1.02w Graph transformations: simple transformations of f(x)

The function f is defined by $$f: x \mapsto |2x - a|, \quad x \in \mathbb{R}$$ where \(a\) is a positive constant.

1. Sketch the graph of \(y = f(x)\), showing the coordinates of the points where the graph cuts the axes. [2]
2. On a separate diagram, sketch the graph of \(y = f(2x)\), showing the coordinates of the points where the graph cuts the axes. [2]
3. Given that a solution of the equation f(x) = \(\frac{1}{2}x\) is \(x = 4\), find the two possible values of \(a\). [4]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = \text{f}(x)\), \(-1 \leq x \leq 3\). The curve touches the \(x\)-axis at the origin \(O\), crosses the \(x\)-axis at the point \(A(2, 0)\) and has a maximum at the point \(B(\frac{4}{3}, 1)\). In separate diagrams, show a sketch of the curve with equation

1. \(y = \text{f}(x + 1)\), [3]
2. \(y = |\text{f}(x)|\), [3]
3. \(y = \text{f}(|x|)\), [4]

marking on each sketch the coordinates of points at which the curve

1. has a turning point,
2. meets the \(x\)-axis.

The function f is defined by \(\text{f}: x \to \frac{3x - 1}{x - 3}\), \(x \in \mathbb{R}\), \(x \neq 3\).

1. Prove that \(\text{f}^{-1}(x) = \text{f}(x)\) for all \(x \in \mathbb{R}\), \(x \neq 3\). [3]
2. Hence find, in terms of \(k\), \(\text{f}f(k)\), where \(x \neq 3\). [2]

\includegraphics{figure_3} Figure 3 shows a sketch of the one-one function g, defined over the domain \(-2 \leq x \leq 2\).

1. Find the value of \(\text{f}g(-2)\). [3]
2. Sketch the graph of the inverse function \(\text{g}^{-1}\) and state its domain. [3]

The function h is defined by \(\text{h}: x \mapsto 2g(x - 1)\).

1. Sketch the graph of the function h and state its range. [3]

\includegraphics{figure_3} Figure 3 shows a sketch of the curve with equation \(y = \text{f}(x)\), \(x \geq 0\). The curve meets the coordinate axes at the points \((0, c)\) and \((d, 0)\). In separate diagrams sketch the curve with equation

1. \(y = \text{f}^{-1}(x)\), [2]
2. \(y = 3\text{f}(2x)\). [3]

Indicate clearly on each sketch the coordinates, in terms of \(c\) or \(d\), of any point where the curve meets the coordinate axes. Given that f is defined by $$\text{f}: x \mapsto 3(2^{-x}) - 1, \quad x \in \mathbb{R}, x \geq 0,$$

1. state
   1. the value of \(c\),
   2. the range of f.
   [3]
2. Find the value of \(d\), giving your answer to 3 decimal places. [3]

The function g is defined by $$\text{g}: x \mapsto \log_2 x, \quad x \in \mathbb{R}, x \geq 1.$$

1. Find fg(x), giving your answer in its simplest form. [3]

\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]

\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]

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]

\includegraphics{figure_4} The diagram shows the curves \(y = \ln x\) and \(y = 2 \ln(x - 6)\). The curves meet at the point \(P\) which has \(x\)-coordinate \(a\). The shaded region is bounded by the curve \(y = 2 \ln(x - 6)\) and the lines \(x = a\) and \(y = 0\).

1. Give details of the pair of transformations which transforms the curve \(y = \ln x\) to the curve \(y = 2 \ln(x - 6)\). [3]
2. Solve an equation to find the value of \(a\). [4]
3. Use Simpson's rule with two strips to find an approximation to the area of the shaded region. [3]

The transformations R, S and T are defined as follows. \begin{align} \text{R} &: \text{ reflection in the } x\text{-axis}
\text{S} &: \text{ stretch in the } x\text{-direction with scale factor 3}
\text{T} &: \text{ translation in the positive } x\text{-direction by 4 units} \end{align}

1. The curve \(y = \ln x\) is transformed by R followed by T. Find the equation of the resulting curve. [2]
2. Find, in terms of S and T, a sequence of transformations that transforms the curve \(y = x^3\) to the curve \(y = \left(\frac{1}{3}x - 4\right)^3\). You should make clear the order of the transformations. [2]

The functions \(f(x)\) and \(g(x)\) are defined for the domain \(x > 0\) as follows: $$f(x) = \ln x, \quad g(x) = x^3.$$ Express the composite function \(fg(x)\) in terms of \(\ln x\). State the transformation which maps the curve \(y = f(x)\) onto the curve \(y = fg(x)\).

Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = \frac{1}{\cos^2 x}\), \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), together with its asymptotes \(x = \frac{1}{2}\pi\) and \(x = -\frac{1}{2}\pi\). \includegraphics{figure_9}

1. Use the quotient rule to show that the derivative of \(\frac{\sin x}{\cos x}\) is \(\frac{1}{\cos^2 x}\). [3]
2. Find the area bounded by the curve \(y = f(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\pi\). [3]

The function \(g(x)\) is defined by \(g(x) = \frac{1}{2}f(x + \frac{1}{4}\pi)\).

1. Verify that the curves \(y = f(x)\) and \(y = g(x)\) cross at \((0, 1)\). [3]
2. State a sequence of two transformations such that the curve \(y = f(x)\) is mapped to the curve \(y = g(x)\). On the copy of Fig. 9, sketch the curve \(y = g(x)\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve. [8]
3. Use your result from part (ii) to write down the area bounded by the curve \(y = g(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = -\frac{1}{4}\pi\). [1]

Each of the graphs of \(y = \text{f}(x)\) and \(y = \text{g}(x)\) below is obtained using a sequence of two transformations applied to the corresponding dashed graph. In each case, state suitable transformations, and hence find expressions for \(\text{f}(x)\) and \(\text{g}(x)\).

1. \includegraphics{figure_5i} [3]
2. \includegraphics{figure_5ii} [3]

Fig. 8 shows parts of the curves \(y = f(x)\) and \(y = g(x)\), where \(f(x) = \tan x\) and \(g(x) = 1 + f(x - \frac{1}{4}\pi)\). \includegraphics{figure_8}

1. Describe a sequence of two transformations which maps the curve \(y = f(x)\) to the curve \(y = g(x)\). [4]

It can be shown that \(g(x) = \frac{2\sin x}{\sin x + \cos x}\).

1. Show that \(g'(x) = \frac{2}{(\sin x + \cos x)^2}\). Hence verify that the gradient of \(y = g(x)\) at the point \((\frac{1}{4}\pi, 1)\) is the same as that of \(y = f(x)\) at the origin. [7]
2. By writing \(\tan x = \frac{\sin x}{\cos x}\) and using the substitution \(u = \cos x\), show that \(\int_0^{\frac{1}{4}\pi} f(x)dx = \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{u}du\). Evaluate this integral exactly. [4]
3. Hence find the exact area of the region enclosed by the curve \(y = g(x)\), the \(x\)-axis and the lines \(x = \frac{1}{4}\pi\) and \(x = \frac{1}{2}\pi\). [2]

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]

Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = e^{2x} + k e^{-2x}\) and \(k\) is a constant greater than 1. The curve crosses the \(y\)-axis at P and has a turning point Q. \includegraphics{figure_9}

1. Find the \(y\)-coordinate of P in terms of \(k\). [1]
2. Show that the \(x\)-coordinate of Q is \(\frac{1}{4}\ln k\), and find the \(y\)-coordinate in its simplest form. [5]
3. Find, in terms of \(k\), the area of the region enclosed by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\ln k\). Give your answer in the form \(ak + b\). [4]

The function \(g(x)\) is defined by \(g(x) = f(x + \frac{1}{4}\ln k)\).

1. 1. Show that \(g(x) = \sqrt{k}(e^{2x} + e^{-2x})\). [3]
   2. Hence show that \(g(x)\) is an even function. [2]
   3. Deduce, with reasons, a geometrical property of the curve \(y = f(x)\). [3]

\includegraphics{figure_1} Figure 1 shows the curve \(y = f(x)\) which has a maximum point at \((-45, 7)\) and a minimum point at \((135, -1)\).

1. Showing the coordinates of any stationary points, sketch on separate diagrams the graphs of
   1. \(y = f(|x|)\),
   2. \(y = 1 + 2f(x)\). [6]

Given that $$f(x) = A + 2\sqrt{2} \cos x^{\circ} - 2\sqrt{2} \sin x^{\circ}, \quad x \in \mathbb{R}, \quad -180 \leq x \leq 180,$$ where \(A\) is a constant,

1. show that f(x) can be expressed in the form $$f(x) = A + R \cos (x + \alpha)^{\circ},$$ where \(R > 0\) and \(0 < \alpha < 90\), [3]
2. state the value of \(A\), [1]
3. find, to \(1\) decimal place, the \(x\)-coordinates of the points where the curve \(y = f(x)\) crosses the \(x\)-axis. [4]

\includegraphics{figure_7} The diagram shows the curve \(y = \text{f}(x)\) which has a maximum point at \((-45, 7)\) and a minimum point at \((135, -1)\).

1. Showing the coordinates of any stationary points, sketch the curve with equation \(y = 1 + 2\text{f}(x)\). [3]

Given that $$\text{f}(x) = A + 2\sqrt{2} \cos x° - 2\sqrt{2} \sin x°, \quad x \in \mathbb{R}, \quad -180 \leq x \leq 180,$$ where \(A\) is a constant,

1. show that f\((x)\) can be expressed in the form $$\text{f}(x) = A + R \cos (x + \alpha)°,$$ where \(R > 0\) and \(0 < \alpha < 90\), [3]
2. state the value of \(A\), [1]
3. find, to 1 decimal place, the \(x\)-coordinates of the points where the curve \(y = \text{f}(x)\) crosses the \(x\)-axis. [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]

Each of the graphs of \(y = \text{f}(x)\) and \(y = \text{g}(x)\) below is obtained using a sequence of two transformations applied to the corresponding dashed graph. In each case, state suitable transformations, and hence find expressions for f(x) and g(x).

1. \includegraphics{figure_3i} [3]
2. \includegraphics{figure_3ii} [3]

Fig. 4 shows the curve \(y = \text{f}(x)\), where \(\text{f}(x) = \sqrt{1 - 9x^2}\), \(-a < x < a\). \includegraphics{figure_4}

1. Find the value of \(a\). [2]
2. Write down the range of f(x). [1]
3. Sketch the curve \(y = \text{f}(\frac{1}{3}x) - 1\). [3]

Fig. 8 shows part of the curve \(y = \text{f}(x)\), where \(\text{f}(x) = e^{-\frac{1}{5}x} \sin x\), for all \(x\). \includegraphics{figure_8}

1. Sketch the graphs of (A) \(y = \text{f}(2x)\), (B) \(y = \text{f}(x + \pi)\). [4]
2. Show that the \(x\)-coordinate of the turning point P satisfies the equation \(\tan x = 5\). Hence find the coordinates of P. [6]
3. Show that \(\text{f}(x + \pi) = -e^{-\frac{1}{5}\pi}\text{f}(x)\). Hence, using the substitution \(u = x - \pi\), show that $$\int_{\pi}^{2\pi} \text{f}(x)\,dx = -e^{-\frac{1}{5}\pi} \int_{0}^{\pi} \text{f}(u)\,du.$$ Interpret this result graphically. [You should not attempt to integrate f(x).] [8]