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

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[0,∞)

[2,1]

14,∞

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Correct option is C

Since co-domain =0,π2∴ for f to be onto, range =0,π2 This is possible only when x2+x+a≥0 The quadratic expression is positive for all real values, if the discriminant is negative∴12−4a≤0⇒a≥14

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Let f:R→0,π2 defined by f(x)=tan−1⁡x2+x+a , then the set of values of a for which/is onto is

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