If f: R rightarrow A,defined by f(x) = cos x | vedclass

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Question

(A) Given the function $f(x) = \cos x + \sqrt{3} \sin x - 1$. We can rewrite the expression as $f(x) = 2(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin

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$[-3, 1]$

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(A) Given the function $f(x) = \cos x + \sqrt{3} \sin x - 1$. We can rewrite the expression as $f(x) = 2(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x) - 1$. Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f(x) = 2 \sin(x + \frac{\pi}{6}) - 1$. Since the range of $\sin(x + \frac{\pi}{6})$ is $[-1, 1]$,the range of $f(x)$ is $2 	imes [-1, 1] - 1 = [-2, 1] - 1 = [-3, 1]$. For $f$ to be an onto function,the codomain $A$ must be equal to the range of the function. Therefore,$A = [-3, 1]$.

If $f: R \rightarrow A$,defined by $f(x) = \cos x + \sqrt{3} \sin x - 1$,is an onto function,then $A =$

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