SPS SPS SM Pure (SPS SM Pure) 2020 October

1. Find $$\int \frac{x}{x^2 + 1} dx$$ [2]
2. Find. $$\int 2\pi(4x + 3)^{10} dx$$ [2]
3. Find. $$\int \frac{2}{e^{4x}} dx$$ [2]

1. Find \(\frac{dy}{dx}\) if \(y = 4\ln(3x)\) [2]
2. Differentiate \(\frac{2x}{\sqrt{3x+1}}\) giving your answer in the form \(\frac{3x+c}{\sqrt{(3x+1)^p}}\), where \(c, p \in \mathbb{N}\) [3]

Expand \((x - 2y)^5\). [3]

What transformations could be used, and in which order, to transform the curve \(y = \sin x\) into the curve \(y = 2 \sin(3x + 30°)\)? [3]

Find the equation of the tangent to the curve $$y = 3x^2(x + 2)^6$$ at the point \((-1, 3)\), giving your answer in the form \(y = mx + c\). [5]

1. Express \(\frac{x}{(x + 1)(x + 2)}\) in partial fractions. [3]
2. Hence find \(\int \frac{x}{(x + 1)(x + 2)} dx\). [2]

1. Express \(24 \sin \theta + 7 \cos \theta\) in the form \(R \sin(\theta + \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). [3]
2. Hence solve the equation \(24 \sin \theta + 7 \cos \theta = 12\) for \(0° < \theta < 360°\). [4]

1. 1. Sketch the graph of \(y = \cos \sec x\) for \(0 < x < 4\pi\). [3]
   2. It is given that \(\cos \sec \alpha = \cos \sec \beta\), where \(\frac{1}{2}\pi < \alpha < \pi\) and \(2\pi < \beta < \frac{5}{2}\pi\). By using your sketch, or otherwise, express \(\beta\) in terms of \(\alpha\). [2]
2. 1. Write down the identity giving \(\tan 2\theta\) in terms of \(\tan \theta\). [1]
   2. Given that \(\cot \phi = 4\), find the exact value of \(\tan \phi \cot 2\phi \tan 4\phi\), showing all your working. [6]

Earth is being added to a pile so that, when the height of the pile is \(h\) metres, its volume is \(V\) cubic metres, where $$V = (h^6 + 16)^2 - 4.$$

1. Find the value of \(\frac{dV}{dh}\) when \(h = 2\). [3]
2. The volume of the pile is increasing at a constant rate of 8 cubic metres per hour. Find the rate, in metres per hour, at which the height of the pile is increasing at the instant when \(h = 2\). Give your answer correct to 2 significant figures. [3]

1. Prove that $$\cos^2(\theta + 45°) - \frac{1}{2}(\cos 2\theta - \sin 2\theta) \equiv \sin^2 \theta.$$ [4]
2. Hence solve the equation $$6\cos^2(\frac{1}{2}\theta + 45°) - 3(\cos \theta - \sin \theta) = 2$$ for \(-90° < \theta < 90°\). [3]