## Hedging for a single step market

I'm trying to understand this method correctly so I wrote my own problem. Can someone check to see that the problem itself makes sense and that I solved it right. (Numbers were chosen pretty much at random but were based kinda on the 'power option' (as is the wording and general layout...))

Consider a single step market with two outcomes and an interest rate of 2%. An option is sold which pays off $(S_T - 1.1K)^2$ if the value of the asset $S$ is above the strike price $K$ at the terminal time $T$. Determine the price of this option if $S_0 = 10$, the market moves up or down by 5% at each time step, and the strike price is 9.2.

---I made this question to better understand this next part---
What is the hedge that should be used over the one step in order to meet the claim? You may assume that the initial value for the riskless asset (i.e. cash/bonds) is 1.

My solution...

The value of the asset if the stock moves up in value is 10*1.05 = 10.5.
The value of the asset if the stock moves down in value is 10*0.95 = 9.5.
(1.1*K = 10.12)

The risk-neutral probability is given by

$p = \frac{(1 + r) - d}{u-d} = \frac{(1.02) - 0.95}{1.05-0.95} = 0.7$

and the price at time 0 with a strike price of 9.2 at time 1 is...

$(1+0.02)^{-1}[0.7 \times (10.5 - 10.12)^2 + 0.3 \times (9.5-10.12)^2 ] = 0.099098$

Now for the hedging part we solve...

$1 \cdot \phi^0 + 10.5 \phi^1 = (10.5-10.12)^2 = 0.1444$
$1 \cdot \phi^0 + 9.5 \phi^1 = 0$

Hence,
$\phi^1 = 0.1444$ and $\phi^0 = -1.3718$.

Now am I right in thinking that $\phi^0$ represents the amount of cash/bonds borrowed (i.e. 1.3718) and $\phi^1$ is the amount of shares bought..?