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If $x, y \in$ $[0, 15]$ , then the number of solutions $(x, y)$ of the equation $3^{\cos e c^{2} x - 1} \times \sqrt{4 y^{2} - 4 y + 2} \leq 1$ is

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answer is D.

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Detailed Solution

We have $3^{\cot^{2} x} \sqrt{{(2 y - 1)}^{2} + 1} \leq 1$ .

But $3^{\cot^{2} x} \geq 1$ and $\sqrt{{(2 y - 1)}^{2}} \geq 1$

$\Rightarrow 3^{\cot^{2} x} = 1$ and $\sqrt{{(2 y - 1)}^{2} + 1} = 1$

$\Rightarrow \cot^{2} x = 0, y = \frac{1}{2}$

$\Rightarrow x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2}, \frac{9 π}{2} (∵ x \in [0, 15])$

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