If [f:R S], defined by [f( x ) = x - 3 x + 1], is onto, then the interval of S is

Question

If $$f:R \to S$$, defined by

$$f\left( x \right) = \sin x - \sqrt 3 \cos x + 1$$,

is onto, then the interval of $S$ is

Accepted Answer

$$f\left( x \right)$$ is onto

$$\therefore$$ $S=$ range of $f(x)$

Now $$f\left( x \right) = \sin \,x - \sqrt 3 \,\cos \,x + 1$$

$$ = 2\sin \left( {x - {\pi  \over 3}} \right) + 1$$

As $$1 -  \le \sin \left( {x - {\pi  \over 3}} \right) \le 1$$

$$ - 1 \le 2\sin \left( {x - {\pi  \over 3}} \right) + 1 \le 3$$

$$\therefore$$ $$f\left( x \right) \in \left[ { - 1,3} \right] = S$$

Solutions · 1