Christmas Streak 56/88: 2018 CSAT (Korean SAT) #30 of Natural Sciences

$\cos(k\pi)=-1$ is trivial. Then let's split the cases into three.

$k+7< t\leq k+8$ and $k+8< t\leq k+9$ are omitted because they're symmetric to $k< t\leq k+1$ and $k-1 \leq t\leq k.$

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$\text{i)}~k-1\leq t\leq k$

$\begin{aligned} g(t) &=\int_{k}^{t+1} (-x+t+1)\cos(\pi x)dx \\ & =\frac{1}{\pi}\left[(-x+t+1)\sin(\pi x)\right]_{k}^{t+1}-\frac{1}{\pi^2}\left[\cos(\pi x)\right]_{k}^{t+1} \\ & =\frac{1}{\pi^2}\cos(t\pi)-\frac{1}{\pi^2}\end{aligned}$

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$\text{ii)}~k<t<k+1$

$\begin{aligned} g(t) &=\int_{k}^{t} (x-t+1)\cos(\pi x)dx +\int_{t}^{t+1} (-x+t+1)\cos(\pi x)dx \\ & =\frac{1}{\pi} \left[(x-t+1)\sin(\pi x)\right]_{k}^{t} +\frac{1}{\pi^2}\left[\cos(\pi x)\right]_{k}^{t} +\frac{1}{\pi}\left[(-x+t+1)\right]_{t}^{t+1}-\frac{1}{\pi^2}\left[\sin(\pi x)\right]_{t}^{t+1} \\ & =\frac{1}{\pi}\sin(t\pi) +\frac{1}{\pi^2}\left(\cos(t\pi)+1\right)-\frac{1}{\pi}\sin(t\pi) + \frac{2}{\pi^2}\cos(t\pi) \\ & =\frac{1}{\pi^2}\left(3\cos(t\pi) + 1\right)\end{aligned}$

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$\text{iii)}~k+1<t<k+7$

$\begin{aligned} g(t) &=\int_{t-1}^{t} (x-t+1)\cos(\pi x)dx + \int_{t}^{t+1} (-x+t+1)\cos(\pi x)dx \\ &=\frac{1}{\pi}\left[(x-t+1)\sin(\pi x)\right]_{t-1}^{t} +\frac{1}{\pi^2}\left[(-x+t+1)\right]_{t}^{t+1} +\frac{1}{\pi}\left[(x-t+1)\sin(\pi x)\right]_{t}^{t+1}-\frac{1}{\pi^2}\left[(-x+t+1)\right]_{t}^{t+1} \\ & =\frac{1}{\pi}\sin(t\pi)+\frac{2}{\pi^2}\cos(t\pi)-\frac{1}{\pi}\sin(t\pi)+\frac{2}{\pi^2}\cos(t\pi) \\ & =\frac{4}{\pi^2}\cos(t\pi)\end{aligned}$

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From above, we can see that $g(t)$ has its local minima at $t=k,~k+2,~k+4,~k+6.$ Also, $t=k+8$ yields a local minimum as well because of symmetry.

$\displaystyle \sum_{i=1}^{m} \alpha_m = k+(k+2)+(k+4)+(k+6)+(k+8)=5k+20=45~\Leftrightarrow~k=5.$

And since the local minima are $\displaystyle -\frac{2}{\pi^2},~-\frac{4}{\pi^2},~-\frac{4}{\pi^2},~-\frac{4}{\pi^2},~-\frac{2}{\pi^2},$ respectively,

we get $\displaystyle k-\sum_{i=1}^{m}g(\alpha_i)=5-\pi^2\cdot\left(-\frac{16}{\pi^2}\right)=\boxed{21}.$