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HARRISON COLLEGE INTERNAL EXAMINATION, APRIL 2018
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION
SCHOOL BASED ASSESSMENT
PURE MATHEMATICS
UNIT 1 β TEST 3
Time: 1 Hour & 20 minutes
This examination paper consists of 3 printed pages.
The paper consists of 7 questions.
The maximum mark for this examination is 60.
INSTRUCTIONS TO CANDIDATES
1. Write your name clearly on each sheet of paper used.
2. Answer ALL questions.
3. Number your questions carefully and do NOT write your solutions to different questions beside one another.
4. Unless otherwise stated in the question, any numerical answer that is not exact, MUST be written correct to three (3) significant figures.
EXAMINATION MATERIALS ALLOWED
1. Mathematical formulae
2. Electronic calculator (non β programmable, non β graphical)
1. (a) Determine
(i) limπ₯β2
π₯3 β 4π₯
π₯ β 2 [3]
(ii) limπ₯β0
sin 3π₯
4π₯ [4]
(b) Find the values of π₯ for which π₯2+1
|3π₯+2|β8 is discontinuous. [4]
(c) A function π(π₯) is defined as
π(π₯) = {π₯ + 6 π₯ β€ 3 π₯2 π₯ > 3
(i) Find limπ₯β3
π(π₯). [2]
(ii) Determine whether π(π₯) is continuous at π₯ = 3. Give a reason for your answer. [4]
(d) Differentiate π(π₯) = cos 2π₯ using first principles. [6]
Total 23 Marks
2. Given that π¦ = 16π₯ +1
π₯, determine the equation of the tangent to the curve at the point where π₯ = 1. [6]
TOTAL 6 Marks
3. The curve πΆ has equation π¦ =π₯
4+π₯2.
(i) Show that ππ¦
ππ₯=
4βπ₯2
(4+π₯2)2 [3]
(ii) Determine the coordinates of the stationary points on πΆ. [4]
Total 7 Marks
4.
Fig. 1 shows an open tank in the shape of a triangular prism. The vertical ends π΄π΅πΈ and π·πΆπΉ are identical
isosceles triangles, angle π΄π΅πΈ = angle π΅π΄πΈ = 30Β°. The length of π΄π· is 40 cm. The tank is fixed in position
with the open top π΄π΅πΆπ· horizontal. Water is poured into the tank at a constant rate of 200 cm3sβ1. The
depth of water, π‘ seconds after filling starts, is β cm (see Fig. 2).
(i) Show that, when the depth of water in the tank is β cm, the volume, π cm3, of water in the tank is
given by π = (40β3)β2. [3]
(ii) Find the rate at which β is increasing when β = 5. [3]
Total 6 Marks
β cm
5. The parametric equations of a curve are given by
π₯ = sin π , π¦ = cos 2π , 0 β€ π β€ 2π
Show that ππ¦
ππ₯= β4 sin π. [4]
Total 4 Marks
6. Use the substitution π’ = sin π₯ + 1 to show that
β« sin π₯ cos π₯ (1 + sin π₯)5 ππ₯ =1
42(1 + sin π₯)6(6 sin π₯ β 1) + π
[8]
Total 8 Marks
7.
The diagram shows part of the curve π₯ =12
π¦2 β 2. The shaded region is bounded by the curve, the π¦ β axis
and the lines π¦ = 1 and π¦ = 2. Find the volume, in terms of π, when the shaded region is rotated through
360Β° about the π¦ β axis. [6]
Total 6 Marks
END OF EXAMINATION
CAPE UNIT 1 TEST 3 MARK SCHEME
1. (i) limπ₯β2
π₯3β4π₯
π₯β2
= limπ₯β2
π₯(π₯2 β 4)
π₯ β 2
= limπ₯β2
π₯(π₯ + 2)(π₯ β 2)
π₯ β 2
= limπ₯β2
π₯(π₯ + 2)
= 2(2 + 2)
= 8
(ii) limπ₯β0
sin 3π₯
4π₯
= limπ₯β0
sin 3π₯
3π₯Γ
3π₯
4π₯
=3
4limπ₯β0
sin 3π₯
3π₯
=3
4Γ 1
=3
4
(b) |3π₯ + 2| β 8 = 0
|3π₯ + 2| = 8
3π₯ + 2 = 8
3π₯ = 6
π₯ = 2
3π₯ + 2 = β8
3π₯ = β10
π₯ = β10
3
1 - Completely factoring π₯3 β 4π₯
1 β cancelling π₯ β 2
1 β C.A.O
1 β Separation of terms
1 β Use of appropriate property of limits (S.O.I)
1 β Use of limπ₯β0
sin π₯
π₯= 1
1 β C.A.O
1 β equating denominator to 0
1 β for 3π₯ + 2 = 8
1 β for 3π₯ + 2 = β8
1 β mark for π₯ = β10
3 and 2
c) (i) limπ₯β3β
π(π₯) = 3 + 6 = 9
limπ₯β3+
π(π₯) = 32 = 9
limπ₯β3
π(π₯) = 9
(ii) limπ₯β3β
π(π₯) = limπ₯β3+
π(π₯)
π(3) = 9
limπ₯β3β
π(π₯) = limπ₯β3+
π(π₯) = π(3)
Therefore, π(π₯) is continuous at π₯ = 3
d) π(π₯) = cos 2π₯
π(π₯ + β) = cos(2π₯ + 2β)
π(π₯ + β) β π(π₯) = cos(2π₯ + 2β) β cos 2π₯
= β2 sin (2π₯ + 2β + 2π₯
2) sin (
2π₯ + 2β β 2π₯
2)
limββ0
β2 sin(2π₯ + β) sin β
β
= limββ0
β2 sin(2π₯ + β)sin β
β
= β2 sin(2π₯ + 0) Γ 1
= β2 sin 2π₯
2. π¦ = 16π₯ +1
π₯= 16π₯ + π₯β1
ππ¦
ππ₯= 16 β π₯β2 = 16 β
1
π₯2
when π₯ = 1
ππ¦
ππ₯= 16 β
1
12= 15
π¦ = 16(1) +1
1= 17
1 β for determining the right hand and left hand limits
1 β stating limπ₯β3
π(π₯) = 9
1 β stating that the right hand limit equals the left hand limit
1 β determining π(3)
1 β stating that limπ₯β3
π(π₯) = π(3)
1 β stating that π is continuous at π₯ = 3
1 β expression for π(π₯ + β) β π(π₯)
1 β use of factor formula
1 β use of limit formula
1 β use of limββ0
sin π₯
π₯= 1
1 β evaluating limit at β = 0
1 β C.A.O
1 β determining ππ¦
ππ₯
1 β substituting π₯ = 1 into ππ¦
ππ₯
1 β evaluating ππ¦
ππ₯ when π₯ = 1
1 β evaluating π¦ when π₯ = 1
π¦ = ππ₯ + π
17 = 15(1) + π
2 = π
π¦ = 15π₯ + 2
3. (i) π¦ =π₯
4+π₯2
ππ¦
ππ₯=
1(4 + π₯2) β π₯(2π₯)
(4 + π₯2)2
=4 + π₯2 β 2π₯2
(4 + π₯2)2
=4 β π₯2
(4 + π₯2)2
(ii)4 β π₯2
(4 + π₯2)2= 0
4 β π₯2 = 0
π₯2 = 4
π₯ = Β±2
π¦ =2
4 + 22=
1
4 (2,
1
4)
π¦ = β2
4 + (β2)2= β
1
4 (β2, β
1
4)
4. (i) tan 30Β° =β
π
π =3β
β3
Volume =1
2(
6β
β3) (β)(40)
=120β2
β3
=120β3β2
3
= 40β3β2
1 β correct use of π¦ = ππ₯ + π
1 β correct equation
1 β correct numerator
1 β correct denominator
1 β simplification of the numerator
1 β use of ππ¦
ππ₯= 0
1 β for π₯ = Β±2
2 β 1 mark for each correct pair of coordinates
1 β correct expression for base of triangle
1 β use of formula for volume of a prism
1 β simplification
(ii) π = 40β3β2
ππ
πβ= 80β3β
πβ
ππ‘
200 = 80β3(5)πβ
ππ‘
200
400β3=
πβ
ππ‘
1
2β3=
πβ
ππ‘
5. π₯ = sin π
ππ₯
ππ= cos π
π¦ = cos 2π
ππ¦
ππ= β2 sin 2π
ππ¦
ππ₯= β
2 sin 2π
cos π
= β2(2 sin π cos π)
cos π
= β4 sin π
6. π’ = sin π₯ + 1 β π’ β 1 = sin π₯
ππ’
ππ₯= cos π₯
β«(π’ β 1) cos π₯ (π’5)ππ’
cos π₯
β« π’6 β π’5 ππ’
=π’7
7β
π’6
6+ π
=6π’7 β 7π’6
42+ π
=π’6(6π’ β 7)
42+ π
1 β determining ππ
πβ
1 β use of parametric differentiation and substituting β = 5
1 β correct value for πβ
ππ‘
1 β for ππ₯
ππ= cos π
1 β for ππ¦
ππ= β2 sin 2π
1 β for ππ¦
ππ₯=
ππ¦
ππΓ
ππ
ππ₯
1 β for use of sin 2π = 2 sin π cos π
1 β differentiating π’
1 β substituting ππ₯ =ππ’
cos π₯
1 β substituting π’ β 1 = sin π₯
1 β simplifying integrand
1 β integrating correctly
1 β combining fraction
1 β correct factoring
=(sin π₯ + 1)6(6(sin π₯ + 1) β 7)
42+ π
=(sin π₯ + 1)6(6 sin π₯ β 1)
42+ π
7. π₯ =12
π¦2 β 2
π₯2 = (12
π¦2β 2) (
12
π¦2β 2)
π₯2 =144
π¦4β
48
π¦2+ 4
π β«(144π¦β4 β 48π¦β2 + 4)
2
1
ππ¦
= π [144π¦β3
β3β
48π¦β1
β1+ 4π¦]
21
= π ([144
β3(2)3+
48
2+ 4(2)] β [
144
β3(1)3+
48
1+ 4(1)])
= 22π
1 β resubstituting π’ = sin π₯ + 1
1 β determining π₯2 =144
π¦4 β48
π¦2 + 4
1 β substituting π₯2 into correct formula for volume of revolution
1 β correct integration
1 β substitution of the upper limit
1 β substitution of the lower limit
1 β C.A.O
HARRISON COLLEGE INTERNAL EXAMINATION, APRIL 2018
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION
SCHOOL BASED ASSESSMENT
PURE MATHEMATICS
UNIT 1 β TEST 3 (PREVIEW)
Time: 1 Hour & 20 minutes
8. (a) Determine
(i) limπ₯β3
π₯3 β 9π₯
π₯ β 3 [18]
(ii) limπ₯β0
sin 5π₯
2π₯ [
5
2]
(b) Find the values of π₯ for which π₯2+1
|2π₯+3|β6 is discontinuous. [β
9
2,
3
2]
(c) A function π(π₯) is defined as
π(π₯) = {π₯ + 2 π₯ β€ 2 π₯2 π₯ > 2
(iii) Find limπ₯β2
π(π₯). [4]
(iv) Determine whether π(π₯) is continuous at π₯ = 2. Give a reason for your answer. [Yes]
(d) Differentiate π(π₯) = sin 2π₯ using first principles. [2 cos 2π₯]
Total 23 Marks
9. Given that π¦ = 8π₯ +1
π₯, determine the equation of the tangent to the curve at the point where π₯ = 1.
[π¦ = 7π₯ + 2]
TOTAL 6 Marks
10. The curve πΆ has equation π¦ =π₯
1+π₯2.
(iii) Show that ππ¦
ππ₯=
1βπ₯2
(1+π₯2)2
(iv) Determine the coordinates of the stationary points on πΆ. [(1,1
2) , (β1, β
1
2)]
Total 7 Marks
11.
Fig. 1 shows an open tank in the shape of a triangular prism. The vertical ends π΄π΅πΈ and π·πΆπΉ are identical
isosceles triangles, angle π΄π΅πΈ = angle π΅π΄πΈ = 30Β°. The length of π΄π· is 40 cm. The tank is fixed in position
with the open top π΄π΅πΆπ· horizontal. Water is poured into the tank at a constant rate of 100 cm3sβ1. The
depth of water, π‘ seconds after filling starts, is β cm (see Fig. 2).
(iii) Show that, when the depth of water in the tank is β cm, the volume, π cm3, of water in the tank is
given by π = (40β3)β2.
(iv) Find the rate at which β is increasing when β = 4. [5β3
48]
Total 6 Marks
12. The parametric equations of a curve are given by
π₯ = cos π , π¦ = sin 2π , 0 β€ π β€ 2π
find ππ¦
ππ₯. [β
2 cos 2π
sin π]
Total 4 Marks
13. Use the substitution π’ = sin π₯ + 2 to show that
β« cos π₯ (2 + sin π₯)6 ππ₯ =(2 + sin π₯)7
7+ π
Total 8 Marks
β cm
14. The diagram below represents the finite region π which is enclosed by the curve π¦ = π₯3 β 1 and the lines
π₯ = 0 and π¦ = 0.
Calculate the volume of the solid that results from rotating π about the π¦ β axis. [3π
5]
Total 6 Marks
15.
16. a