Find $\displaystyle\int \frac{ax+b}{(x+c)(x+d)}\;dx,\;a\neq 0,\;c \neq d $.

Factorise $x^2+cx+d$ into 2 distinct linear factors and then find $\displaystyle \int \frac{ax+b}{x^2+cx+d}\;dx,\;a \neq 0$ using partial fractions or otherwise.

Factorise $x^2+bx+c$ into 2 distinct linear factors and then find $\displaystyle \int \frac{a}{x^2+bx+c }\;dx$ using partial fractions or otherwise.

Direct calculation of low positive and negative powers of complex numbers. Calculations involving a complex conjugate. Powers of $i$. Four parts.

Find the gcd $d$ of two positive integers $a$ and $b$ also find integers $x,y$ such that $ax+by=d$, using the extended Euclidean algorithm.

Given a number evaluate simple power, negative power, to one half.

Given a number evaluate simple power, negative power, to one half.

Given a number evaluate simple power, negative power, to one half.

This is the question for week 5 of the MA100 course at the LSE. It looks at material from chapters 9 and 10.

The following describes how we define our revenue and cost functions for part b of the question.

We have variables c, f, m, h.

The revenue function is R(q) = -c q^2 + 2mf q .
The cost function is C(q) = f q^2 - 2mc q + h .

The "revenue - cost" function is -(c+f) q^2 +2m(c+f) q - h

Differentiating, we see that there is a maximum point at m.

We pick each one of f, m, h randomly from the set {2, .. 6}, and we pick c randomly from {h+1 , ... , h+5}. This ensures that the discriminant of the "revenue - cost" function is positive, meaning there are two real roots, meaning the maximum point lies above the x-axis. I.e. we can actually make a profit.

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.

Solve simple two step linear equations with feedback.

A quartic is given and the student is asked whether the gradient is positive or negative at various values of x.

A quartic graph is given. The question is to determine whether the gradient is positive or negative at various values of x. Non-calculator. Advice is given.

Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

Given a number evaluate simple power, negative power, to one half.

Given a number evaluate simple power, negative power, to one half.

Solve for $x(t)$, $\displaystyle\frac{dx}{dt}=\frac{a}{(x+b)^n},\;x(0)=0$

Direct calculation of low positive and negative powers of complex numbers. Calculations involving a complex conjugate. Powers of $i$. Four parts.

Find modulus and argument of the complex number $z_1$ and find the $n$th roots of $z_1$ where $n=5,\;6$ or $7$.

Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

This question tests a student's ability to raise a positive base to a negative exponent and then take the negative. The base and the exponent have the same absolute value.

This question tests a student's ability to raise a positive base to a negative exponent and then take the negative.

This question tests a student's ability to raise a positive base to a negative exponent.

This question tests a student's ability to raise a positive base to an exponent of -1.

Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

This question tests a student's ability to raise a positive base to a negative exponent.

This question tests a student's ability to raise a positive base to an exponent of -1.

This question tests a student's ability to raise a positive base to a negative exponent and then take the negative. The base and the exponent have the same absolute value.

This question tests a student's ability to raise a positive base to a negative exponent and then take the negative.