If f:R to S defined by f(x) = sin x - sqrt{3} | vedclass

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(A) The function is given by $f(x) = \sin x - \sqrt{3} \cos x + 1$. We know that the expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^

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$[-1, 3]$

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(A) The function is given by $f(x) = \sin x - \sqrt{3} \cos x + 1$. We know that the expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$. Here,$a = 1$ and $b = -\sqrt{3}$. Thus,the range of $\sin x - \sqrt{3} \cos x$ is $[-\sqrt{1^2 + (-\sqrt{3})^2}, \sqrt{1^2 + (-\sqrt{3})^2}] = [-\sqrt{1+3}, \sqrt{1+3}] = [-2, 2]$. Adding $1$ to the entire inequality,we get: $-2 + 1 \le \sin x - \sqrt{3} \cos x + 1 \le 2 + 1$. $-1 \le f(x) \le 3$. Since the function $f$ is onto,the codomain $S$ must be equal to the range of the function. Therefore,$S = [-1, 3]$.

If $f:R \to S$ defined by $f(x) = \sin x - \sqrt{3} \cos x + 1$ is onto,then the interval of $S$ is

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