IB Mathematics SL 2.4 Key features of graphs AA SL Paper 2- Exam Style Questions- New Syllabus
Question
(ii) Determine the period of \( f \).
(b) A point \( P \) is defined by the coordinates \( (1.63, 2.16) \). Determine whether \( P \) is located above, below, or on the graph of \( f \), providing a mathematical justification for your conclusion.
(c) A line \( L_1 \) has the equation \( x – 6y + 11 = 0 \). State the gradient of \( L_1 \).
(d) The line \( L_1 \) is the normal to the graph of \( f \) at the point \( A(1, 2) \). A second line, \( L_2 \), is the tangent to the graph of \( f \) at point \( A \).
(i) Find the gradient of \( L_2 \).
(ii) Hence, find the equation of \( L_2 \).
(f) Calculate the area of the shaded region bounded by the graph of \( f \) and the line \( L_1 \) between points \( A \) and \( B \).
Most-appropriate topic codes (IB Mathematics AA SL):
• SL 5.4: Tangents and normals at a given point, and their equations – parts (c), (d)
• SL 5.11: Areas of a region enclosed by a curve and the x-axis, or between curves –part (f)
• SL 2.4: Finding the point of intersection of two curves or lines using technology – parts (b), (e)
▶️ Answer/Explanation
(a)(i)
Amplitude = \( \frac{2}{\pi} \approx 0.637 \).
(a)(ii)
Period = \( \frac{2\pi}{b} = \frac{2\pi}{3\pi} = \frac{2}{3} \approx 0.667 \).
(b)
Calculate \( f(1.63) \):
\( f(1.63) = \frac{2}{\pi} \sin(3\pi \times 1.63) + 2 \approx 2.2156 \).
Since \( 2.2156 > 2.16 \), the graph is above the point, so point \( P \) lies below the graph.
(c)
Rearranging \( x – 6y + 11 = 0 \):
\( 6y = x + 11 \Rightarrow y = \frac{1}{6}x + \frac{11}{6} \).
Gradient of \( L_1 \) is \( \frac{1}{6} \approx 0.167 \).
(d)(i)
Since the normal \( L_1 \) has gradient \( \frac{1}{6} \), the tangent \( L_2 \) has gradient \( m = -1 \div \frac{1}{6} = -6 \).
(d)(ii)
Using \( y – y_1 = m(x – x_1) \) at \( A(1, 2) \):
\( y – 2 = -6(x – 1) \Rightarrow y = -6x + 8 \).
(e)
Solve \( f(x) = \text{Line } L_1 \) using technology:
\( \frac{2}{\pi} \sin(3\pi x) + 2 = \frac{1}{6}x + \frac{11}{6} \).
For \( x > 1.5 \), \( B \approx (1.65, 2.11) \).
(f)
Area = \( \int_{1}^{1.6486} (f(x) – L_1(x)) \, dx \).
Using GDC integration: Area ≈ 0.253.