If `alpha, beta(alpha,beta)` are the points of discontinuity of the function `f(f(x))`, where `f(x)=1/(1-x)`, then the set of values of a foe which the points `(alpha,beta)` and `(a,a^2)` lie on the same side of the line `x+2y-3=0` , isA. `(-3//2,1)`B. `[-3//2,1]`C. `[1,oo)`D. `(-oo,-3//2]`

If `alpha, beta(alpha,beta)` are the points of discontinuity of the fu

1.	If `alpha, beta(alpha,beta)` are the points of discontinuity of the function `f(f(x))`, where `f(x)=1/(1-x)`, then the set of values of a foe which the points `(alpha,beta)` and `(a,a^2)` lie on the same side of the line `x+2y-3=0` , isA. `(-3//2,1)`B. `[-3//2,1]`C. `[1,oo)`D. `(-oo,-3//2]`
Answer» Correct Answer - A We have `f(x)=(1)/(1-x)` Clearly, f(x) is defined for all `x in R -(1)}` Now, `f(f(x))=f((1)/(1-x))=(1)/(1-(1)/(1-x))=(x-1)/(x)` We find that f(f(x)) is defined for all `x ne 0,1` Now, `f(f(x))=f((x-1)/(x))=(1)/(1-(x-1)/(x))="x for all x"in 0,1` Thus, the set of points of discontinuity is `{0,1}` the line x+2y-3=0. Therefore, `(0+2-3)(a+2a^(2)-3)gt0` `Rightarrow 2a^(2)+a-3 lt 0` `Rightarrow (2a+3) (a-1) lt 0 Rightarrow -3//2 lt a lt 1` `Rightarrow a in (-3//2,1)`