OCR C3 (Core Mathematics 3)

The function f is defined for all real values of \(x\) by $$f(x) = 10 - (x + 3)^2.$$

1. State the range of f. [1]
2. Find the value of ff(-1). [3]

Show that \(\int_2^8 \frac{3}{x} \, dx = \ln 64\). [4]

Find the equation of the tangent to the curve \(y = \sqrt{4x + 1}\) at the point \((2, 3)\). [5]

Find the equation of the tangent to the curve \(y = \frac{2x + 1}{3x - 1}\) at the point \((1, \frac{3}{2})\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [5]

Differentiate each of the following with respect to \(x\).

1. \(x^3(x + 1)^5\) [2]
2. \(\sqrt{3x^4 + 1}\) [3]

Functions f and g are defined for all real values of \(x\) by $$f(x) = x^3 + 4 \quad \text{and} \quad g(x) = 2x - 5.$$ Evaluate

1. fg(1), [2]
2. \(f^{-1}(12)\). [3]

Find the exact solutions of the equation \(|6x - 1| = |x - 1|\). [4]

Solve, for \(0° < \theta < 360°\), the equation \(\sec^2 \theta = 4 \tan \theta - 2\). [5]

Solve the inequality \(|2x - 3| < |x + 1|\). [5]

It is given that \(\theta\) is the acute angle such that \(\sin \theta = \frac{12}{13}\). Find the exact value of

1. \(\cot \theta\), [2]
2. \(\cos 2\theta\). [3]

Solve the inequality \(|4x - 3| < |2x + 1|\). [5]

The sequence defined by $$x_1 = 3, \quad x_{n+1} = \sqrt{31 - \frac{5}{2}x_n}$$ converges to the number \(\alpha\).

1. Find the value of \(\alpha\) correct to 3 decimal places, showing the result of each iteration. [3]
2. Find an equation of the form \(ax^3 + bx + c = 0\), where \(a\), \(b\) and \(c\) are integers, which has \(\alpha\) as a root. [3]

The mass, \(m\) grams, of a substance at time \(t\) years is given by the formula $$m = 180e^{-0.017t}.$$

1. Find the value of \(t\) for which the mass is 25 grams. [3]
2. Find the rate at which the mass is decreasing when \(t = 55\). [3]

1. Differentiate \(x^2(x + 1)^6\) with respect to \(x\). [3]
2. Find the gradient of the curve \(y = \frac{x^2 + 3}{x^2 - 3}\) at the point where \(x = 1\). [3]

The equation \(2x^3 + 4x - 35 = 0\) has one real root.

1. Show by calculation that this real root lies between 2 and 3. [3]
2. Use the iterative formula $$x_{n+1} = \sqrt[3]{17.5 - 2x_n},$$ with a suitable starting value, to find the real root of the equation \(2x^3 + 4x - 35 = 0\) correct to 2 decimal places. You should show the result of each iteration. [3]

1. It is given that \(a\) and \(b\) are positive constants. By sketching graphs of $$y = x^5 \quad \text{and} \quad y = a - bx$$ on the same diagram, show that the equation $$x^5 + bx - a = 0$$ has exactly one real root. [3]
2. Use the iterative formula \(x_{n+1} = \sqrt[5]{53 - 2x_n}\), with a suitable starting value, to find the real root of the equation \(x^5 + 2x - 53 = 0\). Show the result of each iteration, and give the root correct to 3 decimal places. [4]

The function f is defined for all non-negative values of \(x\) by $$f(x) = 3 + \sqrt{x}.$$

1. Evaluate ff(169). [2]
2. Find an expression for \(f^{-1}(x)\) in terms of \(x\). [2]
3. On a single diagram sketch the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\), indicating how the two graphs are related. [3]

1. Solve, for \(0° < \alpha < 180°\), the equation \(\sec \frac{1}{2}\alpha = 4\). [3]
2. Solve, for \(0° < \beta < 180°\), the equation \(\tan \beta = 7 \cot \beta\). [4]

1. \includegraphics{figure_4a} The diagram shows the curve \(y = \frac{2}{\sqrt{x}}\). The region \(R\), shaded in the diagram, is bounded by the curve and by the lines \(x = 1\), \(x = 5\) and \(y = 0\). The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed. [4]
2. Use Simpson's rule, with 4 strips, to find an approximate value for $$\int_1^5 \sqrt{(x^2 + 1)} \, dx,$$ giving your answer correct to 3 decimal places. [4]

\includegraphics{figure_4} The function f is defined by \(f(x) = 2 - \sqrt{x}\) for \(x \geq 0\). The graph of \(y = f(x)\) is shown above.

1. State the range of f. [1]
2. Find the value of ff(4). [2]
3. Given that the equation \(|f(x)| = k\) has two distinct roots, determine the possible values of the constant \(k\). [2]

It is given that \(y = 5^{x-1}\).

1. Show that \(x = 1 + \frac{\ln y}{\ln 5}\). [2]
2. Find an expression for \(\frac{dx}{dy}\) in terms of \(y\). [2]
3. Hence find the exact value of the gradient of the curve \(y = 5^{x-1}\) at the point \((3, 25)\). [2]

1. Given that \(x = (4t + 9)^{\frac{1}{2}}\) and \(y = 6e^{\frac{2t+1}{4}}\), find expressions for \(\frac{dx}{dt}\) and \(\frac{dy}{dx}\). [4]
2. Hence find the value of \(\frac{dy}{dt}\) when \(t = 4\), giving your answer correct to 3 significant figures. [3]

The integral \(I\) is defined by $$I = \int_0^{13} (2x + 1)^{\frac{3}{2}} \, dx.$$

1. Use integration to find the exact value of \(I\). [4]
2. Use Simpson's rule with two strips to find an approximate value for \(I\). Give your answer correct to 3 significant figures. [3]

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 $$V = (h^6 + 16)^{\frac{1}{2}} - 4.$$

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. Express \(3 \sin \theta + 2 \cos \theta\) in the form \(R \sin(\theta + \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). [3]
2. Hence solve the equation \(3 \sin \theta + 2 \cos \theta = \frac{5}{2}\), giving all solutions for which \(0° < \theta < 360°\). [5]

\includegraphics{figure_5} The diagram shows the curves \(y = (1 - 2x)^5\) and \(y = e^{2x-1} - 1\). The curves meet at the point \((\frac{1}{2}, 0)\). Find the exact area of the region (shaded in the diagram) bounded by the \(y\)-axis and by part of each curve. [8]

1. Write down the identity expressing \(\sin 2\theta\) in terms of \(\sin \theta\) and \(\cos \theta\). [1]
2. Given that \(\sin \alpha = \frac{1}{4}\) and \(\alpha\) is acute, show that \(\sin 2\alpha = \frac{1}{8}\sqrt{15}\). [3]
3. Solve, for \(0° < \beta < 90°\), the equation \(5 \sin 2\beta \sec \beta = 3\). [3]

1. Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos(\theta + \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). [3]
2. Hence solve the equation \(4 \cos \theta - \sin \theta = 2\), giving all solutions for which \(-180° < \theta < 180°\). [5]

A substance is decaying in such a way that its mass, \(m\) kg, at a time \(t\) years from now is given by the formula $$m = 240e^{-0.04t}.$$

1. Find the time taken for the substance to halve its mass. [3]
2. Find the value of \(t\) for which the mass is decreasing at a rate of 2.1 kg per year. [4]

1. Find \(\int (3x + 7)^9 \, dx\). [3]
2. \includegraphics{figure_5b} The diagram shows the curve \(y = \frac{1}{2\sqrt{x}}\). The shaded region is bounded by the curve and the lines \(x = 3\), \(x = 6\) and \(y = 0\). The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid produced, simplifying your answer. [5]

1. Find the exact value of the \(x\)-coordinate of the stationary point of the curve \(y = x \ln x\). [4]
2. The equation of a curve is \(y = \frac{4x + c}{4x - c}\), where \(c\) is a non-zero constant. Show by differentiation that this curve has no stationary points. [3]

1. \(t\)	0	10	20
   \(X\)	275	440

   The quantity \(X\) is increasing exponentially with respect to time \(t\). The table above shows values of \(X\) for different values of \(t\). Find the value of \(X\) when \(t = 20\). [3]
2. The quantity \(Y\) is decreasing exponentially with respect to time \(t\) where $$Y = 80e^{-0.02t}.$$
   1. Find the value of \(t\) for which \(Y = 20\), giving your answer correct to 2 significant figures. [3]
   2. Find by differentiation the rate at which \(Y\) is decreasing when \(t = 30\), giving your answer correct to 2 significant figures. [3]

\includegraphics{figure_6} The diagram shows the graph of \(y = f(x)\), where $$f(x) = 2 - x^2, \quad x \leq 0.$$

1. Evaluate ff(-3). [3]
2. Find an expression for \(f^{-1}(x)\). [3]
3. Sketch the graph of \(y = f^{-1}(x)\). Indicate the coordinates of the points where the graph meets the axes. [3]

\includegraphics{figure_6} The diagram shows the curve with equation \(y = \frac{1}{\sqrt{3x + 2}}\). The shaded region is bounded by the curve and the lines \(x = 0\), \(x = 2\) and \(y = 0\).

1. Find the exact area of the shaded region. [4]
2. The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed, simplifying your answer. [5]

1. Given that \(\int_0^a (6e^{2x} + x) \, dx = 42\), show that \(a = \frac{1}{2} \ln(15 - \frac{1}{6}a^2)\). [5]
2. Use an iterative formula, based on the equation in part (i), to find the value of \(a\) correct to 3 decimal places. Use a starting value of 1 and show the result of each iteration. [4]

1. Write down the formula for \(\cos 2x\) in terms of \(\cos x\). [1]
2. Prove the identity \(\frac{4 \cos 2x}{1 + \cos 2x} = 4 - 2 \sec^2 x\). [3]
3. Solve, for \(0 < x < 2\pi\), the equation \(\frac{4 \cos 2x}{1 + \cos 2x} = 3 \tan x - 7\). [5]

\includegraphics{figure_7} The diagram shows the curve with equation \(y = \cos^{-1} x\).

1. Sketch the curve with equation \(y = 3 \cos^{-1}(x - 1)\), showing the coordinates of the points where the curve meets the axes. [3]
2. By drawing an appropriate straight line on your sketch in part (i), show that the equation \(3 \cos^{-1}(x - 1) = x\) has exactly one root. [1]
3. Show by calculation that the root of the equation \(3 \cos^{-1}(x - 1) = x\) lies between 1.8 and 1.9. [2]
4. The sequence defined by $$x_1 = 2, \quad x_{n+1} = 1 + \cos(\frac{1}{3}x_n)$$ converges to a number \(\alpha\). Find the value of \(\alpha\) correct to 2 decimal places and explain why \(\alpha\) is the root of the equation \(3 \cos^{-1}(x - 1) = x\). [5]

1. Find the exact value of \(\int_1^2 \frac{2}{(4x - 1)^2} \, dx\). [4]
2. \includegraphics{figure_7b} The diagram shows part of the curve \(y = \frac{1}{x}\). The point \(P\) has coordinates \((a, \frac{1}{a})\) and the point \(Q\) has coordinates \((2a, \frac{1}{2a})\), where \(a\) is a positive constant. The point \(R\) is such that \(PR\) is parallel to the \(x\)-axis and \(QR\) is parallel to the \(y\)-axis. The region shaded in the diagram is bounded by the curve and by the lines \(PR\) and \(QR\). Show that the area of this shaded region is \(\ln(\frac{4}{e})\). [6]

The curve \(y = \ln x\) is transformed to the curve \(y = \ln(\frac{1}{2}x - a)\) by means of a translation followed by a stretch. It is given that \(a\) is a positive constant.

1. Give full details of the translation and stretch involved. [2]
2. Sketch the graph of \(y = \ln(\frac{1}{2}x - a)\). [2]
3. Sketch, on another diagram, the graph of \(y = |\ln(\frac{1}{2}x - a)|\). [2]
4. State, in terms of \(a\), the set of values of \(x\) for which \(|\ln(\frac{1}{2}x - a)| = -\ln(\frac{1}{2}x - a)\). [2]

1. Sketch the graph of \(y = \sec x\) for \(0 \leq x \leq 2\pi\). [2]
2. Solve the equation \(\sec x = 3\) for \(0 \leq x \leq 2\pi\), giving the roots correct to 3 significant figures. [3]
3. Solve the equation \(\sec \theta = 5 \cos \theta\) for \(0 \leq \theta \leq 2\pi\), giving the roots correct to 3 significant figures. [4]

\includegraphics{figure_8} The diagram shows part of each of the curves \(y = e^{\frac{1}{3}x}\) and \(y = \sqrt[3]{(3x + 8)}\). The curves meet, as shown in the diagram, at the point \(P\). The region \(R\), shaded in the diagram, is bounded by the two curves and by the \(y\)-axis.

1. Show by calculation that the \(x\)-coordinate of \(P\) lies between 5.2 and 5.3. [3]
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac{2}{3} \ln(3x + 8)\). [2]
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(P\) correct to 2 decimal places. [3]
4. Use integration, and your answer to part (iii), to find an approximate value of the area of the region \(R\). [5]

\includegraphics{figure_8} The diagram shows part of the curve \(y = \ln(5 - x^2)\) which meets the \(x\)-axis at the point \(P\) with coordinates \((2, 0)\). The tangent to the curve at \(P\) meets the \(y\)-axis at the point \(Q\). The region \(A\) is bounded by the curve and the lines \(x = 0\) and \(y = 0\). The region \(B\) is bounded by the curve and the lines \(PQ\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\). [5]
2. Use Simpson's Rule with four strips to find an approximation to the area of the region \(A\), giving your answer correct to 3 significant figures. [4]
3. Deduce an approximation to the area of the region \(B\). [2]

1. Express \(5 \cos x + 12 \sin x\) in the form \(R \cos(x - \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). [3]
2. Hence give details of a pair of transformations which transforms the curve \(y = \cos x\) to the curve \(y = 5 \cos x + 12 \sin x\). [3]
3. Solve, for \(0° < x < 360°\), the equation \(5 \cos x + 12 \sin x = 2\), giving your answers correct to the nearest \(0.1°\). [5]

\includegraphics{figure_8} The diagram shows the curve with equation \(y = x^8 e^{-x^2}\). The curve has maximum points at \(P\) and \(Q\). The shaded region \(A\) is bounded by the curve, the line \(y = 0\) and the line through \(Q\) parallel to the \(y\)-axis. The shaded region \(B\) is bounded by the curve, the line \(y = 0\) and the line \(PQ\).

1. Show by differentiation that the \(x\)-coordinate of \(Q\) is 2. [5]
2. Use Simpson's rule with 4 strips to find an approximation to the area of region \(A\). Give your answer correct to 3 decimal places. [4]
3. Deduce an approximation to the area of region \(B\). [2]

1. Given that \(y = \frac{4 \ln x - 3}{4 \ln x + 3}\), show that \(\frac{dy}{dx} = \frac{24}{x(4 \ln x + 3)^2}\). [3]
2. Find the exact value of the gradient of the curve \(y = \frac{4 \ln x - 3}{4 \ln x + 3}\) at the point where it crosses the \(x\)-axis. [4]
3. \includegraphics{figure_8iii} The diagram shows part of the curve with equation $$y = \frac{2}{x^2(4 \ln x + 3)}.$$ The region shaded in the diagram is bounded by the curve and the lines \(x = 1\), \(x = e\) and \(y = 0\). Find the exact volume of the solid produced when this shaded region is rotated completely about the \(x\)-axis. [4]

\includegraphics{figure_9} The function f is defined by \(f(x) = \sqrt{(mx + 7)} - 4\), where \(x \geq -\frac{7}{m}\) and \(m\) is a positive constant. The diagram shows the curve \(y = f(x)\).

1. A sequence of transformations maps the curve \(y = \sqrt{x}\) to the curve \(y = f(x)\). Give details of these transformations. [4]
2. Explain how you can tell that f is a one-one function and find an expression for \(f^{-1}(x)\). [4]
3. It is given that the curves \(y = f(x)\) and \(y = f^{-1}(x)\) do not meet. Explain how it can be deduced that neither curve meets the line \(y = x\), and hence determine the set of possible values of \(m\). [5]

1. By first writing \(\sin 3\theta\) as \(\sin(2\theta + \theta)\), show that $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta.$$ [4]
2. Determine the greatest possible value of $$9 \sin(\frac{10}{3}\alpha) - 12 \sin^3(\frac{10}{3}\alpha),$$ and find the smallest positive value of \(\alpha\) (in degrees) for which that greatest value occurs. [3]
3. Solve, for \(0° < \beta < 90°\), the equation \(3 \sin 6\beta \cos 2\beta = 4\). [6]

\includegraphics{figure_9} The diagram shows the curve with equation \(y = 2 \ln(x - 1)\). The point \(P\) has coordinates \((0, p)\). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 0\), \(y = 0\) and \(y = p\). The units on the axes are centimetres. The region \(R\) is rotated completely about the \(y\)-axis to form a solid.

1. Show that the volume, \(V \text{ cm}^3\), of the solid is given by $$V = \pi(e^p + 4e^{\frac{p}{2}} + p - 5).$$ [8]
2. It is given that the point \(P\) is moving in the positive direction along the \(y\)-axis at a constant rate of \(0.2 \text{ cm min}^{-1}\). Find the rate at which the volume of the solid is increasing at the instant when \(p = 4\), giving your answer correct to 2 significant figures. [5]

Functions f and g are defined by $$f(x) = 2 \sin x \quad \text{for } -\frac{1}{2}\pi \leq x \leq \frac{1}{2}\pi,$$ $$g(x) = 4 - 2x^2 \quad \text{for } x \in \mathbb{R}.$$

1. State the range of f and the range of g. [2]
2. Show that gf(0.5) = 2.16, correct to 3 significant figures, and explain why fg(0.5) is not defined. [4]
3. Find the set of values of \(x\) for which \(f^{-1}g(x)\) is not defined. [6]

1. Prove the identity $$\tan(\theta + 60°) \tan(\theta - 60°) \equiv \frac{\tan^2 \theta - 3}{1 - 3 \tan^2 \theta}.$$ [4]
2. Solve, for \(0° < \theta < 180°\), the equation $$\tan(\theta + 60°) \tan(\theta - 60°) = 4 \sec^2 \theta - 3,$$ giving your answers correct to the nearest \(0.1°\). [5]
3. Show that, for all values of the constant \(k\), the equation $$\tan(\theta + 60°) \tan(\theta - 60°) = k^2$$ has two roots in the interval \(0° < \theta < 180°\). [3]