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Q.
The set of values of $\alpha$ such that $f: R \rightarrow\left[0, \frac{\pi}{2}\right)$ defined by $f(x)=\tan^{-1}\left(x^{2}+x+\alpha^{2}\right)$ is onto is
TS EAMCET 2020
Solution:
Let $A=\left\{x: 0 \leq x < \frac{\pi}{2}\right\}$
Since $f: R \rightarrow A$ is an onto function,
therefore, Range of $f=A$ Range of $f=A$
$\Rightarrow 0 \leq f(x) \leq \frac{\pi}{2} $ for all $x \in R$
$\Rightarrow 0 \leq \tan ^{-1}\left(x^{2}+x+\alpha^{2}\right) \leq \frac{\pi}{2} $ for all $x \in R$
$\Rightarrow 0 \leq x^{2}+x+\alpha^{2} \leq \infty$ for all $x \in R$
$\Rightarrow x^{2}+x+\alpha^{2} \geq 0$ for all $x \in R$
$\Rightarrow 1-4 \alpha^{2} \leq 0 \Rightarrow \alpha^{2} \geq \frac{1}{4}$
$\therefore \left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)$