Flashcards · 9660 · Topic 2

Unit PSM1 – PP1: Pure Maths (International AS) flashcards

18 cards on Topic 2, subtopics PP1.1 to PP1.3. Pick a subtopic or study them all.

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Circle equation, and the centre and radius of x² + y² + px + qy + c = 0: Centre (a, b), radius r: (x − a)² + (y − b)² = r². Expanded form: centre (−p/2, −q/2), r² = p²/4 + q²/4 − c (from completing the square).
Find the centre and radius of x² + y² − 12x + 6y + 9 = 0: (x − 6)² − 36 + (y + 3)² − 9 + 9 = 0 (x − 6)² + (y + 3)² = 36. Centre (6, −3), radius 6.
How do you tell whether a point is inside, on or outside a circle?: Work out (x − a)² + (y − b)² for the point and compare with r²: less means inside, equal means on the circle, greater means outside.
Three circle properties used in PP1.1: The angle in a semicircle is 90°. The perpendicular from the centre to a chord bisects the chord. The tangent is perpendicular to the radius at the point of contact.
Find the tangent to (x − 2)² + (y − 3)² = 52 at (8, 7): Check: 36 + 16 = 52. Radius gradient 4/6 = 2/3, so tangent gradient −3/2. y − 7 = −(3/2)(x − 8), which gives 3x + 2y = 38.
Translating a circle by the vector (s, t): The centre (a, b) moves to (a + s, b + t); the radius is unchanged. Example: (x + 3)² + (y − 1)² = 49 translated by (2, −5) gives (x + 1)² + (y + 4)² = 49.

Sine rule, cosine rule and the area of a triangle: a/sin A = b/sin B = c/sin C. a² = b² + c² − 2bc cos A, or cos A = (b² + c² − a²)/(2bc). Area = ½ab sin C, with C between sides a and b.
a = 5, b = 8, A = 30°. Find the possible values of B.: sin B = 8 sin 30°/5 = 0.8, so B = 53.1° or 180° − 53.1° = 126.9°. Both leave room for a third angle, so there are two triangles (the ambiguous case).
Radians, arc length, sector area and segment area: π rad = 180°. Arc l = rθ; sector area A = ½r²θ (θ in radians). Segment area = ½r²(θ − sin θ). Example: r = 6, θ = 2.5 gives arc 15 and area 45.
Periods of y = sin x, cos x and tan x, and the period of y = cos 4x: sin x and cos x: 360° (2π). tan x: 180° (π), with asymptotes at 90° and 270°. For sin kx, cos kx and tan kx divide the period by k, so cos 4x has period 90°.
cos θ = −12/13 and θ is in the second quadrant. Find sin θ and tan θ.: sin²θ = 1 − 144/169 = 25/169. Sine is positive in the second quadrant, so sin θ = 5/13. tan θ = sin θ/cos θ = −5/12.
Solve 2sin²θ + 5cos θ = 4 for 0° ≤ θ < 360°: Use sin²θ = 1 − cos²θ: 2cos²θ − 5cos θ + 2 = 0. (2cos θ − 1)(cos θ − 2) = 0. cos θ = 2 is impossible, so cos θ = 1/2. θ = 60° or 300°.

Key features of the graph of y = aˣ: Passes through (0, 1); always above the x-axis; asymptote y = 0. Increasing if a > 1; decreasing if 0 < a < 1. y = (1/a)ˣ is y = aˣ reflected in the y-axis.
Link between index form and logarithm form, and two special values: y = aˣ ⇔ x = log_a y. The log is the index. log_a 1 = 0 and log_a a = 1. log_a x is defined only for x > 0.
Laws of logarithms (same base): log_a x + log_a y = log_a(xy) log_a x − log_a y = log_a(x/y) k log_a x = log_a(xᵏ) log(x + y) is not log x + log y.
Solve 2ˣ = 11: Take logs: x log 2 = log 11, so x = log 11/log 2 = 3.46 (3 s.f.). Check: 2³ = 8 and 2⁴ = 16, so x lies between 3 and 4.
Solve log₂(x + 1) + log₂(x − 1) = 3: log₂(x² − 1) = 3, so x² − 1 = 8 and x = ±3. x = −3 makes x − 1 negative, so reject it. Answer x = 3.
Solve 3^(x + 1) = 2^(x + 4): (x + 1) log 3 = (x + 4) log 2 x(log 3 − log 2) = 4 log 2 − log 3 x = (4 log 2 − log 3)/(log 3 − log 2) = 4.13 (3 s.f.).