Question

Let f(x) = |x − 1|. Then,

(a) f(x²) = [f(x)]²
(b) f(x + y) = f(x) f(y)
(c) f(|x| = |f(x)|
(d) None of these

Answer 1

(d) None of these

$$\begin{gathered} f\left(x\right)=\left|x-1\right| \\ \mathrm{Since},\left|x^2-1\right|\ne \left|x-1\right|{}^2, \\ f\left(x^2\right)\ne \left(f\right(x\left)\right){}^2 \\ \mathrm{Thus},\left(\mathrm{i}\right)\mathrm{is}\mathrm{wrong}. \\ \mathrm{Since},\left|x+y-1\right|\ne \left|x-1\right|\left|y-1\right|, \\ f(x+y)\ne f\left(x\right)f\left(y\right) \\ \mathrm{Thus},\left(\mathrm{ii}\right)\mathrm{is}\mathrm{wrong}. \\ \mathrm{Since}\left|\left|x\right|-1\right|\ne \left|\left|x-1\right|\right|=\left|x-1\right|, \\ f\left(\left|x\right|\right)\ne \left|f\left(x\right)\right| \\ \mathrm{Thus},\left(\mathrm{iii}\right)\mathrm{is}\mathrm{wrong}. \\ \mathrm{Hence},\mathrm{none}\mathrm{of}\mathrm{the}\text{given options is the answer.} \end{gathered}$$