Let $$g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$$ and $$f^{\prime \prime}(x)>0$$ for all $$x \in(0,3)$$. If $$g$$ is decreasing in $$(0, \alpha)$$ and increasing in $$(\alpha, 3)$$, then $$8 \alpha$$ is :

Let $$\mathrm{R}$$ be the interior region between the lines $$3 x-y+1=0$$ and $$x+2 y-5=0$$ containing the origin. The set of all values of $$a$$, for which the points $$\left(a^2, a+1\right)$$ lie in $$R$$, is :

Let $$\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$$ and $$\beta=\frac{(5 !) !}{(5 !)^{4 !}}$$. Then :

$$\text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : }$$