First slide

Functions (XII)

Question

$If f (x) = | x - 2 | + 2 x and f^{- 1} (x) = a | x - 4 | + b x + c; (if f (x) is an invertible function),$ $then (a, b, c are real constants) which of the following is/are correct?$

Difficult

A

$a = 1$
B

$a = \frac{- 1}{3}$
C

$b = \frac{2}{3}$
D

$c = \frac{- 2}{3}$

Solution

$W h e n x < 4 f (x) = x - 2 + 2 x = 3 x - 2 a n d f^{- 1} (x) = a (x - 4) + b x + c f \circ f^{- 1} (x) = x 3 (a (x - 4) + b x + c) - 2 = x 3 a + 3 b = 1 a n d - 12 a + 3 c - 2 = 0 a n d w h e n x < 4 f (x) = - x + 2 + 2 x = x + 2 a n d f^{- 1} (x) = a (x - 4) + b x + c f \circ f^{- 1} (x) = x 3 (a (x - 4) + b x + c) - 2 = x 3 a + 3 b = 1 a n d - 12 a + 3 c - 2 = 0 f (x) = x - 2 + 2 x = 3 x - 2 x < 2 = - x + 2 + 2 x = x + 2 x > 2$

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