SPS SPS FM (SPS FM) 2020 October

1. Find the binomial expansion of \((2 + x)^5\), simplifying the terms. [4]
2. Hence find the coefficient of \(y^3\) in the expansion of \((2 + 3y + y^2)^5\). [3]

Let \(a = \log_2 x\), \(b = \log_2 y\) and \(c = \log_2 z\). Express \(\log_2(xy) - \log_2(\frac{z}{y})\) in terms of \(a\), \(b\) and \(c\). [3]

1. Give full details of a sequence of two transformations needed to transform the graph \(y = |x|\) to the graph of \(y = |2(x + 3)|\). [3]
2. Solve \(|x| > |2(x + 3)|\), giving your answer in set notation. [5]

Prove by induction that, for \(n \geq 1\), \(\sum_{r=1}^n r(3r + 1) = n(n + 1)^2\). [5]

\includegraphics{figure_5} The diagram shows triangle \(ABC\), with \(AB = x\) cm, \(AC = (x + 2)\) cm, \(BC = 2\sqrt{7}\) cm and angle \(CAB = 60°\).

1. Find the value of \(x\). [4]
2. Find the area of triangle \(ABC\), giving your answer in an exact form as simply as possible. [2]

Prove by contradiction that \(\sqrt{7}\) is irrational. [5]

A curve has equation \(y = \frac{1}{4}x^4 - x^3 - 2x^2\).

1. Find \(\frac{dy}{dx}\). [1]
2. Hence sketch the gradient function for the curve. [4]
3. Find the equation of the tangent to the curve \(y = \frac{1}{4}x^4 - x^3 - 2x^2\) at \(x = 4\). [2]

The equation of a circle is \(x^2 + y^2 + 6x - 2y - 10 = 0\).

1. Find the centre and radius of the circle. [3]
2. Find the coordinates of any points where the line \(y = 2x - 3\) meets the circle \(x^2 + y^2 + 6x - 2y - 10 = 0\). [4]
3. State what can be deduced from the answer to part ii. about the line \(y = 2x - 3\) and the circle \(x^2 + y^2 + 6x - 2y - 10 = 0\). [1]
4. The point \(A(-1,5)\) lies on the circumference of the circle \(x^2 + y^2 + 6x - 2y - 10 = 0\). Given that \(AB\) is a diameter of the circle, find the coordinates of \(B\). [2]

In this question you must show detailed reasoning. A sequence \(t_1, t_2, t_3 \ldots\) is defined by \(t_n = 5 - 2n\). Use an algebraic method to find the smallest value of \(N\) such that $$\sum_{n=1}^{\infty} 2^{t_n} - \sum_{n=1}^{N} 2^{t_n} < 10^{-8}$$ [8]