11. Find all values of $b$ for which the equation

$$
a^{2-2 x^{2}}+(b+4) a^{1-x^{2}}+3 b+4=0
$$

has no solutions for any $a>1$.
Solution 1. Rewrite the equation as $-\frac{(t+2)^{2}}{t+3}=b$ and consider the function $f(t)=-\frac{(t+2)^{2}}{t+3}$. To find the set of its values, we will investigate this function for monotonicity. Since

$$
f^{\prime}(t)=-\frac{2(t+2)(t+3)-(t+2)^{2}}{(t+3)^{2}}=-\frac{(t+2)(t+4)}{(t+3)^{2}}<0 \text { for } t \geqslant 0
$$

the function is decreasing on the interval $[0 ;+\infty)$. Since $f(0)=-\frac{4}{3}$ and $f(t) \rightarrow-\infty$ as $t \rightarrow+\infty$, the set of its values on the considered interval is the interval $\left(-\infty ;-\frac{4}{3}\right)$. Therefore, the given equation has no positive solutions for $b \geqslant-\frac{4}{3}$.

Answer: $\left[-\frac{4}{3} ;+\infty\right)$.

Topic: "Application of Derivatives for Function Investigation"