Let $f(x)=\left\{\begin{array}{cc}\frac{1}{3}, & x \leq \pi / 2 \\ \frac{\mathrm{~b}(1-\sin x)}{(\pi-2 x)^2}, & x>\pi / 2\end{array}\right.$. If $f$ is continuous at $x=\pi / 2$, then the value of $\int\limits_0^{3 \mathrm{~b}-6}\left|x^2+2 x-3\right| \mathrm{d} x$ is :

Let $\frac{x^2}{f\left(a^2+7 a+3\right)}+\frac{y^2}{f(3 a+15)}=1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbf{R}$. If the set of all possible values of $a$ is $\mathbf{R}-[\alpha, \beta]$, then $\alpha^2+\beta^2$ is equal to :

The sum of squares of all the real solutions of the equation

$\log _{(x+1)}\left(2 x^2+5 x+3\right)=4-\log _{(2 x+3)}\left(x^2+2 x+1\right)$ is equal to $\_\_\_\_$ .

If $\int\limits_{\pi / 6}^{\pi / 4}\left(\cot \left(x-\frac{\pi}{3}\right) \cot \left(x+\frac{\pi}{3}\right)+1\right) d x=\alpha \log _{\mathrm{e}}(\sqrt{3}-1)$, then $9 \alpha^2$ is equal to $\_\_\_\_$ .