GATE CS 2016 Set 2 Q.12 (CS Q.2): Calculus

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 ________.

Your answer

Official answer and solution

Official answer: 9
From: Official GATE 2016 answer key

Solution

Decompose the polynomial into even and odd degree components and determine the degree of the derived difference.

Let $f(x) = \sum_{k=0}^d a_k x^k$ be a polynomial. Then:

f(x) + f(-x) = \sum_{k=0}^d a_k x^k + \sum_{k=0}^d a_k (-1)^k x^k = 2 \sum_{k \text{ even}} a_k x^k

Only the terms with even powers of $x$ survive in $f(x) + f(-x)$ .

We are given that the degree of $(f(x) + f(-x))$ is 10. This implies:

a_{10} \ne 0 \quad \text{and} \quad a_k = 0 \text{ for all even } k > 10

Now consider $g(x) = f'(x) = \sum_{k=1}^d k a_k x^{k-1}$ . Then:

g(-x) = \sum_{k=1}^d k a_k (-1)^{k-1} x^{k-1}

g(x) - g(-x) = \sum_{k=1}^d k a_k \left[1 - (-1)^{k-1}\right] x^{k-1}

Notice that:

When $k$ is odd, $k-1$ is even, so $(-1)^{k-1} = 1$ , and $1 - 1 = 0$ .
When $k$ is even, $k-1$ is odd, so $(-1)^{k-1} = -1$ , and $1 - (-1) = 2$ .

Thus, only even $k$ contribute to $g(x) - g(-x)$ :

g(x) - g(-x) = 2 \sum_{k \text{ even}, k \ge 2} k a_k x^{k-1}

The highest non-zero even index is $k = 10$ with $a_{10} \ne 0$ . Its term in $g(x) - g(-x)$ is:

2 \times 10 \times a_{10} x^{10 - 1} = 20 a_{10} x^9 \ne 0

Since no even power greater than 10 has a non-zero coefficient, the degree of $(g(x) - g(-x))$ is exactly 9.

Final answer

Common trap

Assuming the degree of the derivative is always (degree - 1) without checking whether the leading term cancels.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 2
Question: 12 of 65, printed as CS Q.2
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source