<p>If <strong>f</strong> : ℝ → \(\left[-1, 1\right]\), <i>f</i>(<i>x</i>) = \(\sin\left(\tan^{-1}\left(\frac{x^2 - a}{x^2 + 1}\right)\right)\) is an onto function, the set of values of '<i>a</i>' is</p>
Step-by-Step Solution
Key Concept: For an onto function, the range must equal the codomain. Find the range of the inner function and ensure the full codomain is covered.
<p><strong>Step 1:</strong> For <i>f</i> to be onto, the range of <i>f</i> must equal the codomain $[-1, 1]$.</p><p><strong>Step 2:</strong> Let $u = \frac{x^2 - a}{x^2 + 1}$. Then $f(x) = \sin(\tan^{-1}(u))$.</p><p><strong>Step 3:</strong> We have $\sin(\tan^{-1}(u)) = \frac{u}{\sqrt{1 + u^2}}$.</p><p><strong>Step 4:</strong> The range of <i>u</i> is $[-a, 1)$ as <i>x</i> varies over ℝ. For the range of <i>f</i> to be $[-1, 1]$, we need $-a \geq -1$, so $a \leq 1$.</p><p><strong>Step 5:</strong> However, for <i>f</i> to be onto to $[-1, 1]$, we need the minimum of <i>u</i> to approach values such that the full range is attained. This requires $a > -1$.</p><p>∴ The set of values of <i>a</i> is $(-1, \infty)$.</p>
Correct Answer: C