Answer: (a)(i) \(3\leq x\lt 5\); (a)(ii) \(x=4\); (b) \(\displaystyle f^{-1}g(x)=\ln\left(\frac{2-2x}{2-5x}\right)\).

Answer: (a)(i) \(3\leq x\lt 5\); (a)(ii) \(x=4\); (b) \(\displaystyle f^{-1}g(x)=\ln\left(\frac{2-2x}{2-5x}\right)\).

(a)(i) Since \(x\geq0\), we have \(0\lt \mathrm e^{-x}\leq1\).

Therefore

\(3\leq 5-2\mathrm e^{-x}\lt 5.\)

The range of \(f\) is \(3\leq y\lt 5\), so the domain of \(f^{-1}\) is

\(3\leq x\lt 5.\)

(a)(ii) If \(f^{-1}(x)=\sqrt{5x-4}\), apply \(f\) to both sides:

\(x=f(\sqrt{5x-4}).\)

It is quicker to note that \(f^{-1}(x)=u\) means \(x=f(u)\). The equation given also says \(u=\sqrt{5x-4}\). Thus the mark scheme equation reduces to

\(x=\sqrt{5x-4}.\)

Squaring gives

\(x^2=5x-4,\)

so

\(x^2-5x+4=0.\)

Factorising,

\((x-1)(x-4)=0.\)

The domain of \(f^{-1}\) is \(3\leq x\lt 5\), so reject \(x=1\). Hence

\(x=4.\)

(a)(iii) The graph of \(y=f(x)\) starts at \((0,3)\), increases, and has horizontal asymptote \(y=5\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). It starts at \((3,0)\), increases, and has vertical asymptote \(x=5\).

(b) To find \(f^{-1}\), let

\(y=5-2\mathrm e^{-x}.\)

Then

\(2\mathrm e^{-x}=5-y,\)

so

\(\mathrm e^{-x}=\frac{5-y}{2}.\)

Taking logarithms,

\(-x=\ln\left(\frac{5-y}{2}\right),\)

and hence

\(f^{-1}(y)=\ln\left(\frac{2}{5-y}\right).\)

Now substitute \(g(x)=\frac{3}{1-x}\):

\(f^{-1}g(x)=\ln\left(\frac{2}{5-\frac{3}{1-x}}\right).\)

Simplify the denominator:

Therefore

\(f^{-1}g(x)=\ln\left(\frac{2}{(2-5x)/(1-x)}\right) =\ln\left(\frac{2-2x}{2-5x}\right).\)