= Solution
{c}
<Complete-linkage clustering> defines the dissimilarity between <clusters in cluster analysis> $A,B$ as $d_{\rm CL}(A,B)=\max_{i\in A,j\in B}d_{ij}$. Remove point 6 and first merge 1 with 4 at height $0.67$. The distances from $\{1,4\}$ to 2, 3, 5 are respectively
$$
\max(1.41,1.19)=1.41,\quad
\max(2.68,2.19)=2.68,\quad
\max(3.01,3.32)=3.32.
$$
The next merge therefore attaches 2 at $1.41$. Now the distances from $\{1,2,4\}$ to 3 and 5 are $2.95$ and $3.32$, whereas $d_{35}=2.82$. Merge 3 and 5 at $2.82$. The final distance between the two remaining <clusters in cluster analysis> is the largest cross-entry, $3.32$. Thus the <dendrogram> has merges
$$
\boxed{(1,4):0.67,\quad(\{1,4\},2):1.41,\quad(3,5):2.82,\quad
(\{1,2,4\},\{3,5\}):3.32.}
$$
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/undated/prelims/mathematics-iii-resit-complete-linkage.png]
{title=Complete-linkage dendrogram after removing observation 6}
Cutting strictly between heights $2.82$ and $3.32$ gives the two <clusters in cluster analysis>
$$
\boxed{\{1,2,4\}\quad\text{and}\quad\{3,5\}.}
$$
The figure is an original drawing of these computed merges.
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