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JEE MATHEMATICS MCQ: FUNCTIONS

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Let f:A→Bf:A \to B be a function defined by f(x)=3sin⁡x+cos⁡x+4.f\left( x \right) = \sqrt 3 \sin x + \cos x + 4. If f is invertible, then
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Correct option: A
A=[−2π/3,π/3],B=[2,  6]A = \left[ { - 2\pi /3,\pi /3\left] {,B = } \right[2,\;6} \right]
Option A is correct. To be invertible, the function must be a bijection. By rewriting f(x) as 2sin⁡(x+π/6)+42\sin(x + \pi/6) + 4, we find the domain and range that make it a bijection.

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