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Q.

The equation 4Sin²x + 4Sinx + a² - 3 = 0 has a solution if

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a

$- 2 \leq a \leq 2$

b

$- 1 \leq a \leq 1$

c

$- 3 \leq a \leq 3$

d

$- 4 \leq a \leq 4$

answer is A.

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Detailed Solution

$\sin x = \frac{- 4 \pm \sqrt{16 - 16 (a^{2} - 3)}}{8}$

$- 1 \leq \frac{- 1 \pm \sqrt{4 - a^{2}}}{2} \leq 1$

$- 2 \leq - 1 \pm \sqrt{4 - a^{2}} \leq 2$

$0 \leq \sqrt{4 - a^{2}} \leq 3$

$4 - a^{2} \geq 0 \Rightarrow a \in [- 2, 2]$

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