Flashcards · YMA01 · Topic 2

Unit P2: Pure Mathematics 2 flashcards

47 cards on Topic 2, subtopics P2.1 to P2.8. Pick a subtopic or study them all.

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P2.1 Proof

How do you write a general even integer, a general odd integer and three consecutive integers?: Even: 2n. Odd: 2n + 1. Consecutive: n, n + 1, n + 2, where n is an integer.
Proof by exhaustion: Split the statement into a set of cases that covers every possibility, prove each case, and say that the cases are complete.
Disproof by counter-example: Give one value that meets the stated conditions, work out both sides, and say clearly why the statement fails. One counter-example is enough to disprove; many confirming cases never prove.
Prove that n² + n is even for every integer n: n² + n = n(n + 1). If n is even, n(n + 1) is even × odd = even. If n is odd, n + 1 is even, so n(n + 1) is odd × even = even. Both cases give an even number.
What should the first and last lines of a proof by deduction contain?: First: say what the letters stand for, such as "let n be an integer". Last: a sentence linking back to the claim, such as "so the expression is a multiple of 4". Each step must follow from the one before.

P2.2 Algebra and functions

The remainder theorem: When f(x) is divided by (ax + b), the remainder is f(−b/a). For a divisor (x − a), the remainder is f(a).
The factor theorem: (ax + b) is a factor of f(x) exactly when f(−b/a) = 0.
Factorise f(x) = x³ − 4x² + x + 6 fully: f(−1) = −1 − 4 − 1 + 6 = 0, so (x + 1) is a factor. Dividing gives x² − 5x + 6 = (x − 2)(x − 3). f(x) = (x + 1)(x − 2)(x − 3).
Find the remainder when x³ − 5x + 2 is divided by (x + 3): Use x = −3: f(−3) = −27 + 15 + 2 = −10. The remainder is −10.
f(x) = x³ + kx² − x − 10 has a factor (x − 2). Find k.: f(2) = 8 + 4k − 2 − 10 = 4k − 4 = 0, so k = 1.
Which values should you try when looking for a factor of a cubic?: If the leading coefficient is 1, try the factors of the constant term, positive and negative. Otherwise try fractions p/q with p dividing the constant and q dividing the leading coefficient.

P2.3 Coordinate geometry in the (x, y) plane

Equation of a circle with centre (a, b) and radius r: (x − a)² + (y − b)² = r². Example: centre (−2, 4), radius 3 gives (x + 2)² + (y − 4)² = 9.
Find the centre and radius of x² + y² − 2x + 6y − 6 = 0: Complete the square in x and in y: (x − 1)² − 1 + (y + 3)² − 9 − 6 = 0 (x − 1)² + (y + 3)² = 16. Centre (1, −3), radius 4.
Three circle properties used in coordinate geometry: The angle in a semicircle is 90°. The perpendicular from the centre to a chord bisects the chord. The tangent at a point is perpendicular to the radius there.
How do you find the tangent to a circle at a point P on it?: 1. Check P is on the circle. 2. Find the gradient of the radius from the centre to P. 3. The tangent gradient is its negative reciprocal. 4. Use y − y₁ = m(x − x₁) through P.
How do you decide whether a line meets a circle, touches it, or misses it?: Substitute the line into the circle equation to get a quadratic. Two real roots: meets at two points. One repeated root: tangent. No real roots: misses the circle.
A diameter has endpoints (1, 2) and (7, 10). Find the centre and radius.: Centre = midpoint = (4, 6). Radius = half the length = ½√(6² + 8²) = ½ × 10 = 5.

P2.4 Sequences and series

Arithmetic sequence: nth term and sum of the first n terms: uₙ = a + (n − 1)d. Sₙ = (n/2)(2a + (n − 1)d) = (n/2)(a + l), where l is the last term.
Geometric sequence: nth term and sum of the first n terms: uₙ = arⁿ⁻¹. Sₙ = a(1 − rⁿ)/(1 − r), for r ≠ 1.
Sum to infinity of a geometric series, and when it exists: S∞ = a/(1 − r), valid only for |r| < 1 (a convergent series). Example: a = 9, r = 1/3 gives S∞ = 13.5.
Sum of the first n natural numbers, and Σ (r = 1 to 12) of (2r + 1): 1 + 2 + … + n = n(n + 1)/2. Σ (2r + 1) = 2 × (12 × 13/2) + 12 = 156 + 12 = 168.
Binomial expansion of (a + bx)ⁿ for a positive integer n: (a + bx)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ(bx)ʳ for r = 0 to n, which has n + 1 terms. ⁿCᵣ = n!/(r!(n − r)!). Bracket the whole of bx, including its sign.
Find the coefficient of x² in (1 + 3x)⁵: Term in x²: ⁵C₂ × (3x)² = 10 × 9x² = 90x². The coefficient is 90.

P2.5 Exponentials and logarithms

Key features of the graph of y = aˣ (a > 0, a ≠ 1): Passes through (0, 1). Asymptote y = 0. Range y > 0. Increasing if a > 1; decreasing if 0 < a < 1. aˣ is never zero or negative.
What does log_a x = y mean?: It is the same statement as aʸ = x, for a > 0, a ≠ 1 and x > 0. So log_a a = 1 and log_a 1 = 0.
The three laws of logarithms: log_a(xy) = log_a x + log_a y log_a(x/y) = log_a x − log_a y log_a(xᵏ) = k log_a x Traps: log_a(x + y) is not log_a x + log_a y.
Solve 5^(2x + 1) = 40: (2x + 1) ln 5 = ln 40 2x + 1 = ln 40/ln 5 = 2.2920 x = 0.646 (3 s.f.)
Change of base formula for logarithms: log_a b = log_c b / log_c a. Also log_a b = 1/log_b a. Example: log₄ 8 = log₂ 8 / log₂ 4 = 3/2.
Find the smallest whole n with 1.5ⁿ ≥ 100: n ln 1.5 ≥ ln 100, so n ≥ ln 100/ln 1.5 = 11.36. The smallest whole n is 12. When the base is between 0 and 1 its log is negative, so dividing by it reverses the inequality.

P2.6 Trigonometry

The two identities needed in Unit P2 trigonometry: tan θ = sin θ / cos θ (cos θ ≠ 0). sin²θ + cos²θ = 1, so sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ.
Second solutions in 0° to 360° for sin u = k, cos u = k and tan u = k (principal value α): sin u = k: α and 180° − α, repeating every 360°. cos u = k: α and 360° − α (that is, −α), repeating every 360°. tan u = k: α, then add 180° each time.
Which trig ratios are positive in each quadrant (0° to 360°)?: First (0° to 90°): all. Second (90° to 180°): sine. Third (180° to 270°): tangent. Fourth (270° to 360°): cosine.
Solve cos 3x = 1/2 for 0° ≤ x < 180°: Interval for 3x: 0° ≤ 3x < 540°. cos u = 1/2: u = 60°, 300°, 420°. x = 20°, 100°, 140°.
How do you solve an equation like a cos²x + b sin x + c = 0?: Replace cos²x by 1 − sin²x to get a quadratic in sin x. Factorise, reject any value outside −1 to 1, then find all angles in the interval.
Why should you factorise rather than divide by sin x when solving sin x = k tan x?: Dividing by sin x loses the solutions where sin x = 0. Write tan x = sin x/cos x, collect on one side, take out sin x as a factor, then solve each factor.

P2.7 Differentiation

Stationary point: A point where dy/dx = 0. Find the y-coordinate by substituting the x-value back into the curve.
Classify a stationary point with the second derivative: d²y/dx² > 0: minimum. d²y/dx² < 0: maximum. d²y/dx² = 0: inconclusive, so test the sign of dy/dx on each side.
Classify a stationary point using the sign of dy/dx on each side: − then +: minimum. + then −: maximum. Same sign on both sides: point of inflection.
Increasing and decreasing functions: f is increasing where f′(x) > 0 and decreasing where f′(x) < 0.
Find and classify the stationary points of y = x³ + 3x² − 24x + 1: dy/dx = 3x² + 6x − 24 = 3(x + 4)(x − 2). x = 2: y = −27, d²y/dx² = 6x + 6 = 18 > 0, minimum (2, −27). x = −4: y = 81, d²y/dx² = −18 < 0, maximum (−4, 81).
How do you solve an optimisation problem?: 1. Write the quantity to optimise. 2. Use the constraint to leave one variable. 3. Differentiate and set equal to zero. 4. Prove it is a maximum or minimum. 5. Answer in the words of the question, with units.

P2.8 Integration

Evaluating a definite integral: ∫ from a to b of f(x) dx = F(b) − F(a), where F is an antiderivative. No constant is needed.
Area between a curve and the x-axis when part of the curve is below the axis: A definite integral is signed: a region below the axis gives a negative value. Split the integral at each root, find each part, make every part positive, then add.
Area enclosed between two graphs: Find the intersections, decide which graph is higher, then integrate (upper − lower) between the intersections. Split wherever the graphs cross.
Find the area enclosed by y = x² and y = 4x + 5: x² = 4x + 5 gives x = −1 or 5. ∫ from −1 to 5 of (4x + 5 − x²) dx = [2x² + 5x − x³/3] = 100/3 − (−8/3) = 36.
The trapezium rule: Area ≈ (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)], where h = (b − a)/n. n strips need n + 1 y-values.
When does the trapezium rule over-estimate, and how can it be improved?: It over-estimates when the curve bends upwards (the chords lie above the curve) and under-estimates when it bends downwards. More strips (smaller h) give a better estimate.