OCR C4 (Core Mathematics 4) 2005 June

Find the quotient and the remainder when \(x^4 + 3x^3 + 5x^2 + 4x - 1\) is divided by \(x^2 + x + 1\). [4]

Evaluate \(\int_0^{\frac{\pi}{2}} x \cos x dx\), giving your answer in an exact form. [5]

The line \(L_1\) passes through the points \((2, -3, 1)\) and \((-1, -2, -4)\). The line \(L_2\) passes through the point \((3, 2, -9)\) and is parallel to the vector \(\mathbf{4i} - \mathbf{4j} + \mathbf{5k}\).

1. Find an equation for \(L_1\) in the form \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\). [2]
2. Prove that \(L_1\) and \(L_2\) are skew. [5]

1. Show that the substitution \(x = \tan \theta\) transforms \(\int \frac{1}{(1 + x^2)^2} dx\) to \(\int \cos^2 \theta d\theta\). [3]
2. Hence find the exact value of \(\int_0^1 \frac{1}{(1 + x^2)^2} dx\). [4]

\(ABCD\) is a parallelogram. The position vectors of \(A\), \(B\) and \(C\) are given respectively by $$\mathbf{a} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}, \quad \mathbf{b} = 3\mathbf{i} - 2\mathbf{j}, \quad \mathbf{c} = \mathbf{i} - \mathbf{j} - 2\mathbf{k}.$$

1. Find the position vector of \(D\). [3]
2. Determine, to the nearest degree, the angle \(ABC\). [4]

The equation of a curve is \(xy^2 = 2x + 3y\).

1. Show that \(\frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}\). [5]
2. Show that the curve has no tangents which are parallel to the \(y\)-axis. [3]

A curve is given parametrically by the equations $$x = t^2, \quad y = \frac{1}{t}.$$

1. Find \(\frac{dy}{dx}\) in terms of \(t\), giving your answer in its simplest form. [3]
2. Show that the equation of the tangent at the point \(P\left(4, -\frac{1}{4}\right)\) is \(x - 16y = 12\). [3]
3. Find the value of the parameter at the point where the tangent at \(P\) meets the curve again. [4]

1. Given that \(\frac{3x + 4}{(1 + x)(2 + x)^2} \equiv \frac{A}{1 + x} + \frac{B}{2 + x} + \frac{C}{(2 + x)^2}\), find \(A\), \(B\) and \(C\). [5]
2. Hence or otherwise expand \(\frac{3x + 4}{(1 + x)(2 + x)^2}\) in ascending powers of \(x\), up to and including the term in \(x^2\). [5]
3. State the set of values of \(x\) for which the expansion in part (ii) is valid. [1]

Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of \(20°C\). At time \(t\) minutes later, the temperature of the liquid is \(\theta°C\).

1. Explain how the information above leads to the differential equation $$\frac{d\theta}{dt} = -k(\theta - 20),$$ where \(k\) is a positive constant. [2]
2. The liquid is initially at a temperature of \(100°C\). It takes 5 minutes for the liquid to cool from \(100°C\) to \(68°C\). Show that $$\theta = 20 + 80e^{-(\frac{k}{5} \ln \frac{5}{3})t}.$$ [8]
3. Calculate how much longer it takes for the liquid to cool by a further \(32°C\). [3]