If $\alpha, \beta(\alpha<\beta)$ are the points of discontinuity of the function $f(f(f(x)))$, where $f(x)=\frac{1}{1-x}$, then the set of values of $a$ for which the points $(\alpha, \beta)$ and $\left(a, a^2\right)$ lie on the same side of the line $x+2 y-3=0$, is
Answer & explanation
Correct answer: option 1
We have,
$f(x)=\frac{1}{1-x}$
Clearly, f(x) is defined for all $x \in R-\{1\}$.
Now,
$f(f(x))=f\left(\frac{1}{1-x}\right)=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$
We find that $f(f(x))$ is defined for all $x \neq 0,1$.
Now, $f(f(f(x)))=f\left(\frac{x-1}{x}\right)=\frac{1}{1-\frac{x-1}{x}}=x$ for all $x \neq 0,1$.
Thus, the set of points of discontinuity is $\{0,1\}$.
∴ $\alpha=0$ and $\beta=1$.
It is given that $(\alpha, \beta)=(0,1)$ and $\left(a, a^2\right)$ lie on the same side of the line $x+2 y-3=0$. Therefore,
$(0+2-3)\left(a+2 a^2-3\right)>0$
$\Rightarrow 2 a^2+a-3<0$
$\Rightarrow (2 a+3)(a-1) <0 \Rightarrow-3 / 2<a<1$
$\Rightarrow a \in(-3 / 2,1)$