nice problem. a) Consider $P(x)=x^4-x^3+x-a$ where $a \in \mathbb{R}^{+}$. Note that $P(0)<0$ and $\lim_{x \to \infty} P(x)=\infty$ which implies that by IVT $P(x)$ has a positive root. Suppose $P(x)$ has more than one positive root say $\alpha_1 \leq \alpha_2$. Then by Mean Value Theorem $P'(x)$ will also have one root $\beta$ such that $0<\alpha_1 \leq \beta \leq \alpha_2$. Consider $P'(x)=4x^3-3x^2+1$. Note that $P''(x)=0 \implies x=0,1/2$ and $P'''(1/2)>0$ which means that $P'(x)$ attains a local minima at $x=1/2$. But $P'(1/2)=3/4>0$. It can be shown that $P'(x)$ is decreasing between $0$ and $1/2$ and increasing for $x \geq 1/2$. Since $P'(1/2)>0$ it follows that $P'(x)$ lies above the x-axis when $x$ is positive. Hence $P'(x)$ always lies above the x-axis for $x \in \mathbb{R}^{+}$ and hence cannot have a positive root. b) We claim that the sequence $\{a_n\}$ is bounded above by $1$ and increasing. We prove this by induction on $n$. Note that it suffices to show that $Q_n(1)>0$ where $Q_n(x):= P(x)-a_n$. Indeed, we have that $Q_n(1)=1-a_n$ which is positive by the induction hypothesis. Since $Q_n(0)<0$ for all $n$, it follows by an application of IVT that $a_{n+1}<1$. We now show that the sequence is increasing. Enough to show that $Q_n(a_n)<0$ since our claim will then follow by IVT. Note that $Q_n(a_n)$ is $a_n^3(a_n-1)$ which is negative and hence we are done. Now, we have that the sequence is increasing and bounded above and hence it must have a limit. We call the limit $\ell{}$. Note that we have $\ell{}^4-\ell{}^3+\ell{}-\ell{}=0$ which implies that $\ell{}=1$.

Easy one to begin the contest.