This is the question for week 4 of the MA100 course at the LSE. It looks at material from chapters 7 and 8. The following describes how a polynomial was defined in the question. This may be helpful for anyone who needs to edit this question.

For parts a to c, we used a polynomial defined as m*(x^4 - 2a^2  x^2 + a^4 + b), where the variables "a" and "b" are randomly chosen from a set of reaosnable size, and the variable $m$ is randomly chosen from the set {+1, -1}. We can easily see that this polynomial has stationary points at -a, 0, and a. We introduced the variable "m" so that these stationary points would not always have the same classification. The variable "b" is always positive, and so this ensures that our polynomial does not cross the x-axis. The first and second derivatives; stationary points; the evaluation of the second derivative at the stationary points; the classification of the stationary points; and the axes intercepts can all be easily expressed in terms of the variables "a", "b", and "m". Indeed, this is what we did to mark the student's answers.

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.

Factorise polynomials by identifying common factors. The first expression has a constant common factor; the rest have common factors involving variables.

I feel this question has too many questions inside it, I have since made a question that just asks a single division problem called Decimals: Division (includes rounding the answer).

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.

old question, way too many things in one question! I have made better questions out of each part now.

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.

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.

Evaluate $\int_0^{\,m}e^{ax}\;dx$, $\int_0^{p}\frac{1}{bx+d}\;dx,\;\int_0^{\pi/2} \sin(qx) \;dx$. 

Evaluate $\int_0^{\,m}e^{ax}\;dx$, $\int_0^{p}\frac{1}{bx+d}\;dx,\;\int_0^{\pi/2} \sin(qx) \;dx$. 

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.

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: 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.

Student is given two whole numbers to add, they have to give the sum.

The numbers are in the range -9 to 9.

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.

Statements about the different kinds of averages where you have to select "always", "sometimes", or "never".

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.

Evaluate $\int_0^{\,m}e^{ax}\;dx$, $\int_0^{p}\frac{1}{bx+d}\;dx,\;\int_0^{\pi/2} \sin(qx) \;dx$. 

There are copious comments in the definition of the function eqnline about the voodoo needed to have a JSXGraph diagram interact with the input box for a part.

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.

Factorise polynomials by identifying common factors. The first expression has a constant common factor; the rest have common factors involving variables.

There are copious comments in the definition of the function eqnline about the voodoo needed to have a JSXGraph diagram interact with the input box for a part.

There are copious comments in the definition of the function eqnline about the voodoo needed to have a JSXGraph diagram interact with the input box for a part.

Using Jsxgraph to draw the vector field for a differential equation of the form $\frac{dy}{dx}=f(x,y)=x^2-y^2$, and also by moving the point $(x_0,y_0)$ you can see the solution curves going through that point.

If you want to modify $f(x,y)$ simply change the definition of  $f(x,y)$ and that of the variable  str in the user defined function testfield in Extensions and scripts. You have to use javascript notation for functions and powers in the definition of $f(x,y)$.