Flashcards · YMA01 · Topic 3

Unit P3: Pure Mathematics 3 flashcards

35 cards on Topic 3, subtopics P3.1 to P3.6. Pick a subtopic or study them all.

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One-one and many-one functions, and which ones have an inverse: One-one: each output comes from exactly one input. Many-one: two or more inputs give the same output. Only a one-one function has an inverse, so a many-one function needs its domain restricted first.
What does fg(x) mean?: fg(x) = f(g(x)): apply g first, then f. In general fg ≠ gf.
How do you find f⁻¹(x), and how are f and f⁻¹ related?: Put y = f(x), rearrange for x, then rename y as x. Domain of f⁻¹ = range of f; range of f⁻¹ = domain of f. The graph of f⁻¹ is the reflection of f in y = x.
y = |f(x)| compared with y = f(|x|): y = |f(x)|: reflect every part below the x-axis in the x-axis. y = f(|x|): keep the part for x ≥ 0, delete x < 0, and reflect the kept part in the y-axis.
Solve |2x + 1| = 5: 2x + 1 = 5 gives x = 2; 2x + 1 = −5 gives x = −3. Both check in the original. When the other side is not a constant, test every answer in the original equation.
Divide 2x² + x − 1 by x + 2: 2x² + x − 1 = (x + 2)(2x − 3) + 5, so (2x² + x − 1)/(x + 2) = 2x − 3 + 5/(x + 2).

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
Ranges (outputs) of arcsin x, arccos x and arctan x in radians: arcsin x: −π/2 ≤ y ≤ π/2 (input −1 to 1). arccos x: 0 ≤ y ≤ π (input −1 to 1). arctan x: −π/2 < y < π/2 (any real input).
Identities linking sec² with tan², and cosec² with cot²: sec²θ ≡ 1 + tan²θ cosec²θ ≡ 1 + cot²θ Both come from dividing sin²θ + cos²θ ≡ 1 by cos²θ or by sin²θ.
Compound angle formulae for sin(A ± B), cos(A ± B) and tan(A ± B): sin(A ± B) ≡ sin A cos B ± cos A sin B (signs match). cos(A ± B) ≡ cos A cos B ∓ sin A sin B (signs opposite). tan(A ± B) ≡ (tan A ± tan B)/(1 ∓ tan A tan B).
Double angle formulae: sin 2A ≡ 2 sin A cos A cos 2A ≡ cos²A − sin²A ≡ 2cos²A − 1 ≡ 1 − 2sin²A tan 2A ≡ 2 tan A/(1 − tan²A)
How do you write a cos θ + b sin θ in the form R cos(θ − α)?: Expand: R cos α = a and R sin α = b. Then R = √(a² + b²) and tan α = b/a (α acute when a, b > 0). Greatest value R, least value −R. Example: √3 cos θ + sin θ = 2 cos(θ − π/6).

Key features of y = eˣ and y = ln x: y = eˣ: through (0, 1), range y > 0, asymptote y = 0, increasing. y = ln x: through (1, 0), domain x > 0, asymptote x = 0. Each is the reflection of the other in y = x.
Inverse relations between eˣ and ln x: e^(ln x) = x for x > 0. ln(eˣ) = x for all x.
Solve e^(4 − 3x) = 9 exactly, then to 3 s.f.: 4 − 3x = ln 9, so x = (4 − ln 9)/3 = 0.601 (3 s.f.).
Asymptote and y-intercept of y = e^(ax + b) + c: Asymptote y = c. y-intercept (0, eᵇ + c). The curve is increasing if a > 0 and decreasing if a < 0.
How do you turn y = axⁿ into a straight-line graph?: Take logs: log y = n log x + log a. Plot log y against log x: gradient n, intercept log a, so a = 10^(intercept) with base-10 logs.
How do you turn y = kbˣ into a straight-line graph?: Take logs: log y = (log b)x + log k. Plot log y against x (not log x): gradient log b, intercept log k.

Derivatives of eᵏˣ, ln kx, sin kx, cos kx, tan kx and aˣ (x in radians): eᵏˣ → keᵏˣ ln kx → 1/x sin kx → k cos kx cos kx → −k sin kx tan kx → k sec² kx aˣ → aˣ ln a (not xaˣ⁻¹)
Derivatives of sec x, cosec x and cot x: sec x → sec x tan x cosec x → −cosec x cot x cot x → −cosec²x
The product rule: If y = uv, then dy/dx = u′v + uv′. Example: y = x² cos 3x gives dy/dx = 2x cos 3x − 3x² sin 3x.
The quotient rule: If y = u/v, then dy/dx = (vu′ − uv′)/v².
The chain rule: dy/dx = (dy/du) × (du/dx). Examples: d/dx e^f(x) = f′(x)e^f(x); d/dx ln f(x) = f′(x)/f(x); d/dx (f(x))ⁿ = n(f(x))ⁿ⁻¹ f′(x).
How do you find dy/dx when x is given in terms of y?: Use dy/dx = 1/(dx/dy). Example: x = y³ + 2y gives dx/dy = 3y² + 2, so dy/dx = 1/(3y² + 2).

Integrals of eᵏˣ, 1/x, sin kx and cos kx: ∫ eᵏˣ dx = (1/k)eᵏˣ + c ∫ 1/x dx = ln|x| + c ∫ sin kx dx = −(1/k) cos kx + c ∫ cos kx dx = (1/k) sin kx + c
Integrate f′(x)/f(x): ∫ f′(x)/f(x) dx = ln|f(x)| + c. The numerator must be the derivative of the denominator, or a constant multiple of it.
Find ∫ 5/(2x + 1) dx: The derivative of 2x + 1 is 2, so write 5/(2x + 1) = (5/2) × 2/(2x + 1). Answer: (5/2) ln|2x + 1| + c.
Integrals of tan x and sec² kx: ∫ tan x dx = −ln|cos x| + c = ln|sec x| + c. ∫ sec² kx dx = (1/k) tan kx + c.
How do you integrate sin²x and cos²x?: Use the double angle identities first: sin²x = ½(1 − cos 2x), cos²x = ½(1 + cos 2x). So ∫ sin²x dx = x/2 − (sin 2x)/4 + c and ∫ cos²x dx = x/2 + (sin 2x)/4 + c.
Evaluate ∫ from 0 to π/16 of sec² 4x dx: [(1/4) tan 4x] from 0 to π/16 = (1/4) tan(π/4) − 0 = 1/4.

How do you show that f(x) = 0 has a root between a and b?: Calculate f(a) and f(b). State that f is continuous on [a, b]. If f(a) and f(b) have opposite signs, there is a root between a and b.
When can a change of sign mislead you?: A vertical asymptote (such as tan x between 1 and 2) can give a sign change with no root. A repeated root touches the axis without a sign change. Two roots in one interval give the same sign at both ends.
How do you justify that a root is 2.543 to 3 decimal places?: Show that f(2.5425) and f(2.5435) have opposite signs, and state that f is continuous, so the root lies in [2.5425, 2.5435).
Iteration xₙ₊₁ = g(xₙ): what does the limit of the sequence tell you?: If the terms converge to α, then α = g(α), so α is a root of x = g(x), the rearranged equation. Carry full accuracy between steps and stop when terms agree to the accuracy needed.
What does a staircase or cobweb diagram show?: It is drawn with y = g(x) and y = x. The steps show whether successive iterates move towards the root (convergence) or away from it (divergence) for that rearrangement and starting value.