Try this option:
the rule: if f(-x)=f(x), then the function f(x) is even, if f(-x)=-f(x) then the function is odd.
1. f(x)=-5x⁴-2;
if to substitute x→(-x), then f(-x)=-5*(-x)⁴-2; ⇔ f(-x)=-5x⁴-2, in other words f(x)=f(-x), it means that this function is even.
2. f(x)=x³+2x.
if to substitute x→(-x), then f(-x)=(-x)³+2*(-x); ⇔ f(-x)=-(x³+2x), in other words f(-x)=-f(x), it means that this function is odd.
Answer:
A) JK and AB =
![[tex] \boxed{( \theta) = \f{12}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%5Cf%7B12%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, =》 
=》 
[/tex]
B) AABC
![[tex] \boxed{( \theta) = {1}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%7B1%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, 
C) AABM:
![[tex] e \fbox{: }](https://tex.z-dn.net/?f=%5Btex%5D%20e%20%5Cfbox%7B%3A%20%7D)
The value of k for which 
is 
ans[/tex]
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