The graph of an inverse trigonometric function passes through the point $(0, \frac{\pi}{2})$ with asymptotes at $y = 0$ and $y = \pi$. This graph represents the function:
Show Hint
Always check the domain and boundaries (y-intercepts and asymptotes) of the basic inverse trigonometric functions:
- \(\tan^{-1}(0) = 0\)
- \(\cot^{-1}(0) = \frac{\pi}{2}\)
- \(\sec^{-1}(x)\) and \(\csc^{-1}(x)\) are undefined at \(x = 0\).
Concept:
To identify the correct inverse trigonometric function from geometric or graphical features, we evaluate its principal value branch (range) and domain:
• For \(y = \cot^{-1} x\), the domain is \(\mathbb{R}\) (all real numbers) and the principal value range is \((0, \pi)\).
• Let us evaluate the value of the function at \(x = 0\):
\[
y = \cot^{-1}(0)
\]
Since \(\cot\left(\frac{\pi}{2}\right) = 0\) and \(\frac{\pi}{2} \in (0, \pi)\), it follows that \(\cot^{-1}(0) = \frac{\pi}{2}\). Thus, the graph must cross the y-axis exactly at the point \(\left(0, \frac{\pi}{2}\right)\).
• Furthermore, as \(x \to \infty\), \(\cot^{-1}x \to 0\), and as \(x \to -\infty\), \(\cot^{-1}x \to \pi\), which establishes horizontal asymptotes at \(y = 0\) and \(y = \pi\).
Step 1: Testing alternative options for verification • For option (A), \(y = \sec^{-1}x\), the domain is \((-\infty, -1] \cup [1, \infty)\). The value \(x = 0\) does not lie within the domain, so it cannot have a y-intercept at \((0, \frac{\pi}{2})\).
• For option (C), \(y = \tan^{-1}x\), the principal value range is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). At \(x = 0\), \(\tan^{-1}(0) = 0\), meaning it passes through the origin \((0,0)\), not \((0, \frac{\pi}{2})\).
• For option (D), \(y = \csc^{-1}x\), the domain is \((-\infty, -1] \cup [1, \infty)\). Like \(\sec^{-1}x\), \(x = 0\) is outside the domain.
Hence, the only function that is continuous across all real values and contains the point \((0, \frac{\pi}{2})\) while bounded between $0$ and $\pi$ is \(y = \cot^{-1} x\).
