Pre C13: ROUND 1A
Solve for x from the given equation
1 √(x – 4)3 = 125
ANSWER: x = 29
[(x – 4)3 = (53)2 = (52)3, x – 4 = 52 = 25, x = 29]
2 ∛(x + 7)2 = 9
ANSWER: x = 20, - 34
[(x + 7)2 = (32)3 = (33)2 = 272, x + 7 = ±27, x = -7 ±27, x = 20, -
34]
3 √(2x – 5)3 = 27
ANSWER: x = 7
[(2x – 5)3 = (33)2 = (32)3 , 2x – 5 = 32 = 9, 2x = 9 + 5 = 14, x = 7]
Pre C13 : ROUND 1B
Determine the values of the constants a, b and c given that the
function
y = ax3 + bx2 + cx has gradient
1. 6x2 – 4x + 2 ANSWER: a = 2, b = -2, c = 2
[dy/dx = 3ax + 2bx + c = 6x2 - 4x + 2, 3a = 6, a = 2, 2b = -4, b =
2
-2, c = 2]
2. 9x2 – 12x – 5 ANSWER: a = 3, b = -6, c = -5
[dy/dx = 3ax + 2bx + c = 9x2 - 6x - 4, 3a = 9, a = 3, 2b = -12, b =
2
-6, c = -5]
3. -3x2 + 10x + 10. ANSWER: a = -1, b = 5, c = 4
[dy/dx = 3ax + bx + c = -3x2 + 10x + 4, 3a = -3, a = -1, 2b = 10,
2
b =5, c = 4]
Pre C 13 ROUND 2: SPEED RACE
1 Solve the equation sin2x = ¾ for x in the interval 0 < x < π/2.
(Give answer in radians)
ANSWER: x = π/3
[sinx = ±√3/2, sinx = √3/2, x = π/3, sinx = - √3/2, no solution ]
2 Find dy/dx given that x2 + xy – y2 = 8
ANSWER: dy/dx = (2x + y)/(2y – x) [2x + y + x(dy/dx) –
2y(dy/dx) = 0, (dy/dx)(2y - x) = 2x + y, (dy/dx) = (2x + y)/(2y –
x)]
3 The roots of the equation x2 – 7x + 9 = 0 are α and β. Evaluate
α2 + β2.
ANSWER: 31
[α2 + β2 = (α + β)2 - 2αβ = 72 – 2(9) = 49 – 18 = 31]
Pre C 13 ROUND 5: RIDDLES
1 I am a line segment.
2 You may find me inside a polygon.
3 You may also find me inside a polyhedron.
4 You will not find me in a triangle or a pyramid.
5 In a polygon I connect non-adjacent vertices.
6 In a polyhedron I connect vertices not in the same face.
Who am I
ANSWER: DIAGONAL
Pre C13 ROUND 4A: TRUE OR FALSE
1 (x + 3) is a factor of x2 – 2x – 15
ANSWER: TRUE
2 (x – 3) is a factor of 2x2 – 3x – 9
ANSWER: TRUE
3 (x + 2) is a factor of 2x2 + 3x + 1
ANSWER: FALSE
Pre C 13 ROUND 4B: TRUE OR FALSE
In the following, a, b and c are real numbers.
1 If a > b, then ac > bc
ANSWER: FALSE
2 If a < b, then a – c < b – c
ANSWER: TRUE
3 If ac > bc, then a > b
ANSWER: FALSE
ROUND 3: POTD 4 prelim
ABCD is a cyclic quadrilateral such that AB = 6 cm, BC = 4 cm, CD = 12 cm and
AD = 8 cm. Find cosine of ( angle D). (Hint: Find the length of AC)
B
A 6cm 4cm
C
8cm 12cm
D
SOLUTION:
Apply cosine rule to
AC2 = 82 + 122 – 2(8)12cosD = 64 + 144 – 192cosD
= 208 – 192cosD (2 MARKS)
Apply cosine rule to
AC2 = 62 + 42 – 2(6)4cosB = 36 + 16 – 48cosB
= 52 – 48cosB (2 MARKS)
ABCD is a cyclic quadrilateral, hence opposite angles are supplementary
cosB = cos(180 – D) = -cosD (2 MARKS)
AC2 = 52 + 48cosD
208 – 192cosD = 52 + 48cosD (2 MARKS)
208 – 52 = (192 + 48)cosD, 156 = 240cosD
cosD = 156/240 = 39/60 = 13/20 (2 MARKS)
Pre C 14 ROUND 1A
Find the identity e of the given binary operation ∗ on Z set of
integers.
1a∗b=a+b+3
ANSWER: e = -3
[a ∗ e = a + e + 3 = a, e + 3 = 0, e = -3]
2a∗b=a-b+3
ANSWER: No identity
[operation is not commutative hence no identity]
3 a ∗ b = a + 2b - 3
ANSWER: No identity
[operation is not commutative]
Pre C 14 ROUND 1B
Find the next 3 terms of a sequence defined by the recurrence
relation
1 Un = 2Un – 1 + 3, U1 = - 2 for n > 1.
ANSWER: U2 = -1, U3 = 1, U4 = 5 [U2 = 2U1 + 3 = 2(-2) +
3 = -1,
U3 = 2(U2) + 3 = 2(-1) + 3 = 1, U4 = 2U3 + 3
= 2(1) + 3 = 5]
2 Un = 3 – 2Un – 1, U1 = -2 for n > 1
ANSWER: U2 = 7, U3 = -11, U4 = 25 [U2 =3 – 2U1 = 3 – 2(-2)
= 7,
U3 = 3 - 2U2 = 3 – 2(7) = -11, U4 = 3 – 2U3 = 3 – 2(-11) = 25]
3 Un = 3Un – 1 + 7, U1 = -5 for n > 1
ANSWER: U2 = -8, U3 = -17, U4 = -44 [U2 = 3U1 + 7 = 3(-5) + 7
= -8,
U3 = 3U2 + 7 = 3(-8) + 7 = -17, U4 = 3U3 + 7 = 3(-17) + 7 = -51 +
7 = -44]
Pre C14 ROUND 2: SPEED RACE
1 Find the cartesian equation of a curve defined by x = 2cost and
y = 3sint.
ANSWER: x2/4 + y2/9 = 1, or 9x2 + 4y2 = 36
[cost = x/2, sint = y/3, cos2t + sin2t = 1, x2/4 + y2/9 = 1, 9x2 + 4y2
= 36]
2 Find the coordinates of the point of inflexion of the curve y = x3
+ 3x2 + 5
ANSWER: (-1, 7)
[dy/dx = 3x2 + 6x, d2y/dx2 = 6x + 6 = 0, x = -1, y = -1+ 3 + 5 = 7,
(-1. 7)]
3 Find the inverse of the matrix A =
ANSWER: A-1 = [det A = 4 – 3 = 1, A-1 = ]
Pre C 14 ROUND 5: RIDDLE
1 I am a solid 3 - dimensional figure.
2 I consist of faces, edges and points.
3 I am bounded by polygonal faces.
4 My faces meet in line segments or points.
5 Examples of me are pyramids and cuboids.
6 Others are tetrahedron and octahedron.
Who am I?
ANSWER: POLYHEDRON
Pre C 14 ROUND 4A: TRUE OR FALSE
In the following, x, y and z are variables and k is a constant
1 The relation xy = k is an inverse variation.
ANSWER: TRUE
2 The relation y/xz = k is a joint variation.
ANSWER: TRUE
3 The relation y = k/x is a direct variation.
ANSWER: FALSE
4 The relation 1/y = k/x is an inverse relation.
ANSWER: FALSE
Pre C 14: ROUND 4B: TRUE OR FALSE
1 sin(π/2 + A) = cosA
ANSWER: TRUE
2 sin(π/2 – A) = cosA
ANSWER: TRUE
3 sin(π + A) = sinA
ANSWER: FALSE
4 sin(π – A) = sinA
ANSWER: TRUE
