Answer: (a) \(u_n<5\) for all positive integers \(n\).

(b) \(u_{n+1}>u_n\) for \(n\ge1\).

(a)

Base case: \(u_1<5\), which is given.

Assume that \(u_k<5\) for some positive integer \(k\). Then

\(5-u_{k+1}=5-\frac{6u_k+5}{u_k+2}=\frac{5(u_k+2)-(6u_k+5)}{u_k+2}.\)

So

\(5-u_{k+1}=\frac{5-u_k}{u_k+2}.\)

Since \(u_k<5\), the numerator is positive. Since the sequence is positive, \(u_k+2>0\). Hence \(5-u_{k+1}>0\), so \(u_{k+1}<5\).

Therefore, by induction, \(u_n<5\) for all positive integers \(n\).

(b)

Consider the difference:

\(u_{n+1}-u_n=\frac{6u_n+5}{u_n+2}-u_n.\)

Then

\(u_{n+1}-u_n=\frac{6u_n+5-u_n(u_n+2)}{u_n+2}=\frac{-u_n^2+4u_n+5}{u_n+2}.\)

Factorise the numerator:

\(-u_n^2+4u_n+5=(5-u_n)(u_n+1).\)

So

\(u_{n+1}-u_n=\frac{(5-u_n)(u_n+1)}{u_n+2}.\)

From part (a), \(5-u_n>0\), and since \(u_n>0\), the denominator and \(u_n+1\) are positive. Hence \(u_{n+1}-u_n>0\), so \(u_{n+1}>u_n\).

Mark scheme