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Continuous f on R with f(f(f(x))) = x | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Let ff be a continuous function on R\mathbb R satisfying the relation f(f(f(x)))=x ∀ x∈Rf\left(f\left(f\left(x\right)\right)\right) = x\ \forall\,x \in \mathbb R. Then
Athere exists ff which is non-decreasingcorrect
Bthere exists ff which is non-increasing
Cff has to be differentiable ∀ x∈R\forall\,x \in \mathbb Rcorrect
Dthere exists only one function that satisfies the given conditionscorrect
Solution
Step 1: If f(x)=f(y)f\left(x\right) = f\left(y\right), applying ff twice more gives x=yx = y, so ff is one-one; and every xx is the image of f(f(x))f\left(f\left(x\right)\right), so ff is onto. A continuous one-one function on R\mathbb R is strictly monotonic, by the intermediate value theorem: one that changes direction takes some value twice. Step 2: If ff were strictly decreasing, then f∘ff\circ f would be strictly increasing and f∘f∘ff\circ f\circ f strictly decreasing, but f∘f∘ff\circ f\circ f is the identity, which is increasing. So ff is strictly increasing. Step 3: Let f(x0)>x0f\left(x_0\right)>x_0 for some x0x_0. Applying the increasing ff,
f(f(x0))>f(x0)>x0,f(f(f(x0)))>f(f(x0))>x0f\left(f\left(x_0\right)\right)>f\left(x_0\right)>x_0 ,\qquad f\left(f\left(f\left(x_0\right)\right)\right)>f\left(f\left(x_0\right)\right)>x_0
 i.e. x0>x0x_0>x_0, which is not possible. The same argument with the inequality reversed rules out f(x0)<x0f\left(x_0\right)<x_0. Hence
f(x)=xfor every xf\left(x\right) = x \quad\text{for every } x
 and this is the only continuous solution. Step 4: Testing the four options against f(x)=xf\left(x\right) = x: (1) f(x)=xf\left(x\right) = x is non-decreasing. True. (2) it is strictly increasing, hence not non-increasing; and there is no other candidate. False. (3) f(x)=xf\left(x\right) = x is differentiable everywhere, so any ff meeting the conditions is. True. (4) the solution is unique. True. Answer: (1),(3),(4)\left(1\right),\left(3\right),\left(4\right)
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