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Let E1={x in R : x!=1\ a n dx/(x-1)>0} ...

Let `E_1={x in R : x!=1\ a n dx/(x-1)>0}` and `E_2={x in E_1:sin^(-1)((log)_e(x/(x-1)))\ i s\ a\ r e a l\ nu m b e r}` . Here, the inverse trigonometric function `sin^(-1)x` assumes values in `[pi/2,pi/2]` . Let `f: E_1->R` be the function defined by `f(x)=(log)_e(x/(x-1))` and `g: E_2->R` be the function defined by `g(x)=sin^(-1)((log)_e(x/(x-1)))` . LIST-I LIST-II P. The range of `f` is 1. `(-oo,1/(1-e)]uu[e/(e-1),\ oo)` Q. The range of `g` contains 2. `(0,\ 1)` R. The domain of `f` contains 3. `[1/2,1/2]` S. The domain of `g` is 4. `(-oo,\ 0)uu(0,\ oo)` 5. `(-oo,\ e/(e-1)]` 6. `(-oo,\ 0)uu(1/2, e/(e-1)]` The correct option is: `P->4;\ \ Q->2;\ \ R->1;\ \ S->1` (b) `P->3;\ \ Q->3;\ \ R->6;\ \ S->5` (c) `P->4;\ \ Q->2;\ \ R->1;\ \ S->6` (d) `P->4;\ \ Q->3;\ \ R->6;\ \ S->5`

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