Làm quen toán tiếng Anh

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1. (a) It is given that ξ = {1, 2, 3, 4, 5, ..., k} where k is a positive integer.
P is a subset where P = {x: x is a prime number}.
Q is another subset and Q = {x: the unit digit is 9}
(i) List all the members of Q given that k = 100
(ii) If n(P ∩ Q) = 4, list the elements of the set P ∩ Q and hence state the largest
possible value of k.
(b) It is given that n(ξ) = 23, n(A ∩ B) = x, n(A) = y, n(B) = 2y and n(A' ∩ B') = 7.
(i) Use a Venn diagram and show that 3y = 16 + x
(ii) Hence find the least possible value of y.
2. Find the equation of the perpendicular bisector of AB where A is (3, 10) and B is (7, 2)
3. The roots of the quadratic equation x
2
-
6
x +
x
= 0 are α and β
Without using a calculator, show that
α
1
+
β
1
=
3
4. The tangents TA and TD are drawn from a point T to the circle, centre O. The diameter
DB and the tangent TA when produced meet at E. Given that ∠ ADO = 26
o
, find

(a) ∠ ADT
(b) ∠ ATD
(c) ∠ BAE
(d) ∠ ACD ( C is a point on the cirlce, centre O)

Tôn Nữ Bích Vân giới thiệu

Solution:
1. (a) (i) Q = {9, 19, 29, 39, 49, 59, 69, 79, 89, 99}
(ii) P ∩ Q = {19, 29, 59, 79} ; k = 88
(b)
least value of y = 6 [when x = 2]
2. The mid-point of AB is (5, 6). The gradient of the line AB is
3 - 7
10 - 2
= -2. The
perpendicular is the
line with gradient
2
1
and passing through the
point (5, 6). Equation of the perpendicular of AB is:
y - 6 =
2
1
(x - 5)
y =
2
1
x +

2
7
Alternative method
Let P(x, y) lie on the perpendicular of AB.
Therefore PA = PB
⇒
22
10) -(y 3) -(x
+
=
22
2) -(y 7) -(x
+
⇒ x
2
- 6x + 9 + y
2
- 20y + 100 = x
2
- 14x + 49 + y
2
- 4y + 4
Simplifying, 8x - 16y + 56 = 0
⇒ y =
2
1
x +
2
7

3. The quadratic equation whose roots are α and β is (x - α) (x - β) = 0
x
2
- (α + β)x + αβ = 0
By inspection: α + β =
6
αβ =
2
Now
3
2
611
==
αβ
β+α
=
β
+
α
[Hence shown]
4. (a) ∠ODT = 90
o
(tan ⊥ rad. )
∠ADT = 90
o
- 26
o
= 64
o
(b) TA = TD (tangents from an external point)

∠TDA = ∠TAD (base ∠s of isos. ∆)
∠ATD = 180
o
- 64
o
- 64
o
(∠ sum of ∆)
= 52
o
ξ
A B
7
(y-x) x (2y-x)
26
o
O
E
A
B
C
D
T
(c) ∠BAE = ∠ADB (∠ in alt. segment) = 26
o
(d) ∠ACD = ∠ADT (∠ in alt. segment) = 64
o