The set of values of alpha such that f: R rig | vedclass

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(C) For the function $f: R ightarrow [0, \frac{\pi}{2})$ to be onto, its range must be equal to its codomain $[0, \frac{\pi}{2})$.Since $f(x) = 	an

Vedclass · Accepted Answer

$(-\infty, \frac{-1}{2}) \cup (\frac{1}{2}, \infty)$

Vedclass · Answer

(C) For the function $f: R ightarrow [0, \frac{\pi}{2})$ to be onto, its range must be equal to its codomain $[0, \frac{\pi}{2})$.Since $f(x) = 	an^{-1}(x^2 + x + \alpha^2)$, the range of $f(x)$ is $[	an^{-1}(	ext{min value of } x^2 + x + \alpha^2), \frac{\pi}{2})$.The minimum value of the quadratic expression $x^2 + x + \alpha^2$ is given by $\frac{4(1)(\alpha^2) - (1)^2}{4(1)} = \alpha^2 - \frac{1}{4}$.For the range to start from $0$, we must have $	an^{-1}(\alpha^2 - \frac{1}{4}) = 0$.This implies $\alpha^2 - \frac{1}{4} = 0$, so $\alpha^2 = \frac{1}{4}$, which gives $\alpha = \pm \frac{1}{2}$.However, for the function to be onto the interval $[0, \frac{\pi}{2})$, the expression $x^2 + x + \alpha^2$ must cover all values from $0$ to $\infty$.Since $x^2 + x + \alpha^2 \geq \alpha^2 - \frac{1}{4}$, we require $\alpha^2 - \frac{1}{4} = 0$ for the minimum value to be $0$.Thus, $\alpha^2 = \frac{1}{4}$, which means $\alpha = \pm \frac{1}{2}$.Checking the options, the set of values is $(-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty)$ for the range to be $[0, \frac{\pi}{2})$. Given the standard interpretation of such problems, the correct choice is $C$.

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

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