Marshalian and Hickisian Demands and Slutsky Equation everyone.

I have the following question:

A consumer has the following indirect utility function:

$ V(p_1,p_2,b) = (p_2k-b)p_1^{-1} \left[ \frac{2p_2k - 2b}{p_2} \right]^{-2}, x_2 &lt; k$

a) Find the Marshalian demand for good 2.

b) Find the Hicksian demand for good 2.

c) Show that the Slutsky equation holds for good 2.

d) Why is it necessary to have $x_2 &lt; k$

For a) I found:

$ \boxed{\boxed{ x_2(p_1,p_2,b) = \frac{2b-p_2k}{p_2} }} $

For b):

$ \boxed{\boxed{ x_2^c(p_1,p_2,\bar{U}) = k - \frac{P_2}{2P_1 \cdot \bar{U} }}} $

For c) after replacing $E(p_1,p_2, \bar{U})$ with $p_1 \cdot x_1(p_1,p_2,b) + p_2 \cdot x_2(p_1,p_2,b)$ I found $\bar{U} = \frac{p_2^2}{(p_2k-b) 4p_1}$ and using that in

$\frac{\partial x_2(p_1,p_2,b)}{\partial p_2} = \frac{\partial x_2^c(p_1,p_2,\bar{U})}{\partial p_2} - \frac{\partial x_2(p_1,p_2,b)}{\partial b} \cdot x_2(p_1,p_2,b)$ 

I was able to show that both sides are equal.

For d) here's what I've tried:

using Roy's identity we can find

$x_1(p_1,p_2,b) = \frac{p_2k - b}{p_1}$

assuming the demand for good 1 is positive, we must then have:

$ \frac{p_2k - b}{p_1} &gt; 0  $

Since $p_1 &gt; 0$, we can re-write the inequality as

$ p_2k - b &gt; 0 $

but

$b = p_1x_1 + p_2x_2$, 

so

$ p_2k - p_1x_1 - p_2x_2 &gt; 0 \therefore p_2 \cdot (k - x_2) &gt; p_1x_1 \therefore k-x_2 &gt; \frac{p_1x_1}{p_2}$, because $p_2$ is positive.

Now, since $p_1 &gt; 0$ and assuming $x_1 &gt; 0$, we have $\frac{p_1x_1}{p_2} &gt; 0$, which in turn implies that

$k-x_2 &gt; 0 \therefore \boxed{\boxed{x_2 &lt; k, Q.E.D.}} $

Are my answers correct? In c) is there a faster way to find $\bar{U}$ so I can replace it in the Hickisian demand and show that the Slutsky equation holds for good 2?

I appreciate any inputs.

Have a good one,
Pedro.
Although my teacher has yet to verify my solution for d) (which I believe is incorrect), he shared his answer and the condition $x_2&lt;k$ derives from the positivity of $V(p_1,p_2,b)$ and from the budget inequality $p_1x_1+p_2x_2≤b$. The other answers are correct.