Let E 1 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − ≠ ∈ 0 1 x x and 1 x : R x and E 2 = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ∈ − number real a is 1 x x log sin : E x e 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ π π . 2 , 2 – in values assumes x sin function ric trigonomet inverse the Here, 1 – Let ƒ : E 1 → R be

Question

Let E 1 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ > − ≠ ∈ 0 1 x x and 1 x : R x and E 2 = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ∈ − number real a is 1 x x log sin : E x e 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ π π . 2 , 2 – in values assumes x sin function ric trigonomet inverse the Here, 1 – Let ƒ : E 1 → R be the function defined by ƒ(x) = log e ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 1 x x and g : E 2 → R be the function defined by g(x) = sin –1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 1 x x log e List-I List-II (P) The range of ƒ is (1) ⎥ ⎦ ⎤ ⎜ ⎝ ⎛ − ∞ − e 1 1 , ∪ ⎟ ⎠ ⎞ ⎢ ⎣ ⎡ ∞ − , 1 e e (Q) The range of g contains (2) (0, 1) (R) The domain of ƒ contains (3) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − 2 1 , 2 1 (S) The domain of g is (4) (– ∞ , 0) ∪ (0, ∞ ) (5) ⎥ ⎦ ⎤ ⎜ ⎝ ⎛ − ∞ − 1 e e , (6) (– ∞ , 0) ∪ ⎥ ⎦ ⎤ ⎜ ⎝ ⎛ − 1 e e , 2 1 The correct option is :

Accepted Answer

(P) → (4); (Q) → (2); (R) → (1); (S) → (1)

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