Introduction to Advanced Mathematics (C1)
THURSDAY 7JUNE 2007
Section A (36 marks)
1. Solve the inequality 1 – 2x < 4 + 3x.
Dividing by a
negative number
reverses the direction
of inequality sign
[3]
1 – 2x < 4 + 3x
– 2x < 4 + 3x – 1
– 2x – 3x < 4 – 1
– 5x < 3
3
x>
−5
Subtract 1 from both sides
Subtract 3x from both sides
Divide both sides by – 5
2. Make t the subject of the formula s = 12 at 2 .
[3]
s = 12 at 2
2s = at 2
2s 2
=t
a
2s
t=
a
Multiply both sides by 2
Divide both sides by a
Square root both sides
3. The converse of the statement ‘P ⇒ Q’ is ‘Q ⇒ P’.
Write down the converse of the following statement.
‘n is an odd integer ⇒ 2n is an even integer.’
Show that this converse is false.
[2]
‘If 2n is an even integer, then n is an odd integer’
If 2n = 2, n = 1which is odd
If 2n = 4, n = 2 which is even
There is a counter-example and so the converse is not true.
4. You are given that f(x) = x3 + kx + c. The value of f(0) is 6, and x – 2 is a factor of f(x).
Find the values of k and c.
f(x) = x3 + kx + c
f(0) = 03 + k × 0 + c
∴6=c
f(2) = 23 + k × 2 + c
0 = 8 + 2k + c
0 = 8 + 2k + 6
– 2k = 14
k=–7
The value of f(0) is 6
Using the factor theorem if x – 2
is a factor of f(x) then f(2) = 0
[3]
5. (i) Find a, given that a3 = 64x12y3.
⎛1⎞
(ii) Find the value of ⎜ ⎟
⎝2⎠
[2]
−5
[2]
(i)
a3 = 64x12y3
= 43x12y3
= ( 4x 4 y )
3
4
∴ a = 4x y
⎛1⎞
(ii) ⎜ ⎟
⎝2⎠
−5
= ( 2−1 )
−5
= 2( −1)×( −5)
= 25
= 32
6. Find the coefficient of x3 in the expansion of (3 – 2x)5.
The term will be 5C3 (3)5−3 (−2 x)3
=
[4]
5
(3) 2 (−2 x)3
3× 5 − 3
5 × 4 × 3 × 2 ×1
× 9 × (−8 x3 )
3 × 2 ×1 × 2 ×1
= – 720x3
∴ The coefficient of x3 is – 720
=
7. Solve the equation
4x + 5
= −3 .
2x
Multiply both sides by 2x
Add 6x to both sides
Subtract 5 from both sides
Divide both sides by 10
8. (i) Simplify
(ii) Express
[3]
4x + 5
= −3
2x
4x + 5
2x ×
= −3 × 2x
2x
4 x + 5 = −6 x
10 x + 5 = 0
10 x = −5
5
x=−
10
x=–½
98 − 50 .
6 5
in the form a + b 5 , where a and b are integers.
2+ 5
(i) 98 − 50
= 49 × 2 − 25 × 2
= 49 × 2 − 25 × 2
=7 2 −5 2
=2 2
[2]
[3]
(ii)
6 5
2+ 5
Multiply top and bottom by
2 − 5 to make the bottom a
whole number
6 5 2− 5
×
=
2+ 5 2− 5
=
2×6 5 − 6 5 × 5
2× 2 − 2× 5 + 2× 5 − 5 × 5
12 5 − 30
4−5
12 5 30
=
−
−1
−1
= 30 − 12 5
=
Tidy up
9. (i) A curve has equation y = x2 – 4. Find the x-coordinates of the points on the curve where y = 21. [3]
⎛ 2⎞
(ii) The curve y = x2 – 4 is translated by ⎜ ⎟ .
⎝0⎠
Write down an equation for the translated curve. You need not simplify your answer.
[2]
(i) y = x2 – 4 and
∴ x2 – 4 = 21
x2 = 21 + 4
x2 = 25
x = ±5
(ii) y = x2 – 4
y = ( x − 2) 2 − 4
y = 21
Add 4 to both sides
Replace x with x – 2
10. The triangle shown in the following has height (x + 1) cm and base (2x – 3) cm. Its area is 9 cm2.
Show that 2x2 – x – 21 = 0.
[2]
2
By factorising, solve the equation 2x – x – 21 = 0. Hence find the height and base of the triangle.[3]
Area of a triangle = ½ × base × height
= ½ × (2x – 3) × (x + 1)
2
Its area is 9 cm
∴ ½ (2x – 3) (x + 1) = 9
½ (2x2 – 3x + 2x – 3) = 9
Multiply both sides by 2
2x2 – 3x + 2x – 3 = 18
Tidy up
2x2 – x – 3 = 18
Subtract 18 from both sides
2x2 – x – 21 = 0
2x2 – x – 21 = 0.
(2x – 7)(x + 3) = 0
Either 2x – 7= 0
or x + 3 = 0
7
The solution is x =
or x = – 3
x = – 3 is not exactable
2
7
9
∴ Height x + 1 = + 1 = cm
2
2
7
Base 2x – 3 = 2 × − 3 = 4 cm
2
Section B (36 marks)
11. A circle has centre C(1, 3) and passes through the point A(3, 7) as shown in the following figure.
(i) Show that the equation of the tangent at A is x + 2y = 17.
The line with equation y = 2x – 9 intersects this tangent at the point T.
(ii) Find the coordinates of T.
[4]
[3]
(iii) The equation of the circle is (x – 1)2 + (y – 3)2 = 20.
Show that the line with equation y = 2x – 9 is a tangent to the circle. Give the coordinates of the
point where this tangent touches the circle.
[5]
(i) The gradient of the line joining A(3, 7) and C(1, 3) is:
3−7
−4
=
=2
1− 3
−2
1
2
y – y1 = m (x – x1)
1
y – 7 = − (x – 3)
2
2 × (y – 7) = – (x – 3)
2y – 14 = – x + 3
2y = – x + 17
x + 2y = 17
The gradient of the perpendicular is −
Equation is:
Using A( 3, 7)
Multiply both sides by 2
Add 14 to both sides
Add x to both sides
(ii) x + 2y = 17
& y = 2x – 9
x + 2(2x – 9) = 17
x + 4x – 18 = 17
5x = 35
∴ x=7
y=2×7–9
=5
∴ coordinate of is T (7 , 5)
(iii) (x – 1)2 + (y – 3)2 = 20
(x – 1)2 + (2x – 9 – 3)2 = 20
(x – 1)2 + (2x – 12)2 = 20
x2 – 2x + 1 + 4x2 – 48x + 144 = 20
5x2 – 50x + 125 = 0
x2 – 10x + 25 = 0
(x – 5)( x – 5) = 0
x=5&5
∴ y = 2x – 9 is a tangent to the circle
y = 2 × 5 – 9 = 1 Coordinate (5 , 1)
12. (i) Write 4x2 – 24x + 27 in the form a(x – b)2 + c.
(ii) State the coordinates of the minimum point on the curve y = 4x2 – 24x + 27.
(iii) Solve the equation 4x2 – 24x + 27 = 0.
(iv) Sketch the graph of the curve y = 4x2 – 24x + 27.
(i) 4x2 – 24x + 27
(iii) 4x2 – 24x + 27 = 0
2
= 4(x – 6x) + 27
or 4(x – 3)2 – 9 = 0
2
2
= 4{(x – 3) – 3 } + 27
or 4(x – 3)2 = 9
= 4{(x – 3)2 – 9} + 27
or (x – 3)2 = 9/4
2
= 4(x – 3) – 36 + 27
9
3
or x – 3 = ±
=±
= 4(x – 3)2 – 9
4
2
(ii) minimum point is (3, –9)
∴ x = 3 + 3/2 or 3 – 3/2
= 4.5 or 1.5
(iv)
[4]
[2]
[3]
[3]
( 3, –9)
13. A cubic polynomial is given by f(x) = 2x3 – x2 – 11x – 12.
(i) Show that (x – 3)(2x2 + 5x + 4) = 2x3 – x2 –11x – 12.
Hence show that f(x) = 0 has exactly one real root.
[4]
Show that x = 2 is a root of the equation f(x) = – 22 and find the other roots of this equation. [5]
(iii) Using the results from the previous parts, sketch the graph of y = f(x).
[3]
(i) (x – 3)(2x2 + 5x + 4)
2x2 + 5x + 4 = 0
= x×(2x2 + 5x + 4) – 3 × (2x2 + 5x + 4)
can’t be solved, because
= 2x3 + 5x2 + 4x – 6x2 – 15x – 12
52 – 4 × 2 × 4 < 0
3
2
2
= 2x + 5x – 6x + 4x – 15x – 12
= 2x3 – x2 – 11x – 12
(ii) f(x) = – 22
2x3 – x2 – 11x – 12 = – 22
2x3 – x2 – 11x + 10 = 0
2 x 2 + 3x − 5
x − 2 2 x 3 − x 2 − 11x + 10
2 x3 − 4 x 2
3 x 2 − 11x + 10
3x 2 − 6 x
− 5 x + 10
−5 x + 10
2x2 + 3x – 5
= (2x + 5)(x – 1)
2x3 – x2 – 11x + 10 = (x – 2)(2x + 5)(x – 1)
∴ other roots are x = 1 & x = – 5/2
(iii)