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The function f(x) is continuous and $f(f(x))=1-x$, for $x\in [0,1]$ then,

(A) $f(\frac{1}{8})+f(\frac{7}{8})=3$

(B) $f(\frac{2}{3})+f(\frac{1}{3})=2$

(C) $f(\frac{5}{6})+f(\frac{1}{6})=1$

(D) None of These

My approach is as follow $f\left( {f\left( x \right)} \right) = 1 - x \Rightarrow f'\left( {f\left( x \right)} \right) \times f'\left( x \right) = - 1$

$f'\left( {f\left( 0 \right)} \right) \times f'\left( 0 \right) = - 1\& f'\left( {f\left( 1 \right)} \right) \times f'\left( 1 \right) = - 1$

$f\left( {f\left( 0 \right)} \right) = 1;f\left( {f\left( 1 \right)} \right) = 0$

Not able to procced from here

insipidintegrator
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Samar Imam Zaidi
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1 Answers1

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From $f(f(x)) =1-x$ follows $$1-f(x)=f(f(f(x)))=f(1-x) .$$ Hence, $$f(x) +f(1-x) =1$$ for all $x\in[0, 1]$.

Jochen
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. Instead of answering it, it would be better to look for a good duplicate target, or help the user by posting comments. You can use this. – Sourav Ghosh Aug 12 '22 at 06:27
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    More generally, if $f(f(x))=g(x)$ then $f(g(x))=f(f(f(x)))=g(f(x))$. – J.G. Aug 12 '22 at 07:19
  • $f(x)+f(1-x)=1$ is necessary but not sufficient condition. This answer does not give full solution of question in the title, only solution to test problem in the text. – Ivan Kaznacheyeu Aug 12 '22 at 07:22
  • Yes I agree as $f(x)=x, \frac{a^x}{a^x+\sqrt{a}}$ satisfy $f(x)+f(1-x)=1$ but fail to satisfy $ff(x)=1-x$. – Z Ahmed Aug 12 '22 at 15:31