0580/41 May/Jun 2022, Question 6: Solving Equations by Graphical Methods

Part (ai) · 1 mark

Read the value of f(2) from the graph.

1. 1Read up from x = 2 to the curve
2. 2f(2) = −3

Answer−3

Part (aii) · 3 marks

Find all x in −1.5 ≤ x ≤ 5 where f(x) = 0.

1. 1f(x) = 0 where the curve crosses the x-axis
2. 2x = −1, x ≈ 1.58, x ≈ 4.42

Answer−1, 1.55 to 1.6, 4.4 to 4.45

Part (aiii) · 1 mark

Given f(x) = k has three solutions, find the smallest integer k.

1. 1Three solutions when y = k cuts the curve three times: k above the minimum (≈ −8.2) and below the maximum (≈ 7.05)
2. 2The left end of the curve is at y = −9, so y = −8 still cuts it three times
3. 3Smallest integer k = −8

Answer−8

Part (aiv) · 2 marks

Draw a line y = mx on the grid so f(x) = mx has exactly one solution.

1. 1Draw a ruled line through the origin with a steep negative gradient, e.g. y = −5x
2. 2y = −5x passes through (−1.5, 7.5), (0, 0) and (2, −10)
3. 3It crosses the curve only once, between x = −1.5 and x = 0

AnswerRuled line through origin intersecting curve once

Part (bi) · 3 marks

Differentiate y = 3x² − 12x + 7 and evaluate dy/dx at x = 5.

1. 13x² → 6x, −12x → −12, 7 → 0
2. 2dy/dx = 6x − 12
3. 3At x = 5: 6 × 5 − 12 = 30 − 12 = 18

Answer18

Part (bii) · 2 marks

Find the coordinates where the gradient of y = 3x² − 12x + 7 is 0.

1. 16x − 12 = 0, so x = 2
2. 2y = 3 × 2² − 12 × 2 + 7 = 12 − 24 + 7 = −5
3. 3(2, −5)

Answer(2, −5)

Part (c) · 2 marks

Given dy/dx = 14x⁶ + 6x for y = 2xᵖ + qx², find p and q.

1. 1dy/dx = 2p xᵖ⁻¹ + 2qx
2. 22p xᵖ⁻¹ = 14x⁶: p − 1 = 6, so p = 7 (and 2 × 7 = 14)
3. 32qx = 6x, so q = 3
4. 4p = 7, q = 3

Answerp = 7, q = 3

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