Answer: \(\frac12\ln5-\frac1{10}\).

(a)(i) At the \(y\)-axis, \(x=0\). For the curve,

\(y=\frac1{2(0)+1}=1.\)

So

\(A=(0,1).\)

At the \(x\)-axis, \(y=0\). For the line \(5y=x-1\),

\(0=x-1,\)

so \(x=1\). Hence

\(B=(1,0).\)

(a)(ii) When \(x=2\), the curve gives

\(y=\frac1{2(2)+1}=\frac15.\)

The line gives

\(5y=2-1=1,\)

so

\(y=\frac15.\)

Both equations give the same \(y\)-value, so the \(x\)-coordinate of \(C\) is \(2\).

(b) The shaded area is the area under the curve from \(x=0\) to \(x=2\), minus the triangle under the line from \(x=1\) to \(x=2\).

The area under the curve is

The triangle has base \(1\) and height \(\frac15\), so its area is

\(\frac12\cdot1\cdot\frac15=\frac1{10}.\)

Therefore the shaded area is

\(\frac12\ln5-\frac1{10}.\)