = n 1
Vertical intercept = log10
k
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2. Gradient =13 3
6 1
=10
5
= 2
Therefore, y = 2x2+ c......................
Substitutex2= 1, y = 3 into ,3 = 2(1) + c
c= 1
Hence, the equation is y = 2x2+ 1,that is,y= (2x2+ 1)2.
3. y=px+ qx32
yx
=px
x
+qx
32
x
=p+ qx
= qx +p...................................... Y= mX+ c
q= Gradient
=8 4
4 6
=4
2
= 2
Substitute x = 4,yx
= 8, q= 2 into ,8 = (2)(4) +p
p= 16
Therefore,p= 16 and q= 2.
4. y= 102x+ 3
log10
y= log10
102x+ 3
= (2x+ 3) log10
10
log10
y= 2x+ 3
Y= mX+ c
log10
y-intercept = 3
Gradient = 2
For pointA,4 3
q 0
= 2
1 = 2q
q=12
For pointB,p 31 0
= 2
p 3 = 2
p= 5
Therefore,p= 5, q=12
.
5. y= ax5
log10
y= log10
a+ 5 log10
x
log10
y= 5 log10
x+ log10
a
Y= mX+ c
log10
a= log10
y-intercept
= 4
a= 104
= 10 000
Gradient = 5
k 4
3 0
= 5
k 4 = 15
k= 19
Therefore, a= 10 000 and k= 19.
6. (a) Gradient =5 2
3 0
= 1
Vertical intercept = 2
The equation of the line is
Y= mX+ c
y= log10
x+ 2
(b) x= a10y 1
log10
x= log10
a+ (y 1) log10
10
log10
x= log10
a+ y 1
y= 1 log10
a+ log10
x
1 log10
a= 2
log10
a= 1
a= 0.1
7. (a) Gradient =10 4
3 1
= 3
Equation of the line is log10
y= 3 log10
(x+ 2) + c
Substitute log10
(x+ 2) = 1 and log10
y= 4 into
the equation,
log10
y= 3 log10
(x+ 2) + c
4 = 3(1) + c
c= 1
Therefore, log10
y= 3 log10
(x+ 2) + 1
y= k(x+ 2)n log
10y= log
10k+ nlog
10(x+ 2)
log10
y= nlog10
(x+ 2) + log10
k
Compare to log10
y= 3 log10
(x+ 2) + 1
Hence, n= 3 and log10
k= 1
k= 10
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(b) Wheny= 0.08, y= k(x+ 2)n
0.08 = 10(x+ 2)3
0.008 = (x+ 2)3
x+ 2 = 3 0.008 = 0.2
x= 0.2 2
= 1.8
8.xy
=3x
+ x
1xxy =
1x
3x
+ x
1y
=3
x2
+ 1
1y
= 3 1x2
+ 1 ..........................
Substitute1
x2
= 3 and1y
= kinto ,
1y
= 3 1x2+ 1 k= 3(3) + 1
= 10
Substitute1
x2
= pand1y
= 16 into ,
16 = 3(p) + 1
3p= 15
p= 5
9. (a) y= rx2+ t....................................
Substitutex= 1,y= 4 into , 4 = r(1)2+ t
r+ t= 4 .......................................
Substitutex= 3,y= 36 into, 36 = r(3)2+ t
9r+ t= 36 ...................................
, 8r= 32 r= 4
Substitute r= 4 into , 4 + t= 4
t= 0
(b) y= rx2+ tbecomesy= 4x2
1x
(y) =1x
(4x2)
yx
= 4x
Therefore, the gradient is 4.
ForA(2,p),
p 02 0
= 4
p= 8
ForB(q, 12),
12 0
q 0
= 4
12 = 4q
q= 3
10. (a) The gradient of the straight line =
11 7
4 2 = 2
Gradient =r 7
3 2
= 2
r 7 = 2
r= 9
(b) log10
y= 2x+ c............................
Substitute log10
y= 7,x= 2 into , 7 = 2(2) + c
c= 3
Therefore, log10
y= 2x+ 3
y= 102x+ 3
11. (a) y= px2+ q...................................
Substitutex= 1,y= 4 into , 4 =p(1)2+ q
p+ q= 4......................................
Substitutex= 3,y= 12 into , 12 =p(3)2+ q
9p+ q= 12 ..................................
, 8p= 8 p= 1
Substitutep= 1 into , 1 + q= 4
q= 3
(b) y=px2+ qbecomes
y=x2+ 3
y
x2
= 1 +3
x2
y
x2
= 3 1x2
+ 1
ForA,
y
x2
=4
12
= 4
1
x2
=1
12
= 1
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ForB,
y
x2
=12
32
=12
9
=43
1
x2
=1
32
=19
0 x21
1
A(1, 4)
x2y
, 1943
B
12. x2y=px.......................................
1x2
(x2y) =1
x2
(px)
y=p 1xIn Diagram (b), gradient =p
=14 7
1 12
= 14Substitutep= 14 into , x2y= 14x
ForA, 12 r= 14(1)
r= 14
ForB, t2(7) = 14t
7t2= 14t
7t= 14
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 7.1 7.9
0
4
6
8
2
x
2 4 6 8 10
y2
(b) (i) y2= px+ q
p= Gradient
=6 1
7 0.5
=10
13
q=y2
-intercept = 0.6
(ii) Whenx= 1.5,y2= 1.7
y= 1.3
14. (a)x 1 4 9 16 20 25
yx 4.01 14.02 30.06 52.80 66.0 82.0
0
40
60
80
20
x
10 20 30
yx
(b) (i) y=p
x+ (q+ 1)x
x(y) = x px
+ (q+ 1)x yx =p+ (q+ 1)x yx = (q+ 1)x+ p
p= vertical intercept
= 2
q+ 1 = Gradient
=82 34
25 10
= 3.2 q = 2.2
(ii) Whenx= 10,yx = 34
y10 = 34
y=34
10
= 10.75
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15. (a)log
10x 0 0.30 0.54 0.65 0.699 0.778
log10
y 0.6 1.18 1.69 1.91 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=pxn2
log10
y= log10
p+n2
log10
x
log10
y =n2
log10
x+ log10
p
log10
p= Intercept of log10
y
= 0.6
p= 3.981
n2
= Gradient
=2.2 1.0
0.8 0.2
= 2
n= 4
(c) Incorrect value ofyis 180.
The correct value is log10
y= 2.2,
that is, y= 158.5
16. (a)x 3.0 3.5 4.0 4.5 5.0 5.6
log10
y 1.4 1.53 1.68 1.83 1.98 log10
p
0
1.0
1.5
2.0
0.5
x
log10y
2 4 6
(b) (i) y= k ax 1
log10
y= log10
k+ (x 1) log10
a
= log10
k+ xlog10
a log10
a
log10
y= (log10
a)x+ log10
k log10
a
log10
a= Gradient
=2.1 0.8
5.4 1 = 0.2955
a= 1.975
log10
k log10
a= Intercept of log10
y
= 0.5
log10
k 0.2955 = 0.5
log10
k= 0.7955
k= 6.24
(ii) Whenx= 5.6,
log10
y= 2.15
y= 141.3
Therefore,p= 141.3
(iii) Whenx= 2.0, log10
y= 1.1
y= 12.6
17. (a)T 0 26.8 32.9 38.0 42.9 46.5
l 0 4.5 5.5 6.3 7.1 7.7
0
10
20
30
40
50
10
l
T
2 4
New
Old
6 8
(b) Gradient =30 05 0
= 6
Therefore, T= 6l T 2= 36l
(c) (i) T= 6l 6
(ii) When T= 6.5, l= 2.1 l = 2.12
= 4.41
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Additional Mathematics SPM Chapter 13
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18. (a) y = k(x 1)n
log10
y= log10
k+ nlog10
(x 1)
log10
y = nlog10
(x 1) + log10
k
Y = mX+ c
log10
y 0.7 1.3 1.66 1.79 1.95 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= 100, log10
100 = 2
log10
(x 1) = 0.65
x 1 = 4.467
x= 5.467
(ii) Whenx= 8, log10
7 = 0.85
log10
y= 2.4
y= 251.2
1. y=axn
log2y= log
2(axn)
= log2a+ log
2xn
log2y= nlog
2x+ log
2a
Y= mX+ c
Y= log2y, m= n,X= log
2x, c= log
2a
SubstituteX= 1, Y= 5 into the equation,
5 = m+ c...................
SubstituteX= 3, Y= 11 into the equation,11 = 3m+ c................
, 2m= 6 m= 3
Substitute m= 3 into ,5 = 3 + c
c= 2
n= m
= 3
c= log2a= 2
a= 22
= 4
Therefore, n= 3, a= 4.
2.y
x
2 = ax+
bx
yx2
x = ax+ bxxyx
= ax2+ b
Y = mX+ c
Y=yx , m= a,X= x
2, c= b
Given the gradient =12
a=12
y
x
=1
2
x2+ b
Substitutex2= 4,yx
= 6 into the equation,
6 =12
(4) + b
b = 4
Therefore, a=12
, b= 4
3. (a) y=bx
x+ ab
1y
=x+ abbx
= xbx + abbx
1y
= a 1x+1b
Whenx= 3,y= 1
1x
=13
,1y
=11
= 1
Whenx = 6,y=32
1x
=16
,1y
=1
3
2
=23
,
, 1
0
6 3
1 2
y
1
1
3
1
x
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Additional Mathematics SPM Chapter 13
Penerbitan Pelangi Sdn. Bhd.
(b) a = Gradient
=
1 23
13
16
= 2
Substitute
1
x =
1
3 and
1
y = 1 into the equation,
1y
= a 1x+1b
1 = 2 13+1b
1b
= 1 23
=13
b = 3
Therefore, a= 2, b= 3.
4. (a)yx
7 8 6 4 2 0 2
x 1 2 4 6 8 10 12
0
4
6
8
10
2
2
x
x
y
2 64 8 10 12
(b) (i) Wrongly recorded y = 7, actual value of
y= 9
(ii) y= ax2+ bx
y
x
= ax2+ bx
x
yx
= ax+ b
b =yx
-intercept
= 10
a= Gradient
=10 0
0 10
= 1
(iii) Wheny= 0,x= 10
Therefore, the horizontal distance of the
point Cis 10 m.
(iv) Whenx= 5,
yx
= 5
y5
= 5
y= 25
Therefore, the height of pointBis 25 m.
5. (a)x 0 10 20 30 40 50
y 20 30 40 50 60 70
0
10
20
30
40
50
60
70
x
y
10 3020 40 50
(b) (i) 4k2x= (y c)2
4k2
x =y c y c= 2kx y= 2kx+ c Y= mX+ C
Y= y, m= 2k,X= x, C= c
c=y-intercept
= 20
2k= Gradient
=70 20
50 0
= 1
k= 12
Therefore, c= 20, k=12
.
(ii) Wheny= 55,
x= 35 x= 352
= 1225
(iii) Whenx= 500,
x= 22.4, y= 42.5
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Additional Mathematics SPM Chapter 13
Penerbitan Pelangi Sdn. Bhd.
6.f 2 400 196 100 64 49
1m
100 50 25 16.7 12.5
(a), (c)
0
50
100
150
200
250
300
350
400
m
1
f2
20 6040 80
(c)
(a)
100
(b) (i) When m= 0.05 kg,
1
m
=1
0.05
= 20
f 2= 80
f= 9 oscillations per second
(ii) Whenf = 15,
f 2= 152
= 225
1
m
= 56
m=1
56
= 0.01786 kg
= 18 g
(iii) f 2=1
km
= 1k 1m Gradient =
1k
=400 0
100 0
= 4
k=14
7. (a) 2(y+ 1)2= kx+ t
(y+ 1)2=k
2x+t
2
...................
Y= mX+ c
Y= (y+ 1)2, m=k
2
,X= x, c=t
2
Substitutex= 0, (y+ 1)2
= 10 into , 10 =
t2
t= 20
Substitutex= 5, (y+ 1)2= 0, t= 20 into ,
0 =k
2
(5) +202
5k2
= 10
k=205
= 4
(b) (y+ 1)2 =k
2x+t
2
becomes
(y+ 1)2=42x+
202
y2+ 2y+ 1 = 2x+ 10
y2= 2x 2y+ 10 1
= 2(x y) + 9
Y= mX+ c
Y= y2, m= 2,X= x y, c= 9
Gradient = 2
Intercept on the Y-axis = 9