IB DP Maths- MAA HL Prediction Paper 1 – 2026 Edition
IB DP Math AA HL Prediction Paper 1 – April/May 2026 Exam
IB DP Math AA HL Prediction Paper 1: Prepare for the IB exams with subject-specific Prediction questions, model answers. All topics covered.
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Question
An interior designer designed a built-in cupboard for his client. The built-in cupboard of length \( 9a \) metres, \( a > 0 \), has three equal sections and each section has a semi-elliptical hole in the centre. The designer wants to fit a hollow rectangular compartment, for storage, into each of the elliptical hole. Each rectangular compartment with negligible thickness, has a length of \( l \) metres, where \( l < 2a \), a height of \( y \) metres, and a fixed depth. The cross-section for part of the built-in cupboard is modelled by the equation
\[ \mathrm{f}(x) = \begin{cases} \sqrt{1 – \frac{x^2}{a^2}} & \text{for } -a \leq x \leq a, \\ 0 & \text{for } a \leq x \leq 2a, \end{cases} \]
and \( \mathrm{f}(x + 3a) = \mathrm{f}(x) \) for \( -\frac{9a}{2} \leq x \leq \frac{9a}{2} \), where \( a \) is a real constant.
▶️Answer/Explanation
Part (a):
\[ \mathrm{f}\left(3a + \frac{1}{2}l\right) = \sqrt{1 – \frac{l^2}{4a^2}} \]
Working: Since \( \mathrm{f}(x + 3a) = \mathrm{f}(x) \), evaluate \( \mathrm{f}\left(3a + \frac{1}{2}l\right) = \mathrm{f}\left(\frac{1}{2}l\right) \). For \( -\frac{1}{2}l \leq \frac{1}{2}l \leq a \) (since \( l < 2a \)), use the definition:
\[ \mathrm{f}\left(\frac{1}{2}l\right) = \sqrt{1 – \frac{\left(\frac{1}{2}l\right)^2}{a^2}} = \sqrt{1 – \frac{l^2}{4a^2}} \]
Part (b):
\[ l = \sqrt{2}a, \quad y = \frac{\sqrt{2}}{2} \]
Working: The area of the rectangle’s cross-section is \( A = l \cdot y \), where \( y = \mathrm{f}\left(\frac{1}{2}l\right) = \sqrt{1 – \frac{l^2}{4a^2}} \). Thus:
\[ A = l \sqrt{1 – \frac{l^2}{4a^2}} \]
Let \( k = \frac{l}{a} \), so \( l = ka \), and:
\[ A = (ka) \sqrt{1 – \frac{(ka)^2}{4a^2}} = a^2 k \sqrt{1 – \frac{k^2}{4}} \]
Maximize \( f(k) = k \sqrt{1 – \frac{k^2}{4}} \), for \( 0 < k < 2 \). Compute the derivative:
\[ f(k) = k \left(1 – \frac{k^2}{4}\right)^{1/2} \]
Using the product rule:
\[ f'(k) = \left(1 – \frac{k^2}{4}\right)^{1/2} + k \cdot \frac{1}{2} \left(1 – \frac{k^2}{4}\right)^{-1/2} \cdot \left(-\frac{2k}{4}\right) \]
Simplify:
\[ f'(k) = \sqrt{1 – \frac{k^2}{4}} – \frac{k^2}{4 \sqrt{1 – \frac{k^2}{4}}} \]
Set \( f'(k) = 0 \):
\[ \sqrt{1 – \frac{k^2}{4}} = \frac{k^2}{4 \sqrt{1 – \frac{k^2}{4}}} \]
Multiply both sides by \( \sqrt{1 – \frac{k^2}{4}} \):
\[ 1 – \frac{k^2}{4} = \frac{k^2}{4} \]
Solve:
\[ 1 = \frac{k^2}{4} + \frac{k^2}{4} = \frac{k^2}{2} \implies k^2 = 2 \implies k = \sqrt{2} \]
Thus, \( l = ka = \sqrt{2}a \). The height is:
\[ y = \sqrt{1 – \frac{(\sqrt{2}a)^2}{4a^2}} = \sqrt{1 – \frac{2a^2}{4a^2}} = \sqrt{1 – \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2} \]
Key Concept:
Optimization of a function involving a geometric model requires differentiation and solving for critical points.