If $$2 \tan ^2 \theta-5 \sec \theta=1$$ has exactly 7 solutions in the interval $$\left[0, \frac{n \pi}{2}\right]$$, for the least value of $$n \in \mathbf{N}$$, then $$\sum_\limits{k=1}^n \frac{k}{2^k}$$ is equal to:

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 :