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Q.
Let $f : R \to [0, \frac{\pi}{2})$ defined by $f(x) = tan^{-1}(x^2 + x + 2a)$,then the set of values of ‘$a$' for which $f$ is onto, is
Relations and Functions - Part 2
Solution:
Since for given function co-domain is $0 \le x < \frac{\pi}{2}$,
For onto function, we have
Co-domain = Range = $0 \le x < \frac{\pi}{2}$
This will be possible if $x^2 + x + 2a \ge 0$ Fact: Quadratic function $f(x) \ge 0$
i.e. $Ax^2 + Bx + C \ge 0$ then $D \le 0$ if $ A > 0$
$ \therefore x^2 + x + 2a \ge 0$
$\Rightarrow 1^2 - 4(2a) \le 0$
$\Rightarrow a \ge 1/8$
$\therefore a \in [\frac{1}{8}, \infty)$