For any real number x , let [x] denotes the largest integer less...

Given, f(x)=\left\{$\begin{array}{cc}x-[x] & \text{ if }[x] \text{ is odd } \\ 1+[x]-x & \text{ if }[x] \text{is even }\end{array}$\right.f(x)$ and $\cos \pi x$ both are periodic with period 2 and both are even.\begin{aligned} \therefore \quad& \int_{-10}^{10} f(x) \cos \pix d x \\ & =2 \int_0^{10} f(x) \cos \pix d x \end{aligned}$$<img src="/api/proxy-image?url=https$$\begin{aligned}& =10 \int_0^2 f(x) \cos \pix d x \\ & \text{ Now, } \quad \int_0^1 f(x) \cos \pix d x \\ & =\int_0^1(1-x) \cos \pix d x=-\int_0^1 u \cos \piu d u \\ & \text{\text{ and } } \int_1^2 f(x) \cos \pix d x \\ & =\int_1^2(x-1) \cos \pix d x=-\int_0^1 u \cos \piu d u \\ & \therefore \quad \int_{-10}^{10} f(x) \cos \pix d x \\ & =-20 \int_0^1 u \cos \piu d u=\frac{40}{\pi^2}\\ & \Rightarrow \quad \frac{\pi^2}{10} \int_{-10}^{10}f(x) \cos \pix d x=4 \end{aligned}$$