PYQs / GATE CE / 2022 / Set 1 / Q30 GATE CE 2022 Set 1 — Question 30 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +1 / -0 Hard Half-Range Sine & Cosine Series Partial Differential Equations Engineering Mathematics
Engineering Mathematics → Partial Differential Equations → Half-Range Sine & Cosine Series
Last updated 5 September 2026
Question The Fourier cosine series of a function is given by:
f ( x ) = ∑ n = 0 ∞ f n cos n x f(x) = \sum_{n=0}^{\infty} f_n \cos nx f ( x ) = n = 0 ∑ ∞ f n cos n x For
f ( x ) = cos 4 x f(x) = \cos^4 x f ( x ) = cos 4 x , the numerical value of
( f 4 + f 5 ) (f_4 + f_5) ( f 4 + f 5 ) is
______ . (round off to three decimal places)
Correct answer 0.12 to 0.13
Solution We need to express
f ( x ) = cos 4 x f(x) = \cos^4 x f ( x ) = cos 4 x in terms of multiple angles using trigonometric identities:
cos 2 x = 1 + cos 2 x 2 \cos^2 x = \frac{1 + \cos 2x}{2} cos 2 x = 2 1 + cos 2 x cos 4 x = ( cos 2 x ) 2 = ( 1 + cos 2 x 2 ) 2 = 1 4 ( 1 + 2 cos 2 x + cos 2 2 x ) \cos^4 x = (\cos^2 x)^2 = \left( \frac{1 + \cos 2x}{2} \right)^2 = \frac{1}{4} (1 + 2 \cos 2x + \cos^2 2x) cos 4 x = ( cos 2 x ) 2 = ( 2 1 + cos 2 x ) 2 = 4 1 ( 1 + 2 cos 2 x + cos 2 2 x ) Now, substitute
cos 2 2 x = 1 + cos 4 x 2 \cos^2 2x = \frac{1 + \cos 4x}{2} cos 2 2 x = 2 1 + c o s 4 x :
cos 4 x = 1 4 ( 1 + 2 cos 2 x + 1 + cos 4 x 2 ) \cos^4 x = \frac{1}{4} \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) cos 4 x = 4 1 ( 1 + 2 cos 2 x + 2 1 + cos 4 x ) cos 4 x = 1 4 + 1 2 cos 2 x + 1 8 + 1 8 cos 4 x \cos^4 x = \frac{1}{4} + \frac{1}{2} \cos 2x + \frac{1}{8} + \frac{1}{8} \cos 4x cos 4 x = 4 1 + 2 1 cos 2 x + 8 1 + 8 1 cos 4 x cos 4 x = 3 8 + 1 2 cos 2 x + 1 8 cos 4 x \cos^4 x = \frac{3}{8} + \frac{1}{2} \cos 2x + \frac{1}{8} \cos 4x cos 4 x = 8 3 + 2 1 cos 2 x + 8 1 cos 4 x Comparing this with the Fourier cosine series
f ( x ) = ∑ n = 0 ∞ f n cos n x f(x) = \sum_{n=0}^{\infty} f_n \cos nx f ( x ) = ∑ n = 0 ∞ f n cos n x :
f 0 = 3 8 f_0 = \frac{3}{8} f 0 = 8 3 f 2 = 1 2 f_2 = \frac{1}{2} f 2 = 2 1 f 4 = 1 8 = 0.125 f_4 = \frac{1}{8} = 0.125 f 4 = 8 1 = 0.125 All other f n = 0 f_n = 0 f n = 0 n ≠ 0 , 2 , 4 n \neq 0, 2, 4 n = 0 , 2 , 4
Therefore,
f 5 = 0 f_5 = 0 f 5 = 0 .
f 4 + f 5 = 0.125 + 0 = 0.125 f_4 + f_5 = 0.125 + 0 = 0.125 f 4 + f 5 = 0.125 + 0 = 0.125 .
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Correct answer 0.12 to 0.13
Solution We need to express
f ( x ) = cos 4 x f(x) = \cos^4 x f ( x ) = cos 4 x in terms of multiple angles using trigonometric identities:
cos 2 x = 1 + cos 2 x 2 \cos^2 x = \frac{1 + \cos 2x}{2} cos 2 x = 2 1 + cos 2 x cos 4 x = ( cos 2 x ) 2 = ( 1 + cos 2 x 2 ) 2 = 1 4 ( 1 + 2 cos 2 x + cos 2 2 x ) \cos^4 x = (\cos^2 x)^2 = \left( \frac{1 + \cos 2x}{2} \right)^2 = \frac{1}{4} (1 + 2 \cos 2x + \cos^2 2x) cos 4 x = ( cos 2 x ) 2 = ( 2 1 + cos 2 x ) 2 = 4 1 ( 1 + 2 cos 2 x + cos 2 2 x ) Now, substitute
cos 2 2 x = 1 + cos 4 x 2 \cos^2 2x = \frac{1 + \cos 4x}{2} cos 2 2 x = 2 1 + c o s 4 x :
cos 4 x = 1 4 ( 1 + 2 cos 2 x + 1 + cos 4 x 2 ) \cos^4 x = \frac{1}{4} \left( 1 + 2 \cos 2x + \frac{1 + \cos 4x}{2} \right) cos 4 x = 4 1 ( 1 + 2 cos 2 x + 2 1 + cos 4 x ) cos 4 x = 1 4 + 1 2 cos 2 x + 1 8 + 1 8 cos 4 x \cos^4 x = \frac{1}{4} + \frac{1}{2} \cos 2x + \frac{1}{8} + \frac{1}{8} \cos 4x cos 4 x = 4 1 + 2 1 cos 2 x + 8 1 + 8 1 cos 4 x cos 4 x = 3 8 + 1 2 cos 2 x + 1 8 cos 4 x \cos^4 x = \frac{3}{8} + \frac{1}{2} \cos 2x + \frac{1}{8} \cos 4x cos 4 x = 8 3 + 2 1 cos 2 x + 8 1 cos 4 x Comparing this with the Fourier cosine series
f ( x ) = ∑ n = 0 ∞ f n cos n x f(x) = \sum_{n=0}^{\infty} f_n \cos nx f ( x ) = ∑ n = 0 ∞ f n cos n x :
f 0 = 3 8 f_0 = \frac{3}{8} f 0 = 8 3 f 2 = 1 2 f_2 = \frac{1}{2} f 2 = 2 1 f 4 = 1 8 = 0.125 f_4 = \frac{1}{8} = 0.125 f 4 = 8 1 = 0.125 All other f n = 0 f_n = 0 f n = 0 n ≠ 0 , 2 , 4 n \neq 0, 2, 4 n = 0 , 2 , 4
Therefore,
f 5 = 0 f_5 = 0 f 5 = 0 .
f 4 + f 5 = 0.125 + 0 = 0.125 f_4 + f_5 = 0.125 + 0 = 0.125 f 4 + f 5 = 0.125 + 0 = 0.125 .
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More questions on Partial Differential Equations ← Q29 Full paper Q31 → 