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If \( f(x)=\sqrt{4-x^{2}}+\frac{1}{\sqrt{\sin x \mid-\sin x}} \), then the domain of \( f(x) \) isA) \( [-2,0] \)B) \( (0,2] \)C) \( [-2,2] \) |
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Answer» If |sin x| = sin x Then |sin x| - sin x = 0 but \(\sqrt{|sin\,\text x|-sin \text x}\neq0\) Therefore f(x) is not designed at points where |sin x| = sin x Therefore |sin x| = - sin x Also |sin x| - sin x > 0 ⇒ -2 sin x > 0 (\(\because\) |sin x| = - sin x) ⇒ sin x < 0 (Multiplying both sides by negative number -2) ⇒ x ∈ (-π, 0) ⋃ (π, 2π).........(i) Also 4 - x2 ≥ 0 (\(\because\) domain of √x is [0, 0]) ⇒ x2 ≤ 4 ⇒ x ∈ [-2, 2]......(ii) Since, π = 3.14 >2 and - π < -2 Therefore from equation(i) and (ii), we observed that domain of function f(x) is x ∈ [-2, 2] ∩ (-π, 0) (\(\because\) π = 3.14 >2) ⇒ x ∈ [-2, 0) Hence, domain of given function f(x) is [-2, 0). |
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