Para que una función sea par, ha de verificarse que f(x)=f(-x)
a) f(x) = x2+x
f(-x) = (-x)2-x = x2-x, f(x)≠f(-x)
b) f(x) = x3
f(-x) = (-x)3=-x3, f(x)≠f(-x)
c) f(x) = xcos(x)
f(-x) = (-x)cos(-x) = -xcos(x), f(x)≠f(-x)
d) f(x) = xsen(x)
f(-x) = (-x)sen(-x) = -x(-sen(x)) = xsen(x), f(x)=f(-x)
e) f(x) = x2sen(x)
f(-x) = (-x)2sen(-x) = x2(-sen(x)) = - x2sen(x), f(x)≠f(-x)
La solución correcta es la d)