OCR C3 (Core Mathematics 3) 2013 January

For each of the following curves, find the gradient at the point with \(x\)-coordinate 2.

1. \(y = \frac{3x}{2x + 1}\) [3]
2. \(y = \sqrt{4x^2 + 9}\) [3]

The acute angle \(A\) is such that \(\tan A = 2\).

1. Find the exact value of \(\cosec A\). [2]
2. The angle \(B\) is such that \(\tan (A + B) = 3\). Using an appropriate identity, find the exact value of \(\tan B\). [3]

1. Given that \(|t| = 3\), find the possible values of \(|2t - 1|\). [3]
2. Solve the inequality \(|x - t^2| > |x + 3\sqrt{2}|\). [4]

The mass, \(m\) grams, of a substance is increasing exponentially so that the mass at time \(t\) hours is given by $$m = 250e^{0.02t}.$$

1. Find the time taken for the mass to increase to twice its initial value, and deduce the time taken for the mass to increase to 8 times its initial value. [3]
2. Find the rate at which the mass is increasing at the instant when the mass is 400 grams. [3]

\includegraphics{figure_5} The diagram shows the curve \(y = \frac{6}{\sqrt{3x + 1}}\). The shaded region is bounded by the curve and the lines \(x = 2\), \(x = 9\) and \(y = 0\).

1. Show that the area of the shaded region is \(4\sqrt{7}\) square units. [4]
2. The shaded region is rotated completely about the \(x\)-axis. Show that the volume of the solid produced can be written in the form \(k\ln 2\), where the exact value of the constant \(k\) is to be determined. [5]

1. By sketching the curves \(y = \ln x\) and \(y = 8 - 2x^2\) on a single diagram, show that the equation $$\ln x = 8 - 2x^2$$ has exactly one real root. [3]
2. Explain how your diagram shows that the root is between 1 and 2. [1]
3. Use the iterative formula $$x_{n+1} = \sqrt{4 - \frac{1}{2}\ln x_n},$$ with a suitable starting value, to find the root. Show all your working and give the root correct to 3 decimal places. [4]
4. The curves \(y = \ln x\) and \(y = 8 - 2x^2\) are each translated by 2 units in the positive \(x\)-direction and then stretched by scale factor 4 in the \(y\)-direction. Find the coordinates of the point where the new curves intersect, giving each coordinate correct to 2 decimal places. [3]

\includegraphics{figure_7} The diagram shows the curve with equation $$x = (y + 4)\ln (2y + 3).$$ The curve crosses the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).

1. Find an expression for \(\frac{dx}{dy}\) in terms of \(y\). [3]
2. Find the gradient of the curve at each of the points \(A\) and \(B\), giving each answer correct to 2 decimal places. [5]

The functions f and g are defined for all real values of \(x\) by $$\text{f}(x) = x^2 + 4ax + a^2 \text{ and } \text{g}(x) = 4x - 2a,$$ where \(a\) is a positive constant.

1. Find the range of f in terms of \(a\). [4]
2. Given that fg(3) = 69, find the value of \(a\) and hence find the value of \(x\) such that \(\text{g}^{-1}(x) = x\). [6]

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 $$6\cos^2(\frac{1}{3}\theta + 45°) - 3(\cos \theta - \sin \theta) = 2$$ for \(-90° < \theta < 90°\). [3]
3. It is given that there are two values of \(\theta\), where \(-90° < \theta < 90°\), satisfying the equation $$6\cos^2(\frac{1}{3}\theta + 45°) - 3(\cos \frac{2}{3}\theta - \sin \frac{2}{3}\theta) = k,$$ where \(k\) is a constant. Find the set of possible values of \(k\). [3]