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Best answer
75 votes
75 votes
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 $9.$
edited by
57 votes
57 votes
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 $9$.
edited by
21 votes
21 votes
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.
3 votes
3 votes
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.
Answer:

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