Question
Let $$f\left( x \right) = \left| {x - 1} \right|.$$ Then
A.
$$f\left( {{x^2}} \right) = {\left( {f\left( x \right)} \right)^2}$$
B.
$$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)$$
C.
$$f\left( {\left| x \right|} \right) = \left| {f\left( x \right)} \right|$$
D.
None of these
Answer :
None of these
Solution :
$$f\left( x \right) = \left| {x - 1} \right| = \left\{ {_{x - 1,}^{ - x + 1}} \right.\,_{x \geqslant 1}^{x < 1}$$
Consider $$f\left( {{x^2}} \right) = {\left( {f\left( x \right)} \right)^2}$$
If it is true it should be $$\forall x$$
$$\therefore $$ Put $$x = 2$$
$${\text{LHS}} = f\left( {{2^2}} \right) = \left| {4 - 1} \right| = 3\,{\text{and}}\,{\text{RHS}} = {\left( {f\left( 2 \right)} \right)^2} = 1$$
$$\therefore \left( A \right)$$ is not correct
Consider $$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)$$
Put $$x = 2,y = 5$$ we get $$f\left( 7 \right) = 6;f\left( 2 \right) + f\left( 5 \right) = 1 + 4 = 5$$
$$\therefore \left( B \right)$$ is not correct
Consider $$f\left( {\left| x \right|} \right) = \left| {f\left( x \right)} \right|$$
Put $$x = - 5$$ then $$f\left( {\left| { - 5} \right|} \right) = f\left( 5 \right) = 4$$
$$f\left( {\left| { - 5} \right|} \right) = \left| { - 5 - 1} \right| = 6$$
$$\therefore \left( C \right)$$ is not correct.
Hence (D) is the correct alternative.