If $ f: R \rightarrow A $ defined as $ f(x)=\tan ^{-1}\left(\sqrt{4\left(x^{2}+x+1\right)}\right) $ is surjective, then $ A $ is equal to

Question

If $ f: R \rightarrow A $ defined as $ f(x)=\tan ^{-1}\left(\sqrt{4\left(x^{2}+x+1\right)}\right) $ is surjective, then $ A $ is equal to, $ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $, $ \left[0, \frac{\pi}{2}\right) $, $ \left[\frac{\pi}{3}, \frac{\pi}{2}\right) $, $ \left(0, \frac{\pi}{3}\right] $

MARKS App · Accepted Answer

∵ x 2 + x + 1= x + 1 2 2 + 3 4 ∴ 4 x 2 + x + 1 ≥ 3 ⇒ 4 x 2 + x + 1 ≥ 3 ⇒ tan - 1 4 x 2 + x + 1 ≥ tan - 1 ( 3 ) ⇒ f ( x ) ≥ π 3 ∴ Range of f x is π 3 , π 2

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