First slide

Functions (XII)

Question

$Let g (x) = f (x) - 1 . If f (x) + f (1 - x) = 2 \forall x \in R, then g (x) is symmetrical about$

Moderate

A

the origin
B

$the line x = \frac{1}{2}$
C

$the point (1, 0)$
D

$the point (\frac{1}{2}, 0)$

Solution

$f (x) - 1 + f (1 - x) - 1 = 0$
$So, g (x) + g (1 - x) = 0$
$Replacing x by x + \frac{1}{2}, we get g (\frac{1}{2} + x) + g (\frac{1}{2} - x) = 0$
$So, it is symmetrical about (\frac{1}{2}, 0) .$ $(I f f (a + x) + f (a - x) = 0 t h e n f (x) i s s y m m e t r i c a b o u t (a, 0))$

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