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If a function f:R→Tf: R \to T, defined by f(x)=3sin⁡x+cos⁡xf(x) = \sqrt{3}\sin x + \cos x is an onto function, then the interval of TT is

Solution

✅ Correct Option: 1

Write 3sin⁡x+cos⁡x=2sin⁡(x+π6)\sqrt3\sin x + \cos x = 2\sin\left(x + \frac{\pi}{6}\right), since the amplitude is 3+1=2\sqrt{3+1}=2.

The sine factor covers [−1,1][-1,1], so the range is [−2,2][-2,2]. For the map to be onto, T=[−2,2]T = [-2,2].

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