Pure · Statistics · Mechanics
The notes follow the OCR H240 specification, restricted to the Year 1 (AS) content. Everything else has been deferred or removed. Most sections open with the headline result and then the working that makes it stick.
The A‑level is examined by three 100‑mark papers, each two hours long.
All papers permit a calculator. There is no non‑calculator paper at H240.
The verb at the start of a question tells you what kind of answer earns the marks.
| Word | What it asks for |
|---|---|
| Exact | Unrounded — leave $\pi$, surds, and fractions in symbolic form. |
| Prove | Formal argument, every logical step justified, concise conclusion. |
| Show that | Every step of working visible. The target answer is given, so the marks are for the route, not the destination. |
| Determine | Justification and working needed. |
| Verify | Substitute the given value and check. |
| Find / Solve / Calculate | Working may help but no justification required. |
| Give / State / Write down | No working needed. |
| Hence | The next step must use the result of the previous part. |
| Hence or otherwise | The previous part probably gives the cleanest route, but any valid method scores. |
| Plot | Mark accurate points, possibly join them. |
| Sketch | Show key features: intercepts, turning points, asymptotes. Not to scale. |
| Draw | Draw to a sensible level of accuracy. |
The instruction "In this question you must show detailed reasoning" is a deliberate calculator‑restriction signal. You may still use the calculator to check, but the marks are for analytical working. A correct final answer with no analytical method scores zero on these questions.
Set the calculator to degree mode for Year 1. Radians don't appear in the AS specification, but a calculator stuck in radians will silently give nonsense for every trig question. Check once at the start.
Three methods only: deduction, exhaustion, and disproof by counter‑example. Proof by contradiction (and the two standard proofs of $\sqrt{2}$ being irrational and the infinitude of primes) is Year 2 content — leave it for now.
Proof by deduction. Start from established facts (definitions, axioms, earlier theorems) and proceed by valid logical steps to the conclusion. The most common form. Be explicit at each step about what you are using.
Proof by exhaustion. Check every possible case. The "cases" can be discrete values (prove a result for $n = 1, 2, 3$), or you can partition an infinite domain into a finite number of categories (e.g. prove a result for all integers by proving it for evens and odds separately).
Disproof by counter‑example. A single concrete instance where the claim fails disproves it. You only need one.
Disprove the claim: "if $n$ is prime then $n$ is odd".
$n = 2$ is prime and even. ∎
The symbol $\equiv$ denotes an identity: a relation that holds for all values of the variables involved (not just for some). Use $=$ for equations that hold conditionally, $\equiv$ for things true by definition.
$A \Rightarrow B$ means "$A$ implies $B$": whenever $A$ is true, $B$ is true. It does not mean $B \Rightarrow A$.
$A \Leftrightarrow B$ ("$A$ if and only if $B$", or "iff") means implication in both directions: $A \Rightarrow B$ and $B \Rightarrow A$.
For a result like "$P$ if and only if $Q$" you have to prove both implications separately.
If $A \Rightarrow B$ then $A$ is sufficient for $B$ (knowing $A$ is enough). And $B$ is necessary for $A$ (without $B$, $A$ cannot hold). Both arrows means "necessary and sufficient" — that's $\Leftrightarrow$.
Prove by deduction that the product of any two odd integers is odd.
Let $m$ and $n$ be odd integers. Then $m = 2a + 1$ and $n = 2b + 1$ for some integers $a, b$.
$$mn = (2a+1)(2b+1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1.$$
Since $2ab + a + b$ is an integer, $mn$ is of the form $2k + 1$ and is therefore odd. ∎
A student claims: "For every positive integer $n$, the value $n^2 + n + 41$ is prime." Show that this claim is false.
Take $n = 40$. Then $n^2 + n + 41 = 1600 + 40 + 41 = 1681 = 41^2$, which is not prime.
The trap: checking $n = 1, 2, 3, \ldots, 39$ all return primes (this is a famous expression due to Euler), which makes the claim look true by exhaustion. But exhaustion of finitely many cases proves nothing about an infinite claim. Only one counterexample is needed to disprove. ∎
Prove by exhaustion that every prime number greater than $3$ can be written in the form $6k \pm 1$ for some integer $k$.
Every integer is of exactly one of the forms $6k$, $6k+1$, $6k+2$, $6k+3$, $6k+4$, $6k+5$ for some integer $k$. We consider each case for a number $p > 3$:
Only $p = 6k + 1$ and $p = 6k + 5$ remain possible. Rewriting $6k + 5 = 6(k+1) - 1$, we see every prime greater than $3$ is of the form $6m + 1$ or $6m - 1$ for some integer $m$ — that is, $6k \pm 1$. ∎
Note: the converse is not true. $25 = 6(4) + 1$ is not prime. So this gives a necessary, not sufficient, condition.
The laws of indices, valid for all real $a, b$ (with $x > 0$ where needed for fractional powers):
$$x^a x^b = x^{a+b}, \qquad \frac{x^a}{x^b} = x^{a-b}, \qquad (x^a)^b = x^{ab}$$
$$x^{-a} = \frac{1}{x^a}, \qquad x^{m/n} = \sqrt[n]{x^m}, \qquad x^0 = 1$$
$x^0 = 1$ for all $x \neq 0$. The expression $0^0$ is left undefined for our purposes. Don't write $0^0 = 1$ in a proof.
A surd is an irrational root, like $\sqrt{2}$ or $\sqrt[3]{5}$. The point of working with surds is to keep answers exact.
Rationalising single‑surd denominators. Multiply top and bottom by the surd:
$$\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}.$$
Rationalising binomial surds. Multiply by the conjugate:
$$\frac{a}{b + \sqrt{c}} \cdot \frac{b - \sqrt{c}}{b - \sqrt{c}} = \frac{a(b - \sqrt{c})}{b^2 - c}.$$
The conjugate of $b + \sqrt{c}$ is $b - \sqrt{c}$. The product $(b + \sqrt{c})(b - \sqrt{c}) = b^2 - c$ is rational by the difference of two squares.
If the question says "give your answer in exact form" or "in the form $a + b\sqrt{c}$", surds in the denominator usually lose marks even if the value is correct. Rationalise.
Two linear equations. Eliminate a variable by scaling and adding/subtracting, or rearrange one equation and substitute.
One linear, one quadratic (or one quadratic in disguise — e.g. a circle). Rearrange the linear equation for one variable, substitute into the quadratic, solve.
Solve $x + y = 1$ and $x^2 + y^2 = 1$.
From the first equation, $y = 1 - x$. Substitute:
$$x^2 + (1-x)^2 = 1 \;\Rightarrow\; 2x^2 - 2x = 0 \;\Rightarrow\; x(x-1) = 0.$$
So $x = 0$ or $x = 1$, giving the points $(0, 1)$ and $(1, 0)$.
If the question asks for points of intersection, give coordinates as pairs. Don't leave the answer as just the $x$‑values — substitute back to find each $y$.
Three solution methods, choose by inspection:
For $ax^2 + bx + c = 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$
The discriminant is $\Delta = b^2 - 4ac$:
Completed square form: $a(x + p)^2 + q$ has turning point at $(-p, q)$. Minimum if $a > 0$, maximum if $a < 0$. The line of symmetry is $x=-p$.
Anything quadratic in some expression of $x$ can be solved by the same techniques. The trick is to spot the quadratic structure:
Substitute $y$ for whatever the quadratic is in. Solve for $y$. Then back‑substitute, paying attention to the natural range of the original expression.
Solve $\sin^2 x - \sin x - 1 = 0$ for $0° \le x < 360°$.
Let $y = \sin x$. Then $y^2 - y - 1 = 0$, so by the formula $y = \dfrac{1 \pm \sqrt{5}}{2}$.
Now $\sin x$ lies in $[-1, 1]$. The root $\tfrac{1 + \sqrt{5}}{2} \approx 1.618$ is outside this range and is rejected. The remaining root $\tfrac{1 - \sqrt{5}}{2} \approx -0.618$ is valid.
So $\sin x \approx -0.618$, giving $x \approx 218.2°$ or $x \approx 321.8°$.
After substituting back in a disguised quadratic, always check the natural range of the original expression. $\sin x$ and $\cos x$ must lie in $[-1, 1]$. $e^x$ must be positive. Solutions outside the range are rejected, not carried forward.
Linear inequalities are solved like equations, with one rule: flip the inequality sign when multiplying or dividing both sides by a negative.
Quadratic inequalities. Find the critical values (roots) by setting the expression equal to zero. Sketch the parabola. Read off the region matching the inequality.
For a positive‑leading quadratic, $(x - \alpha)(x - \beta) > 0$ (with $\alpha < \beta$) holds for $x < \alpha$ or $x> \beta$. The other inequality $(x - \alpha)(x - \beta) < 0$ holds for $\alpha < x < \beta$.
Never multiply both sides of an inequality by an expression in $x$ unless you know its sign — the sign might flip the inequality and might not, depending on $x$. Cross‑multiplying $\tfrac{1}{x} < 2$ to get $1 < 2x$ is wrong if $x$ could be negative. Move everything to one side and use a sign diagram.
The interval $a < x < b$ is written $(a, b)$ in interval notation. Round bracket=strict. Square bracket=non‑strict. So $a \le x < b$ is $[a, b)$.
Infinity always takes a round bracket: $x \ge 3$ is $[3, \infty)$.
Set‑builder notation uses curly braces: $\{x : 2 \le x < 3\}$ reads as "the set of $x$ such that $2 \le x < 3$" .
$\cup$ is "or" (union). $\cap$ is "and" (intersection). The empty set is $\emptyset$.
On a graph, shade the accepted region. Use a solid line for $\le$ or $\ge$ (boundary included) and a dotted line for $<$ or $>$ (boundary excluded).
You should be able to expand brackets, collect like terms, factorise, and divide.
For any polynomial $f(x)$:
$$f(a) = 0 \iff (x - a) \text{ is a factor of } f(x).$$
More generally, $f\!\left(\tfrac{b}{a}\right) = 0 \iff (ax - b)$ is a factor.
Polynomial division leaves a quotient and remainder, just like integer division. Write $f(x) = (x - a) q(x) + r$ where $r$ is the remainder (a constant). Substitute $x = a$: $f(a) = r$. So the remainder when $f(x)$ is divided by $(x - a)$ equals $f(a)$. Setting $r = 0$ gives the factor theorem.
You should be able to divide a polynomial by a linear factor (and sometimes by a quadratic). Two methods work:
Factorise $f(x) = x^3 - 4x^2 + x + 6$.
Try small integer values for a root. $f(-1) = -1 - 4 - 1 + 6 = 0$. So $(x + 1)$ is a factor.
Write $x^3 - 4x^2 + x + 6 = (x + 1)(x^2 + Bx + 6)$. Expanding the right gives $x^3 + Bx^2 + 6x + x^2 + Bx + 6 = x^3 + (B+1)x^2 + (B+6)x + 6$. Comparing $x^2$ coefficients: $B + 1 = -4$, so $B = -5$.
So $f(x) = (x + 1)(x^2 - 5x + 6) = (x + 1)(x - 2)(x - 3)$.
If a polynomial with integer coefficients has a rational root $\tfrac{p}{q}$, then $p$ divides the constant term and $q$ divides the leading coefficient. So for $x^3 - 4x^2 + x + 6$, the candidate rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$. Try those first.
Sketching ≠ plotting. A sketch shows the shape and the key features: intercepts with the axes, turning points, asymptotes, and behaviour as $x \to \pm\infty$. It is not drawn to scale.
For a polynomial $f(x)$ given in factorised form:
$y = \dfrac{a}{x}$ has two branches, with the $x$‑axis and $y$‑axis as asymptotes. For $a > 0$, the branches sit in the first and third quadrants; for $a < 0$, the second and fourth.
$y = \dfrac{a}{x^2}$ has two branches both on the same side of the $x$‑axis (above if $a > 0$, below if $a < 0$). Again the axes are asymptotes.
Starting from $y = f(x)$:
Only single transformations are examinable at AS. Combined transformations (like $y = 2f(3x - 1) + 5$) are Year 2 — you'll meet them later.
If asked to describe a transformation, name the type, the direction, and the magnitude. "Translation by the vector $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$" or "stretch parallel to the $y$‑axis, scale factor $3$". Words like "shift" or "moved" cost marks. Mark schemes are pedantic here.
Direct proportion. $y \propto x$ means $y = kx$ for some constant $k$. Graph: straight line through the origin.
Inverse proportion. $y \propto \tfrac{1}{x}$ means $y = \tfrac{k}{x}$. Graph: reciprocal curve.
Other proportional forms. $y \propto x^2$, $y \propto \sqrt{x}$, $y \propto \tfrac{1}{x^2}$ — recognise the graph shape and find $k$ from a single data point.
(i) Express $2x^2 - 8x + 11$ in the form $a(x + p)^2 + q$, stating the values of $a$, $p$, and $q$.
(ii) Hence state the minimum value of $2x^2 - 8x + 11$ and the value of $x$ at which it occurs.
(i) Factor out $2$ from the $x$ terms: $2x^2 - 8x + 11 = 2(x^2 - 4x) + 11 = 2\!\left[(x-2)^2 - 4\right] + 11 = 2(x-2)^2 - 8 + 11 = 2(x-2)^2 + 3$.
So $a = 2$, $p = -2$, $q = 3$.
(ii) The minimum value is $3$, occurring at $x = 2$.
Solve the equation $9^x - 4 \cdot 3^x + 3 = 0$.
Notice $9^x = (3^2)^x = (3^x)^2$. Let $y = 3^x$. Then $y^2 - 4y + 3 = 0$, so $(y-1)(y-3) = 0$ and $y = 1$ or $y = 3$.
Substituting back: $3^x = 1 \Rightarrow x = 0$, and $3^x = 3 \Rightarrow x = 1$.
The substitution $y = 3^x$ requires $y > 0$ throughout (since $3^x$ is always positive). Any negative or zero value of $y$ from the quadratic would be rejected. Always check.
The polynomial $f(x) = x^3 + ax^2 + bx + 18$ has $(x - 3)$ and $(x + 1)$ as factors.
(i) Find the values of $a$ and $b$.
(ii) Fully factorise $f(x)$.
(iii) Hence solve the inequality $f(x) \le 0$, giving your answer in set notation.
(i) By the factor theorem, $f(3) = 0$ and $f(-1) = 0$.
$f(3) = 27 + 9a + 3b + 18 = 0 \Rightarrow 9a + 3b = -45 \Rightarrow 3a + b = -15$.
$f(-1) = -1 + a - b + 18 = 0 \Rightarrow a - b = -17$.
Adding: $4a = -32$, so $a = -8$. Then $b = a + 17 = 9$.
(ii) So $f(x) = x^3 - 8x^2 + 9x + 18$. We know $(x - 3)$ and $(x + 1)$ are factors. The third factor is linear, say $(x - c)$. Comparing constant terms in $(x-3)(x+1)(x-c)$: $(-3)(1)(-c) = 3c = 18$, so $c = 6$.
$$f(x) = (x - 3)(x + 1)(x - 6).$$
(iii) Roots at $x = -1, 3, 6$. The leading coefficient is positive, so the cubic goes from $-\infty$ to $+\infty$. Sign analysis:
| Interval | $x < -1$ | $-1 < x < 3$ | $3 < x < 6$ | $x > 6$ |
|---|---|---|---|---|
| Sign of $f(x)$ | $-$ | $+$ | $-$ | $+$ |
$f(x) \le 0$ on $\{x : x \le -1\} \cup \{x : 3 \le x \le 6\}$.
Three equivalent forms of the equation of a line:
For points $A = (x_1, y_1)$ and $B = (x_2, y_2)$:
For two lines with gradients $m_1$ and $m_2$:
Horizontal lines have gradient $0$; the perpendicular has undefined gradient — it's vertical, with equation $x = $ constant. The rule $m_1 m_2 = -1$ doesn't apply when one gradient is $0$ or undefined. Sketch first; don't blindly multiply.
The equation of a circle with centre $(a, b)$ and radius $r$:
$$(x - a)^2 + (y - b)^2 = r^2.$$
Some questions present the equation expanded: $x^2 + y^2 + Dx + Ey + F = 0$. Convert to standard form by completing the square on the $x$ and $y$ terms separately.
Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$.
Group: $(x^2 - 6x) + (y^2 + 4y) = 12$.
Complete the square: $(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$, so $(x - 3)^2 + (y + 2)^2 = 25$.
Centre $(3, -2)$, radius $5$.
You need these explicitly, by name and by use:
To find the equation of the tangent to a circle at point $P$:
This is a standard exam question. Don't try to differentiate implicitly — that's Year 2 and unnecessary here.
To find where a line and a circle intersect, substitute the line equation into the circle equation. This gives a quadratic in one variable. The discriminant tells you:
The points $A(2, 5)$ and $B(6, 13)$ are given.
(i) Find the equation of the line $AB$ in the form $y = mx + c$.
(ii) Find the equation of the perpendicular to $AB$ that passes through $A$.
(i) Gradient $m = \dfrac{13 - 5}{6 - 2} = \dfrac{8}{4} = 2$. Using $y - 5 = 2(x - 2)$: $y = 2x + 1$.
(ii) Perpendicular gradient $= -\tfrac{1}{2}$. Equation: $y - 5 = -\tfrac{1}{2}(x - 2)$, i.e. $y = -\tfrac{1}{2}x + 6$.
A circle has equation $(x - 3)^2 + (y + 2)^2 = 25$. The line $y = 2x + k$ is a tangent to the circle. Find the two possible values of $k$.
Substitute $y = 2x + k$ into the circle:
$$(x - 3)^2 + (2x + k + 2)^2 = 25.$$
Expand: $x^2 - 6x + 9 + 4x^2 + 4x(k+2) + (k+2)^2 = 25$.
$$5x^2 + \big(4(k+2) - 6\big)x + 9 + (k+2)^2 - 25 = 0$$
$$5x^2 + (4k + 2)x + (k^2 + 4k - 12) = 0.$$
For tangency, the discriminant is zero:
$$(4k+2)^2 - 4(5)(k^2 + 4k - 12) = 0$$
$$16k^2 + 16k + 4 - 20k^2 - 80k + 240 = 0$$
$$-4k^2 - 64k + 244 = 0 \;\Rightarrow\; k^2 + 16k - 61 = 0.$$
By the quadratic formula: $k = \dfrac{-16 \pm \sqrt{256 + 244}}{2} = \dfrac{-16 \pm \sqrt{500}}{2} = -8 \pm 5\sqrt{5}$.
The points $A(1, 2)$ and $B(7, 6)$ are fixed. The point $C(p, q)$ varies, subject to the condition $\angle ACB = 90°$.
(i) Explain why $C$ lies on a circle with $AB$ as a diameter, and find the equation of this circle.
(ii) If, in addition, $C$ lies on the line $y = x - 2$, find all possible coordinates of $C$.
(i) By the converse of the "angle in a semicircle" theorem, if $\angle ACB = 90°$ then $C$ lies on the circle with diameter $AB$.
Centre = midpoint of $AB$ = $(4, 4)$. Radius $= \tfrac{1}{2}|AB| = \tfrac{1}{2}\sqrt{6^2 + 4^2} = \tfrac{1}{2}\sqrt{52} = \sqrt{13}$.
Circle: $(x - 4)^2 + (y - 4)^2 = 13$.
(ii) Substitute $y = x - 2$ into the circle: $(x - 4)^2 + (x - 6)^2 = 13$.
$x^2 - 8x + 16 + x^2 - 12x + 36 = 13 \Rightarrow 2x^2 - 20x + 39 = 0$.
$x = \dfrac{20 \pm \sqrt{400 - 312}}{4} = \dfrac{20 \pm \sqrt{88}}{4} = \dfrac{10 \pm \sqrt{22}}{2} = 5 \pm \tfrac{\sqrt{22}}{2}$.
Corresponding $y$ values: $y = x - 2 = 3 \pm \tfrac{\sqrt{22}}{2}$.
So $C = \left(5 + \tfrac{\sqrt{22}}{2},\, 3 + \tfrac{\sqrt{22}}{2}\right)$ or $C = \left(5 - \tfrac{\sqrt{22}}{2},\, 3 - \tfrac{\sqrt{22}}{2}\right)$.
Only $(a + bx)^n$ for positive integer $n$. The extended binomial expansion for negative or fractional $n$ (with the validity condition $|x| < 1$) is Year 2.
For a positive integer $n$:
$$(a + b)^n = a^n + \binom{n}{1} a^{n-1} b + \binom{n}{2} a^{n-2} b^2 + \cdots + \binom{n}{r} a^{n-r} b^r + \cdots + b^n.$$
The general term is $\binom{n}{r} a^{n-r} b^r$. This is in the formula booklet.
Factorial: $n! = n(n-1)(n-2)\cdots 2 \cdot 1$, with the convention $0! = 1$.
Binomial coefficient: $\displaystyle \binom{n}{r} = \,^n\!C_r = \frac{n!}{r!(n-r)!}$ — the number of ways to choose $r$ objects from $n$ when order doesn't matter. Also the entry in row $n$, position $r$ of Pascal's triangle (counting from $0$).
$(x + y)^4 = (x + y)(x + y)(x + y)(x + y)$. To get a term in $x^3 y$, pick $x$ from three of the four brackets and $y$ from the remaining one. The number of ways to choose which bracket contributes $y$ is $\binom{4}{1} = 4$. So the coefficient of $x^3 y$ is $4$. Generalising gives the formula.
Find the coefficient of $x^3$ in the expansion of $(2 + 3x)^7$.
The general term is $\binom{7}{r}(2)^{7-r}(3x)^r$. For $x^3$, take $r = 3$:
$$\binom{7}{3}(2)^4 (3x)^3 = 35 \cdot 16 \cdot 27 \, x^3 = 15120 \, x^3.$$
Coefficient: $15120$.
For $(a + bx)^n$, students routinely lose marks by forgetting to raise $b$ and $x$ to the appropriate power. The general term has $(bx)^r = b^r x^r$ — both the constant $b$ and the variable $x$ are raised to the power $r$.
Find the first four terms, in ascending powers of $x$, of the expansion of $(1 + 2x)^7$.
General term: $\binom{7}{r}(1)^{7-r}(2x)^r = \binom{7}{r} 2^r x^r$.
$r = 0$: $\binom{7}{0} \cdot 1 = 1$.
$r = 1$: $\binom{7}{1} \cdot 2 = 14$.
$r = 2$: $\binom{7}{2} \cdot 4 = 21 \cdot 4 = 84$.
$r = 3$: $\binom{7}{3} \cdot 8 = 35 \cdot 8 = 280$.
So $(1 + 2x)^7 = 1 + 14x + 84x^2 + 280x^3 + \cdots$.
In the expansion of $(1 + ax)^n$ in ascending powers of $x$, the first three terms are $1 + 24x + 240x^2$. Find the values of $a$ and $n$.
The expansion begins $1 + nax + \binom{n}{2} a^2 x^2 + \cdots$. Comparing:
$$na = 24, \qquad \tfrac{n(n-1)}{2} a^2 = 240 \;\Rightarrow\; n(n-1)a^2 = 480.$$
From the first, $a = \dfrac{24}{n}$. Substitute:
$$n(n-1) \cdot \frac{576}{n^2} = 480 \;\Rightarrow\; \frac{576(n-1)}{n} = 480.$$
$576(n - 1) = 480 n \Rightarrow 96n = 576 \Rightarrow n = 6$.
Then $a = 24/6 = 4$.
Trap to avoid: students often try to solve for $a$ first, treating $n$ as known. The two equations have to be handled simultaneously.
In the expansion of $(1 + ax)(2 + bx)^4$ in ascending powers of $x$, the coefficient of $x^2$ is $-40$ and the coefficient of $x^3$ is $40$. Given that $a > 0$, find the values of $a$ and $b$.
First expand $(2 + bx)^4$:
$$(2 + bx)^4 = 16 + 32 b x + 24 b^2 x^2 + 8 b^3 x^3 + b^4 x^4.$$
Now multiply by $(1 + ax)$ and read off coefficients:
Coefficient of $x^2$: from $1 \cdot 24 b^2 + ax \cdot 32 bx$, giving $24 b^2 + 32 ab = -40$. Divide by $8$: $3 b^2 + 4 a b = -5 \quad (\star)$.
Coefficient of $x^3$: from $1 \cdot 8 b^3 + ax \cdot 24 b^2 x^2$, giving $8 b^3 + 24 a b^2 = 40$. Divide by $8$: $b^3 + 3 a b^2 = 5 \quad (\star\star)$.
From $(\star)$: $a = \dfrac{-5 - 3b^2}{4b}$ (assuming $b \neq 0$). Substitute into $(\star\star)$:
$$b^3 + 3 b^2 \cdot \frac{-5 - 3b^2}{4b} = 5 \;\Rightarrow\; 4 b^3 + 3 b(-5 - 3b^2) = 20.$$
Expanding: $4 b^3 - 15 b - 9 b^3 = 20 \Rightarrow -5 b^3 - 15 b - 20 = 0 \Rightarrow b^3 + 3 b + 4 = 0$.
Factorise: $(b + 1)(b^2 - b + 4) = 0$. The quadratic factor has discriminant $1 - 16 < 0$, so no real roots. Hence $b=-1$.
Then $a = \dfrac{-5 - 3}{-4} = 2$. Since $a > 0$, this is consistent.
So $a = 2$, $b = -1$.
Degrees only. The trig identities are just $\sin^2 x + \cos^2 x \equiv 1$ and $\tan x \equiv \tfrac{\sin x}{\cos x}$. Radians, small‑angle approximations, $\sec/\csc/\cot$, the compound‑angle and double‑angle formulae, and the $R\sin(x+\alpha)$ form are all Year 2.
For a triangle with sides $a, b, c$ opposite to angles $A, B, C$:
Sine rule:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$
Use it when you know an angle and its opposite side, plus one other piece of information.
Cosine rule:
$$a^2 = b^2 + c^2 - 2bc \cos A.$$
Use it when you know either two sides and the included angle (finding the third side), or all three sides (finding an angle). Rearranged for an angle:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}.$$
Area of a triangle:
$$\text{Area} = \tfrac{1}{2} ab \sin C,$$
where $C$ is the angle included between sides $a$ and $b$.
The sine rule can give two valid triangles. If you compute $\sin B = 0.7$, then $B$ could be $44.4°$ or $135.6°$. Both might be geometrically valid given the data. Always check whether the obtuse possibility is ruled out by the other angles summing to less than $180°$. If both work, give both triangles.
For acute angles, $\sin, \cos, \tan$ come from right‑angled triangles (SOHCAHTOA). For arbitrary angles, define them by the coordinates of a point on the unit circle, rotated through angle $\theta$ from the positive $x$‑axis.
This gives:
Mnemonic: CAST, read anticlockwise from the fourth quadrant — Cos, All, Sin, Tan are positive in their respective quadrants.
Memorise:
| $x$ | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
| $\sin x$ | $0$ | $\tfrac{1}{2}$ | $\tfrac{\sqrt{2}}{2}$ | $\tfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos x$ | $1$ | $\tfrac{\sqrt{3}}{2}$ | $\tfrac{\sqrt{2}}{2}$ | $\tfrac{1}{2}$ | $0$ |
| $\tan x$ | $0$ | $\tfrac{\sqrt{3}}{3}$ | $1$ | $\sqrt{3}$ | undef. |
Plus: $\sin 180° = 0$, $\cos 180° = -1$, $\tan 180° = 0$, etc., by symmetry of the graphs.
"Give your answer in exact form" on a trig question means you should leave $\sqrt{2}, \sqrt{3}$ etc. unevaluated. A decimal answer can lose marks even if numerically correct.
Only two for Year 1:
$$\sin^2 x + \cos^2 x \equiv 1, \qquad \tan x \equiv \frac{\sin x}{\cos x}.$$
From SOHCAHTOA: $\sin x = \tfrac{o}{h}$, $\cos x = \tfrac{a}{h}$, so $\tan x = \tfrac{o}{a} = \tfrac{\sin x}{\cos x}$.
From Pythagoras: $o^2 + a^2 = h^2$, dividing by $h^2$ gives $\sin^2 x + \cos^2 x = 1$.
The question always specifies an interval for the solution, like $0° \le x < 360°$.
Procedure for $\sin x = k$ (and similarly $\cos, \tan$):
For $\sin x = k$ in $[0°, 360°)$: solutions are at $x = \sin^{-1}(k)$ and $x = 180° - \sin^{-1}(k)$.
For $\cos x = k$: solutions are at $x = \cos^{-1}(k)$ and $x = 360° - \cos^{-1}(k)$.
For $\tan x = k$: solutions repeat every $180°$, so add $180°$ to the principal value.
An equation like $6\sin^2 x + \cos x - 4 = 0$ involves two different trig functions. Use the identity $\sin^2 x = 1 - \cos^2 x$ to write everything in terms of one function:
$$6(1 - \cos^2 x) + \cos x - 4 = 0 \;\Rightarrow\; -6\cos^2 x + \cos x + 2 = 0 \;\Rightarrow\; 6\cos^2 x - \cos x - 2 = 0.$$
Now this is quadratic in $\cos x$. Solve for $\cos x$, then for $x$.
For $\tan 3x = -1$ in $0° \le x < 360°$:
When the unknown is multiplied by a constant ($\sin 2x$, $\cos \tfrac{x}{2}$, $\tan 3x$), always change the interval first to match. Otherwise you'll miss solutions or include spurious ones. Find all solutions in the new interval, then divide (or multiply) at the end.
Solve $2\sin x + 1 = 0$ for $0° \le x < 360°$.
$\sin x = -\tfrac{1}{2}$. Principal value (calculator): $x = -30°$.
$\sin$ is negative in quadrants 3 and 4, so in $[0°, 360°)$: $x = 180° + 30° = 210°$ and $x = 360° - 30° = 330°$.
In triangle $ABC$, side $a = 12$ cm, side $b = 16$ cm, and angle $A = 40°$. Find the possible values of angle $B$ to one decimal place, and for each, find the corresponding length of side $c$.
Sine rule: $\dfrac{\sin B}{16} = \dfrac{\sin 40°}{12}$, so $\sin B = \dfrac{16 \sin 40°}{12} \approx 0.8571$.
So $B \approx 59.0°$ or $B \approx 180° - 59.0° = 121.0°$. Both must be tested against the constraint $A + B < 180°$:
Both triangles are geometrically valid. The mark scheme expects both to be stated.
The trap: most students only give the acute value $B = 59.0°$ and miss two marks. Sketching both triangles makes the ambiguity obvious.
Solve the equation $6\cos^2 x + 7\sin x - 8 = 0$ for $0° \le x < 360°$. Give exact answers where possible, otherwise to 1 decimal place.
Use $\cos^2 x = 1 - \sin^2 x$:
$$6(1 - \sin^2 x) + 7 \sin x - 8 = 0 \;\Rightarrow\; -6\sin^2 x + 7\sin x - 2 = 0 \;\Rightarrow\; 6\sin^2 x - 7\sin x + 2 = 0.$$
Let $y = \sin x$. Factorise $6y^2 - 7y + 2 = (3y - 2)(2y - 1) = 0$, so $y = \tfrac{2}{3}$ or $y = \tfrac{1}{2}$.
$\sin x = \tfrac{1}{2}$: $x = 30°$ or $x = 150°$ (exact).
$\sin x = \tfrac{2}{3}$: $x = \sin^{-1}(\tfrac{2}{3}) \approx 41.8°$ or $x = 180° - 41.8° = 138.2°$.
All four solutions: $x = 30°,\, 41.8°,\, 138.2°,\, 150°$.
For positive $a$, the graph of $y = a^x$:
The special base $e \approx 2.71828\ldots$ gives the function $y = e^x$, distinguished by:
$$\frac{d}{dx} e^{kx} = k e^{kx}.$$
That is: the gradient of $e^{kx}$ at any point is $k$ times the value there. This is why $e^x$ models any quantity whose rate of change is proportional to itself — radioactive decay, population growth under unlimited resources, Newton's law of cooling, continuously compounded interest.
$\log_a x$ is defined as the inverse of $a^x$ (for $a > 0$, $a \neq 1$, $x > 0$):
$$y = \log_a x \iff a^y = x.$$
So $\log_a (a^x) = x$ and $a^{\log_a x} = x$.
Two consequences from the definition:
The natural logarithm is $\ln x = \log_e x$. Graph of $\ln x$ is the reflection of $e^x$ in the line $y = x$.
For positive $x, y$ and a valid base $a$:
$$\log_a x + \log_a y = \log_a(xy)$$
$$\log_a x - \log_a y = \log_a\!\left(\frac{x}{y}\right)$$
$$\log_a(x^k) = k \log_a x$$
Let $A = \log_a x$ and $B = \log_a y$. Then $a^A = x$ and $a^B = y$. So $xy = a^A \cdot a^B = a^{A+B}$. Taking $\log_a$ of both sides: $\log_a(xy) = A + B = \log_a x + \log_a y$. ∎
For $a^x = b$, take logarithms of both sides:
$$x \log a = \log b \;\Rightarrow\; x = \frac{\log b}{\log a}.$$
Any base will do. The natural log is often cleanest for equations involving $e$.
Solve $2^{x+1} = 5^x$.
Take natural logs: $(x + 1) \ln 2 = x \ln 5$, so $x \ln 2 + \ln 2 = x \ln 5$. Rearrange:
$$x(\ln 5 - \ln 2) = \ln 2 \;\Rightarrow\; x = \frac{\ln 2}{\ln 5 - \ln 2} \approx 0.756.$$
Many real‑world relationships are of the form $y = ax^n$ or $y = kb^x$. Taking logs converts them to linear form in transformed variables — useful for fitting a model to data plotted on log axes.
Power model $y = ax^n$. Take logs:
$$\log y = \log a + n \log x.$$
So a graph of $\log y$ against $\log x$ is a straight line, gradient $n$, intercept $\log a$.
Exponential model $y = kb^x$. Take logs:
$$\log y = \log k + x \log b.$$
So a graph of $\log y$ against $x$ is a straight line, gradient $\log b$, intercept $\log k$.
If a question gives a table of values and the graph appears curved, try plotting:
Then read $a, n$ (or $k, b$) from the gradient and intercept.
The general form is $P = P_0 e^{kt}$ (or $P = P_0 b^t$):
Always check: state the assumptions ("the growth rate is constant", "no external factors limit growth"), and comment on limitations ("the model predicts unbounded growth, which is unrealistic in practice for a finite resource").
Solve $\log_2(x + 3) + \log_2(x - 4) = 3$.
Combine using the product law: $\log_2\big((x+3)(x-4)\big) = 3$, so $(x+3)(x-4) = 2^3 = 8$.
Expand: $x^2 - x - 12 = 8 \Rightarrow x^2 - x - 20 = 0 \Rightarrow (x - 5)(x + 4) = 0$.
So $x = 5$ or $x = -4$. Check both in the original equation: $\log_2$ requires its argument to be positive. For $x = -4$: $\log_2(-1)$ is undefined — reject.
So $x = 5$.
Trap: forgetting to check the domain of the logarithm. Quadratics from log equations frequently have one valid root and one extraneous root.
The mass $M$ grams of a radioactive isotope is modelled by $M = M_0 e^{-kt}$, where $t$ is the time in days and $M_0$ is the initial mass. After 5 days, the mass has fallen to $60\%$ of its initial value.
(i) Find the value of $k$, giving your answer to 3 decimal places.
(ii) Find the half‑life of the isotope.
(i) At $t = 5$: $0.6 M_0 = M_0 e^{-5k}$, so $e^{-5k} = 0.6$.
$-5k = \ln 0.6 \Rightarrow k = -\tfrac{1}{5}\ln 0.6 = \tfrac{1}{5}\ln\tfrac{5}{3} \approx 0.102$.
(ii) Half‑life $T$ satisfies $0.5 M_0 = M_0 e^{-kT}$:
$$T = \frac{\ln 2}{k} = \frac{\ln 2}{\tfrac{1}{5}\ln\tfrac{5}{3}} = \frac{5 \ln 2}{\ln 5 - \ln 3} \approx 6.78 \text{ days.}$$
A scientist suspects two variables $x$ and $y$ are related either by $y = a x^n$ or by $y = k b^x$. She plots two graphs of her experimental data:
(i) State, with reasoning, which model fits the data.
(ii) Find the relationship between $y$ and $x$ in the form $y = ax^n$, giving exact values for $a$ and $n$.
(i) Taking logs of $y = ax^n$ gives $\ln y = \ln a + n \ln x$, which is linear in $\ln x$. Taking logs of $y = k b^x$ gives $\ln y = \ln k + x \ln b$, which is linear in $x$. Since the plot of $\ln y$ vs $\ln x$ (Graph A) is straight but the plot of $\ln y$ vs $x$ (Graph B) is not, the data fits the power model $y = ax^n$.
(ii) From Graph A: $n$ is the gradient, so $n = -3$. The line passes through $(\ln 2, 5)$:
$$5 = \ln a + (-3) \ln 2 \;\Rightarrow\; \ln a = 5 + 3\ln 2 = 5 + \ln 8.$$
So $a = e^{5 + \ln 8} = 8 e^5$.
Therefore $y = 8 e^5 \cdot x^{-3}$, or equivalently $y = \dfrac{8 e^5}{x^3}$.
Differentiate $x^n$ for rational $n$, together with sums, differences, and constant multiples. First principles is restricted to positive integer powers. Differentiating $\sin x, \cos x, e^x, \ln x$ and the product, quotient, chain rules are all Year 2.
The derivative of $f(x)$ at a point gives:
Notation: $\dfrac{dy}{dx}$ (Leibniz) and $f'(x)$ (Lagrange) mean the same thing.
The formal definition:
$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$$
This is in the formula booklet.
Differentiate $f(x) = x^3$ from first principles.
$$f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}.$$
Expand: $(x + h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3$. So:
$$f'(x) = \lim_{h \to 0} \frac{3x^2 h + 3x h^2 + h^3}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2.$$
First principles questions ask for positive integer powers of $x$ (typically $x^2$ or $x^3$). The mark scheme wants to see:
Skipping any of these typically costs marks even if the final answer is correct.
For any rational $n$:
$$\frac{d}{dx}(x^n) = n x^{n-1}.$$
This applies to all rational exponents — fractional powers like $x^{1/2}$ (rewrite as $\sqrt{x}$), and negative powers like $x^{-1}$ (rewrite as $\tfrac{1}{x}$). Rewrite first, then differentiate.
Linearity: the derivative of a sum is the sum of derivatives, and a constant multiple comes outside:
$$\frac{d}{dx}\big(af(x) + bg(x)\big) = a f'(x) + b g'(x).$$
At the point $(x_1, y_1)$ on a curve:
Equation of either line: $y - y_1 = m(x - x_1)$.
A stationary point is where $\dfrac{dy}{dx} = 0$. To classify:
A function is increasing on an interval where $\dfrac{dy}{dx} \ge 0$ and decreasing where $\dfrac{dy}{dx} \le 0$. ("Strictly" replaces $\ge$ with $>$.)
To find the intervals: differentiate, set $\dfrac{dy}{dx} = 0$ to locate the stationary points, then check signs of $\dfrac{dy}{dx}$ in each region.
$\dfrac{d^2 y}{dx^2}$ is the rate of change of the gradient. Geometrically:
Given $y = 3x^4 - \dfrac{2}{x^2} + 5\sqrt{x}$, find $\dfrac{dy}{dx}$.
Rewrite using index notation: $y = 3x^4 - 2x^{-2} + 5 x^{1/2}$.
Apply the power rule term by term:
$$\frac{dy}{dx} = 12 x^3 + 4 x^{-3} + \tfrac{5}{2} x^{-1/2} = 12 x^3 + \frac{4}{x^3} + \frac{5}{2\sqrt{x}}.$$
The curve $C$ has equation $y = x^3 - 4x + 2$.
(i) Find the equation of the tangent to $C$ at the point $P$ where $x = 2$.
(ii) The tangent meets the curve $C$ again at point $Q$. Find the coordinates of $Q$.
(i) At $x = 2$: $y = 8 - 8 + 2 = 2$, so $P = (2, 2)$. Then $\dfrac{dy}{dx} = 3x^2 - 4$, giving gradient $= 12 - 4 = 8$.
Tangent: $y - 2 = 8(x - 2)$, i.e. $y = 8x - 14$.
(ii) Set tangent equal to curve: $x^3 - 4x + 2 = 8x - 14 \Rightarrow x^3 - 12x + 16 = 0$.
Since the tangent touches the curve at $x = 2$, $(x - 2)$ is a repeated factor of this cubic. Factorise:
$$x^3 - 12x + 16 = (x - 2)^2 (x + 4).$$
(Check by expansion.) So the other root is $x = -4$. Substitute into the tangent: $y = 8(-4) - 14 = -46$. So $Q = (-4, -46)$.
Key idea: a tangent touches the curve with multiplicity 2 at the point of tangency, so $(x - 2)^2$ must factorise from the difference between curve and tangent. This guarantees a clean third root.
A closed rectangular box has a square base of side $x$ cm and height $h$ cm. The total surface area is $300$ cm$^2$.
(i) Show that $h = \dfrac{300 - 2x^2}{4x}$.
(ii) Show that the volume $V$ of the box is given by $V = 75x - \tfrac{1}{2} x^3$.
(iii) Find the value of $x$ that maximises $V$, and find this maximum volume, giving exact answers.
(i) Surface area $= 2x^2 + 4xh = 300$, so $4xh = 300 - 2x^2$, giving $h = \dfrac{300 - 2x^2}{4x}$.
(ii) $V = x^2 h = x^2 \cdot \dfrac{300 - 2x^2}{4x} = \dfrac{x(300 - 2x^2)}{4} = \dfrac{300x - 2x^3}{4} = 75x - \tfrac{1}{2}x^3$.
(iii) $\dfrac{dV}{dx} = 75 - \tfrac{3}{2}x^2$. Set $= 0$: $x^2 = 50$, so $x = 5\sqrt{2}$ (positive root).
Second derivative: $\dfrac{d^2 V}{dx^2} = -3x < 0$ for $x> 0$. So this is a maximum.
Maximum volume:
$$V = 75(5\sqrt{2}) - \tfrac{1}{2}(5\sqrt{2})^3 = 375\sqrt{2} - \tfrac{1}{2}(250\sqrt{2}) = 375\sqrt{2} - 125\sqrt{2} = 250\sqrt{2} \text{ cm}^3.$$
Integrate $x^n$ for $n \neq -1$, plus sums, differences, and constant multiples. Definite integrals and areas. Integration by substitution, by parts, of $\sin x, \cos x, e^x, \tfrac{1}{x}$, and differential equations are all Year 2.
Integration is the reverse of differentiation. If $F'(x) = f(x)$, then:
$$\int f(x)\, dx = F(x) + c.$$
The "$+ c$" is the constant of integration — because the derivative of a constant is zero, there's a whole family of functions with the same derivative.
For $n \neq -1$:
$$\int x^n \, dx = \frac{x^{n+1}}{n + 1} + c.$$
Add one to the power, divide by the new power.
Forgetting $+ c$ on an indefinite integral costs marks every single time, on every question. The mark scheme is non‑negotiable on this. Train yourself to write it as soon as you write the integral sign.
If you're told the curve passes through a specific point, substitute that point's coordinates after integrating to determine $c$.
A curve has $\dfrac{dy}{dx} = 3x^2 - 4x + 1$ and passes through $(1, 5)$. Find the equation of the curve.
Integrate: $y = x^3 - 2x^2 + x + c$. Substitute $(1, 5)$:
$$5 = 1 - 2 + 1 + c \;\Rightarrow\; c = 5.$$
So $y = x^3 - 2x^2 + x + 5$.
$$\int_a^b f(x)\, dx = F(b) - F(a),$$
where $F$ is any antiderivative of $f$. No constant of integration appears, because it cancels.
Geometrically, $\int_a^b f(x)\, dx$ is the signed area between the curve $y = f(x)$ and the $x$‑axis from $x = a$ to $x = b$:
If a question asks for the area bounded by a curve and the $x$‑axis (not the value of an integral), and the curve crosses the axis in the interval:
If you blindly integrate from $a$ to $b$ across a sign change, the answer is the net signed area — positive bits minus negative bits — which is not usually what "find the area" means. Find the roots first.
The area between $y = f(x)$ and $y = g(x)$, with $f(x) \ge g(x)$ on $[a, b]$:
$$\int_a^b \big(f(x) - g(x)\big)\, dx.$$
Always subtract the lower curve from the upper. If they cross, split the interval.
(i) Find $\displaystyle \int (3x^2 - 4x + 5)\, dx$.
(ii) Hence evaluate $\displaystyle \int_0^2 (3x^2 - 4x + 5)\, dx$.
(i) $\displaystyle \int (3x^2 - 4x + 5)\, dx = x^3 - 2x^2 + 5x + c$.
(ii) $\Big[x^3 - 2x^2 + 5x\Big]_0^2 = (8 - 8 + 10) - 0 = 10$.
Find the total area enclosed between the curve $y = x^3 - 4x$ and the $x$‑axis between $x = -2$ and $x = 2$.
Roots: $x^3 - 4x = x(x^2 - 4) = 0$ gives $x = -2, 0, 2$. Check the sign on each sub‑interval:
$\displaystyle \int_{-2}^0 (x^3 - 4x)\, dx = \Big[\tfrac{x^4}{4} - 2x^2\Big]_{-2}^0 = 0 - (4 - 8) = 4$.
$\displaystyle \int_0^2 (x^3 - 4x)\, dx = \Big[\tfrac{x^4}{4} - 2x^2\Big]_0^2 = (4 - 8) - 0 = -4$.
Total area $= |4| + |-4| = 8$.
Trap: $\displaystyle\int_{-2}^2 (x^3 - 4x)\, dx = 0$ (by symmetry — odd function), but the area is $8$, not $0$. Always split at roots when "area" is asked for.
The curves $y = 6x - x^2$ and $y = x^2 - 2x$ intersect at two points. Find the exact area of the region enclosed between them.
Find intersections: $6x - x^2 = x^2 - 2x \Rightarrow 2x^2 - 8x = 0 \Rightarrow 2x(x - 4) = 0$. So $x = 0$ and $x = 4$.
Which curve is on top? At $x = 2$: $6(2) - 4 = 8$ vs $4 - 4 = 0$. So $y = 6x - x^2$ is above on $(0, 4)$.
Area $= \displaystyle \int_0^4 \big[(6x - x^2) - (x^2 - 2x)\big]\, dx = \int_0^4 (8x - 2x^2)\, dx$.
$$= \Big[4x^2 - \tfrac{2}{3}x^3\Big]_0^4 = 64 - \tfrac{128}{3} = \tfrac{192 - 128}{3} = \tfrac{64}{3}.$$
Two‑dimensional vectors only. Three‑dimensional vectors (and the corresponding distance formula) are Year 2.
A scalar has magnitude only (mass, time, temperature, speed).
A vector has both magnitude and direction (displacement, velocity, force).
Notation in 2D — three equivalent forms:
$$\mathbf{a} = x\mathbf{i} + y\mathbf{j} \quad=\quad \begin{pmatrix} x \\ y \end{pmatrix} \quad=\quad \overrightarrow{AB} \text{ from origin to } (x, y).$$
$\mathbf{i}$ and $\mathbf{j}$ are unit vectors along the $x$‑ and $y$‑axes. In print, vectors are bold; in handwriting, underline: $\underline{a}$.
For $\mathbf{a} = x\mathbf{i} + y\mathbf{j}$:
$$|\mathbf{a}| = \sqrt{x^2 + y^2} \quad \text{(Pythagoras).}$$
The direction is the angle measured anticlockwise from the positive $x$‑axis, in $[0°, 360°)$. Compute via $\tan^{-1}(y/x)$, but check the quadrant:
Always sketch the vector first. The calculator's $\tan^{-1}$ only returns values in $(-90°, 90°)$, which is just one of two possible quadrants. The sketch resolves the ambiguity in seconds.
Addition by components, or diagrammatically tip‑to‑tail:
$$\begin{pmatrix} a \\ b \end{pmatrix} + \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} a + c \\ b + d \end{pmatrix}.$$
Subtraction: $\mathbf{a} - \mathbf{b}$ is $\mathbf{a}$ plus the negative of $\mathbf{b}$.
Scalar multiplication scales the magnitude (and reverses direction if the scalar is negative):
$$k \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} ka \\ kb \end{pmatrix}.$$
The position vector of point $P$ is $\overrightarrow{OP}$ — written $\mathbf{p}$.
The displacement from $A$ to $B$ is:
$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}.$$
(Memorise: "to minus from".)
The distance between points $A$ and $B$ is the magnitude of the displacement:
$$|AB| = |\mathbf{b} - \mathbf{a}| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}.$$
Forces are vectors. You can add them (resultant force), test for equilibrium (resultant is zero), and resolve them into horizontal and vertical components using a sketch and basic trigonometry. See the mechanics section for the worked physics.
Given $\mathbf{a} = 3\mathbf{i} - 4\mathbf{j}$ and $\mathbf{b} = -2\mathbf{i} + \mathbf{j}$:
(i) Find $|\mathbf{a}|$.
(ii) Find $\mathbf{a} + 2\mathbf{b}$ and its magnitude.
(iii) Find a unit vector in the direction of $\mathbf{a}$.
(i) $|\mathbf{a}| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$.
(ii) $\mathbf{a} + 2\mathbf{b} = (3 - 4)\mathbf{i} + (-4 + 2)\mathbf{j} = -\mathbf{i} - 2\mathbf{j}$. Magnitude: $\sqrt{1 + 4} = \sqrt{5}$.
(iii) $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|} = \tfrac{1}{5}(3\mathbf{i} - 4\mathbf{j}) = \tfrac{3}{5}\mathbf{i} - \tfrac{4}{5}\mathbf{j}$.
The points $A$, $B$, $C$ have position vectors $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$, $\begin{pmatrix} 5 \\ 7 \end{pmatrix}$, $\begin{pmatrix} 11 \\ 15 \end{pmatrix}$ respectively.
(i) Find $\overrightarrow{AB}$ and $\overrightarrow{BC}$.
(ii) Show that $A$, $B$, $C$ are collinear and find the ratio $AB : BC$.
(iii) Find the position vector of the point $D$ such that $ABCD$ (in this order) is a parallelogram.
(i) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$. $\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$.
(ii) $\overrightarrow{BC} = 2\overrightarrow{AB}$, so $\overrightarrow{BC}$ is parallel to $\overrightarrow{AB}$. Since they share point $B$, the three points are collinear. $|AB| = 5$, $|BC| = 10$, ratio $AB:BC = 1:2$.
(iii) Trap: since $A, B, C$ are collinear, $ABCD$ cannot be a (non‑degenerate) parallelogram — three collinear vertices means $D$ would have to lie on the same line, giving a degenerate (zero‑area) figure. The question has no valid answer; recognising this is the point.
Always check the geometric setup before grinding through algebra.
$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. The point $M$ is the midpoint of $AB$. The line $OM$ extended meets the line $BC$ extended at the point $P$.
(i) Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.
(ii) Hence find $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.
(i) Since $OABC$ is a parallelogram, $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$. $M$ is the midpoint of $AB$:
$$\overrightarrow{OM} = \tfrac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB}) = \tfrac{1}{2}(\mathbf{a} + \mathbf{a} + \mathbf{c}) = \mathbf{a} + \tfrac{1}{2}\mathbf{c}.$$
(ii) Parametrise line $OM$: $\overrightarrow{OP} = t\,\overrightarrow{OM} = t\mathbf{a} + \tfrac{t}{2}\mathbf{c}$ for some scalar $t$.
Parametrise line $BC$: starting at $B = \mathbf{a} + \mathbf{c}$ with direction $\overrightarrow{BC} = -\mathbf{a}$, giving $\overrightarrow{OP} = (\mathbf{a} + \mathbf{c}) + s(-\mathbf{a}) = (1 - s)\mathbf{a} + \mathbf{c}$.
Equate coefficients of $\mathbf{a}$ and $\mathbf{c}$ (linearly independent):
$$t = 1 - s, \qquad \tfrac{t}{2} = 1.$$
The second equation gives $t = 2$. Then $s = -1$ (so $P$ is the reflection of $B$ in $C$ along the line $BC$).
$$\overrightarrow{OP} = 2\mathbf{a} + \mathbf{c}.$$
The population is the whole set of items you want to make a conclusion about. A sample is a subset of the population, used as the basis for that conclusion. Different samples can give different conclusions about the same population.
A census records data from every member of the population. Often impossible, expensive, or destructive (testing every battery means draining every battery).
Simple random sampling. Assign each member of the population a unique number; use a random number generator to pick a set; sample those members.
Opportunity sampling. Sample whoever is conveniently available — the first people who walk past, the patients in a single clinic.
Systematic sampling. Sample every $k$th member of an ordered list.
Stratified sampling. Divide the population into strata (e.g. by age, gender, location); take a random sample within each stratum, with sample sizes proportional to the size of the stratum.
Cluster sampling. Split the population into naturally occurring clusters (e.g. schools, neighbourhoods). Randomly select some clusters; sample everyone in them.
Quota sampling. Choose a target number from each category and then sample (often opportunistically) until the quotas are met.
You only need to be able to carry out simple random sampling and opportunity sampling. For the others, recognise them, give advantages and disadvantages, and critique their use in context.
Describe how to take a simple random sample of $30$ students from a school of $600$.
Method:
A researcher wants to investigate the average commuting time for workers in a large city. He stands outside the central train station between 8 am and 9 am on a Tuesday and asks the first $100$ people who emerge.
(i) Identify the sampling method used.
(ii) State two specific reasons why this method may give a biased estimate of the city's average commuting time.
(i) Opportunity sampling.
(ii) Two of the following (any two clearly distinct points):
OCR trap: generic answers like "the sample isn't random" lose marks. The mark scheme rewards specific, contextual reasons.
A school has $1000$ students: $400$ in Years 7–9, $350$ in Years 10–11, and $250$ in Years 12–13. The head wishes to take a sample of size $50$, using stratified sampling by year group.
(i) Calculate the number of students to be sampled from each stratum. Identify and resolve any practical difficulty.
(ii) Within Years 12–13, the students are split: $100$ at Site A, $80$ at Site B, and $70$ at Site C. The head wonders whether to take a simple random sample within Years 12–13, or to further stratify by site. Give one specific advantage and one specific disadvantage of further stratifying by site.
(i) Proportional allocation:
Practical difficulty: two of these are not whole numbers. The head must round; one common rule is to round half-integers in opposite directions so the total stays at $50$ (e.g. $18$ and $12$, or $17$ and $13$). Whatever convention is used, it should be stated.
(ii) Advantage: stratifying by site guarantees representation from all three sites in proportion to their size. A simple random sample of only $12$ or $13$ from Years 12–13 could by chance under-represent or even miss an entire site.
Disadvantage: with only $12$–$13$ students from Years 12–13, the sub-strata would be very small (about $5$, $4$, $3$ students). Further stratification adds administrative complexity for very little additional precision; smaller strata are also harder to randomise from cleanly.
You should be able to interpret each of these. Drawing is rarely required in the exam, but you should recognise the conventions.
If the question gives class widths that differ, the height of each bar is not the frequency. It is frequency density. Confusing height with frequency is the single most common mistake on histogram questions.
A scatter diagram shows pairs of observations. Look for:
A regression line is the line of best fit. You only need to interpret one at Year 1, not compute one.
Strong correlation between two variables does not mean one causes the other. Both could be caused by a third variable; the relationship might be coincidence. Mark schemes accept "correlation does not imply causation" as the standard caveat — say it.
Mean: $\bar{x} = \dfrac{\sum x}{n}$. Sensitive to outliers.
Median: the middle value when the data is in order. Resistant to outliers.
Mode: the most frequent value. There may be no mode, or more than one.
Quartiles ($Q_1$, $Q_2 = $ median, $Q_3$) divide ordered data into four equal parts. The interquartile range is $\text{IQR} = Q_3 - Q_1$, a measure of spread resistant to outliers.
Percentiles: the $k$th percentile is the value at the $k\%$ position in the ordered data. The median is the $50$th percentile; quartiles are the $25$th and $75$th.
Variance:
$$\sigma^2 = \frac{\sum(x - \bar{x})^2}{n} = \frac{\sum x^2}{n} - \bar{x}^2.$$
Standard deviation: $\sigma = \sqrt{\sigma^2}$. Has the same units as the data — easier to interpret than variance.
For grouped data, treat each data point as the class midpoint and multiply by the frequency. This gives an estimate of the mean and standard deviation.
Use your calculator's statistics mode for mean and standard deviation. Enter the values once, then read off $\bar{x}$ and $\sigma_x$ directly. Computing by hand wastes time and invites arithmetic slips.
Standard deviation is the root‑mean‑square deviation from the mean. We can't average the raw deviations $(x - \bar{x})$ because they sum to zero by definition. We could take absolute values, but $|x - \bar{x}|$ is awkward to work with algebraically. Squaring gives a positive quantity that's calculus‑friendly. The square root at the end puts the result back in the original units.
When asked to compare two distributions, comment on:
Always frame the comparison in context. "Group A has a higher median test score than Group B" beats "the median of A is bigger".
Two definitions of outlier are on the spec:
These can disagree. If the question doesn't specify, use the relevant one for context (the IQR rule alongside box plots, the SD rule alongside means).
An identified outlier might be:
Never silently delete outliers. Always state what you've done and why.
The number of pets owned by $8$ families is: $0, 1, 2, 1, 3, 0, 2, 1$. Find the mean and the median.
Mean: sum $= 0 + 1 + 2 + 1 + 3 + 0 + 2 + 1 = 10$. So $\bar{x} = \tfrac{10}{8} = 1.25$.
Median: sort the data — $0, 0, 1, 1, 1, 2, 2, 3$. With $n = 8$ values, the median is the average of the 4th and 5th: $\tfrac{1 + 1}{2} = 1$.
The heights (in cm) of $40$ plants are summarised below.
| Height $h$ (cm) | Frequency |
|---|---|
| $0 \le h < 10$ | 6 |
| $10 \le h < 15$ | 14 |
| $15 \le h < 25$ | 12 |
| $25 \le h < 50$ | 8 |
(i) Find the frequency density for each class.
(ii) On a histogram of this data, the bar for $10 \le h < 15$ has height $2.8$ cm. Find the height of the bar for $25 \le h < 50$.
(i) Frequency density = frequency ÷ class width:
(ii) On the histogram, the bar for $10 \le h < 15$ has height $2.8$ cm and represents a frequency density of $2.8$. So the scale is $1$ cm per unit of frequency density. The bar for $25 \le h < 50$ has frequency density $0.32$, so its height is $\mathbf{0.32}$ cm.
OCR trap: students sometimes use the raw frequency ($8$) instead of the frequency density ($0.32$) to find the bar height — that would give $8$ cm, an absurd answer. On a histogram with unequal widths, bar height represents frequency density, not frequency. The area gives the frequency.
A data set $\{x_1, x_2, \ldots, x_n\}$ has mean $\bar{x}$ and standard deviation $\sigma$. Each value is transformed by $y_i = a x_i + b$ for constants $a > 0$ and $b$.
(i) Show that the mean of the transformed data is $a\bar{x} + b$ and the standard deviation is $a\sigma$.
(ii) A set of exam marks has mean $56$ and standard deviation $12$. The marks are scaled by the formula $y = 1.2 x + 4$. State the mean and standard deviation of the moderated marks.
(iii) A student receives a moderated mark of $76$. Find their original mark and show that their standardised position (number of standard deviations above the mean) is the same before and after moderation.
(i) Mean: $\bar{y} = \tfrac{1}{n}\sum y_i = \tfrac{1}{n}\sum (a x_i + b) = \tfrac{a}{n}\sum x_i + \tfrac{nb}{n} = a\bar{x} + b$.
Variance: $\sigma_y^2 = \tfrac{1}{n}\sum (y_i - \bar{y})^2 = \tfrac{1}{n}\sum (a x_i + b - a\bar{x} - b)^2 = \tfrac{1}{n}\sum a^2(x_i - \bar{x})^2 = a^2 \sigma^2$.
So $\sigma_y = a \sigma$ (since $a > 0$). ∎
(ii) New mean: $1.2 \times 56 + 4 = 71.2$. New SD: $1.2 \times 12 = 14.4$.
(iii) $76 = 1.2x + 4 \Rightarrow x = 60$.
Original z-score: $\dfrac{60 - 56}{12} = \dfrac{4}{12} = \dfrac{1}{3}$.
Moderated z-score: $\dfrac{76 - 71.2}{14.4} = \dfrac{4.8}{14.4} = \dfrac{1}{3}$.
Both equal $\tfrac{1}{3}$, so the standardised position is unchanged. This is the defining feature of a linear scaling: it preserves z-scores, hence preserves the relative standing of every student in the cohort.
Mutually exclusive events cannot both occur:
$$P(A \cap B) = 0.$$
It follows that $P(A \cup B) = P(A) + P(B)$.
Independent events do not affect each other's probabilities:
$$P(A \cap B) = P(A) \cdot P(B).$$
Mutually exclusive and independent are different — not synonyms, not opposites. Two events can be both, but only if at least one has probability zero. In practice, "mutually exclusive" means the outcomes don't overlap; "independent" means the outcomes don't influence each other.
$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$
Reason: adding $P(A)$ and $P(B)$ counts the overlap twice; subtract it once to correct.
Venn diagrams: circles for events, overlap for $A \cap B$. Useful when several events are involved.
Sample space diagrams: grids showing every possible outcome (two dice, for example). Each cell has equal probability.
Tree diagrams: branches show possible outcomes; multiply along a branch, add between branches. Especially useful for sequences of events ("with replacement" vs "without replacement").
"With replacement" — probabilities stay the same at each stage (independent trials). "Without replacement" — probabilities change at each stage, because the population is smaller. On a tree diagram, write the conditional probabilities on the second set of branches.
$P(A \mid B)$ is the probability of $A$ given that $B$ has occurred. The basic interpretation:
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$$
If $A$ and $B$ are independent, $P(A \mid B) = P(A)$ — knowing $B$ tells you nothing extra about $A$.
"$A$ given $B$" restricts attention to the outcomes where $B$ occurs. Within those, we ask what fraction also have $A$. So divide $P(A \cap B)$ (both happen) by $P(B)$ (the new total).
Two fair six-sided dice are rolled. Find the probability that the sum of the scores is greater than $9$.
Out of $36$ equally likely outcomes:
Total: $6$ favourable outcomes. So $P(\text{sum} > 9) = \tfrac{6}{36} = \tfrac{1}{6}$.
Events $A$ and $B$ satisfy $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cup B) = 0.7$.
(i) Find $P(A \cap B)$.
(ii) Determine whether $A$ and $B$ are independent, justifying your answer.
(iii) Determine whether $A$ and $B$ are mutually exclusive, justifying your answer.
(i) Using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$$0.7 = 0.4 + 0.5 - P(A \cap B) \;\Rightarrow\; P(A \cap B) = 0.2.$$
(ii) Independent if $P(A \cap B) = P(A) \cdot P(B)$. Check: $0.4 \times 0.5 = 0.2$ ✓. Yes, $A$ and $B$ are independent.
(iii) Mutually exclusive if $P(A \cap B) = 0$. Here $P(A \cap B) = 0.2 \neq 0$. No, $A$ and $B$ are not mutually exclusive.
OCR trap: "independent" and "mutually exclusive" are often confused, but they describe different things and are almost always mutually exclusive themselves (a non-trivial pair can be one or the other, not usually both). Always test each property against its own definition.
A bag contains $5$ red, $4$ blue, and $3$ green balls.
(i) Two balls are drawn at random without replacement. Find the probability that both balls are the same colour.
(ii) Given that both balls are the same colour, find the probability that they are both red.
(iii) The balls are returned to the bag, and the experiment is repeated, but this time three balls are drawn without replacement. Find the probability that all three are different colours.
Total $12$ balls.
(i)
$$P(\text{same colour}) = \tfrac{20 + 12 + 6}{132} = \tfrac{38}{132} = \tfrac{19}{66}.$$
(ii) By the definition of conditional probability:
$$P(\text{both red} \mid \text{same colour}) = \dfrac{P(\text{both red})}{P(\text{same colour})} = \dfrac{20/132}{38/132} = \dfrac{20}{38} = \dfrac{10}{19}.$$
(iii) Want one of each colour out of three drawn. The three balls can come out in any of $3! = 6$ orderings — e.g. R then B then G, or B then G then R, etc. The probability of one specific ordering (e.g. R, B, G):
$$\tfrac{5}{12} \cdot \tfrac{4}{11} \cdot \tfrac{3}{10} = \tfrac{60}{1320}.$$
By symmetry of the multiplication, every ordering has the same probability. So
$$P(\text{all different}) = 6 \cdot \tfrac{60}{1320} = \tfrac{360}{1320} = \tfrac{3}{11}.$$
A random variable $X$ is binomially distributed if the following all hold:
Notation: $X \sim B(n, p)$, with the random variable $X$ being the number of successes in $n$ trials.
$$P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x}, \qquad x = 0, 1, \ldots, n.$$
This is in the formula booklet. In practice, almost every question uses the calculator's built‑in binomial functions:
For $P(X \ge x)$, use $1 - P(X \le x - 1)$. For $P(a \le X \le b)$, use $P(X \le b) - P(X \le a - 1)$.
The cumulative function $P(X \le x)$ includes $x$. So $P(X \ge 5) = 1 - P(X \le 4)$, not $1 - P(X \le 5)$. Mistaking the boundary is the single most common error on binomial questions. Write out which values are included before you press a calculator button.
Questions often ask: "Is $B(n, p)$ a sensible model here?" To answer:
The random variable $X \sim B(10, 0.3)$. Find $P(X = 4)$.
Using the formula:
$$P(X = 4) = \binom{10}{4}(0.3)^4(0.7)^6 = 210 \cdot 0.0081 \cdot 0.117649 \approx 0.2001.$$
So $P(X = 4) \approx 0.200$. (Use the calculator's binomial PD function for speed.)
$X \sim B(20, 0.35)$. Find $P(X \ge 8)$.
The calculator gives cumulative probabilities $P(X \le k)$. To convert "at least $8$" to a CDF query:
$$P(X \ge 8) = 1 - P(X \le 7).$$
Note carefully: the boundary is $X \le 7$, not $X \le 8$. From the calculator: $P(X \le 7) \approx 0.6010$, so
$$P(X \ge 8) \approx 1 - 0.6010 = 0.3990.$$
OCR trap: using $1 - P(X \le 8)$ would give the probability $P(X \ge 9)$, missing the value $X = 8$ from the answer. Always write down the boundary explicitly — "I want $X \ge 8$ which means $X = 8, 9, 10, \ldots$, so I need $1 - P(X \le 7)$" — to avoid the off-by-one error.
In a particular production process, the probability that any item is defective is $0.04$, and items are independent. Items are packaged in boxes of $50$. A box is rejected by quality control if it contains more than $4$ defective items.
(i) Find the probability that a randomly chosen box contains exactly $2$ defective items.
(ii) Find the probability that a randomly chosen box is rejected.
(iii) A retailer purchases $20$ boxes. Find the probability that at most $1$ of these boxes is rejected.
Let $X$ = number of defective items in a box. Then $X \sim B(50, 0.04)$.
(i) $P(X = 2) = \binom{50}{2}(0.04)^2(0.96)^{48} \approx 0.2762$.
(ii) A box is rejected if $X > 4$:
$$p_{\text{rej}} = P(X > 4) = 1 - P(X \le 4) \approx 1 - 0.9509 = 0.0491.$$
(iii) Now let $Y$ = number of boxes rejected out of $20$. The trial is "select a box, observe whether it is rejected", which is binomial with $n = 20$ and probability $p_{\text{rej}} \approx 0.0491$:
$$Y \sim B(20, 0.0491).$$
$P(Y \le 1) = P(Y = 0) + P(Y = 1)$:
$$P(Y = 0) = (0.9509)^{20} \approx 0.3651.$$
$$P(Y = 1) = 20 \cdot (0.0491)(0.9509)^{19} \approx 0.3771.$$
$$P(Y \le 1) \approx 0.7422.$$
(Calculator binomial CD with $n = 20$, $p = 0.0491$, $k = 1$ confirms this.)
What makes this hard: noticing that $p_{\text{rej}}$ from part (ii) becomes the parameter for a second binomial. The two-stage structure (items in a box → boxes in a shipment) trips up many candidates.
Hypothesis tests for a binomial proportion only. Tests on the mean of a normal distribution and Pearson's correlation coefficient are Year 2.
The mark scheme wants language like:
"There is sufficient evidence at the $5\%$ significance level to reject $H_0$. The data suggests that the proportion of [context] is greater than $p_0$."
Wrong: "We have proved $p > p_0$." (A test doesn't prove anything; it gathers evidence.) Wrong: "We accept $H_1$." (You don't accept the alternative; you reject or fail to reject the null.) Wrong: a conclusion that doesn't mention the context — bare numbers without nouns lose marks.
For $H_1: p \neq p_0$, split the significance level: $\alpha/2$ in each tail. The critical region has two parts (one in each tail), and the $p$‑value is doubled compared to a one‑tailed test.
Because $X$ is discrete, you can't always hit exactly $5\%$ in the tail. The actual significance level is the largest probability less than or equal to $\alpha$ that the critical region achieves. Past papers often ask for this explicitly.
A bag contains a large number of coloured balls. A claim is made that the proportion of red balls is $0.4$. A sample of $20$ balls is taken with replacement, and $4$ are red. Test, at the $5\%$ significance level, whether the proportion of red balls is less than $0.4$.
Let $p$ = proportion of red balls.
$H_0: p = 0.4$, $H_1: p < 0.4$. One-tailed test at $5\%$.
Under $H_0$, $X \sim B(20, 0.4)$, where $X$ is the number of red balls in the sample.
Observed: $X = 4$. Compute the $p$-value:
$$P(X \le 4) \approx 0.0510.$$
Compare to $0.05$: $0.0510 > 0.05$. Do not reject $H_0$.
Conclusion: there is insufficient evidence at the $5\%$ significance level to suggest that the proportion of red balls is less than $0.4$.
Sara claims that a coin is biased towards heads. She tosses it $30$ times and observes $20$ heads. Test her claim at the $5\%$ significance level.
Let $p$ = probability of heads on a single toss.
$H_0: p = 0.5$ (coin is fair), $H_1: p > 0.5$ (Sara's claim — biased towards heads). One-tailed test at $5\%$.
Under $H_0$, $X \sim B(30, 0.5)$.
Observed: $X = 20$. Compute:
$$P(X \ge 20) = 1 - P(X \le 19) \approx 1 - 0.9506 = 0.0494.$$
Compare to $0.05$: $0.0494 < 0.05$. Reject $H_0$.
Conclusion: there is sufficient evidence at the $5\%$ significance level to support Sara's claim that the coin is biased towards heads.
OCR trap: this is a marginal call ($p$-value $= 0.0494$ vs threshold $0.05$). Two common slips: using $P(X \le 20)$ (wrong direction — that's nearly $1$) or using $1 - P(X \le 20)$ (excludes the value $20$ from the "at least $20$" event). Showing the boundary explicitly is essential to scoring full marks.
A drug manufacturer claims that $60\%$ of patients respond to a new drug. A doctor is sceptical and conducts a trial on $25$ patients, using a two-tailed test at the $10\%$ significance level.
(i) Find the critical region for the test.
(ii) State the actual significance level of the test.
(iii) In the trial, $11$ patients respond to the drug. Carry out the test, stating your conclusion in context.
(iv) The doctor concludes "the manufacturer is wrong". Comment on this conclusion.
Let $p$ = proportion of patients responding to the drug.
$H_0: p = 0.6$, $H_1: p \neq 0.6$. Two-tailed at $10\%$, so $5\%$ in each tail.
Under $H_0$, $X \sim B(25, 0.6)$.
(i) Lower tail: find largest $c$ with $P(X \le c) \le 0.05$.
So lower CR: $X \le 9$.
Upper tail: find smallest $c$ with $P(X \ge c) \le 0.05$, i.e. $P(X \le c - 1) \ge 0.95$.
So upper CR: $X \ge 19$.
Critical region: $X \le 9$ or $X \ge 19$.
(ii) Actual significance level:
$$P(X \le 9) + P(X \ge 19) \approx 0.0294 + (1 - 0.9577) = 0.0294 + 0.0423 = 0.0717,$$
or about $7.17\%$.
(iii) Observed $X = 11$. Since $9 < 11 < 19$, the observation is not in the critical region. Do not reject $H_0$. Conclusion: there is insufficient evidence at the $10\%$ significance level to dispute the manufacturer's claim that $60\%$ of patients respond.
(iv) The doctor's conclusion overstates the evidence. Failing to reject $H_0$ does not prove $H_0$ is true; it only means the observed data is consistent with $H_0$. The true proportion could be slightly different from $60\%$, but the sample size may be too small to detect it, or the doctor's $10\%$ significance level may have been too stringent given the small sample. A more accurate phrasing: "the trial provides no evidence against the manufacturer's claim at the $10\%$ significance level".
OCR's prescribed Large Data Set (LDS) is pre‑released material common to all candidates. Statistics questions on Paper 2 assume you are familiar with it — the context, the variables, the sources of bias, and the sensible techniques. You are not expected to memorise specific numerical values.
The LDS contains two topics, each given for two census years (2001 and 2011), split by Local Authority District (LAD) / Unitary Authority (UA):
The LDS is available on the OCR website (in "Pre‑release Materials" for H240). Open it once at the start of the course, then again a few times during revision. Looking at the actual columns and rows takes ten minutes and consistently pays off in the exam.
The OCR Large Data Set covers "method of travel to work" for each Local Authority District (LAD).
(i) Name two pieces of information given for each LAD on this topic.
(ii) State one limitation of the data.
(i) Any two of:
(ii) Any one of:
A student wants to compare the proportion of commuters who travel to work by train in two contrasting districts using the LDS.
(i) State what data should be extracted from the LDS.
(ii) State what calculation should be performed.
(iii) Give two reasons why a direct comparison of the train-commuter numbers between the two districts could be misleading.
(i) For each district, extract the number of people commuting by train, and the total number of people in work (the sum of all method-of-travel categories, excluding unemployed).
(ii) For each district, calculate the proportion: (train commuters) ÷ (total in work). Compare these proportions, not the raw numbers.
(iii) Any two of:
A researcher wants to test whether the proportion of car commuters in a particular LAD changed significantly between 2001 and 2011.
(i) Identify what data she would need to extract from the LDS to set up such a test.
(ii) Explain whether this can be tested using the binomial proportion test from H240, or whether different methods would be needed.
(iii) Suggest one potential source of bias that could affect the comparison even if the LDS values are fully accurate.
(i) For each year (2001 and 2011), the researcher needs the number of car commuters in that LAD and the total number of people in work in that LAD. From these she can compute a proportion for each year.
(ii) The H240 binomial proportion test is a one-sample test: it compares one observed proportion against a fixed hypothesised value. Here, the researcher wants to compare two observed proportions — a two-sample test — which is not in the H240 specification. (It would typically use a two-proportion $z$-test from a wider statistics syllabus.)
Within H240, the most she can do is:
(iii) Any one of:
Three SI base quantities are used in mechanics, and they are mutually independent — no one of them is defined in terms of the others.
| Quantity | SI unit | Symbol |
|---|---|---|
| Length | metre | m |
| Time | second | s |
| Mass | kilogram | kg |
Everything else in mechanics is derived from these three:
| Quantity | Unit |
|---|---|
| Velocity | m s$^{-1}$ |
| Acceleration | m s$^{-2}$ |
| Force, Weight | newton (N) $= $ kg m s$^{-2}$ |
Always include units in your final answer. A number with no units rarely scores full marks for the question. Convert mixed units (km, hours, grams) into SI base units before substituting into formulae — leaving them mixed is a common, costly slip.
The first marks on most mechanics questions go to getting the vocabulary right. Vectors and scalars are not interchangeable.
A particle that runs $10$ m to the right and $10$ m back has distance travelled $= 20$ m but displacement $= 0$. The two values are different, and the question wording chooses which one the answer should be.
| Graph | Gradient gives | Area gives |
|---|---|---|
| Displacement–time | Velocity | — |
| Velocity–time | Acceleration | Displacement |
| Acceleration–time | — | Change in velocity |
"Area" means signed area: regions below the $x$‑axis count negatively.
For motion in a straight line with constant acceleration, $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $s$ = displacement, $t$ = time:
$$\begin{aligned} v &= u + at \\ s &= ut + \tfrac{1}{2} a t^2 \\ s &= \tfrac{1}{2}(u + v) t \\ v^2 &= u^2 + 2as \\ s &= v t - \tfrac{1}{2} a t^2 \end{aligned}$$
All five are in the formula booklet.
Take $a$ as constant. Integrating $a$ with respect to $t$ gives velocity: $v = u + at$. Integrating velocity gives displacement: $s = ut + \tfrac{1}{2} a t^2$.
For $v^2 = u^2 + 2as$, eliminate $t$ between the first two: from the first, $t = (v - u)/a$; substitute into the second and simplify.
The third comes from average velocity times time: $s = \tfrac{1}{2}(u + v) t$ holds whenever acceleration is constant, because velocity is then linear in time and its average is the arithmetic mean of start and end.
Each equation involves four of the five SUVAT quantities. List what's known and what's wanted, then pick the equation that contains exactly those.
| Equation | Missing |
|---|---|
| $v = u + at$ | $s$ |
| $s = ut + \tfrac{1}{2}at^2$ | $v$ |
| $s = \tfrac{1}{2}(u + v)t$ | $a$ |
| $v^2 = u^2 + 2as$ | $t$ |
| $s = vt - \tfrac{1}{2}at^2$ | $u$ |
Pick a positive direction at the start, and stick with it. Quantities pointing the other way carry a negative sign. The most common case: take "up" as positive, so $g$ (gravitational acceleration) is $-9.8$ m s$^{-2}$.
A projectile is modelled as moving only under gravity, with $a = -g = -9.8$ m s$^{-2}$ taking upwards as positive. For Year 1, this is one‑dimensional motion — a ball thrown straight up or dropped.
"State one assumption" is a recurring one‑mark question. Have one of these ready.
The spec uses $g = 9.8$ m s$^{-2}$ unless told otherwise. Don't slip to $9.81$ or $10$ unless the question specifies. And remember $g$ is not a universal constant — it varies with location. A question may ask you to comment on this.
When acceleration is not constant, SUVAT no longer applies. Use calculus instead:
$$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2 s}{dt^2}.$$
$$s = \int v \, dt, \qquad v = \int a \, dt.$$
The constants of integration are determined by initial conditions (typical wording: "starts from rest" means $v = 0$ at $t = 0$; "from the origin" means $s = 0$ at $t = 0$).
If $a$ is given as a function of $t$, SUVAT is invalid. Recognising this is examined — questions that look like SUVAT can ambush you by giving acceleration as $a = 2t$, say. Read the question. If $a$ depends on $t$, switch to calculus.
A car accelerates uniformly from rest at $2$ m s$^{-2}$ for $8$ seconds. Find its final velocity and the distance it travels.
$u = 0$, $a = 2$, $t = 8$.
$v = u + at = 0 + 2 \times 8 = 16$ m s$^{-1}$.
$s = ut + \tfrac{1}{2} a t^2 = 0 + \tfrac{1}{2}(2)(64) = 64$ m.
A stone is thrown vertically upwards from the top of a cliff $30$ m above the sea, with initial speed $20$ m s$^{-1}$. Take $g = 9.8$ m s$^{-2}$ and upwards as positive.
(i) Find the maximum height of the stone above the sea.
(ii) Find the speed at which the stone hits the sea.
(iii) Find the total time the stone is in the air.
Set the launch point as $s = 0$, so the sea is at $s = -30$. With upwards positive: $u = +20$, $a = -9.8$.
(i) At maximum height, $v = 0$. Use $v^2 = u^2 + 2as$:
$$0 = 400 + 2(-9.8)s \Rightarrow s = \tfrac{400}{19.6} \approx 20.4 \text{ m above the cliff}.$$
Maximum height above the sea: $30 + 20.4 = 50.4$ m.
(ii) When the stone hits the sea, $s = -30$. Use $v^2 = u^2 + 2as$:
$$v^2 = 400 + 2(-9.8)(-30) = 400 + 588 = 988 \Rightarrow v \approx \pm 31.4.$$
The stone is moving downwards, so $v \approx -31.4$ m s$^{-1}$. Speed: $\mathbf{31.4}$ m s$^{-1}$.
(iii) Use $s = ut + \tfrac{1}{2}at^2$ with $s = -30$:
$$-30 = 20t - 4.9 t^2 \Rightarrow 4.9 t^2 - 20t - 30 = 0.$$
Solve: $t = \dfrac{20 \pm \sqrt{400 + 588}}{9.8} = \dfrac{20 \pm \sqrt{988}}{9.8} \approx \dfrac{20 \pm 31.4}{9.8}$.
Take the positive root: $t \approx 5.24$ s.
OCR trap: sign convention. With "upwards positive", $u$ is positive, $a = -g$ is negative, and the final position below the launch point gives $s$ negative. Mixing up signs is the most common error here. State the convention at the start and stick to it.
A particle moves in a straight line. Its acceleration at time $t$ seconds is $a = (6 - 4t)$ m s$^{-2}$ for $0 \le t \le 4$. At $t = 0$, the particle is at the origin and at rest.
(i) Find expressions for the velocity $v$ and displacement $s$ at time $t$.
(ii) Find the time(s) within the interval at which the particle is at rest, and the displacement at each such time.
(iii) Find the total distance travelled in the interval $0 \le t \le 4$.
(i) Integrate: $v = \int (6 - 4t) \, dt = 6t - 2t^2 + C_1$. At $t = 0$, $v = 0$, so $C_1 = 0$. So
$$v = 6t - 2t^2 = 2t(3 - t).$$
Integrate again: $s = \int (6t - 2t^2) \, dt = 3t^2 - \tfrac{2}{3}t^3 + C_2$. At $t = 0$, $s = 0$, so $C_2 = 0$. So
$$s = 3t^2 - \tfrac{2}{3}t^3.$$
(ii) $v = 0$ when $2t(3 - t) = 0$, giving $t = 0$ (initial) or $t = 3$ s.
At $t = 3$: $s = 27 - \tfrac{2}{3}(27) = 27 - 18 = 9$ m.
(iii) On $[0, 3]$: $v = 2t(3 - t) \ge 0$, so particle moves in the positive direction. Distance covered = $s(3) - s(0) = 9$ m.
On $[3, 4]$: $v \le 0$, so particle moves in the negative direction. At $t = 4$: $s = 48 - \tfrac{128}{3} = \tfrac{144 - 128}{3} = \tfrac{16}{3}$ m.
Displacement change on $[3, 4]$: $\tfrac{16}{3} - 9 = -\tfrac{11}{3}$. Distance covered (absolute value): $\tfrac{11}{3}$ m.
Total distance: $9 + \tfrac{11}{3} = \tfrac{38}{3} \approx 12.67$ m.
OCR trap: total distance ≠ |net displacement|. When the particle changes direction, you must sum the absolute distances of each leg separately. A simple $|s(4) - s(0)| = |\tfrac{16}{3}| \approx 5.33$ m would be wrong — that's the net displacement, not the total distance.
Newton's three laws in 1D and 2D, with smooth contact (no friction coefficient), connected particles, and simple pulleys. Friction with $F \le \mu R$, projectiles (with both horizontal and vertical components of $g$), and moments / statics of rigid bodies are all Year 2.
A force has both magnitude and direction. On a diagram, draw forces as arrows from the point of application; for a particle, all forces act at a single point.
A free body diagram shows a single body with all the forces acting on it. Include weight (downwards), any normal reactions, tensions or thrusts in strings/rods, and any applied forces. Do not include forces acting on other bodies (those go on a separate diagram for that body).
You should be ready to state and to critique these standard assumptions:
"State one modelling assumption" is a routine one‑mark question. The mark scheme wants both the assumption and a brief justification or implication. Better: "We model the rope as light and inextensible, so the tension is the same throughout its length and we can ignore its weight."
A body remains at rest, or moves with constant velocity in a straight line, unless acted on by a non‑zero resultant force.
Equivalently: if the resultant force on a body is zero, the body is in equilibrium — meaning either at rest, or moving with constant velocity.
For a particle in equilibrium under several forces, the forces must "close up" into a polygon, or their components in any chosen direction must sum to zero.
$$\mathbf{F} = m \mathbf{a}.$$
$\mathbf{F}$ is the resultant force on the body, $m$ is its mass, $\mathbf{a}$ is its acceleration. The equation applies in either scalar form (motion in a straight line) or vector form (2D motion, with $\mathbf{F}$ and $\mathbf{a}$ given as $\mathbf{i}, \mathbf{j}$ components or columns).
$F = ma$ uses the net force. If multiple forces act on a body, sum them (with signs!) before applying the equation. The single most common error in mechanics is forgetting to include weight, or forgetting the normal reaction, when computing the resultant.
$$W = mg.$$
Weight is the gravitational force on a mass $m$ near the Earth's surface. It acts vertically downwards. At our specified $g = 9.8$ m s$^{-2}$, a $5$ kg mass has weight $49$ N.
Every action has an equal and opposite reaction. More precisely: if body $A$ exerts a force on body $B$, then body $B$ exerts a force of equal magnitude and opposite direction on body $A$.
Crucially: these two forces act on different bodies. They do not cancel; they cannot cancel, because they don't act on the same thing.
When a body rests on a surface, the surface pushes back on the body with a normal reaction force ($R$ or $N$), perpendicular to the surface. The body's weight is opposed by this normal reaction (and any other vertical force) so that — if there is no vertical acceleration — the net vertical force is zero.
Contact is lost when the normal reaction reaches zero. (Example: a passenger in a falling lift, momentarily.)
At Year 1, contact surfaces are typically modelled as smooth — no friction. The only force at a smooth contact is the normal reaction. The limitations of this model — that real surfaces do exhibit friction — should be acknowledged in context.
Two bodies connected by a light, inextensible string or rod move with the same acceleration (in the direction the string can pull, or the rod can push or pull).
Two equally valid approaches:
The "system" approach gives the acceleration quickly. The "separate" approach is needed to find the internal forces (tension in the string, force on the connecting bar).
An engine of mass $4000$ kg pulls a carriage of mass $2000$ kg along a horizontal track with a driving force of $9000$ N. Resistive forces are $1500$ N on the engine and $500$ N on the carriage. Find the acceleration of the train and the tension in the coupling.
System approach. Total mass $= 6000$ kg. Net force $= 9000 - 1500 - 500 = 7000$ N. So $a = \tfrac{7000}{6000} = \tfrac{7}{6} \approx 1.17$ m s$^{-2}$.
Tension. Apply $F = ma$ to the carriage alone: $T - 500 = 2000 \cdot \tfrac{7}{6}$, giving $T \approx 2833$ N.
For two particles connected by a light inextensible string over a smooth, light pulley:
Treat each particle separately with $F = ma$, taking the direction of motion for each as positive, and solve the resulting pair of equations.
Draw the system. Mark the direction of motion with an arrow on each particle. Label the tension $T$ acting up on each (the pulley turns the direction). Write $F = ma$ for each particle in its own direction of motion. Two equations, two unknowns ($T$ and $a$).
In 2D, a single force $F$ at angle $\theta$ to a chosen axis splits into components parallel and perpendicular to that axis:
$$F_{\parallel} = F \cos\theta, \qquad F_{\perp} = F \sin\theta.$$
For equilibrium of a particle in 2D, the sum of components in any chosen direction is zero. Picking convenient directions — e.g. parallel and perpendicular to a slope — usually simplifies the algebra dramatically.
A horizontal force of $30$ N is applied to a $5$ kg block at rest on a smooth horizontal surface. Find the acceleration of the block.
$F = ma$. So $a = \dfrac{F}{m} = \dfrac{30}{5} = 6$ m s$^{-2}$.
Two blocks $A$ (mass $3$ kg) and $B$ (mass $5$ kg) are connected by a light inextensible string and lie at rest on a smooth horizontal surface. A horizontal force of $40$ N is applied to $A$, in the direction away from $B$, pulling the system into motion.
(i) Find the acceleration of the system.
(ii) Find the tension in the string.
(i) Treat the system as a single body. Total mass $= 8$ kg, resultant horizontal force $= 40$ N. So
$$a = \dfrac{40}{8} = 5 \text{ m s}^{-2}.$$
(ii) To find the internal tension, apply $F = ma$ to one block in isolation. Block $B$ has only one horizontal force on it — the tension $T$ pulling it forward:
$$T = m_B \cdot a = 5 \times 5 = 25 \text{ N}.$$
Check using block $A$: net force $= 40 - T = 40 - 25 = 15$ N, which equals $m_A \cdot a = 3 \times 5 = 15$ N ✓.
OCR trap: students sometimes apply $F = ma$ to the applied force ($40$ N) and one of the masses (e.g. $3$ kg or $5$ kg) alone, which is wrong because that mass is not the only thing the $40$ N is accelerating. Use the system to find $a$; use a single body to find the internal force.
Two particles $P$ (mass $4$ kg) and $Q$ (mass $3$ kg) are connected by a light inextensible string passing over a smooth, light, fixed pulley. Both particles hang vertically. The system is released from rest. Take $g = 9.8$ m s$^{-2}$.
(i) Find the acceleration of the system and the tension in the string.
(ii) After $1.5$ seconds, the string breaks. Find the maximum additional height above its position at the break that $Q$ reaches, stating any assumption you make.
(iii) The ground is $5$ m below $P$'s position at the moment the string breaks. Find the speed with which $P$ hits the ground.
(i) Take downwards positive for $P$ (heavier, moves down), upwards positive for $Q$. Both share the same magnitude of acceleration $a$.
For $P$: $4g - T = 4a \quad (\star)$
For $Q$: $T - 3g = 3a \quad (\dagger)$
Add: $4g - 3g = 7a \Rightarrow a = \dfrac{g}{7} = \dfrac{9.8}{7} = 1.4$ m s$^{-2}$.
From $(\dagger)$: $T = 3a + 3g = 3(1.4) + 3(9.8) = 4.2 + 29.4 = \mathbf{33.6}$ N.
(ii) At the moment of the break, both particles have speed $v = u + at = 0 + 1.4 \times 1.5 = 2.1$ m s$^{-1}$. $Q$ is moving upwards.
After the break, $Q$ is in free fall under gravity alone (assuming air resistance is negligible). Take upwards positive: $u = 2.1$, $a = -9.8$. At maximum height, $v = 0$. Use $v^2 = u^2 + 2as$:
$$0 = (2.1)^2 + 2(-9.8) s \Rightarrow s = \dfrac{4.41}{19.6} = 0.225 \text{ m}.$$
So $Q$ rises an additional $\mathbf{0.225}$ m above its position at the break.
Assumption: air resistance is negligible. (Other valid assumptions: $Q$ is modelled as a particle; $g$ is constant.)
(iii) For $P$ after the break: $u = 2.1$ m s$^{-1}$ downwards, $a = 9.8$ m s$^{-2}$ downwards. With downwards positive, $P$ falls $5$ m. Use $v^2 = u^2 + 2as$:
$$v^2 = (2.1)^2 + 2(9.8)(5) = 4.41 + 98 = 102.41 \Rightarrow v \approx \mathbf{10.12} \text{ m s}^{-1}.$$
What makes this hard: three distinct phases (constant-tension system, free fall up for $Q$, free fall down for $P$), each with its own setup. Recognising that both particles have the same speed at the break — and that $Q$'s direction is upwards while $P$'s is downwards — is the conceptual crux.
OCR A‑Level Mathematics A (H240) — Year 1 / AS Stage notes. Specification reference 1.01–1.10 (Pure), 2.01–2.05 (Statistics), 3.01–3.03 (Mechanics), restricted to AS content.