The graph of $y = f (x)$ is symmetrical about the line $x = 1,$ then

f (- x) = f (x)

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f (1 + x) = f (1 - x)

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f (x + 1) = f (x - 1)

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None of these

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Solution

The correct option is C $f (x + 1) = f (x - 1)$
If $y = f (x)$ is symmetrical about the line $x = a$ then

$f (x + α) = f (x - α)$

$∴ f (x + 1) = f (x - 1)$

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Similar questions

f (x) = x - \frac{1}{x}

f (\frac{1}{x}) =

f (x)

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- f (x)

\frac{1}{f (x)}

f (x) = \frac{4^{x}}{4^{x} + 2} for all x \in R .

If $f : R \to R$ be given by $f (x) = \frac{4^{x}}{4^{x} + 2}$ for all $x ϵ R$ . Then,

If $f (x) = \frac{x - 1}{x + 1}$ , then show that
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(ii) $f (- \frac{1}{x}) = \frac{- 1}{f (x)}$

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