IBDP Maths SL 2.4 Key features of graphs AA HL Paper 1- Exam Style Questions- New Syllabus
Question
Consider the curve \( y = x^2 – x – 1 \) and the line \( y = mx – 3 \), where \( m \in \mathbb{R} \).
(a) Show that the curve and the line meet at points satisfying \( x^2 – (m + 1)x + 2 = 0 \).
(b) Hence, determine the values of \( m \) for which the line is tangent to the curve.
Most-appropriate topic codes (IB Mathematics: Analysis and Approaches HL 2025):
• SL 2.4: Intersection of curves and lines — part (a)
• SL 2.7: Discriminant and nature of roots — part (b)
• SL 2.7: Discriminant and nature of roots — part (b)
▶️ Answer/Explanation
(a)
Set the equations equal to find intersection:
\( x^2 – x – 1 = mx – 3 \)
Rearrange:
\( x^2 – x – 1 – mx + 3 = 0 \)
\( x^2 – (m + 1)x + 2 = 0 \)
(b)
The line is tangent to the curve when the quadratic \( x^2 – (m + 1)x + 2 = 0 \) has exactly one solution (repeated root). This occurs when the discriminant is zero.
Discriminant \( \Delta = [-(m+1)]^2 – 4(1)(2) = (m+1)^2 – 8 \).
Set \( \Delta = 0 \):
\( (m+1)^2 – 8 = 0 \)
\( (m+1)^2 = 8 \)
\( m+1 = \pm 2\sqrt{2} \)
\( m = -1 \pm 2\sqrt{2} \)
\(\boxed{m = -1 – 2\sqrt{2},\quad -1 + 2\sqrt{2}}\)