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1

No, risk neutrality is not assumed. Risk preferences are given by the concavity of $u_i$ which is arbitrary in this setup.


2

Hint: This suggests $y=\dfrac{U}{\sqrt{x}}$ so draw the three curves in the usual way $y=\dfrac{10}{\sqrt{x}}$; three of the points are $(1,10)$ and $(4,5)$ and $(9,3.333)$ $y=\dfrac{15}{\sqrt{x}}$; three of the points are $(1,15)$ and $(4,7.5)$ and $(9,5)$ $y=\dfrac{20}{\sqrt{x}}$; three of the points are $(1,20)$ and $(4,10)$ and $(9,6.667)$


0

My math is a bit rusty, but I believe you can show that, for a well-behaved (e.g., open and connected) feasible set $T$ and a consideration set $T'$, $x \in S \subseteq T' \subseteq T$, e.g. Sen's property $\alpha$ holds. That doesn't seem like a very interesting question/result though, as it just follows directly from the definition of "well-behaved". ...


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