Function satisfying some constraints

Question

I am in a hunt for a continuous function $Q : [0,1] \to \mathbb R$ that satisfies the following criteria:

1. $Q(0) = 0$

2. $Q(1) = 1$

3. $Q'(x) \geq 0$ for all $x \in [0,1]$

4. $\int_0^1 P(x)Q(x)= 0.7$, where $P(x)$ is the standard beta distribution $$P(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}$$

I understand that there might be many possibilities for a $Q(x)$ with only these constraints, Since $Q(x)$ seems to resemble a CDF (of the distribution my interest: a beta distribution), I was considering that also the as probable candidate with appropriates scaling. But it seems like the incomplete beta function is giving me a hard time to compute this reasonably well. Can someone suggest an elegant $Q(x)$ that satisfies all conditions in the range $0\leq x\leq1$. Please correct me if any of the conditions I described above makes it impossible to define a $Q(x)$.

Answer 1

It's arguably not elegant, because it is only differentiable on $(0,1)$, but this works:

1. $Q(0) = 0$
2. $Q(1) = 1$
3. $Q(x) = 0.7$ for all $x \in (0,1)$

Answer 2

$$Q(x)=x, \alpha=7, \beta=3$$ gives: $$\frac{\int_{0}^{1} x~ x^7 (1-x)^3 dx}{\int_{0}^{1} x^7 (1-x)^3 dx}=\frac{B(8,3)}{B(7,3)}=\frac{\Gamma(8)\Gamma(3) \Gamma(10)}{\Gamma(7) \Gamma(3)\Gamma(11)}=\frac{7!~2!~9!}{6!~3!~10!}=\frac{7}{10}$$