Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
Find the limit L = limx→∞ (√(9x2 + 3x) − 3x).
Find the limit L = limx→0 (1 − cos 5x)/(2x²), with x in radians.
Let f(x) = x2·e−x. Find f′(1).
The tangent line to f(x) = 0.5x2 + 2x − 2 at x = 3 crosses the y-axis at y = b. Find b.
An open box (no lid) with a square base must hold 13500 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 1650 tickets per show at €20 each. Market research shows that every €1 increase in the ticket price means 50 fewer tickets sold (and every €1 decrease, 50 more). Find the maximum revenue R per show.
The region under y = 0.5x from x = 0 to x = 1 is rotated about the x-axis. Find the volume V of the solid (a cone).
The region between the line y = 4x and the parabola y = 3x² is rotated about the x-axis. Find the volume V of the solid.
Answer: L = 0.5
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 = 6.25
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′(1) = 0.3679
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: b = −6.5
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: S = 2700 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 + 108000/x³ > 0, so this is a minimum.
The best open box is half as tall as it is wide.
Answer: R = 35113 €
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) = −100 < 0, so this is a maximum.
Answer: V = 0.2618
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.
Check: a cone of radius 0.5 and height 1 has (1/3)πr²h = 0.2618.
Answer: V = 15.89
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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