Question:

Let \[ f(x) = \begin{cases} 1 + \dfrac{2x}{\pi}, & -\pi \le x \le 0
1 - \dfrac{2x}{\pi}, & 0<x \le \pi \end{cases} \] The constant term of the Fourier series of \(f(x)\) is:

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To find the constant term in a Fourier series, always compute the average value over the interval using \( \dfrac{1}{\pi} \int_{-\pi}^{\pi} f(x)\, dx \).
Updated On: May 26, 2025
  • \(0\)
  • \(\dfrac{2}{\pi}\)
  • \(-\dfrac{2}{\pi}\)
  • \(\dfrac{\pi}{2}\)
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The Correct Option is A

Solution and Explanation

The constant term in the Fourier series is given by: \[ a_0 = \dfrac{1}{\pi} \int_{-\pi}^{\pi} f(x)\, dx \] Split the integral: \[ a_0 = \dfrac{1}{\pi} \left( \int_{-\pi}^{0} \left(1 + \dfrac{2x}{\pi} \right) dx + \int_{0}^{\pi} \left(1 - \dfrac{2x}{\pi} \right) dx \right) \] Compute each part: - First integral: \[ \int_{-\pi}^{0} \left(1 + \dfrac{2x}{\pi} \right) dx = \left[x + \dfrac{x^2}{\pi}\right]_{-\pi}^{0} = (0 + 0) - (-\pi + \pi) = 0 \] - Second integral: \[ \int_{0}^{\pi} \left(1 - \dfrac{2x}{\pi} \right) dx = \left[x - \dfrac{x^2}{\pi}\right]_{0}^{\pi} = (\pi - \pi) - (0 - 0) = 0 \] So, \[ a_0 = \dfrac{1}{\pi}(0 + 0) = 0 \]
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