GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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MATHEMATICS (H2) Paper 1 Suggested Solutions
9740/01 October/November 2010
1. Topic: Vectorsin Three Dimensions(Points & Lines)
(i) Given a = οΏ½2π3π6π
οΏ½, b = οΏ½1
β22
οΏ½, where p > 0.
|π| = pβ22 + 32 + 62 = 7p
|π| = οΏ½12 + (β2)2 + (2)2 = 3
Since|π| = |π| 7p = 3
β΄ p = ππ
(ii) a = 37
οΏ½236
οΏ½ =
β
ββ
6797
187 β
ββ
(a +b)β(a β b) =
β£β’β’β’β‘
β
ββ
6797
187 β
ββ
+ οΏ½1
β22
οΏ½
β¦β₯β₯β₯β€
β
β£β’β’β’β‘
β
ββ
6797
187 β
ββ
β οΏ½1
β22
οΏ½
β¦β₯β₯β₯β€
=
β
ββ
137
β57
327 β
ββ
β
β
ββ
β17
23747 β
ββ
= β1349
β 11549
+ 12849
= 0
β΄ (a +b) β(a β b) = 0 (shown)
ALTERNATIVE APPROACH
(a + b) β(a β b) = aβa βaβb + aβbβbβb = |π|2 β |π|2 = |π|2 β |π|2(since |π| = |π|) = 0
2. Topic: Maclaurinβs Series
(i) eπ₯(1 + sin 2π₯)
= οΏ½1 + π₯ + 12
π₯2 + β― οΏ½ [1 + 2π₯ β β― ]
= 1 + 2π₯ + π₯ + 2π₯2 + 12
π₯2 + β―
= π + ππ + ππ
ππ + β―β¦β¦β¦β¦.β¦β¦β¦β¦β¦β¦β¦ (1)
(ii) (1 + 43
π₯)π
= 1 + π1
(43
π₯)1 + π(πβ1)2!
(43
π₯)2 + β―
= 1 + 43
ππ₯ + 89
π(π β 1)π₯2 + β― β¦β¦β¦β¦β¦β¦β¦ (2)
Comparing coefficients of x in (1) and (2)
43
π = 3
π = ππ
β΄ Third term in series (2)
= ππ
οΏ½πποΏ½ οΏ½π
πβ ποΏ½ ππ
= ππ
ππ
= Third term in series (1) (shown)
eπ₯ = 1 + π₯ +π₯2
2! +π₯3
3! + β― +π₯π
π! + β―
sin π₯ = π₯ βπ₯3
3! +π₯5
5! β β―(β1)ππ₯2π+1
(2π + 1)! + β―
From MF15:
π β π = |π|2
From MF15: (1 + π₯)π = 1 + ππ₯ + π(πβ1)
2!π₯2 + β―
Sub 2x into sin x expansion
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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3. Topic: Arithmetic Progression, Recurrence (i) Given Sn = n(2n + c) un = Sn β Snβ1
= n (2n + c) β (n β 1) [2(n β 1) + c]
= 2n2 + nc β 2(n β 1)2 β c (n β 1)
= 2n2 + nc β 2 (n2 β 2n + 1) β cn + c
= 4n + c β 2
ALTERNATIVE APPROACH
S1 = u1 = 1[2(1) + c] = 2 + c
S2 = u2 + u1 = 2[4 + c] = 8 + 2c
β u2 = S2 β S1 = 8 + 2c β 2 β c = 6 + c
S3 = u3 + u2 + u1 = 3(6 + c) = 18 + 3c
β u3 = S3 β S2 = 18 + 3c β 8 β 2c = 10 + c
S4 = u4 + u3 + u2 + u1 = 4[8 + 2c] = 32 + 4c
β u4 = S4 β S3 = 32 + 4c β 18 β 3c = 14 + c
Since u2 β u1 = u3 β u2 = u4 β u3 = 4
β un is an arithmetic progression (A. P.).
β΄un = 2 + c + (n β 1) 4
= 2 + c + 4n β 4
= 4n + cβ 2
(ii) un = 4n + c β 2
un+1 = 4(n + 1) + c β 2
= 4n + 4 + c β 2
= (4n + c β 2) + 4
= un + 4
ALTERNATIVE APPROACH
From un in (i):
u1 = 2 + c
u2 = 6 + c = 4 + (2 + c) = 4 + u1
u3 = 10 + c = 4 + (6 + c) = 4 + u2
u4 = 14 + c = 4 + (10 + c) = 4 + u3
β΄un+1 = 4 + un
1st term, a = 2 + c Common difference, d = 4
nth term in an A.P.: Tn = a + (n β 1) d
f(un) = 4 + un
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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4. Topic: Differentiation (Tangents &Normals)
(i) Given x2 β y2 + 2xy + 4 = 0β¦β¦β¦β¦β¦β¦β¦ (1)
Using implicit differentiation,
2π₯ β 2π¦ dπ¦dπ₯
+ οΏ½2π¦ + 2π₯ dπ¦dπ₯
οΏ½ = 0
π₯ + π¦ = (π¦ β π₯) dπ¦dπ₯
β΄ππππ
= π+ππβπ
(ii) For the tangent of the curve to be parallel to the x-axis,
dπ¦dπ₯
= 0
β π₯+π¦π¦βπ₯
= 0
y = β x β¦β¦β¦β¦β¦β¦ (2) Sub (2) into (1) x2 β (β x)2 + 2x(β x) + 4 = 0 x2 β x2 β 2x2 + 4 = 0 2x2 = 4
x = Β±β2
Sub x = β β2 into (2)
y = β2
Sub x= β2 into (2)
y = β β2
The coordinates are (ββπ, βπ) and (βπ, ββπ).
5. Topic: Functions, Graphs (Transformations)
(i) Given y = x3 = f(x) y = (x β 2)3 = g(x)
y = 12(x β 2)3 = h(x)
y = ππ(x β 2)3β 6
When x = 0, y = 12(β2)3 β 6
= β 10
When y = 0, 0 = 12(x β 2)3 β 6
(x β 2)3 = 12
x = 2 + β123
β΄Curve crosses the y-axis at (0, β10) and the x-axis at (2 +βπππ , 0).
f(x β 2) Translate 2 units in the positive x- direction
12g(x)
Stretch with scale factor 12 parallel to the y- axis
h(x)β 6 Translate 6 units in the negative y -direction
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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(ii) Using G. C. (refer to Appendix for detailed steps),
For y = fβ1(x), when y = 0, x = β10 x = 0, y = 2 + βπππ
6. Topic: Integration
(i) Using G.C. for y = x3 β 3x + 1 (refer to Appendix for detailed steps and various methods that can be used),
Ξ² = 0.3473, Ξ³ = 1.5321 β΄Ξ² = 0.347 andΞ³= 1.532(3d.p.)
(ii) Using G.C. (refer to Appendix for detailed steps and various methods used),
Area ofthe region bounded by the curve and the x- axis (between x = π½ and x = πΎ) = οΏ½β« (π₯3 β 3π₯ + 1)πΎ
π½ dπ₯οΏ½ = | (β0.7814168)|
β 0.781 units2(3sig.fig.)
The graph of y = fβ1(x) is obtained by reflecting the graph of y = f(x) in the line y = x.
Remember to obtain the absolute value for the answer as we are finding the area.
The x-coordinate in f(x) becomes the y-coordinate in fβ1(x) and vice versa.
y = x y = f(x)
y = fβ1(x)
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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(iii) Obtain the x-coordinates of the intersection points for y = x3 β 3x + 1 β¦β¦β¦β¦β¦β¦β¦ (1) y = 1 β¦β¦β¦β¦β¦β¦β¦ (2) (1) = (2) x3β 3x + 1 = 1 x(x2 β 3) = 0
x = 0, x = Β±β3
From the diagram, the lower and upper limit of integration to find the region is ββ3 and 0 respectively. Hence, area of region (shaded region A) between the curve and the line
= β« (π₯3 β 3π₯ + 1)dπ₯0ββ3 β οΏ½1 Γ β3οΏ½
= οΏ½14
π₯4 β 32
π₯2 + π₯οΏ½ββ3
0β β3
= οΏ½0 β οΏ½14
Γ 9 β 32
(3) β β3οΏ½οΏ½ β β3
= β 94
+ 92
+ β3 β β3
= ππ units2
(iv)
Sincex = β1 and x = 1 are stationary points, When x = β1, ymax = (β1)3 β 3(β1) + 1 = 3 When x = 1, ymin = (1)3 β 3(1) + 1 = β1 For x3 β 3x + 1 = k to have three real distinct roots,
ymin<k<ymax β΄Set of values of k = {kπ β: β1<k <3} Area of rectangle
Check final answer with G. C. (time permitting)
Note: β’ Final answer expressed in set notation as
question asks for set of values of k. β’ Three real distinct roots β no equal sign
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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7. Topic: Differential Equations
Given dπdπ‘
β (20 β π)
βdπdπ‘
= π(20 β π), where k is a constantβ¦β¦β¦β¦β¦β¦β¦ (1)
Whent = 0, sub ΞΈ = 10 and dπdπ‘
= 1 into (1)
1 = k (20 β 10)
k = 120β10
= 110
βdπdπ‘
= 110
(20 β π)
β« 120βπ
dπ = β« 110
dπ‘
β ln|20 β π| = 110
π‘ + π
When t = 0,ΞΈ = 10 β ln|20 β 10| = 0 + c c = β ln 10
β β ln|20 β π| = 110
π‘ β ln 10
ln(20 β π) = β 110
+ ln 10
20 β ΞΈ = eβ 110π‘ β eln 10
20 β ΞΈ = 10eβ 110π‘
β΄ΞΈ = ππ β πππβ ππππ (shown)β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (2)
Sub ΞΈ = 15 into (2), 15 = 20 β 10eβ 1
10π‘
10eβ 110π‘ = 5
eβ 110π‘ = 1
2
t = β10 ln 12
= 10 π₯π§ π mins
From (2), ΞΈ = 20 β 10eβ 110π‘
As tββ, eβ 110π‘ β 0 β ΞΈ β 20
β΄ΞΈ approaches 20Β°C for large values oft.
Using G.C. (refer to Appendix for detailed steps),
t 0
20
ΞΈ
10
ΞΈ = 20 β 10eβ 110π‘
eln π₯ = π₯
οΏ½1π₯
dπ₯ = ln|π₯| + π
οΏ½ g(π¦) dπ¦ = οΏ½ f(π₯) dπ₯
Variable separable form
t
ΞΈ
10
0
20 ΞΈ is always β€ 20
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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8. Topic: Complex Numbers (Polar Form, Loci & Argand Diagrams)
(i) z1 = 1 + iβ3
|π§1| = οΏ½12 + οΏ½β3οΏ½2 = 2
ΞΈ1 = tanβ1 οΏ½β31
οΏ½ = Ο3
β΄z1 = π οΏ½ππ¨π¬ π π
+ π’ π¬π’π§ π π
οΏ½
z2 = β1 β i
|π§2| = οΏ½(β1)2 + (β1)2 = β2
ΞΈ2 = βΟ + tanβ1 οΏ½β1β1
οΏ½ = β 3Ο4
β΄z2 = βπ οΏ½ππ¨π¬ οΏ½β πππ
οΏ½ + π’ π¬π’π§ οΏ½β πππ
οΏ½οΏ½
(ii) z1 = 2 οΏ½cos Ο3
+ i sin Ο3
οΏ½ = 2eiπ3
z2 = β2(cos οΏ½β 3Ο4
οΏ½ + i sin οΏ½β 3Ο4
οΏ½ = β2eiοΏ½β3Ο4 οΏ½
π§1π§2
= 2eiΟ3
β2eiοΏ½β3Ο4 οΏ½
= 2β2
eiοΏ½Ο3βοΏ½β3Ο
4 οΏ½οΏ½
= β2eiοΏ½1312ΟοΏ½
= β2eiοΏ½β1112ΟοΏ½
= β2 οΏ½cos οΏ½β 1112
ΟοΏ½ + i sin οΏ½β 1112
ΟοΏ½οΏ½
β΄οΏ½π§1π§2
οΏ½β = β2 οΏ½cos οΏ½β 11
12ΟοΏ½ β i sin οΏ½β 11
12ΟοΏ½οΏ½
= βπ οΏ½ππ¨π¬ οΏ½ππππ
ποΏ½ + π’ π¬π’π§ οΏ½ππππ
ποΏ½οΏ½
ALTERNATIVE APPROACH
οΏ½οΏ½π§1π§2
οΏ½βοΏ½ = οΏ½π§1
π§2οΏ½
= |π§1||π§2|
= 2β2
= β2
argοΏ½π§1π§2
οΏ½β = βargοΏ½π§1
π§2οΏ½
= β οΏ½π3
β οΏ½β 3π4
οΏ½οΏ½
= β 13π12
= 11π12
β΄οΏ½π§1π§2
οΏ½β = βπ οΏ½ππ¨π¬ οΏ½ππ
ππποΏ½ + π’ π¬π’π§ οΏ½ππ
ππποΏ½οΏ½
Complex conjugate z = a + ib z* = aβib
Arg οΏ½π§1
π§2οΏ½ =
13Ο12
β 2Ο = β11Ο12
Principal argument (βΟ < ΞΈ β€ Ο)
arg (z*) = β arg (z)
|π§β| = |π§|
Arg οΏ½π§1
π§2οΏ½
β= β
13Ο12
+ 2Ο =11Ο12
Principal argument (βΟ < ΞΈ β€ Ο)
cos(βπ)= cos π sin(βπ)= βsin π
π§ = π₯ + iπ¦ = π(cos π + i sin π) Polar Form:
where π = |π§| = οΏ½π₯2 + π¦2 π = tanβ1 π¦
π₯
1
Im
Re β1
z2 β1 ΞΈ2
β3 z1
ΞΈ1
π§ = π(cos π + i sin π) = πeiπ Exponential Form:
argοΏ½π§1π§2
οΏ½ = arg (z1) β arg (z2)
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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(iii) (a) |π§ β π§1| = 2
οΏ½π§ β (1 + i β3)οΏ½ = 2
(b) arg (z β z2) = π4
arg[z β (β1 β i)] = π4
(iv) CO = CA = 2 β β COA is isosceles. CMβ₯OA β OM = MA = 1. β΄OA = OM + MA = 1 +1 = 2.
β΄The locus |π β ππ| = 2 meets the positive real axis at (2, 0).
ALTERNATIVE APPROACH
Cartesian equation of |π§ β π§1| = 2 loci:
(x β 1)2 + (yββ3)2 = 22
Sub y = 0, (x β 1)2 + 3 = 22 (x β 1)2 = 1 x β 1 = Β±1 x = 0 (reject), 2
β΄The locus |π β ππ| = 2 meets the positive real axis at (2, 0).
0
C
Im(z)
2
β3
Re(z) 1
1
β1
β1
|π§ β π§1| = 2 arg (z β z2) = π4
Ο4
O
C
M Re(z)
Im(z)
A
2
2
1
β3
β circle with center C (1, β3) and radius 2 units.
βstraight line beginning at (β1, β1) with an angle of π
4 rad with respect
to the Re(z) axis.
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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9. Topic: Differentiation (Maxima & Minima)
(i) Volume of box = x(3x)y 300 = 3x2y
y = 100π₯2 cm β¦β¦β¦β¦β¦β¦β¦β¦ (1)
External surface area of box, A1 = x(3x) + 2yx + 2y(3x) = (3x2 + 8xy) cm2
External surface area of lid, A2 = x(3x) + 2(ky)x + 2(ky)(3x) = (3x2 + 8kxy) cm2
Total external surface area of the box and the lid, A = A1 + A2 = (3x2 + 8xy) + (3x2 + 8kxy) = 6x2 + 8(1 + k) xy
= 6π₯2 + 8(1 + π)π₯ οΏ½100π₯2 οΏ½
A = 6π₯2 + 800(1 + π)π₯
dπ΄dπ₯
= 12π₯ β 800(1 + π)π₯2 β¦β¦β¦β¦β¦β¦β¦β¦ (2)
d2π΄dπ₯2 = 12 + 1600(1 + π)
π₯3 β¦β¦β¦β¦β¦β¦β¦β¦ (3)
Stationary point of A occurs when dπ΄dπ₯
= 0.
From (2), 12x β 800(1+π)π₯2 = 0
12x = 800(1+π)π₯2
x3 = 200(1+π)3
x = οΏ½200(1+π)3
οΏ½13
Using second derivative test, sub x = οΏ½200(1+π)3
οΏ½13into (3),
d2π΄dπ₯2 = 12 + 1600(1+π)
οΏ½οΏ½200(1+π)3 οΏ½
13οΏ½
3
= 12 + 1600(1 + π) Γ 3200(1+π)
= 36 >0
β Ais minimum when x = οΏ½πππ(π+π)π
οΏ½ππ
(ii) From (1) y = 100π₯2
π¦π₯ = 100
π₯3
= 100
οΏ½οΏ½200(1+π)3 οΏ½
13οΏ½
3
= 100 οΏ½ 3200(1+π)
οΏ½
= ππ(π+π)
(iii) Given 0 < k β€ 1 1 < 1 + k β€ 2 2 < 2(1 + k) β€ 4
14 β€ 1
2(1+π) < 1
2
34 β€ 3
2(1+π) < 3
2
β΄ ππ β€ π
π < π
π
Sub x = οΏ½200(1+π)3
οΏ½13
Manipulate k to get 3
2(1+π) = π
π
Sub y = 100π₯2
Second Derivative Test:
Sign of πππ
πππ Nature of stationary point
β Maximum + Minimum 0 Point of inflexion
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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ALTERNATIVE APPROACH π¦π₯ = 3
2(1+π) β Let Y = π¦
π₯ and X = k β Y = 3
2(1+X)
Using G.C. to obtain range of Y for 0 < X β€ 1 (refer to Appendix for detailed steps),
Given 0 < k β€ 1
β΄ ππ β€ π
π < π
π
(iv) For the box to have square ends, y = x.
From (ii), π¦π₯ = 3
2(1+π)
1 = 32(1+π)
1 + k = 32
k = ππ
10.
Topic: Vectors in Three Dimensions (Lines &Planes)
(i) l: π₯β10β3
= π¦+16
= π§+39
β r = οΏ½10β1β3
οΏ½ + π οΏ½β369
οΏ½ , π β ββ¦β¦β¦β¦β¦. (1)
r = οΏ½10β1β3
οΏ½ + π οΏ½β123
οΏ½ , where π = 3π
β direction vector of l, d = οΏ½β123
οΏ½
p: x β 2y β 3z = 0
β π« β οΏ½1
β2β3
οΏ½ = 0 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (2)
β normal of plane p, n = οΏ½1
β2β3
οΏ½ = βοΏ½β123
οΏ½
Since n = β d, l is parallel to normal, n.
β΄l is perpendicular to plane p. (shown)
(ii) From (1), r = οΏ½10 β π
β1 + 2πβ3 + 3π
οΏ½β¦β¦β¦ (3)
At intersection point, sub (3) into (2),
οΏ½10 β π
β1 + 2πβ3 + 3π
οΏ½ β οΏ½1
β2β3
οΏ½ = 0
10 β π + 2 β 4π + 9 β 9π = 0 21β 14π = 0
π = 32
π₯ β π1
π1=
π¦ β π2
π2=
π§ β π3
π3
r = οΏ½π1π2π3
οΏ½ + π οΏ½π1π2π3
οΏ½ , Ξ» β β
Equation of Line:
Cartesian form
Vector form
π1π₯ + π2π¦ + π3π§ = π·
r β οΏ½π1π2π3
οΏ½ = π·
Equation of Plane:
Cartesian form
Standard form
0 X 1
0.75
1.5
Y = 32(1+X)
Y
Inclusive (β€)
Exclusive (<)
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Sub π = 32 into (2),
r =
β
ββ
10 β 32
β1 + 2 οΏ½32οΏ½
β3 + 3 οΏ½32οΏ½β
ββ
= οΏ½
172232
οΏ½
β΄ Coordinates of the point of intersection of l and p is (πππ
, π, ππ ).
(iii) Sub A (β2, 23, 33) into (1),
οΏ½β22333
οΏ½ = οΏ½10β1β3
οΏ½ + π οΏ½β123
οΏ½
π οΏ½β123
οΏ½ = οΏ½β122436
οΏ½
π οΏ½β123
οΏ½ = 12 οΏ½β123
οΏ½
π = 12
Since there is a valid solution forΒ΅, point A lies on l.
From (2), r β οΏ½1
β2β3
οΏ½=0 β Origin O lies on plane p.
Using Midpoint Theorem,
πποΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ΄οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β +ππ΅οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
2
ππ΅οΏ½οΏ½οΏ½οΏ½οΏ½β = 2πποΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β β ππ΄οΏ½οΏ½οΏ½οΏ½οΏ½β
= 2οΏ½
172232
οΏ½ β οΏ½β22333
οΏ½
= οΏ½19
β19β30
οΏ½
β΄ Coordinates of the point B is (19, β19, β 30).
M(172
, 2,32) O
A (β2, 23, 33)
p
B
l
Midpoint Theorem:
If M is the midpoint of AB, then π¦ = π+π
2
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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(iv) Area of βOAB = 12
οΏ½ππ΄οΏ½οΏ½οΏ½οΏ½οΏ½β Γ ππ΅οΏ½οΏ½οΏ½οΏ½οΏ½β οΏ½
= 12
οΏ½οΏ½β22333
οΏ½ Γ οΏ½19
β19β30
οΏ½οΏ½
= 12
οΏ½οΏ½β63567
β399οΏ½οΏ½
= 12
οΏ½(β63)2 + (567)2 + (β399)2
= 348.087 β 348 unit2
11. Topic:Differentiation, Parametric Equations & Graphs
(i) Given x = t + 1π‘ and y = t β 1
π‘
dπ₯dπ‘
= 1 β 1π‘2 dπ¦
dπ‘ = 1 + 1
π‘2
= π‘2β1π‘2 = π‘2+1
π‘2
βdπ¦dπ₯
= dπ¦dπ‘dπ₯dπ‘
= π‘2+1π‘2β1
At point P,t = p,
x = π + 1π and y = π β 1
π
dπ¦dπ₯
= π2+1π2β1
Equation of tangent at point P:
π¦ β οΏ½π β 1π
οΏ½ = π2+1π2β1
οΏ½π₯ β οΏ½π + 1π
οΏ½οΏ½
π¦ β οΏ½π2β1π
οΏ½ = π2+1π2β1
(π₯) β οΏ½π2+1π2β1
οΏ½ οΏ½π2+1π
οΏ½
p(p2 β 1) y β (p2 β 1)(p2 β 1) = p (p2 + 1) x β (p2 + 1)(p2 + 1)
p(p2 β 1) y β p4 + 2p2 β 1 = p (p2 + 1) x β p4 β 2p2 β 1
p(p2 β 1) y = p (p2 + 1) x β 4p2
(p2 β 1)y = (p2 + 1) x β 4p (p2 + 1) x β (p2 β 1)y = 4p (shown)
Gradient of tangent at point (x1, y1) = π¦βπ¦1
π₯βπ₯1
οΏ½π1π2π3
οΏ½ Γ οΏ½π1π2π3
οΏ½ = οΏ½π2π3 β π3π2π3π1 β π1π3π1π2 β π2π1
οΏ½
Vector Product:
dπ¦dπ‘
=dπ¦dπ₯
Γdπ₯dπ‘
Chain Rule:
Gradient of tangent at point P.
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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(ii)
Equation of tangent at P, (p2 + 1) x β (p2 β 1) y = 4p β¦β¦β¦β¦β¦β¦β¦β¦(1) At point A, sub y = x into (1) [(π2 + 1) β (π2 β 1)]π₯ = 4p x = 2p β y = 2p
βA (2p, 2p)
At point B, sub y = β x into (1)
[π2 + 1 + π2 β 1]π₯ = 4p 2p2x = 4p
x = 2π β y = β 2
π
βπ΅ οΏ½2π
, β 2π
οΏ½
Area of βOAB = 12
|ππ΄||ππ΅|
= 12
οΏ½οΏ½(2π)2 + (2π)2οΏ½ οΏ½οΏ½οΏ½2π
οΏ½2
+ οΏ½β2π
οΏ½2
οΏ½
= 12
οΏ½πβ8οΏ½ οΏ½1π β8οΏ½
= 12
(8) = 4 units2
β΄Area of βOAB is independent of p. (shown)
(iii) x = t + 1π‘ β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(2)
y = t β 1π‘ β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(3)
(2) + (3) x + y = 2t
t = π₯+π¦2
β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(4)
Sub (4) into (2)
x = π₯+π¦2
+ 2π₯+π¦
2x(x + y) = (x + y)2 + 4 2x2 + 2xy = x2 + 2xy + y2 + 4 4 + y2 = x2 x2 β y2 = 4
ππ
ππ β ππ
ππ = 1
O
A
B
x
y
y = x
y = βx
(p2 + 1) x β (p2 β 1)y = 4p (arbitrary sketch for illustrative purpose)
Eliminate t, so that only x and y are left in the equation.
Graph of Hyperbola:
Equation: (π₯ββ)2
π2 β (π¦βπ)2
π2 = 1 Center: (h, k) Oblique Asymptotes: π¦ = π Β± π
π(π₯ β β)
OAβ₯OB
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Using G.C. (refer to Appendix for detailed steps),
Asymptotes: y = x and y = β x x-intercepts: (β2, 0) and (2, 0)
y = x y = βx y
x 0
(2, 0) (β2, 0)
πΆ:π₯2
22 βπ¦2
22 = 1
y = x
y = βx
(β2, 0) (2, 0)
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Appendix: Detailed G. C. Steps (for those still trapped in G. C. limbo)
Q5 (ii): Sketching the Inverse of a Function
β Ensure G. C. is in FUNC mode.
β Plot f(x)
β Plot fβ1(x)
β Plot f(x)
[SHIFT] [EXIT] [F6]
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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β Plot fβ1(x)
Q11: Sketching A Hyperbola
β Enter Conics. β Enter HYPERBOLA.
[F4]
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Q6 (i): Finding the Roots of a Function
Method I:Using Zero Values of Graphs
β Define approximate bounds of x value to left/right of Ξ². β Repeat steps for root Ξ³.
Method II: Using Solver
βEnter Solver (last item on MATH menu) β Enter equation.
β Enter suitable approximate value of x (i.e. x = 0) for root Ξ². β Enter suitable approximate value of x (i.e. x = 2) for root Ξ³.
Method III: Using Poly Root Finder Application
β Enter PlySmlt2
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Method I: Find Roots in Graph
Method II: Using Polynomials in Equation
Q6 (ii): Finding Region Bounded by Curve
Method I:Using βfnInt(β Function
β Key in lower/upper limits, expression and the integrating
variable (x).
Method II: Using Integration Function in Graph
β Enter Ξ² and Ξ³ as the values for lower and upper limit
respectively.
[F5]
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Method I:Using Integration Function in Graph
Method II: Using Integration Function in MATHInput Mode β Key in lower/upper limits and the expression.
[F5]
[SHIFT][MENU]
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Q7: Graph Sketching
β Ensure G. C. is in FUNC mode.
[SHIFT] [EXIT] [F6]
GCE βAβ Level October/November 2010 Suggested Solutions
Mathematics H2 (9740/01) version 2.1
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Q9 (iii): Finding the Range of Values in a Function
β Enter expression
β Enter X = 0 and X = 1to obtain the corresponding values of Y.
β Enter X = 0 and X = 1to obtain the corresponding values of Y.
[F5]