Question

If  \[f\left( x \right) = \frac{1}{1 - x}\] , then the set of points discontinuity of the function f (f(f(x))) is

पर्याय

• {1}

• {0, 1} 

• {−1, 1}

• none of these 

Answer 1

 {0, 1} 

Given: 

\[f\left( x \right) = \frac{1}{1 - x}\]

Clearly, 

\[f: R - \left\{ 1 \right\} \to R\]

Now, 

\[f\left( f\left( x \right) \right) = f\left( \frac{1}{1 - x} \right) = \left( \frac{1}{1 - \left( \frac{1}{1 - x} \right)} \right) = \left( \frac{1 - x}{- x} \right) = \left( \frac{x - 1}{x} \right)\]

∴  \[fof:\]

\[R - \left\{ 0, 1 \right\} \to R\]

Now,

\[f\left( f\left( f\left( x \right) \right) \right) = f\left( \frac{x - 1}{x} \right) = \left( \frac{1}{1 - \left( \frac{x - 1}{x} \right)} \right) = x\]

∴ \[fofof:\]

\[R - \left\{ 0, 1 \right\} \to R\]

Thus, ​

\[f\left( f\left( f\left( x \right) \right) \right)\]  is not defined at 

 

\[x = 0, 1\] .

Hence, ​

\[f\left( f\left( f\left( x \right) \right) \right)\]  is discontinuous at {0, 1}.