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SL AA · Paper 1 (No Calculator)

IB Math AA SL Paper 1 practice questions (non-calculator)

Paper 1 rewards clean algebra, fluent trig identities, and confident derivatives — all without a calculator. Two original SL AA questions below, plus a focused skills briefing so you know exactly what to master and why each technique wins marks.

Total: 25 marks · Recommended: 40 minutes · No calculator

What Paper 1 actually tests — and how to improve fast

Paper 1 (non-calculator) at SL AA is deliberately narrow. Under exam pressure and no calculator, the mistakes that cost marks are almost always mechanical: bad algebra, wrong sign, or reaching for a shortcut that isn't in your toolkit.

The five skills you must have automatic

  • Exact-value trigonometry. sin,cos,tan of 0,π6,π4,π3,π2 (and their multiples across all four quadrants) should come out of your fingers in under 3 seconds. The identity sin2⁡θ+cos2⁡θ=1 must be used both forwards and backwards.
  • Log rules under pressure. loga⁡(xy)=loga⁡x+loga⁡y, loga⁡xn=nloga⁡x, change of base. Practise combining and expanding in both directions. Never leave log2⁡8 as anything but 3.
  • Quadratics without formula-panic. Completing the square, discriminant b2−4ac, sum and product of roots (−b/a and c/a), and knowing when a substitution turns a quartic into a quadratic.
  • Derivative fluency. Chain, product, quotient rules and ddx[sin,cos,ex,ln⁡x]. Reasonably clean simplification is expected — leaving unsimplified f′(x)g(x)−f(x)g′(x)[g(x)]2 often loses A marks.
  • Integration by inspection. Reversing the chain rule on ∫sin⁡(ax+b)dx, ∫(2x+3)5dx, ∫1x−2dx. Substitution appears in Paper 1 but expect the substitution to be either obvious or hinted.

How to actually improve

  1. Twenty-minute mechanical fluency reps. Every day, 20 minutes of pure calculation — no notes, no calculator. Not full past papers; just the five mechanical skills above rotated across the week.
  2. Redo your worst mistake three times. Whenever you drop a mark on Paper 1, write the error, then redo the exact question a day later, a week later, and a month later. Never let the same mistake happen twice.
  3. Show every method line. M marks are the biggest hidden points reservoir. Write the substitution, the rule name, the identity used — even one word ("chain") next to a derivative earns method credit.
  4. Time yourself on part (a). Paper 1 has short parts (2-4 marks) that must be done at ~1 min per mark. If part (a) drags past 4 minutes, move on. This is a Paper 1-specific pacing skill.
  5. Practise "reverse engineering" answers. Given an integral, differentiate it back. Given a quadratic's roots, verify by expansion. Fast self-checking on Paper 1 is where the top grades win.

Question 1 · Composite functions and trig identity

Skills: composition, inverse trig, exact values · 12 marks

(a) [3 marks]

Given f(x)=2sin⁡x+1 and g(x)=x−12, show that (g∘f)(x)=sin⁡x.

Reveal worked solution
(g∘f)(x)=g(f(x))=g(2sin⁡x+1)=(2sin⁡x+1)−12=2sin⁡x2=sin⁡x.

(M1 for substitution · A1 for simplification · AG1 for showing equality)

(b) [3 marks]

Hence solve f(x)=2 exactly for x∈[0,2π].

Reveal worked solution

Step 1. 2sin⁡x+1=2⇒sin⁡x=12. (M1)

Step 2. Reference angle =π6. In [0,2π], sin is positive in Q1 and Q2.

x=π6 or π−π6=5π6. (A1 · A1)

(c) [3 marks]

Prove the identity sin2⁡x1−cos⁡x=1+cos⁡x for cos⁡x≠1.

Reveal worked solution
sin2⁡x1−cos⁡x=1−cos2⁡x1−cos⁡x=(1−cos⁡x)(1+cos⁡x)1−cos⁡x=1+cos⁡x.

(M1 for sin2⁡x=1−cos2⁡x · M1 for factorising · AG1)

(d) [3 marks]

Hence find the exact value of sin2⁡(5π6)1−cos⁡(5π6).

Reveal worked solution

Using (c): value =1+cos⁡(5π6)=1+(−32)=2−32. (M1 · A1 · A1)

Question 2 · Turning points and definite integral

Skills: derivatives, stationary points, integration · 13 marks

(a) [2 marks]

Let f(x)=x3−6x2+9x+2. Find f′(x).

Reveal worked solution
f′(x)=3x2−12x+9. (M1 · A1)

(b) [3 marks]

Find the exact x-coordinates of the stationary points and classify each as a local maximum or minimum.

Reveal worked solution

f′(x)=0⇒3(x2−4x+3)=0⇒3(x−1)(x−3)=0⇒x=1 or x=3. (M1 · A1)

f″(x)=6x−12. f″(1)=−6<0 ⇒ local max. f″(3)=6>0 ⇒ local min. (A1)

(c) [3 marks]

Find the exact y-values at each stationary point.

Reveal worked solution
f(1)=1−6+9+2=6,f(3)=27−54+27+2=2. (M1 · A1 · A1)

(d) [5 marks]

Find the exact value of ∫13f(x)dx and interpret geometrically.

Reveal worked solution
∫(x3−6x2+9x+2)dx=x44−2x3+9x22+2x+C. (M1 · A1)
At x=3: 814−54+812+6=81+1624−48=2434−48=514.
At x=1: 14−2+92+2=1+184+0=194.
∫13f(x)dx=514−194=324=8. (M1 · A1)

Geometric interpretation: the area under f between its local max (x=1,y=6) and local min (x=3,y=2) is 8. Since f>0 throughout, no sign change. (A1)