1.02v Inverse and composite functions: graphs and conditions for existence

The functions f and g are defined for all real values of \(x\) by $$f(x) = 4x^2 - 12x \quad \text{and} \quad g(x) = ax + b,$$ where \(a\) and \(b\) are non-zero constants.

1. Find the range of f. [3]
2. Explain why the function f has no inverse. [2]
3. Given that \(g^{-1}(x) = g(x)\) for all values of \(x\), show that \(a = -1\). [4]
4. Given further that gf\((x) < 5\) for all values of \(x\), find the set of possible values of \(b\). [4]

The function \(f(x) = \ln(1 + x^2)\) has domain \(-3 \leq x \leq 3\). Fig. 9 shows the graph of \(y = f(x)\). \includegraphics{figure_9}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve. [3]
2. Find the gradient of the curve at the point P\((2, \ln 5)\). [4]
3. Explain why the function does not have an inverse for the domain \(-3 \leq x \leq 3\). [1]

The domain of \(f(x)\) is now restricted to \(0 \leq x \leq 3\). The inverse of \(f(x)\) is the function \(g(x)\).

1. Sketch the curves \(y = f(x)\) and \(y = g(x)\) on the same axes. State the domain of the function \(g(x)\). Show that \(g(x) = \sqrt{e^x - 1}\). [6]
2. Differentiate \(g(x)\). Hence verify that \(g'(\ln 5) = \frac{1}{4}\). Explain the connection between this result and your answer to part (ii). [5]

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. 9 shows the curves \(y = \text{f}(x)\) and \(y = \text{g}(x)\). The function \(y = \text{f}(x)\) is given by $$\text{f}(x) = \ln \left( \frac{2x}{1+x} \right), \quad x > 0.$$ The curve \(y = \text{f}(x)\) crosses the \(x\)-axis at P, and the line \(x = 2\) at Q. \includegraphics{figure_9}

1. Verify that the \(x\)-coordinate of P is 1. Find the exact \(y\)-coordinate of Q. [2]
2. Find the gradient of the curve at P. [Hint: use \(\frac{a}{b} = \ln a - \ln b\).] [4]

The function \(\text{g}(x)\) is given by $$\text{g}(x) = \frac{e^x}{2-e^x}, \quad x < \ln 2.$$ The curve \(y = \text{g}(x)\) crosses the \(y\)-axis at the point R.

1. Show that \(\text{g}(x)\) is the inverse function of \(\text{f}(x)\). Write down the gradient of \(y = \text{g}(x)\) at R. [5]
2. Show, using the substitution \(u = 2 - e^x\) or otherwise, that \(\int_0^{\ln \frac{4}{3}} \text{g}(x) dx = \ln \frac{3}{2}\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac{32}{27}\). [Hint: consider its reflection in \(y = x\).] [7]

Fig. 9 shows the line \(y = x\) and the curve \(y = f(x)\), where \(f(x) = \frac{1}{2}(e^x - 1)\). The line and the curve intersect at the origin and at the point P\((a, a)\). \includegraphics{figure_9}

1. Show that \(e^a = 1 + 2a\). [1]
2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac{1}{2}a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\). [6]
3. Show that the inverse function of f\((x)\) is g\((x)\), where g\((x) = \ln(1 + 2x)\). Add a sketch of \(y = g(x)\) to the copy of Fig. 9. [5]
4. Find the derivatives of f\((x)\) and g\((x)\). Hence verify that \(g'(a) = \frac{1}{f'(a)}\). Give a geometrical interpretation of this result. [7]

Given that \(f(x) = 2\ln x\) and \(g(x) = e^x\), find the composite function \(gf(x)\), expressing your answer as simply as possible. [3]

Fig. 9 shows the curve \(y = f(x)\). The endpoints of the curve are P \((-\pi, 1)\) and Q \((\pi, 3)\), and \(f(x) = a + \sin bx\), where \(a\) and \(b\) are constants. \includegraphics{figure_9}

1. Using Fig. 9, show that \(a = 2\) and \(b = \frac{1}{2}\). [3]
2. Find the gradient of the curve \(y = f(x)\) at the point \((0, 2)\). Show that there is no point on the curve at which the gradient is greater than this. [5]
3. Find \(f^{-1}(x)\), and state its domain and range. Write down the gradient of \(y = f^{-1}(x)\) at the point \((2, 0)\). [6]
4. Find the area enclosed by the curve \(y = f(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\). [4]

Fig. 4 shows the curve \(y = f(x)\), where $$f(x) = a + \cos bx, \quad 0 \leq x \leq 2\pi,$$ and \(a\) and \(b\) are positive constants. The curve has stationary points at \((0, 3)\) and \((2\pi, 1)\). \includegraphics{figure_4}

1. Find \(a\) and \(b\). [2]
2. Find \(f^{-1}(x)\), and state its domain and range. [5]

The functions \(f(x)\) and \(g(x)\) are defined by \(f(x) = \ln x\) and \(g(x) = 2 + e^x\), for \(x > 0\). Find the exact value of \(x\), given that \(fg(x) = 2x\). [5]

The functions f(x) and g(x) are defined by $$f(x) = x^2, \quad g(x) = 2x - 1,$$ for all real values of \(x\).

1. State the ranges of f(x) and g(x). Explain why f(x) has no inverse. [3]
2. Find an expression for the inverse function g\(^{-1}\)(x) in terms of \(x\). Sketch the graphs of \(y = g(x)\) and \(y = g^{-1}(x)\) on the same axes. [4]
3. Find expressions for gf(x) and fg(x). [2]
4. Solve the equation gf(x) = fg(x). Sketch the graphs of \(y = gf(x)\) and \(y = fg(x)\) on the same axes to illustrate your answer. [4]
5. Show that the equation f(x + a) = g\(^{-1}\)(x) has no solution if \(a > \frac{1}{4}\). [5]

The function f is defined by $$f : x \to 3e^{x-1}, \quad x \in \mathbb{R}.$$

1. State the range of f. [1]
2. Find an expression for \(f^{-1}(x)\) and state its domain. [4]

The function g is defined by $$g : x \to 5x - 2, \quad x \in \mathbb{R}.$$ Find, in terms of e,

1. the value of gf(ln 2), [3]
2. the solution of the equation $$f^{-1}g(x) = 4.$$ [4]

\(f(x) \equiv \frac{2x-3}{x-2}\), \(x \in \mathbb{R}\), \(x > 2\).

1. Find the range of \(f\). [2]
2. Show that \(f(f(x) = x\) for all \(x > 2\). [3]
3. Hence, write down an expression for \(f^{-1}(x)\). [1]

The function f is defined by $$\text{f}(x) \equiv 2 + \ln (3x - 2), \quad x \in \mathbb{R}, \quad x > \frac{2}{3}.$$

1. Find the exact value of \(\text{f}(1)\). [2]
2. Find an equation for the tangent to the curve \(y = \text{f}(x)\) at the point where \(x = 1\). [4]
3. Find an expression for \(\text{f}^{-1}(x)\). [2]

The function f is defined by $$\text{f}(x) \equiv 3 - x^2, \quad x \in \mathbb{R}, \quad x \geq 0.$$

1. State the range of f. [1]
2. Sketch the graphs of \(y = \text{f}(x)\) and \(y = \text{f}^{-1}(x)\) on the same diagram. [3]
3. Find an expression for f\(^{-1}(x)\) and state its domain. [3]

The function g is defined by $$\text{g}(x) \equiv \frac{8}{3-x}, \quad x \in \mathbb{R}, \quad x \neq 3.$$

1. Evaluate fg\((-3)\). [2]
2. Solve the equation $$\text{f}^{-1}(x) = \text{g}(x).$$ [3]

The functions f(x) and g(x) are defined as follows. $$\text{f}(x) = \ln x, \quad x > 0$$ $$\text{g}(x) = 1 + x^2, \quad x \in \mathbb{R}$$ Write down the functions fg(x) and gf(x), and state whether these functions are odd, even or neither. [4]

Fig. 9 shows the curve \(y = f(x)\), where $$f(x) = (e^x - 2)^2 - 1, x \in \mathbb{R}.$$ The curve crosses the x-axis at O and P, and has a turning point at Q. \includegraphics{figure_9}

1. Find the exact x-coordinate of P. [2]
2. Show that the x-coordinate of Q is \(\ln 2\) and find its y-coordinate. [4]
3. Find the exact area of the region enclosed by the curve and the x-axis. [5]

The domain of f(x) is now restricted to \(x \geqslant \ln 2\).

1. Find the inverse function \(f^{-1}(x)\). Write down its domain and range, and sketch its graph on the copy of Fig. 9. [7]

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]

Given that \(f(x) = 2\ln x\) and \(g(x) = e^x\), find the composite function gf(x), expressing your answer as simply as possible. [3]

Write down the conditions for f(x) to be an odd function and for g(x) to be an even function. Hence prove that, if f(x) is odd and g(x) is even, then the composite function gf(x) is even. [4]

The function f(x) is defined by $$f(x) = 1 + 2\sin 3x, \quad -\frac{\pi}{6} \leqslant x \leqslant \frac{\pi}{6}.$$ You are given that this function has an inverse, \(f^{-1}(x)\). Find \(f^{-1}(x)\) and its domain. [6]

Given that \(f(x) = \frac{1}{2}\ln(x - 1)\) and \(g(x) = 1 + e^{2x}\), show that g(x) is the inverse of f(x). [3]

Fig. 8 shows the line \(y = x\) and parts of the curves \(y = f(x)\) and \(y = g(x)\), where $$f(x) = e^{x-1}, \quad g(x) = 1 + \ln x.$$ The curves intersect the axes at the points A and B, as shown. The curves and the line \(y = x\) meet at the point C. \includegraphics{figure_8}

1. Find the exact coordinates of A and B. Verify that the coordinates of C are \((1, 1)\). [5]
2. Prove algebraically that \(g(x)\) is the inverse of \(f(x)\). [2]
3. Evaluate \(\int_0^1 f(x) \, dx\), giving your answer in terms of \(e\). [3]
4. Use integration by parts to find \(\int \ln x \, dx\). Hence show that \(\int_{e^{-1}}^1 g(x) \, dx = \frac{1}{e}\). [6]
5. Find the area of the region enclosed by the lines OA and OB, and the arcs AC and BC. [2]

The function f is defined by \(\text{f} : x \mapsto \frac{1}{2-2x}\) for \(x \in \mathbb{R}, x \neq 0, x \neq \frac{1}{2}, x \neq 1\). The function g is defined by \(\text{g}(x) = \text{ff}(x)\).

1. Show that \(\text{g}(x) = \frac{1-x}{1-2x}\) and that \(\text{gg}(x) = x\). [4]

It is given that f and g are elements of a group \(K\) under the operation of composition of functions. The element e is the identity, where \(\text{e} : x \mapsto x\) for \(x \in \mathbb{R}, x \neq 0, x \neq \frac{1}{2}, x \neq 1\).

1. State the orders of the elements f and g. [2]
2. The inverse of the element f is denoted by h. Find \(\text{h}(x)\). [2]
3. Construct the operation table for the elements e, f, g, h of the group \(K\). [4]

$$f(x) = \frac{ax + b}{x + 2}; \quad x \in \mathbb{R}, x \neq -2,$$ where \(a\) and \(b\) are constants and \(b > 0\).

1. Find \(f^{-1}(x)\). [2]
2. Hence, or otherwise, find the value of \(a\) so that \(f(x) = x\). [2]

The curve \(C\) has equation \(y = f(x)\) and \(f(x)\) satisfies \(f(x) = x\).

1. On separate axes sketch
   1. \(y = f(x)\), [3]
   2. \(y = f(x - 2) + 2\). [3]

On each sketch you should indicate the equations of any asymptotes and the coordinates, in terms of \(b\), of any intersections with the axes. The normal to \(C\) at the point \(P\) has equation \(y = 4x - 39\). The normal to \(C\) at the point \(Q\) has equation \(y = 4x + k\), where \(k\) is a constant.

1. By considering the images of the normals to \(C\) on the curve with equation \(y = f(x - 2) + 2\), or otherwise, find the value of \(k\). [5]