FunctionMedium
Question
A function whose graph is symmetrical about the origin is given by -
Options
A.f(x) = ex + e-x
B.f(x) = sin(sin(cos(sin x)))
C.f(x + y) = f(x) + f(y)
D.sin x + sin | x |
Solution
Put y = - x, we get f(x) = -x also f(0) = 0
f(x + y) = f(x) + f(y) is an odd function so it is symmetric about origin.
f(x + y) = f(x) + f(y) is an odd function so it is symmetric about origin.
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