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Pure Mathematics 1 flashcards

53 cards on Topic 1, subtopics 1.1 to 1.8. Pick a subtopic or study them all.

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1.1 Quadratics

ax² + bx + c = a(x + p)² + q: what are the vertex and the line of symmetry?: Vertex (−p, q); line of symmetry x = −p. If a > 0, q is the minimum value; if a < 0, q is the maximum value.
Write 2x² − 12x + 5 in completed-square form and state its minimum point.: 2(x² − 6x) + 5 = 2[(x − 3)² − 9] + 5 = 2(x − 3)² − 13. Minimum point (3, −13). The classic slip is not multiplying the −9 by 2.
The quadratic formula for ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) / (2a) It always works, even when the quadratic does not factorise.
What does the discriminant b² − 4ac tell you about the roots?: > 0: two distinct real roots. = 0: one repeated root (the curve touches the x-axis). < 0: no real roots.
A line y = mx + k and a curve: how do you find k for which the line is a tangent?: 1. Substitute the line into the curve to get one quadratic. 2. Set its discriminant b² − 4ac = 0 and solve for k. (Two points of intersection: > 0. No intersection: < 0.)
Which two conditions make ax² + bx + c positive for all x?: a > 0 and b² − 4ac < 0. Both are needed; giving only the discriminant loses a mark.
Solve (x − 2)(x − 5) < 0 and (x − 2)(x − 5) > 0.: < 0: 2 < x < 5 (below the axis, between the roots). > 0: x < 2 or x > 5 (above the axis, outside the roots). Always sketch the parabola first.
Solve x − 7√x + 12 = 0.: Let u = √x: u² − 7u + 12 = 0, so u = 3 or u = 4 (both positive, so both valid). x = 9 or x = 16. Always substitute back, and reject u < 0 when u = √x.

1.2 Functions

One-one function, and when does f⁻¹ exist?: A one-one function gives different outputs for different inputs. f⁻¹ exists only when f is one-one on its domain (test: a horizontal line meets the graph at most once).
When can the composite gf be formed, and what does gf(x) mean?: Only when the range of f lies within the domain of g. gf(x) = g(f(x)): apply f first, then g.
How are the domain and range of f⁻¹ related to f?: Domain of f⁻¹ = range of f. Range of f⁻¹ = domain of f.
How do you find f⁻¹(x)?: 1. Check f is one-one. 2. Write y = f(x) and make x the subject (choose the square-root sign that matches the domain of f). 3. Swap letters to give f⁻¹(x). 4. State its domain: the range of f.
How is the graph of y = f⁻¹(x) related to y = f(x)?: It is the reflection of y = f(x) in the line y = x. A point (a, b) on f becomes (b, a) on f⁻¹. Draw and label y = x on the sketch.
Describe the transformations y = f(x) + a, y = f(x + a), y = af(x) and y = f(ax).: f(x) + a: translation by the vector (0, a). f(x + a): translation by the vector (−a, 0). af(x): stretch parallel to the y-axis, scale factor a. f(ax): stretch parallel to the x-axis, scale factor 1/a.
(2, 5) lies on y = f(x). Where does it move to on y = 3f(2x) − 1?: Halve the x-coordinate; multiply the y-coordinate by 3, then subtract 1. The image is (1, 14).
f(x) = 2x² − 12x + 7 for x ≥ 4. Find the range of f.: f(x) = 2(x − 3)² − 11. The vertex x = 3 is outside the domain and f is increasing for x ≥ 4, so the least value is f(4) = −9. Range: f(x) ≥ −9.

1.3 Coordinate geometry

Gradient, midpoint and length of the line joining (x₁, y₁) and (x₂, y₂): Gradient m = (y₂ − y₁)/(x₂ − x₁). Midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2). Length √((x₂ − x₁)² + (y₂ − y₁)²).
Gradients of parallel and perpendicular lines: Parallel: m₁ = m₂. Perpendicular: m₁m₂ = −1, so m₂ = −1/m₁ (the negative reciprocal, not just −m₁).
Equation of a circle with centre (a, b) and radius r, and the centre and radius of x² + y² + 2gx + 2fy + c = 0: (x − a)² + (y − b)² = r². Expanded form: centre (−g, −f), radius √(g² + f² − c). If g² + f² − c ≤ 0 it is not a real circle.
Find the centre and radius of x² + y² − 10x + 4y + 13 = 0.: (x − 5)² − 25 + (y + 2)² − 4 + 13 = 0, so (x − 5)² + (y + 2)² = 16. Centre (5, −2), radius 4.
How do you find the tangent to a circle at a point P on it?: 1. Find the centre C. 2. Find the gradient of the radius CP. 3. Tangent gradient = −1/(gradient of CP), because a tangent is perpendicular to the radius. 4. Use y − y₁ = m(x − x₁) through P.
How do you find the perpendicular bisector of AB?: 1. Find the midpoint M of AB. 2. Find the gradient of AB, then the perpendicular gradient −1/m. 3. Write the line through M with that gradient.
A line and a curve give a quadratic after substitution. What do "two distinct points", "meets" and "touches" each need?: Two distinct points: b² − 4ac > 0. Meets (at least once): b² − 4ac ≥ 0. Touches (tangent): b² − 4ac = 0.

1.4 Circular measure

Radian: The angle at the centre of a circle subtended by an arc equal in length to the radius. π radians = 180°, so 1 radian ≈ 57.3°.
How do you convert between degrees and radians?: Degrees to radians: × π/180 (for example 72° = 2π/5). Radians to degrees: × 180/π (for example 1.5 rad ≈ 85.9°).
Arc length and sector area for radius r and angle θ: Arc length s = rθ. Sector area A = ½r²θ. θ must be in radians.
Area of a segment cut off by a chord that subtends θ at the centre: Segment = sector − triangle = ½r²θ − ½r² sin θ = ½r²(θ − sin θ), with θ in radians (and the calculator in radian mode).
Length of a chord that subtends θ at the centre of a circle of radius r: 2r sin(θ/2), or from the cosine rule: AB² = r² + r² − 2r² cos θ.
A sector has radius 12 cm and angle 0.75 rad. Find its arc length, area and perimeter.: Arc = 12 × 0.75 = 9 cm. Area = ½ × 144 × 0.75 = 54 cm². Perimeter = 2 × 12 + 9 = 33 cm (include both radii).

1.5 Trigonometry

Exact values of sin, cos and tan for π/6, π/4 and π/3: sin: 1/2, √2/2, √3/2. cos: √3/2, √2/2, 1/2. tan: 1/√3 (= √3/3), 1, √3.
Principal value ranges of sin⁻¹x, cos⁻¹x and tan⁻¹x: sin⁻¹x: −π/2 ≤ y ≤ π/2. cos⁻¹x: 0 ≤ y ≤ π. tan⁻¹x: −π/2 < y < π/2.
After the calculator value x₁, how do you find the second solution of sin x = k, cos x = k and tan x = k?: sin x = k: 180° − x₁ (π − x₁). cos x = k: 360° − x₁ (2π − x₁). tan x = k: x₁ + 180° (x₁ + π). Then add or subtract 360° (2π) to find every value in the interval.
The two identities in Pure Mathematics 1 trigonometry: tan θ ≡ sin θ / cos θ sin²θ + cos²θ ≡ 1
Greatest value, least value and period of y = a sin bx + c (a, b > 0): Greatest c + a; least c − a; period 360°/b (2π/b). For tan bx the period is 180°/b.
Why must you not cancel sin x in 7 sin x cos x = 2 sin x?: Cancelling loses the solutions of sin x = 0. Factorise instead: sin x(7 cos x − 2) = 0. For 0° ≤ x ≤ 360° this gives x = 0°, 180°, 360°, 73.4° and 286.6°: five solutions.

1.6 Series

Binomial coefficient ⁿCᵣ, and the general term of (a + b)ⁿ: ⁿCᵣ = n!/(r!(n − r)!). The general term ⁿCᵣ aⁿ⁻ʳ bʳ is the (r + 1)th term, so the third term uses ⁿC₂.
Find the coefficient of x³ in (1 + 2x)⁶.: ⁶C₃ × 2³ = 20 × 8 = 160. Raise the whole of 2x to the power, not just x.
nth term and sum of an arithmetic progression (first term a, common difference d): uₙ = a + (n − 1)d Sₙ = ½n(a + l) = ½n{2a + (n − 1)d}, where l is the last term.
nth term, sum and sum to infinity of a geometric progression (first term a, ratio r): uₙ = arⁿ⁻¹ Sₙ = a(1 − rⁿ)/(1 − r), r ≠ 1 S∞ = a/(1 − r), only when |r| < 1.
Three-term tests: when are a, b, c in AP, and when in GP?: AP: 2b = a + c. GP: b² = ac.
Show that the GP 12, 8, 16/3, … converges and find its sum to infinity.: r = 8/12 = 2/3 and |2/3| < 1, so it converges. S∞ = 12/(1 − 2/3) = 36.

1.7 Differentiation

d/dx (xⁿ), and what must you do first with terms like 6/x⁴ or ∛x?: d/dx (xⁿ) = nxⁿ⁻¹ for rational n. Rewrite every term as a single power first: 6/x⁴ = 6x⁻⁴, ∛x = x^(1/3).
Chain rule, with an example: dy/dx = (dy/du) × (du/dx). d/dx (x² + 1)⁴ = 4(x² + 1)³ × 2x = 8x(x² + 1)³.
Gradient of the tangent and of the normal at x = a: Tangent: m = dy/dx at x = a. Normal: −1/m. Then use y − y₁ = m(x − x₁).
How do you find and classify stationary points?: 1. Solve dy/dx = 0 and find each y. 2. Find d²y/dx² at each point: negative → maximum, positive → minimum. 3. If d²y/dx² = 0, test the sign of dy/dx either side.
Connected rates of change: the rule, and the side of a square, x cm, grows at 0.2 cm s⁻¹. Rate of change of area at x = 6?: dA/dt = (dA/dx) × (dx/dt). A = x², dA/dx = 2x = 12, so dA/dt = 12 × 0.2 = 2.4 cm² s⁻¹. Differentiate before substituting the value.
Increasing and decreasing functions: f is increasing where f′(x) > 0 and decreasing where f′(x) < 0. To show f is increasing, prove f′(x) > 0 for all x in the domain (often by completing the square) and say so.

1.8 Integration

∫ xⁿ dx and ∫ (ax + b)ⁿ dx: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, n ≠ −1. ∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/(a(n + 1)) + c, n ≠ −1: divide by a as well as by the new power.
How do you find a curve from its gradient function and a point on it?: 1. Integrate dy/dx, including + c. 2. Substitute the given point to find c. 3. Write y = … in full.
Area between two graphs: ∫ (upper − lower) dx between the x-coordinates of the intersection points. Sketch first to see which graph is on top.
Volume of revolution about the x-axis and about the y-axis: About the x-axis: V = π ∫ y² dx (x-limits). About the y-axis: V = π ∫ x² dy (y-limits). With a hole: π ∫ (outer² − inner²), never π ∫ (outer − inner)².
Evaluate ∫₁^∞ 2/x² dx.: [−2/x]₁^∞. As x → ∞, −2/x → 0. Value = 0 − (−2) = 2. State the limit in words; do not substitute ∞ as a number.
Why is ∫ y dx not always the area between a curve and the x-axis?: Parts of the curve below the x-axis give negative values. Split the integral at each root, find each part, and add their sizes. An area is never negative.