WMA11/01 Oct 2024, Question 6: Lines Meeting Curves: the Discriminant

Part (a) · 3 marks

Sketch C for y = 1/(2 − x), x ≠ 2, labelling the vertical asymptote equation and the y-axis intersection coordinates.

1. 1y = 1/(2 − x) = −1/(x − 2): the graph of y = 1/x translated 2 right and reflected in the x-axis
2. 2Vertical asymptote where 2 − x = 0: x = 2; horizontal asymptote y = 0
3. 3At x = 0, y = ½, so it crosses the y-axis at (0, ½)
4. 4For x < 2, y > 0 (top-left branch); for x > 2, y < 0 (bottom-right branch)

AnswerTwo-branch reciprocal curve: vertical asymptote x = 2 and horizontal asymptote y = 0; for x < 2 the branch is above the x-axis, crossing the y-axis at (0, ½) and rising to +∞ as x → 2 from the left; for x > 2 the branch is below the x-axis, coming up from −∞ just right of x = 2 and approaching y = 0 from below

Part (bi) · 4 marks

Given line l with y = kx − 4 cuts C at least once, show that k² − 5k + 4 ≥ 0.

1. 1Set kx − 4 = 1/(2 − x), so (kx − 4)(2 − x) = 1
2. 22kx − kx² − 8 + 4x = 1, so kx² + (−4 − 2k)x + 9 = 0
3. 3At least one intersection needs b² − 4ac ≥ 0: (−4 − 2k)² − 4 × k × 9 ≥ 0
4. 44k² + 16k + 16 − 36k ≥ 0, so 4k² − 20k + 16 ≥ 0
5. 5Divide by 4: k² − 5k + 4 ≥ 0

AnswerShown: (kx − 4)(2 − x) = 1 ⇒ kx² − (4 + 2k)x + 9 = 0; b² − 4ac ≥ 0 ⇒ (4 + 2k)² − 36k ≥ 0 ⇒ 4k² − 20k + 16 ≥ 0 ⇒ k² − 5k + 4 ≥ 0

Part (bii) · 2 marks

Find the range of possible values of k.

1. 1k² − 5k + 4 = (k − 1)(k − 4)
2. 2Critical values k = 1 and k = 4
3. 3(k − 1)(k − 4) ≥ 0 outside the roots: k ≤ 1 or k ≥ 4

Answerk ≤ 1 or k ≥ 4

More past-paper questions on lines meeting curves: the discriminant

• WMA11/01 Jun 2024 Q4 · 6 marks
• WMA11/01 Jan 2023 Q7 · 10 marks
• WMA11/01 Oct 2023 Q6 · 6 marks
• WMA11/01 Jan 2022 Q10 · 9 marks
• WMA11/01 Oct 2021 Q8 · 10 marks
• WMA11/01 Jan 2019 Q9 · 7 marks