Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
Find the limit L = limx→∞ (−5x3 + 5x − 3)/(5x3 + 5x + 8).
Find the limit L = limx→0 ln(1 + 3x)/sin(2x), with x in radians.
Let f(x) = ln(2x2 + 1). Find f′(1).
Let f(x) = x3 − 13.5x2 + 54x − 2. Find the x-coordinate x of the local maximum.
An open box (no lid) with a square base must hold 108000 cm³. Use the least possible material, that is the smallest area of the base plus the four sides. Find the minimum surface area S.
A theatre sells 390 tickets per show at €15 each. Market research shows that every €1 increase in the ticket price means 10 fewer tickets sold (and every €1 decrease, 10 more). Find the ticket price p that maximises revenue.
Evaluate I = ∫01 2x·(x² + 1)3 dx.
The region between the line y = 2x and the parabola y = x² is rotated about the x-axis. Find the volume V of the solid.
Answer: L = −1
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
For equal degrees the limit is the ratio of the leading coefficients.
Answer: L = 1.5
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Answer: f′(1) = 1.333
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Answer: x = 3
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
f″(x) = 6x − 27: negative at a maximum, positive at a minimum.
Answer: S = 10800 cm²
Optimization: write the quantity to optimise as a function of one variable, using the constraint to eliminate the other. Set the derivative to zero and solve. Check that the critical point is a maximum or minimum (second derivative or sign change of the first derivative) and that it lies inside the allowed range; an endpoint such as a zero side length is not the answer.
S″(x) = 2 + 864000/x³ > 0, so this is a minimum.
The best open box is half as tall as it is wide.
Answer: p = 27 €
Optimization: write the quantity to optimise as a function of one variable, using the constraint to eliminate the other. Set the derivative to zero and solve. Check that the critical point is a maximum or minimum (second derivative or sign change of the first derivative) and that it lies inside the allowed range; an endpoint such as a zero side length is not the answer.
R″(p) = −20 < 0, so this is a maximum.
Answer: I = 3.75
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
Answer: V = 13.4
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
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