1. Week 4 Point C (2,4) lies on a circle x2 + y2 - 8x - 6y + 20 = 0.
Find the equation of the tangent to the circle at C
Find the first 3 terms in the expansion of
(2 – 3x)8
Given that (x+3) is a factor of f(x) = x3 + 6x2 + 5x -12 factorise f(x) and sketch the graph of y = f(x)
𝑂𝐴 =−2𝑖+3𝑗 𝑂𝐵 =4𝑖+6𝑗
Calculate | 𝐴𝐵 |
20 g of a substance is dropped into a solution. The mass not dissolved after t seconds is modelled by M = Ae-kt
After 5 seconds 10g remains. Find the value of A and k (3 s.f.)

2. 1
Point C (2,4) lies on a circle x2 + y2 - 8x - 6y + 20 = 0.
Find the equation of the tangent to the circle at C
(x - 4)2 + (y – 3)2 = 5
Centre (4,3)
Gradient of normal = - ½
Gradient of the tangent = 2
y – 4 = 2(x - 2)
y = 2x
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SOLUTION
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3. 2
Find the first 3 terms in the expansion of
(2 – 3x)8
× 27 × (-3x) + 28 × 26 × (-3x)2
256 – 3072x x2
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SOLUTION
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4. 3
Given that (x+3) is a factor of f(x) = x3 + 6x2 + 5x -12 factorise f(x) and sketch the graph of y = f(x)
f(x) = (x + 3)(x + 4)(x – 1)
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SOLUTION
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5. 4 𝐴𝐵 = 6i + 3j 𝐴𝐵 = 6 2 + 3 2 = 3 5 𝑂𝐴 =−2𝑖+3𝑗 𝑂𝐵 =4𝑖+6𝑗
𝑂𝐴 =−2𝑖+3𝑗 𝑂𝐵 =4𝑖+6𝑗
Calculate | 𝐴𝐵 |
𝐴𝐵 = 6i + 3j
𝐴𝐵 =
= 3 5
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SOLUTION
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6. 5
20 g of a substance is dropped into a solution. The mass not dissolved after t seconds is modelled by M = Ae-kt
After 5 seconds 10g remains. Find the value of A and k.
(Give k correct to 3 significant figures)
𝑡= =𝐴 𝑒 0
A = 20
t = 5 M = 10
10 = 20e-5k
e-5k = 0.5
-5k = ln (0.5) k = 0.139
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SOLUTION
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SOLUTION

7. Week 4 Point C (2,4) lies on a circle x2 + y2 - 8x - 6y + 20 = 0.
Find the equation of the tangent to the circle at C
y = 2x
Find the first 3 terms in the expansion of
(2 – 3x)8
256 – 3072x x2
Given that (x+3) is a factor of f(x) = x3 + 6x2 + 5x -12 factorise f(x) and sketch the graph of y = f(x)
(x + 3)(x + 4)(x – 1)
𝑂𝐴 =−2𝑖+3𝑗 𝑂𝐵 =4𝑖+6𝑗
Calculate | 𝐴𝐵 |
3 5
20 g of a substance is dropped into a solution. The mass not dissolved after t seconds is modelled by M = Ae-kt
After 5 seconds 10g remains. Find the value of A and k (3 s.f.)
A = 20 k = 0.139