Markdown Editor

Exam with Answer Key Template

Exam paper with point allocation, multiple choice, short answer and problems, KaTeX worked solutions and marking guidance

ÉducationStandardexamanswerkeymath

Mode d'emploi: Use to set a written exam and its marking scheme together. Build the point allocation from learning outcomes first, then write each solution with marks per step. Main pitfall: printing the student version without removing the answer key section.

Aperçu

Exam: [Course code] [Exam name] with Answer Key

Exam paper template with a separate answer key: candidate instructions, multiple-choice, short-answer and extended problem sections, a point allocation table, worked solutions written in KaTeX math, and marking guidance. The sample questions use introductory calculus and algebra so the format is concrete; replace them with your own. Based on common assessment practice.

FieldValue
Course[MATH 110 Calculus I]
Exam[Midterm exam 1]
Date / timeYYYY-MM-DD, [09:00-10:30]
Duration[90 minutes]
Total points[50]
Permitted materials[Non-programmable calculator; one A4 formula sheet]
Examiner[@name]
Moderator[@name]
Version[A]

1. Instructions to candidates

  1. Write your name and student ID on every page.
  2. Answer all questions in Sections A, B and C.
  3. In Section A, circle one answer per question. No working is required.
  4. In Sections B and C, show all working. Correct answers without working may receive limited credit.
  5. Give exact answers unless a question asks for a decimal approximation.
  6. If you believe a question contains an error, state your assumption and answer the question as you interpret it.
  7. Points for each question are shown in brackets.
NameStudent IDSeat
[ ][ ][ ]

2. Point allocation

SectionQuestionsPoints eachSection totalLearning outcomes
A. Multiple choice1-5210LO1, LO2
B. Short answer6-8412LO2, LO3
C. Problems9-101428LO2, LO3, LO4
Total1050

Check: 10 + 12 + 28 = 50.

3. Section A: Multiple choice (10 points)

Q1. [2] What is ddx(x3)\dfrac{d}{dx}\left(x^3\right)?

(a) x2x^2   (b) 3x23x^2   (c) 3x33x^3   (d) x44\dfrac{x^4}{4}

Q2. [2] Evaluate ∫032x dx\displaystyle\int_0^3 2x \, dx.

(a) 3   (b) 6   (c) 9   (d) 18

Q3. [2] What is lim⁡x→0sin⁡xx\displaystyle\lim_{x \to 0} \frac{\sin x}{x}?

(a) 0   (b) 1   (c) ∞\infty   (d) The limit does not exist

Q4. [2] What is ddx(e2x)\dfrac{d}{dx}\left(e^{2x}\right)?

(a) e2xe^{2x}   (b) 2ex2e^{x}   (c) 2e2x2e^{2x}   (d) 2x e2x−12x\,e^{2x-1}

Q5. [2] What is the slope of the tangent to y=x2y = x^2 at x=3x = 3?

(a) 3   (b) 6   (c) 9   (d) 2

4. Section B: Short answer (12 points)

Q6. [4] Use the product rule to differentiate f(x)=x2sin⁡xf(x) = x^2 \sin x.

Q7. [4] Find the critical points of f(x)=x3−3xf(x) = x^3 - 3x and classify each as a local maximum or local minimum.

Q8. [4] Solve 2x2−8x+6=02x^2 - 8x + 6 = 0.

5. Section C: Problems (28 points)

Q9. [14] Let f(x)=(3x2+1)4f(x) = \left(3x^2 + 1\right)^4.

(a) [6] Find f′(x)f'(x).

(b) [3] Evaluate f′(1)f'(1).

(c) [5] Find the equation of the tangent line to y=f(x)y = f(x) at x=0x = 0.

Q10. [14]

(a) [8] Find the area of the region enclosed by y=x2y = x^2 and y=2xy = 2x.

(b) [6] Solve the system of equations

{2x+y=7x−y=2\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}

6. Answer key

6.1 Section A

QuestionAnswerCommon wrong answer and the error it reveals
Q1(b) 3x23x^2(d): integrated instead of differentiated
Q2(c) 9(d): forgot to divide by 2 when integrating
Q3(b) 1(d): treated 0/00/0 as undefined without taking the limit
Q4(c) 2e2x2e^{2x}(a): omitted the chain rule factor
Q5(b) 6(c): gave the yy-value instead of the slope

Worked note for Q2:

∫032x dx=[x2]03=9−0=9\int_0^3 2x\,dx = \left[x^2\right]_0^3 = 9 - 0 = 9

6.2 Section B

Q6. Product rule (uv)′=u′v+uv′(uv)' = u'v + uv' with u=x2u = x^2, v=sin⁡xv = \sin x:

f′(x)=2xsin⁡x+x2cos⁡xf'(x) = 2x\sin x + x^2\cos x

Marks: rule stated or applied correctly [2]; each term correct [1 + 1].

Q7. Differentiate and set to zero:

f′(x)=3x2−3=3(x−1)(x+1)=0  ⟹  x=−1 or x=1f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1) = 0 \implies x = -1 \text{ or } x = 1

Second derivative f′′(x)=6xf''(x) = 6x. At x=1x = 1, f′′(1)=6>0f''(1) = 6 > 0: local minimum, f(1)=−2f(1) = -2. At x=−1x = -1, f′′(−1)=−6<0f''(-1) = -6 < 0: local maximum, f(−1)=2f(-1) = 2.

Marks: f′(x)f'(x) [1]; both critical points [1]; correct classification of each with justification [1 + 1]. Accept a sign chart of f′f' instead of f′′f''.

Q8. Divide by 2 and factor:

x2−4x+3=(x−1)(x−3)=0  ⟹  x=1 or x=3x^2 - 4x + 3 = (x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3

Marks: simplification or correct use of the quadratic formula [2]; each root [1 + 1]. Check: 2(1)2−8(1)+6=02(1)^2 - 8(1) + 6 = 0 and 2(9)−24+6=02(9) - 24 + 6 = 0.

6.3 Section C

Q9 (a). Chain rule with outer function u4u^4 and inner u=3x2+1u = 3x^2 + 1:

f′(x)=4(3x2+1)3⋅6x=24x(3x2+1)3f'(x) = 4\left(3x^2 + 1\right)^3 \cdot 6x = 24x\left(3x^2 + 1\right)^3

Marks: derivative of outer function [2]; derivative of inner function [2]; correct product and simplification [2].

Q9 (b).

f′(1)=24(1)(3+1)3=24×64=1536f'(1) = 24(1)\left(3 + 1\right)^3 = 24 \times 64 = 1536

Marks: substitution [1]; arithmetic [2]. Follow-through: award full marks for correct evaluation of an incorrect f′(x)f'(x) from part (a).

Q9 (c). f(0)=(0+1)4=1f(0) = (0 + 1)^4 = 1 and f′(0)=0f'(0) = 0, so the tangent line is

y−1=0 (x−0)  ⟹  y=1y - 1 = 0\,(x - 0) \implies y = 1

Marks: point f(0)f(0) [1]; slope f′(0)f'(0) [2]; line equation [2].

Q10 (a). Intersections: x2=2x  ⟹  x(x−2)=0  ⟹  x=0,2x^2 = 2x \implies x(x - 2) = 0 \implies x = 0, 2. On [0,2][0, 2], 2x≥x22x \ge x^2, so

A=∫02(2x−x2)dx=[x2−x33]02=4−83=43A = \int_0^2 \left(2x - x^2\right) dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}

Marks: limits of integration [2]; correct integrand with the upper curve first [1]; antiderivative [3]; final value [2]. Accept ≈1.33\approx 1.33.

Q10 (b). Adding the equations eliminates yy:

3x=9  ⟹  x=3,y=7−2(3)=13x = 9 \implies x = 3, \qquad y = 7 - 2(3) = 1

Marks: valid elimination or substitution [3]; xx [1]; yy [1]; check in both equations [1].

7. Marking guidance

  • Award method marks for correct reasoning even when an arithmetic slip leads to a wrong final answer.
  • Apply follow-through: an error in one part is penalised once, and later parts are marked on the student's own earlier value.
  • Accept any correct equivalent form (for example 2xsin⁡x+x2cos⁡x2x\sin x + x^2\cos x or x(2sin⁡x+xcos⁡x)x(2\sin x + x\cos x)).
  • Do not award marks for a correct answer obtained by clearly invalid working.
  • Use half-marks only where the scheme shows them.
  • Second-mark a sample of [10%] of scripts, including all scripts within [2] points of a grade boundary.
DateQuestionUnexpected answerDecision
Example: YYYY-MM-DDQ7Classified points using a graph sketch only1 of 2 classification marks: justification incomplete
[Date][Question][Answer][Decision]

Results summary

StatisticValue
Scripts marked[n]
Mean (SD)[ ]
Questions with mean below 50%[ ]
Action for next exam or teaching[ ]

Standard and licence note

Based on common assessment practice for written examinations and marking schemes. Original text written for this collection; sample questions and solutions are illustrative and were checked for correctness. Template licence: MIT.

Modèles associés