A set of MCQ designed to help Level 2 Engineering students prepare/practice for the on-line GOLA test that is used to assess the C&G 2850, Level 2 Engineering, Unit 202: Engineering Principles.

Harder implicit differentiation requiring both chain rule and product rule.

Given a statement in words of a relationship between two variables, write down a corresponding formula. It's marked correct if values satisfying the relationship satisfy the formula, and values not satisfying the relationship don't.

Then, given a value for one of the variables, work out the value of the other one. They're substituted into the formula given in the first part and marked correct if it's satisfied.

Both parts use custom marking algorithms.

Straightforward question: student must find the general solution to a second order constant coefficient ODE. Uses custom marking algorithm to check that both roots appear and that the solution is in the correct form (e.g. two arbitrary constants are present). Arbitrary constants can be any non space-separated string of characters. The algorithm also allows for the use of $e^x$ rather than $\exp(x)$.

Unit tests are also included, to check whether the algorithm accurately marks when the solution is correct; when it's correct but deviates from the 'answer'; when one or more roots is incorrect; or when the roots are correct but constants of integration have been forgotten.

Manipulate fractions in order to add and subtract them. The difficulty escalates through the inclusion of a whole integer and a decimal, which both need to be converted into a fraction before the addition/subtraction can take place. 

Manipulate fractions in order to add and subtract them. The difficulty escalates through the inclusion of a whole integer and a decimal, which both need to be converted into a fraction before the addition/subtraction can take place. 

Solve for $x$: $\displaystyle ax ^ 2 + bx + c=0$.

Entering the correct roots in any order is marked as correct. However, entering one correct and the other incorrect gives feedback stating that both are incorrect.

Straightforward question: student must find the general solution to a second order constant coefficient ODE. Uses custom marking algorithm to check that both roots appear and that the solution is in the correct form (e.g. two arbitrary constants are present). Arbitrary constants can be any non space-separated string of characters. The algorithm also allows for the use of $e^x$ rather than $\exp(x)$.

Unit tests are also included, to check whether the algorithm accurately marks when the solution is correct; when it's correct but deviates from the 'answer'; when one or more roots is incorrect; or when the roots are correct but constants of integration have been forgotten.

This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

Two questions testing the application of the Cosine Rule when given two sides and an angle. In these questions, the triangle is always acute and both of the given side lengths are adjacent to the given angle.

Two questions testing the application of the Cosine Rule when given two sides and an angle. In these questions, the triangle is always acute and both of the given side lengths are adjacent to the given angle.

Two questions testing the application of the Sine Rule when given two angles and a side. In this question the triangle is obtuse. In one question, the two given angles are both acute. In the second, one of the angles is obtuse.

Straightforward question: student must find the general solution to a second order constant coefficient ODE. Uses custom marking algorithm to check that both roots appear and that the solution is in the correct form (e.g. two arbitrary constants are present). Arbitrary constants can be any non space-separated string of characters. The algorithm also allows for the use of $e^x$ rather than $\exp(x)$.

Unit tests are also included, to check whether the algorithm accurately marks when the solution is correct; when it's correct but deviates from the 'answer'; when one or more roots is incorrect; or when the roots are correct but constants of integration have been forgotten.

This is the question for week 9 of the MA100 course at the LSE. It looks at material from chapters 17 and 18.

Description of variables for part b:
For part b we want to have four functions such that the derivative of one of them, evaluated at 0, gives 0; but for the rest we do not get 0. We also want two of the ones that do not give 0, to be such that the derivative of their sum, evaluated at 0, gives 0; but when we do this for any other sum of two of our functions, we do not get 0. Ultimately this part of the question will show that even if two functions are not in a vector space (the space of functions with derivate equal to 0 when evaluated at 0), then their sum could nonetheless be in that vector space. We want variables which statisfy:

a,b,c,d,f,g,h,j,k,l,m,n are variables satisfying

Function 1: x^2 + ax + b sin(cx)
Function 2: x^2 + dx + f sin(gx)
Function 3: x^2 + hx + j sin(kx)
Function 4: x^2 + lx + m sin(nx)

u,v,w,r are variables satifying
u=a+bc
v=d+fg
w=h+jk
r=l+mn

The derivatives of each function, evaluated at zero, are:
Function 1: u
Function 2: v
Function 3: w
Function 4: r

So we will define
u as random(-5..5 except(0))
v as -u
w as 0
r as random(-5..5 except(0) except(u) except(-u))

Then the derivative of function 3, evaluated at 0, gives 0. The other functions give non-zero.
Also, the derivative of function 1 + function 2 gives 0. The other combinations of two functions give nonzero.

We now take b,c,f,g,j,k,m,n to be defined as \random(-3..3 except(0)).
We then define a,d,h,l to satisfy
u=a+bc
v=d+fg
w=h+jk
r=l+mn

Description for variables of part e:

Please look at the description of each variable for part e in the variables section, first.
As described, the vectors V3_1 , V3_2 , V3_3 are linearly independent. We will simply write v1 , v2 , v3 here.
In part e we ask the student to determine which of the following sets span, are linearly independent, are both, are neither:

both: v1,v2,v3
span: v1,v1+v2,v1+v2+v3, v1+v2+v3,2*v1+v2+v3
lin ind: v1+v2+v3
neither: v2+v3 , 2*v2 + 2*v3
neither:v1+v3,v1-2*v3,2*v1-v3
neither: v1+v2,v1-v2,v1-2*v2,2*v1-v2

This is the question for week 3 of the MA100 course at the LSE. It looks at material from chapters 5 and 6. The following describes how two polynomials were defined in the question. This may be helpful for anyone who needs to edit this question.

In part a we have a polynomial. We wanted it to have two stationary points. To create the polynomial we first created the two stationary points as variables, called StationaryPoint1 and StationaryPoint2 which we will simply write as s1 ans s2 here. s2 was defined to be larger than s1. This means that the derivative of our polynomial must be of the form a(x-s1)(x-s2) for some constant a. The constant "a" is a variable called PolynomialScalarMult, and it is defined to be a multiple of 6 so that when we integrate the derivative a(x-s1)(x-s2) we only have integer coefficients. Its possible values include positive and negative values, so that the first stationary point is not always a max (and the second always a min). Finally, we have a variable called ConstantTerm which is the constant term that we take when we integrate the derivative derivative a(x-s1)(x-s2). Hence, we can now create a randomised polynomial with integers coefficients, for which the stationary points are s1 and s2; namely (the integral of a(x-s1)(x-s2)) plus ConstantTerm.

In part e we created a more complicated polynomial. It is defined as -2x^3 + 3(s1 + s2)x^2 -(6*s1*s2) x + YIntercept on the domain [0,35]. One can easily calculate that the stationary points of this polynomials are s1 and s2. Furthermore, they are chosen so that both are in the domain and so that s1 is smaller than s2. This means that s1 is a min and s2 is a max. Hence, the maximum point of the function will occur either at 0 or s2 (The function is descreasing after s2). Furthermore, one can see that when we evaluate the function at s2 we get (s2)^2 (s2 -3*s1) + YIntercept. In particular, this is larger than YIntercept if s2 > 3 *s1, and smaller otherwise. Possible values of s2 include values which are larger than 3*s1 and values which are smaller than 3*s1. Hence, the max of the function maybe be at 0 or at s2, dependent on s2. This gives the question a good amount of randomisation.

A set of MCQ designed to help Level 2 Engineering students prepare/practice for the on-line GOLA test that is used to assess the C&G 2850, Level 2 Engineering, Unit 202: Engineering Principles.

A set of MCQ designed to help Level 2 Engineering students prepare/practice for the on-line GOLA test that is used to assess the C&G 2850, Level 2 Engineering, Unit 202: Engineering Principles.

This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

Straightforward question: student must find the general solution to a second order constant coefficient ODE. Uses custom marking algorithm to check that both roots appear and that the solution is in the correct form (e.g. two arbitrary constants are present). Arbitrary constants can be any non space-separated string of characters. The algorithm also allows for the use of $e^x$ rather than $\exp(x)$.

Unit tests are also included, to check whether the algorithm accurately marks when the solution is correct; when it's correct but deviates from the 'answer'; when one or more roots is incorrect; or when the roots are correct but constants of integration have been forgotten.

This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve. 

Simple procedures are given and student is asked to carry them out or un-do them.

Version 1: bi and bii have the same answer. biii and biv both have two answers.

Version 2: bi and bii have different answers. biii has two answers, biv has one answer.

Version 3: bi and bii have different answer. biii has one answer, biv has two answers.

Version 4: bi and bii have the same answer. biii has one answer, biv has two answers.

Simple procedures are given and student is asked to carry them out or un-do them.

Version 1: bi and bii have the same answer. biii and biv both have two answers.

Version 2: bi and bii have different answers. biii has two answers, biv has one answer.

Version 3: bi and bii have different answer. biii has one answer, biv has two answers.

Version 4: bi and bii have the same answer. biii has one answer, biv has two answers.

Simple procedures are given and student is asked to carry them out or un-do them.

Version 1: bi and bii have the same answer. biii and biv both have two answers.

Version 2: bi and bii have different answers. biii has two answers, biv has one answer.

Version 3: bi and bii have different answer. biii has one answer, biv has two answers.

Version 4: bi and bii have the same answer. biii has one answer, biv has two answers.

Simple procedures are given and student is asked to carry them out or un-do them.

Version 1: bi and bii have the same answer. biii and biv both have two answers.

Version 2: bi and bii have different answers. biii has two answers, biv has one answer.

Version 3: bi and bii have different answer. biii has one answer, biv has two answers.

Version 4: bi and bii have the same answer. biii has one answer, biv has two answers.

Simple procedures are given and student is asked to carry them out or un-do them.

Version 1: i and ii have the same answer. iii and iv both have two answers.

Version 2: i and ii have different answers. iii has two answers,biv has one answer.

Version 3: i and ii have different answer. iii has one answer, iv has two answers.

Version 4: i and ii have the same answer. iii has one answer, iv has two answers.

Simple procedures are given and student is asked to carry them out or un-do them.

Version 1: i and ii have the same answer. iii and iv both have two answers.

Version 2: i and ii have different answers. iii has two answers,biv has one answer.

Version 3: i and ii have different answer. iii has one answer, iv has two answers.

Version 4: i and ii have the same answer. iii has one answer, iv has two answers.