Question: I am looking for a method to prove that the following function is quasi-concave in $\alpha$ (or find conditions under which it is true):

$ \pi=F(-k)(f(0)^2-f(h(1-\alpha))^2)+ \frac{1}{2}-\frac{1}{2}f(0)^2-\frac{1}{2}f(-k^2)(2F(h(1-\alpha))-1)^2 $,

where:

$F$ - is a CDF function of Normal Distribution, $N[0,\sigma^2]$;

$f$ - is a PDF function of Normal Distribution, $N[0,\sigma^2]$;

$k$ - exogenous patameter, k $\in [0,+∞)$;

$h$ - exogenous patameter, h $\in [0,+∞)$.

Any help would be appreciated.

**UPD**

**I have forgotten to mention that I am trying to prove quasi-concavity on the interval $\alpha \in [0,1]$**

Answer: 
Your conjecture seems to be contradicted, at least for small values of $\sigma$. 
You can draw the function with the following R-code: 

```
qq_f = function(x,k,h,sig){   
-pnorm(-k, sd=sig)*( (dnorm(h*(1-x), sd=sig))^2 ) - 0.5*dnorm(-k^2, sd=sig)*( 2*pnorm(h*(1-x), sd=sig) -1 )^2  
}  
curve(qq_f(x,k=0,h=1,sig=0.5),col='blue',xlim=c(-1,3),type='l',main="A candidate quasi-concave function")

```

[![enter image description here](https://i.stack.imgur.com/Bxa7A.png)](https://i.stack.imgur.com/Bxa7A.png)
After a while, I have not found conditions or proof the function is quasi-concave at $\alpha \in [0,1]$.

However, the ultimate goal was to show that there is an intermediate solution to the equation. It was indeed possible to condition on the second derivatives w.r.t. to $\alpha$ and $h$ under which: 
- the function equals to zero and increases at $0$ 
- the function equals to 0 and decreases at $1$. 

Those conditions grant that there is at least one inflection point somewhere in $[0,1]$

I can provide more details if somebody is interested.