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 the above
Solution
Given, f(x) = |x -1|
∴ f(x2) = | x2 -1| and {f(x)}2 = (x - 1)2
⇒ f(x)2 ≠ ∀ (f(x))2, hence (a) is false.
Also, f(x + y) =| x + y - 1| and f(x) = | x - 1|, f(y) = | y - 1|
⇒ f(x + y) ≠ f(x) + f( y), hence (b) is false.
f(| x |) = || X | - 1| and | f(x) |=|| x - 1||=| x - 1|
∴ f(| x |) ≠ | f(x) |, hence (c) is false.
∴ f(x2) = | x2 -1| and {f(x)}2 = (x - 1)2
⇒ f(x)2 ≠ ∀ (f(x))2, hence (a) is false.
Also, f(x + y) =| x + y - 1| and f(x) = | x - 1|, f(y) = | y - 1|
⇒ f(x + y) ≠ f(x) + f( y), hence (b) is false.
f(| x |) = || X | - 1| and | f(x) |=|| x - 1||=| x - 1|
∴ f(| x |) ≠ | f(x) |, hence (c) is false.
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