IB Math AA Assessment Example (SL + HL Extensions)
Topic Focus: Functions, Algebra, and Modeling
Time: 35–45 minutes (can be shortened to 25)
Calculator: Allowed
Total: 30 marks
IB Mathematics: Analysis & Approaches (AA) Assessment
Section A: Non-Calculator (Recommended for first 10 minutes)
(No calculator recommended, but you can allow it if you want.)
1. Solve and justify (4 marks)
Solve the equation:
2x2−5x−3=02x^2 - 5x - 3 = 02x2−5x−3=0
(a) Solve using factorization. (2 marks)
(b) Verify your solutions by substitution. (2 marks)
2. Functions and reasoning (5 marks)
A function is defined by
f(x)=x−2f(x) = \sqrt{x-2}f(x)=x−2
(a) State the domain of fff. (2 marks)
(b) State the range of fff. (2 marks)
(c) Explain why f(x)f(x)f(x) cannot take negative values. (1 mark)
Section B: Calculator Allowed
3. Modeling with a quadratic (7 marks)
A student throws a ball upward from a height of 1.5 meters. The height hhh, in meters, after ttt seconds is modeled by:
h(t)=−4.9t2+6t+1.5h(t) = -4.9t^2 + 6t + 1.5h(t)=−4.9t2+6t+1.5
(a) State the height of the ball when t=0t=0t=0. (1 mark)
(b) Find the maximum height reached by the ball. (3 marks)
(c) Find the time when the ball hits the ground. (3 marks)
4. Interpreting functions (6 marks)
Consider the functions:
g(x)=x2−4x+3g(x) = x^2 - 4x + 3g(x)=x2−4x+3 p(x)=g(x+2)p(x) = g(x+2)p(x)=g(x+2)
(a) Expand and simplify p(x)p(x)p(x). (2 marks)
(b) Describe the transformation from g(x)g(x)g(x) to p(x)p(x)p(x). (2 marks)
(c) Find the x-intercepts of p(x)p(x)p(x). (2 marks)
5. A combined reasoning question (8 marks)
A function is defined by:
k(x)=x+1x−2k(x)=\frac{x+1}{x-2}k(x)=x−2x+1
(a) State the value(s) of xxx for which k(x)k(x)k(x) is undefined. (1 mark)
(b) Solve:
k(x)=3k(x)=3k(x)=3
(3 marks)
(c) Explain whether the function has a horizontal asymptote and state its equation. (2 marks)
(d) The function is restricted to x>2x>2x>2. Describe the behavior of k(x)k(x)k(x) as x→2+x\to 2^+x→2+. (2 marks)
✅ HL Extension (Optional Add-On: +6 marks)
6. HL Challenge: Proof / Generalization (6 marks)
The sequence is defined by
u1=3,un+1=2un−1u_1 = 3,\quad u_{n+1} = 2u_n - 1u1=3,un+1=2un−1
(a) Find u2u_2u2, u3u_3u3, and u4u_4u4. (2 marks)
(b) Conjecture a formula for unu_nun. (2 marks)
(c) Prove your conjecture using mathematical induction. (2 marks)
Answer Key (Quick Teacher Version)
1.
(a) (2x+1)(x−3)=0⇒x=3, x=−12(2x+1)(x-3)=0\Rightarrow x=3,\; x=-\frac12(2x+1)(x−3)=0⇒x=3,x=−21
(b) Sub works
2.
Domain: x≥2x\ge2x≥2
Range: f(x)≥0f(x)\ge0f(x)≥0
Negative not possible due to square root output
3.
(a) h(0)=1.5h(0)=1.5h(0)=1.5
(b) Max at t=−b2a=69.8=0.612...t=-\frac{b}{2a}=\frac{6}{9.8}=0.612...t=−2ab=9.86=0.612...
Sub to get max height
(c) Solve h(t)=0h(t)=0h(t)=0 (quadratic formula or calculator)
4.
(a) p(x)=(x+2)2−4(x+2)+3=x2−1p(x)=(x+2)^2-4(x+2)+3=x^2-1p(x)=(x+2)2−4(x+2)+3=x2−1
(b) shift left 2
(c) x2−1=0⇒x=±1x^2-1=0\Rightarrow x=\pm1x2−1=0⇒x=±1
5.
(a) x≠2x\ne2x=2
(b) x+1x−2=3⇒x+1=3x−6⇒x=72\frac{x+1}{x-2}=3\Rightarrow x+1=3x-6\Rightarrow x=\frac{7}{2}x−2x+1=3⇒x+1=3x−6⇒x=27
(c) Horizontal asymptote: y=1y=1y=1
(d) As x→2+x\to2^+x→2+, denominator tiny positive → function →+∞\to +\infty→+∞
HL 6.
(a) 5, 9, 17
(b) un=2n+1u_n=2^n+1un=2n+1
(c) induction proof
