Question

The domain of the function
\(f(x)=\tan ^{-1}\left(\sqrt{x^{2}+x-2}\right)+\sec ^{-1}\left(x^{2}-5 x+3\right)\) is :

• A. \((-\infty,-2] \cup[1, \infty)\)
• B. \([1,4] \cup\left[\frac{5+\sqrt{17}}{2}, \infty\right)\)
• C. \(\left(-\infty, \frac{5-\sqrt{17}}{2}\right] \cup[1,4] \cup\left[\frac{5+\sqrt{17}}{2}, \infty\right)\)
• D. \((-\infty,-2] \cup[1,4] \cup\left[\frac{5+\sqrt{17}}{2}, \infty\right)\)

Hint:

The square root needs x^2 + x - 2 >= 0, and sec inverse needs |x^2 - 5x + 3| >= 1; intersect the two solution sets.

Answer 1

The Correct Option is D

Condition 1 (square root): \(x^2+x-2\ge0 \Rightarrow (x+2)(x-1)\ge0 \Rightarrow x\le-2\) or \(x\ge1\). (\(\tan^{-1}\) is defined for all reals.)
Condition 2 (\(\sec^{-1}\)): \(|x^2-5x+3|\ge1\).
Case \(x^2-5x+3\ge1\): \(x^2-5x+2\ge0 \Rightarrow x\le\frac{5-\sqrt{17}}{2}\) or \(x\ge\frac{5+\sqrt{17}}{2}\).
Case \(x^2-5x+3\le-1\): \(x^2-5x+4\le0 \Rightarrow 1\le x\le4\).
Here \(\frac{5-\sqrt{17}}{2}\approx0.44\) and \(\frac{5+\sqrt{17}}{2}\approx4.56\).
Intersect with Condition 1: for \(x\le-2\) the first case holds, giving \((-\infty,-2]\); for \(x\ge1\) we get \([1,4]\) and \(\left[\frac{5+\sqrt{17}}{2},\infty\right)\). The part \(x\le0.44\) does not meet \(x\ge1\).
Domain \(=(-\infty,-2]\cup[1,4]\cup\left[\frac{5+\sqrt{17}}{2},\infty\right)\).
Answer: (4) \((-\infty,-2] \cup[1,4] \cup\left[\frac{5+\sqrt{17}}{2}, \infty\right)\)