If a function \( f:R\to T \), defined by \( f(x)=\sqrt{3}\sin x+\cos x \) is an onto function, then the interval of \( T \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → [−2, 2]
For f to be onto, T must be exactly the range of f.
Write f in the form R sin(x + α), with \( R=\sqrt{(\sqrt{3})^{2}+1^{2}}=2 \):
\( \displaystyle f(x)=2\left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right) \)
\( \displaystyle =2\left(\sin x\cos\frac{\pi}{6}+\cos x\sin\frac{\pi}{6}\right) \)
\( \displaystyle =2\sin\left(x+\frac{\pi}{6}\right) \)
As x runs over R, \( \sin\left(x+\frac{\pi}{6}\right) \) takes every value in [−1, 1], so f takes every value in [−2, 2].
\( T=[-2,\ 2] \)
Tip: The range of a sin x + b cos x is always \( \left[-\sqrt{a^{2}+b^{2}},\ \sqrt{a^{2}+b^{2}}\right] \); remember it as a standard result.