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Question

If $$r> 0, -\pi \leq \theta \leq \pi $$ and r, $$\theta $$ satisfy $$r\sin \theta =3$$ and $$r=4\left ( 1+\sin \theta  \right )$$ then the number of possible solutions of the pair $$\left ( r, \theta  \right )$$ is


A
2
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B
4
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C
0
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D
infinite
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Solution

The correct option is B 2
$$ r sin\theta = 3 $$ and $$ r = 4( 1 + sin\theta)$$
using these equation,
$$4(1 + sin\theta) sin\theta = 3$$
$$ \Rightarrow 4  sin^2\theta + 4 sin\theta - 3 =0$$
$$\Rightarrow (2 sin\theta - 1) (2 sin\theta + 3) = 0$$
but $$-1 \leq sin\theta \leq 1$$,so
$$ \Rightarrow sin\theta = 1/2$$
Therefore solution in $$ -\pi \leq \theta \leq \pi$$ is,
$$ \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$$
and $$r = 6$$
Thus, number of possible pair $$(r, \theta)$$ is 2.
Hence, option 'A' is correct.

Maths

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