PYQs / GATE CE / 2025 / Set 1 / Q13 GATE CE 2025 Set 1 — Question 13 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Medium Fourier Series Partial Differential Equations Engineering Mathematics
Engineering Mathematics → Partial Differential Equations → Fourier Series
Last updated 5 September 2026
Question Which one of the following options is the correct Fourier series of the periodic function
f ( x ) f(x) f ( x ) described below:
f ( x ) = { 0 if − 2 < x < − 1 2 k if − 1 < x < 1 0 if 1 < x < 2 ; period = 4 f(x) = \begin{cases} 0 & \text{if } - 2 < x < -1 \\ 2k & \text{if } - 1 < x < 1 \\ 0 & \text{if } 1 < x < 2 \end{cases} ; \text{period} = 4 f ( x ) = ⎩ ⎨ ⎧ 0 2 k 0 if − 2 < x < − 1 if − 1 < x < 1 if 1 < x < 2 ; period = 4 Correct answer (C) f(x) = k + (4k)/(π) (cos (π)/(2)x - (1)/(3) cos (3π)/(2)x + (1)/(5) cos (5π)/(2)x - + …)
Solution The function
f ( x ) f(x) f ( x ) is an even function over the interval
( − 2 , 2 ) (-2, 2) ( − 2 , 2 ) , so its Fourier series will only contain cosine terms (
b n = 0 b_n = 0 b n = 0 ). The period is
T = 4 T = 4 T = 4 , so
L = T / 2 = 2 L = T/2 = 2 L = T /2 = 2 .
1. Constant term a 0 / 2 a_0/2 a 0 /2 a 0 = 1 L ∫ − L L f ( x ) d x = 1 2 ∫ − 1 1 2 k d x = 1 2 [ 2 k x ] − 1 1 = 1 2 ( 2 k − ( − 2 k ) ) = 2 k a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx = \frac{1}{2} \int_{-1}^{1} 2k dx = \frac{1}{2} [2kx]_{-1}^{1} = \frac{1}{2} (2k - (-2k)) = 2k a 0 = L 1 ∫ − L L f ( x ) d x = 2 1 ∫ − 1 1 2 k d x = 2 1 [ 2 k x ] − 1 1 = 2 1 ( 2 k − ( − 2 k )) = 2 k So, the constant term is
a 0 / 2 = k a_0/2 = k a 0 /2 = k .
2. Cosine coefficients a n a_n a n a n = 1 L ∫ − L L f ( x ) cos ( n π x L ) d x = 1 2 ∫ − 1 1 2 k cos ( n π x 2 ) d x = k [ 2 n π sin ( n π x 2 ) ] − 1 1 a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx = \frac{1}{2} \int_{-1}^{1} 2k \cos\left(\frac{n\pi x}{2}\right) dx = k \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_{-1}^{1} a n = L 1 ∫ − L L f ( x ) cos ( L nπ x ) d x = 2 1 ∫ − 1 1 2 k cos ( 2 nπ x ) d x = k [ nπ 2 sin ( 2 nπ x ) ] − 1 1 a n = 2 k n π ( sin n π 2 − sin ( − n π 2 ) ) = 4 k n π sin n π 2 a_n = \frac{2k}{n\pi} \left( \sin\frac{n\pi}{2} - \sin\left(-\frac{n\pi}{2}\right) \right) = \frac{4k}{n\pi} \sin\frac{n\pi}{2} a n = nπ 2 k ( sin 2 nπ − sin ( − 2 nπ ) ) = nπ 4 k sin 2 nπ For
n = 1 , a 1 = 4 k π n=1, a_1 = \frac{4k}{\pi} n = 1 , a 1 = π 4 k ; for
n = 2 , a 2 = 0 n=2, a_2 = 0 n = 2 , a 2 = 0 ; for
n = 3 , a 3 = − 4 k 3 π n=3, a_3 = -\frac{4k}{3\pi} n = 3 , a 3 = − 3 π 4 k ; for
n = 5 , a 5 = 4 k 5 π n=5, a_5 = \frac{4k}{5\pi} n = 5 , a 5 = 5 π 4 k .
The series is:
f ( x ) = k + 4 k π cos π x 2 − 4 k 3 π cos 3 π x 2 + 4 k 5 π cos 5 π x 2 − … f(x) = k + \frac{4k}{\pi} \cos\frac{\pi x}{2} - \frac{4k}{3\pi} \cos\frac{3\pi x}{2} + \frac{4k}{5\pi} \cos\frac{5\pi x}{2} - \dots f ( x ) = k + π 4 k cos 2 π x − 3 π 4 k cos 2 3 π x + 5 π 4 k cos 2 5 π x − … This matches option (C).
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Correct answer (C) f(x) = k + (4k)/(π) (cos (π)/(2)x - (1)/(3) cos (3π)/(2)x + (1)/(5) cos (5π)/(2)x - + …)
Solution The function
f ( x ) f(x) f ( x ) is an even function over the interval
( − 2 , 2 ) (-2, 2) ( − 2 , 2 ) , so its Fourier series will only contain cosine terms (
b n = 0 b_n = 0 b n = 0 ). The period is
T = 4 T = 4 T = 4 , so
L = T / 2 = 2 L = T/2 = 2 L = T /2 = 2 .
1. Constant term a 0 / 2 a_0/2 a 0 /2 a 0 = 1 L ∫ − L L f ( x ) d x = 1 2 ∫ − 1 1 2 k d x = 1 2 [ 2 k x ] − 1 1 = 1 2 ( 2 k − ( − 2 k ) ) = 2 k a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx = \frac{1}{2} \int_{-1}^{1} 2k dx = \frac{1}{2} [2kx]_{-1}^{1} = \frac{1}{2} (2k - (-2k)) = 2k a 0 = L 1 ∫ − L L f ( x ) d x = 2 1 ∫ − 1 1 2 k d x = 2 1 [ 2 k x ] − 1 1 = 2 1 ( 2 k − ( − 2 k )) = 2 k So, the constant term is
a 0 / 2 = k a_0/2 = k a 0 /2 = k .
2. Cosine coefficients a n a_n a n a n = 1 L ∫ − L L f ( x ) cos ( n π x L ) d x = 1 2 ∫ − 1 1 2 k cos ( n π x 2 ) d x = k [ 2 n π sin ( n π x 2 ) ] − 1 1 a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx = \frac{1}{2} \int_{-1}^{1} 2k \cos\left(\frac{n\pi x}{2}\right) dx = k \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_{-1}^{1} a n = L 1 ∫ − L L f ( x ) cos ( L nπ x ) d x = 2 1 ∫ − 1 1 2 k cos ( 2 nπ x ) d x = k [ nπ 2 sin ( 2 nπ x ) ] − 1 1 a n = 2 k n π ( sin n π 2 − sin ( − n π 2 ) ) = 4 k n π sin n π 2 a_n = \frac{2k}{n\pi} \left( \sin\frac{n\pi}{2} - \sin\left(-\frac{n\pi}{2}\right) \right) = \frac{4k}{n\pi} \sin\frac{n\pi}{2} a n = nπ 2 k ( sin 2 nπ − sin ( − 2 nπ ) ) = nπ 4 k sin 2 nπ For
n = 1 , a 1 = 4 k π n=1, a_1 = \frac{4k}{\pi} n = 1 , a 1 = π 4 k ; for
n = 2 , a 2 = 0 n=2, a_2 = 0 n = 2 , a 2 = 0 ; for
n = 3 , a 3 = − 4 k 3 π n=3, a_3 = -\frac{4k}{3\pi} n = 3 , a 3 = − 3 π 4 k ; for
n = 5 , a 5 = 4 k 5 π n=5, a_5 = \frac{4k}{5\pi} n = 5 , a 5 = 5 π 4 k .
The series is:
f ( x ) = k + 4 k π cos π x 2 − 4 k 3 π cos 3 π x 2 + 4 k 5 π cos 5 π x 2 − … f(x) = k + \frac{4k}{\pi} \cos\frac{\pi x}{2} - \frac{4k}{3\pi} \cos\frac{3\pi x}{2} + \frac{4k}{5\pi} \cos\frac{5\pi x}{2} - \dots f ( x ) = k + π 4 k cos 2 π x − 3 π 4 k cos 2 3 π x + 5 π 4 k cos 2 5 π x − … This matches option (C).
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