2019 May TZ1 Paper 2 - Q4

(a)

For 0 \le x < \dfrac \pi 2, 0 < \cos x \le 1, so \sec x \ge 1. So f(x) = \sec x + 2 \ge 3. So the range of f is [3, \infty).

(b)

We have:

x = \sec (f^{-1}(x)) + 2

So:

x - 2 = \sec(f^{-1}(x))

giving:

\displaystyle \frac 1 {x - 2} = \cos(f^{-1}(x))

Since 0 \le f^{-1}(x) < \dfrac \pi 2 (the domain of f), we have:

\displaystyle \arccos \left(\frac 1 {x - 2}\right) = f^{-1}(x)

The domain of f^{-1} is the range of f, so x \ge 3, or [3, \infty).

[such a function exists because f is injective]