Question

All x satisfying the inequality \((\cot^{-1}x)^2 - 7(\cot^{-1}x) + 10 \gt 0\), lie in the interval :

• A. \((\cot 2, \infty)\)
• B. \((-\infty, \cot 5) \cup (\cot 2, \infty)\)
• C. \((\cot 5, \cot 4)\)
• D. \((-\infty, \cot 5) \cup (\cot 4, \cot 2)\)

Hint:

Put t = cot^{-1}x, which lies in (0, pi). Solve the quadratic inequality in t and convert back using that cot^{-1} is decreasing.

Answer 1

The Correct Option is A

Let \(t = \cot^{-1}x\), so \(t \in (0, \pi)\) and \(t^2 - 7t + 10 \gt 0\), i.e. \((t-2)(t-5) \gt 0\).
So \(t \lt 2\) or \(t \gt 5\).
Since \(\pi \approx 3.14 \lt 5\), the case \(t \gt 5\) is impossible. Hence \(0 \lt t \lt 2\).
\(\cot^{-1}x\) is a decreasing function of x, so \(\cot^{-1}x \lt 2\) gives \(x \gt \cot 2\).
Thus \(x \in (\cot 2, \infty)\).
Answer: (1) \((\cot 2, \infty)\)