2015 17 From previous KVPY papers!

Let $a_0 = 1$ , $a_1 = A$ , $a_2 = B$ , $a_3 = C$ , $a_4 = D$ and $a_5 = E$ , then:

$\begin{aligned} \Rightarrow f(x) & = \sum_{m=0}^5 a_{5-m} \cos{mx} \\ \Rightarrow T & = \sum_{k=0}^9 (-1)^k f\left( \frac{k\pi}{5} \right) \\ & = \sum_{m=0}^5 \color{#3D99F6}{\sum_{k=0}^9 (-1)^k} a_{5-m} \color{#3D99F6} {\cos{\left(\frac{km\pi}{5}\right)} } \end{aligned}$

Consider $\displaystyle \color{#3D99F6}{\sum_{k=0}^9 (-1)^k \cos{\left(\frac{km\pi}{5}\right)} }$

For $m = 0$ :

$\begin{aligned} \sum_{k=0}^9 (-1)^k \cos{\left(\frac{k(0)\pi}{5}\right)} & = \sum_{k=0}^9 (-1)^k \cos{0} = \sum_{k=0}^9 (-1)^k = 0 \end{aligned}$

For $1\le m \le 4$ , we note that $\cos{\left(\frac{km\pi}{5}\right)} = - \cos{\left(\frac{(5+km)\pi}{5}\right)}$ .

$\begin{aligned} \small \sum_{k=0}^9 (-1)^k \cos{\left(\frac{km\pi}{5}\right)} & \small = \sum_{k=0}^4 (-1)^k \cos{\left(\frac{km\pi}{5}\right)} + \sum_{k=5}^9 (-1)^k \cos{\left(\frac{km\pi}{5}\right)} \\ & \small = \sum_{k=0}^4 (-1)^k \cos{\left(\frac{km\pi}{5}\right)} + \sum_{k=0}^4 (-1)^k \cos{\left(\frac{(5+km)\pi}{5}\right)} \\ & \small = \sum_{k=0}^4 (-1)^k \cos{\left(\frac{km\pi}{5}\right)} - \sum_{k=0}^4 (-1)^k \cos{\left(\frac{km\pi}{5}\right)} \\ & \small = 0 \end{aligned}$

For $m=5$ :

$\begin{aligned} \sum_{k=0}^9 (-1)^k \cos{\left(\frac{5k\pi}{5}\right)} & = \sum_{k=0}^9 (-1)^k \cos{k\pi} = \sum_{k=0}^9 (-1)^k (-1)^k = \sum_{k=0}^9 1 = 10 \end{aligned}$

Therefore, $T = 10$ , which is $\boxed{\text{independent of A, B, C, D and E}}$ .