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$Let f : R \to (0, \frac{π}{2}]$ o. defined by $f (x) = \tan^{- 1} (x^{2} + x + a) .$
Then the set of values of a for which f 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 =co domain=0,π2This is possible only when x2+x+a is perfect square.a-¼=0∴ 1-4a=0 or a=14

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Let f:R→0,π2 o. defined by fx=tan-1x2+x+a.Then the set of values of a for which f is onto is

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