Mark the correct alternative in each of the following:
The function f : R → R defined by f(x) = 6x + 6|x| is
Given that function f : R → R defined by f(x) = 6x + 6|x|
Let f(x) = f(y)
⇒ 6x + 6|x|=6y + 6|y|
Only when x = y
So, f is one-one.
Now for y=f(x)
y can never be negative which means for no x ϵ R y is negative.
So, f is not onto but into.