Question

The constant term of the Fourier coefficients of the periodic function
\[ f(x)= \begin{cases} -k, & -\pi<x<0 \\ k, & 0<x<\pi \end{cases} \quad\text{and}\quad f(x+2\pi)=f(x),k=\text{constant} \] is

• A. $k$
• B. $2k$
• C. $2\pi$
• D. $0$

Hint:

A function symmetric about the origin with equal positive and negative areas has zero average value, giving a zero constant Fourier coefficient.

Answer 1

The Correct Option is D

The constant Fourier coefficient $a_0$ for a $2\pi$–periodic function is given by:
\[ a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\,dx \]
Split the integral across the two intervals where $f(x)$ is defined:
\[ \int_{-\pi}^{\pi} f(x)\,dx = \int_{-\pi}^{0} (-k)\,dx + \int_{0}^{\pi} k\,dx \]
Evaluate the first integral:
\[ \int_{-\pi}^{0} (-k)\,dx = -k(0 - (-\pi)) = -k(\pi) \]
Evaluate the second integral:
\[ \int_{0}^{\pi} k\,dx = k(\pi - 0) = k\pi \]
Add both contributions:
\[ -k\pi + k\pi = 0 \]
Thus the total integral is zero, giving:
\[ a_0 = \frac{1}{2\pi}(0) = 0 \]
Therefore the constant term of the Fourier series is zero.
Final Answer: 0