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Evaluation of definite integrals

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Question

$Let a non constant polynomial f satisfies the relation f (f (f (x))) + (1 - p) f (x) = 3$
$\forall x \in R, where p is some real number. If leading coefficient of f (x) is negative and$ $f (0) = 4 then \int_{- 1}^{1} f^{- 1} (x) dx is equal to$

Difficult

Solution

Degree is 1
$\Rightarrow f (x) must be linear$
$\begin{array}{l} Let f (x) = ax + 4 \Rightarrow f (f (f (x))) = a^{3} x + 4 a^{2} + 4 a + 4 \\ \Rightarrow a^{3} x + 4 a^{2} + 4 a + 4 + (1 - p) (ax + 4) = 3, \forall x \in R \\ \Rightarrow a^{3} + (1 - p) a = 0 \Rightarrow a^{2} = p - 1 \\ 4 a^{2} + 4 a + 4 + 4 (1 - p) = 3 \\ - 4 (1 - p) + 4 a + 4 + 4 (1 - p) = 3 \\ 4 a = - 1 \Rightarrow a = - \frac{1}{4} \\ f (x) = - \frac{1}{4} x + 4 \\ f^{- 1} (x) = 4 (4 - x) \end{array}$

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Similar Questions

Let $f (x) = \sin x + \sum_{r = 0}^{2} \cos (x + \frac{2 r π}{3})$

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