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