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Consider f(x) = arcsin([e^x]) + arcsin([e^(-x)]), where [.] denotes the greatest integer function. Which of the following are correct? (A) The domain of f(x) is (-ln2, ln2). (B) The range of f(x) is {pi}. (C) f(x) has a removable discontinuity at x = 0. (D) The equation f(x) = arccos(x) has exactly one solution.

1. (A) Domain of f(x) is (-ln2, ln2).
2. (B) Range of f(x) is {pi}.
3. (C) f(x) has a removable discontinuity at x = 0.
4. (D) f(x) = arccos(x) has exactly one solution.

Correct answer: (A) Domain of f(x) is (-ln2, ln2).

Solution

For f to be defined: [e^x] and [e^(-x)] must both lie in [-1,1]. Since e^x > 0, [e^x] >= 0, so we need [e^x] in {0,1}, giving x < ln2. Similarly [e^(-x)] in {0,1} gives x > -ln2. Domain = (-ln2, ln2). On this domain: f = pi/2 for x != 0 and f = pi at x=0. Range = {pi/2, pi}. Since lim_(x->0) f = pi/2 but f(0)=pi, there is a removable discontinuity at x=0. For f = arccos(x): neither pi/2 nor pi lies in the range of arccos for x in [-1,1] such that a solution exists in (-ln2,ln2). So A and C are correct.

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