I1.4 Given that C is a root of the equation kx
9 I1 a
1 I2 R
I3 S
10 I4 k
1
25
1
b S
1 R 30 r
2
2
1
2 C
10 T
1 T 6 s (= )
2
2
9
5
7
2 Group Events
- D –2 W P w
G1 k
1 G2 w
45 G3 r 2006 G4 R 12 3
16
7 B z –13 x (=1.75) S
8
15
4
1
1
1 C s z
30 T
4
4
2
15
a
1 t
14 R (= 3.75) W 2013021
4 Individual Event 1 I1.1 If a is a real number satisfying log ( x + 3) – log ( x + 1) = 1, find the value of a .
2
2
3 log = log
- x
2
2
2
1
- x
x + 3 = 2x + 2 ⇒ x = 1 a = 1
I1.2 In Figure 1, O is the centre of the circle with radius 1 cm. If the A
length of the arc AB is equal to a cm and the area of the shaded 1 cm
2
sector OAB is equal to b cm , find the value of b. (Take = 3)
π O a cm
1
1
1 1 cm
b = rs =
1 1 =
⋅
2
2
2 B
Figure 1 I1.3 An interior angle of a regular C-sided polygon is 288b , find the value of C.
°
Each interior angle = 288b = 144
° ° o
360 Each exterior angle = 36 =
° C C = 10
2 I1.4 Given that C is a root of the equation kx + 2x + 5 = 0, where k is a constant.
If D is another root, find the value of D.
1
⇒
100k + 20 + 5 = 0 k =
−
4
2 C + D = sum of roots =
− k
⇒
10 + D = 8 D = –2
I2.1 Given that a, b and c are three numbers not equal to 0 and a : b : c = 6 : 3 : 1.
2 b
3 If R = , find the value of R.
2
- 2 a bc
- I2.2 Given that . If S = , find the value of S.
- R
- k = 0 k = –
- −
- k
- k R
- −
- = cos
- cos
- y x T
- y x
- x
- x – 1 = 0 or x – x + 1 = 0
- 1
- 2
- −
- y
- 2
- 2 x
- If P = 3 , find the value of P.
- 2
- 2
- 2
- a =
- =
- 2
- If r = |xy|, find the value of r.
- which is a sum of two non-negative numbers.
+ +
- G1.2 Given that x = x and y = y satisfy the system of equations . If B = , find the
- = = 1 − = − x = y
- Sub. x = y into the first equation: = 1 x = y = B =
- ⇒ ⇒
- – f (–6) = 3 p
- (1 + x )(1 + y ) = 55 1 + x = 5, 1 + y = 11 or 1 + x = 11, 1 + y = 5
- y +5y G3.2 Given that 6 = 36 and 6 = 216, find the value of x.
- tan y tan x tan x + tan y + 1 =
- tan x tan y
- G4.2 Given that a and b are positive real numbers and a b = 2. If S = a b ,
- a b a b
- G4.3 Let 2 = 7 = 196. If T = , find the value of T.
- 1
- T = = =
- =
- G4.4 If W = 2006 – 2005 + 2004 – 2003 … + 4 – 3 + 2 – 1 , find the value of W.
- = 2006
2
3 3 k
9
( ) Let a = 6k, b = 3k, c = k, then R = = 2 +
2 6 k 3 k k
25
( ) ( ) k R k
2 R
=
2 k R
9
9
⇒
25
25
9
18
2 R
25
25 S = = = 1
18
9
2
25
25 I2.3 Given that T = sin 50 (S + 3 tan 10 ), find the value of T.
°× × ° o
sin
10 T = sin 50 (1 + 3 )
°× ⋅ o o cos
10 sin
5 o o
10 3 sin
10 o ⋅ cos
( )
10 o 2 sin
5
1
3 o o +
= cos o 10 sin
10 cos
10
2
2 o 2 sin 5 o o o o
= cos 60 cos 10 sin 60 sin
10 o ⋅ ⋅
( )
10 o o o 2 sin 5 cos 50 sin 100
=
1 o o = = cos 10 cos
10
T
y
= I2.4 Given that x and y are real numbers satisfying the system of equations . x
= If W = x + y , find the value of W.
1
y =
x
1
=
1
1 ..........(*)
= x
1
1 = x + 1 or = –x + 1
x x
2
2
1 = x + x or 1 = –x + x
2
2 x
− 1 ±
5 x
= or no solution
2 − 1 −
5 Check: sub. x = into (*)
2
2 LHS = < 0, RHS > 0 (rejected)
1
5 − − − +
1
5
2
5
1 −
1
5
5
1 + + +
When x = ; LHS = = ; RHS = = (accepted)
2
2
2
1
5 −
1
5
1
5
1
5
= ⇒ W = x + y = = 5
2
2
2
2 x
3 A B
−
2 I3.1 Given that ,where A and B are constants. If S= A + B , find the value of S =
2 x x
1 x x −
−
3 Ax B x
1 − −
( ) =
2 x x
1 x x −
− ( ) A + B = 2, –B = –3 A = –1, B = 3
2
2 S = (–1) + 3 = 10
I3.2 In Figure 1, ABCD is an inscribed rectangle, AB = (S – 2) cm and D C AD = (S – 4) cm. If the circumference of the circle is R cm, find the value of R. (Take = 3)
π AB = 8 cm, CD = 6 cm AC = 10 cm (Pythagoras’ theorem) A B
R = 10 = 30 π
Figure 1 R I3.3 Given that x and y are integers satisfying the equation xy = 21x + 20y – 13.
2 If T = xy, find the value of T. 15xy = 21x + 20y – 13 (3x – 4)(5y – 7) = 15
3 x
4
1 3 x
4
3 3 x
4
5 3 x
4
15 3 x
4
1
− = − = − = − = − = − or or or or or
5 y
7
15 5 y
7
5 5 y
7
3 5 y
7
1 5 y
7
15
− = − = − = − = − = −
3 x
4
3 3 x
4
5 3 x
4
15
− = − − = − − = − or or
5 y
7
5 5 y
7
3 5 y
7
1
− = − − = − − = −
For integral solution: 3x – 4 = 5, 5y – 7 = 3
⇒ x = 3, y = 2 T = 3 2 = 6
×
2 I3.4 Let a be the positive root of the equation x – 2x – T = 0.
T
T
T
T
a
2 x – 2x – 6 = 0
1
7
6
2
2 ⇒ ⇒ a – 2a – 6 = 0 a = 2a + 6 a = 2 + a
T
6 T
6 T
6
⇒ ⇒
2 + = 2 + = a
2 2 = a 2 = 2 + = a
T T a a a a
2
2
T a
a
6
6 P =
3
1 2 = 1+ a = 1 +
1 7 =
2
7
= + + + + + a a
k
1
1
1
1
1
1
1
1
1
1
I4.1 Let
1 1 , find the value of k .
= × − + + + + × + + + +
4
2
3
2
3
4
2
3
4
2
3 Reference: 1995 FG6.2
k
1
1
1
1
1
1
⇒
Let x = 1 , y = 1 , then = x(y – 1) – y(x – 1) = –x + y = k = 1
2
3
2
3
4
4
4
2 I4.2 Let x and y be real numbers satisfying the equation y + 4y + 4 + x y k = 0.
Reference: 2005 FI4.1, 2009 FG1.4, 2011 FI4.3, 2013 FI1.4, 2015 HG4, 2015 FI1.1
2 x y k
The equation is equivalent to (y + 2) = 0
⇒
y + 2 = 0 and x + y + 1 = 0
y = –2 and x – 2 + 1 = 0, x = 1 r = |–2 1| = 2 × rd
I4.3 In Figure 1, there are eight positive numbers in series. Starting from the 3 number, each
th1 number is the product of the previous two numbers. Given that the 5 number is and the
r
1
th 8 number is . If the first number is s, find the value of s.
4 r
1
1
s
4 r r nd
Let the 2 number be a, then other numbers are as follows:
1
8
13
1
2
3
2
3
5
5
8 s a s a = s a sa sa = s a s a
4 r r
1
2
3 s a = ................. (1) r
1
8
13 s a
= .............. (2)
4 r
4
(2) (1) : a = 1
÷
1
1
1
2 ⇒
Sub. a = 1 into (1): s = ; s > 0 s =
= r
2
2 I4.4 Let [x] be the largest integer not greater than x. For example, [2.5] = 2.
2
4 6 2n Let w = 1 + [10 s ] + [10 s ] + [10 s ] + + [10 s ] + , find the value of w.
× × × … × …
1
1
1
1
w
= 1 + [10 × ] + [10 × ] + [10 × ] + … + [10 × ] + …
n
2
4
8
2 = 1 + 5 + 2 + 1 + 0 + 0 + = 9
…
2
2
G1.1 Given that k is a real number. If x + 2kx – 3k can be divisible by x – 1, find the greatest value of k.2
2 By factor theorem, 1 + 2k – 3k = 0
2 3k – 2k – 1 = 0 (3k + 1)(k – 1) = 0
1
k = – or 1
3 Greatest value of k = 1
x y
= +
1
1
1
3
5
x y x y
= +
1
5
3 value of B . x y x y x x y y
⇒ ⇒
3
5
5
3
3
5
3
5 x x
15
16 ⇒ ⇒
3
5
8
15
2 G1.3 x 3 x x Given that = 2 + is a root of the equation – (tan + cot ) + 1 = 0.
α α C C If = sin cos , find the value of .
α × α ⇒ y
3 y y
3 Let the other root be , (2 + ) = product of roots = 1 = 2 –
3
3
tan + cot = sum of roots = 2 + + 2 – = 4
α α
sin cos
1
α α
1
C =
4 = 4 = sin α × cos α =
4
cos sin sin cos
α α α α
G1.4 Let a be an integer. If the inequality | x + 1| < a – 1.5 has no integral solution, find the greatest
value of a .| x + 1| 0, In order that the equation has no integral solution, it is sufficient that a – 1.5 < 0
Q ≥ a < 1.5
Greatest integral value of a = 1
Group Event 2 G2.1 In Figure 1, PRS is a straight line, PQ = PR = QS and Q QPR = 30 . If RQS = w , find the value of w .
∠ ° ∠ ° QPR = QSP = 30 (base s isos. )
∠ ∠ ° ∠ ∆
30 °
PQS = 120 ( s sum of ) ∠ ° ∠ ∆ P S R
圖圖 PQR = PRQ =(180 –30 ) 2 = 75 ( s sum of isos. )
∠ ∠ ° ° ÷ ° ∠ ∆ Figure 1
⇒ RQS = 120 – 75 = 45 w = 45
∠ ° ° °
7
3 G2.2 Let + + f ( x ) = px qx rx – 5, where p , q and r are real numbers.
f z f z (Reference: 1995 FI1.3) If (–6) = 3 and = (6) Find the value of .
7
3 ⇒
6 – q r 6 – 6 – 5 = 3
× ×
7
3
7
3
f (6) = p6 q + – 6 + 6 r – 5 = –(– p 6 q 6 – 6 r – 5) – 10 = –3 – 10 = –13
× × × × n 1
20 G2.3 If n 0 and s = , find the value of s .
≠
- 4 2 2 2 n n +
2 n n 1
2 1
1
20
20 s = =
=
16
2
4 2 ⋅ ⋅ ⋅ +
20 2
4 2 n n n 2 2 , find the value of + + + G2.4 Given that x and y are positive integers and x y xy = 54. If t = x y t .
x y xy
1 + = 55 +
x = 4, y = 10 or x = 10, y = 4 t = 14
8
2 − G3.1 Given that r = 2006 , find the value of r.
×
2 It is easy to show that r = 2006. x x
x + y = 2 ............. (1) x + 5y = 3 ........... (2)
7
5(1) – (2): 4x = 7 ⇒ x =
4 Given that tan + G3.3 x + tan y + 1 = cot x + cot y = 6. If z = tan( x y ), find the value of z .
6
= x y
tan tan
5 tan x + tan y = 5; tan x tan y =
6
5 tan( + x y ) = = 30
= 5
1 tan x tan y
1
− − 6 A G3.4 In Figure 1, ABCD is a rectangle, F is the midpoint of CD and D
2 BE : EC = 1 : 3. If the area of the rectangle ABCD is 12 cm and
2 F the area of BEFD is R cm , find the value of R.
2 Area of BCD = 6 cm ∆ B C E
3
1
9
2
2 Area of CEF =
6 cm = cm
∆ ⋅ ⋅ Figure 1
4
2
4
9
15
2
2 Area of BEFD = (6 – ) cm = cm
4
4
2 cm G4.1
In Figure 1, ABCD is a parallelogram, BE CD , BF AD, ⊥ ⊥ 1 cm D C E CE = 2 cm, DF = 1 cm and EBF = 60 .
∠ °
2 F If the area of the parallelogram ABCD is R cm , find the value of R. 60
° EDF = 360 – 90 – 90 – 60 = 120 ( s sum of polygon)
∠ ° ° ° ° ° ∠ BAD = BCD = 180 – 120 = 60 (int. s //-lines)
∠ ∠ ° ° ° ∠ A B Figure 1
2 BC = cm = 4 cm = AD o cos
60 BE = 2 tan 60 = 2 3 cm
° AF = (4 – 1) cm = 3 cm
3 AB = cm = 6 cm o cos
60
2
2 Area of ABCD = AB BE = 6 2 3 cm = 12 3 cm × × 2 2
1
1
a b the minimum value S.
2 ⇒ ⇒ ⇒ a b a + b – 2 ab
1 ab 1 ab ............. (1)
− ≥ ≥ ≥ ≥ ( )
2
2
2 a + b = (a + b) – 2ab = 4 – 2ab 4 – 2 = 2 ...............(2)
≥
1
1
⇒ ≥
1 ≥ 1 ......... (3) 2 2
ab a b 2 2
1
1 2 2
1
1 S a b a b = =
- 2 2
4
a b a b 2 2 2 2
=
4
- 2 2 a b
2 2
1
a b
=
1
4
( ) 2 2
a b
2 (1 + 1) + 4 = 8 (by (2) and (3))
≥ × x y
1
1
x y Reference: 2001 HI1, 2003 FG2.2, 2004 FG4.3, 2005 HI9 x log 2 = y log 7 = log 196
log 196 log 196
x = , y =
log 2 log
7
1
1 log 2 log 7 log
14
= 2 x y log 196 log
14
2 Method 2 (provided by Denny) x 1 1 y 2 = 196 , 7 = 196 1 1 1 1 x y x y +
2 7 = 14 = 196 196 = 196
× ×
1
1
1
x y
2
2
2
2
2
2
2
2
2
W = (2006 + 2005)(2006–2005) + (2004 + 2003)(2004–2003) + + (4+3)(4–3) + (2+1)(2–1) …
= 2006 + 2005 + 2004 + + 4 + 3 + 2 + 1
…
2006
1 = 1003 2007 = 2013021
× ( )
2