GATE 2022 IN Q.13: solution and answer

Official answer: BSource: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. The period is $2L=2$, so $L=1$. The Fourier series is therefore built from the constant $a_0$, $\cos (n\pi x)$ and $\sin (n\pi x)$.

2. Check parity. Take $0<x\le 1$, so $-x\in [-1,0)$. Then $f(-x)=-1-(-x)=-(1-x)=-f(x)$. So $f$ is odd on $[-1,1]$. The value at the single point $x=0$ does not affect any coefficient.

3. For an odd $f$, the product $f(x)\cos (n\pi x)$ is odd, so $a_0=\frac{1}{2}{\int }_{-1}^{1}f(x)dx=0$ and $a_n={\int }_{-1}^{1}f(x)\cos (n\pi x)dx=0$. There is no constant term and no cosine term.

4. Since $f(x)\sin (n\pi x)$ is even, the sine coefficients are $$b_n={\int }_{-1}^{1}f(x)\sin (n\pi x)dx=2{\int }_0^1(1-x)\sin (n\pi x)dx$$

5. Integrate by parts with $u=1-x$ and $dv=\sin (n\pi x)dx$: $${\int }_0^1(1-x)\sin (n\pi x)dx={\left[-\frac{(1-x)\cos (n\pi x)}{n\pi }\right]}_0^1-{\int }_0^1\frac{\cos (n\pi x)}{n\pi }dx=\frac{1}{n\pi }-\frac{\sin (n\pi )}{(n\pi {)}^2}=\frac{1}{n\pi }$$

6. So $b_n=\frac{2}{n\pi }\ne 0$ for every $n=1,2,3,\dots$, and $$f(x)=\frac{2}{\pi }\sum _{n=1}^{\infty }\frac{\sin (n\pi x)}{n}$$

7. The series contains only $\sin (n\pi x)$ terms, and it contains one for every $n$. The answer is (B).