SL AA · Practice Paper 1 · Interactive
IB Maths Analysis & Approaches — Interactive Practice Paper 1
Paper 1 is the no-calculator paper — algebraic manipulation, exact values and short-answer proof. Two modes: instant per-part feedback, or full timed exam with a countdown.
30 marks · 5 questions · No GDC · Exact-value answers
Question 1 6 marks
Sequences & Series
The first three terms of an arithmetic sequence are $u_1 = 5$, $u_2 = 8$, $u_3 = 11$.
(a) Find the common difference $d$. [1]
(b) Find the 20th term $u_{20}$. [2]
(c) Find the sum of the first 20 terms, $S_{20}$. [3]
Show worked solution ▾
(a) $d = u_2 - u_1 = 8 - 5 = \mathbf{3}$.(b) $u_{20} = u_1 + 19d = 5 + 19(3) = 5 + 57 = \mathbf{62}$.(c) $S_{20} = \tfrac{20}{2}(u_1 + u_{20}) = 10(5 + 62) = 10 \cdot 67 = \mathbf{670}$.
Question 2 5 marks
Algebra — logarithms
Given that $\log_2 x = 5$, find the exact value of:
(a) $x$. [1]
(b) $\log_2(4x)$. [2]
(c) $\log_2\!\left(\dfrac{x^2}{8}\right)$. [2]
Show worked solution ▾
(a) $x = 2^5 = \mathbf{32}$.(b) $\log_2(4x) = \log_2 4 + \log_2 x = 2 + 5 = \mathbf{7}$.(c) $\log_2\!\left(\tfrac{x^2}{8}\right) = 2\log_2 x - \log_2 8 = 2(5) - 3 = \mathbf{7}$.
Question 3 6 marks
Trigonometry — exact values
Given that $\theta$ is acute and $\sin\theta = \tfrac{3}{5}$, find the exact value of:
(a) $\cos\theta$. [2]
(b) $\tan\theta$. [2]
(c) $\sin 2\theta$. [2]
Show worked solution ▾
(a) $\cos^2\theta = 1 - \sin^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}$. Since $\theta$ acute, $\cos\theta = \mathbf{\tfrac{4}{5}}$.(b) $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{4/5} = \mathbf{\tfrac{3}{4}}$.(c) $\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \tfrac{3}{5} \cdot \tfrac{4}{5} = \mathbf{\tfrac{24}{25}}$.
Question 4 7 marks
Calculus — differentiation
Let $f(x) = x^3 - 6x^2 + 9x + 1$.
(a) Find $f'(x)$. Type your answer in the form $ax^2 + bx + c$. [2]
(b) Find the $x$-coordinates of the two stationary points. Type the smaller first. [3]
(c) Classify the stationary point at $x = 1$ as a local maximum or minimum. [2]
Show worked solution ▾
(a) $f'(x) = \mathbf{3x^2 - 12x + 9}$.(b) $f'(x) = 0 \Rightarrow 3(x^2 - 4x + 3) = 0 \Rightarrow 3(x-1)(x-3) = 0$, so $x = \mathbf{1}$ and $x = \mathbf{3}$.(c) $f''(x) = 6x - 12$; at $x=1$, $f''(1) = -6 < 0$, so it is a local maximum .
Question 5 6 marks
Functions — composite & inverse
Let $f(x) = 2x - 3$ and $g(x) = x^2 + 1$, both with domain $x \in \mathbb{R}$.
(a) Find $(f \circ g)(2)$. [2]
(b) Find $f^{-1}(x)$. Type in the form $\dfrac{x+a}{b}$. [2]
(c) Solve $f(x) = 0$. [2]
Show worked solution ▾
(a) $g(2) = 4 + 1 = 5$; then $f(5) = 2(5) - 3 = \mathbf{7}$.(b) Let $y = 2x - 3 \Rightarrow x = \dfrac{y+3}{2}$, so $f^{-1}(x) = \mathbf{\dfrac{x+3}{2}}$.(c) $2x - 3 = 0 \Rightarrow x = \mathbf{\tfrac{3}{2}}$.