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Calculus

GATE CSE 2016 Set 2 · Q2

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Let f(x)f(x) be a polynomial and g(x)=f′(x)g(x) = f^{\prime}(x) be its derivative. If the degree of (f(x)+f(−x))(f(x) + f(-x)) is 10, then the degree of (g(x)−g(−x))(g(x) - g(-x)) is ________.

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Official answer: 9

Calculus

Any polynomial f(x)f(x) can be decomposed into the sum of an even polynomial fe(x)f_e(x) and an odd polynomial fo(x)f_o(x):

f(x)=fe(x)+fo(x)f(x) = f_e(x) + f_o(x)

Then:

f(x)+f(−x)=[fe(x)+fo(x)]+[fe(−x)+fo(−x)]=[fe(x)+fo(x)]+[fe(x)−fo(x)]=2fe(x)f(x) + f(-x) = [f_e(x) + f_o(x)] + [f_e(-x) + f_o(-x)] = [f_e(x) + f_o(x)] + [f_e(x) - f_o(x)] = 2f_e(x)

Given that the degree of (f(x)+f(−x))=2fe(x)(f(x) + f(-x)) = 2f_e(x) is 1010, the highest degree term of the even part fe(x)f_e(x) is of power 1010.

Now, g(x)=f′(x)=fe′(x)+fo′(x)g(x) = f^{\prime}(x) = f_e^{\prime}(x) + f_o^{\prime}(x). Since the derivative of an even function is odd and the derivative of an odd function is even:

  • fe′(−x)=−fe′(x)f_e^{\prime}(-x) = -f_e^{\prime}(x)

  • fo′(−x)=fo′(x)f_o^{\prime}(-x) = f_o^{\prime}(x)

Therefore:

g(x)−g(−x)=[fe′(x)+fo′(x)]−[−fe′(x)+fo′(x)]=2fe′(x)g(x) - g(-x) = [f_e^{\prime}(x) + f_o^{\prime}(x)] - [-f_e^{\prime}(x) + f_o^{\prime}(x)] = 2f_e^{\prime}(x)

Since fe(x)f_e(x) has degree 1010, its derivative fe′(x)f_e^{\prime}(x) has degree:

degree(fe′(x))=10−1=9\text{degree}(f_e^{\prime}(x)) = 10 - 1 = 9

Thus, the degree of $(g(x) - g(-x))$ is9.

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This Calculus previous year question is Q2 of the GATE CSE 2016 Set 2 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

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Calculus in GATE CSE

Calculus has 40 GATE CSE PYQs across 25 papers from 2011 to 2026. 11 of them are 1-mark NAT questions like this one, and it is the only one in GATE CSE 2016 Set 2.

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GateAI solution cardGATE CSE 2016 Set 2 · Q2 · NAT

Let f(x) be a polynomial and g(x) = f'(x) be its derivative. If the degree of (f(x) + f(-x)) is 10, then the degree of (g(x) - g(-x)) is …

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