IB DP Math AA SL Prediction Paper 1 for 2025 Exams
IB DP Math AA SL Prediction Paper 1 – April/May 2025 Exam
IIB DP Math AA SL Prediction Paper 1: Prepare for the IB exams with subject-specific Prediction questions, model answers. All topics covered.
Prepared by IB teachers: Access our IB DP Math AA SL Prediction Paper 1 with model answer. Students: Practice with exam-style papers for IB DP Math AA HL Exam
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) \(\sqrt{5}\), \(-\sqrt{5}\)
Question 2
The diagram shows two curves. One curve has the equation \( y = \sin x \) and the other curve has the equation \( y = f(x) \).
(a) In order to transform the curve \( y = \sin x \) to the curve \( y = f(x) \), the curve \( y = \sin x \) is first reflected in the \( x \)-axis.
Describe fully a sequence of two further transformations which are required.
(b) Find \( f(x) \) in terms of \( \sin x \).
▶️ Answer/Explanation
Solution:
2(a)
Stretch by a factor of \( 3 \) in the \( y \)-direction, followed by a translation of \[ \begin{pmatrix} 0 \\ -2 \end{pmatrix}. \]
Alternative Method for Question 2(a)
Translation of \[ \begin{pmatrix} 0 \\ -\frac{2}{3} \end{pmatrix}, \] followed by a stretch by a factor of \( 3 \) in the \( y \)-direction.
2(b)
The equation of \( f(x) \) is: \[ f(x) = -3 \sin x – 2. \]
Question 3
The coefficient of $x^3$ in the expansion of $(3+ax)^6$ is 160.
(a) Find the value of the constant $a$.
(b) Hence find the coefficient of $x^3$ in the expansion of $(3+ax)^6(1-2x)$.
▶️Answer/Explanation
(a) Find the value of \(a\)
For \((3 + ax)^6\), use the binomial expansion:
\[(3 + ax)^6 = \sum_{k=0}^{6} \binom{6}{k} 3^{6-k} (ax)^k\]
The \(x^3\) term occurs when \(k = 3\):
Coefficient = \(\binom{6}{3} 3^{6-3} a^3 = 20 \cdot 3^3 \cdot a^3\)
\(3^3 = 27\), so coefficient = \(20 \cdot 27 \cdot a^3 = 540a^3\)
Given it equals 160:
\[ 540a^3 = 160 \]
\[ a^3 = \frac{160}{540} = \frac{8}{27} \]
\[ a = \left(\frac{8}{27}\right)^{\frac{1}{3}} = \frac{2}{3} \]
Answer: \(a = \frac{2}{3}\)
(b) Coefficient of \(x^3\) in \((3 + ax)^6 (1 – 2x)\) with \(a = \frac{2}{3}\)
First, \(a = \frac{2}{3}\), so \((3 + ax)^6 = \left(3 + \frac{2}{3}x\right)^6\).
\(x^3\) coefficient in \(\left(3 + \frac{2}{3}x\right)^6\) is \(540 \left(\frac{2}{3}\right)^3 = 540 \cdot \frac{8}{27} = 160\) (from part (a)).
For \((3 + \frac{2}{3}x)^6 (1 – 2x)\), the \(x^3\) term comes from:
\(x^3\) from \((3 + \frac{2}{3}x)^6\) × constant from \((1 – 2x)\): \(160 \cdot 1 = 160\)
\(x^2\) from \((3 + \frac{2}{3}x)^6\) × \(-2x\) from \((1 – 2x)\):
\(x^2\) coefficient: \(\binom{6}{2} 3^4 \left(\frac{2}{3}\right)^2 = 15 \cdot 81 \cdot \frac{4}{9} = 15 \cdot 36 = 540\)
Total = \(540 \cdot (-2) = -1080\)
Combine:
\[ 160 + (-1080) = -920 \]
Answer: \(-920\)
Final Answers:
(a) \(\frac{2}{3}\)
(b) \(-920\)