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Question:

Let E1 = {x ∈ R: x ≠ 1 and xx > 0} and E2 = {x ∈ E1: sin-1(loge(xx)) is a real number}. (Here, the inverse trigonometric function sin-1x assumes values in [-π/2, π/2]). Let f: E1 → R be the function defined by f(x) = loge(xx) and g: E2 → R be the function defined by g(x) = sin-1(loge(xx)).
LIST - I LIST - II
P. The range of f is 1. (-∞, 1/1-e] ∪ [ee, ∞)
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. (-∞, 0) ∪ (0, ∞)
5. (-∞, ee]
6. (-∞, 0) ∪ (1/2, ee]
The correct option is
P → 4; Q → 2; R → 1; S → 1
P → 3; Q → 3; R → 6; S → 5
P → 4; Q → 3; R → 6; S → 5
P → 4; Q → 2; R → 1; S → 6

P → 4; Q → 3; R → 6; S → 5

P → 4; Q → 2; R → 1; S → 1

P → 3; Q → 3; R → 6; S → 5

P → 4; Q → 2; R → 1; S → 6

Solution:

E1: xx > 0 ⇒ E1: x ∈ (-∞, 0) ∪ (1, ∞)
E2: -π/2 ≤ ln(xx) ≤ π/2
1/e ≤ xx ≤ e
Now xx ≥ 1/e ≥ 0 ⇒ (ex)1/x ≥ 0 ⇒ x ∈ (-∞, 1/1-e] ∪ (1, ∞)
also xx ≤ e
So, E2: (-∞, 1/1-e] ∪ [ee, ∞)
as Range of xx is R+ - {1} ⇒ Range of f is R - {0} or (-∞, 0) ∪ (0, ∞)
Range of g is [-π/2, π/2] - {0} or [-π/2, 0) ∪ (0, π/2]
Now P → 4; Q → 2; R → 1; S → 1
Hence A is correct.