Let f:R→0,π2 defined by f(x)=tan−1x2+x+a , then the set of values of a for which/is onto is
[0,∞)
[2,1]
14,∞
None of these
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