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If (f:\mathbb{R}\to\mathbb{R}), (f(x)=\sin x), on which interval is (f) one-one?

Answer and explanation

Correct answer: \([-\frac{\pi}{2},\frac{\pi}{2}]\) / ([-\frac{\pi}{2},\frac{\pi}{2}])

Step 1: (\sin x) is periodic on the real line, so it is not one-one on all real numbers. Step 2: On ([-\frac{\pi}{2},\frac{\pi}{2}]), (\sin x) is strictly increasing. Step 3: For trigonometric functions, choose a proper restricted interval to make them one-one.

Related tags

Sine FunctionOne-OneRestricted DomainTrigonometryOne One FunctionRelations And FunctionsMathematicsClass 12 Mcq

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