Mathematics AA: Standard Level
Calculus — Topic 5
- 1.
Let g(x) = x³ − 6x² + 9x. Find the coordinates of all stationary points and classify each as a local maximum or local minimum.
[6 marks] · no calculator - 2.
Marking analysis: A learner attempts the following task: “Let g(x) = x³ − 6x² + 9x. Find the coordinates of all stationary points and classify each as a local maximum or local minimum.” Their response addresses only this point: “Differentiates to obtain g′(x) = 3x² − 12x + 9.” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[6 marks] · no calculator - 3.
Evaluate the definite integral from 0 to 2 of (3x² + 2) dx. Interpret your answer as the signed area between the curve and the x-axis on this interval.
[3 marks] · no calculator - 4.
Marking analysis: A learner attempts the following task: “Evaluate the definite integral from 0 to 2 of (3x² + 2) dx. Interpret your answer as the signed area between the curve and the x-axis on this interval.” Their response addresses only this point: “Finds an antiderivative x³ + 2x.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator - 5.
Find the equation of the tangent to the curve y = x² − 3x + 5 at the point where x = 2.
[5 marks] · no calculator - 6.
Marking analysis: A learner attempts the following task: “Find the equation of the tangent to the curve y = x² − 3x + 5 at the point where x = 2.” Their response addresses only this point: “Differentiates to obtain dy/dx = 2x − 3.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator - 7.
Differentiate y = (3x + 1)⁴ using the chain rule. Hence find the gradient of the curve at x = 0.
[4 marks] · no calculator - 8.
Marking analysis: A learner attempts the following task: “Differentiate y = (3x + 1)⁴ using the chain rule. Hence find the gradient of the curve at x = 0.” Their response addresses only this point: “Applies the chain rule: dy/dx = 4(3x + 1)³ × 3.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator - 9.
Find the equation of the normal to the curve y = x³ at the point (1, 1).
[5 marks] · no calculator - 10.
Marking analysis: A learner attempts the following task: “Find the equation of the normal to the curve y = x³ at the point (1, 1).” Their response addresses only this point: “Differentiates to obtain dy/dx = 3x².” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator - 11.
A particle moves along a line so that its displacement is s(t) = t³ − 6t² + 9t metres, t ≥ 0 seconds. Find the particle's velocity function and determine when the particle is at rest.
[3 marks] · no calculator - 12.
Marking analysis: A learner attempts the following task: “A particle moves along a line so that its displacement is s(t) = t³ − 6t² + 9t metres, t ≥ 0 seconds. Find the particle's velocity function and determine when the particle is at rest.” Their response addresses only this point: “Differentiates displacement to obtain v(t) = 3t² − 12t + 9.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator - 13.
Find ∫(4x³ − 6x + 2) dx.
[3 marks] · no calculator - 14.
Marking analysis: A learner attempts the following task: “Find ∫(4x³ − 6x + 2) dx.” Their response addresses only this point: “Integrates each term using the power rule.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator - 15.
For the curve y = 12/x, x ≠ 0, find the equation of the tangent at x = 3.
[4 marks] - 16.
Marking analysis: A learner attempts the following task: “For the curve y = 12/x, x ≠ 0, find the equation of the tangent at x = 3.” Their response addresses only this point: “Differentiates to obtain dy/dx = −12/x².” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 17.
Determine the interval(s) over which the function h(x) = x³ − 3x is increasing.
[5 marks] · no calculator - 18.
Marking analysis: A learner attempts the following task: “Determine the interval(s) over which the function h(x) = x³ − 3x is increasing.” Their response addresses only this point: “Differentiates to obtain h′(x) = 3x² − 3.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator - 19.
Water flows into a tank at a rate given by R(t) = 6t − t² litres per minute, for 0 ≤ t ≤ 6. Find the total volume of water that flows into the tank during this time.
[4 marks] - 20.
Marking analysis: A learner attempts the following task: “Water flows into a tank at a rate given by R(t) = 6t − t² litres per minute, for 0 ≤ t ≤ 6. Find the total volume of water that flows into the tank during this time.” Their response addresses only this point: “Recognises that total volume is the definite integral of the rate function from 0 to 6.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 21.
A farmer has 40 m of fencing to enclose a rectangular field, using an existing wall as one side (so fencing is only needed for the other three sides). Find the dimensions that maximise the enclosed area, and calculate this maximum area.
[5 marks] - 22.
Marking analysis: A learner attempts the following task: “A farmer has 40 m of fencing to enclose a rectangular field, using an existing wall as one side (so fencing is only needed for the other three sides). Find the dimensions that maximise the enclosed area, and calculate this maximum area.” Their response addresses only this point: “Sets up the constraint x + 2y = 40, where x is the side parallel to the wall and y are the two perpendicular sides, and expresses x = 40 − 2y.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] - 23.
A particle's displacement is given by s(t) = t³ − 9t² + 24t, for t ≥ 0 (metres, seconds). Find the particle's acceleration function, and determine the value of t at which the particle's acceleration is zero.
[4 marks] - 24.
Marking analysis: A learner attempts the following task: “A particle's displacement is given by s(t) = t³ − 9t² + 24t, for t ≥ 0 (metres, seconds). Find the particle's acceleration function, and determine the value of t at which the particle's acceleration is zero.” Their response addresses only this point: “Differentiates displacement twice: v(t) = 3t² − 18t + 24, then a(t) = 6t − 18.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 25.
Find the area of the region enclosed between the curves y = x² and y = 2x.
[5 marks] - 26.
Marking analysis: A learner attempts the following task: “Find the area of the region enclosed between the curves y = x² and y = 2x.” Their response addresses only this point: “Finds the points of intersection by solving x² = 2x: x = 0 and x = 2.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] - 27.
For the curve y = x³ − 3x² + 2, find the coordinates of the point of inflection, using the second derivative.
[4 marks] - 28.
Marking analysis: A learner attempts the following task: “For the curve y = x³ − 3x² + 2, find the coordinates of the point of inflection, using the second derivative.” Their response addresses only this point: “Finds y′ = 3x² − 6x and y″ = 6x − 6.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 29.
The area of a circular oil spill is increasing at a rate of 8 m² s⁻¹. Find the rate at which the radius is increasing at the instant when the radius is 5 m. (A = πr²)
[4 marks] - 30.
Marking analysis: A learner attempts the following task: “The area of a circular oil spill is increasing at a rate of 8 m² s⁻¹. Find the rate at which the radius is increasing at the instant when the radius is 5 m. (A = πr²)” Their response addresses only this point: “Differentiates A = πr² with respect to t, using the chain rule: dA/dt = 2πr(dr/dt).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] - 31.
For f(x) = x^2 e^(-x), with real x, find the stationary points and classify each using the sign of the derivative.
[5 marks] · no calculator - 32.
Marking analysis: A learner attempts the following task: “For f(x) = x^2 e^(-x), with real x, find the stationary points and classify each using the sign of the derivative.” Their response addresses only this point: “f'(x) = e^(-x)(2x - x^2) = e^(-x)x(2 - x).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator - 33.
Find the intersections of y = 2x and y = x^2, then calculate the exact area enclosed between the curves.
[4 marks] · no calculator - 34.
Marking analysis: A learner attempts the following task: “Find the intersections of y = 2x and y = x^2, then calculate the exact area enclosed between the curves.” Their response addresses only this point: “x^2 = 2x gives x = 0 and x = 2, at (0,0) and (2,4).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator - 35.
Find the tangent to y = ln x at x = e. Prove that ln x <= x/e for every x > 0, stating when equality holds.
[5 marks] · no calculator - 36.
Marking analysis: A learner attempts the following task: “Find the tangent to y = ln x at x = e. Prove that ln x <= x/e for every x > 0, stating when equality holds.” Their response addresses only this point: “At x = e, y = 1 and dy/dx = 1/e.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator - 37.
f(x) = x^3 - 3kx, where k > 0. The vertical separation between its local maximum and local minimum is 32. Determine k.
[5 marks] · no calculator - 38.
Marking analysis: A learner attempts the following task: “f(x) = x^3 - 3kx, where k > 0. The vertical separation between its local maximum and local minimum is 32. Determine k.” Their response addresses only this point: “f'(x) = 3x^2 - 3k, so stationary inputs are +/-sqrt(k).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator - 39.
A temperature is modelled by T(t) = 20 + 60e^(-0.15t), where t is minutes after cooling begins. Find when T first reaches 35 degrees, the initial rate of temperature change, and whether the model reaches 20 degrees at a finite time.
[5 marks] - 40.
Marking analysis: A learner attempts the following task: “A temperature is modelled by T(t) = 20 + 60e^(-0.15t), where t is minutes after cooling begins. Find when T first reaches 35 degrees, the initial rate of temperature change, and whether the model reaches 20 degrees at a finite time.” Their response addresses only this point: “35 = 20 + 60e^(-0.15t) gives e^(-0.15t) = 1/4.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]