Given the function $f (x) = \frac{1}{1 - x}$ , the number points of discontinuity of the composite function $y = f^{3 n} (x)$ , where, $f^{n} (x) = f o f . . . o f (n t i m e s)$ are $(n \in N)$

0, 1

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2 n

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3 n

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2 n + 1

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Solution

The correct option is A $0, 1$
$f (x) = \frac{1}{x}$
$f^{2} (x) = f o f (x) = \frac{1}{1 - (\frac{1}{1 - x})} = \frac{1 - x}{- x} = \frac{x - 1}{x}$
$f^{3} (x) = f o f o f (x) = \frac{(\frac{1}{1 - x}) - 1}{(\frac{1}{1 - x})} = \frac{x}{1}$
$f^{4} (x) = f o f o f o f (x) = \frac{1}{1 - x} = f (x)$
$f^{3 n - 2} (x) = \frac{1}{1 - x}$
$f^{3 n - 1} (x) = \frac{x - 1}{x}$
$f^{3 n} (x) = x$

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Similar questions

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

y = f^{3 n} (x), w h e r e f^{n} (x) = f o f o f . . . (n t i m e s) (n ϵ N) a r e 0, 1.

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

f^{3 n} (x)

f^{n} (x) = f o f o f . . . o f (n

)

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

y = f^{3 n} (x),

f^{n} (x) = f o f \dots o f (n times) (x), (n \in N)

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

y = [f {f (x)}]

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

y = f {f (x)}

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