Question

The set of values of $\alpha$ such that $f:R\to \left[0,\frac{\pi }{2}\right)$ defined by $f(x)={\tan }^{-1}\left(x^2+x+{\alpha }^2\right)$ is onto is

• A. $\left(-\frac{1}{2},\frac{1}{2}\right)$
• B. $\left(-\frac{1}{4},\frac{1}{4}\right)$
• C. $\left(-\infty ,-\frac{1}{2}\right)\cup \left(\frac{1}{2},\infty \right)$
• D. $\left(-\infty ,-\frac{1}{4}\right)\cup \left(\frac{1}{4},\infty \right)$

Answer 1

Answer: (C)

Let $A=\left\{x:0\le x<\frac{\pi }{2}\right\}$
Since $f:R\to A$ is an onto function,
therefore, Range of $f=A$ Range of $f=A$
$\Rightarrow 0\le f(x)\le \frac{\pi }{2}$ for all $x\in R$
$\Rightarrow 0\le {\tan }^{-1}\left(x^2+x+{\alpha }^2\right)\le \frac{\pi }{2}$ for all $x\in R$
$\Rightarrow 0\le x^2+x+{\alpha }^2\le \infty$ for all $x\in R$
$\Rightarrow x^2+x+{\alpha }^2\ge 0$ for all $x\in R$
$\Rightarrow 1-4{\alpha }^2\le 0\Rightarrow {\alpha }^2\ge \frac{1}{4}$
$\therefore \left(-\infty ,-\frac{1}{2}\right)\cup \left(\frac{1}{2},\infty \right)$