Given the first three terms of a sequence, give a formula for the $n^\text{th}$ term.

In the first sequence, $d$ is positive. In the second sequence, $d$ is negative.

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multiple choice testing sin, cos, tan of angles that are negative or greater than 360 degrees that result in nice exact values. 

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.

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Using the chain rule with polynomials and negative powers

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.

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.

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.

Find modulus and argument of two complex numbers. Then use De Moivre's Theorem to find negative powers of the complex numbers.

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

testing sin, cos, tan of angles that are negative or greater than 2pi radians that result in nice exact values. 

Find the $x$ and $y$ components of a force which is applied at an angle to a particle. Resolve using $F \cos \theta$. The force is applied in the negative $x$ direction but the positive $y$. 

multiple choice testing sin, cos, tan of angles that are negative or greater than 360 degrees that result in nice exact values. 

Find a unit vector orthogonal to two others.

Uses $\wedge$ for the cross product. The interim calculations should all be displayed to enough dps, not 3,  to ensure accuracy to 3 dps. If the cross product has a negative x component then it is not explained that the negative of the cross product is taken for the unit vector.

Scores (including negatives)  in 20 games and frequency given -- calculate mean.

rebelmaths

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 the first three terms of a sequence, give a formula for the $n^\text{th}$ term.

In the first sequence, $d$ is positive. In the second sequence, $d$ is negative.

Find $\displaystyle \int \frac{a}{(bx+c)^n}\;dx$

Find $\displaystyle\int \frac{ax^3-ax+b}{1-x^2}\;dx$. Input constant of integration as $C$.

Find $\displaystyle \int \frac{a}{(bx+c)^n}\;dx$

Find $\displaystyle \int ae ^ {bx}+ c\sin(dx) + px ^ {q}\;dx$.

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

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.

Testing addition, multiplication and division involving negative numbers.

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.

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Asks the student to add two single-digit numbers, one of which might be negative.