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प्रश्न
Let f(x) = |x − 1|. Then,
पर्याय
(a) f(x2) = [f(x)]2
(b) f(x + y) = f(x) f(y)
(c) f(|x| = |f(x)|
(d) None of these
उत्तर
(d) None of these
\[\text{ Since,} \left| x^2 - 1 \right| \neq \left| x - 1 \right| {}^2 , \]
\[f( x^2 ) \neq (f(x)) {}^2 \]
\[\text { Thus, (i) is wrong} . \]
\[\text{ Since, } \left| x + y - 1 \right| \neq \left| x - 1 \right|\left| y - 1 \right|, \]
\[f(x + y) \neq f(x) f(y)\]
\[\text{ Thus, (ii) is wrong} . \]
\[\text{ Since } \left| \left| x \right| - 1 \right| \neq \left| \left| x - 1 \right| \right| = \left| x - 1 \right|, \]
\[f\left( \left| x \right| \right) \neq \left| f(x) \right|\]
\[\text{ Thus, (iii) is wrong } . \]
\[\text{ Hence, none of the given options is the answer} .\]
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