CSEC ADDITIONAL MATHEMATICS MAY 2014
SECTION I
1. (a)
Data: f : x ® 1 - x 2 , x ÎÂ .
Required To Prove: f is not one to one
Proof:
(i)
f : x ® 1 - x 2 , x ÎÂ
as
s
m
at
hs
.
co
m
A sketch is made of f : x ® 1 - x 2 , x ÎÂ
There can be more than one values of x that are mapped onto the same
value of f ( x ) . In fact, with the exception of x = 0, there are always two
sp
values of x that are mapped onto the same value of f ( x ) .
For example,
.fa
f (1) = 0
w
w
w
f ( -1) = 0
Hence, f is not one to one.
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Page 1 of 32
1
x - 3, x ÎÂ
2
Required To Find: fg ( x ) and state its domain.
Solution:
f : x ® 1 - x 2 , x ÎÂ
1
g : x ® x - 3, x ÎÂ
2
2
æ1
ö
\ fg ( x ) = 1 - ç x - 3 ÷ , x ÎÂ
è2
ø
2
æ x 3x 3x
ö
= 1- ç - - + 9÷
è 4 2 2
ø
a)
x2
4
x2
fg ( x ) = 3x - 8 4
at
hs
.
= 3x - 8 -
co
m
Data: g : x ®
(ii)
Domain of g ( x ) is x Î Â .
m
Domain of f ( x ) is x Î Â .
as
s
\ Domain of fg ( x ) is Â Ç Â , which is  .
x2
, x ÎÂ .
4
Required To Determine: g -1 and to sketch g and g -1 on the
same axes.
Solution:
1
g ( x) = x - 3
2
1
Let y = x - 3
2
1
\y +3= x
2
x = 2 ( y + 3)
Replace y by x to obtain:
g -1 ( x ) = 2 ( x + 3)
w
w
w
.fa
b)
sp
\ fg ( x ) = 3x - 8 -
= 2x + 6
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1
x - 3 and g -1 ( x ) = 2 x + 6.
2
Reflection in
Recall that g ( x ) ¾¾¾¾®
g -1 ( x ) .
y=x
Hence, g ( x ) =
Data: When 2 x3 + ax 2 - 5 x - 2 is divided by 2 x - 1, the remainder is – 3.5.
Required To Determine: The value of the constant, a.
Solution:
We first recall the remainder and factor theorem.
If f ( x ) is any polynomial and f ( x ) is divided by ( ax + b ) the remainder is
w
.fa
(b)
sp
as
s
m
at
hs
.
co
m
g ( x ) and g -1 ( x ) are sketched below.
w
w
æ bö
æ bö
f ç - ÷ . If f ç - ÷ = 0 , then ( ax + b ) is a factor of f ( x ) .
è aø
è aø
Let f ( x ) = 2 x3 + ax 2 - 5x - 2
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æ ( -1) ö
1
\ f ç÷ = -3
2 ø
2
è
1
æ1ö
f ç ÷ = -3
2
è2ø
3
2
Data: A kitchen of length y m and width x m.
1
Length of kitchen = (Square of the width) and
2
Perimeter of the kitchen = 48 m.
Required To Calculate: The value of x and of y
Calculation:
w
.fa
sp
as
s
m
(c)
at
hs
.
1
1
1
1
a = -3 + 2 + 2 4
2
2
4
1
3
a=
4
4
a=3
co
m
1
æ1ö
æ1ö
æ1ö
2 ç ÷ + a ç ÷ - 5 ç ÷ - 2 = -3
2
è2ø
è2ø
è2ø
1 1
1
1
+ a - 2 - 2 = -3
4 4
2
2
w
Length of kitchen =
1
(square of the width of the kitchen)
2
1 2
x
2
Perimeter = 48 m
w
\y =
\ y + x + y + x = 48
2 y + 2 x = 48
÷2
y + x = 24
Let
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1 2
x
2
y + x = 24
y=
…u
…v
Substituting y = 24 - x into equation u
1
24 - x = x 2
2
´2
48 - 2 x = x 2
x 2 + 2 x - 48 = 0
at
hs
.
( x + 8)( x - 6 ) = 0
co
m
From v
y = 24 - x
x = -8 or 6
= 24 - 6
sp
= 18
as
s
When x = 6
y = 24 - x
m
x ¹ negative, since the width of the kitchen cannot be a negative value.
\ x = 6 only
w
w
w
.fa
\ The rectangular kitchen is 18 m by 6 m.
2. (a)
Data: f ( x ) = -2 x2 - 12 x - 9
(i)
Required To Express: f ( x ) in the form k + a ( x + h ) , where k, a and h
are integers.
Solution:
-2 x 2 - 12 - 9 = -9 - 2 x 2 + 12 x
2
(
)
= -9 - 2 ( x + 6 x )
2
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One half the coefficient of 6 x =
1
( 6)
2
=3
-2 x2 - 12 x - 9 = * - 2 ( x + 3) , where * is a number to be found.
2
-2 ( x + 3) = -2 ( x + 3)( x + 3)
2
co
m
= -2 ( x 2 + 6 x + 9 )
= -2 x 2 - 12 x - 18
Hence,
* + ( -18) = -9
at
hs
.
\* = 9
\-2 x2 -12 x - 9 = 9 - 2 ( x + 3)
2
and which is of the form k + a ( x + h ) , where k = 9Î Z , a = -2Î Z and
h = 3Î Z .
m
2
as
s
Alternative Method:
-2 x 2 - 12 x - 9 = k + a ( x + h )
2
= k + a ( x 2 + 2hx + h 2 )
.fa
sp
= ax 2 + 2ahx + ( ah 2 + k )
w
w
w
Equating coefficients:
a = -2
2ah = -12
2 ( -2 ) h = -12
h=3
ah 2 + k = -9
-2 ( 3) + k = -9
2
k =9
\ f ( x ) = -2 x2 - 12 x - 9 º 9 - 2 ( x + 3) and is of the form k + a ( x + h ) ,
2
2
where k = 9Î Z , a = -2Î Z and h = 3Î Z .
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Required To State: The maximum value of f ( x ) .
Solution:
(ii)
f ( x ) = 9 - 2 ( x + 3)
2
¯
³ 0 "x
co
m
\ The maximum value of f ( x ) = 9 - 0
=9
Alternative Method:
f ( x ) = -2 x 2 - 12 - 9
f ¢ ( x ) = -2 ( 2 x ) - 12
at
hs
.
= -4 x - 12
Let f ¢ ( x ) = 0
-4 x - 12 = 0
m
\ x = -3
as
s
The stationary value occurs when f ¢ ( x ) = 0, that is at x = -3 .
f ( -3) = -2 ( -3) - 12 ( -3) - 9
2
sp
= -18 + 36 - 9
=9
.fa
f ¢¢ ( x ) = -4 ( < 0) Þ f ( x ) = 9 is the maximum value of f ( x ) .
Required To Determine: The value of x for which f ( x ) is a maximum.
Solution:
f ( x ) = -2 x 2 - 12 x - 9
w
w
w
(iii)
= 9 - 2 ( x + 3)
2
³ 0 "x
f ( x ) has a maximum of 9 - 0 and this occurs when -2 ( x + 3) = 0
2
and x = -3
OR
AS SHOWN BEFORE
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The stationary value of f ( x ) occurs at x = -3 and f ( -3) gives a
maximum value of 9.
Hence the maximum value of f ( x ) occurs at x = -3 .
Required To Solve: For x in 3 + 5 x - 2 x 2 £ 0 .
co
m
(b)
Solution:
3 + 5x - 2 x2 £ 0
´ -1
at
hs
.
2 x2 - 5x - 3 ³ 0
( 2 x + 1)( x - 3) ³ 0
1
and 3.
2
The coefficient of x 2 > 0 Þ the quadratic curve has a minimum point.
w
w
.fa
sp
as
s
m
Hence, y = 2 x 2 - 5x - 3 cuts the horizontal axis at -
w
The solution set of x for which 3 + 5 x - 2 x 2 £ 0 is the same for 2 x 2 - 5 x - 3 ³ 0 .
1ü
ì
The range of values is { x : x ³ 3} È í x : x £ - ý as illustrated in the diagram
2þ
î
above.
(c)
Data: Series is given by 0.2 + 0.02 + 0.002 + 0.0002 + ...
(i)
Required To Prove: That the given series is geometric.
Proof:
Let us look at the terms of the series and the number of each of the terms.
No. of
1
2
3
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4
…
Page 8 of 32
term
Term
0.2
0.02
= 0.2 ´10
-1
0.002
= 0.2 ´10-2
0.0002
= 0.2 ´10-3
…
This is of the form:
ar
a
ar 2
ar 3
…
where a = 0.2 and r = 10-1 .
co
m
Hence, Tn , the nth term is of the form Tn = ar n - 1 , where n is the number of
the term.
at
hs
.
Hence, the series is a geometric progression with 1st term, a = 0.2 and the
common ratio, r = 10-1 .
Required To Calculate: The sum to infinity of the series.
Calculation:
a
Recall, the sum to infinity of the geometric series, S¥ =
, r < 1.
1- r
m
(ii)
0.2
1 - 10-1
0.2
=
1 - 0.1
0.2
=
0.9
2
= (as a fraction in exact form)
9
(i)
w
w
3. (a)
w
.fa
sp
as
s
\ S¥ =
SECTION II
Data: The lines with equations x + 3 y = 6 and kx + 2 y = 12 are
perpendicular.
Required To Calculate: The value of k
Calculation:
Let us consider the line with equation
x + 3y = 6
3y = -x + 6
1
1
y = - x + 2 and is of the form y = mx + c , where m = - is the
3
3
gradient.
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kx + 2 y = 12
2 y = -kx + 12
k
k
y = - x + 6 and is of the form y = mx + c , where m = - is the
2
2
gradient.
co
m
Hence,
1
k
- ´ - = -1 (the product of the gradients of perpendicular lines = -1 )
3
2
k
= -1
6
k = -6
Data: The center of a circle is at the point of intersection of the lines given
in 3(a)(i).
The radius of the circle is 5 cm.
Note: The radius should be given as 5 units.
at
hs
.
(ii)
as
s
m
Required To Determine: The equation of the circle
Solution:
Let
…u
x + 3y = 6
.fa
sp
-6 x + 2 y = 12 …v
From v
-6 x + 2 y = 12
2 y = 12 + 6 x
y = 6 + 3x
w
w
w
Substitute y = 6 + 3x into u
x + 3 ( 6 + 3x ) = 6
x + 18 + 9 x = 6
10 x = -12
x = -1
When x = -1
1
5
1
5
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æ 1ö
y = 6 + 3 ç -1 ÷
è 5ø
3
= 6-3
5
2
=2
5
at
hs
.
co
m
æ 1 2ö
\ The center of the circle is ç -1 , 2 ÷ .
è 5 5ø
( x - a) + ( y - b)
2
= r2 .
as
s
2
m
The equation of the circle with center ( a, b ) and radius r, is
\ Equation of the circle is
!
#
( x – (-1 )) 2 + ( x – 2 ) 2 = (5)2
"
sp
"
$
.fa
= ( x + ") 2 + ( x –
!#
"
%$
(5)2
#'
!''
x + #" + y2 -
"
y + #" = 25
w
= x2 +
!# 2
) =
"
w
x 25
w
25 x2 + 25y2 +60 x – 120y + 36 + 144 = 625
25 x2 + 25y2 +60 x – 120y = 445
25 x2 + 25y2 + 60 x –120y – 445 = 0
(b)
æ 1ö
æ 3ö
æ 4ö
Data: OR = ç ÷ , OS = ç ÷ and OR = ç ÷ .
è 1ø
è -1ø
è 4ø
ˆ = 90°.
(i)
Required To Prove: TRS
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at
hs
.
co
m
Proof:
as
s
a . b = a b cos q
m
ˆ = 90°, then RT . RS = 0 by the dot product law.
If TRS
If q = 90° , then a . b = 0.
sp
RT = RO + OT
w
w
w
.fa
æ 1ö æ 4 ö
= -ç ÷ + ç ÷
è 1ø è 4 ø
æ 3ö
=ç ÷
è 3ø
RS = RO + OS
æ 1ö æ 3 ö
= -ç ÷ + ç ÷
è1ø è -1 ø
æ 2ö
=ç ÷
è -2 ø
æ 3ö æ 2 ö
RT . RS = ç ÷ . ç ÷
è 3 ø è -2 ø
= ( 3 ´ 2 ) + ( 3 ´ -2 )
=0
Q.E.D.
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Required To Calculate: The length of the hypotenuse of triangle TRS.
Calculation:
at
hs
.
co
m
(ii)
as
s
RT =
m
æ 3ö
RT = ç ÷
è 3ø
æ 2ö
RS = ç ÷
è -2 ø
( 3) + ( 3)
2
2
sp
= 18
( 2) + ( -2 )
2
2
.fa
RS =
w
w
w
= 8
TS 2 =
( 18 ) + ( 8 )
2
2
Pythagoras' Theorem
TS 2 = 18 + 8
= 26
TS = 26 units
4. (a)
Data: Diagram showing a sector OAB of a circle with center O, radius 9 cm and
angle at O = 0.7 radians.
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Required To Calculate: The area of sector OAB.
Calculation:
1 2
Area of sector OAB = ( 9 ) ´ 0.7
2
at
hs
.
= 28.35 cm2
co
m
(i)
()
*
m
Required To Calculate: The area of the shaded region, H.
Calculation:
Angle AOC = 900
(The angle made by the tangent to a circle and a radius, at the point of
contact, is a right angle
= tan (0.7 radians)
as
s
(ii)
sp
AC = 7.580 cm
w
w
.fa
Area of the triangle OAB = ½ (9 x 7.580) cm2
= 34.11 cm2
\ Area of the shaded region, H = Area of triangle OAC – Area of sector
OAB
2
= (34.11- 28.35) cm
= 5.76 cm2
p 1
p
3
Data: sin = , cos =
6 2
6
2
pö 1
æ
Required To Prove: cos ç x + ÷ =
3 cos x - sin x , where x is acute.
6ø 2
è
w
(b)
(
)
Proof: By the compound angle formula
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pö
p
p
æ
cos ç x + ÷ = cos x cos - sin x sin
6ø
6
6
è
= ( cos x )
(
3 cos x - sin x
)
Q.E.D.
Required To Prove:
tan q sin q
1
º 1+
1 - cos q
cos q
m
Proof:
Left hand side:
sin q
.sin q
tan q sin q cos q
=
1 - cos q
1 - cos q
sin 2 q
=
cos q (1 - cos q )
Recall: sin 2 q + cos 2 q = 1
\ sin 2 q = 1 - cos 2 q
co
m
(c)
1
2
at
hs
.
=
3
1
- ( sin x )
2
2
as
s
= (1 - cos q )(1 + cos q )
sp
Hence,
tan q sin q (1 + cos q )(1 - cos q )
=
1 - cos q
cos q (1 - cos q )
1 + cos q
cos q
1
cos q
=
+
cos q cos q
1
=
+ 1 (Right hand side)
cos q
w
w
w
.fa
=
5. (a)
Q.E.D.
SECTION III
Data: A curve has equation y = 3 + 4 x - x 2 and the point P ( 3, 6 ) lies on the
curve.
Required To Find: The equation of the tangent to the curve at P
Solution:
y = 3 + 4x - x2
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dy
= 4 - 2x
dx
\ The gradient of the tangent at P = 4 - 2 ( 3)
Gradient function,
= -2
Hence, the equation of the tangent at P is
y-6
= -2
x -3
y - 6 = -2 ( x - 3)
Data: f ( x ) = 2 x3 - 9 x 2 - 24 x + 7
at
hs
.
(b)
co
m
y - 6 = -2 x + 6
y = -2 x + 12
2 x + y - 12 = 0 which is of the form ax + by + c = 0, where a = 2Î Z , b = 1Î Z
and c = -12Î Z .
Required To Find: All the stationary points of f ( x ) .
Solution:
(i)
m
y = 2 x3 - 9 x 2 - 24 x + 7
dy
\ Gradient function,
= 2 ( 3x3 - 1 ) - 9 ( 2 x ) - 24
dx
as
s
= 6 x 2 - 18 x - 24
sp
At a stationary point,
.fa
Let
6 x 2 - 18 x - 24 = 0
dy
= 0.
dx
÷6
x 2 - 3x - 4 = 0
w
w
w
( x + 1)( x - 4 ) = 0
\ Stationary points occur at x = -1 and x = 4 .
f ( -1) = 2 ( -1) - 9 ( -1) - 24 ( -1) + 7
3
2
= -2 - 9 + 24 + 7
= 20
f ( 4 ) = 2 ( 4 ) - 9 ( 4 ) - 24 ( 4 ) + 7
3
2
= 128 - 144 - 96 + 7
= -105
\ Stationary points are ( -1, 20 ) and ( 4, - 105) .
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(ii)
Required To Determine: The nature of each of the stationary points of
f ( x).
When x = -1
d2y
= 12 ( -1) - 18
dx 2
= -30 (Negative)
at
hs
.
\ ( -1, 20 ) is a maximum point.
co
m
Solution:
d2y
= 6 ( 2 x ) - 18
dx 2
= 12 x - 18
m
When x = 4
d2y
= 12 ( 4 ) - 18
dx 2
= 30 (Positive)
Required To Evaluate:
Solution:
ò {x ( x
4
2
2
}
- 2 ) dx = ò
4
2
(x
3
2
- 2 ) dx
- 2 x ) dx
.fa
2
ò x(x
4
sp
6. (a)
as
s
\( 4, -105) is a minimum point.
4
w
é x4
ù
= ê - x2 ú
ë4
û2
The constant of integration is omitted in a definite integral as it cancels out.
ìï ( 4 )4
ü ì 4
ü
4
2 ï ï ( 2)
2ï
2
x
x
2
dx
=
4
2
(
)
(
)
(
)
í
ý
í
ý
ò2
îï 4
þï îï 4
þï
w
w
{
}
= ( 64 - 16 ) - ( 4 - 4 )
= 64 - 16
= 48
p
(b)
Required To Evaluate:
ò ( 4 cos x + 2sin x ) dx
3
0
Solution:
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p
p
ò ( 4 cos x + 2sin x ) dx = éë4 ( sin x ) + 2 ( - cos x )ùû 03
3
0
p
= [ 4sin x - 2 cos x ]03
(
)
= 2 3 - 1 - ( -2 )
= 2 3 +1
Data: A curve with gradient function
dy
= 6 x 2 - 1 passes through the point
dx
at
hs
.
(c)
co
m
æ
æp ö
æ p öö
= ç 4sin ç ÷ - 2 cos ç ÷ ÷ - ( 4sin ( 0 ) - 2 cos ( 0 ) )
è3ø
è 3 øø
è
æ
3
1ö
= çç 4 ´
- 2 ´ ÷÷ - ( 4 ´ 0 - 2 ´1)
2
2ø
è
P ( 2, - 5) .
Required To Determine: The equation of the curve.
Solution:
dy
= 6x2 -1
dx
\ Equation of the curve is 6 x2 -1 dx
m
(i)
)
as
s
ò(
3
6x
- x + C (where C is a constant)
3
y = 2 x3 - x + C
sp
y=
.fa
Since ( 2, - 5 ) lies on the curve
-5 = 2 ( 2 ) - 2 + C
3
w
-5 = 16 - 2 + C
w
w
C = -19
\ Equation of the curve is y = 2 x3 - x - 19 .
(ii)
Required To Calculate: The area bounded by the curve, the x – axis and
the verticals x = 3 and x = 4 .
Calculation:
The region whose area is required lies entirely above the x – axis.
Area required = ò
4
3
( 2x
3
- x - 19 ) dx
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4
é 2 x4 x2
ù
=ê
- - 19 x ú
2
ë 4
û3
ìï 2 ( 4 )4 ( 4 )2
üï ìï 2 ( 3)4 ( 3)2
üï
=í
- 19 ( 4 ) ý - í
- 19 ( 3) ý
2
2
ïî 4
ïþ ïî 4
ïþ
1
ì 1
ü
= {128 - 8 - 76} - í40 - 4 - 57 ý
2
î 2
þ
= 65 square units
SECTION IV
Data: A class has 60 students. 27 study Mathematics, 20 study Biology and 22 do
not study either Mathematics or Biology.
(i)
Required To Calculate: The probability that a student chosen at random
studies both Mathematics and Biology.
Calculation:
Creating a Venn diagram to illustrate the data:
m
at
hs
.
7. (a)
co
m
= 44 - ( -21)
Let:
as
s
M = {Students who study Mathematics}
w
w
w
.fa
sp
B = {Students who study Biology}
Let x be the number of students who study both Mathematics and Biology.
Hence,
( 27 - x ) + x + ( 20 - x ) + 22 = 60
69 - x = 60
x=9
\ 9 students study both Mathematics and Biology.
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Page 19 of 32
co
m
at
hs
.
The probability that a student chosen at random studies both Mathematics
and Biology
No. of students studying both Mathematics and Biology
=
Total no. of students
9
60
3
=
20
as
s
m
=
Required To Calculate: The probability that a student chosen at random
studies Biology only.
Calculation:
The probability that a student chosen at random studies Biology only
No. of students studying Biology only
=
Total no. of students
.fa
sp
(ii)
Data: Two ordinary 6 sided dice are tossed.
S represents the sum of their scores.
An incomplete sample space diagram for S is given.
w
w
(b)
11
60
w
=
(i)
Required To Complete: The sample space diagram.
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Page 20 of 32
Solution:
6 +1 = 7
6+2 =8
6+3=9
6 + 4 = 10
6 + 5 = 11
6 + 6 = 12
5
5 +1 = 6
5+ 2 = 7
5+3 = 8
5+ 4 = 9
5 + 5 = 10
5 + 6 = 11
4
4 +1 = 5
4+2 = 6
4+3= 7
4+4 =8
4+5 = 9
4 + 6 = 10
3
3 +1 = 4
3+ 2 = 5
3+3 = 6
3+ 4 = 7
3+5 = 8
3+ 6 = 9
2
2 +1 = 3
2+2 = 4
2+3=5
2+4 = 6
2+5 = 7
2+6 =8
1
1+1 = 2
1+ 2 = 3
1+ 3 = 4
1+ 4 = 5
1+ 5 = 6
1+ 6 = 7
1
2
3
4
5
6
Score
on Die 1
6
7
8
5
6
7
at
hs
.
co
m
6
4
5
6
2
11
12
8
9
10
11
7
8
9
10
m
10
4
5
6
7
8
9
3
4
5
6
7
8
2
3
4
5
6
7
1
2
3
4
5
6
sp
1
9
as
s
3
Sum, S
.fa
Score
on Die 2
a)
w
w
w
(ii)
Required To Find: P ( S > 9)
Solution:
P ( S > 9) =
Number of scores >9
Total number of scores
1+ 2 + 3
6´ 6
6
=
36
1
=
6
=
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Page 21 of 32
5–6
4–5
3–4
2–3
1–2
as
s
6–5
5–4
4–3
3–2
2–1
at
hs
.
Data: D is the difference between the scores on the dice. An incomplete
table for d and P ( D = d ) is given.
Required To Find: The value of a, b and c in the table, that is, to
complete the table.
Solution:
For d = 1 the scores on the dice could be
m
(iii)
co
m
Required To Find: P ( S £ 4 )
Solution:
Number of scores ≤ 4
𝑃(𝑆 ≤ 4) =
Total number of scores
3+2+1
=
6×6
6
=
36
1
=
6
b)
.fa
sp
Total number of scores in which the difference is 1 is 10
10
\ P ( D = d = 1) =
36
5
=
18
w
w
w
For d = 3 the scores on the dice could be
6–3
5–2
4–1
3–6
2–5
1–4
Total number of scores in which the difference is 3 is 6
6
36
1
=
6
\ P ( D = d = 3) =
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Page 22 of 32
For d = 5 the scores on the dice could be
1–6
6–1
Total number of scores in which the difference is 5 is 2
2
36
1
=
18
5
1
1
, b = and c = .
18
6
18
Data: Stem and leaf diagram showing the scores of 51 people in an aptitude test.
(i)
Required To Calculate: The median and the quartiles for the data.
Calculation:
There are 51 scores in the data.
The middle score of 51 scores is the 26th score.
From the diagram, the 26th score in ascending or descending degree of
magnitude is 71.
Hence, the median score is 71.
sp
as
s
m
(c)
at
hs
.
\a =
co
m
\ P ( D = d = 5) =
w
w
w
.fa
1
3
of 51 = 12
4
4
Hence, the 13th value is the lower quartile, Q1 .
(obtained from the diagram)
Q1 = 56
3
1
of 51 = 38
4
4
Hence, the 39th value is the upper quartile, Q3 .
Q3 = 79
The interquartile range is Q3 - Q 1 = 79 - 56
= 23
The semi-interquartile range is
1
1
( Q3 - Q1 ) = ( 23)
2
2
= 11.5
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Page 23 of 32
Required To Construct: A box and whisker plot to illustrate the data and
to comment on the data.
Solution:
The median of the data is the 26th value, which is 71.
There are 25 scores before the median.
The 13th score is the lower median, which is 56.
There are 25 scores after the median score.
26 + 13 = 39 and hence the 39th score is the upper median. This is 79.
as
s
m
at
hs
.
co
m
(ii)
w
w
w
.fa
sp
Comments
1
of the scores were less than 56.
4
1
of the scores were more than 79.
4
1
of the scores were between 56 and 71.
4
1
of the scores were between 71 and 79.
4
Interquartile range = 23
1
1
Q1 - 1 I.Q.R = 56 - 1 ( 23)
2
2
= 21.5
1
Q3 + 1 I.Q.R. = 113.5
2
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Page 24 of 32
There are no values < 21.5 or > 113.5
Hence, there are no outliers.
The whisker to the left of the box is longer than the whisker to the right of
the box. Hence, there are more extreme values towards the negative end
and so the distribution is said to be negatively skewed.
Data: A velocity – time graph showing the motion of a car travelling 1410 m
from A to B.
sp
as
s
m
at
hs
.
co
m
8. (a)
Required To Calculate: The distance covered by the car before it
decelerates.
Note: A car cannot travel. It is better to have said ‘the distance
covered by the car’.
Calculation:
The ‘branches’ of the graph are named PQ, QR and RS for convenience
and are as shown on the diagram.
The car moves at a constant 25 ms-1 for 30 seconds before starting to
decelerate.
w
w
w
.fa
(i)
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Page 25 of 32
co
m
at
hs
.
\ Distance covered = Area under the ‘branch’
= 30 ´ 25
m
= 750 m
as
s
OR
Distance = Constant Speed ´ Time
=25 ms -1 ´ 30 s
sp
= 750 m
Required To Calculate: The deceleration of the car as it goes from 25
ms-1 to 10 ms-1.
Calculation:
w
w
w
.fa
(ii)
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Page 26 of 32
The regions A1 , A2 and A3 are as shown on the diagram
The distance covered by the car is the area under the graph.
From t = 30 to t = 90 the distance can be obtained by the sum of the area
of the regions A2 and A3 .
Area of A1 = 750
1
( 25 + 10 ) ´ ( t - 30 )
2
Area of A3 = ( 90 - t ) ´10
at
hs
.
Area of A2 ==
co
m
\ Area of ( A2 + A3 ) = 1410 - 750
= 660
w
w
w
.fa
sp
as
s
m
Hence,
1
( 35)( t - 30 ) + 10 ( 90 - t ) = 660
2
1
17 t - 525 + 900 - 10t = 660
2
1
7 t = 285
2
t = 38
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Page 27 of 32
10 - 25
38 - 30
15
= - ms -2
8
co
m
Hence, the acceleration of the car is the gradient of the branch QR.
15
ms -2 .
8
m
\ The deceleration is
at
hs
.
Acceleration =
as
s
OR
sp
From Q to R:
Initial velocity, u = 25 ms-1
Final velocity, v = 10 ms-1
Time = 8 s
The acceleration, a, is constant
v = u + at
.fa
\10 = 25 + a ( 8 )
w
w
w
a=-
(iii)
15
8
\ The deceleration is
15
ms -2 .
8
Required To Find: The period of time for which the car maintained the
speed of 10 ms-1.
Solution:
The car maintains the speed of 10 ms-1 from R to S, that is, from t = 38 to
t = 90 .
\ Car maintains a speed of 10 ms-1 for 90 - 38 = 52 s .
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Page 28 of 32
(iv)
Data: The car comes to rest at C, decelerating uniformly from B and
covering a further 30 m.
Required To Calculate: The average velocity of the car from A to C
as
s
m
at
hs
.
co
m
Calculation:
w
w
w
.fa
sp
In the branch from S to C, the area of triangle SBC = 30 (since 30 m is
covered).
SB = 10
BC ´ 10
\
= 30
2
BC = 6
\ Time at C = 96
OR
u = 10 ms-1
v = 0 ms-1
s = 30 m
v 2 = u 2 + 2as
0 = 10 2 + 2a ( 30 )
100
60
5
=3
\a = -
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Page 29 of 32
v = u + at
æ 5 ö
0 = 10 + ç - t ÷
è 3 ø
t =6
89:;< =>?:;@AB A9CBDB=
89:;< :>EB :;FB@
a=
at
hs
.
Data: A particle starts from rest, at P, and travels in a straight line. The particle
comes to rest at Q, 10 seconds later. Velocity is v ms-1.
0£t £5
v = 0.72t 2 - 0.096t 3
2
5 £ t £ 10
v = 2.4t - 0.24t
(i)
Required To Calculate: The time when velocity is at a maximum.
Calculation:
Let acceleration at time, t, by a ms-2.
dv
dt
m
(b)
1410 + 30
96
!''G
= *$ ms-1
=15 ms-1
=
co
m
From A to C:
0£t £5
sp
as
s
dv
= 0.72 ( 2t ) - 0.096 ( 3t 2 )
dt
= 1.44t - 0.288t 2
dv
=0
dt
t (1.44 - 0.288t ) = 0
t = 0 or 5
When t = 0 the particle is at rest.
\ t = 5 , when a = 0 .
w
w
w
.fa
Let
OR
dv
= 2.4 - 0.24 ( 2t )
dt
5 £ t £ 10
dv
=a=0
dt
Maximum velocity occurs when the acceleration is 0
When
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Page 30 of 32
2.4 - 0.24 ( 2t ) = 0
2.4
0.48
=5
t=
\ Maximum velocity occurs at t = 5 .
Required To Calculate: The maximum velocity.
Calculation:
When t = 5
v = 0.72 ( 5) - 0.096 ( 5 )
2
3
0£t £5
at
hs
.
= 6 ms-1
co
m
(ii)
OR
v = 2.4 ( 5 ) - 0.24 ( 5 )
Required To Calculate: The distance moved by the particle from P to
where the particle attains its maximum velocity.
Calculation:
The distance required is the distance moved by the particle from t = 0 to
t = 5.
sp
as
s
(iii)
5 £ t £ 10
-1
m
= 6 ms
2
v = 0.72t 2 - 0.096t 3
0£t £5
.fa
Let s be the distance from O after time, t.
s = ò 0.72t 2 - 0.096t 3 dt
w
(
w
w
= ( 0.72 )
)
t 3 0.096t 4
+C
3
4
(where C is the constant of integration)
s = 0 at t = 0 ,
\C = 0
\ s = 0.24t 3 - 0.024t 4
When v = 0
0.72t 2 - 0.096t 3 = 0
t 2 ( 0.72 - 0.096t ) = 0
t = 0 or 7
1
2
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Page 31 of 32
Hence, the particle does not come to instantaneous rest in the period t = 0
to t = 5 .
\ From t = 0 to t = 5 , distance covered = 0.24 ( 5) - 0.024 ( 5)
= 15 m
4
w
w
w
.fa
sp
as
s
m
at
hs
.
co
m
3
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Page 32 of 32