8a) Because is monic and of degree 4, the leading term is .

is also even, so no terms of odd degree can be present.

Therefore our expression is .

8b) We know that ,

and .

From this, you can substitute into the general expression from 8a) and obtain a pair of simultaneous equations which can be solved for and to obtain the entire polynomial.

After the original polynomial is obtained, 8c) is trivial.

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10a) First note that the constant terms in each expression can be equated, so .

Normally these these types of questions, you can fully expand the second equation, and factorise to separate the terms out, which you can directly equate with the original equation.

(if the expression is too complicated to expand, you can repeatedly differentiate the equations and equate the constant terms each time)

But when I tried both methods on that question, I got some conflicting equations so I'm not sure about that. Are you sure you copied the question correctly? Maybe it was a +7 instead of a -7?

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11)

Start by factorising the fraction.

Note that and (always be careful to check if the value of is restricted, this is where people lose marks).

Start by finding the critical points, i.e. points where the denominator is zero or the two sides would be equal.

So our critical points are -4, -3, 6 and 7. These points are where the inequality changes states from true to false. Pick one test point between each critical point to check for whether the inequality holds.

- True

- False

- True

- False

- True

Remember that and

So, , ,

This can be verified by examining the graph of

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