Flashcards · 9709 · Topic 3

Pure Mathematics 3 flashcards

51 cards on Topic 3, subtopics 3.1 to 3.9. Pick a subtopic or study them all.

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3.1 Algebra

Partial fractions: what form do you use for a linear factor, a repeated factor (cx + d)² and a quadratic factor cx² + d?: Linear (ax + b): A/(ax + b). Repeated: B/(cx + d) + C/(cx + d)². Quadratic: (Bx + C)/(cx² + d). The numerator must have lower degree than the denominator.
Express 1/((x + 1)(x + 2)) in partial fractions.: 1/(x + 1) − 1/(x + 2). Check: ((x + 2) − (x + 1))/((x + 1)(x + 2)) = 1/((x + 1)(x + 2)).
Binomial expansion of (1 + x)ⁿ for rational n, and when is it valid?: (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + n(n − 1)(n − 2)x³/3! + … Valid for |x| < 1.
Expand √(1 + x) up to the term in x².: (1 + x)^(1/2) = 1 + ½x + (½)(−½)x²/2 + … = 1 + ½x − ⅛x² + …, for |x| < 1.
How do you expand (a + bx)ⁿ when n is not a positive integer?: Take out aⁿ: (a + bx)ⁿ = aⁿ(1 + bx/a)ⁿ, then expand. Valid for |bx/a| < 1, that is |x| < |a/b|.

3.2 Logarithmic and exponential functions

Solve 3^(2x) − 5 × 3ˣ + 6 = 0.: Let u = 3ˣ: u² − 5u + 6 = 0, so u = 2 or u = 3. 3ˣ = 2 gives x = ln 2/ln 3 = 0.631 (3 s.f.); 3ˣ = 3 gives x = 1.
How do you solve an equation with the unknown in an index, such as 5^(2x − 1) = 3^(x + 2)?: 1. Take ln of both sides. 2. Bring the indices down: (2x − 1) ln 5 = (x + 2) ln 3. 3. Expand, collect the x terms, factorise out x and divide.
A graph of ln y against x is a line with gradient 0.2 and intercept 1.5, for y = k(aˣ). Find a and k.: ln y = ln k + x ln a, so ln a = 0.2 and ln k = 1.5. a = e^0.2 = 1.22, k = e^1.5 = 4.48 (3 s.f.).
In a hidden quadratic in eˣ, why must you reject a root such as eˣ = −3?: eˣ > 0 for every real x (and aˣ > 0 for a > 0), so eˣ = −3 has no solution. Say why you reject it.
Sketch y = e^(kx) for k > 0 and for k < 0.: Both pass through (0, 1) with asymptote y = 0 and lie above the x-axis. k > 0: growth (steeper for larger k). k < 0: decay towards the x-axis.

3.3 Trigonometry

sec x vs cos⁻¹ x: sec x = 1/cos x; cos⁻¹ x is the inverse function.
Find the exact value of tan 75°.: tan(45° + 30°) = (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1) = 2 + √3.
tan(A ± B) and tan 2A: tan(A ± B) ≡ (tan A ± tan B)/(1 ∓ tan A tan B) tan 2A ≡ 2 tan A/(1 − tan²A)
cos²A and sin²A in terms of cos 2A: cos²A ≡ ½(1 + cos 2A) sin²A ≡ ½(1 − cos 2A)
Key features of y = sec θ, y = cosec θ and y = cot θ: sec and cosec: |y| ≥ 1, period 360°; asymptotes where cos θ = 0 (sec) or sin θ = 0 (cosec). cot: period 180°, asymptotes where sin θ = 0.
To solve R sin(θ + α) = c for 0° ≤ θ ≤ 360°, what must you do before finding the angles?: Shift the interval: 0° + α ≤ θ + α ≤ 360° + α. Find every value of θ + α in that range, then subtract α.

3.4 Differentiation

d/dx tan⁻¹x, and d/dx tan⁻¹(2x): d/dx tan⁻¹x = 1/(1 + x²). d/dx tan⁻¹(2x) = 2/(1 + 4x²): do not forget the inner derivative 2.
d/dx ln(f(x)) and d/dx e^(f(x)): d/dx ln(f(x)) = f′(x)/f(x) d/dx e^(f(x)) = f′(x) e^(f(x))
For y = eˣ/(x − 2), find the stationary point.: Quotient rule: dy/dx = ((x − 2)eˣ − eˣ)/(x − 2)² = eˣ(x − 3)/(x − 2)². eˣ > 0, so dy/dx = 0 only at x = 3. Stationary point (3, e³).
x = t² + 1, y = t³ − 3t. Where is the tangent horizontal, and where vertical?: dx/dt = 2t, dy/dt = 3t² − 3. Horizontal: t = ±1, at (2, −2) and (2, 2). Vertical: t = 0, at (1, 0).
What must you substitute to find the gradient of an implicit curve at a point?: Both the x- and the y-coordinate of the point, after making dy/dx the subject.

3.5 Integration

ln|x| vs ln x: The modulus is needed when x may be negative.
∫ f′(x)/f(x) dx: ln|f(x)| + c. Use it when the numerator is a constant multiple of the derivative of the denominator, e.g. ∫ 2x/(x² + 1) dx = ln(x² + 1) + c.
Integration by parts: the formula and how to choose u: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. Choose u as the part that becomes simpler when differentiated (a power of x, or ln x).
Find ∫ x eˣ dx and ∫ ln x dx.: ∫ x eˣ dx = x eˣ − eˣ + c (u = x). ∫ ln x dx = x ln x − x + c (u = ln x, dv/dx = 1).
∫ 1/(x² + a²) dx: (1/a) tan⁻¹(x/a) + c
Integration by substitution: what must change besides the integrand?: Replace dx using du/dx, and change the limits to values of the new variable (or substitute back before using x-limits).

3.6 Numerical solution of equations

Show that 3 sin x = x has a root between 2 and 3.: f(x) = 3 sin x − x (radians). f(2) = 0.728 > 0, f(3) = −2.58 < 0. Sign change and f is continuous, so a root lies between 2 and 3.
Iterating to a given accuracy: three rules for the working: Use the full stored value (Ans) each time, not a rounded one. Write each iterate to at least one more decimal place than the answer needs. Stop when two successive iterates round to the same value.
How do you show the limit α of xₙ₊₁ = F(xₙ) is a root of a given equation?: Put xₙ₊₁ = xₙ = α, so α = F(α), then rearrange this to the given equation, showing every step.
How can you tell from the numbers that an iteration is failing?: The iterates move away from the root, or oscillate with growing size, instead of settling down. A different rearrangement of the same equation may converge.

3.7 Vectors

Parallel vs skew: Parallel lines have direction vectors that are multiples of each other; skew lines are not parallel and do not meet.
Scalar product and the angle between two vectors: a.b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ, so cos θ = a.b/(|a||b|). a.b = 0 means the vectors are perpendicular.
Vector equation of a line: r = a + tb, where a is the position vector of a point on the line, b is a direction vector and t is a scalar parameter.
How do you find the angle between two lines?: Use the scalar product of their direction vectors only, not the position vectors.
How do you decide whether two lines in 3D intersect?: 1. Equate the general points: a + sb = c + td. 2. Solve two of the three component equations for s and t. 3. Check the third. If it holds, they meet; if not (and they are not parallel), they are skew.
Magnitude and unit vector of a = xi + yj + zk: |a| = √(x² + y² + z²). Unit vector â = a/|a|.
How do you find the foot of the perpendicular from a point C to a line?: 1. Write a general point P on the line in terms of t. 2. Form CP. 3. Solve CP . (direction vector) = 0 for t. 4. Substitute t to get P; |CP| is the shortest distance.

3.8 Differential equations

How do you solve a first-order differential equation by separating the variables?: 1. Rearrange so all y terms are with dy and all x terms with dx. 2. Integrate both sides, with one constant c. 3. Use the given condition to find c. 4. Rearrange if asked, and interpret in context.
Translate "the rate of decrease of x is proportional to x" into a differential equation.: dx/dt = −kx, where k is a positive constant.
Solve dy/dx = ky.: ∫ (1/y) dy = ∫ k dx, so ln|y| = kx + c, giving y = Ae^(kx) (A a constant).
Solve dy/dx = 2xy given y = 1 when x = 0.: ∫ (1/y) dy = ∫ 2x dx, so ln y = x² + c. At x = 0, y = 1: c = 0. y = e^(x²).
When must the constant of integration be added?: At the integration step, before you take exponentials or rearrange. Adding it afterwards gives the wrong general solution.

3.9 Complex numbers

|z − a| = k vs |z − a| = |z − b|: A circle vs a perpendicular bisector.
arg(z − a) = α: A half-line from a, not a full line.
Complex conjugate of z = x + iy, and zz*: z* = x − iy. zz* = x² + y² = |z|², which is real.
Express (3 + 4i)/(1 + 2i) in the form x + iy.: Multiply top and bottom by 1 − 2i: (3 − 6i + 4i + 8)/5 = (11 − 2i)/5 = 2.2 − 0.4i.
Modulus-argument (polar) form, and the range of the principal argument: z = r(cos θ + i sin θ) = re^(iθ), where r = |z| and θ = arg z. Principal argument: −π < θ ≤ π. Place z in the correct quadrant; do not just use tan⁻¹(y/x).
Multiplying and dividing in polar form: |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg z₁ + arg z₂. |z₁/z₂| = |z₁|/|z₂| and arg(z₁/z₂) = arg z₁ − arg z₂. Adjust the argument into −π < θ ≤ π if needed.
How do you find the square roots of a complex number z?: Set (a + ib)² = z. Compare real parts (a² − b² = Re z) and imaginary parts (2ab = Im z), then solve for real a and b.
Non-real roots of a polynomial equation with real coefficients: They occur in conjugate pairs: if a + ib is a root, so is a − ib.