Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
Find the limit L = limx→∞ (√(x2 − 9x) − x).
Find the limit L = limx→0 ln(1 + 5x)/sin(4x), with x in radians.
Let f(x) = ln(3x2 + 5). Find f′(3).
Let f(x) = x3 − 7.5x2 + 18x + 2. Find the x-coordinate x of the local minimum.
An open box (no lid) with a square base must hold 62500 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 company sells q units per week. The price at which it can sell them is p = 100 − 0.5q euros per unit, and the cost is C(q) = 300 + 70q euros. Find the price p that maximises profit.
Evaluate I = ∫₀2 x·sin x dx (x in radians).
Evaluate I = ∫₀1 x²·eˣ dx.
Answer: L = −4.5
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: L = 1.25
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′(3) = 0.5625
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 − 15: negative at a maximum, positive at a minimum.
Answer: S = 7500 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 + 500000/x³ > 0, so this is a minimum.
The best open box is half as tall as it is wide.
Answer: p = 85 €
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.
P″(q) = −1 < 0, so this is a maximum.
Answer: I = 1.742
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Answer: I = 0.7183
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
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