University of Malta
Junior College
Subject:
Intermediate Pure Mathematics
Date:
June 2009
Time:
09.00 - 12.00
End of Year Test
Worked Solutions
Question 1
a) To write 27 +
21
in the form k 3 .
3
27 +
21
21 3
= 9!3+
3 3
3
21 3
3
=3 3+7 3
= 9 3+
= 10 3.
b) To express
8! 5
in the form a + b 5 .
3+ 5
8 ! 5 8 ! 5 3! 5
=
"
3+ 5 3+ 5 3! 5
(8 ! 5 )(3 ! 5 )
9!5
24 ! 8 5 ! 3 5 + 5
=
4
29 ! 11 5
=
4
29 11
=
!
5.
4
4
=
c) To express
2x ! 7
into partial fractions.
(x + 1)(2x ! 1)
2x ! 7
A
B
"
+
(x + 1)(2x ! 1) (x + 1) 2x ! 1
2x ! 7 " A(2x ! 1) + B(x + 1)
Case 1 when x = !1
Case 2 when x =
2x ! 7 = A(2x ! 1) + B(x + 1)
2x ! 7 = A(2x ! 1) + B(x + 1)
2(!1) ! 7 = A(2(!1) ! 1) + B(!1 + 1)
!9 = !3A
3= A
Answer:
1
2
(
)
2 ( 12 ) ! 7 = A 2 ( 12 ) ! 1 + B ( 12 + 1)
1! 7 = B
3
2
!4 = B
2x ! 7
3
4
=
!
.
(x + 1)(2x ! 1) (x + 1) 2x ! 1
2
Question 2
a) Given that a polynomial is given by f (x) ! x 3 " 2x 2 + 4x + a .
To find the remainder R, when f (x) is divided by (x ! 3) by using factor theorem:
f (x) ! x 3 " 2x 2 + 4x + a
f (3) ! (3)3 " 2(3)2 + 4(3) + a
! 27 " 18 + 12 + a
! 21 + a
Hence the remainder is R = (21 + a) .
To find the remainder S, when f (x) is divided by (x + 1) by using factor theorem:
f (x) ! x 3 " 2x 2 + 4x + a
f ("1) ! ("1)3 " 2("1)2 + 4("1) + a
! "1 " 2 " 4 + a
!a"7
Hence the remainder is S = (a ! 7) .
To find the value of a, given that R is three times S.
R = 3S
21 + a = 3( a ! 7 )
21 + a = 3a ! 21
42 = 2a
21 = a
Answer a = 21 .
b) To simplify 43 log 2 81 + 2 log 2 2 ! 12 log 2 9 ! 2 log 2 6 .
3
4
( )
( )
log 2 81 + 2 log 2 2 ! 12 log 2 9 ! 2 log 2 6 = 43 log 2 34 + 2 ! 12 log 2 32 ! 2 log 2 ( 2 " 3)
= 3log 2 3 + 2 ! log 2 3 ! 2 ( log 2 2 + log 2 3)
= 3log 2 3 + 2 ! log 2 3 ! 2 log 2 2 ! 2 log 2 3
(but log 2 2 = 1)
= ( 3 ! 1 ! 2 ) log 2 3 + 2 ! 2 log 2 2
= 0 log 2 3 + 2 ! 2
=0
Answer 0.
3
Question 3
a) Let ! and ! be the roots of the equation 5x 2 + 3x ! 1 = 0 .
b
a
3
=#
5
Sum of roots ! ! + " = #
Product of roots ! !" =
c
a
=#
1
5
The roots are one less, i.e. the roots are
(! " 1)
and ( ! " 1) .
Sum of new roots ! (! " 1) + ( # " 1) = ! + # " 2
3
=" "2
5
13
="
5
Product of new roots !
(! " 1) ( # " 1) = !# " (! + # ) + 1
="
=
1 3
+ +1
5 5
7
5
New equation is x 2 ! ( Sum of new roots ) x + ( Product of new roots ) = 0 .
x 2 ! ( Sum of new roots ) x + ( Product of new roots ) = 0
" 13 %
" 7%
x2 ! $ ! ' x + $ ' = 0
# 5&
# 5&
13
7
x+ =0
5
5
2
5x + 13x + 7 = 0
x2 +
Answer 5x 2 + 13x + 7 = 0 .
4
b) To expand ( 2 + x ) by using Pascal’s triangle.
4
Using the triagnle shown on the right we have that :
( 2 + x )4 = 1( 2 4 ) + 4 ( 2 3 )( x ) + 6 ( 2 2 ) ( x 2 ) + 4 ( 2 ) ( x 3 ) + 1( x 4 )
= 16 + 32x + 24x 2 + 8x 3 + x 4
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
We next use the above to evaluate ( 2.02 ) by putting x = 0.02 .
4
( 2.02 )4 = ( 2 + 0.02 )4
2
3
4
= 16 + 32 ( 0.02 ) + 24 ( 0.02 ) + 8 ( 0.02 ) + ( 0.02 )
= 16 + 0.64 + 0.0096 + 0.000064 + 0.00000016
= 16.64966416
Question 4
Given that the points A and B have coordinates ( 2, 5 ) and ( 8, ! 1) . Given also that C is the
midpoint of AB.
! x + x2 y1 + y2 $
! 2 + 8 5 ' 1$
= #
i) The coordiantes of C are # 1
,
,
&
&
" 2
" 2
2 %
2 %
! 10 4 $
=# ,
&
" 2
2%
= ( 5, 2 )
ii) The gradient of AB is m1 =
y2 ! y1 5 ! ( !1) 5 + 1 6
=
=
=
= !1
x2 ! x1
2!8
!6
!6
iii) To find the equation of line which passes through C and which is perpendicular to AB.
Let the equation of this line be y = m2 x + c .
First we find the value of the gradient m2 . Since this line is perpendicular to line AB then,
m1 ! m2 = "1
"1 ! m2 = "1
m2 = 1
Next we find the value of c.
The line y = x + c passes through ( 5, 2 ) . Hence: 2 = 5 + c
!3 = c
Answer: y = x ! 3 .
5
iv) To find the value of k if the line y = x ! 3 passes through the point ( k + 1, 2k ) .
y= x!3
2k = ( k + 1) ! 3
2k = k ! 2
k = !2
Answer k = !2 .
Question 5
a) To show that 3! = 2 tan ! .
Given that OA = OB = 2cm
shaded area = Half area of sector OAB.
!
# " r2
2"
!
=
#"4
2"
= 2! .....(i)
Area of Sector OAB =
AT
OA
AT
=
2
In triangle OAT, we have that tan ! =
Hence, the length of AT = 2 tan ! .
Area of shaded region = Half area of sector OAB =! from (i).
( base )( perpendicular height )
= 12 ( 2 ) ( 2 tan " )
Area of !OAT =
1
2
= 2 tan "
Hence we have that, Area of !OAT = Area of shaded region + Area of sector OAB
2 tan " = " + 2"
2 tan " = 3"
b) From standard theory we know that sin 2 ! + cos 2 ! = 1 .
2
! 5$
By substitution, we get # & + cos 2 ' = 1
" 13 %
2
144
! 5$
cos ' = 1 ( # & =
" 13 %
169
12
cos' = ±
13
2
6
Since x is obtuse - the angle x is in the 2nd quadrant then we say that cos! = "
Hence, tan x =
5
sin x
5
= 1312 = !
cos x ! 13
12
Question 6
The domain is ( !1, 1)
The range is [0, 0.632) .
Question 7
2x 2 + 5x > 3 .
i) To find the range of values of
2x 2 + 5x ! 3 > 0
( 2x ! 1) ( x + 3) > 0
y
x
1
!3
!3
2
From the diagram we see that the inquality region is x < !3 and x > 12 .
ii) To find the range of values of ( x ! 1) ( x + 1) ( x ! 2 ) < 0 .
A sketch of y = ( x ! 1) ( x + 1) ( x ! 2 ) is shown below:
y
2
!1
1
The range of values of x are x < !1 and 1 < x < 2 .
7
2
x
12
13
Question 8
To differentiate the following with respect to x.
(
i) To differentiate y = x 3 + 4
(
)
5
dy
= 5 x3 + 4
dx
)
4
(
! 3x 2
)
4
= 15x 2 x 3 + 4 .
ii) To differentitate y = e2 x cos 2x by using the product rule:
u = e2 x
Let
and
du
= 2e2 x
dx
Then
dy
= !2e2 x sin 2x + 2e2 x cos 2x
dx
= 2e2 x ( cos 2x ! sin 2x ) .
iii) To differentiate y =
ln ( 4x )
by using the quotient rule:
4x ! 3
u = ln ( 4x )
Let
4
1
du
=
=
dx 4x x
Then
v = cos 2x
dv
= !2 sin 2x
dx
dy
=
dx
1
x
and
v = 4x ! 3
dv
=4
dx
( 4x ! 3) ! 4 ln ( 4x )
.
( 4x ! 3)2
End of Exam
8