If tan-1 x = y, then :(a) −1 < y < 1(b) \(\frac{-\pi}{2}\) ≤ y �

1.	If tan-1 x = y, then :(a) −1 < y < 1(b) \(\frac{-\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)(c) \(\frac{-\pi}{2}\) < y < \(\frac{\pi}{2}\)(d) y ∈ {\(\frac{-\pi}{2}\),\(\frac{\pi}{2}\)}
Answer» Option : (c) Since, principal value branch of \(tan^{-1}x\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\). \(\therefore -\frac{\pi}{2}<tan^{-1} x< \frac{\pi}{2}\) \(\Rightarrow \frac{-\pi}{2}<y<\frac{\pi}{2}\) \((\because tan^{-1}x =y)\)

If tan-1 x = y, then :(a) −1 < y < 1(b) \(\frac{-\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)(c) \(\frac{-\pi}{2}\) < y < \(\frac{\pi}{2}\)(d) y ∈ {\(\frac{-\pi}{2}\),\(\frac{\pi}{2}\)}

Answer»

Option : (c)

Since, principal value branch of \(tan^{-1}x\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

\(\therefore -\frac{\pi}{2}<tan^{-1} x< \frac{\pi}{2}\)

\(\Rightarrow \frac{-\pi}{2}<y<\frac{\pi}{2}\) \((\because tan^{-1}x =y)\)

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If tan-1 x = y, then :(a) −1 &lt; y &lt; 1(b) \(\frac{-\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)(c) \(\frac{-\pi}{2}\) &lt; y &lt; \(\frac{\pi}{2}\)(d) y ∈ {\(\frac{-\pi}{2}\),\(\frac{\pi}{2}\)}

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If tan-1 x = y, then :(a) −1 < y < 1(b) \(\frac{-\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)(c) \(\frac{-\pi}{2}\) < y < \(\frac{\pi}{2}\)(d) y ∈ {\(\frac{-\pi}{2}\),\(\frac{\pi}{2}\)}