OCR MEI C3 (Core Mathematics 3)

1. The function f(x) is defined by $$f(x) = \frac{1-x}{1+x}, x \neq -1.$$ Show that f(f(x)) = x. Hence write down \(f^{-1}(x)\). [3]
2. The function g(x) is defined for all real x by $$g(x) = \frac{1-x^2}{1+x^2}.$$ Prove that g(x) is even. Interpret this result in terms of the graph of \(y = g(x)\). [3]

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]

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