SL AA · Paper 1 (No Calculator)

SL AA Paper 1 practice — 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 $0, \tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}, \tfrac{\pi}{2}$ (and their multiples across all four quadrants) should come out of your fingers in under 3 seconds. The identity $\sin^2\theta + \cos^2\theta = 1$ must be used both forwards and backwards.
  • Log rules under pressure. $\log_a(xy) = \log_a x + \log_a y$, $\log_a x^n = n\log_a x$, change of base. Practise combining and expanding in both directions. Never leave $\log_2 8$ as anything but $3$.
  • Quadratics without formula-panic. Completing the square, discriminant $b^2 - 4ac$, 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 $\dfrac{d}{dx}[\sin, \cos, e^x, \ln x]$. Reasonably clean simplification is expected — leaving unsimplified $\dfrac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ often loses A marks.
  • Integration by inspection. Reversing the chain rule on $\int \sin(ax+b)\,dx$, $\int (2x+3)^5\,dx$, $\int \tfrac{1}{x-2}\,dx$. 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 $f(x) = 2\sin x + 1$ and $g(x) = \dfrac{x-1}{2}$, show that $(g \circ f)(x) = \sin x$.

Reveal worked solution
$(g \circ f)(x) = g(f(x)) = g(2\sin x + 1) = \dfrac{(2\sin x + 1) - 1}{2} = \dfrac{2\sin x}{2} = \sin x.$

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

(b) [3 marks]

Hence solve $f(x) = 2$ exactly for $x \in [0, 2\pi]$.

Reveal worked solution

Step 1. $2\sin x + 1 = 2 \Rightarrow \sin x = \tfrac12.$ (M1)

Step 2. Reference angle $= \tfrac{\pi}{6}$. In $[0, 2\pi]$, $\sin$ is positive in Q1 and Q2.

$x = \tfrac{\pi}{6} \text{ or } \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}.$ (A1 · A1)

(c) [3 marks]

Prove the identity $\dfrac{\sin^2 x}{1 - \cos x} = 1 + \cos x$ for $\cos x \ne 1$.

Reveal worked solution
$\dfrac{\sin^2 x}{1 - \cos x} = \dfrac{1 - \cos^2 x}{1 - \cos x} = \dfrac{(1 - \cos x)(1 + \cos x)}{1 - \cos x} = 1 + \cos x.$

(M1 for $\sin^2 x = 1 - \cos^2 x$ · M1 for factorising · AG1)

(d) [3 marks]

Hence find the exact value of $\dfrac{\sin^2(\tfrac{5\pi}{6})}{1 - \cos(\tfrac{5\pi}{6})}$.

Reveal worked solution

Using (c): value $= 1 + \cos(\tfrac{5\pi}{6}) = 1 + \left(-\tfrac{\sqrt 3}{2}\right) = \tfrac{2 - \sqrt 3}{2}.$ (M1 · A1 · A1)

Question 2 · Turning points and definite integral

Skills: derivatives, stationary points, integration · 13 marks

(a) [2 marks]

Let $f(x) = x^3 - 6x^2 + 9x + 2$. Find $f'(x)$.

Reveal worked solution
$f'(x) = 3x^2 - 12x + 9.$ (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

$f'(x) = 0 \Rightarrow 3(x^2 - 4x + 3) = 0 \Rightarrow 3(x-1)(x-3) = 0 \Rightarrow x = 1 \text{ or } x = 3.$ (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
$f(1) = 1 - 6 + 9 + 2 = 6, \quad f(3) = 27 - 54 + 27 + 2 = 2.$ (M1 · A1 · A1)

(d) [5 marks]

Find the exact value of $\displaystyle\int_1^3 f(x) \, dx$ and interpret geometrically.

Reveal worked solution
$\int (x^3 - 6x^2 + 9x + 2)\,dx = \tfrac{x^4}{4} - 2x^3 + \tfrac{9x^2}{2} + 2x + C.$ (M1 · A1)
At $x = 3$: $\tfrac{81}{4} - 54 + \tfrac{81}{2} + 6 = \tfrac{81 + 162}{4} - 48 = \tfrac{243}{4} - 48 = \tfrac{51}{4}.$
At $x = 1$: $\tfrac{1}{4} - 2 + \tfrac{9}{2} + 2 = \tfrac{1 + 18}{4} + 0 = \tfrac{19}{4}.$
$\int_1^3 f(x)\,dx = \tfrac{51}{4} - \tfrac{19}{4} = \tfrac{32}{4} = 8.$ (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)