5. Find all values of the parameter $b$, for which for any value of the parameter $a \in[-1 ; 1]$ the inequality $x^{2}+6 x+2(a+b+1) \sqrt{-x^{2}-6 x-5}+8<a^{2}+b^{2}+2 a$ is not satisfied for at least one value of $x \in[-5 ;-1]$.

(20 points)
Solution: Let's make the substitution $y=\sqrt{-x^{2}-6 x-5}=\sqrt{4-(x+3)^{2}}, \quad y \in[0 ; 2]$. We get $y^{2}-2(a+b+1) y+a^{2}+b^{2}+2 a-3>0$. We need to determine for which values of $a$ and $b$ the inequality $y^{2}-2(a+b+1) y+a^{2}+b^{2}+2 a-3>0$ holds for any $y \in[0 ; 2]$. Consider the function $f(y)=y^{2}-2(a+b+1) y+a^{2}+b^{2}+2 a-3$. Its graph is a parabola with branches directed upwards, with the vertex at the point with abscissa $y_{0}=a+b+1$.

1) $\left\{\begin{array}{c}a+b+1 \leq 0, \\ f(0)>0,\end{array} \quad\left\{\begin{array}{c}a+b \leq-1, \\ (a+1)^{2}+b^{2}>4 .\end{array}\right.\right.$
2) $\left\{\begin{array}{l}00,\end{array}\left\{\begin{array}{l}-10,\end{array} \quad\left\{\begin{array}{c}a+b \geq 1, \\ (a-1)^{2}+(b-2)^{2}>8 .\end{array}\right.\right.$

![](https://cdn.mathpix.com/cropped/2024_05_06_9a8ac98263d65b486910g-3.jpg?height=317&width=263&top_left_y=784&top_left_x=1299)

On the coordinate plane $O a b$, we will plot the set of points $(a ; b)$ that satisfy conditions 1-3. The points that do not satisfy conditions 1-3 are the points for which the inequality $y^{2}-2(a+b+1) y+a^{2}+b^{2}+2 a-3>0$ does not hold for at least one $y \in[0 ; 2]$. The intersection points of the hyperbola $(a+1) b=-2$ and the line $a+b=-1$ are found by solving the equation $(a+1)^{2}=2$. We get the points ( $-1 \pm \sqrt{2} ; \mp \sqrt{2}$ ). The circle $(a+1)^{2}+b^{2}=4$ intersects the line $a+b=-1$ at the same points (this can be verified by substitution). Similar reasoning is applied to the second circle and the line. In the end, the points for which the inequality $y^{2}-2(a+b) y+a^{2}+b^{2}-3>0$ does not hold for at least one $y \in[0 ; 2]$ form a closed region, the boundary of which consists of the graphs of two circles and a hyperbola, with the boundary included. To solve the problem, we need to find such values of $b$ for which the points $(a ; b)$ fall into the resulting region for any $a \in[-1 ; 1]$. These values of $b$ form the interval $\left[b_{1} ; b_{2}\right]$. The lower boundary $b_{1}$ is found by substituting $a=1$ into the hyperbola equation. We have $b_{1}=-1$. The upper boundary $b_{2}$ is found by substituting $a=-1$ into the circle equation $(a-1)^{2}+(b-2)^{2}=8$. We have $b_{2}=4$.

Answer: $[-1 ; 4]$.