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Let $f : R \to [0, \frac{π}{2})$ defined by $f (x) = \tan^{- 1} (x^{2} + x + λ)$ , then the set of values of $λ$ for which f is onto, is

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[0, 4)

[-2, 1]

$\frac{1}{4}$

[0,1]

answer is C.

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Detailed Solution

$∵$ f is onto Hence this is possible only when

$x^{2} + x + λ \geq 0 \Rightarrow ∆ = 0$ Hence required answer will $λ$ = $\frac{1}{4}$

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