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1. Find ππππ
ππππ when y = x5 + x2 (1)
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2. Find ππππππππ
when y = 6x3 + 5x-2 (1) __________________________________________________________________________________________
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3. Find ππππ
ππππ when y = 3x-1 β 5π₯π₯β
32 (1)
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4. Find ππππ
ππππ when y = (x + 1)(x + 6) (2)
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5. Find ππππ
ππππ when y = βπ₯π₯(π₯π₯ β 4) (2)
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6. Find ππππ
ππππ when y = 5+βππ
ππ2 (2)
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A-Level Starter Activity
Topic: Differentiation Chapter Reference: Pure 1, Chapter 12
8 minutes
Solutions
1. y = x5 + x2 ππππππππ
= 5x4 + 2x M1
2.
y = 6x3 + 5x-2 ππππππππ
= 12x2 β 10x-3 M1
3.
y = 3x-1 β 5π₯π₯β32
ππππππππ
= -3x-2 + 152π₯π₯β
52
M1
4.
y = (x + 1)(x + 6) y = x2 + 7x + 6 M1 ππππππππ
= 2x + 7 M1 5.
y = βπ₯π₯(π₯π₯ β 4) y = x1.5 β 4x0.5 M1 ππππππππ
= 1.5x0.5 β 4x-1.5 M1 6.
y = 5+βππππ2
y = 5x-2 + x-1.5 M1
ππππππππ
= β10x-3 -1.5x-2.5 M1
1. Find the value of the gradient of the curve y = 3x2 + x β 5 at the point x = 1 (2) __________________________________________________________________________________________
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2. Find the value of the gradient of the curve y = (x + 1)2 at the point (4, 25) (2) __________________________________________________________________________________________
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3. Find the value of the gradient of the curve y = 8π₯π₯+π₯π₯
3
4βπ₯π₯ (2)
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4. Find the equation of the tangent to the curve y = 3 β x2 at the point (-3, -6) (4) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Differentiation Chapter Reference: Pure 1, Chapter 10
10 minutes
Solutions
1. y = 3x2 + x β 5 πππππππ₯π₯
= 6π₯π₯ + 1 M1
At x = 1, πππππππ₯π₯
= 6(1) + 1 = 7 M1
2.
y = (x + 1)2 = (x + 1)(x + 1) y = x2 + 2x + 1 M1 πππππππ₯π₯
= 2π₯π₯ + 2 M1 When x = 4, πππππππ₯π₯
= 2(4) + 2 = 10 M1
3.
y = 8π₯π₯+π₯π₯3
4βπ₯π₯ = 8π₯π₯
4π₯π₯12
+ π₯π₯3
4π₯π₯12 = 2x0.5 + 1
4π₯π₯2.5 M1
πππππππ₯π₯
= x-0.5 + 58x1.5 M1
4.
y = 3 β x2
πππππππ₯π₯
= β2π₯π₯ M1
When x = -3, πππππππ₯π₯
= β2(β3) = 6 M1
Equation of line crossing (-3, -6) -6 = 6(-3) + c -6 = -18 + c c = 12
M1
y = 6x + 12 M1
1. Find ππππ
ππππ of the curve, y = x2 + 4x + 6 (1)
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2. Find ππππ
ππππ of the curve, y = 2x2 - 5x - 2 (1)
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3. Find ππππ
ππππ of the curve, y = (x + 5)(4x - 2) (2)
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4. Find ππππ
ππππ of the curve, y = (-9x + 3)(2x - 4) (2)
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5. Find the values of x for which ππππππππ
= 0 when y = (x - 3)(x + 5) (4)
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A-Level Starter Activity
Topic: Differentiating Quadratics Chapter Reference: Pure 1, Chapter 12
6 minutes
Solutions
1. y = x2 + 4x + 6 ππππππππ
= 2π₯π₯ + 4 M1
2.
y = 2x2 - 5x β 2 ππππππππ
= 4π₯π₯ β 5 M1
3.
y = (x + 5)(4x - 2) y = x2 + 18x - 10 M1 ππππππππ
= 2π₯π₯ + 18 M1 4.
y = (-9x + 3)(2x - 4) y = -18x2 + 42x - 12 M1 ππππππππ
= β36π₯π₯ + 42 M1 5.
π¦π¦ = (π₯π₯ β 3)(π₯π₯ + 5) y = x2 + 2x β 15 M1 ππππππππ
= 2π₯π₯ + 2 M1 2x + 2 = 0 2x = -2 M1
x = -1 M1
1. A curve has the equation y = 2 + 4
π₯π₯.
a. Find an equation of the normal to the curve at the point M (4, 3). The normal to the curve at M intersects the curve again at the point N. b. Find the coordinates of the point N. (3) __________________________________________________________________________________________
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2. A curve has the equation y = (x + 2)(x β 5) a. Find an equation of the normal to the curve at the point P(2, -12). (3) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Differentiation and Normals Chapter Reference: Pure 1, Chapter 12
7 minutes
Solutions
1a. y = 2 + 4
π₯π₯
πππππππ₯π₯
= β4π₯π₯-2 M1
Gradient at M = -14
Gradient of normal = 4 M1
y β 3 = 4(x β 4) y = 4x β 13 M1
1b.
4x β 13 = 2 + 4π₯π₯
4x2 β 15x β 4 = 0 M1
(4x + 1)(x β 4) = 0 M1 x = 4 (at M) x = β1
4
Therefore, N(-14
,β14) M1
2.
y = (x + 2)(x β 5) y = x2 -3x - 10 M1 πππππππ₯π₯
= 2π₯π₯ β 3 M1 At x = 2, πππππππ₯π₯
= 2(2) β 3 = 1 Therefore, gradient of normal = -1
M1
y - -12 = -1(x β 2) y + 12 = -x + 2 y = -x β 10
M1
1. The point A lies on the curve y = 12
π₯π₯2 and the x-coordinate of A is 2.
a. Find an equation of the tangent to the curve at A. Give your answer in the form ax + by + c = 0, where a, b and c, are integers. (3) b. Verify that the points where the tangent at A intersects the curve again has the coordinates (-1, 12). (3) __________________________________________________________________________________________
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2. The curve C has the equation y = x β 3π₯π₯β
12 + 3 and passes through the point P (4, 1). Show that the tangent to C
at P passes through the origin. (3) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Differentiation and Tangents Chapter Reference: Pure 1, Chapter 12
7 minutes
Solutions
1a. πππππππ₯π₯
= β24π₯π₯-3 M1 At A, y = 3, gradient = -3 M1 y β 3 = -3(x β 2) 3x + y β 9 = 0 M1
1b.
Tangent: x = -1, -3 + y β 0 = 0 y = 12
M1
For the curve, x = -1, y = 12
1 = 12 M1
Therefore, tangent intersects curve at (-1, 12) M1 2.
πππππππ₯π₯
= 1 β 32π₯π₯β
12
Gradient at P = 14
M1
y β 1 = 14
(π₯π₯ β 4) M1 y = ΒΌ x which passes through (0,0) M1
1. Find the set of values of x for which f(x) is increasing when, f(x) = 2x2
+ 2x + 1 (3) __________________________________________________________________________________________
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2. Find the set of values of x for which f(x) is decreasing when f(x) = x(x β 6)2 (4) __________________________________________________________________________________________
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3. f(x) = x3 + kx2 + 3 Given that (x + 1) is a factor of f(x) a. Find the values of the constant k, (2) b. Find the set of values of x for which f(x) is increasing. (3) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Increasing and Decreasing Functions Chapter Reference: Pure 1, Chapter 12
8 minutes
Solutions
1. fβ(x) = 4x + 2 M1 4x + 2 β₯ 0 M1 x β₯ -1
2 M1
2.
f(x) = x(x β 6)2
f(x) = x(x2 β 12x +36) f(x) = x3 β 12x2 + 36x
M1
fβ(x) = 3x2 β 24x + 36 M1 3x2 β 24x + 36 β€ 0 x2 β 8x + 12 β€ 0 M1
(x β 6)(x β 2) β€ 0 2 β€ x β€ 6 M1
3a.
As (x + 1) is a factor, f(-1) = 0 M1 Therefore, -1 + k + 3 = 0 k = -2
M1
3b.
fβ(x) = 3x2 β 4x M1 3x2 β 4x β₯ 0 x(3x β 4) β₯ 0 M1
x β€ 0 and x β₯ 4
3 M1
1. The graph shows the height, h cm of letters on a website advert t seconds after the advert appears on the screen.
For t in the interval 0 β€ t β€ 2, h is given by the equation,
H = 2t4 β 8t3 + 8t2 + 1
For larger values of t, the variation of h over this interval is repeated every 2 seconds. a. Find ππβ
ππππ for t in the interval 0 β€ t β€ 2 (1)
b. Find the rate at which the maximum height of the letters is increasing when t = 0.25. (1) c. Find the maximum height of the letters. (3) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Modelling with Differentiation Chapter Reference: Pure 1, Chapter 12
5 minutes
Solutions
1a. ππβππππ
= 8π‘π‘3 β 24t2 + 16t M1 1b.
When t = 0.25, ππβππππ
= 2.625 cm per second. M1
1c.
Stationary point: 8t3 β 24t2 + 16t = 0 M1
8t(t β 1)(t β 2) = 0 t = 0 t = 1 t = 2
M1
From graph, maximum occurs when t = 1, Therefore, maximum height is 3 cm. M1
1. Find ππ
2π¦π¦πππ₯π₯2
when y = x3 β 6x2 β 36x + 15 (2) __________________________________________________________________________________________
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2. Find the stationary points of y = π₯π₯
4+162π₯π₯2
and state if it is a maximum or minimum. (6) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Second Order Derivatives Chapter Reference: Pure 1, Chapter 12
8 minutes
Solutions
1a. πππ¦π¦πππ₯π₯
= 3π₯π₯2 β 12x β 36 M1 ππ2π¦π¦πππ₯π₯2
= 6x β 12 M1 1b.
y = 12π₯π₯2 + 8π₯π₯-2 M1
πππ¦π¦πππ₯π₯
= π₯π₯ β 16π₯π₯-3
At stationary point, πππ¦π¦πππ₯π₯
= 0 M1
π₯π₯ β 16π₯π₯-3 = 0 x4 = 16 x = Β± 2
M1
ππ2π¦π¦πππ₯π₯2
= 1 + 48x-4 M1
At (-2, 4), ππ2π¦π¦
πππ₯π₯2 = 1 + 48(-2)-4 = 4
Therefore, minimum point M1
At (2, 4), ππ2π¦π¦
πππ₯π₯2 = 1 + 48(2)-4 = 4
Therefore, minimum point M1
1. Complete the table:
Curve π π π π π π π π
y = x5
f(x) = 2x3
f(x) = x-3
y = 5x-8
y = 4x3 + 3x-4
f(x) = 2x + 13π₯π₯6
f(x) = 7 + π₯π₯β45
y = 3x-1 - 5π₯π₯β32
f(x) = 2 β 7x-1 + π₯π₯β83
y = 14π₯π₯β 1
π₯π₯2
(10)
A-Level Starter Activity
Topic: Simple Differentiation Chapter Reference: Pure 1, Chapter 12
7 minutes
Solutions
Curve π π π π π π π π
y = x5 πππππππ₯π₯
= 5π₯π₯4 M1
f(x) = 2x3 fβ(x) = 6x2 M1
f(x) = x-3 fβ(x) -3x-4 M1
y = 5x-8 πππππππ₯π₯
= -40x-9 M1
y = 4x3 + 3x-4 πππππππ₯π₯
= 12x2 β 12x-5 M1
f(x) = 2x + 13π₯π₯6 fβ(x) = 2 + 2x5 M1
f(x) = 7 + π₯π₯β45 fβ(x) = -4
5π₯π₯β1.8 M1
y = 3x-1 - 5π₯π₯β32
πππππππ₯π₯
= β3π₯π₯-2 + 152π₯π₯-2.5 M1
f(x) = 2 β 7x-1 + π₯π₯β83 fβ(x) = 7x-2 - 8
3π₯π₯β
113 M1
y = 14π₯π₯β 1
π₯π₯2 ππππ
πππ₯π₯= -4x-2 + 2x-3 M1
1. The curve y = x3 β 3x + 1 is stationary at the points P and Q. a. Find the coordinates of the points P and Q. (4) b. Find the length of PQ in the form kβ5 (2) __________________________________________________________________________________________
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2a. Find the coordinates of the stationary points on the curve, y = 2 + 9x + 3x2 β x3 (4) b. Determine whether each stationary point is a maximum or minimum point. (3) __________________________________________________________________________________________
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A-Level Starter Activity
Topic: Stationary Points Chapter Reference: Pure 1, Chapter 12
8 minutes
Solutions
1a. ππππππππ
= 3π₯π₯2 β 3
At stationary point, ππππππππ
= 0 M1
3π₯π₯2 β 3 = 0 x2 = 1 x = Β± 1
M1
When x = 1, y = -1 M1
When x = -1, y = 3 M1
1b.
PQ2 = 22 + 42 = 20 M1 PQ = β20 = 2β5 M1
2a.
ππππππππ
= 9 + 6π₯π₯ β 3x2
At stationary point, ππππππππ
= 0 M1
9 + 6π₯π₯ β 3x2 = 0 -3(x+1)(x β 3) = 0 x = -1, 3
M1
When x = -1, y = -3 M1
When x = 3 y = 29 M1
(-1, -3) and (3, 29) 2b.
ππ2ππππππ2
= 6 β 6π₯π₯ M1 At (-1, -3), ππ2ππππππ2
= 6 β 6(β1) = 12 Therefore, minimum point
M1
At (3, 29), ππ2ππππππ2
= 6 β 6(3) = -12 Therefore, maximum point
M1
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