Flashcards · 9709 · Topic 2

Pure Mathematics 2 flashcards

35 cards on Topic 2, subtopics 2.1 to 2.6. Pick a subtopic or study them all.

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Rewrite |x − a| < b and |x − a| > b without the modulus (b > 0).: |x − a| < b ⇔ a − b < x < a + b. |x − a| > b ⇔ x < a − b or x > a + b.
How do you solve |a| = |b| when both sides are moduli?: Square both sides: |a| = |b| ⇔ a² = b². When one side is not a modulus, solve the two cases separately and check each answer instead.
Remainder theorem: The remainder when p(x) is divided by (x − a) is p(a). When divided by (ax + b) it is p(−b/a).
Factor theorem: (ax + b) is a factor of p(x) ⇔ p(−b/a) = 0. For (3x + 1), substitute x = −1/3.
What form does the remainder take when you divide by a linear polynomial, and by a quadratic?: Linear divisor: a constant. Quadratic divisor: Ax + B (A or B may be zero). Identity: p(x) ≡ divisor × quotient + remainder.
Find the remainder when 9x³ − 3x + 5 is divided by (3x + 1).: p(−1/3) = 9(−1/27) − 3(−1/3) + 5 = −1/3 + 1 + 5 = 17/3.

Definition of a logarithm: logₐ b = x ⇔ aˣ = b, with a > 0, a ≠ 1 and b > 0. lg x = log₁₀ x; ln x = logₑ x.
The three laws of logarithms: log(xy) = log x + log y log(x/y) = log x − log y log(xⁿ) = n log x log(x + y) cannot be split.
eˣ and ln x as inverses: the key facts and how their graphs are related: e^(ln x) = x for x > 0; ln(eˣ) = x for all x. y = eˣ passes through (0, 1), asymptote y = 0; y = ln x passes through (1, 0), asymptote x = 0. They are reflections in y = x.
Solve 6ˣ = 11.: Take logs: x ln 6 = ln 11, so x = ln 11/ln 6 = 1.34 (3 s.f.).
Which graph is a straight line for y = kxⁿ, and which for y = k(aˣ)?: y = kxⁿ: ln y = ln k + n ln x, so plot ln y against ln x (gradient n). y = k(aˣ): ln y = ln k + x ln a, so plot ln y against x (gradient ln a). In both, the intercept is ln k, so k = e^(intercept).
Solving an inequality such as 0.6ⁿ < 0.01 with logs: what is the trap?: n ln 0.6 < ln 0.01. ln 0.6 is negative, so dividing by it reverses the inequality: n > ln 0.01/ln 0.6 = 9.02, so the least integer is n = 10.

Define sec θ, cosec θ and cot θ.: sec θ = 1/cos θ cosec θ = 1/sin θ cot θ = 1/tan θ = cos θ/sin θ Each has the same sign as its parent function.
The two Pythagorean identities involving sec, cosec, tan and cot: 1 + tan²θ ≡ sec²θ 1 + cot²θ ≡ cosec²θ
Compound angle formulae for sin(A ± B) and cos(A ± B): sin(A ± B) ≡ sin A cos B ± cos A sin B cos(A ± B) ≡ cos A cos B ∓ sin A sin B (the sign changes).
sin 2A and the three forms of cos 2A: sin 2A ≡ 2 sin A cos A cos 2A ≡ cos²A − sin²A ≡ 2cos²A − 1 ≡ 1 − 2sin²A
How do you write a sin θ + b cos θ as R sin(θ + α)?: 1. Expand: R sin θ cos α + R cos θ sin α. 2. Match: R cos α = a, R sin α = b. 3. R = √(a² + b²), tan α = b/a. Greatest value R, least value −R.
Express 5 sin θ + 2 cos θ in the form R sin(θ + α), 0° < α < 90°.: R = √(25 + 4) = √29; tan α = 2/5, α = 21.80°. 5 sin θ + 2 cos θ = √29 sin(θ + 21.80°).

Derivatives of eˣ, ln x, sin x, cos x and tan x: eˣ → eˣ ln x → 1/x sin x → cos x cos x → −sin x tan x → sec²x (x in radians)
Product rule, with an example: d/dx (uv) = u dv/dx + v du/dx. d/dx (x e⁻ˣ) = −x e⁻ˣ + e⁻ˣ = e⁻ˣ(1 − x).
Quotient rule: d/dx (u/v) = (v du/dx − u dv/dx)/v². The minus sign makes the order matter: v du/dx comes first.
Gradient of a parametric curve x = f(t), y = g(t): dy/dx = (dy/dt) ÷ (dx/dt), usually left in terms of t. Horizontal tangent where dy/dt = 0; vertical tangent where dx/dt = 0.
Implicit differentiation: how do you differentiate y² and xy with respect to x?: d/dx (y²) = 2y dy/dx. d/dx (xy) = x dy/dx + y (product rule). Then collect the dy/dx terms on one side and factorise.
Differentiate e^(kx), ln(ax + b) and sin(kx).: e^(kx) → k e^(kx) ln(ax + b) → a/(ax + b) sin(kx) → k cos(kx)

∫ e^(ax + b) dx and ∫ 1/(ax + b) dx: ∫ e^(ax + b) dx = (1/a) e^(ax + b) + c ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + c
∫ sin(ax + b) dx, ∫ cos(ax + b) dx and ∫ sec²(ax + b) dx: −(1/a) cos(ax + b) + c (1/a) sin(ax + b) + c (1/a) tan(ax + b) + c (radians)
How do you integrate sin²x and cos²x?: Use the double-angle formula: sin²x = ½(1 − cos 2x), integral ½x − ¼ sin 2x + c. cos²x = ½(1 + cos 2x), integral ½x + ¼ sin 2x + c.
The trapezium rule: ∫ₐᵇ y dx ≈ ½h{y₀ + 2(y₁ + … + yₙ₋₁) + yₙ}, where h = (b − a)/n. n strips need n + 1 ordinates. It is not printed in MF19.
When does the trapezium rule over-estimate, and when does it under-estimate?: Curve bends upward (chords above the curve, e.g. eˣ): over-estimate. Curve bends downward (chords below the curve, e.g. ln x, √x): under-estimate. Give the reason from a sketch.
Find the exact value of ∫₀² 3/(4 − x) dx.: [−3 ln|4 − x|]₀² = −3 ln 2 + 3 ln 4 = 3 ln 2. Here a = −1, so the minus sign comes out in front.

How do you show that f(x) = 0 has a root between a and b?: Work out f(a) and f(b) and show they have opposite signs. State: "sign change and f is continuous, so a root lies between a and b".
When can the sign-change method mislead you?: A repeated (touching) root, or two roots close together, gives no sign change. A discontinuity, such as an asymptote of 1/x or tan x, can give a sign change with no root.
An iteration xₙ₊₁ = F(xₙ) converges to α. What equation does α satisfy?: α = F(α), so α is a root of x = F(x). To find α exactly, replace xₙ₊₁ and xₙ by α and solve.
When do you stop iterating, and how can you confirm a root is 2.66 to 2 d.p.?: Stop when two successive iterates round to the same value (use the stored calculator value each time). Confirm by a sign change of f on [2.655, 2.665].
xₙ₊₁ = 6/(xₙ + 1), x₁ = 1, converges to α. Find α exactly.: α = 6/(α + 1) gives α² + α − 6 = 0, so (α + 3)(α − 2) = 0. All iterates are positive, so α = 2.