Flashcards · 9660 · Topic 5

Unit P2: Pure Maths (International A-level) flashcards

57 cards on Topic 5, subtopics P2.1 to P2.10. Pick a subtopic or study them all.

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P2.1 Algebra and functions

When does a function have an inverse, and how are f and f⁻¹ related?: Only when f is one-one (restrict the domain if needed). Domain of f⁻¹ = range of f; range of f⁻¹ = domain of f. y = f⁻¹(x) is the reflection of y = f(x) in y = x.
f(x) = 3x − 5 and g(x) = x² + 2. Find fg(x).: fg(x) = f(g(x)): g acts first. fg(x) = 3(x² + 2) − 5 = 3x² + 1. (gf(x) = (3x − 5)² + 2 is different.)
Find the inverse of f(x) = (4x + 1)/(x − 2), x ≠ 2: y(x − 2) = 4x + 1, so x(y − 4) = 2y + 1. f⁻¹(x) = (2x + 1)/(x − 4), x ≠ 4.
Solve |x − 4| = 2x + 1: x − 4 = 2x + 1 gives x = −5, but then the right side is −9: reject. 4 − x = 2x + 1 gives x = 1: both sides equal 3. Answer x = 1.
(3, 4) lies on y = f(x). Find its image on y = 2f(x − 1) + 5.: Inside: solve x − 1 = 3, so x = 4. Outside: y = 2 × 4 + 5 = 13. Image (4, 13).
Express (3x² − 14x − 1)/((x + 1)(x − 3)²) in partial fractions: Form A/(x + 1) + B/(x − 3) + C/(x − 3)². x = −1: 16 = 16A, A = 1. x = 3: −16 = 4C, C = −4. x² terms: 3 = A + B, B = 2. Result: 1/(x + 1) + 2/(x − 3) − 4/(x − 3)².

P2.2 Sequences and series

The binomial series for (1 + x)ⁿ with n rational: (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + … Valid for |x| < 1 when n is negative or a fraction. For a positive integer n the series stops and holds for all x.
How do you expand (a + bx)ⁿ and state its validity?: Take out aⁿ: aⁿ(1 + (b/a)x)ⁿ. Expand with u = (b/a)x, then multiply by aⁿ. Valid for |u| < 1, that is |x| < |a/b|.
Series for (1 + u)^(−1/2) up to u³: 1 − u/2 + 3u²/8 − 5u³/16 + …, valid for |u| < 1. With u = 4x: 1 − 2x + 6x² − 20x³.
Expand (7 − x)/((1 − x)(1 + 2x)) up to x² and state where it is valid: = 2/(1 − x) + 5/(1 + 2x) = 2(1 + x + x²) + 5(1 − 2x + 4x²) + … = 7 − 8x + 22x² + …, valid for |x| < 1/2 (the narrower range).
State where the expansions of (2 + x)^(1/2) and (3 − 2x)⁻² are valid: (2 + x)^(1/2) = √2(1 + x/2)^(1/2): valid for |x| < 2. (3 − 2x)⁻² = (1/9)(1 − 2x/3)⁻²: valid for |x| < 3/2.

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

How do you convert parametric equations to a Cartesian equation?: Rearrange the easier equation for the parameter and substitute; or, for trig, use sin²θ + cos²θ = 1 or a double angle identity. Then state any restriction on x or y.
Cartesian equations of x = t², y = 2t and of x = a cos θ, y = b sin θ: x = t², y = 2t: y² = 4x (a parabola). x = a cos θ, y = b sin θ: x²/a² + y²/b² = 1 (an ellipse; a circle if a = b).
Cartesian equations of x = 1/t, y = 3t and of x = t + 1/t, y = t − 1/t: x = 1/t, y = 3t: xy = 3 (a rectangular hyperbola, t ≠ 0). x = t + 1/t, y = t − 1/t: x² − y² = 4 (for t > 0, only x ≥ 2).
Parametric equations for the circle (x − p)² + (y − q)² = r²: x = p + r cos θ, y = q + r sin θ.
Find the Cartesian equation of x = 3 sin θ, y = 1 − cos 2θ: cos 2θ = 1 − 2sin²θ, so y = 2sin²θ = 2(x/3)² = 2x²/9, with −3 ≤ x ≤ 3.

P2.4 Trigonometry

sec x, cosec x and cot x: definitions and where each is undefined: sec x = 1/cos x, undefined at x = π/2 + nπ. cosec x = 1/sin x, undefined at x = nπ. cot x = cos x/sin x, undefined at x = nπ. |sec x| ≥ 1 and |cosec x| ≥ 1.
The two identities that come from sin²x + cos²x = 1: Divide by cos²x: tan²x + 1 = sec²x. Divide by sin²x: 1 + cot²x = cosec²x.
Domains and ranges of sin⁻¹x, cos⁻¹x and tan⁻¹x: sin⁻¹x: domain [−1, 1], range [−π/2, π/2]. cos⁻¹x: domain [−1, 1], range [0, π]. tan⁻¹x: all real x, range (−π/2, π/2).
Compound angle formulae (sin, cos and tan of A ± B): sin(A ± B) = sin A cos B ± cos A sin B cos(A ± B) = cos A cos B ∓ sin A sin B tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)
Double angle formulae, including all three forms of cos 2x: sin 2x = 2 sin x cos x cos 2x = cos²x − sin²x = 2cos²x − 1 = 1 − 2sin²x tan 2x = 2 tan x/(1 − tan²x)
Write 5 cos θ − 3 sin θ in the form r cos(θ + α): r cos α = 5 and r sin α = 3, so r = √34 and tan α = 3/5, α = 31.0°. 5 cos θ − 3 sin θ = √34 cos(θ + 31.0°).

P2.5 Exponentials and logarithms

Compare the graphs of y = eˣ and y = ln x: eˣ: all real x, y > 0, through (0, 1) and (1, e), asymptote y = 0. ln x: x > 0 only, all real y, through (1, 0) and (e, 1), asymptote x = 0. Each is the reflection of the other in y = x.
Inverse identities for e and ln: ln(eˣ) = x for all x; e^(ln x) = x for x > 0. ln 1 = 0 and ln e = 1.
Solve e²ˣ − 7eˣ + 10 = 0: Let u = eˣ: u² − 7u + 10 = 0, (u − 2)(u − 5) = 0. eˣ = 2 or 5, so x = ln 2 or x = ln 5.
Exponential growth and decay models: doubling time and half-life: Growth N = Ae^(kt), k > 0; decay N = Ae^(−kt), k > 0. A is the value at t = 0. Doubling time and half-life are both ln 2/k.
Find the inverse of f(x) = 3eˣ + 2: y = 3eˣ + 2, so x = ln((y − 2)/3). f⁻¹(x) = ln((x − 2)/3), with domain x > 2 (the range of f).
Asymptote and y-intercept of y = 2eˣ − 5: Asymptote y = −5; y-intercept (0, −3).

P2.6 Differentiation

Derivatives of e^(kx), ln x, ln(ax + b), sin kx, cos kx and tan kx: e^(kx) → ke^(kx) ln x → 1/x ln(ax + b) → a/(ax + b) sin kx → k cos kx cos kx → −k sin kx tan kx → k sec²kx (x in radians)
The chain, product and quotient rules: Chain: dy/dx = (dy/du)(du/dx). Product: y = uv gives dy/dx = uv′ + vu′. Quotient: y = u/v gives dy/dx = (vu′ − uv′)/v².
Differentiate y = (5 − x)/(2x + 3): dy/dx = [(−1)(2x + 3) − (5 − x)(2)]/(2x + 3)² = −13/(2x + 3)².
Find the gradient of x³ − 2xy + y² = 4 at (1, 3): 3x² − 2y − 2x dy/dx + 2y dy/dx = 0 dy/dx = (2y − 3x²)/(2y − 2x) At (1, 3): (6 − 3)/(6 − 2) = 3/4.
x = 1 + 2t², y = t³ − 6t. Find dy/dx and the point when t = 2.: dy/dx = (dy/dt)/(dx/dt) = (3t² − 6)/(4t). At t = 2: 6/8 = 3/4, at the point (9, −4).
x = y² + 7y. Find dy/dx where y = 2.: dx/dy = 2y + 7 = 11, so dy/dx = 1/(dx/dy) = 1/11.

P2.7 Integration

Integrals of e^(kx), 1/x, 1/(ax + b), sin kx and cos kx: e^(kx) → e^(kx)/k 1/x → ln|x| 1/(ax + b) → (1/a) ln|ax + b| sin kx → −cos(kx)/k cos kx → sin(kx)/k Add + c in each case.
Evaluate ∫ from 0 to 2 of 3x²/(x³ + 1) dx: The numerator is the derivative of the denominator, so the integral is ln(x³ + 1). [ln(x³ + 1)] from 0 to 2 = ln 9 − ln 1 = ln 9 = 2 ln 3.
Integration by substitution for a definite integral: Choose u, find du/dx and replace dx. Replace any remaining x using u. Change the limits to u-values at the same time, integrate in u, and evaluate in u (no need to return to x).
Integration by parts, and the choice of u for x e^(4x) and for x ln x: ∫ u v′ dx = uv − ∫ u′ v dx. x e^(4x): u = x, giving ¼x e^(4x) − (1/16)e^(4x) + c. x ln x: u = ln x, giving (x²/2) ln x − x²/4 + c.
Find ∫ 7/((x − 2)(x + 5)) dx: 7/((x − 2)(x + 5)) = 1/(x − 2) − 1/(x + 5). Integral: ln|x − 2| − ln|x + 5| + c = ln|(x − 2)/(x + 5)| + c.
Volumes of revolution about the x-axis and about the y-axis: About the x-axis: V = π ∫ y² dx with x-limits. About the y-axis: V = π ∫ x² dy with y-limits. Example: y = x² from y = 0 to 4 about the y-axis: π ∫ y dy = 8π.

P2.8 Differential equations

Turn into differential equations: "decreasing at a rate proportional to the amount m", and "cooling of T in surroundings at S": dm/dt = −km, k > 0. dT/dt = −k(T − S), k > 0. State what k is and match the sign to the words.
How do you solve a separable differential equation?: Write dy/dx = f(x)g(y), separate to (1/g(y)) dy = f(x) dx, integrate each side with a single + c, use the initial condition for c, then rearrange for y if asked.
General solutions of dy/dt = ky and of dT/dt = −k(T − S): dy/dt = ky: y = Ae^(kt), A = y at t = 0. dT/dt = −k(T − S): T − S = Be^(−kt), so T tends to S.
Solve dy/dx = 2x/(y + 1) with y = 1 when x = 0 (y > −1): (y + 1) dy = 2x dx, so (y + 1)²/2 = x² + c. At x = 0, y = 1: c = 2. (y + 1)² = 2x² + 4, so y = −1 + √(2x² + 4).
A body at 70 °C in a room at 18 °C reaches 46 °C after 10 minutes. Find k and the temperature at 20 minutes.: T − 18 = 52e^(−kt). 28 = 52e^(−10k), so k = (1/10) ln(52/28) = 0.0619. At t = 20: T = 18 + 52e^(−20k) = 33.1 °C (3 s.f.).

P2.9 Numerical methods

Locating a root by a change of sign: If f is continuous on [a, b] and f(a) and f(b) differ in sign, there is at least one root in (a, b). State the values, the signs, continuity and the conclusion.
Iteration xₙ₊₁ = g(xₙ): when does it converge to the root α?: Rearrange f(x) = 0 as x = g(x). The iteration converges near α when |g′(α)| < 1 and the start value is close enough. Different rearrangements may converge or diverge.
Staircase compared with cobweb diagrams: Staircase: 0 < g′(α) < 1, values approach α from one side. Cobweb: −1 < g′(α) < 0, values alternate either side of α.
The mid-ordinate rule, and whether it over- or under-estimates: ∫ from a to b of y dx ≈ h × (sum of the heights at the n strip midpoints), h = (b − a)/n. Convex curve: too low. Concave curve: too high.
Simpson's rule: (h/3)[first + last + 4(odd ordinates) + 2(even interior ordinates)], with an even number of strips. It is exact for polynomials up to cubic.
When can a change of sign mislead you?: An asymptote in the interval can give a sign change with no root; a repeated root (touching the axis) gives no sign change; two roots in one interval can leave the end signs equal.

P2.10 Vectors

Magnitude of (x, y, z), the vector AB, and the mid-point of AB: |(x, y, z)| = √(x² + y² + z²). AB = OB − OA ("end minus start"). Mid-point: (OA + OB)/2. Example: |(3, 4, 12)| = 13.
Vector equation of a line: r = a + tb, where a is a point on the line, b is a direction vector and t is any real number. Through A and B: a = OA, b = AB.
How do you test whether two lines in 3D intersect?: Use different parameters. If the directions are parallel, check for a common point. Otherwise equate components, solve two equations, and test the third: if it holds they intersect, if not they are skew.
Scalar product: The number xa + yb + zc found from two vectors (x, y, z) and (a, b, c); also called the dot product. Geometric form: u.v = |u||v| cos θ, so u.v = 0 for perpendicular non-zero vectors.
Skew lines: Lines in three dimensions that are neither parallel nor intersecting. They still have an angle between them.
Foot of the perpendicular from P to a line, and the perpendicular distance: The foot X is the point on the line nearest to P, where PX meets the line at a right angle: solve PX.d = 0 for t. The perpendicular distance is |PX|, the shortest distance from P to the line.