If $x \in [0, 3 π]$ and $r \in R$ , then the number of pairs of $(r, x)$ satisfying $2 sin x = r^{4} - 2 r^{2} + 3$ is

2

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3

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4

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6

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Solution

The correct option is C $4$
$2 sin x = r^{4} - 2 r^{2} + 3 \Rightarrow 2 sin x = (r^{2} - 1)^{2} + 2 \Rightarrow 2 sin x = (r^{2} - 1)^{2} + 2$
We know that,
$2 sin x \leq 2 and (r^{2} - 1)^{2} + 2 \geq 2$

So, both are equal only when
$2 sin x = 2 \Rightarrow sin x = 1 \Rightarrow x = \frac{π}{2}, \frac{5 π}{2}$
And $(r^{2} - 1)^{2} + 2 = 2$
$\Rightarrow (r^{2} - 1)^{2} = 0 \Rightarrow r = \pm 1$

$∴$ The required ordered pairs are
$(1, \frac{π}{2}), (- 1, \frac{π}{2}), (1, \frac{5 π}{2}), (- 1, \frac{5 π}{2})$

Hence, there are $4$ ordered pairs.

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