GCE 'A' Level October/November 2008 Suggested Solutions. Mathematics H2 (
9740/01) version 1.1. For tuition, exam papers & Last-Minute Buddha FootΒ ...
GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01)
version 1.1
MATHEMATICS (H2)
9740/01
Paper 1 Suggested Solutions
October/November 2008
2.
Topic: Summation of Series (Mathematical Induction) 1
Let Pn denotes the statement βππ = π(π + 1)(4π + 5), β π β β€+ β. 6
When n = 1, 1.
Topic: Definite Integrals
L.H.S.
= S1 = u1 = 3
From the diagram:
R.H.S.
= (1)(1+1)(4+5) = 3
2 2 β«1 π¦ dπ₯ = β«1 π₯ 2 dπ₯β¦β¦ 4 4 β«π π₯ dπ¦ = β«π οΏ½π¦ dπ¦
Area A = Area B =
=
Equating (1) & (2): 2
4 1 β«π π¦ 2 dπ¦β¦β¦
(2)3 3
2
π₯3
οΏ½ οΏ½
β 8 3
3 1
(1)3 3
β
1 3
3 2
3 4 2
2
=οΏ½ π¦ οΏ½ 3
π
3
2
=
π =
16 3
9
(1)
β΄ P1 is true. 1
Assume Pk is true i.e. ππ = π(π + 1)(4π + 5), for some π β β€+ 6
(2)
2
3
2
β π 3
3 2
3
1
To show that Pk+1 is also true i.e. ππ+1 = (π + 1)(π + 2)(4π + 9), L.H.S. = Sk+1 1
= k (k + 1) (4k + 5) + (k + 1) [2(k+1) + 1]
π₯ π+1 οΏ½ π₯ dπ₯ = +π π+1
6
π
1
= (k + 1) [k (4k + 5) + 6(2k + 3)] 6
1
β΄a = ( )3
1
= (k + 1) (4k2 + 5k + 12k + 18) Bring out the (k + 1) 6 factor since itβs found on the R.H.S.
2 9 2
6
1 ππ = π(π + 1)(4π + 5) 6
= Sk + uk+1
= (4)2 β (π)2 3
6
β L.H.S. = R.H.S.
1
4
β«1 π₯ 2 dπ₯ = β«π π¦ 2 dπ¦
1
2
6
uk+1 = (k + 1) [2(k+1) + 1]
1
= (k + 1) (4k2 + 17k + 18) 6 1
= (k + 1) (k + 2) (4k + 9) 6
β 2.73 (3 sig. fig.)
= R. H. S. β΄Pk+1 is also true if Pk is true. Since P1 is true and Pk+1 is true if Pk is true, by mathematical induction, Pn is true β π β β€+
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01) 3.
Topic: Vectors
= =
1 5 οΏ½ 4 οΏ½βοΏ½β1οΏ½ β3 0 οΏ½12 +4 2 +(β3)2 οΏ½52 +(β1)2 +02 (1)(5)+(4)(β1)+(β3)(0) 26
β26β26
4.
Topics: Differentiation, Differential Equations (i)
dπ¦ dπ₯
=
3π₯
π₯2+ 1
3π₯
Express in
β« dπ¦ = β« π₯ 2+ 1 dπ₯ 3
y =
O
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β ππ΄ β οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β ππ΅ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βοΏ½οΏ½ππ΅ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βοΏ½ οΏ½ππ΄
1
P
οΏ½οΏ½οΏ½οΏ½οΏ½β = π΄π΄ οΏ½οΏ½οΏ½οΏ½οΏ½β ππ β΅ same magnitude & direction
1 5 = οΏ½ 4 οΏ½ + οΏ½β1οΏ½ β3 0 π =οΏ½ π οΏ½ βπ
=
B
Equal Vectors:
οΏ½οΏ½οΏ½οΏ½οΏ½β + οΏ½οΏ½οΏ½οΏ½οΏ½β (i) οΏ½οΏ½οΏ½οΏ½οΏ½β ππ΄ = ππ΄ π΄π΄ οΏ½οΏ½οΏ½οΏ½οΏ½β οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ΄ + ππ
(ii) cos β AOB =
version 1.1
A Scalar Product of two vectors a and b: π β π = |π||π| cos π b
ΞΈ
a
1
2
π
=
π
2π₯
β« π₯ 2+ 1 dπ₯
π₯π§|ππ + π| + π
3
οΏ½
fβ² (π₯) f(π₯)
f β² (π₯) dπ₯ = ln|f(π₯)| + π f(π₯)
(ii) Sub y = 2, x = 0 into π¦ = ln|π₯ 2 + 1| + π, 3
2
2 = ln|02 + 1| + π β c = 2 2
π
β΄ π = π₯π§|ππ + π| + π π
(iii) As x β +β, As x β ββ,
dπ¦
dπ₯ dπ¦ dπ₯
β 0+
β 0β
β΄the gradient of every solution curve tends to be horizontal as x β Β± β.
β AOB = cos β1 ( ) 26
= 87.795Β° β 87.8Β° (3 sig. fig.)
(iv)
(iii) Area of parallelogram OAPB οΏ½οΏ½οΏ½οΏ½οΏ½β Γ οΏ½οΏ½οΏ½οΏ½οΏ½β = οΏ½ππ΄ ππ οΏ½ 1
5
= οΏ½οΏ½ 4 οΏ½ Γ οΏ½β1οΏ½οΏ½ β3
β3
= οΏ½οΏ½β 15οΏ½ οΏ½
0
β21
π’ π£ οΏ½1 4 5 β1
π€ 4 β3 1 β3 1 4 οΏ½π’ β οΏ½ οΏ½π£+ οΏ½ οΏ½π€ β3οΏ½ = οΏ½ β1 0 5 0 5 β1 0 = β3π’ β 15π£ β 21π€
TI-84 Plus
= οΏ½(β3)2 + (β15)2 + (β21)2 = β675
= ππβπ units2
Casio fx-9860G
N.B. Final answer expressed in surd form as question asks for exact area.
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01) 5.
version 1.1
Topic: Integration 1 β3
(i) β«0
1
1+9π₯ 2
6. 1 β3
dπ₯ = β«0 = = = = =
1 3
1
1
dπ₯
1+(3π₯)2 1
Express in
3
β«0β3 1+(3π₯)2 dπ₯
οΏ½
1 β3
[tanβ1 (3π₯)]0 3 1 β1 3 3 1 3 Ο
π
12
οΏ½tan ( ) β tanβ1 (0)οΏ½ Ο
οΏ½ β0οΏ½ 3
β3
e
π₯ π+1
e
e π₯ π+1
ILATE/LIATE Rule: Sub u = ln π₯ (L) dπ’ 1 = dπ₯ π₯
Sub
dπ£
dπ₯
= π₯ π (A)
π₯ π+1 π£= π+1
= οΏ½ln e β οΏ½ = = = =
eπ+1 π+1
eπ+1 π+1 1
β β
(π+1)2
π+1 1
π+1
1
1
οΏ½ β ln 1 β οΏ½
οΏ½
π₯ π+1
e
οΏ½
π+1 1
(π+1)2
1π+1 π+1
(a) By Cosine Rule, cos β ABC =
fβ²(π₯)
1+[f(π₯)]2
cos ΞΈ =
β²
f (π₯) dπ₯ = tanβ1 [f(π₯)] + π + [f(π₯)]2 Integration by parts: dπ£ dπ’ οΏ½ π’ dπ₯ = π’π’ β οΏ½ π£ dπ₯ dπ₯ dπ₯ 1
(ii) β«1 π₯ π ln π₯ dπ₯ = οΏ½ln π₯ β οΏ½ οΏ½οΏ½ β β«1 οΏ½ οΏ½ οΏ½ οΏ½ dπ₯ π+1 π+1 π₯ eπ+1
Topic: Maclaurinβs Series
οΏ½οΏ½ β
e π₯π β«1 οΏ½π+1οΏ½ dπ₯
= 2
6
1
1 2
β 4 οΏ½1 +
2 3 οΏ½ ΞΈ2 οΏ½ & 4
β 2 [1 + οΏ½ ΞΈ2 οΏ½ + β¦]
above
β2 β2
2 4 3 [1 + ΞΈ2 8 3 2 + ΞΈ 4 π
β΄ a = 2, b =
π
1
+ β¦]
1
B
Small Angle Approximation: 1 cos π β 1 β π 2 2
Express in (1 + x)n since n is not a positive integer
β f(0) = tan ( Ο) = 1
(b) f(x) = tan (2x + Ο) 4
1
3 2 2 ΞΈ οΏ½ 4
1 3
ΞΈ
A
β΄ AC β οΏ½4 + 3ΞΈ2 οΏ½2 (Shown)
Maclaurinβs expansion: (1 + π₯)π = 1 + ππ + β―
1
1
f β²(x) = 2 sec2 (2x + Ο)
(π+π)π
4
4
1
β f β²(0) = 2sec2 ( Ο) = 4
1
1
4
1
f β³(x) = 2 β
2 sec(2x + Ο) [2 sec(2x + Ο) tan (2x + Ο)] 2
1
4
1
= 8 sec (2x + Ο) tan (2x + Ο) 4 4 β f β³(0) = 8(2)(1) = 16 f(x) = f(0) + xf β²(0) + π₯2
π₯2 2!
f β³(0) + β¦
= 1 + x(4) + οΏ½ οΏ½16 + β¦ 2!
= 1 + 4x + 8x2 + β¦
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10β π΄πΆ 2
2
ignored since ΞΈ is small
πππ+π +π
3
2 (1)(3)
β 10 β 6 + 3ΞΈ 2 β 4 + 3ΞΈ 2
x term i.e.
[(π + 1)eπ+1 β eπ+1 + 1]
C
2 (π΄π΅)(π΅πΆ)
12 + 32 β π΄πΆ 2
β AC = 10 β 6 cos ΞΈ 1 β 10 β 6 (1 β ΞΈ 2)
2
(eπ+1 β 1π+1 )
π΄π΅2 + π΅πΆ 2 β π΄πΆ 2
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4
4
d sec π [f(π₯)] dπ₯
= π sec πβ1 [f(π₯)] β
d sec[f(π₯)] dπ₯
d sec[f(π₯)] dπ₯ = f β² (π₯) β sec[f(π₯)] tan[f(π₯)]
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01) 7.
version 1.1
Topic: Differentiation
8. π₯
Total time taken: 180 = (2y + x)(3) + ΟοΏ½ οΏ½ (9) 2
π₯
Total time taken = length of straight part Γ 3 hrs/m + length of semicircular part Γ 9 hrs/m
60 = 2y + x + ΟοΏ½ οΏ½ (3) 2y = 60 β x β π₯
y = 30 β β 2
2 3Οπ₯ 2 3Οπ₯ 4
β¦β¦β¦ (1)
1 2
π₯ 2
= 30π₯ β
When A is maximum,
π₯+
2
= π₯ οΏ½30 β β
Sub y expression from (1)
β 30 β π₯ β
π₯
10Οπ₯
8 10Οπ₯ 8
= 30π₯ β
dπ΄ dπ₯
= 0.
2
π₯2 2 π₯2 2
β β
3Οπ₯ 4
4 5Οπ₯ 2
π₯2
1
8
+
4
2
Οπ₯ 2 8
=0
= 30
x =
30
1+
10Ο 8
β 6.0889 β 6.09 (3 sig. fig.) Second derivative test:
d2 π΄ dπ₯ 2
= β1 β
5Ο 4
< 0 β A is maximum when x = 6.09
6.0889
3Ο(6.0889)
Sub x β 6.0889 into (1): y = 30 β β 2 β 12.608 β 12.6 (3 sig. fig.)
Alternative Method 2 3 Ο = 1 + 3β3i + 3οΏ½β3iοΏ½ + οΏ½β3iοΏ½ π§ = 1 + β3i = 2ei 3 Ο 3 Ο = 1 + 3β3i β 3(3) β 3β3i π§ 3 = οΏ½2ei 3 οΏ½ = 23 ei 3 = β8 = β8 (ii) Given that 1 + β3i is a root, sub π§ = 1 + β3i into 2π§ 3 + ππ§ 2 + ππ§ + 4 = 0: 2(β8) + ποΏ½1 + β3iοΏ½ + ποΏ½1 + β3iοΏ½ + 4 = 0 β16 + ποΏ½1 + 2β3i β 3οΏ½ + π + πβ3i + 4 = 0 z3= β8 β16 + 2πβ3i β 2π + π + πβ3i + 4 = 0 from β12 β 2π + π + 2πβ3i + πβ3i = 0 Comparing real parts: β12 β 2π + π = 0 β2π + π = 12 π = 12 + 2π β¦ β¦ β¦ (1) Comparing imaginary parts:
οΏ½ + ΟοΏ½ οΏ½
3Οπ₯ 2
3
(i) z13 = οΏ½1 + β3iοΏ½
2
Total area of rectangular & semicircular parts: A = π₯π¦ + Ο οΏ½ οΏ½
Binomial Expansion: (1 + π₯)3 = 1 + 3π₯ + 3π₯ 2 + π₯ 3
Topic: Complex Numbers
4
Given 1 + β3i is a root β 1 β β3i is also a root. β
Non-real roots occur in conjugate pairs in polynomial equations οΏ½π§ β οΏ½1 + β3iοΏ½οΏ½οΏ½π§ β οΏ½1 β β3iοΏ½οΏ½ = π§ 2 β 2π§ + 4 with real coefficients.
A quadratic factor of the equation is
β 2π§ 3 β 3π§ 2 + 6π§ + 4 β‘ (π§ 2 β 2π§ + 4)(2π§ + 1) = 0
β΄x = 6.09 & y = 12.6 gives the flower-bed a maximum area.
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2πβ3 + πβ3 = 0 π = β2π β¦ β¦ β¦ . β¦ (2) Equating (1) & (2): 12 + 2π = β2π 4π = β12 β π = βπ π = β2(β3) = π (iii) Since the equation 2π§ 3 β 3π§ 2 + 6π§ + 4 = 0 has real coefficients,
β π = π + βππ’, π β βππ’, β facebook.com/JossSticksTuition
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π π
By inspection or long division.
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01)
version 1.1
9.
ALTERNATE APPROACH (ii) Since the equation 2π§ 3 + ππ§ 2 + ππ§ + 4 = 0 has real coefficients, Given 1 + β3i is a root β 1 β β3i is also a root. β A quadratic factor of the equation is
οΏ½π§ β οΏ½1 + β3iοΏ½οΏ½οΏ½π§ β οΏ½1 β β3iοΏ½οΏ½ = π§ 2 β 2π§ + 4
Non-real roots occur in conjugate pairs in polynomial equations with real coefficients.
β 2π§ 3 + ππ§ 2 + ππ§ + 4 β‘ (π§ 2 β 2π§ + 4)(π΄π§ β π) where A, B β β
Comparing coefficients of z3, A = 2 Comparing coefficients of constant, 4 = β4B β B = β1 Comparing coefficients of z2, a = βB β 2A = β(β1) β 2(2) = β3 Comparing coefficients of z, b = β2(βB) + 4A = β2[β(β1)] + 4(2) = 6
Topic: Graphing Techniques (i)
f(π₯) =
Quotient Rule: π’ β² π£π’β² β π’π’β² (ππ₯ (ππ₯ + π)(π) β + π)(π) οΏ½ οΏ½ = f β² (π₯) = π£ π£2 (ππ₯ + π)2 πππ₯ + ππ β πππ₯ β ππ = (ππ₯ + π)2 ππ β ππ β 0, βπ₯ β΅ ππ β ππ β 0 (Given) = (ππ₯ + π)2
β΄By differentiation, fβ²(x) β 0, βx β the graph of y = f(x) has no turning points. π π
π
= +
2π§ 3 β 3π§ 2 + 6π§ + 4 β‘ οΏ½π§ β οΏ½1 + β3iοΏ½οΏ½οΏ½π§ β οΏ½1 β β3iοΏ½οΏ½(2π§ + 1) = 0 π
β3 π§3 = β 1 β 2
1 2
0
π§1 = 1 + β3i
1
ππ₯+π
ππβππ
π (ππ₯+π)
π
π(ππ₯+π)
π
By long division:
ππ π
cx + d ax + b ax +
ππβππ
π
When ad β bc = 0, π¦ = f(π₯) = , π₯ β β π
π π
a c
ad c ad bβ c
π
π
β΄ The graph is a horizontal line at π = , but undefined at π = β π π y
Re (z)
(iii) π¦ =
π π
3π₯β7
2π₯+1
Sub a = 3, b = β7, c = 2, d = 1 into f β²(x) in (i):
ππ + π , ππ + π ππ = ππ
π¦=
π x π β΄ the graph has a positive gradient at all points since (2x + 1)2 > 0, x β βππ.
π§2 = 1 β β3i
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πβ
π
= β
β ββ3
ππ₯+π
(ii) f(x) = +
(iii) Sub a = β3, b = 6, A = 2, B = β1 in (ii), β π = π + βππ’, π β βππ’, β π Im (z)
ππ₯+π
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dπ¦ dπ₯
=
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3(1)β (β7)(2) (2π₯+1)2
=
17
(2π₯+1)2
>0
β
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01)
version 1.1
10. Topic: Arithmetic & Geometric Series
(iv)
(i) Since the student saves $3 more than the previous month in each subsequent month, the monthly savings = 10, 13, 16, 19, β¦ β Arithmetic series with the 1st term, a = 10, and common difference, d = 3. To save over $2000 in total, ππ > 2000 π [2π + (π β 1)π] > 2000 2 Sub a = 10, d = 3 into Sn:
Casio fx-9860G TI-84 Plus
(a) y
π¦= 1 π₯=β 2
2
3π₯ β 7 π¦= 2π₯ + 1
3 2
x
7 οΏ½ , 0οΏ½ 3
(0, β7)
π
Sum of A. P.: ππ π = [2π + (π β 1)π] 2
3 π¦=οΏ½ 2 π₯=β
1 2
7
οΏ½3, 0οΏ½
π¦2 =
Casio fx-9860G
n > 33.79 or n < β39.46 (reject) β n = 34 (earliest no. of months since 1 Jan 2009)
β΄ she will first have saved over $2000 on the 1st of October 2011. (ii) (a) Taking into account only the original $10 deposit:
(b) y
[2(10) + (π β 1)(3)] > 2000 π [17 + 3π] > 2000 2 17π + 3π2 > 4000 3π2 + 17π β 4000 > 0
3π₯ β 7 2π₯ + 1
Month 1 2 3
Balance at Start of Month ($) 10 10+(0.02)(10) = (1.02)(10) (1.02)(10)+(0.02)(1.02)(10) = (1.02)(1.02)(10)
Interest Earned End of Month ($) (0.02)(10) (0.02)(1.02)(10) (0.02)(1.02)(1.02)(10)
n
(1.02)nβ1(10)
(0.02)(10)(1.02)nβ1
Total compound interest earned after n months from the original $10 x
= (10) (0.02) + (10) (0.02) [1.02 + 1.022 + β¦ + 1.02n-1] = (10)(0.02) + (10)(0.02) οΏ½
3 π¦ = βοΏ½ 2
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= (10)(0.02) οΏ½1 + = 10(1.02π β 1) facebook.com/JossSticksTuition
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1.02πβ1.02 0.02
οΏ½
1.02οΏ½1.02πβ1 β1οΏ½ 1.02β1
οΏ½
Sum of G.P. where r > |1|: π(π π β 1) ππ = πβ1
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01)
version 1.1
β΄total interest earned in 2 years (24 months) from original $10 = 10(1.0224 β 1) β $6.08 (3 sig. fig.)
(b) Account balance including $10 deposited monthly: Month
Balance at Start of Month ($)
Balance at End of Month ($)
1
10
(1.02)(10)
2
(1.02)(10) + 10
3
1.022(10) + 1.02(10) + 10
1.02[1.02(10) + 10] = 1.022(10) + 1.02(10) 1.02[1.022(10) + 1.02(10) + 10] = 1.023(10) + 1.022(10) + 1.02(10) (1.02 + 1.022 + β¦ + 1.02n)10
n
11. Topic: Three-Dimensional Geometry π₯ Let οΏ½π¦οΏ½be the point of intersection of p1, p2, p3. Solving for p1, p2 and p3, π§ π Cartesian equation of plane β ππ π₯ π1 2 β5 3 3 π₯ β πβ οΏ½π¦ οΏ½ = οΏ½3 2 β5οΏ½ β5 οΏ½ = ββ ππβ π« β π§ = οΏ½π¦οΏ½ β οΏ½π2 οΏ½ = π: π§ π3 5 β20.9 17 16.6 π π§ π1 π₯ + π2 π¦ + π3 π§ = π β ππ β
1.02(1.02π β1)
= 10 οΏ½ οΏ½ 1.02β1 = 510(1.02π β 1) β¦β¦β¦ (1)
Total in account at end of 2 years (24 months) = 510(1.0224 β 1) Sub n = 24 into (1) β 310.3029972 β $310 (3 sig. fig.)
TI-84 Plus
(c) Let n be the number of complete months taken for the total in the account to first exceed $2000. Using (1) from (b)(ii), 510(1.02π β 1) > 2000 1.02π > 4.9215 π ln 1.02 > ln 4.9215 π > 80.475 β΄ n = 81 months.
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01)
version 1.1
(ii) Given that l lies on p3 β all points on l lies on p3.
(i) Since l lies on p1 and p2 β solve for p1 and p2, 2π₯ β 5π¦ + 3π§ = 3β¦β¦β¦β¦ (1) 3π₯ + 2π¦ β 5π§ = β5 β¦β¦β¦ (2)
Eliminate y (1) Γ 2 + (2) Γ 5: 19π₯ β 19π§ = β19 β π§ = π₯ + 1
Eliminate x (1) Γ 3 β (2) Γ 2: β19π¦ + 19π§ = 19 β π§ = π¦ + 1 β
π₯+1 1
=
π¦+1 1
=
π§
1
Cartesian equation of π1 π1 line οΏ½π2 οΏ½ + π οΏ½π2 οΏ½: π3 π3
π₯ β π1 π¦ + π2 π§ β π3 = = π1 π2 π3
βπ π β΄vector equation of l is: π« = οΏ½βποΏ½ + π οΏ½ποΏ½ , π β β π π ALTERNATE APPROACH
Picking two points on l from the vector equation of l in (i), β1 β1 5 Let s = 0: π« = οΏ½β1οΏ½ β οΏ½β1οΏ½ β οΏ½ π οΏ½ = π 0 0 17 β5 β π = π β¦β¦β¦β¦β¦β¦ (1)
0 0 5 Let s = 1: π« = οΏ½0οΏ½ β οΏ½0οΏ½ β οΏ½ π οΏ½ = π 1 1 17 π = 17 β¦β¦β¦.β¦β¦β¦ (2)
Sub (2) into (1): β5 β π = 17 β π = β22 β΄Ξ» = β22, Β΅ = 17
Direction vector of l, ποΏ½π = π§π1 Γ π§π2 (where π§π1 and π§π2 are the normal vectors of p1 and p2 respectively)
(iii) For the three planes to have no point in common, a.
2 3 1 = οΏ½β5οΏ½ Γ οΏ½ 2 οΏ½ = 19 οΏ½1οΏ½ 3 β5 1
3
Using the common point obtained earlier, π
β ππ π β πβ vector equation of l: π« = ββ ππβ + π οΏ½ποΏ½ , π β β π π π’ β ππ β
b.
β π₯ β π§ = β1 β π¦ β π§ = β1
2 3 β5 3 2 β5 =οΏ½ οΏ½π’ β οΏ½ οΏ½π£ + οΏ½ οΏ½π€ 3 β5 2 β5 3 2 = 19π’ + 19π£ + 19π€
1 β ποΏ½π = οΏ½1οΏ½ 1
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5 1 οΏ½ π οΏ½ β οΏ½1οΏ½ = 0 1 17 5 + π + 17 = 0 π = β22
p3 must not contain l β al β
π§π3 β π
π£ π€ ππ = οΏ½2 β5 3 οΏ½ 3 2 β5
TI-84 Plus
p3 must be parallel to l β π§π β₯ πΜπ direction vector of π
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β1 5 β οΏ½β1οΏ½ β οΏ½β22οΏ½ β π 0 17 π β 17
np3 dl
p1 p2 p3
aβ₯bβaβ
b=0 al: Pick a point on l obtained from (i)
β΄Given that the three planes have no point in common, Ξ» = β22 and Β΅ can be any real number but not 17.
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GCE βAβ Level October/November 2008 Suggested Solutions
Mathematics H2 (9740/01)
version 1.1
(iv) Let p and q be two distinct direction vectors that lie on the plane (required to get the equation of the plane). n p q
l
(-1, -1, 0)
O
(1, -1, 3)
1 Since plane contains l β π© = οΏ½1οΏ½ 1
Direction vector of l, ποΏ½π obtained in (i).
1 Since plane contains the point (1, β1, 3) β πͺ = οΏ½β1οΏ½ β ππ 3 1 β1 2 Position vector of point on l, = οΏ½ οΏ½ β οΏ½ οΏ½ = οΏ½ β1 β1 0οΏ½ al obtained in (i). 3 0 3 2 1 3 Normal vector to the plane, π§ = π© Γ πͺ = οΏ½1οΏ½ Γ οΏ½0οΏ½ = οΏ½β1οΏ½ 1 3 β2 1 3 Vector equation of plane: π« β π§ = οΏ½β1οΏ½ β οΏ½β1οΏ½ 3 β2 π₯ 3 οΏ½π¦οΏ½ β οΏ½β1οΏ½ = β2 π§ β2 β΄ Cartesian equation of plane: ππ β π β ππ = βπ
Cartesian equation of plane π1 π₯ π« β π§ = οΏ½π¦οΏ½ β οΏ½π2 οΏ½ = π: π3 π§ π1 π₯ + π2 π¦ + π3 π§ = π
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