GATE CSE 2016 Set 2 | Question: 02

Question

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 __________.

Answer 1

Let $f(x)=x^{10}$   Degree$=10.$

$f(x)+f(-x) =x^{10}+(-x)^{10}$
$\quad\quad=x^{10}+x^{10}$
$\quad\quad=2.x^{10}$

$g(x)-g(-x)=10.x^{9} - \{-10x^{9}\}$
$\quad\quad=20.x^{9}$

So, answer is $9.$

Comments

@gmrishikumar @mehul vaidya 

He does not write the entire polynomial, he just writes the maximum degree term.

What is the problem with this solution$?$

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Some better explanations are given below. Check them out.

The solution is right, but I think the explanation could be better.

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Yes, you are right.

Answer 2

Whole question is based on observation.

Given that Degree of $f(x)+f(-x)$ is $10$ from here we have to observe what can be possible degree of $f(x)$ , first we can say that as Degree of $f(x)+f(-x)$ is $10$ so degree of $f(x)$ would be at least $10$.

Possibility (1) $f(x)=x^{10} +\ldots$      // for simplicity I'm taking coefficients as $1$

Now observe if greater than $10$ degree possible ? if  degree of $f(x)$ is $11$ than still we can see our condition "Degree of $f(x)+f(-x)$ is $10$" is yet satisfied

Possibility (2) $f(x)=x^{11}+x^{10}+\ldots$   

other possibilities $f(x)=x^{19}+x^{15}+x^{13}+x^{10} \ldots$

If we observe bit more we can know that degree of $f(x)$ can be anything but in $f(x)$ the maximum even power of $x$ must be $10$, in that way only the given condition can be satisfied.

Now when we calculate $g(x)-g(-x)$ we can observe that all odd powers of $f(x)$ would be canceled out but even power won't be and the largest even power of $f(x)$ which is $x^{10}$ will result into $x^{9}$ when applying the derivative $g(x)-g(-x)$. Hence answer will be $9$.

Comments

Most Correct answer

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-g(-x) is basically the derivative of f(-x)

$\frac{\mathrm{d} F(-x)}{\mathrm{d} x}$=F'(-x)$\frac{\mathrm{d} (-x)}{\mathrm{d} x}$=-g(-x)

so g(x)-g(-x)=F'(x)+F'(-x)

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Great explanation. I am too naive to think like this. What I did was, took an odd function with power10 and applied on the function above.

Answer 3

9

F is some function where the largest even degree term is having degree  10. no restiction on odd degree terms.

since f(x)+f(-x)= degree 10

even power gets converted to odd in derivative.

then the the degrre of required expression =9.

the odd powers in F will become even in derivative and G(X)-G(-X) retains only odd powers.

Answer 4

f(x) can be either an even function or an odd function.

If f(x) was an odd function, then f(x) + f(-x) = 0, but here it has been given that it has degree 10. So, it must be an even function.

Therefore, f(x) = f(-x) => f'(x) = -f'(-x)

Also, it has been mentioned that g(x) is the derivative of f(x).

So, g(x) = f'(x) and g(-x) = -f'(-x) => g(x) - g(-x) = f'(x) - (-f'(-x)) => g(x) - g(-x) = f'(x) + f'(x) => g(x) - g(-x) = 2 * f'(x) But, f'(x) will have degree 9.

Comments

Function can neither be even or odd.

See this function $f(x)=x^3+x^2+x$ is neither even or odd.

Since $f(-x)=-x^3+x^2-x$

So your argument is incorrect though the final answer is correct.