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Let $f : [\frac{- π}{3}, \frac{π}{6}] \to B$ be defined by $f (x) = 2 \cos^{2} x + \sqrt{3} \sin 2 x + 1,$ where $B = [a, b]$ such that $f^{- 1} (x)$ exists, then the quadratic equation having the roots $a, b$ is

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$x^{2} + 4 x = 0$

$x^{2} + 4 x + 2 = 0$

$x^{2} - 4 x - 2 = 0$

$x^{2} - 4 x = 0$

answer is D.

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Detailed Solution

$f (x) = 1 + \cos 2 x + \sqrt{3} \sin 2 x + 1 = 2 + 2 \sin (2 x + \frac{π}{6})$

Since $- \frac{π}{3} \leq x \leq \frac{π}{6} \Rightarrow \frac{- π}{2} \leq (2 x + \frac{π}{6}) \leq \frac{π}{2}$

Range $[0, 4] = B$

The quadratic equation having roots $0, 4$ is $x^{2} - 4 x = 0$

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