= Solution
First address the positivity request in part (a), which has no separate Solution heading in the stub. For $x\in[0,1]$, each $x^k(1-x)^{n-k}$ and each coefficient $c_k$ is nonnegative. Their sum is therefore nonnegative, including at the endpoints:
$$
\boxed{f(x)\ge0\quad(0\le x\le1).}
$$
This is the elementary direction of the <positive Bernstein coefficients for a strictly positive polynomial> criterion; the converse developed later needs strict positivity.
For completeness, the convergence fact listed as (i) also has a short proof. If $K\sim\operatorname{Bin}(n,x)$, the <Bernstein polynomial> is $B_n(f)(x)=\mathbb E[f(K/n)]$. Its mean argument is $x$ and $\operatorname{Var}(K/n)=x(1-x)/n\le1/(4n)$. Given $\epsilon>0$, <uniform continuity> of $f$ on the compact interval supplies $\delta>0$ such that $|f(s)-f(t)|<\epsilon$ whenever $|s-t|\le\delta$. By <Chebyshev's inequality>, uniformly in $x$,
$$
|B_n(f)(x)-f(x)|\le\epsilon+2\|f\|_\infty\Pr(|K/n-x|>\delta)
\le\epsilon+\frac{\|f\|_\infty}{2n\delta^2}.
$$
Letting $n\to\infty$ and then $\epsilon\downarrow0$ proves $\boxed{\|B_n(f)-f\|_\infty\to0}$. The two numbered items in the source are assumptions supplied for later parts, not additional exam questions; their stub sections contain these supporting derivations.
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