FunctionHard
Question
Let f(x) = |x - 1|, then
Options
A.f(x2) = (f(x))2
B.f(x + y) = f(x) + f(y)
C.f(|x|) = |f(x)|
D.none of these
Solution
f(x) = |x - 1|
f(x2) = |x2 - 1| and f2(x) = |x - 1|2 ⇒ f(x2) ≠f2(x)
f(x + y) = |x + y - 1| and f(x) + f(y) = |x - 1| + |y - 1|
⇒ f(x + y) ≠ f(x) + f(y)
f(|x|) = ||x| - 1| and |f(x)| = ||x - 1||
f(x2) = |x2 - 1| and f2(x) = |x - 1|2 ⇒ f(x2) ≠f2(x)
f(x + y) = |x + y - 1| and f(x) + f(y) = |x - 1| + |y - 1|
⇒ f(x + y) ≠ f(x) + f(y)
f(|x|) = ||x| - 1| and |f(x)| = ||x - 1||
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