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$Let f_{1} (x) = \{\begin{cases} x, 0 \leq x \leq 1 \\ 1, x > 1 \\ 0, otherwise \end{cases}$
$f_{2} (x) = f_{1} (- x) for all x$
$f_{3} (x) = - f_{2} (x) for all x$
$f_{4} (x) = f_{3} (- x) for all x$
which of the following is necessarily true?

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f4(x)=f1(x) for all x

f1(x)=−f3(−x) for all x

f2(−x)=f4(x) for all x

f1(x)+f3(x)=0 for all x

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detailed solution

Correct option is B

f2(x)=f1(-x) f3(x)=-f2(x) f3(x)=-f1(-x) replace x by -x f3(-x)=-f1(x) ⇒f1(x)=-f3(-x)

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Let f1(x)=x,0≤x≤11, x>10, otherwise f2(x)=f1(−x) for all xf3(x)=−f2(x) for all xf4(x)=f3(−x) for all xwhich of the following is necessarily true?

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$Let f_{1} (x) = \{\begin{cases} x, 0 \leq x \leq 1 \\ 1, x > 1 \\ 0, otherwise \end{cases}$
$f_{2} (x) = f_{1} (- x) for all x$
$f_{3} (x) = - f_{2} (x) for all x$
$f_{4} (x) = f_{3} (- x) for all x$
which of the following is necessarily true?