IB DP Maths- MAA SL Prediction Paper 1 – 2026 Edition
IB DP Math AA SL Prediction Paper 1 – April/May 2026 Exam
IIB DP Math AA SL Prediction Paper 1: Prepare for the IB exams with subject-specific Prediction questions, model answers. All topics covered.
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Question 1
(a) Express ${3y^2 – 12y – 15}$ in the form ${3(y+a)^2 + b}$, where $a$ and $b$ are constants.
(b) Hence find the exact solutions of the equation $3x^4 – 12x^2 – 15 = 0$.
▶️Answer/Explanation
(a) Express \(3y^2 – 12y – 15\) as \(3(y + a)^2 + b\)
Factor out 3 from the first two terms:
\[ 3y^2 – 12y – 15 = 3(y^2 – 4y) – 15 \]
Complete the square inside:
\[ y^2 – 4y = (y – 2)^2 – 4 \]
\[ 3(y^2 – 4y) = 3[(y – 2)^2 – 4] = 3(y – 2)^2 – 12 \]
So:
\[ 3(y^2 – 4y) – 15 = 3(y – 2)^2 – 12 – 15 = 3(y – 2)^2 – 27 \]
Thus, \(a = -2\), \(b = -27\).
Answer: \(3(y – 2)^2 – 27\)
(b) Solve \(3x^4 – 12x^2 – 15 = 0\)
Let \(u = x^2\), so the equation becomes:
\[ 3u^2 – 12u – 15 = 0 \]
From (a), this is:
\[ 3(u – 2)^2 – 27 = 0 \]
\[ 3(u – 2)^2 = 27 \]
\[ (u – 2)^2 = 9 \]
\[ u – 2 = \pm 3 \]
\[ u = 2 + 3 = 5 \quad \text{or} \quad u = 2 – 3 = -1 \]
Since \(u = x^2\):
\(x^2 = 5\) → \(x = \pm \sqrt{5}\)
\(x^2 = -1\) → no real solutions (discard)
Answer: \(x = \sqrt{5}\), \(x = -\sqrt{5}\)
Final Answers:
(a) \(3(y – 2)^2 – 27\)
(b) \(\sqrt{5}\), \(-\sqrt{5}\)