JEE MAIN - Mathematics (2004 - No. 3)
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
$$f\left( x \right) = \sin x - \sqrt 3 \cos x + 1$$,
is onto, then the interval of $$S$$ is
[-1, 3]
[-1, 1]
[0, 1]
[0, 3]
Explanation
$$f\left( x \right)$$ 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 $$1 - \le \sin \left( {x - {\pi \over 3}} \right) \le 1$$
$$ - 1 \le 2\sin \left( {x - {\pi \over 3}} \right) + 1 \le 3$$
$$\therefore$$ $$f\left( x \right) \in \left[ { - 1,3} \right] = S$$
$$\therefore$$ $$S=$$ 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 $$1 - \le \sin \left( {x - {\pi \over 3}} \right) \le 1$$
$$ - 1 \le 2\sin \left( {x - {\pi \over 3}} \right) + 1 \le 3$$
$$\therefore$$ $$f\left( x \right) \in \left[ { - 1,3} \right] = S$$
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