Let y = f(x) be a differentiable function \forall x \in R Such that f(f(x)) = x \textrm{ } \forall \textrm{ } x \in [0, 1]. If f(0) = 1, then which of the following is/are correct?

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

Question

Let y = f(x) be a differentiable function $\forall x \in R$ Such that f(f(x)) = $x \forall x \in$ [0, 1]. If f(0) = 1, then which of the following is/are correct?

$\int_{0}^{1} {(x - f (x))}^{2} dx = \frac{1}{2}$

f(x) = 0 has a real root

f '(x) + 1 = 0 has a real root in (0, 1)

$\int_{0}^{1} {(x - f (x))}^{2} dx = \frac{1}{3}$

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f(0) = 1 and f(f(x)) = x ⇒ f(1) = 0

$f' (x) = \frac{f (0) - f (1)}{0 - 1} = - 1$

for atleast one x $\in$ (0, 1)

$I = \int_{0}^{1} {(x - f (x))}^{2} dx$ Put x = f(t)

$I = \int_{1}^{0} {(f (t) - f (f (t)))}^{2} \cdot f' (t) dt = - \int_{0}^{1} {(f (x) - x)}^{2} f' (x) dx$

$I = \int_{0}^{1} {(x - f (x))}^{2} (1 - f' (x)) dx - I$

$2 I = {\frac{{(x - f (x))}^{3}}{3} |}_{0}^{1} = \frac{1}{3} + \frac{1}{3} \Rightarrow I = \frac{1}{3}$