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
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a
[0,∞)
b
[2,1]
c
14,∞
d
None of these
answer is C.
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Detailed Solution
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