Question 1199004
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Think of 5 as 5*1 = 5*x^0


The equation for choice (A) 
f(x) = 3x^3 + 5
is equivalent to
f(x) = 3x^3 + 5x^0


We have an odd exponent 3 and an even exponent 0.
This polynomial function is neither even nor odd.
Even polynomial functions must have all exponents even.
Odd polynomial functions must have all exponents odd.


Choice (B) is an odd function since x = x^1 has an odd exponent.


Choice (C) appears to be an even function since cos(x) is even
But we have a phase shift of pi/2 units to the left, which means we really have a sine function here. Therefore, the function in choice (C) is odd.


You can use the identity
cos(A+B) = cos(A)cos(B)-sin(A)sin(B)
to find that
cos(x+pi/2) = -sin(x)


One common way to prove a function f(x) is odd is to show the following
f(-x) = -f(x)


Or you could use a visual approach to note how we have symmetry about the origin.
{{{drawing(400,400,-4,6,-5,5,
graph(400,400,-4,6,-5,5,-100,cos(x+pi/2)),
locate(1,3,matrix(3,1,y=cos(x+pi/2),"aka",y=-sin(x)))
)}}}
Any point (x,y) on the green curve has a corresponding mirror point at (-x,-y).


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Answer: <font color=red>Choice (A)</font> is not an odd function. 
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