The range of the function $f (x) = ({sin}^{2} x - 2 sin x)$ is

(- \infty, - 1]

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[1, 2]

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[- 1, 3]

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(1, 2)

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Solution

The correct option is C $[- 1, 3]$
given,

$f (x) = {sin}^{2} x - 2 sin x$

${lim}_{x \to 2 π + 2 π n -} ({sin}^{2} (x) - 2 sin (x)) = 0$

$T h e i n t e r v a l h a s a m i n i m u m p o i n t a t x = \frac{π}{2} + 2 π n + 2 π n w i t h v a l u e f (\frac{π}{2} + 2 π n + 2 π n) = - 1$

$T h e i n t e r v a l h a s a m a x i m u m p o i n t a t x = \frac{3 π}{2} + 2 π n + 2 π n w i t h v a l u e f (\frac{3 π}{2} + 2 π n + 2 π n) = 3$

$M i n i m u m f u n c t i o n v a l u e a t t h e d o m a i n i n t e r v a l - \infty < x < \infty i s - 1$

$M a x i m u m f u n c t i o n v a l u e a t t h e d o m a i n i n t e r v a l - \infty < x < \infty i s 3$

$- 1 \leq f (x) \leq 3$

$[- 1, 3]$ is the range

Suggest Corrections

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