11.10 a) Find the reduced quadratic trinomial $P(x)$, for which the graph $\mathrm{y}=P(x)$ is symmetric with respect to the Oy axis and is tangent to the line $y=x$.

b) Let $P(x)$ be a quadratic trinomial with a positive leading coefficient. Prove that if the equation $P(x)=x$ has a unique solution, then the equation $P(P(x))=x$ also has a unique solution.
Answer: a) $P(x)=x^{2}+1 / 4$.

Solution. a) The condition of symmetry with respect to the Oy axis means that $P(x)$ has the form $P(x)=$ $x^{2}+a$. The condition of tangency to the line $y=x$ means that the quadratic equation $x^{2}+a=x$ has a unique solution, i.e., its discriminant $1-4 a$ is equal to zero.

b) Since the equation $P(x)=x$ has a unique solution, say $x_{0}$, the graph of $y=P(x)$ is located above the line $y=x$ everywhere, except for the single point $\left(x_{0}, x_{0}\right)$ - the point of intersection. Therefore, for $x \neq x_{0}$, we have $P(P(x)) \geq P(x)>x$, i.e., the equation $P(P(x))=x$ has no other roots except $x_{0}$.