If f:R→S is given by f(x)=sinx-3cosx+1 is on to, then the interval of S is

We have-a2+b2≤asinx+bcosx≤a2+b2So -1+3≤sinx-3cosx≤1+3⇒ -2≤sinx-3cosx≤2⇒ −2+1≤sin⁡x−3cos⁡x+1≤2+1⇒ −1≤f(x)≤3.

If f:R→S is given by f(x)=sinx-3cosx+1 is on to, then the interval of S is

We have-a2+b2≤asinx+bcosx≤a2+b2So -1+3≤sinx-3cosx≤1+3⇒ -2≤sinx-3cosx≤2⇒ −2+1≤sin⁡x−3cos⁡x+1≤2+1⇒ −1≤f(x)≤3.

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

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$[0, 3]$

$[- 1, 1]$

$[- 1, 3]$

$[0, 1]$

answer is D.

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Detailed Solution

We have

$- \sqrt{a^{2} + b^{2}} \leq a \sin x + b \cos x \leq \sqrt{a^{2} + b^{2}}$

So $- \sqrt{1 + 3} \leq \sin x - \sqrt{3} \cos x \leq \sqrt{1 + 3}$

$\Rightarrow - 2 \leq \sin x - \sqrt{3} \cos x \leq 2$

$\begin{array}{l} \Rightarrow - 2 + 1 \leq \sin x - \sqrt{3} \cos x + 1 \leq 2 + 1 \\ \Rightarrow - 1 \leq f (x) \leq 3 . \end{array}$

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If f:R→S is given by f(x)=sinx-3cosx+1 is on to, then the interval of S is

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