Flashcards · YMA01 · Topic 1

Unit P1: Pure Mathematics 1 flashcards

28 cards on Topic 1, subtopics P1.1 to P1.5. Pick a subtopic or study them all.

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P1.1 Algebra and functions

Laws of indices, and the meaning of a⁰, a⁻ⁿ and a^(m/n): aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ (same base). a⁰ = 1 (a ≠ 0). a⁻ⁿ = 1/aⁿ, a reciprocal, not a negative number. a^(m/n) = (ⁿ√a)ᵐ: root first, then power. Example: 16^(3/4) = 2³ = 8.
How do you rationalise the denominator of k/(p + q√r)?: Multiply top and bottom by the conjugate (p − q√r). The denominator becomes p² − q²r, a rational number. Example: 10/(√7 − √2) = 10(√7 + √2)/5 = 2√7 + 2√2.
Completed-square form of ax² + bx + c, and the turning point of a(x + p)² + q: ax² + bx + c = a(x + b/(2a))² + (c − b²/(4a)). The turning point of a(x + p)² + q is (−p, q): a minimum if a > 0, a maximum if a < 0. Example: x² − 12x + 29 = (x − 6)² − 7, minimum (6, −7).
The discriminant of ax² + bx + c = 0 and what each sign means: b² − 4ac > 0: two different real roots. = 0: one repeated root (graph touches the x-axis). < 0: no real roots. ax² + bx + c is positive for all x only if a > 0 and b² − 4ac < 0 (both conditions are needed).
Solve x² + 4x − 21 > 0: (x + 7)(x − 3) > 0. The parabola opens upwards, so the region is outside the roots: x < −7 or x > 3. (For "< 0" with a positive x² coefficient, the answer is between the roots.)
Describe y = f(x + a), y = f(x) + a, y = af(x) and y = f(ax): f(x + a): translation a to the left. f(x) + a: translation a up. af(x): stretch in the y-direction, scale factor a. f(ax): stretch in the x-direction, scale factor 1/a. Inside the bracket acts in reverse; outside acts as written.

P1.2 Coordinate geometry in the (x, y) plane

Gradient of the line through (x₁, y₁) and (x₂, y₂), and the equation of a line through (x₁, y₁) with gradient m: m = (y₂ − y₁)/(x₂ − x₁). Line: y − y₁ = m(x − x₁).
Gradient conditions for parallel and perpendicular lines: Parallel lines have equal gradients. Perpendicular gradients multiply to −1, so the new gradient is the negative reciprocal: −1/m.
Distance between two points and the midpoint of the segment joining them: Distance = √((x₂ − x₁)² + (y₂ − y₁)²). Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2).
Gradient and y-intercept of the line ax + by + c = 0: Rearrange to y = −(a/b)x − c/b. Gradient −a/b, y-intercept −c/b.
Find the line perpendicular to 2x + 3y = 6 through (2, 1), in the form ax + by + c = 0: Given gradient −2/3, so the perpendicular gradient is 3/2. y − 1 = (3/2)(x − 2) → 2y − 2 = 3x − 6 → 3x − 2y − 4 = 0. Check (2, 1): 6 − 2 − 4 = 0.

P1.3 Trigonometry

The sine rule, and the ambiguous case: a/sin A = b/sin B = c/sin C. Use it when a side and its opposite angle are known. When finding an angle, sin B = k also gives 180° − B (since sin(180° − θ) = sin θ). Keep the obtuse value only if the angles still total less than 180°.
The cosine rule for a side and for an angle: a² = b² + c² − 2bc cos A (two sides and the included angle). cos A = (b² + c² − a²)/(2bc) (all three sides).
Area of a triangle from two sides and an angle: Area = ½ab sin C, where C is the angle between sides a and b (the included angle).
What is one radian, and how do you convert between degrees and radians?: One radian is the angle at the centre when the arc length equals the radius (about 57.3°). π radians = 180°. Degrees to radians: × π/180. Radians to degrees: × 180/π.
Arc length, sector area and segment area (θ in radians): Arc s = rθ. Sector area = ½r²θ; sector perimeter = 2r + rθ. Segment area = sector − triangle = ½r²(θ − sin θ). These need θ in radians.
Exact values of sin, cos and tan at π/6, π/4 and π/3: π/6 (30°): sin ½, cos √3/2, tan 1/√3. π/4 (45°): sin √2/2, cos √2/2, tan 1. π/3 (60°): sin √3/2, cos ½, tan √3.

P1.4 Differentiation

Differentiation from first principles: the limit definition of f′(x): f′(x) = lim (h → 0) [f(x + h) − f(x)]/h. The mark is for writing the limit notation, not just the algebra.
Differentiate axⁿ: d/dx (axⁿ) = anxⁿ⁻¹, for any rational n. A constant differentiates to 0. Differentiate sums term by term.
Differentiate y = (x² − 1)/x: Split first: y = x − x⁻¹. dy/dx = 1 + x⁻² = 1 + 1/x². Always rewrite roots, reciprocals and single-term fractions as powers before differentiating.
How do you find the tangent and normal to y = f(x) at x = a?: 1. Find y at x = a. 2. Find m = dy/dx at x = a. 3. Tangent: y − y₁ = m(x − a). 4. Normal: gradient −1/m through the same point.
Find the tangent to y = x³ − 5x at x = 2: y = 8 − 10 = −2. dy/dx = 3x² − 5 = 7. y + 2 = 7(x − 2), so y = 7x − 16.
What does the second derivative d²y/dx² measure?: The rate of change of the gradient: the derivative of dy/dx. Example: y = x⁴/2 − 3x² + x gives dy/dx = 2x³ − 6x + 1 and d²y/dx² = 6x² − 6.

P1.5 Integration

Integrate xⁿ (indefinite): ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, for n ≠ −1. Raise the power by one and divide by the new power. The + c is a mark.
Find ∫ (4x³ − 6√x + 2/x²) dx: Write as 4x³ − 6x^(1/2) + 2x⁻². = x⁴ − 6x^(3/2)/(3/2) + 2x⁻¹/(−1) + c = x⁴ − 4x^(3/2) − 2/x + c.
How do you find the equation of a curve from f′(x) and one point on it?: 1. Rewrite f′(x) as powers of x. 2. Integrate term by term and add + c. 3. Substitute the point to find c. 4. State y = f(x) in full.
f′(x) = 12x³ − 6x^(−1/2) and the curve passes through (1, 5). Find f(x).: f(x) = 3x⁴ − 12x^(1/2) + c. At (1, 5): 3 − 12 + c = 5, so c = 14. f(x) = 3x⁴ − 12√x + 14.
How can you check an indefinite integral?: Differentiate your answer. You should get back the original integrand exactly. If not, look for a slip such as dividing by the old power instead of the new one.