IB Mathematics SL 5.6 Local maximum and minimum points AI HL Paper 1- Exam Style Questions- New Syllabus

Question

The following diagram shows the graph of a function \(f\). There is a local minimum point at \(A\), where \(x > 0\).

Graph of function f with local minimum at A

The derivative of \(f\) is given by \(f'(x) = 3x^2 – 8x – 3\).

a. Find the \(x\)-coordinate of \(A\). [5]

b. The \(y\)-intercept of the graph is at \((0,6)\). Find an expression for \(f(x)\).

The graph of a function \(g\) is obtained by reflecting the graph of \(f\) in the \(y\)-axis, followed by a translation of \(\left(\begin{array}{c}m\\ n\end{array}\right)\). [6]

c. Find the \(x\)-coordinate of the local minimum point on the graph of \(g\). [3]

▶️ Answer/Explanation

Solution a

Set \(f'(x) = 0\): \(3x^2 – 8x – 3 = 0\) (M1)

Solve: \((3x + 1)(x – 3) = 0\) (M1A1)

Solutions: \(x = -\frac{1}{3}\) or \(x = 3\)

Since \(x > 0\): \(\boxed{x = 3}\) (A2 N3)

Solution b

Integrate \(f'(x)\): \(\int (3x^2 – 8x – 3)dx = x^3 – 4x^2 – 3x + C\) (M1A1A1)

Using \((0,6)\): \(6 = 0 – 0 – 0 + C \Rightarrow C = 6\) (A1)

\(\boxed{f(x) = x^3 – 4x^2 – 3x + 6}\) (A1 N6)

Solution c

Reflect \(f\): \(g(x) = f(-x + m) + n\) (A1)

Minimum was at \(x=3\), now at \(x=-3\) after reflection (M1)

After translation: \(\boxed{x = -3 + m}\) (A1 N3)

—Markscheme—

a. [5 marks]

Setting derivative to zero (M1)

Solving quadratic (M1A1)

Correct solution (A2)

b. [6 marks]

Correct integration (3 marks) (M1A1A1)

Finding constant (A1)

Final answer (A1 N6)

c. [3 marks]

Reflection (A1)

Understanding transformation (M1)

Final coordinate (A1 N3)

Total [14 marks]

Question

A function \( f \) is of the form \( f(t) = p e^{q \cos(rt)} \), \( p, q, r \in \mathbb{R}^+ \). Part of the graph of \( f \) is shown.

Graph of f(t)

The points A and B have coordinates A(0, 6.5) and B(5.2, 0.2), and lie on \( f \). The point A is a local maximum and the point B is a local minimum.

Find the value of \( p \), of \( q \), and of \( r \).

▶️ Answer/Explanation

Markscheme

Substitute coordinates of A: \( f(0) = p e^{q \cos(0)} = p e^q = 6.5 \) (M1)
Substitute coordinates of B: \( f(5.2) = p e^{q \cos(5.2r)} = 0.2 \) (M1)
EITHER: Differentiate: \( f'(t) = -p q r \sin(rt) e^{q \cos(rt)} \)
At minimum B, set \( f'(5.2) = 0 \): \( -p q r \sin(5.2r) e^{q \cos(5.2r)} = 0 \), so \( \sin(5.2r) = 0 \)
Solve: \( 5.2r = \pi \), \( r = \frac{\pi}{5.2} \approx 0.604152 \) (A1)
OR: At minimum B, \( \cos(5.2r) = -1 \), so \( 5.2r = \pi \), \( r = \frac{\pi}{5.2} \approx 0.604152 \) (A1)
OR: Period = \( 2 \times 5.2 = 10.4 \), so \( r = \frac{2\pi}{10.4} = \frac{\pi}{5.2} \approx 0.604152 \) (A1)
At B: \( \cos(5.2r) = \cos \pi = -1 \), so \( f(5.2) = p e^{-q} = 0.2 \)
Eliminate \( p \): \( \frac{p e^q}{p e^{-q}} = \frac{6.5}{0.2} \), \( e^{2q} = 32.5 \) (M1)
Solve: \( 2q = \ln 32.5 \approx 3.481240 \), \( q \approx 1.740620 \approx 1.74 \) (A1)
Solve for \( p \): \( p = \frac{6.5}{e^q} \), \( e^q \approx e^{1.740620} \approx 5.697 \), \( p \approx \frac{6.5}{5.697} \approx 1.14017 \approx 1.10 \) (A1)
Result: \( p \approx 1.10 \), \( q \approx 1.74 \), \( r \approx 0.604 \) [6]

IB Mathematics SL 5.6 Local maximum and minimum points AI HL Paper 1- Exam Style Questions- New Syllabus

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