Exams › JEE Main › Maths

Find the domain of the function f(x)=sin⁻¹{log2(1/2 x²)}.

1. [-2,-1)∪(1,2)
2. (-2,-1]∪[1,2]
3. [-2,-1)∪[1,2]
4. (-2,-1)∪(1,2)

Correct answer: (-2,-1]∪[1,2]

Solution

For arcsin to be defined: -1 <= log2(x^2/2) <= 1, i.e. 1/2 <= x^2/2 <= 2, giving 1 <= x^2 <= 4, so 1 <= |x| <= 2. Including the endpoints, the domain is [-2,-1] U [1,2].

Related JEE Main Maths questions

• For the function f(x) = log(x + √(x² + 1)), which of the following describes it correctly?
• For the function f(x) = log((1 + x)/(1 − x)), which of the following relations is satisfied?
• For which values of x does the inequality sin⁻¹(sin 5) > x² − 4x hold?
• Determine the range of the function f(x) = sin⁻¹x + tan⁻¹x + sec⁻¹x.
• If A = tan⁻¹ ((x√(3))/(2K-x)) and B = tan⁻¹ ((2x-K)/(K√(3))), then what is the value of A - B?
• If the inequality |cos⁻¹ ((1-x²)/(1+x²)) |<(π)/(3) holds, then which of the following is true?

⚔️ Practice JEE Main Maths free + battle 1v1 →