Integrate Twice, Find Two Constants — Sequencing a Two-Stage Problem

~45 min · WMA11 · 5.2

WMA11 · 5.2 · 45 min

Going from $f''(x)$ to $f(x)$ crosses the same boundary — differentiation reversed — twice in a row, and each crossing hands you a brand-new unknown constant that the other crossing knows nothing about. Three independently-verified WMA11 series converge on the same finding: candidates who get every piece of algebra right still lose marks here, not on the integrating, but on the bookkeeping — using the wrong condition on the wrong function, or trying to pay for both constants with one number found only at the very end. This lesson is built around exactly that sequencing, not around the integration itself.

Key terms in this lesson

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Where this question type sits in the spec, and why it needs two constants, not one

Spec 4.1 gives second-order derivatives official standing at P1: "the derivative of f(x) as the gradient of the tangent to the graph of $y = f(x)$ at a point... interpretation as a rate of change; second-order derivatives (notation $f'(x)$ and $f''(x)$ may be used)." $f''(x)$ is not a separate topic bolted onto differentiation — it is what you get by differentiating $f'(x)$ a second time, and by the same logic, undoing it takes integrating twice.

Spec 5.1 states the constant of integration as a requirement, not a footnote: "indefinite integration as the reverse of differentiation (a constant of integration is required)." That sentence does not say a constant is required once per question — it is required every time the reversal happens. Integrating $f''(x)$ to reach $f'(x)$ is one reversal; integrating $f'(x)$ to reach $f(x)$ is a second, separate one. Each earns its own constant, because each is, on its own, exactly the situation spec 5.1 is describing.

Spec 5.2 names this exact question type directly: "Integration of $x^n$ and related sums, differences and constant multiples... given $f'(x)$ and a point on the curve, finding an equation of the curve in the form $y = f(x)$ ." The two-stage version in this lesson is the same mechanism run twice — first from $f''(x)$ to $f'(x)$ using a gradient condition, then from $f'(x)$ to $f(x)$ using a point — and three independently-verified series (Oct 2021 Q10, Jan 2023 Q11(b), Jan 2025 Q6(b)) confirm it is a real, recurring WMA11 question type, not a hypothetical extension of the spec.

The rule you need for the integration itself is short: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$ , $n \neq -1$ . It is not in the exam formula booklet — the spec states plainly that formulae students are expected to know 'are given below and will not appear in the booklet,' and this is one of them — so it has to come from memory, the same as the differentiation rule it reverses.

Mechanism

A gradient condition and a point are statements about two different functions

Everything in this lesson follows from one distinction, and the distinction is definitional, not a rule to memorise. Spec 4.1 defines $f'(x)$ AS the gradient of the to $y = f(x)$ — so a condition written as " $f'(a) = m$ " is a sentence about the STEEPNESS of the curve at $x = a$ , and it can only be checked once $f'(x)$ itself exists as an expression. A condition written as "the curve $y = f(x)$ passes through $(p, q)$ " is a sentence about the HEIGHT of the curve at $x = p$ — it says $f(p) = q$ , and it can only be checked once $f(x)$ itself exists. These are not two versions of the same fact arriving in different notation. $f'(4) = 3$ and $f(4) = 3$ would be two completely different, unrelated statements about the same curve — one about its slope at $x=4$ , one about its height there — and mixing them up is not a small error, because there is no algebra that recovers the right answer once the wrong function has been evaluated. This is exactly the mechanism behind the trap a real Jan 2023 examiner report names: candidates who had correctly integrated once, reaching $f'(x)$ , then reached for the point $(p,q)$ — a fact about $f$ , not about $f'$ — because it was the only other piece of given information left in the question, not because it belonged there.

Mechanism

Why the two constants can't be paid for with one number at the end

The second reason sequencing matters is what integrating twice actually does to a constant algebraically. Suppose $f'(x) = 3x^2 + c_1$ . Integrating again does not just carry $c_1$ along unchanged — it turns it into a genuine TERM of $f(x)$ : $f(x) = \int (3x^2 + c_1)\,dx = x^3 + c_1 x + c_2$ . Notice that $c_1$ now multiplies $x$ inside $f(x)$ ; it did not vanish, and it did not merge with $c_2$ . A candidate who never separately finds $c_1$ — who integrates straight through to $f(x) = x^3 + c$ with one constant standing in for both — has thrown away the $c_1 x$ term entirely, which is a different, incomplete expression, not the same one written differently. That is the mechanism behind the second confirmed trap: a real Jan 2023 examiner report records that "it was very common to see candidates integrate twice and apply a constant only after their second integration, thus missing earlier marks" — the marks missed are not just for tidiness, they are for a term of $f(x)$ that a single deferred constant cannot produce. The safe habit is mechanical, not just procedural: the instant you finish an integration, look at what condition is available for THAT function specifically, and use it before integrating again.

Diagram — One curve, two conditions that describe two different things about it

x-axis: x · y-axis: y = f(x)

y = x³ − 45x + 36: The finished curve from the worked chain below, found from f''(x) = 6x using the real conditions f'(4) = 3 and the point (1, −8). It rises from lower-left, turns down near x ≈ −3.9, falls through the marked point (1, −8), reaches a minimum near x ≈ 3.9, then rises again through x = 4 and beyond.

(1, −8) — a POINT on the curve: This is a height: f(1) = −8. It is what the second integration's constant, c₂, was fixed with — because it is a fact about f itself, and c₂ is the constant that only exists once f(x) exists.
x = 4 — where f′(4) = 3, NOT where f(4) = 3: The condition used here was about the GRADIENT at x = 4, not the height. The curve's actual height at x = 4 is f(4) = −80 (plotted), a completely different number — the gradient of 3 describes how gently the curve is climbing at that x, just past its local minimum, not where the curve sits vertically.

Common error: Reading the marked point (1, −8) as though −8 were a gradient value, or reading "f′(4) = 3" as though the curve passed through the point (4, 3).

Correct: (1, −8) is a height: f(1) = −8. "f′(4) = 3" is a gradient, a slope, not a height — the curve's real height at x = 4 is f(4) = −80, shown on the same plot. Confusing the two is exactly the substitution error a real examiner report names.

examiner-report · Jan 2023 · Q11(b)

Worked, in full

f″(x) = 6x, with f′(4) = 3 and the curve through (1, −8): find f(x) — fixing each constant right after its own integration. (f′(4) = 3 and the point (1, −8) are the real, verified numbers from Jan 2023 Q11(b) and Oct 2021 Q10 respectively, verified directly against the primary source. f″(x) = 6x is VERIDIAN-original, chosen only so these two real conditions resolve cleanly, since neither source document preserves the original polynomial.)

01
Read the two conditions before integrating anything, and decide which function each one belongs to. " $f'(4) = 3$ " is a statement about the gradient — it belongs with $f'(x)$ , once it exists. "The curve passes through $(1, -8)$ " is a statement that $f(1) = -8$ — a height, so it belongs with $f(x)$ , once THAT exists. Nothing gets substituted yet; this is planning, not working.
Earns: No mark of its own, but this is the step whose absence produces both confirmed traps. Skipping it is how a real script ends up reaching for whichever condition is left over, rather than the one that actually matches the function just reached.
02
Integrate $f''(x) = 6x$ once: $f'(x) = 3x^2 + c_1$ .
Earns: M1 — the method mark for integration is earned by the power of at least one term increasing by 1 ( $x \to x^2$ ), which has happened here. This mark does not depend on $c_1$ 's value being found yet.
03
Use $f'(4) = 3$ immediately, because it is a gradient condition and $f'(x)$ now exists: $3(4)^2 + c_1 = 3$ , so $48 + c_1 = 3$ , giving $c_1 = -45$ . So $f'(x) = 3x^2 - 45$ .
Earns: dM1 A1 — dM1 for substituting the condition and solving for $c_1$ , dependent on the first M1 (there must be an $f'(x)$ with a constant in it to substitute into); A1 for the resulting $f'(x) = 3x^2 - 45$ . A real Jan 2023 mark scheme (Q11(b)) splits finding a constant exactly this way — a dependent method mark for the substitute-and-solve step, a separate accuracy mark for the value it produces — rather than one combined mark. Deferring this substitution to after the second integration is exactly the trap a real Jan 2023 examiner report records candidates falling into, and it is unrecoverable later: once $f'(x)$ is folded into $f(x)$ , $c_1$ is no longer a standalone number you can solve for by itself.
04
Integrate $f'(x) = 3x^2 - 45$ a second time: $f(x) = x^3 - 45x + c_2$ .
Earns: dM1 — the second method mark for integration, dependent on the FIRST M1 only (stage 2) — not on stage 3's substitution. A real Jan 2023 mark scheme (Q11(b)) tags this exact step 'dM1: dependent upon first M1 only', because reaching the right cubic-family shape only needs some $f'(x)$ -shaped expression to integrate again, not a correct $c_1$ : the power of both terms has increased by 1 ( $x^2 \to x^3$ and $x^1 \to x^2$ , i.e. the constant $-45x$ term), which is what this mark checks for, separately from whether $c_1$ 's value above was correct.
05
Use the point $(1, -8)$ now, because it is a value of $f$ and $f(x)$ now exists: $f(1) = 1 - 45 + c_2 = -44 + c_2$ , and this equals $-8$ , so $c_2 = 36$ . So $f(x) = x^3 - 45x + 36$ .
Earns: dddM1 A1 — dddM1 for substituting the point and solving for $c_2$ , dependent on ALL three previous method marks (stage 2's integration, stage 3's substitution, stage 4's integration) — a real Jan 2023 mark scheme (Q11(b)) uses this exact triple-dependent code for the analogous step. A1 for the final correct expression. Unlike the method marks above, this A1 genuinely does depend on $c_1$ being right: $c_2$ is solved from $f(1) = 1 - 45 + c_2$ , and a wrong $c_1$ changes that $-45$ term. Check both conditions against the finished answer, in one line each: $f'(x) = 3x^2 - 45$ , so $f'(4) = 3(16) - 45 = 48 - 45 = 3$ ✓. $f(1) = 1 - 45 + 36 = -8$ ✓.

Source — Examiner report, Jan 2023

"It was very common to see candidates integrate twice and apply a constant only after their second integration, thus missing earlier marks"

In your own words

In one sentence: why can't $c_1$ be found correctly if you wait until after the second integration to use the gradient condition?

Complete it yourself

Complete the chain — f″(x) = 12x − 6, with f′(1) = 2 and the curve through (2, 5)

01
$f''(x) = 12x - 6$ . Two integrations are needed to reach $f(x)$ , and each one earns its own, separate constant — they are not the same number.
02
Sort the two given facts by which function they describe before touching the integral. $f'(1) = 2$ is a gradient condition — it belongs with $f'(x)$ , the first thing you reach. The curve passing through $(2, 5)$ means $f(2) = 5$ , a height — it belongs with $f(x)$ , the second thing you reach.

Marked, line by line

$f''(x) = 6 - 6x$ . Given that $f'(3) = -2$ , and that the curve with equation $y = f(x)$ passes through the point $(1, 4)$ , find $f(x)$ . (6) — VERIDIAN-original question, built to demonstrate both confirmed sequencing traps side by side; not a reproduction of any past-paper question. The per-line mark allocations and dependency codes (dM1, dddM1) are modelled directly on a real Jan 2023 mark scheme (Q11(b), Publications Code WMA11_01_MS_2301) for the same 'two integrations, two constants' structure, not copied from a real scheme for THIS question, which for an original question does not exist.

6 marks available

01
$f''(x) = 6 - 6x \Rightarrow f'(x) = 6x - 3x^2 + c_1$
Method mark for the first integration: power of at least one term increased by 1 (x⁰ → x¹, and x¹ → x²). Earned regardless of what c₁ turns out to be.
M1
02
$f'(3) = -2$ : $6(3) - 3(3)^2 + c_1 = 18 - 27 + c_1 = -9 + c_1 = -2$ , so $c_1 = 7$ .
Dependent method mark for substituting the gradient condition and solving for c₁ — dependent on the first M1 only (there must be an f'(x) with a constant in it to substitute into). A real Jan 2023 mark scheme (Q11(b)) tags this exact substitute-and-solve step 'dM1', separately from the M1 for the integration and the A1 below for the value it produces.
dM1
03
$f'(x) = 6x - 3x^2 + 7$
Accuracy mark for the correct c₁ and the resulting f'(x), only available now that f'(x) exists, since f'(3) = −2 is a statement about the gradient function.
A1
04
$f'(x) = 6x - 3x^2 + 7 \Rightarrow f(x) = 3x^2 - x^3 + 7x + c_2$
Second method mark for integration, dependent on the FIRST M1 only — not on the dM1/A1 above. A real Jan 2023 mark scheme (Q11(b)) tags this exact second-integration step 'dM1: dependent upon first M1 only', because reaching the right cubic-family shape only needs some f'(x)-shaped expression to integrate again, not a correct c₁: the method itself is being credited, not the number.
dM1
05
The curve passes through $(1, 4)$ : $f(1) = 3 - 1 + 7 + c_2 = 9 + c_2 = 4$ , so $c_2 = -5$ .
Dependent method mark for substituting the point and solving for c₂ — dependent on all three previous method marks (both integrations and the first substitution). A real Jan 2023 mark scheme (Q11(b)) uses this exact triple-dependent code for the analogous step.
dddM1
06
$f(x) = -x^3 + 3x^2 + 7x - 5$
Correct answer only. Unlike the method marks above, this mark does depend on c₁ being correct even though it is a different constant: the 7x term came from the c₁ found earlier, so an error there would carry through even with flawless method on this line. Check: f'(x) = 6x − 3x², so f'(3) = 18 − 27 + 7 = −2 ✓. f(1) = −1 + 3 + 7 − 5 = 4 ✓.
A1

Named traps

gradient-condition-applied-at-the-wrong-stage: Confirmed directly on a real two-stage integration question: candidates who had correctly reached f'(x) "were often unsuccessful because they substituted the coordinates for P... rather than realising f'(x) is the gradient function so that f'(4) = 3 was needed" (Jan 2023, Q11(b)). The point belongs to f, once f(x) exists — using it while only f'(x) exists attaches a fact about the curve's height to a function that is describing its slope.
both-constants-deferred-to-the-end: Confirmed on the same question: "It was very common to see candidates integrate twice and apply a constant only after their second integration, thus missing earlier marks" (Jan 2023, Q11(b)). Integrating f'(x) turns its constant into a genuine term of f(x) (c₁ multiplying x); a single constant tacked on only after both integrations has nowhere to put that term, and the marks for finding c₁ correctly, at its own line, are gone before the deferred constant is ever reached.
point-coordinates-transposed-or-sign-flipped: Confirmed on a separate, independent series: an examiner report on a two-stage integration question records "slips such as using the point (1, 8) or even (−8, 1)" in place of the correct point (Oct 2021, Q10). Both wrong versions are the same real point with a coordinate's sign or position moved — a copying error, not a conceptual one, but one the paper still records candidates making under time pressure. Write the point down exactly as given before substituting, not from memory a line later.
constant-of-integration-omitted-entirely: The single most common 'silly' final-mark loss on integration questions across the whole WMA11 corpus reviewed for this course, confirmed as a repeated finding across four separate series: Oct 2021 Q1, Oct 2020 Q9, Jan 2023 Q3, and Jan 2025 Q6(b). In a two-stage problem there are two separate places to make this same omission, not one — dropping either constant, not just both, loses the marks attached to that specific integration.

Retrieval — with feedback on every choice

Question 1

3 marks

$f''(x) = 6x$ and $f'(2) = 1$ . What is $f'(x)$ ?

Question 2

1 mark

The curve $y = f(x)$ passes through the point $(1, -8)$ . Which equation should this fact be written as, to find a constant in $f(x)$ ?

Question 3

1 mark

Why does the mark scheme lose marks for integrating $f''(x)$ all the way to $f(x)$ before finding either constant, even when the final answer eventually comes out right?

Question 4

4 marks

$f''(x) = 4 - 6x$ . Given that $f'(1) = 3$ and that the curve $y = f(x)$ passes through $(1, 2)$ , find $f(x)$ .

Reference — not a study method, a lookup

f″(x) → integrate → f′(x) + c₁. Fix c₁ NOW using a GRADIENT condition, f′(a) = m — before integrating again.
f′(x) → integrate → f(x) + c₂. Fix c₂ using a POINT on the curve, f(p) = q — never the gradient condition.
Two integrations, two constants, two different functions. A point belongs to f(x); a gradient value belongs to f′(x).
Method mark for integration: power of at least one term increased by 1 (xⁿ → xⁿ⁺¹). A constant is required every time — spec 5.1.
∫xⁿ dx = xⁿ⁺¹/(n+1) + c, n ≠ −1 — not in the formula booklet. Memorise it.
Finding each constant is really TWO marks, not one: a dependent method mark for substituting the condition and solving, then a separate accuracy mark for the value it produces. The second integration's own method mark survives even a wrong or missing first constant — but the SECOND constant cannot be found at all without the first substitution having been attempted.

Not affiliated with or endorsed by Pearson Edexcel. The worked-chain example's two given conditions — f'(4) = 3 and the point (1, -8) — are real, verified numbers, read directly from the primary source (Jan 2023 Q11(b) and Oct 2021 Q10 respectively; the Oct 2021 point is stated directly in that question's own text, not merely inferred from the examiner report). The function f''(x) = 6x built around them is VERIDIAN-original, since neither source question's own f''(x) survives as a clean, checkable expression once only these two conditions and no others are kept (Jan 2023 Q11's real f''(x) = 4x + 1/√x). Every other question in this lesson — prequestion, chain-drill, marked-solution and MCQ alike — is VERIDIAN-original wording, inspired by the two confirmed real failure modes (Jan 2023 Q11(b)'s wrong-stage substitution and deferred-constant error; Oct 2021 Q10's transposed point) but never a reproduction of a real Pearson question. Their per-line mark codes, including the dM1/dddM1 dependency labels, are modelled directly on Jan 2023 Q11(b)'s own real mark scheme (Publications Code WMA11_01_MS_2301) for the identical two-integration, two-constant structure — verified by a fresh pdftotext re-extraction of that mark scheme, not transcribed from any earlier pass's summary. Every quotation attributed to an examiner report was checked directly against primary Pearson PDF text — this course's own research pass flagged the one case where a web-fetch tool fabricated content on a genuine PDF, and noted this unit's narrower series coverage compared to the Economics/Business precedent.

Question 13 marks

$f''(x) = 6x$ and $f'(2) = 1$ . What is $f'(x)$ ?

$f'(x) = 3x^2 - 11$
Correct. $\int 6x\,dx = 3x^2 + c$ . Using $f'(2) = 1$ : $3(4) + c = 12 + c = 1$ , so $c = -11$ . Check: $3(2)^2 - 11 = 12 - 11 = 1$ . ✓
B $f'(x) = 3x^2$
The integration itself is right and the constant is missing entirely. Spec 5.1 requires a constant of integration every time the reversal is performed — this f'(x) is one specific curve's gradient function, not the general family, and without c it cannot be checked against f'(2) = 1 at all: 3(4) = 12 ≠ 1.
C $f'(x) = 3x^2 + 11$
The size of c is right and the sign is flipped. $12 + c = 1$ rearranges to $c = 1 - 12 = -11$ , not $c = 1 + 12$ . Check the wrong version against the condition: 3(4) + 11 = 23 ≠ 1, which is the giveaway that a sign has gone wrong in the rearrangement, not the integration.
D $f'(x) = 6x^2 - 23$
The power of x was increased correctly (x → x²), which is all the method mark for integration actually checks — but the coefficient was never divided by the new power. $\int 6x\,dx = 6 \cdot \frac{x^2}{2} = 3x^2$ , not $6x^2$ . Carrying that uncorrected $6x^2$ into the condition gives $6(4) + c = 1$ , so $c = -23$ — a different, wrong constant, not just a wrong leading coefficient. This is exactly the accuracy step the method mark alone does not guarantee.

Traps tested: Constant of integration omitted entirely · Sign error solving for c · Power not divided by new index

Question 21 mark

The curve $y = f(x)$ passes through the point $(1, -8)$ . Which equation should this fact be written as, to find a constant in $f(x)$ ?

$f(1) = -8$
Correct. A point $(p, q)$ on the curve $y = f(x)$ means exactly this: the function's value at $x = p$ is $q$ .
B $f(1) = 8$
The y-coordinate's sign has been dropped. A real examiner report on this exact point records candidates making precisely this slip in practice — using (1, 8) where (1, −8) was given (Oct 2021, Q10).
C $f(-8) = 1$
The x- and y-coordinates have been swapped — this reads the point as $(-8, 1)$ rather than $(1, -8)$ . The same real examiner report records this exact transposed version among the wrong substitutions candidates made (Oct 2021, Q10).
D $f'(1) = -8$
A point on the curve $y = f(x)$ is a statement about $f$ , the height, not about $f'$ , the gradient. Writing it as a gradient condition attaches a fact about the curve's position to the wrong function entirely.

Traps tested: Point coordinates transposed or sign flipped · Gradient condition applied at the wrong stage

Question 31 mark

Why does the mark scheme lose marks for integrating $f''(x)$ all the way to $f(x)$ before finding either constant, even when the final answer eventually comes out right?

Because c₁ has to appear as a genuine term inside f(x) (multiplying x) — a single constant added only at the end has no way to produce that term, so a correctly-found final answer this way is rare, not just unrewarded early
Correct. This is the actual mechanism, not just a marking convention: $\int (3x^2 + c_1)\,dx = x^3 + c_1x + c_2$ — the c₁ term survives the second integration as $c_1 x$ . Skip finding c₁ first, and that term is simply missing from whatever f(x) is written down.
BBecause the final value of f(x) is genuinely different depending on the order the constants are found in
Not true when both constants are correctly kept and correctly matched to their conditions — the final f(x) is the same expression whichever order you solve for c₁ and c₂ in, as long as each is matched to the right function. The problem is specifically what happens when they get collapsed into one, not the order of solving two correctly-kept ones.
CBecause WMA11 is a no-calculator paper, so all constants must be found by hand in a fixed sequence
WMA11 does allow calculators generally, with specific sub-parts overridden — but that convention is unrelated to this question. The reason sequencing matters here is structural to what integration does to a constant, not a rule about calculator use.
DBecause c₁ and c₂ must always turn out to be numerically equal, and finding them separately is the only way to guarantee that
There is no such requirement — in the worked chain above, c₁ = −45 and c₂ = 36, two different numbers, and that is entirely normal. Nothing about the topic requires the two constants of integration to match.

Traps tested: Order confused with conflation · Calculator rule misapplied to sequencing · Constants assumed equal

Question 44 marks

$f''(x) = 4 - 6x$ . Given that $f'(1) = 3$ and that the curve $y = f(x)$ passes through $(1, 2)$ , find $f(x)$ .

$f(x) = 2x^2 - x^3 + 2x - 1$
Correct. $f'(x) = 4x - 3x^2 + c_1$ ; $f'(1) = 4 - 3 + c_1 = 1 + c_1 = 3$ , so $c_1 = 2$ and $f'(x) = 4x - 3x^2 + 2$ . Then $f(x) = 2x^2 - x^3 + 2x + c_2$ ; $f(1) = 2 - 1 + 2 + c_2 = 3 + c_2 = 2$ , so $c_2 = -1$ . Check: $f'(1) = 4 - 3 + 2 = 3$ ✓, $f(1) = 2 - 1 + 2 - 1 = 2$ ✓.
B $f(x) = 2x^2 - x^3 + x + 1$
This comes from swapping which condition attaches to which stage — using the point's y-value (2) as the gradient condition on f'(x), and the gradient value (3) as though it were a height on f(x). Both x-values happen to be 1 here, which makes this error easy to miss: it is the FUNCTION that matters, not the x-value.
C $f(x) = 2x^2 - x^3 + 1$
This comes from integrating straight through with a single constant (omitting c₁ separately, so no +2x term appears) and using only the point at the end. It satisfies neither condition properly: differentiating this f(x) gives f'(x) = 4x − 3x², so f'(1) = 4 − 3 = 1, not 3.
D $f(x) = 2x^2 - x^3$
Both constants have been omitted entirely. This is the un-parameterised general antiderivative of f''(x), with neither given condition used at all.

Traps tested: Gradient condition applied at the wrong stage · Both constants deferred to the end · Constant of integration omitted entirely

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2023 · Q11(b) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA11.

Pure Mathematics 1 · progress saved in this browser · sign in to sync across devices

Up next

Sine, Cosine and Tangent Graphs — Period vs Domain Confusion

On a real paper, more than half of candidates got a one-mark question wrong when asked to simply state the period of \tan x — and it isn't because the period is hard to compute. It's because "state the period" and "state the domain" sound like they want the same kind of answer, and they don't: a period is a single number, the length you travel along the x-axis before the graph repeats itself exactly; a domain is a set, every x-value the function is even defined at. \tan x has a domain with gaps in it, and a documented, frequently-given wrong answer on the real question was an interval — -\frac{\pi}{2} < x < \frac{\pi}{2} — that correctly describes something about the graph, just not the thing the question asked for.

50 min