The first three terms of an arithmetic sequence are 𝑢 1 = 5 , 𝑢 2 = 8 , 𝑢 3 = 11 .

(a) d=u2−u1=8−5=3. (b) u20=u1+19d=5+19(3)=5+57=62. (c) S20=202(u1+u20)=10(5+62)=10⋅67=670.

Given that log 2 ⁡ 𝑥 = 5 , find the exact value of:

(a) x=25=32. (b) log2⁡(4x)=log2⁡4+log2⁡x=2+5=7. (c) log2(x28)=2log2⁡x−log2⁡8=2(5)−3=7.

Given that 𝜃 is acute and sin ⁡ 𝜃 = 3 5 , find the exact value of:

(a) cos2⁡θ=1−sin2⁡θ=1−925=1625. Since θ acute, cos⁡θ=45. (b) tan⁡θ=sin⁡θcos⁡θ=3/54/5=34. (c) sin⁡2θ=2sin⁡θcos⁡θ=2⋅35⋅45=2425.

Let 𝑓 ( 𝑥 ) = 𝑥 3 − 6 𝑥 2 + 9 𝑥 + 1 .

(a) f′(x)=3x2−12x+9. (b) f′(x)=0⇒3(x2−4x+3)=0⇒3(x−1)(x−3)=0, so x=1 and x=3. (c) f″(x)=6x−12; at x=1, f″(1)=−6<0, so it is a local maximum.

Let 𝑓 ( 𝑥 ) = 2 𝑥 − 3 and 𝑔 ( 𝑥 ) = 𝑥 2 + 1 , both with domain 𝑥 ∈ 𝑅 .

(a) g(2)=4+1=5; then f(5)=2(5)−3=7. (b) Let y=2x−3⇒x=y+32, so f−1(x)=x+32. (c) 2x−3=0⇒x=32.

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

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Progress: 0 / 16 correct

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]

d =

(b) Find the 20^th term $u_{20}$ . [2]

u_{20} =

(c) Find the sum of the first 20 terms, $S_{20}$ . [3]

S_{20} =

(a)

d = u_{2} - u_{1} = 8 - 5 = 3

.
(b)

u_{20} = u_{1} + 19 d = 5 + 19 (3) = 5 + 57 = 62

.
(c)

S_{20} = \frac{20}{2} (u_{1} + u_{20}) = 10 (5 + 62) = 10 \cdot 67 = 670

Question 2

5 marks

Algebra — logarithms

Given that $\log_{2} x = 5$ , find the exact value of:

(a) $x$ . [1]

x =

(b) $\log_{2} (4 x)$ . [2]

\log_{2} (4 x) =

(c) $\log_{2} (\frac{x^{2}}{8})$ . [2]

value

=

(a)

x = 2^{5} = 32

.
(b)

\log_{2} (4 x) = \log_{2} 4 + \log_{2} x = 2 + 5 = 7

.
(c)

\log_{2} (\frac{x^{2}}{8}) = 2 \log_{2} x - \log_{2} 8 = 2 (5) - 3 = 7

Question 3

6 marks

Trigonometry — exact values

Given that $θ$ is acute and $\sin θ = \frac{3}{5}$ , find the exact value of:

(a) $\cos θ$ . [2]

\cos θ =

(b) $\tan θ$ . [2]

\tan θ =

(c) $\sin 2 θ$ . [2]

\sin 2 θ =

(a)

\cos^{2} θ = 1 - \sin^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}

. Since

θ

acute,

\cos θ = \frac{4}{5}

.
(b)

\tan θ = \frac{\sin θ}{\cos θ} = \frac{3 / 5}{4 / 5} = \frac{3}{4}

.
(c)

\sin 2 θ = 2 \sin θ \cos θ = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}

Question 4

7 marks

Calculus — differentiation

Let $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ .

(a) Find $f^{'} (x)$ . Type your answer in the form $a x^{2} + b x + c$ . [2]

f^{'} (x) =

Whitespace ignored. Accepted: 3x^2-12x+9.

(b) Find the $x$ -coordinates of the two stationary points. Type the smaller first. [3]

smaller

x =

larger

x =

(c) Classify the stationary point at $x = 1$ as a local maximum or minimum. [2]

type:

(a)

f^{'} (x) = 3 x^{2} - 12 x + 9

.
(b)

f^{'} (x) = 0 \Rightarrow 3 (x^{2} - 4 x + 3) = 0 \Rightarrow 3 (x - 1) (x - 3) = 0

, so

x = 1

and

x = 3

.
(c)

f^{″} (x) = 6 x - 12

; at

x = 1

f^{″} (1) = - 6 < 0

, so it is a local maximum.

Question 5

6 marks

Functions — composite & inverse

Let $f (x) = 2 x - 3$ and $g (x) = x^{2} + 1$ , both with domain $x \in R$ .

(a) Find $(f \circ g) (2)$ . [2]

(f \circ g) (2) =

(b) Find $f^{- 1} (x)$ . Type in the form $\frac{x + a}{b}$ . [2]

f^{- 1} (x) =

(c) Solve $f (x) = 0$ . [2]

x =

(a)

g (2) = 4 + 1 = 5

; then

f (5) = 2 (5) - 3 = 7

.
(b) Let

y = 2 x - 3 \Rightarrow x = \frac{y + 3}{2}

, so

f^{- 1} (x) = \frac{x + 3}{2}

.
(c)

2 x - 3 = 0 \Rightarrow x = \frac{3}{2}

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