Flashcards · 9660 · Topic 1

Unit P1: Pure Maths (International AS) flashcards

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Index laws, and the meaning of a⁰, a⁻ⁿ and a^(m/n): Add powers to multiply, subtract to divide, multiply for a power of a power. a⁰ = 1, a⁻ⁿ = 1/aⁿ, a^(m/n) = (ⁿ√a)ᵐ. Example: (1/4)^(−3/2) = 4^(3/2) = (√4)³ = 8.
Rationalise 6/(2 + √3): Multiply top and bottom by the conjugate 2 − √3: 6(2 − √3)/(4 − 3) = 12 − 6√3.
Complete the square on x² + 14x + 10 and state the vertex: x² + 14x + 10 = (x + 7)² − 39. Vertex (−7, −39), line of symmetry x = −7, minimum value −39. For a(x + p)² + q the vertex is (−p, q).
Discriminant conditions, and the values of k for which x² − 4x + k = 0 has equal roots: b² − 4ac > 0: two distinct real roots; = 0: equal roots; < 0: no real roots. Here 16 − 4k: equal roots when k = 4, distinct when k < 4, none when k > 4.
The Remainder Theorem and the Factor Theorem: Remainder Theorem: dividing f(x) by (x − a) leaves remainder f(a). Factor Theorem: (x − a) is a factor of f(x) exactly when f(a) = 0. For a divisor (x + 3), test f(−3).
Describe the transformations y = f(x) + a, y = f(x + a), y = af(x), y = f(ax), y = −f(x), y = f(−x): f(x) + a: translation by vector (0, a). f(x + a): translation by (−a, 0). af(x): stretch parallel to the y-axis, factor a. f(ax): stretch parallel to the x-axis, factor 1/a. −f(x): reflection in the x-axis. f(−x): reflection in the y-axis.

Gradient, mid-point and distance for the points (x₁, y₁) and (x₂, y₂): m = (y₂ − y₁)/(x₂ − x₁). Mid-point ((x₁ + x₂)/2, (y₁ + y₂)/2). Distance √[(x₂ − x₁)² + (y₂ − y₁)²].
Gradient conditions for parallel and perpendicular lines, and the gradient of ax + by + c = 0: Parallel: m₁ = m₂. Perpendicular: m₁m₂ = −1. ax + by + c = 0 has gradient −a/b and y-intercept −c/b.
Find the perpendicular bisector of A(2, −3) and B(10, 3): Gradient AB = 6/8 = 3/4; mid-point (6, 0). Perpendicular gradient −4/3: y = −(4/3)(x − 6). So 4x + 3y − 24 = 0.
Substituting a line into a curve gives a quadratic. What does its discriminant tell you?: b² − 4ac > 0: two points of intersection. = 0: one point; the line is a tangent. < 0: the line and curve do not meet.
The line y = 5x + c touches y = x² − x + 6. Find c and the point of contact.: x² − 6x + (6 − c) = 0. Tangent: 36 − 4(6 − c) = 0, so 12 + 4c = 0 and c = −3. Equal roots at x = 3, y = 5(3) − 3 = 12. Contact point (3, 12).

What does the derivative dy/dx represent?: The gradient of the tangent to y = f(x) at that x-value: the rate at which y changes per unit of x. Its unit is (unit of y) per (unit of x).
Differentiate y = (x³ + 4)/x: Rewrite first: y = x² + 4x⁻¹. dy/dx = 2x − 4x⁻² = 2x − 4/x². Power rule: axⁿ → anxⁿ⁻¹ for rational n.
Find the tangent and normal to y = x² − 6x at x = 5: y = −5 and dy/dx = 2x − 6 = 4. Tangent: y + 5 = 4(x − 5), so y = 4x − 25. Normal: gradient −1/4, y + 5 = −(1/4)(x − 5), so x + 4y + 15 = 0.
How do you find and classify stationary points?: 1. Solve dy/dx = 0. 2. Substitute each x into y for the coordinates. 3. d²y/dx² < 0: maximum; > 0: minimum. 4. If d²y/dx² = 0, test the sign of dy/dx on each side.
Increasing and decreasing functions: f is increasing where f′(x) > 0 and decreasing where f′(x) < 0. To show f is increasing everywhere, show f′(x) > 0 for all x (for a quadratic f′, complete the square).
Steps for an optimisation problem: 1. Define variables and write the constraint. 2. Write the quantity as a function of one variable. 3. Differentiate, set to 0, solve. 4. Prove maximum or minimum. 5. Answer the question with units.

Integrate xⁿ: ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c, for rational n ≠ −1. Integrate term by term; constants multiply through.
Find ∫ (x + 2)/x^(1/2) dx: Split: x^(1/2) + 2x^(−1/2). Integral: (2/3)x^(3/2) + 4x^(1/2) + c.
Evaluate ∫ from 1 to 8 of x^(−1/3) dx: [(3/2)x^(2/3)] from 1 to 8 = (3/2)(4) − (3/2)(1) = 9/2. A definite integral is F(b) − F(a), with no + c.
Find the area between y = x³ − 4x and the x-axis for 0 ≤ x ≤ 2: ∫ from 0 to 2 of (x³ − 4x) dx = [x⁴/4 − 2x²] = 4 − 8 = −4. The region is below the axis, so the integral is negative and the area is 4.
Find the area enclosed by y = 3x² and y = 12: They meet where 3x² = 12, x = ±2. The line is above. ∫ from −2 to 2 of (12 − 3x²) dx = [12x − x³] = 16 − (−16) = 32.
The trapezium rule, and when it over-estimates: (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)], h = (b − a)/n, using n + 1 ordinates. Curve bends upwards (convex): chords above, over-estimate. Bends downwards (concave): under-estimate.

Arithmetic series: nth term and sum of n terms: nth term a + (n − 1)d. S(n) = n/2 × (2a + (n − 1)d) = n/2 × (a + l), where l is the last term.
Geometric series: nth term, sum of n terms and sum to infinity: nth term arⁿ⁻¹. S(n) = a(1 − rⁿ)/(1 − r), r ≠ 1. S∞ = a/(1 − r), only when −1 < r < 1.
Find the limit of the sequence x(n + 1) = 0.75x(n) + 7: Replace both terms by L: L = 0.75L + 7, so 0.25L = 7 and L = 28.
A geometric sequence has 3rd term 20 and 6th term 160. Find r, a and S(8).: ar² = 20 and ar⁵ = 160, so r³ = 8 and r = 2; a = 5. S(8) = 5(2⁸ − 1)/(2 − 1) = 1275.
Binomial expansion of (a + b)ⁿ for a positive integer n, and the expansion of (x − 2)⁴: (a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳbʳ, r = 0 to n, with ⁿCᵣ = n!/(r!(n − r)!). (x − 2)⁴ = x⁴ − 8x³ + 24x² − 32x + 16.
Evaluate Σ from r = 1 to 10 of (3r + 2): 3Σr + 2 × 10 = 3 × 55 + 20 = 185, using Σr = n(n + 1)/2.