1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs

Given that \(\arcsin x = \frac{1}{6}\pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).

Fig. 7 shows the curve \(y = f(x)\), where \(f(x) = 1 + 2 \arctan x\), \(x \in \mathbb{R}\). The scales on the \(x\)- and \(y\)-axes are the same. \includegraphics{figure_7}

1. Find the range of f, giving your answer in terms of \(\pi\). [3]
2. Find \(f^{-1}(x)\), and add a sketch of the curve \(y = f^{-1}(x)\) to the copy of Fig. 7. [5]

Fig. 6 shows part of the curve \(\sin 2y = x - 1\). P is the point with coordinates \((1.5, \frac{1}{12}\pi)\) on the curve. \includegraphics{figure_6}

1. Find \(\frac{dy}{dx}\) in terms of \(y\). Hence find the exact gradient of the curve \(\sin 2y = x - 1\) at the point P. [4]

The part of the curve shown is the image of the curve \(y = \arcsin x\) under a sequence of two geometrical transformations.

1. Find \(y\) in terms of \(x\) for the curve \(\sin 2y = x - 1\). Hence describe fully the sequence of transformations. [4]

$$\text{f}(x) = x^2 + 5x - 2 \sec x, \quad x \in \mathbb{R}, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}.$$

1. Show that the equation \(\text{f}(x) = 0\) has a root, \(\alpha\), such that \(1 < \alpha < 1.5\) [2]
2. Show that a suitable rearrangement of the equation \(\text{f}(x) = 0\) leads to the iterative formula $$x_{n+1} = \cos^{-1} \left( \frac{2}{x_n^2 + 5x_n} \right).$$ [3]
3. Use the iterative formula in part (ii) with a starting value of 1.25 to find \(\alpha\) correct to 3 decimal places. You should show the result of each iteration. [3]

1. Sketch on the same diagram the graphs of $$y = \sin^{-1} x, \quad -1 \leq x \leq 1$$ and $$y = \cos^{-1} (2x), \quad -\frac{1}{2} \leq x \leq \frac{1}{2}.$$ [3]

Given that the graphs intersect at the point with coordinates \((a, b)\),

1. show that \(\tan b = \frac{1}{2}\), [3]
2. find the value of \(a\) in the form \(k\sqrt{5}\). [4]

The function \(\text{f}(x) = \frac{\sin x}{2 - \cos x}\) has domain \(-\pi \leqslant x \leqslant \pi\). Fig. 8 shows the graph of \(y = \text{f}(x)\) for \(0 \leqslant x \leqslant \pi\). \includegraphics{figure_6}

1. Find \(\text{f}(-x)\) in terms of \(\text{f}(x)\). Hence sketch the graph of \(y = \text{f}(x)\) for the complete domain \(-\pi \leqslant x \leqslant \pi\). [3]
2. Show that \(\text{f}'(x) = \frac{2\cos x - 1}{(2 - \cos x)^2}\). Hence find the exact coordinates of the turning point P. State the range of the function \(\text{f}(x)\), giving your answer exactly. [8]
3. Using the substitution \(u = 2 - \cos x\) or otherwise, find the exact value of \(\int_0^\pi \frac{\sin x}{2 - \cos x} dx\). [4]
4. Sketch the graph of \(y = \text{f}(2x)\). [1]
5. Using your answers to parts (iii) and (iv), write down the exact value of \(\int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 - \cos 2x} dx\). [2]

Fig. 3 shows the curve defined by the equation \(y = \arcsin(x - 1)\), for \(0 \leqslant x \leqslant 2\). \includegraphics{figure_7}

1. Find \(x\) in terms of \(y\), and show that \(\frac{dx}{dy} = \cos y\). [3]
2. Hence find the exact gradient of the curve at the point where \(x = 1.5\). [4]

Fig. 7 shows the curve \(y = f(x)\), where \(f(x) = 1 + 2\arctan x, x \in \mathbb{R}\). The scales on the x- and y-axes are the same. \includegraphics{figure_7}

1. Find the range of f, giving your answer in terms of \(\pi\). [3]
2. Find \(f^{-1}(x)\), and add a sketch of the curve \(y = f^{-1}(x)\) to the copy of Fig. 7. [5]

Sketch the curve \(y = 2\arccos x\) for \(-1 \leqslant x \leqslant 1\). [3]

It is given that \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\).

1. By writing \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\) as \(2\tan\theta = 3x\), use implicit differentiation to show that $$\frac{d\theta}{dx} = \frac{k}{4 + 9x^2}$$, where \(k\) is an integer. [3 marks]
2. Hence solve the differential equation $$9y(4 + 9x^2)\frac{dy}{dx} = \cosec 3y$$ given that \(x = 0\) when \(y = \frac{\pi}{3}\). Give your answer in the form \(\mathbf{g}(y) = \mathbf{h}(x)\). [7 marks]

It is given that \(f(x) = \tan^{-1}(1 + x)\).

1. Find \(f(0)\) and \(f'(0)\), and show that \(f''(0) = -\frac{1}{2}\). [4]
2. Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^2\). [2]

By first completing the square in the denominator, find the exact value of $$\int_{\frac{1}{2}}^{\frac{1}{2}} \frac{1}{4x^2 - 4x + 5} dx.$$ [5]

1. Prove that the derivative of \(\cos^{-1} x\) is \(-\frac{1}{\sqrt{1 - x^2}}\). [3]

A curve has equation \(y = \cos^{-1}(1 - x^2)\), for \(0 < x < \sqrt{2}\).

1. Find and simplify \(\frac{dy}{dx}\), and hence show that $$(2 - x^2)\frac{d^2y}{dx^2} = x\frac{dy}{dx}.$$ [5]

\includegraphics{figure_1} The circle, with centre \(C\) and radius \(r\), touches the \(y\)-axis at \((0, 4)\) and also touches the line with equation \(4y - 3x = 0\), as shown in Fig. 1.

1. 1. Find the value of \(r\).
   2. Show that \(\arctan \left(\frac{4}{3}\right) + 2 \arctan \left(\frac{1}{2}\right) = \frac{1}{2} \pi\). [8]

The line with equation \(4x + 3y = q\), \(q > 12\), is a tangent to the circle.

1. Find the value of \(q\). [4]

The equation \(6\arcsin(2x - 1) - x^2 = 0\) has exactly one real root.

1. Show by calculation that the root lies between 0.5 and 0.6. [2]

In order to find the root, the iterative formula \(x_{n+1} = p + q\sin(rx_n^2)\), with initial value \(x_0 = 0.5\), is to be used.

1. Determine the values of the constants \(p\), \(q\) and \(r\). [2]
2. Hence find the root correct to 4 significant figures. Show the result of each step of the iteration process. [2]

One of the diagrams below shows the graph of \(y = \arccos x\) Identify the graph of \(y = \arccos x\) Tick \((\checkmark)\) one box. [1 mark] \includegraphics{figure_4}

Which one of the following equations has no real solutions? Tick (\(\checkmark\)) one box. \(\cot x = 0\) \(\ln x = 0\) \(|x + 1| = 0\) \(\sec x = 0\) [1 mark]

\(f(x) = \arcsin x\) State the maximum possible domain of \(f\) Tick \((\checkmark)\) one box. [1 mark] \(\{x \in \mathbb{R} : -1 \leq x \leq 1\}\) \(\left\{x \in \mathbb{R} : -\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\right\}\) \(\{x \in \mathbb{R} : -\pi \leq x \leq \pi\}\) \(\{x \in \mathbb{R} : -90 \leq x \leq 90\}\)