Mathematics

Let \(f(x) = x^3\). The following diagram shows part of the graph of \(f\).

The point \(\mathrm{P} (a, f(a))\), where \(a > 0\), lies on the graph of \(f\). The tangent at \(\mathrm{P}\) crosses the \(x\)-axis at the point \(\mathrm{Q} \left( \frac{2}{3}, 0 \right)\). This tangent intersects the graph of \(f\) at the point \(\mathrm{R}(-2,-8)\).

1. 1. Show that the gradient of [PQ] is 𝑎3𝑎−23.

   2. Find \(f'(a)\).

   3. Hence show that \(a = 1\).

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The equation of the tangent at \(\mathrm{P}\) is \(y = 3x-2\). Let \(T\) be the region enclosed by the graph of \(f\), the tangent \([\mathrm{PR}]\) and the line \(x = k\), between \(x = -2\) and \(x = k\) where \(-2 < k <1\). This is shown in the diagram below.

2. Given that the area of \(T\) is \(2k +4\), show that \(k\) satisfies the equation \(k^4 - 6k^2 +8=0\).[9]