02134020/CAPE/MS/SPEC
CARIBBEAN EXAMINATIONS COUNCIL
ADVANCED PROFICIENCY EXAMINATION
PURE MATHEMATICS
UNIT 1
ALGEBRA, GEOMETRY AND CALCULUS
SPECIMEN PAPER
PAPER 02
SOLUTIONS AND MARK SCHEMES
2
SECTION A
(MODULE 1)
Question 1
(a)
p
T
T
F
F
q
T
F
T
F
1 mark
~p
F
F
T
T
1 mark
p→q
T
F
T
T
1 mark
~p  q
T
F
T
T
T = true F = false
[1 may be used for T and 0 for F]
(ii)
(iii)
→
[3 marks]
and ~p  q are logically equivalent since columns 4 and 5 are identical.
[2 marks]
p ∧ (p → q) = p ∧ (~p ∨ q)
= (p ∧ ~ ) ∨ (p ∧ )… distribute ∧ over ∨
= f ∨ (p ∧ ~ )
= (p ∧ )
(1 mark)
(1 mark)
(1 mark)
[3 marks]
Specific objectives: (A) 2, 4
(b)
(i)
(ii)
x * y = x + y −1,  x, y in
x,y ∈ ⇒ +  (sum of 2 real numbers)
⇒ + − 1  (difference of 2 real numbers)
⇒ ∗ 
⇒ ∗ is closed in real numbers
x * y = x + y 1= y + x 1 (addition is commutative)
⇒ y *x
⇒ * is commutative in
(1 mark)
(1 mark)
(1 mark)
[3 marks]
(1 mark)
(1 mark)
[2 marks]
3
(b)
(iii)
( ∗ )∗ =( +
− 1) ∗ for x, y, 
= ( + − 1) + − 1
= x + y + z 1 1
= + + −2
∗ ( ∗ ) = ∗ ( + − 1)
= + ( + − 1)1
+ + −2
⇒ ( ∗ ) ∗ = ∗ ( ∗ ) for all , , 
⇒ ∗ is associative in
[2 marks]
[1 mark]
[1 mark]
Specific Objectives (B) 2.
(c)
(i) y =
+ 4) = 2
(1 mark)
For x real, (−2) − 4 (4 ) ≥ 0
⇒ 4y2  1 ≤ 0
⇒ (2y  1)(2y + 1) ≤ 0
⇒  ≤ ≤
(1 mark)
(1 mark)
(1 mark)
(1 mark)
[5 marks]
⇒
(ii)
⇒ (
−2 +4 =0
| | < 2 ⇒ −2 ≤
≤2
[3 marks]
Total 25 marks
Specific Objectives (C) 5; (F) 2.
4
Question 2
(a)
3
2
Let f ( x)  2x  px  qx 2
(i)
f ( 1)  0   2  p  q  2  0  p  q
(1 mark)
1
p q
1
f    0     2  0  p  2q   9
4 4 2
2
 pq   3
(2 mark)
(1 mark)
f ( x )  (2 x  1) ( x  1) ( x  k )
(ii)
(1 mark)
 2 x 3  px 2 qx  2
k 2
 the remaining root is 2
Alternatively
(a)
(1 mark)
Let d be the third root of f (x) = 0
(i)
Then 1  
(ii)
1
= 2 ⇒ −1 + + 2 = − ⇒ = −3
2
2
And  − 2 + 1 = ⇒ = −3
=− ⇒
=1⇒
[3 marks]
=2
[4 marks]
n
Let Pn be the statement  (6r + 5) = n (3n + 8)
r 1
For n = 1, L.H.S. of P1, is 6 + 5 = 11 and R.H.S. of P1
= 1(3 + 8)
= 11
So Pn is true for n = 1
Assume that Pn is true for n = k, i.e
11 + 17 + … . . + (6 + 5) = (3 + 8)
Then, we need to prove Pn is true for n = k + 1
k 1
Now  (6r  5)  11 + 17 + … + (6
r 1
+ 5) + [6(
(1 mark)
+ 1) + 5]
= (3 + 8) + [6( + 1) + 5]
= 3 + 8 + 6 + 6 + 5
= 3k2 + 14k + 11
= (3k + 11) (k + 1)
= (k + 1) [3 (k + 1) + 8]
Thus, if Pn is true when n = k, it is also true with n = (k + 1);
n
i.e.  (6r + 5) = n (3n + 8)  n  .
r 1
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
[10 marks]
5
(c)
e
+ 2e
Alternatively Let
⇒
⇒
⇒
⇒
⇒
=3 ⇒ e
+
=3
(e ) − 3e + 2 = 0
(e − 2)(e − 1) = 0
e = 2 or e = 1
2 = ln2 or 2 = 0
=
= e , giving
ln2 or
=0
− 3 + 2 = 0 etc.
Specific Objectives (B) 5; (D) 4,7; (G) 1, 2
(1 mark)
(1 mark)
(2 mark)
(2 mark)
(2 mark)
[8 marks]
Total 25 marks
6
SECTION B
(MODULE 2)
Question 3
(a)
(i)
LHS 
sin 3  sin 
cos 3  cos
( 1 mark)
 3   
 (3   ) 
2 sin
 cos 

2
 2 



 3   ) 
 3   
2 cos
 cos 

 2 
 2 
sin 2

cos 2
 tan 2
 RHS
(1 mark)
(1 mark)
(1 mark)
[4 marks]
(ii)
 3  
2 sin 
 2



 3   
 cos 
 + sin 2 = 0

 2 
2 sin 2 cos  + sin 2 = 0
sin 2 (2 cos  + 1) = 0
 sin 2 = 0, that is  = 0,

(1 mark)
(1 mark)
(1 mark)
, ,
2
2 4
1
,
cos  =
, that is  =
3 3
2
3
, or
2
(3 marks)
(1 mark)
[7 marks]
(b)
r=
(i)
6
2
 3
 8 2 = 10, x = tan -1   = 36.9o
4

o
 f (  ) = 10 cos(   36.9 )
(2 marks)
(1 mark)
[3 marks]
(ii)
g( ) =
10
10  10 cos(   36 .9 o )
(1 mark)
Minimum value of g (  ) occurs when denominator has maximum value i.e.
cos (  − 36.9 ) = 1.
(1 mark)
10
1
Min g(  ) =
= occurs
10  10 2
(when   36.9  0 ), when  = 36.9.
(2 marks)
[4 marks]
7
(c)
(i)
(ii)
=2 −2
[2 marks]
= ( + + ). (2 − 2 ) = 2 + 0 − 2 = 0
(1 mark)
BC = 2 − 2 and
(1 mark)
. BC = ( + + ). (2 − 2 ) = 0 + 2 − 2 = 0
.
So n is perpendicular to the plane through A, B and C.
[2 marks]
(iii)
Let
=
+
+
with r.n = d
(1 mark)
represent the plane through A, B and C.
At the point A, = 2 so .
⇒ (2 ). ( + + ) =
⇒
=2
Hence the Cartesian equation of the plane is (
⇒
+
(1 mark)
=
+ =2
+
+
). ( + + ) = 2
(1 mark)
[3 marks]
[Total 25 marks]
Specific Objectives: (A) 3, 5, 7, 9, 10; (C) 7, 10
8
Question 4
(a)
L: x + 2y = 7 is a tangent to
(i)
+
at 2 coincident points
Now, = 7 − 2
⇒ (7 − 2 ) − 4(7 − 2 ) +
⇒
−4 +4=0
− 4 = 0 if
+ 2 = 7 if L touches the circle
(1 mark)
(1 mark)
(1 mark)
−1=0
(1 mark)
(1 mark)
⇒ ( − 2) = 0
⇒
= 2(twice)
⇒ when
= 2,
(1 mark)
(1 mark)
=3
So L touches the circle at (3,2)
(1 mark)
[8 marks]
(ii)
a)
Let Q  point diametrically opposite to (3, 2).
The centre of C is (2,0)
3 x
 2, x  1
so
2
2 y
 0, y  2
and,
2
Q  (1,2)
Tangent M at Q: y + 2 =
1
( x  1)
2
(1 mark)
(1 mark)
(2 marks)
(1 mark)
2y + x + 3 = 0
[5 marks]
b)
The equation of the diameter is
This meets C where
(2 − 2 ) +
⇒4−
⇒5
⇒
+4
−5=0
= ±1
+ 2
= 2 + 0=2
− 4(2 − 2 ) − 1 = 0
+
−8+
−1=0
Coordinates of points of intersection are (1, 0), (1, 4)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(2 marks)
[6 marks]
9
(b)
y(1+t) = t2
x(1+t) = t

y(1  t ) t 2
=
t
x(1  t )
y
=t
x

y

x

x

y
x
y
1
(1 mark)
(1 mark)
(1 mark)
x
y
x y
x2
1 x
(1 mark)
(2 marks)
[6 marks]
Total 25 marks
Specific Objectives: (B) 1, 2, 3, 4
10
SECTION C
(MODULE 3)
Question 5
h
lim
h 0 xh 
(a)
x

lim
h

h( xh 
xh 
x

x)
xh  x
h ( x  h  x)
= lim
h 0
x  h   x


lim h x  h  x
h0
h
= lim
xh  x
h 0
=

=
x  x
= 2 x
 (1 mark)
(1 mark)
(1 mark)

(1 mark)
(1 mark)
(1 mark)
[6 marks]
(b)
⇒
⇒
Now
f
(x) =18x + 4
f 1 ( x)  9 x 2  4 x  c
f ( x)  3x 3  2 x 2  cx  d
f (2) = 14 ⇒
⇒
f (3) = 74 ⇒
From (i) + (ii),
So
⇒
(1 mark)
(2 marks)
32 + 2c + d = 14
2c + d = 18…. (i)
99 + 3c + d = 74
(1 mark)
3c + d = 25 …..(ii)
c = 7
d = 4
(1 mark)
(1 mark)
(1 mark)
f (4)  3(43 )  2(42 )  7(4)  4
 192  32  32
= 192
(1 mark)
[8 marks]
11
(c) (i)
=
dy (1  x 2 )1  x(2 x )

dx
(1  x 2 ) 2
=

1 x
2
(1 mark)
2 2
(1  x )
1
x2

(1  x 2 ) 2 (1  x 2 ) 2
1
 x 

=

2 2
(1  x )  1  x 2 

(2 marks)
2
1
 y2
(1  x 2 ) 2
(1 mark)
(1 mark)
[5 marks]
(ii)
d 2 y (1  x 2 )2 ( 2 x )  (1  x 2 )2(1  x 2 )(2 x )

dx 2
(1  x 2 )4
(3 marks)
=
(1  x 2 )(2 x ) [(1  x 2 )  2  2 x 2 ]
(1  x 2 ) 4
(2 marks)
=
2 y ( x 2  3)
(1  x 2 ) 2
(1 mark)
[6 marks]
Total 25 marks
Specific Objectives: (A) 3, 4, 5, 6; (B) 8, 9, 16, (C) 1, 2, 3, 5, 9
12
Question 6
(a) (i)
Finding the equation of the tangent PQ.
dy
 3x 2
dx
a)
 dy 
2
   3( 3)  27
 dx  x  3
(1 mark)
(1 mark)
Equation of tangent:
y  27  27( x  3)
(1 mark)
y  27 x  54
(1 mark)
[4 marks]
b) Q has coordinates (2, 0)
(ii)
a)
(1 mark)
[1 mark]
1
y dx  (3  2)(27)
2
3
27
1

= x4
4
2
Area =

3
0
(2 marks)
(2 marks)
0
81 27

4
2
27
=
units2
4

(1 mark)
[5 marks]
3
Required Volume =   y 2 d x  Volume of the cone with radius 27 units and height
0
1 unit.
(1 mark)
3
1
   x 6 d x   ( 27 ) 2
0
3

x7
7
3
0
(1 mark)
1
  (36 )
3
3  1
      (36 )
7 3

= 3 7  7(35 ) 
7

= 2  35  units3
7
(1 mark)
7
(1 mark)
(1 mark)
[5 marks]
13
(b)
(i)
dy
= 3x2 – 8x + 5
dx
y = x3 – 4x2 + 5x + C
substituting; y = 3 at x = 0
C=3
3
2
y = x – 4x + 5x + 3
(1 mark)
(1 mark)
(1 mark)
[3 marks]
(ii)
dy
= 0 ⇒ 3x3 – 8x + 5 = 0
dx
(3x – 5) (x – 1) = 0
5
x= ;1
3
131
y=
,5
27
 5 131 
co-ordinates are  ,
, (1, 5)
 3 27 
d2 y
= 6x – 8
dx 2
(1 mark)
(1 mark)
(1 mark)
(1 mark)
(1 mark)
 d2 y 
 5 131 
 2  > 0 ⇒  ,

 3 27  max
 dx  x  5 3
 d2 y 
 2  < 0 ⇒ (1, 5) min
 dx  x 1
(1 mark)
(1 mark)
[7 marks]
Total 25 marks
Specific Objective(s): (B) 11, 13, 14, 15, 16, 17;
(C) 8 (i), (ii)
END OF TEST