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Official PYQ · 2016

GATE CS 2016 Set 2 Q02 — Calculus: Limits, Continuity, Differentiability

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Let f(x)f(x) be a polynomial and g(x)=f′(x)g(x)=f'(x) be its derivative. If the degree of (f(x)+f(−x))(f(x)+f(-x)) is 1010, then the degree of (g(x)−g(−x))(g(x) - g(-x)) is __________.
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Answer: 9

Any polynomial f(x)f(x) can be uniquely decomposed into its even part P(x)P(x) and odd part Q(x)Q(x):

f(x)=P(x)+Q(x)f(x) = P(x) + Q(x)

where P(x)=f(x)+f(−x)2P(x) = \frac{f(x) + f(-x)}{2} is an even polynomial, and Q(x)=f(x)−f(−x)2Q(x) = \frac{f(x) - f(-x)}{2} is an odd polynomial.

We are given that the degree of f(x)+f(−x)f(x) + f(-x) is 1010. Since f(x)+f(−x)=2P(x)f(x) + f(-x) = 2P(x), the degree of the even polynomial P(x)P(x) is 1010.

Now, let us find the derivative g(x)=f′(x)g(x) = f'(x):

g(x)=P′(x)+Q′(x)g(x) = P'(x) + Q'(x)

Using the properties of derivatives of even and odd functions:

  • Since P(x)P(x) is even, its derivative P′(x)P'(x) is an odd function, meaning P′(−x)=−P′(x)P'(-x) = -P'(x).
  • Since Q(x)Q(x) is odd, its derivative Q′(x)Q'(x) is an even function, meaning Q′(−x)=Q′(x)Q'(-x) = Q'(x).

Thus, we can compute g(x)−g(−x)g(x) - g(-x) as follows:

g(x)−g(−x)=(P′(x)+Q′(x))−(P′(−x)+Q′(−x))=(P′(x)+Q′(x))−(−P′(x)+Q′(x))=2P′(x)\begin{aligned} g(x) - g(-x) &= \left(P'(x) + Q'(x)\right) - \left(P'(-x) + Q'(-x)\right) \\ &= \left(P'(x) + Q'(x)\right) - \left(-P'(x) + Q'(x)\right) \\ &= 2P'(x) \end{aligned}

Since P(x)P(x) is a polynomial of degree 1010, its derivative P′(x)P'(x) is a polynomial of degree 10−1=910 - 1 = 9. Therefore, the degree of g(x)−g(−x)g(x) - g(-x) is 9.

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